Topics in Finite Geometry: Ovals, Ovoids and Generalized ...

Topics in Finite Geometry: Ovals, Ovoids and
Generalized Quadrangles
S. E. Payne
Edition of 16 May 2007

Contents
1 Synthetic Axioms for Geometry 15
1.1 Linear Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.2 Axioms for Projective Space . . . . . . . . . . . . . . . . . . . 17
1.3 The Axioms of Affine Space . . . . . . . . . . . . . . . . . . . 20
1.4 Planes in Affine Space . . . . . . . . . . . . . . . . . . . . . . 25
1.5 Buekenhout’s Characterization of Affine Space . . . . . . . . . 28
1.6 Embedding Affine Space in Projective Space . . . . . . . . . . 43
1.7 The Principle of Duality . . . . . . . . . . . . . . . . . . . . . 48
1.8 Structure of Projective Geometry . . . . . . . . . . . . . . . . 49
1.9 Quotient Geometries: a First Look . . . . . . . . . . . . . . . 56
1.10 Combinatorics of Finite Projective Spaces . . . . . . . . . . . 57
2 Analytic Geometry 63
2.1 The Projective Space P(V ) . . . . . . . . . . . . . . . . . . . 64
2.2 The Theorems of Desargues and Pappus . . . . . . . . . . . . 67
2.3 Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
2.4 Semilinear Maps: A First Glance . . . . . . . . . . . . . . . . 80
2.5 Some Combinatorics of Finite Geometries . . . . . . . . . . . 81
2.6 Some Geometry in P G(2, q) . . . . . . . . . . . . . . . . . . . 90
2.7 Some Geometry in P G(3, q) . . . . . . . . . . . . . . . . . . . 92
2.8 Polarities, Etc. . . . . . . . . . . . . . . . . . . . . . . . . . . 101
2.9 The Projective Line - Cross Ratio . . . . . . . . . . . . . . . . 105
2.10 Theorem of the Complete Quadrilateral . . . . . . . . . . . . . 110
2.11 Involutions of the Projective Line . . . . . . . . . . . . . . . . 112
2.12 The Hyperbolic Quadric of P G(3, q) - A First Look . . . . . . 115
2.13 3-Dimensional Projective Spaces Are Desarguesian . . . . . . . 118
3

4 CONTENTS
3 The Fundamental Theorem for P G(n, q) 121
3.1 Central Collineations . . . . . . . . . . . . . . . . . . . . . . . 121
3.2 The Group of Translations . . . . . . . . . . . . . . . . . . . . 128
3.3 The Division Ring . . . . . . . . . . . . . . . . . . . . . . . . 133
3.4 The Representation Theorems . . . . . . . . . . . . . . . . . . 138
3.5 The Fundamental Theorem . . . . . . . . . . . . . . . . . . . 140
3.6 Projective Collineations . . . . . . . . . . . . . . . . . . . . . 148
3.7 A Collection of Collineations . . . . . . . . . . . . . . . . . . . 154
4 The Ubiquitous Oval 157
4.1 Arcs and Ovals . . . . . . . . . . . . . . . . . . . . . . . . . . 157
4.2 Conics: The Classical Examples . . . . . . . . . . . . . . . . . 160
4.3 The Group of the Conic . . . . . . . . . . . . . . . . . . . . . 165
4.4 Segre’s Theorem for q Odd . . . . . . . . . . . . . . . . . . . . 174
4.5 o- Polynomials & the Criterion of Glynn . . . . . . . . . . . . 179
4.6 Arcs in Ovals . . . . . . . . . . . . . . . . . . . . . . . . . . . 191
4.7 What is a translation oval? . . . . . . . . . . . . . . . . . . . . 197
4.8 Linear Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198
4.9 Additive Maps on Fq . . . . . . . . . . . . . . . . . . . . . . . 202
4.10 A Characterization of Translation Ovals . . . . . . . . . . . . 206
4.11 Stabilizers of Arcs . . . . . . . . . . . . . . . . . . . . . . . . . 207
4.12 Monomial Hyperovals . . . . . . . . . . . . . . . . . . . . . . . 211
4.13 The Trace Map . . . . . . . . . . . . . . . . . . . . . . . . . . 218
4.14 Translation Ovals and the External Lines Lemma . . . . . . . 222
4.15 Oval Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . 229
4.16 Translation Ovals Again . . . . . . . . . . . . . . . . . . . . . 233
5 Quadratic Forms 237
5.1 Basic Definitions and Results . . . . . . . . . . . . . . . . . . 237
5.2 Structure of Quadrics . . . . . . . . . . . . . . . . . . . . . . . 243
5.3 Change of Basis . . . . . . . . . . . . . . . . . . . . . . . . . . 249
5.4 Canonical Forms over Finite Fields . . . . . . . . . . . . . . . 253
5.5 The Discriminant in Odd Characteristic . . . . . . . . . . . . 262
5.6 Action of the Orthogonal Group . . . . . . . . . . . . . . . . . 265
5.7 The Cartan-Dieudonné Theorem . . . . . . . . . . . . . . . . . 272

CONTENTS 5
6 Quadrics in P G(3, q) 277
6.1 Hyperbolic quadrics in P G(3, q) . . . . . . . . . . . . . . . . . 277
6.2 Quadratic cones in P G(3, q) . . . . . . . . . . . . . . . . . . . 289
6.3 Elliptic quadrics in P G(3, q) . . . . . . . . . . . . . . . . . . . 295
6.4 The (Elliptic) Orthogonal Group . . . . . . . . . . . . . . . . 297
6.5 Elliptic Quadrics Containing a Conic . . . . . . . . . . . . . . 299
6.6 A Condition that Determines a Quadric . . . . . . . . . . . . 311
6.7 Nonsingular Quadrics in PG(3,q) with q odd . . . . . . . . . . 320
7 Ovoids in 3-Dimensional Space 329
7.1 Lines and Planes related to an Ovoid . . . . . . . . . . . . . . 329
7.2 Ovoids as Caps and with many Conics . . . . . . . . . . . . . 330
7.3 Null polarities and skew-symmetric matrices . . . . . . . . . . 336
7.4 Ovoids Yield Null Polarities in Characteristic 2 . . . . . . . . 337
7.5 Symplectic Geometry and Normalized Coordinates . . . . . . 339
7.5.2 Coordinates for the known examples . . . . . . . . . . 341
7.6 The plane representation theorem . . . . . . . . . . . . . . . . 346
7.7 Semi-ovals and Semi-ovoids . . . . . . . . . . . . . . . . . . . 350
8 Trace Equations 353
8.1 Cyclotomic Cosets . . . . . . . . . . . . . . . . . . . . . . . . 354
8.2 More Basic Trace Equations . . . . . . . . . . . . . . . . . . . 355
8.3 Trace Equations by Glynn . . . . . . . . . . . . . . . . . . . . 359
8.4 Ovoids with a pencil of conics . . . . . . . . . . . . . . . . . . 367
8.5 More Trace Equations . . . . . . . . . . . . . . . . . . . . . . 368
9 Finite Generalized Quadrangles 375
9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375
9.2 Axioms and Preliminary Definitions . . . . . . . . . . . . . . . 377
9.3 Basic Combinatorics of GQ . . . . . . . . . . . . . . . . . . . 380
9.4 The Symplectic GQ W (q) . . . . . . . . . . . . . . . . . . . . 384
9.5 Some GQ derived from Quadratic Forms . . . . . . . . . . . . 392
9.6 Triads with Restricted Perp Size . . . . . . . . . . . . . . . . . 396
9.7 The Incidence Matrix . . . . . . . . . . . . . . . . . . . . . . . 401
9.8 Regularity: Variations on a Theme . . . . . . . . . . . . . . . 408
9.9 Bagchi, Brouwer and Wilbrink . . . . . . . . . . . . . . . . . . 413
9.10 Ovoids, Spreads and Polarities . . . . . . . . . . . . . . . . . . 416
9.11 Subquadrangles . . . . . . . . . . . . . . . . . . . . . . . . . . 419

6 CONTENTS
9.12 Collineations . . . . . . . . . . . . . . . . . . . . . . . . . . . 422
9.13 A Coset Geometry . . . . . . . . . . . . . . . . . . . . . . . . 427
9.14 Weak 4-Gonal Families . . . . . . . . . . . . . . . . . . . . . . 431
9.15 Elation Generalized Quadrangles . . . . . . . . . . . . . . . . 432
9.16 Elation Groups as p-Groups . . . . . . . . . . . . . . . . . . . 438
9.17 A Model for STGQ . . . . . . . . . . . . . . . . . . . . . . . . 440
10 GQ Associated with an Oval or Ovoid 445
10.1 Expansion about a Regular Point . . . . . . . . . . . . . . . . 445
10.2 The Construction of T2(O) . . . . . . . . . . . . . . . . . . . . 446
10.3 T2(Oα) is Self-dual when α is Additive . . . . . . . . . . . . . 452
10.4 The isomorphism from T2(C) to W (q) . . . . . . . . . . . . . . 453
10.5 Coordinates for the general T2(O) . . . . . . . . . . . . . . . . 456
10.6 Expanding T2(O) about a Regular Element . . . . . . . . . . . 461
10.6.1 P(T2(O), π) . . . . . . . . . . . . . . . . . . . . . . . . 461
10.6.2 P(T2(O), L) . . . . . . . . . . . . . . . . . . . . . . . . 461
10.7 Constricting about a Regular Spread . . . . . . . . . . . . . . 463
10.8 Characterizing T ∗ 2 (O+ ) with (0, 2)-Sets . . . . . . . . . . . . . 465
10.9 The Basic Assumptions . . . . . . . . . . . . . . . . . . . . . . 468
10.10Parallelism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475
10.10.1 Parallelism in the set of affine planes of type 1 . . . . . 475
10.11A Characterization of T ∗ 2
(O), s + 1 even . . . . . . . . . . . . 480
10.12Regular Ovoids in P(T2(O), L) . . . . . . . . . . . . . . . . . 481
10.13 A Regular Ovoid in GQ(q + 1, q − 1) . . . . . . . . . . . . . . 482
10.14Constricting about a Regular Ovoid . . . . . . . . . . . . . . . 485
10.15Quivers - and A3 . . . . . . . . . . . . . . . . . . . . . . . . . 491
10.16A1, A2, A3 and A4 . . . . . . . . . . . . . . . . . . . . . . . . . 494
10.17Pseudolines: A1, · · · , A6 . . . . . . . . . . . . . . . . . . . . 495
10.18The Construction of T3(Ω) . . . . . . . . . . . . . . . . . . . . 502
10.19The Point-Line Dual of T3(Ω) . . . . . . . . . . . . . . . . . . 502
11 Classifiying Ovoids - Part 2 509
11.1 Ovoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509
11.2 A restriction on plane sections of an ovoid . . . . . . . . . . . 513
11.3 Ovoids of P G(3, q) containing a conic . . . . . . . . . . . . . . 516
11.4 Restrictions on a secant plane section of Ω . . . . . . . . . . . 517
11.5 Sparse and spare o-polynomials . . . . . . . . . . . . . . . . . 522
11.6 The classification of ovoids containing a conic . . . . . . . . . 530

CONTENTS 7
11.7 Translation ovals again . . . . . . . . . . . . . . . . . . . . . . 531
12 The Klein Correspondence 537
12.1 Plücker Coordinates . . . . . . . . . . . . . . . . . . . . . . . 537
12.2 Time out for H5 . . . . . . . . . . . . . . . . . . . . . . . . . . 541
12.3 Return to the Klein Correspondence . . . . . . . . . . . . . . 545
12.4 Dual Coordinates for Lines . . . . . . . . . . . . . . . . . . . . 548
12.5 Linear Congruences . . . . . . . . . . . . . . . . . . . . . . . . 554
12.6 The null system ν . . . . . . . . . . . . . . . . . . . . . . . . . 557
12.7 A polarity π of W (q) when q = 2 2m+1 . . . . . . . . . . . . . . 565
12.8 The ovoid of J. Tits . . . . . . . . . . . . . . . . . . . . . . . . 569
12.9 The Hermitian geometry H(3, q 2 ) . . . . . . . . . . . . . . . . 576
12.10A Closer Look at the Klein Correspondence . . . . . . . . . . 579
12.11Spreads from Flocks: The Thas-Walker Construction . . . . . 584
13 q-Clans and Their Geometries 587
13.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 587
13.2 q-Clans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589
13.3 Flocks of a Quadratic Cone . . . . . . . . . . . . . . . . . . . 590
13.4 4-Gonal Families from q-Clans . . . . . . . . . . . . . . . . . . 593
13.5 Some Families of q-Clans . . . . . . . . . . . . . . . . . . . . . 595
13.5.1 The Classical q-Clans . . . . . . . . . . . . . . . . . . . 595
13.5.2 W.M. Kantor [Ka80]; J.C. Fisher and J.A. Thas [FT79];
M. Walker [Wa76] . . . . . . . . . . . . . . . . . . . . . 596
13.5.3 W.M Kantor [Ka86] . . . . . . . . . . . . . . . . . . . 596
13.5.4 W.M. Kantor [Ka86] for q odd; S.E. Payne [Pa85] for
q even . . . . . . . . . . . . . . . . . . . . . . . . . . . 597
13.5.5 H. Gevaert and N.L. Johnson [GJ88]; q = 5 e . . . . . . 597
13.5.6 H. Gevaert and N.L. Johnson [GJ88]; q = 3 e . . . . . . 597
13.6 q-Clans, Flocks & Spreads of PG(3,q) . . . . . . . . . . . . . . 598
13.7 The Classical Examples . . . . . . . . . . . . . . . . . . . . . 606
13.8 The Bose-Barlotti ∆-Planes . . . . . . . . . . . . . . . . . . . 607
13.9 Property (G); some projective planes . . . . . . . . . . . . . . 608
13.10The Fundamental Lemma . . . . . . . . . . . . . . . . . . . . 611
13.11The Fundamental Theorem . . . . . . . . . . . . . . . . . . . 617
13.12Recoordinatization: Shifts, Flips
& Scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 621
13.13Additive q-Clans and Dual TGQ . . . . . . . . . . . . . . . . 624

8 CONTENTS
13.14Property (G) at a second point of ∞ ⊥ . . . . . . . . . . . . . 626
13.15An additional regular pair of points . . . . . . . . . . . . . . . 628
13.16Flocks Whose Planes Contain a Common Point . . . . . . . . 628
14 Finite Circle Geometries 635
14.1 The Circle Geometries . . . . . . . . . . . . . . . . . . . . . . 635
14.2 Finite Inversive Planes . . . . . . . . . . . . . . . . . . . . . . 638
14.3 Miquelian Inversive Planes of Odd Order . . . . . . . . . . . . 646
14.4 Dembowski’s Theorem . . . . . . . . . . . . . . . . . . . . . . 667
14.5 Finite Laguerre Planes . . . . . . . . . . . . . . . . . . . . . . 669
14.6 Finite Minkowski Planes . . . . . . . . . . . . . . . . . . . . . 674
15 Property G 681
15.1 Property (G) in GQ(s, s 2 ) . . . . . . . . . . . . . . . . . . . . 681
15.2 Property (G) and Subquadrangles . . . . . . . . . . . . . . . . 685
15.3 Tetradic Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 689
15.4 Tetradic Sets of Elliptic Quadrics . . . . . . . . . . . . . . . . 692
15.5 Flocks and Tetradic Sets . . . . . . . . . . . . . . . . . . . . . 697
15.6 Coordinates for the AG(3, q) from Flock GQ . . . . . . . . . 700
15.7 Prop.(G) & Tetradic Sets of Quadrics . . . . . . . . . . . . . . 703
15.8 The ⊲⊳ Equivalence Relation . . . . . . . . . . . . . . . . . . . 709
15.9 Prop.(G) & Tetradic Sets of Ovoids . . . . . . . . . . . . . . . 721
15.10Using Tetradic Sets to Characterize GQ . . . . . . . . . . . . 726
15.11Dual Tetradic Sets with q = 2 e . . . . . . . . . . . . . . . . . . 727
16 Laguerre Geometries and GQ 731
16.1 Laguerre points . . . . . . . . . . . . . . . . . . . . . . . . . . 731
16.2 Ovoids and Laguerre Geometries . . . . . . . . . . . . . . . . 734
16.3 Internal Structure at a Point . . . . . . . . . . . . . . . . . . . 735
16.4 Laguerre Geometries and GQ . . . . . . . . . . . . . . . . . . 740
16.5 Property (G) . . . . . . . . . . . . . . . . . . . . . . . . . . . 743
16.6 Laguerre Sets of Ovoids with q even . . . . . . . . . . . . . . . 746
17 Translation Oval Cones 757
17.1 Representations of Flocks . . . . . . . . . . . . . . . . . . . . 757
17.2 Normalized α-Cones . . . . . . . . . . . . . . . . . . . . . . . 759
17.3 α-Flocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 761
17.4 α-Clans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 765

CONTENTS 9
17.5 Some Families of α-Clans . . . . . . . . . . . . . . . . . . . . . 766
17.6 Herds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 771
17.7 Flippable α-Clans . . . . . . . . . . . . . . . . . . . . . . . . . 773
17.8 A Special α-Clan . . . . . . . . . . . . . . . . . . . . . . . . . 776
17.9 The ovals σ + γ of Glynn . . . . . . . . . . . . . . . . . . . . 780
17.10Cherowitzo Hyperovals . . . . . . . . . . . . . . . . . . . . . . 785
18 Generalized Fans and Spreads of T2(O)) 793
18.1 Generalized f-Fans . . . . . . . . . . . . . . . . . . . . . . . . 793
18.2 Spreads of T2(Of) . . . . . . . . . . . . . . . . . . . . . . . . . 794
19 Appendix 1: Some Elementary Number Theory 797
19.1 Basic Facts about Congruences . . . . . . . . . . . . . . . . . 797
19.2 A Theorem of Lucas . . . . . . . . . . . . . . . . . . . . . . . 799
19.3 The Chinese Remainder Theorem . . . . . . . . . . . . . . . . 800
19.4 Solving Congruences: The Case of a Prime Modulus . . . . . . 803
19.5 The Case of a Prime Power Modulus, etc. . . . . . . . . . . . 805
19.6 Power Residues . . . . . . . . . . . . . . . . . . . . . . . . . . 807
19.7 The 4-Square Theorem of Lagrange . . . . . . . . . . . . . . . 809
19.8 Congruence of Symmetric Matrices . . . . . . . . . . . . . . . 811
19.9 The Bruck-Ryser Theorem . . . . . . . . . . . . . . . . . . . . 812
19.10Number Theoretic Functions . . . . . . . . . . . . . . . . . . . 815
20 Appendix 2: Algebra in Fq[x] 819
20.1 The Galois Field GF (q) = Fq . . . . . . . . . . . . . . . . . . 819
20.2 Relative Trace and Norm . . . . . . . . . . . . . . . . . . . . . 822
20.3 Polynomials over Fq . . . . . . . . . . . . . . . . . . . . . . . . 824
20.4 Symmetric Polynomials . . . . . . . . . . . . . . . . . . . . . . 826
20.5 Newton’s Formulas . . . . . . . . . . . . . . . . . . . . . . . . 828
20.6 The Resultant of Two Polynomials . . . . . . . . . . . . . . . 830
20.7 The Discriminant of a Polynomial . . . . . . . . . . . . . . . . 832
20.8 Reverse Polynomials . . . . . . . . . . . . . . . . . . . . . . . 834
20.9 Roots of Irreducible Polynomials in Zp[x] . . . . . . . . . . . . 835
20.10Stickelberger’s theorem . . . . . . . . . . . . . . . . . . . . . . 836
20.11Factoring x q − 1 in Zp[x] . . . . . . . . . . . . . . . . . . . . . 838
20.12Quadratic Reciprocity . . . . . . . . . . . . . . . . . . . . . . 839
20.13Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . 841
20.14Lifting Roots Modulo Powers of p . . . . . . . . . . . . . . . . 842

10 CONTENTS
20.15Resultants as Determinants . . . . . . . . . . . . . . . . . . . 846
20.16Bezout’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 852
20.17The Hermite-Dickson Criterion . . . . . . . . . . . . . . . . . 855
20.18Greatest Common Divisors of a m ± 1 and a n ± 1 . . . . . . . . 857
20.18.1 Basic Identities . . . . . . . . . . . . . . . . . . . . . . 857
20.18.6 gcd(a m + 1, a n + 1), a > 1 . . . . . . . . . . . . . . . . 859
20.18.13An Application to Finite Fields of Characteristic 2 . . 863
20.18.14Application to Finite Fields of Odd Characteristic . . . 864
20.19Linear Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . 865
20.20K-Subspaces of F as Linearized Polynomials . . . . . . . . . . 870
20.21Additive Maps on Fq . . . . . . . . . . . . . . . . . . . . . . . 874
20.22The Number of Roots of a Special Polynomial . . . . . . . . . 880
20.23Equations Involving Squares . . . . . . . . . . . . . . . . . . . 882
20.24Vandermonde Determinant . . . . . . . . . . . . . . . . . . . . 884
20.25A Theorem of Blokhuis . . . . . . . . . . . . . . . . . . . . . . 885
20.26Theorems of Bruen, Levinger and Carlitz . . . . . . . . . . . . 889
20.27Wedderburn’s Theorem . . . . . . . . . . . . . . . . . . . . . . 901
21 Appendix 3: Review of Sesquilinear Forms 903
21.1 Sesquilinear and Quadratic Forms . . . . . . . . . . . . . . . . 903
21.2 Perps and Radicals . . . . . . . . . . . . . . . . . . . . . . . . 908
21.3 Quotient spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 909
21.4 Polar Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 910
21.5 Hyperbolic Pairs . . . . . . . . . . . . . . . . . . . . . . . . . 914
22 Appendix 4: Interlacing of Eigenvalues 917
22.1 Real Symmetric Matrices: A Review . . . . . . . . . . . . . . 917

CONTENTS 11
Prolegomena
One of the most challenging open problems in finite geometry is the determination
of all ovoids in finite three dimensional projective space P G(3, q). In
1955 A. Barlotti [Ba55] and G. Panella [Pa55] (independently) showed that
if each plane section of an ovoid Ω is a conic, then Ω must be an elliptic
quadric. Since B. Segre had just proved his famous theorem that each oval
in the Desarguesian plane P G(2, q) with q odd must be a conic, it was known
that each ovoid of P G(3, q) with q odd must be an elliptic quadric. Not long
afterwards J. Tits [Ti62] discovered the examples known as the Tits ovoids
(or sometiimes the Suzuki-Tits ovoids because of their connection with the
Suziki groups) for q = 2 e , e odd. These two families remain the only known
ovoids in P G(3, q), and many researchers believe they will turn out to be
the only ones. However, additional improvements on the characterization
theorems were slow in coming. In 1977 Prohaska and Walker [PW77] proved
that an ovoid with even one bundle of conics must be an elliptic quadric.
Several years later D. Glynn [Gl84] succeeded in showing that an ovoid with
a pencil of conics must be an elliptic quadric. The computations in [Gl84]
have remained the central inspiration for several of the more recent papers by
T. Penttila and his coterie of co-authors, in which truly impressive progress
has been made. These papers have become, in our opinion, increasingly challenging
to read, especially because there is so much dependence on earlier
work with many details left to the reader. Possibly the biggest breakthrough
in recent years was the paper [Br00a] by M. Brown showing that even one
conic among the plane sections of an ovoid forces it to be an elliptic quadric.
One major goal of the present treatment is to give a self-contained exposition
of most of this material.
As progress on the problem of determining the ovoids in P G(3, q) was
being made, the study of ovals naturally played a hugh role, so that we
must include a major part of the theory of ovals in P G(2, q). However,
the theory of generalized quadrangles has also become increasingly central
to the study of ovoids. The appearance in 1984 of the monograph [PT84]
contributed to the growing interest in GQ, and although it continues to
provide a significant introduction to the theory, it has become quite out
of date. There are a few detailed surveys (see especially the chapter on
generalized polygons by J. A. Thas in [Bu95]) and the survey [JP97]. The
recent book by H. van Maldeghem [HVM98] is a good general reference, but
it does not really contribute much to the study of finite GQ. The book [KT04]

12 CONTENTS
by K. Thas does give proofs of many new results on finite GQ, but that book
gives relatively few results that relate to the problems that are central to
the study of ovoids. The manuscript [CP02] by Cardinali and Payne does
give a fairly complete treatment of flock GQ of even order, but it is written
for readers who are already somewhat familiar with finite GQ. Since many
readers of this book are likely to be new to the theory of GQ, and there are a
great many pertinent results that have not yet appeared with complete proofs
in book form, we have decided to present a fairly extensive introduction to
the theory of finite GQ.
It was our original expectation that the reader of this book would have
already studied the basics of finite projective and affine spaces and that our
primary contribution should be to present a careful and thorough treatment
of the many results that have appeared during the past twenty years and
have not yet found their way into the available texts. However, in using
this book with our students we have found it expedient to include a considerable
amount of fairly standard material. Moreover, at certain places
it is expedient to rely on a theorem of F. Buekenhout [Bu69] that gives a
synthetic axiomatic approach to finite affine spaces. This material is not
readily available in English (as far as we are aware) and relies on an earlier
axiomatization of affine space, so to be complete we have decided to include
a complete treatment of this material also.
So what are the prerequisites for reading this book? The answer is that,
for the reader who will make use of the appendices to become familiar with
some nearly classical material that is needed in the main chapters, only some
basic abstract algebra, linear algebra, and mathematical maturity are needed.
We have included appendices on basic number theory, equations over finite
fields, and sesquilinear forms. One of the theorems that plays a role in finite
geometry is the theorem of Wedderburn that characterizes finite skew-fields
as fields. In the chapter on finite fields we give Witt’s well known proof.
Chapters 1, 2 and 3 give a synthetic approach to finite projective and and
affine geometry and establish many of the standard results of finite projective
and affiine spaces, the so-called Galois geometries. These chapters might be
viewed as giving an entire first course on these topics. Then Chapter 4 gives
a fairly thorough introduction to the theory of ovals in P G(2, q). Chapter 5
presents the usual treatment of quadrics in finite projective spaces. Much of
the material in these first five chapters (and in the later chapter on the Klein
correspondence) is included in the major treatise by Hirschfeld and Thas
([Hi98], [Hi85] and [HT91]. However, we offer a more thorough and leisurely

CONTENTS 13
treatment of just that part of the general theory that seems relevant to our
overall goals.
In the other chapters we offer complete proofs of many results that were
discovered after the appearance of the references mentioned above. Perhaps
most notable is the inclusion of the complete solution of the problem of
showing that a GQ of order (q 2 , q) with Property (G) at a point must be a
flock GQ. The final step in this solution was given by M. R. Brown, inspired
by the deep and amazing work by J. A. Thas. We believe that the treatment
by Brown places the work of Thas within the grasp of a larger group of
readers.
Many of the very technical computations that were initially undertaken
in the study of ovoids have turned out to be useful in the study of finite
GQ. Our treatment of the (not completely solved) problem of determining
all spreads of T2(O) appears here for the first time in book form with fairly
complete proofs. Although great progress has been made, we hope that the
inclusion of this material will inspire a great deal more.
The present book grew out of notes written for a course by the same
name taught by the author during the Academic Year 2004 – 2005. I would
like to thank the following students and colleague whose constant vigilance
and enthusiastic participation contributed greatly to the current form of this
work: Art Busch, Bill Cherowitzo, Steve Flink, Oscar Jenkins, and Ryan
Pedersen. During these years that immediately follow the original seminar,
other students continue to make their own contributions to this work, and it
continues to evolve .
I would also like to thank my partner in life Angelika Adamic-Payne who
maintains a wonderful nurturing atmosphere in our home throughout the
times when I seem to be living off in my own imaginary world.
Stanley E. Payne

14 CONTENTS

Chapter 1
Synthetic Axioms for Geometry
In the usual synthetic approaches to Galois geometry (affine or projective
space over a finite field) there seems to be a tendency to treat projective
space synthetically and then derive affine space from it without offering a
corresponding synthetic foundation for affine space. In our study of generalized
quadrangles we have on occasion found it convenient to quote the
theorem of Buekenhout [Bu69] characterizing affine space with at least three
points on each line by means of a simple set of axioms. His proof is based
on a paper by Lenz [Le54] which is not quite self-contained. Hence we have
decided to offer here a more nearly complete synthetic treatment of both
affine and projective space. Our treatment owes much to the two references
just mentioned, along with the article [Ha72] by M. Hall, Jr., and the books
[BR98] by Beutelspacher and Rosenbaum (with its reference to [Tam72]) and
[WM69] by Murtha and Willard.
1.1 Linear Space
Definition 1.1.1. A linear space L is a set of points, along with a set of
distinguished subsets of these points, called lines, such that (1) each pair P, Q
of distinct points belongs to exactly one line, denoted P Q, and (2) each line
contains at least two points.
Definition 1.1.2. A set U of points of a linear space L is said to be a
subspace of L provided that for each pair P, Q of distinct points of U, each
point of the line P Q is contained in U. We may express this by saying
P Q ⊆ U.
15

16 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
The empty set, a single point, a line, and the entire space L are examples
of subspaces. The following lemma is self-evident.
Lemma 1.1.3. The intersection of any collection of subspaces is a subspace.
Definition 1.1.4. Let A1, A2, . . . , Ak be arbitrary sets of points. Then the
intersection of all subspaces containing the union of these sets is denoted
〈A1, A2, . . . , Ak〉 and is sometimes called the span or hull of A1, A2, . . . , Ak,
and can be thought of as the smallest subspace containing A1, A2, . . . , Ak. In
an abuse of notation, for a pointset A and a single point P , we let 〈A, P 〉
denote the span of A, {P }.
Note that a line P Q of a linear space is equal to 〈P, Q〉.
Definition 1.1.5. A subspace U of a linear space L is called a plane if there
exist three noncollinear (hence pairwise distinct) points P, Q, R ∈ L such that
U = 〈P, Q, R〉.
Definition 1.1.6. A projective plane is a linear space P having the following
properties: (1) every two lines meet in at least one point, and (2) there are
at least four points, no three collinear.
Exercise 1.1.6.1. Show that a projective plane is a plane.
Exercise 1.1.6.2. Show that a projective plane has at least three points on
each line.
Definition 1.1.7. An affine plane is a linear space πA having the properties:
(1) for each line q and each point P not on q, there exsits a unique line
containing P and disjoint from q, and (2) there are at least three noncollinear
points.
The affine plane having four points is shown below. It is known as the
affine plane of order 2 because it has two points on each line. One can
deduce from the definition of a plane, that the affine plane of order 2 is not
a plane.

1.2. AXIOMS FOR PROJECTIVE SPACE 17
It can be observed in the affine plane of order 2, that every set consisting
of a pair of distinct points is a subspace.
Exercise 1.1.7.1. Show if an affine plane πA has at least three points on
each line, then πA is a plane.
1.2 Axioms for Projective Space
Our axioms for projective space are written in terms of two types of objects
called ”points” and ”lines,” respectively, and prescribe rules for the ”incidence”
relation between them.
Definition 1.2.1. A projective space P is a set of objects of two types. An
object of the first type is called a point; one of the second type is called a line.
A point and a line may or may not be incident, and the incidence relation is
subject to the following axioms.
P1. Given two distinct points A and B there is a unique line ℓ incident with
both. We may write ℓ = AB or ℓ = 〈A, B〉, or ℓ = A + B.
P2. Let A, B, C, D be four distinct points. If there is a point incident with
both lines AB and CD, then there is a point incident with both lines
AC and BD.
Often Axiom P2 is expressed as follows: If A, B, C are distinct points
not on a line and if D = A is a point of AB, and if E = A is a point of
AC, then there is a point F common to DE and BC.

18 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
P3. Each line is incident with at least three points.
The next axiom is sometimes called the ”space axiom” since it rules out
the possiblity that the entire space is merely a plane. We will alert the
reader on those occasions when we want to assume this axiom, but often
we will not need it.
P4. (Space Axiom) There exist two lines that are incident with no point in
common.
When we do not wish to exclude the possibility that the entire space
is just a projective plane, we use the following weaker version of the
preceding axiom.
P5. There exist at least two lines.
Often it is the case that a line is considered to be the set of points incident
with it. However, there will be many times when we wish to construct pointline
incidence geometries starting with objects for which it is not convenient
to take this point of view. Usually we let I denote the incidence relation
and write AIℓ to indicate that the point A and the line ℓ are incident. Alternatively,
the pair (A, ℓ) is said to be a flag provided AIℓ. Later we will
generalize this notion of flag.
Definition 1.2.2. A set U of points of P is said to be a subspace of P
provided that for each pair (A, B) of distinct points of U, each point of the
line AB is contained in U. We may express this by saying AB ⊆ U. In
spite of our remark above about a line sometimes being distinct from the set
of points incident with it, in general a line, considered as a subspace of P, is
indeed identified with the set of points incident with it.
The following lemma is self-evident.
Lemma 1.2.3. The intersection of any collection of subspaces is a subspace.
Definition 1.2.4. If {A1, A2, . . . , Ak} is any set of pointsets, the intersection
of all subspaces containing this set is denoted A1, A2, . . . , Ak, or 〈A1, . . . , Ak〉
or A1 + A2 + · · · + Ak. It is the subspace spanned by these points and is
sometimes called the hull of the set of pointsets or the simply their span.

1.2. AXIOMS FOR PROJECTIVE SPACE 19
Lemma 1.2.5. Let U be a subspace of P and P a point not in U. The span
UP is the set of points incident with the lines UP for all U ∈ U. Moreover,
each line of UP meets U.
Proof. Let A and B be any two points of UP . We need to show that AB ⊆
UP . This is clear if AP = BP , so suppose that this is not the case. There
must be points C and D in U such that CP = AP and DP = BP . By
hypothesis any point of CD is in U. Let E be any point of AB \ {A, B}. We
need to show that P E contains a point of U. To do this we apply Axiom 2
three times.
CA ∩ DB = P =⇒ CD ∩ AB = F ∈ U.
CD ∩ AE = F =⇒ CA ∩ DE = G.
CP ∩ DE = G =⇒ CD ∩ P E = H ∈ U.
By now it is clear that each line of UP meets U.
Corollary 1.2.6. Let U be a subspace of a projective space P, and let P be
a point of P that is not contained in U. For any point Q ∈ U, the span
〈U, P 〉 of U and P consists precisely of the points in the planes 〈g, P 〉, where
g runs over all lines of U through Q.
Proof. Easy exercise.
Corollary 1.2.7. Let g1 and g2 be two skew lines of a projective space P.
Let X1, Y1 be points on g1, and X2, Y2 be points on g2. Let X be a point
on 〈X1, X2〉 and Y a point on 〈Y1, Y2〉 such that 〈X, Y 〉 is skew to g1 and g2.
Then each point Z on 〈X, Y 〉 is on some line (transversal) that intersects g1
and g2.
Proof. Exercise. (Hint: Show that the line g2 intersects the plane π :=
〈g1, Z〉.)
Corollary 1.2.8. (The ’Join theorem’) Let U and V be nonempty subspaces
of P. then the span 〈U, V〉 can be described as follows:
Proof. Exercise.
〈U, V〉 = ∪{〈P, Q〉 : p ∈ U, Q ∈ V, P = Q}.

20 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
1.3 The Axioms of Affine Space
Affine space is linear space in which certain parallel lines exist. The following
axiomatization of affine space is due to Lenz.
Definition 1.3.1. An affine space A is a set of points in which there are
distinguished subsets called lines and an equivalence relation denoted defined
on the set of lines, in such a way that the following axioms are satisfied.
A1. Each pair P, Q of distinct points is contained in a unique line P Q.
A2. For each point P and each line ℓ there is a unique line ℓ ′ such that P ∈ ℓ ′
and ℓℓ ′ .
Of course if the point P is incident with ℓ, then ℓ itself is the unique
line through P parallel to ℓ. This means that two lines that meet in a
single point cannot be parallel.
A3. (Trapezoid axiom) Let P Q and RS be distinct parallel lines and let T
be a point of P R \ {P, R}. Then there must be a point incident with
P Q and T S.
P
R
T
Q
S
Another way to express the Trapezoid axiom is the following: If P , R
and S are three noncollinear points, and T is a third point on the line
P R, then the line through P parallel to RS meets the third side T S.
A4. (Parallelogram axiom) If no line of A has more than two points, and if
P , Q and R are three distinct points, then the line through R parallel
to P Q must have a point in common with the line through P parallel
to QR. See the following diagram.
S
P
T
Q
R

1.3. THE AXIOMS OF AFFINE SPACE 21
Q
P
R
Remark: If the lines all contain at least three points, the Trapezoid
axiom and the Parallelogram axiom are equivalent. Hence in this case
we omit the Parallelogram axiom.
A5. Each line contains at least two points.
A6. (Space axiom) There exist two disjoint lines ℓ and ℓ ′ such that ℓ ℓ ′ .
This axiom excludes affine planes from the affine spaces.
A7. (Triangle axiom) Let P, Q, R be three noncollinear points and let P ′ , Q ′
be points such that P QP ′ Q ′ . If ℓ is the line through P ′ parallel to
P R and ℓ ′ is the line through Q ′ parallel to QR, then ℓ intersects ℓ ′ in
a point R ′ . This axiom is due to O. Tamaschke and may be used to
replace both trapezoid and parallelogram axioms.
P Q P ′
R
Exercise 1.3.1.1. Show the triangle axiom holds if and only if the trapezoid
and parallelogram axioms hold.
ℓ
ℓ ′
R ′
Q ′

22 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
Lemma 1.3.2. All lines in an affine space are incident with the same number
of points.
Proof. If no line has more than two points we are done. Otherwise, let g be
a line with three distinct points A, B and C and let D be an arbitrary point
not on g. By the trapezoid axiom the line AD must meet the line through
C parallel to the line BD in a point E. By varying the point C on AB we
see that AD has at least as many points as AB. Interchanging the roles of
AB and AD shows that any two lines that meet have the same number of
points. If lines g and h are parallel and A is a point of g, B is a point of h,
then both g and h have the same number of points as AB.
Lemma 1.3.3. (The parallelogram theorem) Let A, B and C be three distinct
points. Then the line through C parallel to AB has at least one point in
common with the line through B parallel to AC.
Proof. If A, B and C are on one line, the proof is obvious. So suppose this
is not the case. If each line has exactly two points, the result follows from
the parallelogram axiom. Otherwise, by Lemma 1.3.2 the line AB has a
third point P . By the Trapezoid axiom, the line P C meets the line through
B parallel to AC in a point D. Then DB must meet the parallel to P B
through C.
Lemma 1.3.4. If the lines AB and CD have a point in common or are
parallel, then the lines AC and BD have a point in common or are parallel.
Proof. If three of the points A, B, C and D lie on a line then result is trivial.
Suppose otherwise.
(i) ABCD. According to Lemma 1.3.3, the line CD has a point E in
common with the line through B parallel to the line AC. If E = D we
are done since that gives ACBD. Otherwise by the Trapezoid axiom,
the lines AC and BD have a point in common.
(ii) AB ∩ CD = {S}. By the Trapezoid axiom the line through B parallel
to AC has a point E in common with the line SC = CD. If E = D,
then ACBC. Otherwise, the line through C parallel to BE (i.e., the
line AC) has a point in common with BD.

1.3. THE AXIOMS OF AFFINE SPACE 23
Lemma 1.3.5. Let U be a nonempty subspace; let P a point not in U; and
let O be a point of U. The subspace 〈U, P 〉 consists of exactly the points lying
on lines that are parallel to OP and are incident with a point of U.
Proof. let M be the set of points on parallels to OP through points of U.
We will first show that M is a subspace. Let U1, U2, . . . be points of U.
(i) Each line contains at least three points. Let A1, A2 be distinct points
of M with A1U1A2U2OP . By Lemma 1.3.4 either A1A2 and U1U2
are parallel or they meet in a point S. Let A be a third point of A1A2.
Let ℓ be the line through A parallel to A1U1 so parallel to OP. If A1A2
and U1U2 meet in a point S, which must be in the subspace U, then
the line ℓ meets one side of the triangle (A1, U1, S) and is parallel to
a second, so it meets the third, i.e., ℓ meets SU1 in a point U2 which
must be in U. So A is in M by definition.
If A1A2U1U2, then by Lemma 1.3.3 ℓ meets U1U2, putting A in M.
(ii) Every line has exactly two points. Let A, B, C, D be any four points
(so automatically no three are on a line). Suppose that ABCD. Then
ACBD and ADBC. For example, by the Parallelogram axiom the
line through A parallel to BD must meet the line CD. But it cannot
meet at D, since it is parallel to BD and contains the point A not on
BD. Hence it must meet at C, so ACBD.
Suppose that A1U1A2U2A3U3OP and A3AA1A2. Then U1U2A1A2A3A.
There must be a point U ∈ U such that U3UU1U2, which forces
U3UA3A and hence AUA3U3. This says A ∈ M.
The remaining case is that U1A1OP and U2AU3A1, implying A1AU3U2.
There must be a point U ∈ U such that UU1||U2U3||AA1. It follows that
UAU1A1OP , completing the proof.
We have shown that M is a subspace and clearly it contains the points of
U and {P } so that we know 〈U, P 〉 ⊆ M. We must argue that M ⊆ 〈U, P 〉.
Any point of a line OP , O ∈ U is in 〈U, P 〉 by definition of a subspace. Now
consider a point X of a parallel line ℓ to OP where ℓ meets U in some point
U. If U has only one point, then there are no such X’s and we are done. If
U has only two points, then it must be that each line has exactly two points
and we have a problem. Otherwise everything’s ok.

24 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
***********************************************
Lemma 1.3.6. Let U be a nonempty subspace of an affine space A that has
at least three points on each line; let P a point not in U; and let O be a point
of U. The subspace 〈U, P 〉 consists of exactly the points lying on lines that
are parallel to OP and are incident with a point of U.
Proof. let M be the set of points on parallels to OP which contain a point
O of U. We will first show that M is a subspace. Let A1, A2 be distinct
points of M. Let A be a point of A1A2 and we wish to show that A ∈ M.
Since A1, A2 ∈ M, there exist points O1, O2 ∈ U such that A1O1 and A2O2
are both parallel to OP . By Lemma 1.3.4, either A1A2 and O1O2 are parallel
or they meet in a point S which must be in U since O1O2 ⊆ U. Let ℓ be the
line through A parallel to A1O1 so parallel to OP. If A1A2 and O1O2 meet
in the point S, then the line ℓ meets one side of the triangle (A1, O1, S) and
is parallel to a second, so it meets the third; i.e., ℓ meets O1S in a point O3
which must be in U. So A is in M by definition.
P
O
A1
A2
O1 O2 O3
If A1A2O1O2, then by Lemma 1.3.3, ℓ meets O1O2, putting A in M.
Now we must show that M is the smallest subspace containing U and
P . We will show that every point X of M is necessarily in every subspace
containing U and P . Fix a point O of U and let O ′ be the point where the
parallel line to P O through X meets U. Consider a third point Q of P O.
Then by the trapezoid axiom, QX meets OO ′ in a point O ′′ which must be
in U. Thus the line QO ′′ , which contains X, is in every subspace containing
P and U.
U
ℓ
A
S

1.4. PLANES IN AFFINE SPACE 25
In the case that an affine space A has two points on each line, we make
an observation. Let A, B, C, D be any four points (so automatically no
three are on a line). Suppose that ABCD. Then ACBD and ADBC.
For example, by the Parallelogram axiom the line through A parallel to BD
must meet the line CD. But it cannot meet at D, since it is parallel to BD
and contains the point A not on BD. Hence it must meet at C, so ACBD.
********************************************
1.4 Planes in Affine Space
Exercise 1.4.0.1. Show disjointness of lines in a plane is transitive.
Exercise 1.4.0.2. In an affine space A, show two distinct lines ℓ1, ℓ2 of A
satisfy ℓ1ℓ2 if and only if (ℓ1 ∩ ℓ2 = ∅ and ℓ1, ℓ2 coplanar).
Exercise 1.4.0.3. Consider an affine space A. Show that if the lines have
more than two points, then the plane generated by three noncollinear points
of A is an affine plane.
Including the space axiom among the axioms for affine space would preclude
the entire space being an affine plane. However for now we now we
want to study subspaces that are planes, so we omit the space axiom. There
are several ways of constructing (or generating) a plane as a subspace of an
affine space, but we choose the following after M. Hall, Jr. [Ha72]. Note that
it will turn out that a plane given by this construction is an affine plane
according to Def. 1.1.7.
Definition 1.4.1. Given a line m and a point P not on m, a plane π =
π(P, m) is the set of points on lines P S where S is a point of m together with
the points on the line k through P and parallel to m.
We will show that π(P, m) is a subspace independent of the particular
point-line pair used to define it.
Lemma 1.4.2. If Q ∈ π(P, m) and Q ∈ m, then π(Q, m) = π(P, m).
Proof. There are two cases to be considered: (i) Q is on a line P S, S ∈ m,
and (ii) Q is on k, the paralle to m through P . Of course, if P = Q there is
nothing to prove.

26 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
(i) Suppose Q is on a line P S with S ∈ m. Let W be a point on a line QT
with T ∈ m. If T = S, then W is a point on QP S and so is a point
of π(P, m). If T = S, then Q, T, S are three points not on a line, and
by Lemma 1.2.5 the line P W either intersects the line ST , which is m,
in a point R or is the parallel to m throughP . In either case W is a
point of π(P, m). Thus all points on lines QT with T ∈ m are points
of π(P, m). Consider a point Y = Q on the parallel to m through Q.
P , Q and Y are three noncollinear points and T = S is a point of QP
different from Q. Then m is the line through T = Q parallel to QY , so
it meets the third side Y P of the triangle P QY in a point E (by A3).
As E ∈ m, Y (as a point of P E) is a point of π(P, m). Hence each
point of π(Q, m) is a point of π(P, m).
(ii) Suppose Q is on k, the parallel to m through P . Here, of course, every
point of k is a point of π(P, m). Let T be a point on a line QT S with
S ∈ m. Then P, Q, T are three points not on a line, and by A3, m (the
parallel to P Q through S) intersects P T in a point W . Hence T is a
point of the line P W with W ∈ m so is a point of π(P, m). So also in
this case every point of π(P, m) is a point of π(P, m).
If Q is a point of π(P, m) and Q ∈ m, it is immediate that P is a point
of π(Q, m), and P ∈ m. Thus interchanging the roles of P and Q, the above
argument shows that every point of π(P, m) is a point of π(Q, m). Hence
π(Q, m) = π(P, m), as claimed by the lemma.
Lemma 1.4.3. If R = S are two distinct points of π(P, m), then every point
of the line RS is a point of π(P, m).
Proof. If both R and S are on m, then the line RS is the line m and the
conclusion holds. If, say, R ∈ m, then by Lemma 1.4.2, π(P, m) = π(R, m).
With S ∈ π(R, m), either S is on a line RT with T ∈ m or RS is the parallel
to m through R. In both cases all points of RS are points of π(R, m) =
π(P, m).
Lemma 1.4.4. If k is a line of π(P, m) intersecting m in a point A, then
there is a point Q of π(P, m) not on m and not on k.
Proof. If P ∈ k, take Q = P . If P ∈ k, take Q = P on the parallel to m
through P .

1.4. PLANES IN AFFINE SPACE 27
Lemma 1.4.5. Let k be a line of π(P, m) intersecting m in a point B. Let
Q be a point of π(P, m) not on m and not on k. Then π(P, m) = π(Q, m) =
π(Q, k).
Proof. As Q ∈ m, π(P, m) = π(Q, m) from Lemma 1.4.2. Let C be a point on
k different from B. Since C ∈ π(Q, m), either QC intersects m in a point S or
QC is the parallel to m through Q. Hence by axiom A3 or Lemma 1.3.3(the
parallelogram theorem) the parallel to k through Q intersects m in a point
T . Thus by Lemma 1.4.3 all points of QT are points of π(Q, m). Similarly,
all points on lines QS with S ∈ k are points of π(Q, m). Thus all points of
π(Q, m) are points of π(Q, m). By the same argument, all points of π(Q, m)
are points of π(Q, k).
Theorem 1.4.6. If k is a line of π(P, m) and Q is a point of π(P, m) not
on k, then π(P, m) = π(Q, k).
Proof. If k and m intersect in a point B, then by Lemma 1.4.4 there is a
point T not on k and not on m. Hence by Lemmas 1.4.2 and 1.4.5 we have
π(P, m) = π(T, m) = π(T, k) = π(Q, k). If k and m do not intersect, choose a
point B on m and a point C on k, and let BC be a line r. Let S be a point not
on m or r, and let V be a point not on r or k. Then by Lemmas 1.4.2 and 1.4.5
we have π(P, m) = π(S, m) = π(S, r) = π(V, r) = π(V, k) = π(Q, k).
From here on we may speak of a plane π without reference to the particular
point-line pair (P, m) used to define it. Such a plane is an affine plane.
It should be clear that if π is a subspace of an affine space which is an affine
plane by Def. 1.1.7, then π = π(P, m) for any point-line pair (P, m) with P
not on m.
Theorem 1.4.7. If k and m are two lines of a plane π then either k and m
intersect or k and m are parallel.
Proof. Choose two distinct points P and Q on k. If either of these is on m,
then k and m intersect. If not, then π = π(P, m). If Q is on the parallel to
m through P , then P Q = k is parallel to m. Otherwise Q is on a line P S
with S ∈ m and k = P Q = P S intersects m in S.
It is worth noting that there is one and only one affine plane containing
two intersecting lines and one and only one affine plane containing two parallel
lines. For if we call one of the lines m and take P as a point of the other
line but not on m, then both lines lie in π(P, m).

28 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
The primary reason we needed to include a synthetic approach to affine
space in this book is that on various occasions we make use of a famous
result of F. Buekenhout [Bu69] giving an alternative approach to those affine
spaces in which each line has at least four points. Hence we devote the next
section to this result.
1.5 Buekenhout’s Characterization of Affine
Space
This section is based on the paper [Bu69] by F. Buekenhout. It is wellknown
(implicitly given in Veblen and Young [VY38]) that the projective
spaces can be characterized among the linear spaces by the fact that each of
their planes is a projective plane. One can ask whether or not an analogous
result holds for affine spaces, i.e., is a linear space all of whose planes are
affine planes necessarily an affine space? Marshall Hall, Jr. [Ha60] gave an
example of a Steiner triple system on 81 points whose triangles generate an
affine plane of of 3 2 = 9 points but which is not an affine space. (These Hall
Triple Systems have been greatly generalized and thoroughly studied. See
[Be80].) However, the main theorem of this section is a celebrated result of
F. Buekenhout [Bu69] showing that the answer to the question just posed
is affirmative when the lines of the space considered have more than three
points. Our proof assumes that the order of the underlying space is finite.
Theorem 1.5.1. Let L be a linear space such that
(i) each plane of L is an affine plane,
(ii) L has at least three noncollinear points, and
(iii) each line of L has at least four points.
Then L is an affine space.
Proof. We will first define a relation on the set of lines of L and then
prove that it is an equivalence relation. Once that is accomplished we will
show that L satisfies the axioms of affine space given our hypotheses and
equivalance relation .
Define the parallel relation by two lines ℓ1, ℓ2 ∈ L are parallel and we
write ℓ1ℓ2 if and only if either (ℓ1 ∩ ℓ2 = ∅ and ℓ1, ℓ2 coplanar) or (ℓ1 = ℓ2).

1.5. BUEKENHOUT’S CHARACTERIZATION OF AFFINE SPACE 29
That the relation is reflexive and symmetric follows immediately from the
definition of . We must show that is transitive on the set of lines of L. So
suppose that ℓ1ℓ2 and ℓ2ℓ3. We have several possibilities to investigate.
I. If ℓ1 = ℓ2 and ℓ2 = ℓ3, then clearly ℓ1 = ℓ3, implying ℓ1ℓ3.
II. If ℓ1 = ℓ2 and (ℓ2 ∩ ℓ3 = ∅ and ℓ2, ℓ3 coplanar), then clearly (ℓ1 ∩ ℓ3 = ∅
and ℓ1, ℓ3 coplanar), implying ℓ1ℓ3.
III. If (ℓ1 ∩ ℓ2 = ∅ and ℓ1, ℓ2 coplanar) and ℓ2 = ℓ3, then clearly (ℓ1 ∩ ℓ3 = ∅
and ℓ1, ℓ3 coplanar), implying ℓ1ℓ3.
IV. For the case of (ℓ1 ∩ ℓ2 = ∅ and ℓ1, ℓ2 coplanar) and (ℓ2 ∩ ℓ3 = ∅ and
ℓ2, ℓ3 coplanar), we have two subcases.
A. Suppose ℓ1, ℓ2, and ℓ3 all reside in the same plane. For sake of
contradiction, suppose that ℓ1 ∩ ℓ3 = ∅. If ℓ1 = ℓ3 then we are done
so we may suppose by the definition of linear space that ℓ1 ∩ ℓ3 =
{P } for some point P . Then we contradict the uniqueness of the
line through P disjoint from ℓ2. Therefore, ℓ1 ∩ ℓ3 = ∅, implying
that ℓ1ℓ3. Transitivity of disjointness of lines in an affine plane is
important: we shall call it Property I and cite it often.
ℓ2
ℓ1
ℓ3
B. Now suppose ℓ1, ℓ2, and ℓ3 are not coplanar. Here we need to prove
a strong result of a certain structure to proceed.
Let π be a plane of L and let ℓ be a line of L that contains a unique point
R ∈ π. Let V be the union of the planes that contain both (1) ℓ and (2) a
line of π passing through R. We will show that V is a subspace of L.
P

30 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
π
V
R
Note that it is not clear that a plane containing ℓ meets π in a line so we
must proceed with care. First we introduce some notation. For each point
P ∈ V such that P ∈ ℓ, we denote by ℓ(P ) the point of intersection of ℓ and
the parallel m to the line n = 〈ℓ, P 〉 ∩ π through P , where 〈ℓ, P 〉 denotes
the plane generated by ℓ and P . Since each plane is an affine plane in L,
we know that this line m exists. Furthermore, m meets ℓ since otherwise m
coplanar and disjoint to both ℓ and n implies by Property I that ℓ ∩ n = ∅, a
contradiction. If Q ∈ ℓ, R = Q = ℓ(P ), then the line P Q meets n in a point
designated by Q(P ). Clearly Q(P ) = R, else P ∈ ℓ.
π
n
R
ℓ
P ℓ(P )
ℓ
Q
m
Q(P )

1.5. BUEKENHOUT’S CHARACTERIZATION OF AFFINE SPACE 31
To show that V is a subspace, take arbitrary distinct points X and Y of
V and we will show that XY ⊆ V. We will case on whether or not XY is
coplanar with ℓ.
1. Suppose that X, Y , and ℓ are coplanar in some plane π ′ . Since X and Y
are in V, they have corresponding planes πX and πY that contain ℓ and
meet π in a line containing R. Now since X and ℓ span a unique plane
πX, πX = π ′ . Similarly, πY = π ′ . Thus every point of XY is in π ′ which
is a plane of V since it contains ℓ and a line of π through R.
2. Now suppose that X, Y , and ℓ are not coplanar. Since each line has
at least four points, ℓ has some point Q with Q = ℓ(X), Q = ℓ(Y ),
and Q = R. Construct the points Q(X) and Q(Y ) as discussed in the
notation above. Now Q(X) = Q(Y ) else ℓ and XY would be coplanar.
Furthermore, we have the plane 〈X, Y, Q〉 = 〈Q, Q(X), Q(Y )〉.
Y
π
X
R
ℓ
Q
Q(X)
Q(Y )
Note that XY and Q(X)Q(Y ) are coplanar. We case on whether they
meet or not.
a. Suppose XY ∩ Q(X)Q(Y ) = ∅. Take arbitrary point Z ∈ XY . We will
show that Z ∈ V. Now ZQ is in the plane 〈X, Y, Q〉. Thus ZQ either
meets Q(X)Q(Y ) or is the line through Q disjoint from Q(X)Q(Y ) which
cannot be the case from Property I. Therefore ZQ meets Q(X)Q(Y ) in the
point Q(Z). Now R Q(Z) is a line in π contained by the plane 〈ℓ, Q(Z)〉
and this plane also containes Z. Therefore, Z ∈ V.

32 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
Y
π
Z
X
R
ℓ
Q
Q(X)
Q(Y )
Q(Z)
b. Suppose XY is not disjoint to Q(x)Q(Y ) so that they meet in a point
S ∈ π (since Q(x)Q(Y ) ⊆ π). S = R since otherwise XY and ℓ would
be coplanar. Let q be the parallel line of Q(X)Q(Y ) passing through Q.
Since XY and Q(X)Q(Y ) meet, so do q and XY in some point T . We
will first show that if a point of XY does not belong to V, it must be the
point T . Then we will show that T ∈ V.
π
X
Y
T
S
ℓ
R
q
Q
Q(X)
Q(Y )
Consider Z ∈ XY such that Z = T . Then ZQ intersects Q(X)Q(Y ) in
some point Q(Z). As Q, Q(Z), and ℓ are coplanar in a plane containing

1.5. BUEKENHOUT’S CHARACTERIZATION OF AFFINE SPACE 33
the line R Q(Z) ⊆ π, we have Z ∈ V. Now the difficult work consists of
showing that T ∈ V . We case on two configurations: whether ℓ(X) equals
ℓ(Y ) or not.
i. Suppose that ℓ(X) = ℓ(Y ). First we will show that for any Z ∈ XY
such that Z ∈ V and Z = S, that ℓ(Z) = ℓ(X) = ℓ(Y ). For sake of
contradiction, suppose ℓ(Z) = ℓ(X). Consider the line ℓ(X) Z. Since
Z ∈ V, there is some plane πZ containing Z, ℓ, and a line b of π. Note
that both ℓ(X) Z and ℓ(Z) Z are lines of πZ. By definition of ℓ(Z),
ℓ(Z) Z is the unique line in πZ parallel to b so that ℓ(X) Z must meet b
(and hence π) in some point Z ′ .
π
Z ′
X
b
Y
Z
S
ℓ
ℓ(X) = ℓ(Y )
ℓ(Z)
R
Q(Y )
Q(X)
Note Z ′ = S else XY and ℓ coplanar. Now examine the plane 〈ℓ(X), Z, S〉
and see that it must contain a line of π, namely SZ ′ because both S and
Z ′ are in this plane and both on π. Now the plane 〈ℓ(X), Z, S〉 also
contains X and Y since they are on the line ZS. Thus at least one of
X ℓ(X), Y ℓ(X) must meet the line SZ ′ , in particular in a point of π.

34 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
(Now here we will assume that not both X and Y lie in π; if they did,
then XY is trivially within V. Therefore, since ℓ(X) = ℓ(Y ), neither
of X, Y lie in π.) Suppose the line that meets SZ ′ is X ℓ(X). But this
contradicts the definition of ℓ(X) since this line is parallel to the line
of the plane of V within π and X does not lie in π. Similarily, Y ℓ(X)
does not meet SZ ′ . So we have shown that ℓ(X) = ℓ(Y ) = ℓ(Z) for
Z ∈ V ∩ XY whenever ℓ(X) = ℓ(Y ) and Z = S.
We now examine the plane 〈X, Y, ℓ(X)〉 and note that if it contained
two points of π, then it would contain a line of π; and one of the lines
X ℓ(X), Y ℓ(X) would again have to meet this line of π, a contradiction
as before. So we conclude that the plane 〈X, Y, ℓ(X)〉 contains the single
point S of π, which is where the line XY meets π.
Let X ′ ∈ X ℓ(X) with X ′ = X and X ′ = ℓ(X). We wish to show that
X ′ Y ⊆ V. Suppose not, then X ′ Y is not coplanar to ℓ. Also, we have
that X ′ Y meets π in a point S ′ of Q(X ′ )Q(Y ). Also, it must be that
S ′ = S since 〈X, Y, ℓ(X)〉 ∩ π = {S} and the preceding paragraph. But
this implies X = X ′ , an impossibility. Therefore X ′ Y ⊆ V.
π
X
X ′
Y
ℓ
ℓ(X) = ℓ(Y )
R
S
S ′ ∈ π
Q(Y )
Q(X)
There are at least two distinct points X ′ , X ′′ of X ℓ(X) different from X
and ℓ(X). Thus both X ′ Y and X ′′ Y are contained in V ∩ 〈X, Y, ℓ(X)〉
and neither line contains ℓ(X). If W is a point on one of these lines (so

1.5. BUEKENHOUT’S CHARACTERIZATION OF AFFINE SPACE 35
W ∈ V), then ℓ(X) W is contained in V, this being the unique plane of
V corresponding to W .
Each line of the plane 〈X, Y, ℓ(X)〉 passing through ℓ(X) meets one of
the lines X ′ Y , X ′′ Y (by Property I) in a point W . Since W is in V,
this line through ℓ containing W is also in V, lying in W ′ s plane of V.
Recall that the only point of XY possibly not in V is T . Consider the
line T ℓ(X) ⊂ 〈X, Y, ℓ(X)〉. T is on one of these lines through ℓ meeting
X ′ Y or X ′′ Y . Thus T ∈ V and we are done with this case.
ii. Suppose that ℓ(X) = ℓ(Y ). We will first show that for each point Q ′
of ℓ such that Q ′ = R, there is some point Z ′ ∈ XY such that Z ′ ∈ V,
ℓ(Z ′ ) = Q ′ , and Z ′ = S. XY has at least four points. We now argue that
at most three of these points will not be satisfactory for the selection of
Q ′ . First, the only point of XY that might not be in V, is T . Second,
we may not use S. Third, given the choice of the remaining two points
(there may be more) on XY , at most one of them, call it X ′ , in this case
may have the property that ℓ(X ′ ) = Q.
Then Q(Z ′ ) is defined in π where Z ′ ∈ XY , Z ′ ∈ V, ℓ(Z ′ ) = Q and
Z ′ = S. Note from the argument in the previous paragraph that it is
entirely possible for Z ′ to be either X or Y . Consider the line c = S Q(Z ′ )
which is the intersection of π and 〈Q, X, Y 〉, i.e., c = Q(X)Q(Y ). In
〈Q, X, Y 〉, by Property I, the parallel to XY passing through Q meets c
in a point Q ′ .

36 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
S
c
Z ′
Y
X
Q ′
ℓ
R
ℓ(X)
ℓ(Y )
Q
Q(X)
Q(Z ′ )
Q(Y )
Consider any line d of π passing through R. There are three possibilities
for d: (1) d meets c in a point other than Q ′ , (2) d meets c in Q ′ , or (3)
d is disjoint to c. We are interested in the planes 〈d, ℓ〉. We will study
these planes to see if they meet XY . In particular, if any one contains
T (implying T ∈ V), then we have XY ∈ V.
(1) Suppose d meets c in a point distinct from Q ′ . Note that the lines
(d ∩ c) Q and XY lie in the plane 〈X, Y, Q〉 so that they meet in a unique
point of XY (using Property I and the fact that Q ′ is where the parallel
to XY meets c). Thus the plane 〈ℓ, d〉 contains a unique point of XY ,
namely (d ∩ c) Q ∩ XY . Therefore this type of plane always meets XY .
π

1.5. BUEKENHOUT’S CHARACTERIZATION OF AFFINE SPACE 37
〈ℓ, d〉 ∩ XY
S
Y
c
Z ′
X
Q ′
ℓ
R
Q(X)
Q(Y )
Q(Z ′ d ∩ c
)
d
(2) Denote by α(Q) the plane 〈d, ℓ〉 where d meets c in the point Q ′ . Note
that this plane and 〈Q, XY 〉 are distinct planes (otherwise ℓ and XY
coplanar) whose intersection is the line QQ ′ . A point in common to the
plane 〈d, ℓ〉 and to XY must belong to the line QQ ′ which is disjoint to
XY . Thus the plane α(Q) does not meet XY .
π

38 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
S
Y
c
Z ′
d
X
Q ′
ℓ
R
Q(X)
Q(Z ′ )
Q(Y )
(3) Denote by β(Q) the plane 〈ℓ, d〉 where d ∩ c = ∅. If β(Q) has a point
W on XY , then we are going to show that W is necessarily the point T ;
then XY ⊆ V. For, if W is not on the line disjoint to c passing through
Q in plane 〈X, Y, Q〉, then the line QW meets c in a point J. This point
J is common to β(Q) so J ∈ β(Q) ∩ π implying J ∈ d. But in this case,
d and c are disjoint so we have a contradiction. Therefore, if β(Q) has
a point W on XY , then it is necessarily the point T . If this is the case,
then we have shown a construction for T putting it in V.
Q
π

1.5. BUEKENHOUT’S CHARACTERIZATION OF AFFINE SPACE 39
Y
Z ′
d
X
Q(X)
Q(Z ′ )
Q(Y )
c
S π
We now have the tools to show that XY ∈ V. For suppose XY is not
contained in V. Then our point T is not in mV . Then we do not have a
plane of type β(Q) meeting XY . We now establish a contradiction. Let Q1,
Q2, and Q3 be three distinct points of ℓ \ {R}. We know that the planes
α(Qi), i = 1, 2, 3 are disjoint from XY and we must assume that the planes
β(Qi), i = 1, 2, 3 are also disjoint from XY (else T in V). Since we are
assuming that the α and β planes of Q1 and Q2 do not meet XY , it must
be that these planes equal each other in one of the following two fashions.
(Recall that all other planes of Q1 and Q2 meet XY as described in Case (1)
above.)
(1) α(Q1) = α(Q2) and β(Q1) = β(Q2). Note that by Property I, β(Q1) =
β(Q2) implies Q1(X)Q1(Y ) and Q2(X)Q2(Y ) are parallel in π. Since
both Q1(X)Q1(Y ) and Q2(X)Q2(Y ) meet π in the point S, they must
be the same line. A couple of contradictions are immediate: (1) D and
A would then be coplanar, or (2) Q1 would have to equal Q2.
Q ′
ℓ
R
Q

40 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
Q ′ 2
Q ′ 1
X
Y
ℓ
α(Q1) = α(Q2)
β(Q1) = β(Q2)
Q1(X)Q1(Y )
Q2(X)Q2(Y )
(2) α(Q1) = β(Q2) and β(Q1) = α(Q2). Consider point Q3. By the previous
case we have α(Q3) = β(Q1) and α(Q3) = β(Q2), implying β(Q1) =
β(Q2), a contradiction.
Therefore, some plane β(Qi), i = 1, 2, 3, meets XY . So the point T on
XY is constructed and an element of V. Finally, we see that V as constructed
above must be a subspace of L.
Recall that we have one last case to show concerning the transitivity of .
We are assuming that ℓ1, ℓ2 coplanar, ℓ2, ℓ3 coplanar, not all three coplanar,
ℓ1 ∩ ℓ2 = ∅, and ℓ2 ∩ ℓ3 = ∅. We need to show that ℓ1 ∩ ℓ3 = ∅ and ℓ1, ℓ3
coplanar.
π

1.5. BUEKENHOUT’S CHARACTERIZATION OF AFFINE SPACE 41
A
ℓ1
ℓ2 = ℓ
C
P
B
ℓ3
π = 〈A, B, C〉
For sake of contradiction, suppose that ℓ1 and ℓ3 meet in some point P .
Choose points A ∈ ℓ1, C ∈ ℓ2, and B ∈ ℓ3 such that A = P , B = P , and
C = P . To form our subspace V, let π = 〈A, B, C〉, ℓ = ℓ2, and note that C
is the point where ℓ meets π. Now 〈A, C, P 〉 is a plane containing the line
AC of π, the point C of π, and the line ℓ. Therefore we know P ∈ V. But
P is in precisely one plane of V containing ℓ, the point C, and a line of π
through C. Therefore since P is also in the plane 〈B, C, P 〉 which meets this
criterian, the planes 〈A, C, P 〉 and 〈B, C, P 〉 must be equal, a contradiction of
the hypothesis that not all three of ℓ1, ℓ2, ℓ3 are in the same plane. Therefore
ℓ1 and ℓ3 are disjoint.
Now we will show that ℓ1, ℓ3 coplanar. Suppose that they are not. Choose
a point R of ℓ3. Since ℓ1 ∩ ℓ3 = ∅, we can let π be the plane 〈ℓ1, R〉. For this
configuration we know that V is a subspace. Now ℓ2 is coplanar to ℓ and is in
the plane 〈ℓ2, ℓ〉 of V. In particular, there is a line g of π where 〈ℓ2, ℓ〉 meets
π. By Property I, some point of ℓ2, call it X, must lie on this line g; else g
is disjoint to ℓ2 in 〈ℓ1, ℓ〉 and hence disjoint to ℓ, contradicting the fact that
g contains R.

42 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
g
ℓ2
ℓ3 = ℓ
X π
Note that the plane π contains ℓ1 as well as the point X of ℓ2. We conclude
that ℓ2 must reside in π, else ℓ1, ℓ2 are not coplanar (ℓ1 and any point of ℓ2
will generate the plane that they share since they are disjoint in this plane).
Therefore π is 〈ℓ2, ℓ〉, containing all three of ℓ1, ℓ2, ℓ3. This contradicts the
hypothesis that ℓ1 and ℓ3 are not coplanar finishing this argument. (Although
this seems like a direct proof, it is not. It is not necessary that all three of
ℓ1, ℓ2, ℓ3 lie in the same plane.)
We have shown that is an equivalence relation on the lines of L. Now
show that the axioms of an affine space are satisfied.
A1. By the definition of a linear space we have that each pair P, Q of distinct
points of L is contained in a unique line P Q.
A2. By our definition of the equivalence relation on the lines of V, only
lines in the same plane may be equivalent. In the plane containing ℓ and
P , by the definition of affine plane, there is exactly one line ℓ ′ through P
that is disjoint from ℓ. Thus, applying the definitin of , there is exactly
one line ℓ ′ through P such that ℓ ′ ℓ.
A3. (Trapezoid axiom) Let P Q and RS be distinct parallel lines and let T be
a point of P R \ {P, R}. By the definition of , P Q and RS are coplanar
and disjoint. Since T S meets RS, it cannot be the parallel line of P Q
through T by Property I. Hence T S meets P Q.
A4. (Parallelogram axiom) Since each of our lines has more than two points,
the parallelogram axiom is satisfied vacuously.
R
ℓ1

1.6. EMBEDDING AFFINE SPACE IN PROJECTIVE SPACE 43
A5. By our definition of a linear space, we know that each line contains at
least two points.
A6. (Space axiom) If L has only one plane, then the theorem is trivial. So
suppose that L has a plane π and at least one point P not on π. Take
two points Q, R of π and we have another plane 〈P, Q, R〉. Take another
point S of π that is not on QR. Inspection shows that P Q RS since
they are not coplanar (〈P, Q, S〉 = 〈P, Q, R〉).
π
P
Q S R
For ease of application we state Buekenhout’s result in terms of the most
basic notions.
Theorem 1.5.2. Let L = (P, B) be a linear space with point-set P and
line-set B. This means that each pair of distinct points belongs to a unique
line, and we suppose that each line has at least 4 points and that there are at
least 3 noncollinear points. For each three points P, Q, R not on a line, the
subspace π = 〈P, Q, R〉, which is the intersection of all subspaces containing
{P, Q, R}, must be an affine plane. This means that if S is any point of π
and m is any line of π with S ∈ m, there is a unique line ℓ of π through S
which is disjoint from m. Then L is an affine space.
P Q
1.6 Embedding Affine Space in Projective Space
We start with a given affine space (possibly an affine plane) and adjoin certain
additional points and lines to the points and lines of our given affine
RS

44 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
geometry and then show that the enlarged system of points and lines is a
projective geometry. We shall speak of the points and lines of the original
affine geometry as “finite” points and lines and the new points and lines as
“infinite” points and lines.
Definition 1.6.1. For each parallel class of lines, an infinite point is adjoined
to each of these lines and to no other finite lines.
Definition 1.6.2. If π is an affine plane, the set of infinite points adjoined
to the lines of π is an infinite line.
Having adjoined infinite points to our geometry and defined lines in the
enlarged geometry, we must now show that the axioms P1, P2 and P3 hold
in this enlarged geometry.
Since a finite line contains at least two distinct finite points, and an
infinite point has been added to it, in the enlarged geometry each finite line
contains at least 3 points. In a plane π(P, m), if R and S are two finite points
on m, then through P there are at least the three distinct lines P R, P S and
the parallel to m. Hence there are at least three different infinite points on
the infinite line line of infinite points adjoined to lines of π. Thus axiom P3
holds in the enlarged geometry.
If two distinct points P and Q are both finite, then there is one and only
one finite line joining them, and no infinite line contains any finite point.
Thus, in the enlarged geometry, P and Q are contained in one and only one
line. If one of the points P is finite and the other Q is infinite, let m be a
finite line to which Q has been adjoined. By axiom A1, if P is on m, then
P Q is m, and if P is not on m, then P Q is the unique parallel to m through
P . Thus again there is one and only one line through P and Q. We shall
speak somewhat loosely of infinite points as determining “directions” so that
here we might speak of the line through P in the direction Q. It remains to
show that axiom P1 holds if both P and Q are infinite points. This is not
trivial and we use for the first time the transitive property of the parallelism
relation.
Lemma 1.6.3. Let π1 and π2 be two finite planes whose ideal lines both
contain two distinct ideal points X and Y . Then the ideal lines of π1 and π2
are identical.
Proof. First note that if π1 and π2 have a finite point R in common, then π1
and π2 are identical. For the lines RS and RY are intersecting finite lines

1.6. EMBEDDING AFFINE SPACE IN PROJECTIVE SPACE 45
common to π1 and π2 from which it follows that π1 and π2 are identical,
forcing their infinite lines also to be identical. Hence we may assume that π1
and π2 have no finite point in common.
Let Z be any infinite point different from X and Y on the ideal line of
π1. Let Q be any finite point of π1. By assumption, some line of π1 contains
the infinite point X, so this line or its unique parallel through Q is a line
of π1, which we shall designate as m. Similarly, QY is a line t of π1. If P
is a finite point of t different from Q, the line P Z is a line of π1 different
from t and not parallel to m, so that P Z must intersect m in a finite point
S different from Q. Let R be any finite point of π2, and let RX be the line
h of π2 and RY the line k of π2. Now Q and R are distinct points, since we
are assuming that π1 and π2 have no finite point in common. The lines k
and t are parallel, so by Lemma 1.2.2 the parallel to QR through P meets k
in a finite point M. Similarly, since m and h are parallel, the parallel to QR
through S meets h in a finite point T . Since P M and ST are both parallel to
QR, it follows that P M and ST are parallel to each other. Hence P M and
ST lie in a unique plane π3. By Theorem 1.4.7 the line P S and MT either
intersect in a finite point W or they are parallel. But if they intersect in a
finite point W , this is a finite point common to π1 and π2, contrary to oour
assumption. Hence P S and MT are parallel, implying that MT contains the
ideal point Z of P S. This means that Z is also a point of the infinite line
of π2. Thus every infinite point of π1 is an infinite point of π2. In the same
way, each infinite point of π2 is an infinite point of π1.
Note that Lemma 1.6.3 completes a proof that axiom P1 holds in the
enlarged geometry.
Now we begin to show that Axiom P2 holds in the enlarged geometry. So
let A, B, C be distinct points
not on a line. Let D = A be a point of AB and E = A a point of
AC. We must show that there is a point F common to DE and BC. The
proof must be divided into a number of cases that depend on which of the
given points are finite and which are infinite. First note that if D = B
or E = C, then the conclusion is trivial. So assume D = B and E = C.
Also we may interchange D and E and simultaneously interchange B and
C without affecting the axiom. Similarly, we may interchange B and D and
simultaneously interchange E and C.
Case 1 All of A, B, C, D, E are finite.

46 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
This first part of Lemma 1.2.5 gives the existence of F , either as a finite
point or as the infinite point on the parallel lines BC and DE.
Case 2 A is infinite; B, C, D, are finite.
The second part of Lemma 1.2.5 gives the existence of F either as a finite
point, or as the infinite pont on the parallel lines BC and DE.
Case 3 A is finite and exactly one of B, C, D, E is infinite, which we may
take to be E.
We are going to relabel the points of this set and interpret the desired
result as a consequence of Axiom A3. Relabel points as:
A ↦→ A ′
B ↦→ P ′
C ↦→ B ′
D ↦→ C ′
Finally, let D ′ be any finite point of DE different from D. So C ′ D ′ is
parallel to C ′ D ′ . Also P ′ is on A ′ C ′ but different from A ′ or C ′ . So by Axiom
A3, P ′ B ′ must meet C ′ D ′ in a (finite) point F ′ = F . This says BC meets
DE, as desired.
Case 4 A is finite and exactly two of B, C, D, E are infinite. If D and E
are infinite, then DE is the infinite line of the plane ABC and the conclusion
is merely that the infinite point of the finite line BC lies on the infinite line
of the plane. The reason is the same if B and C are infinite points. Since a
lione containing two distinct infinite points cannot contain any finite points,
we cannot have A finite and both B and D infinite (or alternatively, both
C and E infinite. But we can have A finite and opposite vertices of the
quadrilateral B, C, D, E infinite, and we shall take these to be C and D.
Let k be the finite line BC, i.e., k is the line through B parallel to AE.
Let ℓ be the finite line DE, i.e., ℓ is the line through E parallel to AB. Then
k meets ℓ in a finite point F by Lemma 1.2.2, i.e., BC meets DE as desired.
Case 5 B and D are finite; A. C and E are infinite. Here A is the infinite
point on the finite line BD. If P = B is a finite point on the line joining
B to the infinite point C, then the line BD and the point P determine a
plane π whose infinite line contains both A and C and hence by Lemma 1.6.3

1.6. EMBEDDING AFFINE SPACE IN PROJECTIVE SPACE 47
also the infinite point E. The lines BC and DE are finite lines of π, but
containing the different infinite points C and E are not parallel. Thus BC
and DE intersect in a finite point F .
If as many as three of B, C, D, E are infinite, then A is also infinite,
implying that all of A, B, C, D, E are infinite.
Case 6 All of A, B, C, D, E are infinite. Choose P as any finite point,
and A1, B1, C1 finite points different from P on respectively P A,, P B, P C.
Thus AB is the infinite line of the finite plane P A1B1; AC is the infinite line
of the finite plane P A1C1; and BC is the infinite line of the plane P B1C1. In
the plane P A1B1 the line P D intersects A1B1 in a point D1 different from
A1 and B1. Here D1 may be a finite point, but if D1 is an infinite point, then
D1 = D. Similarly, in the plane P A1C1, P E intersects A1C1 in a point E1
different from A1 and C1, and either E1 is finite or E1 = E. In any event DE
is the infinite line of the plane P D1E1. The lines D1E1 and B1C1 lie in the
plane A1B1C1 and so intersect in a point F ∗ which may be finite or infinite.
It F ∗ is infinite, it is the unique infinite point on B1C1. Let the point F be
defined as the (necessarily unique) infinite point on the line P F ∗ . Since F ∗
is on the line B1C1, it is a point of the plane P B1C1, and therefore, F is on
BC, the line of infinite points in this plane. Similarly, F ∗ is on the line D1E1
and so is in the plane P D1E1. Thus F is a point of DE, the line of infinite
points in this plane. Hence F is the intersection of BC and DE, and so P2
is satisfied in this case.
This last case now shows that in every instance the points and lines of the
enlarged geometry satisfy the axioms P1, P2 and P3 of projective geometry.
We state this as a general theorem.
Theorem 1.6.4. Given an affine geometry satisfying axioms A1, A2, A3,
A4 and A5, we may define an affine plane π = π(P, m) in terms of a line
m and a point P not on m as the set of points on lines P S where S is on
m along with the points of the line through P parallel to m. If we adjoin an
infinite point to each line, adjoining the same infinite point to each line of a
parallel class but to no others, and if we define infinite lines as the set of ideal
points on the lines of a plane π, then the enlarged geometry is a projective
geometry satisfying axioms P1, P2 and P3.

48 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
1.7 The Principle of Duality
An important principle for projective planes, and indeed for all projective
spaces, is the so-called principle of duality . We first develop it for the
plane case and later for more general projective spaces. First suppose that
A is a proposition concerning point-line incidence geometries. We get the
proposition A D dual to A by interchanging the words “point” and “block.”
Example: If A is the proposition ‘there exist four points no three of which
are incident with a common line,’ then A D is the proposition ‘there exist four
lines no three of which meet at a common point.’
Let G be a point-line geometry with point-set P and line-set B, and
incidence relation I. Then the geometry G D dual to G has point-set B, lineset
P , and two objects of G D are incident if and only if they are incident in
G. Clearly (G D ) D = G.
Theorem 1.7.1. (Principle of Duality) Let K be a class of point-line geometries.
Suppose that K has the following property: If K contains the geometry
G, then it also contains the dual geometry G D . Then the following assertion
is true: If A is a proposition that is true for all geometries in K, then A D is
also true for all geometries in K.
Proof. Let G be an arbitrary in K. For G ′ := G D we have that G ′D = G.
Siince G ′ is a geometry in K, the proposition A holds for G ′ . Hence G ′D = G
satisfies the assertion A D .
Let P be a projective plane. According to Definitions 1.1.1, 1.1.6 and
Ex. 1.1.6.2, we see that P is characterized by the following:
PP1 Each two distinct points of P are incident with a unique line.
PP2 Each line is incident with at least three points.
PP3 Each two distinct lines of P meet in a unique point.
PP4 There are four points of P with no three of them collinear.
In order to prove the principle of duality for projective planes we need to
show that any projective plane P must satisfy the propositions that are dual
to PP1, ..., PP4. Clearly PP1 and PP3 are dual to each other. The dual to
PP2 is:

1.8. STRUCTURE OF PROJECTIVE GEOMETRY 49
PP2 D : Each point is incident with at least three lines.
The dual to PP4 is:
PP4 D : There are four lines of P with no three of them concurrent.
So we need to show that each projective plane P satisfies PP2 D and
PP4 D . First let P be an arbitrary point of P. It easily follows from PP1
through PP4 that there is some line m not incident with P . By PP2 there
must be at least three distinct point P1, P2 and P3 incident with m. The
P P1, P P2 and P P3 are three distinct lines through P .
It is also easy to verify that PP4 D follows from PP4. Hence each projective
plane P satisfies the propositions that are the duals of the axioms that
characterize projective planes.
Theorem 1.7.2. (Principle of duality for projective planes) If a proposition
A is true for all projective planes, then the dual propositioin A D also holds
for all projective planes.
Proof. If A is true for all projective planes P, then A D holds for all geometries
of the form P D , hence for all dual projective planes. Furthermore we can
represent any projective plane P as a dual projective plane: P = (P D ) D .
NOTE: A projective plane P may or may not be isomorphic to its dual
P D .
1.8 Structure of Projective Geometry
Lemma 1.2.5 says that all subspaces of a projective space are determined by
their sets of points and lines. This means that the entire projective geometry
is determined by its point-line geometry. We now continue to develop the
structure of projective space, especially the notions of basis and dimension
of a projective space. The following exchange property is fundamental.
Theorem 1.8.1. (Exchange property) Let U be a linear set of the projective
space P, and let P be a point of P that does not lie in U. Then the following
implication is true:
If Q ∈ UP , then P ∈ UQ, hence also UP = UQ.
One can also express this by saying that any two distinct subspaces through
U that are spanned by U and a point outside U intersect only in points of U.

50 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
Proof. Since Q ∈ UP \ U, Lemma 1.2.5, there is a point Q ′ in U such that
P ∈ QQ ′ ⊆ UQ. It then follows easily that the two subspaces in question
are equal: viz, from U ⊆ UQ and P ∈ UQ it follows that UP ⊆ UQ. The
other inclusion follows similarly.
Definition 1.8.2. A set B of points in the projective space P is called independent
provided that for any subset set B ′ ⊆ B and any point P ∈ B \B ′ , we
have P /∈ 〈B ′ 〉. This is the same as saying that for each P ∈ B, P /∈ 〈B\{P }〉.
We say that B is dependent provided there is some point P ∈ B for which
P ∈ 〈B \ {P }〉. A basis for P is an independent set B for which 〈B〉 = P,
i.e., B spans P.
Note: The empty set of points is independent; a single point forms an
independent set; two distinct points form an independent set; three points
form an independent set provided they are not incident with a common line.
Clearly every subset of an independent set is independent.
Theorem 1.8.3. A set B of points of P is a basis of P provided B is a
minimal spanning set of P, that is, 〈B〉 = P, but no proper subset of B spans
P.
Proof. Easy exercise.
Definition 1.8.4. The projective space P is said to be finitely generated
provided there is a finite set B that spans P.
Note: For the remainder of this chapter we let P denote a finitely generated
projective space.
Eventually we will show that each finitely generated projective has a finite
basis, and that any two bases have the same finite number of elements.
Theorem 1.8.5. Let E be a finite spanning set of P. Then there exists a
basis B of P such that B ⊆ E, implying that P has a finite basis.
Proof. Define E0 := E. If E0 is a minimal generating set, then by the preceding
theorem E0 is a basis. Otherwise there is a proper subset E1 of E0 that
spans P also. Since E is finite, after a finite number of steps we must obtain
a minimal spanning set B = En ⊆ E, which is a basis.
Lemma 1.8.6. Let B be an independent set of points of P, and let B1, B2
be subsets of B. If B is finite, then
〈B1 ∩ B2〉 = 〈B1〉 ∩ 〈B2〉.

1.8. STRUCTURE OF PROJECTIVE GEOMETRY 51
Proof. Since B1 ∩ B2 ⊆ B1, B2, clearly 〈B1 ∩ B2〉 ⊆ 〈B1〉 ∩ 〈B2〉. Inclusion in
the other direction is more difficult and will be proved by induction on the
size of B1.
If B1 = ∅, then the assertion is clearly true. Suppose that the statement of
the Lemma holds for all B1 with |B1| = k −1, and suppose that |B1| = k ≥ 1.
We may suppose that B1 is not a subset of B2, since in that case the result
clearly holds. So we may suppose that there is a point P ∈ B1 \ B2. By the
induction hypothesis, the assertion is true for the set B ′ 1 := B1 \ {P }. By
way of contradiction, suppose that there is a point X such that
X ∈ (〈B1〉 ∩ 〈B2〉) \ 〈B1 ∩ B2〉.
If X were a point of 〈B ′ 1 〉, then it would be contained in 〈B′ 1 〉 ∩ 〈B2〉, so
by the induction hypothesis X ∈ 〈B ′ 1 ∩ B2〉, and hence X ∈ 〈B1 ∩ B2〉, giving
the desired contradiction. Hence the hypothetical point X must be a point
of 〈B1〉 \ B ′ 1 〉 = 〈B′ 1 , P 〉 \ 〈B′ 1 〉. Using the exchange property, we see that
P ∈ 〈B ′ 1, X〉. Since X also lies in 〈B2〉, we obtain
P ∈ 〈B ′ 1 , X〉 ⊆ 〈B′ 1 , 〈B2〉〉 = 〈B ′ 1 , B2〉 = 〈B ′ 1 ∪ B2〉.
By the choice of P we have P /∈ B2. Moreover, P ∈ B ′ 1. Therefore we
get a contradiction to the independence of B (for B0 := B ′ 1 ∪ B2 ⊆ B satisfies
P ∈ 〈B0〉 but P ∈ B0).
The next theorem is the Steinitz exchange theorem for finitely generated
projective spaces. We handle the most important special case first.
Lemma 1.8.7. (Exchange lemma) Let B be a finite basis of P, and let P be
a point of P. Then there is a point Q in B with the property that the set
(B \ {Q}) ∪ {P }
is also a basis of P.
Such a point Q can be found in the following way. There is a subset B ′
of B such that P ∈ 〈B ′ 〉, but P ∈ 〈B ′′ 〉 for any proper subset B ′′ of B ′ . Then
for any point Q ∈ B ′ the set (B \ {Q}) ∪ {P } is a basis of P.
Proof. Since B is a finite set there exists a subset B ′ of B with the following
property:
P ∈ 〈B ′ 〉, but P ∈ 〈B ′′ 〉 for any proper subset B ′′ of B ′ .

52 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
Let Q be an arbitrary point of B ′ . We claim that the set B1 := (B \
{Q}) ∪ {P } is a basis of P.
We first show that P ∈ 〈B \ {Q}〉. So suppose that P ∈ 〈B \ {Q}〉. Then
by Lemma 1.8.6 we have
P ∈ 〈B \ {Q}〉 ∩ 〈B ′ 〉 = 〈(B \ {Q}) ∩ B ′ 〉 = 〈B ′ \ {Q}〉,
contradicting the minimality of B ′ .
Using the Exchange property (Theorem 1.8.1) we obtain
〈B1〉 = 〈B \ {Q}, P 〉 = 〈B \ {Q}Q〉 = 〈B〉 = P.
Thus B1 spans P.
It remains to show that B1 is independent. So suppose there is some point
X ∈ B1 for which X ∈ 〈B1 \ {X}〉. If X = P , then P ∈ 〈B1 \ {P }〉. Since
B1 \ {P } = B \ {Q}, and P ∈ 〈B ′ 〉, we get
P ∈ 〈B \ {Q}〉 ∩ 〈B ′ 〉 = 〈(B \ {Q}) ∩ B ′ 〉 = 〈B ′ \ {Q}〉,
contradicting the minimality of B ′ .
So X = P . From the definition of B1, we have X ∈ B \ {Q}, and
X ∈ 〈B1 \ {S}〉 = 〈B \ {Q, X}, P 〉. Since B \ {Q} is independent, it follows
that X ∈ 〈B \ {Q, X}〉. By the exchange property we get
Hence
P ∈ 〈〈B \ {Q, X}〉, X〉 = 〈B \ {Q, X}, X〉 = 〈B \ {Q}〉.
P ∈ 〈B \ {Q}〉 ∩ 〈B ′ \ {Q}〉.
But this contradicts the minimality of B ′ (w.r.t. P ). This shows that B1 is
independent, and hence is a basis for P.
Theorem 1.8.8. (Steinitz exchange theorem for projective spaces) Let B be
a finite basis of P, and let r := |B|. If C is an independent set having s
points, then we have:
(a) s ≤ r;
(b) There is a subset B ∗ of B with |B ∗ | = r − s such that C ∪ B ∗ is a basis
of P.

1.8. STRUCTURE OF PROJECTIVE GEOMETRY 53
Proof. The proof is by induction on the size s of C. If s = 0, the result is
clearly true. If s = 1, the result is true by the Exchange Lemma. Suppose
s > 1 and assume the result is true for s − 1. Let P be an arbitrary point
of C. Then C ′ := C \ {P } is an independent set with |C ′ | = s − 1. By the
induction hypothesis we know that
(i) s − 1 ≤ r;
(ii) There is a subset B ′ of B with |B ′ | = r − (s − 1) such that C ′ ∪ B ′ is
a basis of P .
Claim: We also have s ≤ r. Otherwise, s − 1 = r, so B ′ = ∅. Thus C ′
would be a basis of P. In particular, P ∈ 〈C ′ 〉, contradicting the independence
of C. Hence part (a) is proved.
For part (b), since B is finite there exists a subset B ′′ of C ′ ∪ B ′ such that
P ∈ 〈B ′′ 〉, but P is not contained in the span of any proper subset of B ′′ .
Then B ′′ ∩ B ′ = ∅, since otherwise using Lemma 1.8.6 we have
P ∈ 〈B ′′ 〉 ∩ 〈(C ′ ∪ B ′ )〉 = 〈B ′′ ∩ (C ′ ∪ B ′ )〉 ⊆ 〈B ′′ ∩ C ′ 〉 ⊆ 〈C ′ 〉 = 〈C \ {P }〉.
By the Exchange Lemma, for each Q ∈ B ′′ ∩B ′ the set (C ′ ∪(B ′ \{Q}))∪{P }
is a basis. Since
(C ′ ∪ −(B ′ \ {Q})) ∪ {P } = (C ′ ∪ {P }) ∪ (B ′ \ {Q}) = C ∪ (B ′ \ {Q}),
part (b) follows with B ∗ defined by B ∗ = B ′ \ {Q}.
Finally we consider the case that C is an infinite set. Then C would
contain a finite subset having exactly s ′ = s + 1 elements, and by the above
arguments we would get
r + 1 = s ′ ≤ r,
an impossibility that completes the proof.
The preceding results have an immediate corollary that is quite important.
Corollary 1.8.9. (Basis extension theorem) Let P be a finitely generated
projective space. Then any independent set (in particular the empty set, or
any basis of some subspace) can be extended to a basis of P. Moreover, any
two bases of P have the same number of elements.
Definition 1.8.10. Let P be a finitely generated projective space. If each
basis of P has exactly d + 1 elements, we say that d is the dimension of the
space and write d = dim(P).

54 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
Lemma 1.8.11. Let U be a subspace of the finitely generated projective space
P. Then the followiing are true:
(a) dim(U) ≤ dim(P)
(b) dim(U) = dim(P) if and only if U = P.
Proof. Easy exercise.
Definition 1.8.12. Let P be a projective space of finite dimension d. the
subspaces of dimension 2 are called planes, and the subspaces of dimension
d − 1 are called hyperplanes of P. The set of all subspaces of P is denoted
by U(P). The set U(P) together with the subset relation ⊆ is called the
projective geometry belonging to the projective space P. The empty set and
the whole space are called the trivial subspaces. The set of all nontrivial
subspaces is denoted by U ∗ (P).
In practice we do not usually distinguish between a ’projective space’
and its corresponding ”projective geometry,’ since each uniquely determines
the other. In the projective geometry (U(P), ⊆) we have the following more
general concept of flag. A flag of (U(P), ⊆) is a set of (distinct) elements of
U(P) that are mutually incident. A flag F is called maximal provided there
is no element x ∈ U(P) \ F such that F ∪ {x} is also a flag.
Lemma 1.8.13. Let P be a d-dimensional projective space, and let U be
a t-dimensional subspace of P (with −1 ≤ t ≤ d). Then there exist d − t
hyperplanes of P such that U is the intersection of these hyperplanes.
Proof. Let {P0, P1, . . . , Pt} be a basis of U. By Theorem 1.8.8 we can extend
it to a basis B = {P0, P1, . . . , Pt, Pt+1, . . . , Pd} of P. If we define
Hi := 〈B \ {Pt+i}〉, (i = 1, . . . , d − t),
then we obtain d − t hyperplanes H1, . . . , Hd−t. By Lemma 1.8.6 their intersection
can be computed as follows:
H1 ∩ · · · ∩ Hd−t = 〈{P0, P1, . . . , Pt, Pt+2, . . . , Pd}〉 ∩ . . .
∩〈{P0, P1, . . . , Pt, Pt+1, . . . , Pd−1}〉
= 〈{P0, P1, . . . , Pt, Pt+2, . . . , Pd} ∩ . . .
∩{P0, . . . , Pt, Pt+1, . . . , Pd−1}〉
= 〈{P0, . . . , Pt, Pt}〉 = U.

1.8. STRUCTURE OF PROJECTIVE GEOMETRY 55
Theorem 1.8.14. (dimension formula) Let U and W be subspacers of P.
Then
dim(〈U, W〉) = dim(U) + dim(W) − dim(U ∩ W).
Proof. Choose a basis A = {P1, . . . , Ps} of U ∩ W and extend it to a basis B
of U and also to a basis C of W:
B = {P1, . . . , Ps, Ps+1, . . . , Ps+t}, C = {P1, . . . , Ps, Qs+1, . . . , Qs+t ′}.
If we show that B ∪ C is a b asis of 〈U, W〉, then
dim(〈U, W〉) = |B ∪ C| − 1
= s + t + s + t ′ − s − 1
= s + t − 1 + s + t ′ − 1 − (s − 1)
= dim(U) + dim(W) − dim(U ∩ W).
Because 〈U, W〉 = 〈B, C〉, the set B ∪C certainly spans 〈U, W〉. So it remains
only to show that B ∪ C is an independent set.
Claim: U ∩ 〈C \ A〉 = ∅.
By way of contradiction, suppose that there is a point X ∈ U ∩ 〈C \ A〉.
Since 〈C \ A〉 ⊆ W, the point X would be contained in W and hence in
U ∩ W. Using Lemma 1.8.6 we get the following contradiction:
X ∈ (U ∩ W) ∩ 〈C \ A〉 = 〈A〉 ∩ 〈C \ A = 〈∅〉 = ∅.
So the claim is established. Now suppose that B ∪ C is not independent.
Then there is a point P ∈ B ∪ C with P ∈ 〈(B ∪ C) \ {P }〉. WOLG we may
assume that P ∈ B. It follows that
P ∈ 〈(B ∪ C) \ {P }〉 = 〈B \ {P }, C \ A〉 = 〈〈B \ {P }〉, 〈C \ A〉〉.
The ’Join theorem’ (Cor. 1.2.8) says that there are points T ∈ 〈C \ A〉
and S ∈ 〈B \ {P }〉 ⊆ U such that P ∈ 〈S, T 〉.
We claim that the line 〈S, T 〉 intersects U in a unique point, i.e., the point
S. For if there were two points of U on 〈S, T 〉, then each point of 〈S, T 〉 would
be contained in U. In particular, T would be in U, contradicting the fact that
U and C \A are disjoint. But then, since the point P satisfies P ∈ U ∩〈S, T 〉,
we have P = S ∈ 〈B \ {P }〉, contradicting the independence of B.

56 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
Corollary 1.8.15. Let P be a projective space and let H be a hyperplane of
P. Then for any subspace U of P we have the following:
Either U ⊆ H, or
dim(U ∩ H) = dim(U) − 1.
In particular, any line that is not contained in H intersects H in a unique
point.
1.9 Quotient Geometries: a First Look
In this section we only consider quotient geometries that arise as a ”space
modulo a point.”
Definition 1.9.1. Let Q be a point of P. The ”point-line” geometry whose
points are the lines of P through Q, whose lines are the planes of P through
Q, and whose incidence is the incidence induced by that in P, is called the
quotient geometry of Q, or the quotient geometry P modulo Q.
We shall show that a quotient geometry of a point is a projective space.
In order to do this we need to discuss the concept of isomorphism.
Definition 1.9.2. Let G = (P, B, I) and G ′ = (P ′ , B ′ , I ′ ) be point-line geometries
with point-sets P and P ′ , respectively, and line-sets B and B ′ , respectively.
We say that G and G ′ are isomorphic provided there are bijections
α : P → P ′ and β : B → B ′ such that for all P ∈ P and all B ∈ B the
following equivalence is true:
P IB ⇐⇒ α(P )I ′ α(B).
Such a pair (α, β) is called an isomorphism from G onto G ′ . Often the
symbol α is used to represent both the map α on points and the map β on
lines.
Theorem 1.9.3. Let P be a d-dimensional projective space, and let Q be a
point of P. Then the quotient geometry P/Q of P modulo Q is a projective
space of dimension d − 1.
Proof. The idea of the proof is to show that there is a hyperplane H of P
not through Q, and then to show that P/Q is isomorphic to H.

1.10. COMBINATORICS OF FINITE PROJECTIVE SPACES 57
First, extend {Q} to a basis {Q, P1, . . . , Pd} of P. The subspace H =
〈P1, . . . , Pd〉 is spanned by d independent points, so that H had dimension
d − 1 and is a hyperplane. Since the set {Q, P1, . . . , Pd} is independent, it
follows that Q ∈ H.
Now let H be any hyperplane of P not containing Q. Define the map α
from the points and lines of P/Q to the points and lines of H as follows:
If g is a point of P/Q (i.e., a line of P through Q), put
α : g ↦→ g ∩ H.
If π is a line of P/Q (i.e., a plane of P through Q), put
α : π ↦→ π ∩ H.
It is almost trivial to show that α is bijective and preserves incidence.
For future reference we note that following fact whose proof is contained
in the preceding proof.
Corollary 1.9.4. Let P be a d-dimensional projective space, and let Q be a
point of P. Then there is a hyperplane of P that does not contain Q.
1.10 Combinatorics of Finite Projective Spaces
In this section we count the points, lines and hyperplanes of finite projective
spaces. Later on we will revisit this topic and count ”flags” of various types.
First we need the following lemma.
Lemma 1.10.1. Let g1 and g2 be two lines of a projective space P. Then
there is a bijective map
φ : (g1) → (g2)
from the set (g1) of points on g1 onto the set (g2) of points on g2.
Proof. WOLG we may assume that g1 = g2.
Case 1. The lines g1 and g2 intersect in a point S.
Let P1 be a point on g1 and P2 a point on g2 such that P1 = S = P2. By
axiom P3 there is a third point P on the liine 〈P1, P2〉. by definition, P is
neither on g1 nor on g2. By axiom P2 any liine through P that contains a

58 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
point X = S of g1 intersects the line g2 in a uniquely determined point φ(X)
with φ(X) = S. Hence the map φ defined by
φ : X ↦→ 〈X, P 〉 ∩ g2
is a bijection form (g1) \ {S} onto (g2) \ {S}. Extend this to a bijection from
(g1) to (g2) by defining φ : S ↦→ S.
Case 2. The lines g1 and g2 have no point in common. Let h be a line
connecting some point of g1 to some point of g2. By the first case, there are
bijections
φ1 : (g1) → (h) and φ2 : (h) → (g2).
Then φ := φ2 ◦ φ1 is a bijective map from (g1) onto (g2).
A projective space is said to be finite provided its point-set is a finite
set, in which case it is clear that there could only be finitely many lines.
Conversely, if there were only finitely many lines, and the dimension of the
space was at least 2, there would be a line g with infinitely many points and
a point P not on g. The lines joining P to points of g could not be finite
in number, so if dim(P) is at least 2, then P has finitely many points if and
only if it has finitely many lines.
In this book we are really only interested in finite projective (and affine)
spaces.
Let P be a finite projective space. We now know that each line of P is
incident with the same number of points. Let q be the positive integer for
which each line is incident with q + 1 points. By axiom P3 we have q ≥ 2.
This integer q is called the order of the finite projective space P.
Lemma 1.10.2. Let P be a finite projective space of dimension d ≥ 2 and
order q. Then for each point Q of P, the quotient geometry P/Q has order
q.
Proof. Since P/Q is isomorphic to any hyperplane H not through Q, and
there is such a hyperplane which must have order q since all subspaces of
a projective space of order q clearly have order q. In particular, P/Q must
have order q.
Theorem 1.10.3. Let P be a finite projective space of dimension d and
order q, and let U be a t-dimensional subspace of P (1 ≤ t ≤ d). Then the
following are true:

1.10. COMBINATORICS OF FINITE PROJECTIVE SPACES 59
(a) The number of points of U is
q t + q t−1 + · · · + q + 1 = qt+1 − 1
q − 1 .
In particular, P has exactly q d + q d−1 + · · · + q + 1 points.
(b) The number of lines of U through a fixed point of U equals
q t−1 + · · · + q + 1.
(c) The total number of lines of U equals
(qt + qt−1 + · · · + q + 1)(qt−1 + · · · + q + 1)
.
q + 1
Proof. We can prove (a) and (b) simultaneously by induction on t. If t = 1
then (a) and (b) are rather trivially true. Suppose that (a) and (b) are true
for projective spaces of dimension t−1 ≥ 1, and let Q be an arbitrary point of
U. Since the quotient geometry U/Q is a (t−1)-dimensional projective space
of order q, by the induction hypothesis, its number of points is q t−1 +· · ·+q+1.
Since by the definition of U/Q this is the number of lines of U through Q,
we already have (b). Since each line through Q has exactly q points different
from Q, and since each point R = Q of U is on exactly one of these lines
(i.e., on 〈Q, r〉), U contains exactly
points, proving (a).
1 + (q t−1 + · · · + q + 1)(q = q t + q t−1 + · · · + q + 1
(c) Let b be the number of all lines of U. Since U has exactly q t +· · ·+q+1
points and any point of U is on exactly q t−1 + · · · + q + 1 lines of U, and each
line is on q + 1 points, it follows that
b = (qt + · · · + q + 1)(qt−1 + · · · + q + 1)
.
q + 1
Theorem 1.10.4. Let P be a finite projective space of dimension d and order
q. Then

60 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY
(a) the number of hyperplanes of P is exactly
q d + q d−1 + · · · + q + 1;
(b) the number of hyperplanes of P through a fixed point of P equals
q d−1 + · · · + q + 1.
Proof. First consider part (a). For d = 1 the theorem says that each line has
q + 1 points, which is clearly true. For d = 2, a hyperplane is a line, so part
(a) says that the number of lines of a projective plane is q 2 + q + 1. This is
true by part (c) of Theorem 1.10.3. So suppose that the statement of part (a)
is true for projective spaces of dimension d−1 ≥ 1. Consider a hyperplane H
of P. By the dimension formula (Theorem 1.8.14) every hyperplane different
from H intersects H in a subspace of dimension d − 2. Thus any hyperplane
of P different from H is spanned by a (d − 2)-dimensional subspace of H and
a point outside H.
For each (d − 2)-dimensional subspace U of H and each point P ∈ P \ H,
the subspace 〈U, P〉 is a hyperplane, which contains exactly
(q d−1 + · · · + 1) − (q d−2 + · · · + q + 1) = q d−1
points outside H. Since there exist q d points of P outside H, there are
q hyperplanes different from H through U. By the induction hypothesis
there are exactly q d−1 + · · · + q + 1 hyperplanes of H, which are subspaces
of dimension d − 1 of P. Thus the total number of hyperplanes of P is
q · (q d−1 + · · · + q + 1) + 1. This completes the proof of part (a).
For part (b), let P be a point of P and let H be a hyperplane not through
P . Then any hyperplane of P through P intersects H in a hyperplane of H.
Thus by (a), there are exactly q d−1 + · · · + q + 1 such hyperplanes.
Corollary 1.10.5. Let P be a finite projective plane. Then there exists an
integer q ≥ 2 such that any line of P has exactly q + 1 points, each point is
on q + 1 lines, and the total number of points (resp., lines) is q 2 + q + 1.
For each q that is a power of a single prime there is known at least one
projective plane of order q, and no plane with order not a prime power has
ever been found. The theorem of Bruck-Ryser says the following: If q is a
positive integer of the form q = 4n+1 or q = 4n+2, n a positive integer, and
if there is a projective plane of order q, then q is the sum of two squares, one

1.10. COMBINATORICS OF FINITE PROJECTIVE SPACES 61
of which might be 0. Alternatively, this says exactly that each odd prime
dividing the square-free part of q must be congruent to 1 modulo 4. We
have included a proof of this celebrated theorem in the appendix on number
theory. The Bruck-Ryser theorem implies that there is no projective plane
of order q = 8n + 6, in particular there is no plane of order 6. The only
other integer q that has been excluded as the possible order of a projective
plane is q = 10. This result required many hours on several computers with
the assistance of several mathematicians and a significant body of theory
developed over a period of several years. See [La91].

62 CHAPTER 1. SYNTHETIC AXIOMS FOR GEOMETRY

Chapter 2
Analytic Geometry
Projective (and affine) spaces of dimension at least 3 with order at least 4
can all be ”coordinatized” using the elements of a skewfield F . (Recall that
a skewfield - or division ring - is just a field without the assumption that
multiplication is commutative.) Then certain problems that seem intractable
(or very awkward) using only the synthetic approach can sometimes be solved
in a routine fashion using algebra. In the appendix 20 we give a proof of
the celebrated theorem of Wedderburn that says that any finite skewfield is a
field. Since we are considering only finite geometries, at first glance it might
seem that we need not bother with (noncommutative) skewfields. On the
other hand, they do facilitate our general treatment, so we include them to
begin with.
Vector spaces over skewfields are defined in the same way as vector spaces
over fields. When we say “V is a left vector space over the skewfield F” we
mean that for each element (vector) v ∈ V and each a ∈ F , the “scalar
product” av is a uniquely defined vector in V . We shall deal only with left
vector spaces in this book and refer to them simply as vector spaces. (Note:
When the skewfield F is actually a field, the scalar a may be written either on
the left or on the right of the vector it is multiplying without any confusion.)
The basic theory of vector spaces over skewfields can be developed in the
same way as for vector spaces over fields. The usual elementary properties
of independence, basis, dimension, subspaces, linear maps, etc., hold in the
more general setting. We do not develop these ideas here, partly because
we assume that the reader has a thorough background in linear algebra and
because ultimately we will deal only with vector spaces over fields.
63

64 CHAPTER 2. ANALYTIC GEOMETRY
2.1 The Projective Space P(V )
Definition 2.1.1. Let V be a vector space of dimension d + 1 ≥ 3 over a
skewfield F . We define the point-line geometry P(V ) as follows:
• The points of P(V ) are the 1-dimensional subspaces of V ;
• The lines of P(V ) are the 2-dimensional subspaces of V ;
• The incidence relation in P(V ) is set-theoretical containment.
We shall have to distinguish between the dimension of a vector space V
and the dimension of the associated projective space P(V ). For the time
being we will denote the dimension of a vector space V by dimV . Our first
goal is to show that the geometry P(V ) is a projective space. Later we will
show that if a projective space has dimension 3 or greater, then it is essentially
P(V ) for some vector space V . There are many projective planes that are
not of this type, but in this book we will mention them only occasionally in
passing.
If v, w ∈ V , the subspace of V spanned by {v, w} will be denoted by
〈v, w〉. If 〈v〉 denotes the subspace spanned by v, it is also the subspace of
P(V ) spanned by the point 〈v〉, which might have denoted by 〈〈v〉〉. However,
we write 〈v〉 for either the subspace of V or the point of P(V ), since they
really are the same thing. Then 〈v, w〉
Theorem 2.1.2. Let V be a vector space of dimension d + 1 ≥ 3 over a
skewfield F . Then P(V ) is a projective space called the projective space
coordinatized by F .
Proof. We have four axioms to verify. For P1 suppose that P and Q are
distinct points of P(V ). So there are nonzero vectors v, w ∈ V with P =
〈v〉 = 〈w〉 = Q. It is easy to check that 〈v, w〉 is the unique line of P(V )
containing both v and w.
For P2, let g1 = 〈P0P1〉 and g2 = 〈P0, P2〉 be lines through a common
point P0 that are intersected by the line h = 〈P1, P2〉. Also, let Q = P1, P2
be a point on h and Q1 = P0, P1 a point on g1. We need to show that 〈Q, Q1〉
intersects the line g2 in some point.
If g1 = g2, the desired result is clearly true. So suppose that g1 = g2 and
choose vectors v0, v1, v2 ∈ V so that P0 = 〈v0〉, P1 = 〈v1〉 and P2 = 〈v2〉.

2.1. THE PROJECTIVE SPACE P(V ) 65
Since the points P0, P1 and P2 are noncollinear, the vectors v0, v1, v2 form
an independent set.
Consider the subspace V ′ of V that is generated by v0, v1 and v2. Choose
w1 so that Q1 = 〈w1〉, so w1 ∈ 〈v0, v1〉 ⊆ V ′ . Moreover, Q = 〈w〉 ⊆
〈v0, v1, v2〉 = V ′ . The line 〈Q, Q1〉 is a 2-dimensional subspace of V ′ . Since
Q ∈ g2 and 〈g2, 〈Q, Q1〉〉 ⊆ V ′ , it follows that V ′ = 〈g2, 〈Q, Q1〉〉.
Since g2 is also a 2-dimensional subspace of V ′ , by the dimension formula
for vector spaces,
dimV (g2 ∩ 〈Q, Q1〉〉) = dimV (g2) + dimV (〈Q, Q1〉) − dimV (〈g2, 〈Q, Q1〉〉
= 2 + 2 − 3 = 1.
Hence g2 intersects the line 〈Q, Q1〉 in a 1-dimensional subspace of V , i.e., in
a point of P(V ).
For axiom P3, an arbitrary line 〈v1, v2〉 contains at least the three distinct
points 〈v1〉, 〈v2〉, 〈v2〉.
Finally, for axiom P5 (since we do not wish to exclude the case d =
2), since V has dimension at least 3, there exist three linearly independent
vectors v0, v1, v2 ∈ V . Then 〈v0, v1〈 and 〈v0, v2〉 are two distinct lines of
P(V ). At the same time we notice that if d + 1 ≥ 4, then there are four
nonzero vectors with {v0, v1, v2, v3} an independent set. Then 〈v0, v1〉 and
〈v2, v3〉 are two disjoint lines.
Lemma 2.1.3. The vector subspaces of V are in bijective correspondence
with the geometric subspaces of P(V ), that is:
(a) If V ′ is a subspace of the vector space V , then P(V ′ ) is a subspace of
P(V ).
(b) If U is a subspace of the projective space P(V ), then there is a vector
subspace V ′ of V for such P(V ′ ) = U.
Proof. For (a) we need to show that the set of 1-dimensional subspaces contained
in V ′ forms a linear set (a subspace) of P(V ). Let 〈v〉, 〈w〉 ⊆ V ′ .
Then v, w ∈ V , so clearly 〈v, w〉 ⊆ V ′ .
For (b), let {P0, . . . , Pt} be a basis of U. Then there are vectors v0, . . . , vt
for which Pi = 〈vi〉, 0 ≤ i ≤ t. Define V ′ by
V ′ = 〈v0, v1, . . . , vt〉 ⊆ V.
We claim that P(V ′ ) = U. We show this by induction on t. For t = 0
the claim is trivially true. Also consider separately the case t = 1. Let

66 CHAPTER 2. ANALYTIC GEOMETRY
g = P0P1 be a line of P(V ) with P0 = 〈v0〉 and P1 = 〈v1〉. Since the vector
subspace V ′ = 〈v0, v1〉 is the unique line through the points P0 and P1, and its
1-dimensional subspaces are exactly the points of P(V ′ ), clearly P(V ′ ) = g.
Assume now that the claim is true for t ≥ 1. Let
U = 〈P0, P1, . . . , Pt+1〉
be a subspace of P(V ). By the induction hypothesis there is a vector subspace
V ′′ of V such that P(V ′′ ) = 〈P0, P1, . . . , Pt〉. Thus
U = 〈〈P0, . . . , Pt〉, Pt+1〉 = 〈P(V ′′ ), Pt+1〉.
Let vt+1 be a vector for which Pt+1 = 〈vt+1〉, and let V ′ be the vector
subspace spanned by V ′′ and vt+1.
First we show the inclusion P(V ′ ) ⊆ U. Let P = 〈v〉 be a point of P(V ′ ).
Then there is a vector v ′′ ∈ V ′′ such that
v = av ′′ + bvt+1 (a, b ∈ F ).
Therefore the point P is incident with the line of U that passes through
the points 〈v ′′ 〉 and 〈vt+1〉 of U. Thus any point of P(V ′ ) lies in U.
Conversely, let Q be a point of U. Then by Lemma 1.2.5 Q is incident
with a line XPt+1, where X is a point of P(V ′′ ). If v ′′ is a vector with
X = 〈v ′′ 〉, then Q is on the line through 〈v ′′ 〉 and 〈Vt+1〉. By the case with
t = 1 there exist a, b ∈ F with
v = av ′′ + bvt+1 ∈ V ′′
and Q = 〈v〉. Hence Q is a point of P(V ′ ).
Corollary 2.1.4. The projective space P(V ) has dimension d (when V has
vector space dimension dimV (V ) = d + 1).
Proof. If dimP(V ) ≥ d + 1, there would exist a flag
U0 ⊂ U1 ⊂ · · · ⊂ Ud
of nontrivial subspaces of P(V ). By Lemma 2.1.3 this corresponds to a chain
of d nontrivial subspaces of V that are properly contained in each other. This
is impossible since dimV (V ) = d + 1. Conversely, since dimV (V ) = d + 1,
there is a chain of d + 1 nontrivial subspaces of V properly contained in each
other. These subspaces correspond to a flag of d + 1 nontrivial subspaces of
P(V ).

2.2. THE THEOREMS OF DESARGUES AND PAPPUS 67
Lemma 2.1.5. The line of P(V ) through the points 〈v〉 and 〈w〉 consists of
the point 〈w〉 and the points 〈v + aw〉, a ∈ F . Inparticular, if F is a finite
field with q elements, then any line of P(v) has exactly q +1 point, i.e., P(V )
has order q.
Proof. Each point 〈u〉 of the line through 〈v〉 and 〈w〉 can be written as
〈u〉 = 〈av + bw〉 with a, b ∈ F.
If a = 0, then b = 0, so that 〈u〉 = 〈bw〉 = 〈w〉. If a = 0, then
〈u〉 = 〈av + bw〉 = 〈v + ba −1 w〉.
Conversely, any point of the form 〈v + aw〉 is a point on the line through 〈v〉
and 〈w〉.
See the appendix Chapter 20 for a reasonably thorough introduction to
arithmetic in finite fields. If V is a (d + 1)-dimensional vector space over the
skewfield F , then the projective space P(V ) is denoted by P G(d, F ). Here
P(V ) is the d-dimensional projective space over the skewfield F . If F is a
finite field of order q, then P(V ) is also denoted by P G(d, q). This makes
sense since up to isomorphism there is only one finite field with q elements
( for each prime power q), and any two vector spaces of the same finite
dimension of a skewfield are isomorphic.
2.2 The Theorems of Desargues and Pappus
Definition 2.2.1. Let P be a projective space. We say that the Theorem of
Desargues holds in P provided the following statement is true:
Let A1, A2, A3, B1, B2, B3 be any points with the following properties:
• Ai, Bi and P are three distinct collinear points, i = 1, 2, 3.
• No three of the points A1, A2, A3, P and no three of the points B1, B2, B3, P
are collinear.
• The line Bi−1, Bi+1 meets the line Ai−1, Ai+1 at the point Ci, i = 1, 2, 3.
Here the subscripts are taken modulo 3 to be one of 1,2,3.
Then the three points C1, C2, C3 lie on a common line.

68 CHAPTER 2. ANALYTIC GEOMETRY
C3
A1
B1
A2
B2
A3
B3
C1
Figure 2.1: Desargues’ Configuration
In the above configuration we say that triangles A1A2A3 and B1B2B3
are in perspective from the point P . When the three points C1, C2, C3 lie
on a common line ℓ, we say the two triangles A1A2A3 and B1B2B3 are in
perspective from the line ℓ. When the two triangles are in perspective from
a point P and in perspective from a line ℓ, the configuration is called a
Desargues’ Configuration with vertex P and axis ℓ.
We are going to show that the Theorem of Desargues does hold in projective
spaces of dimension at least 3. However, it may or may not hold
in a projective plane. Note that there are ten distinct points in a Desargues’
Configuration (See Fig. 2.1). Since a projective plane of order 2 has
1 + 2 + 2 2 = 7 points, clearly the statement does not even make sense if the
coordinatizing field has only order 2. But there are no such configurations
where the C1, C2, C3 are not collinear, so we could say the Theorem of Desargues
holds vacuously. In this sense, it has been checked (by computer in
case q = 8) that for each projective plane of order q ≤ 8 the Theorem of
Desargues does hold.
P
C2

2.2. THE THEOREMS OF DESARGUES AND PAPPUS 69
Theorem 2.2.2. Let π be a finite projective plane of order q ≥ 4. Then π
contains ”many” Desarguesian configurations.
Proof. Take any point V , any line ℓ with V ∈ ℓ. Take any three lines p, q, r
through the point V , and let
p ∩ ℓ = X, q ∩ ℓ = Y, r ∩ ℓ = Z.
Let N, L be points of ℓ distinct from X, Y, Z. This is possible since ℓ has at
least 5 points. Define a map σ : A → L, where
as follows:
A = p \ {V, X}, L = ℓ \ {N, LO, Y, X},
• Let A ∈ A, and let NA ∩ V Z = B. So B = V, Z.
• Put C = BL ∩ q., so C = Y .
• Put M = AC ∩ ℓ, so M ∈ {N, L, Y, X}.
• Define σ : A → L by A σ = M.
Now A has q − 1 elements while L has q − 3 elements. Therefore σ is not
one-to-one. Hence there exist distinct point A, A ′ ∈ A with σ(A) = σ(A ′ ).
Let B ′ = NA ′ and C ′ = LB ′ ∩ q. Then triangles ABC and A ′ B ′ C ′ are in
perspective from the point V and from the line ℓ.
Theorem 2.2.3. Desargues’ Theorem is valid in a projective space of dimension
d, d ≥ 3, for any two non-coplanar triangles ABC and A ′ BC ′ in
perspective from a point V .
Proof. Let triangles ABC and A ′ B ′ C ′ belong to distinct planes π, α, respectively.
We first note that the two triangles belong to a projective space Σ of
dimension 3, because Σ = 〈π, V 〉 = 〈α, V 〉. Therefore π ∩ α is a line by the
Dimension Theorem. As lines AB and A ′ B ′ lie in the plane 〈V, A, B〉, they
must intersect in a point. Similarly, BC and B ′ C ′ intersect in a point and so
do the lines AC and A ′ C ′ . It follows that the ponts AB ∩ A ′ B ′ , BC ∩ B ′ C ′
and CA ∩ C ′ A ′ belong to the line π ∩ α.
PUT NEW FIGURE HERE (3.5.5 FROM CASSE)

70 CHAPTER 2. ANALYTIC GEOMETRY
Theorem 2.2.4. Let π be a projective plane of order at least 3 embedded
(as a subspace) in a projective space P with dimension d ≥ 3. Then π is
Desarguesian.
Proof. Let π be a projective plane of order at least 3 embedded in a projective
space P with dimension d ≥ 3. Let ABC, A ′ B ′ C ′ be two triangles of π in
perspective from a point X. Take any line ℓ ⊆ π through X. There are at
least two other points (distinct from X) on ℓ. Let O, O ′ be two such points.
Then A ′ O, O ′ A intersect in a point since they are distinct line of the plane
〈O, X, A〉. Let A ′ O ∩O ′ A = A1. Similarly, define the points B1 and C1. Note
that the points A1, B1 and C1 (∈ π) are not collinear.
Denote the plane 〈A1, B1, C1〉 by π1. Clearly π = π1. Note that triangles
ABC and A1B1C1 are in perspective from O ′ . From the proof of Theorem
2.2.3, A1C1 ∩ AC ∈ π ∩ π1. Let A1C1 ∩ π = P . Similarly, triangles
A ′ B ′ C ′ and A1B1C1 are in perspective from O, so A1C1 ∩ A ′ C ′ ∈ π ∩ π1.
Hence AC ∩ A ′ C ′ = P lies on the line π ∩ π1. In other words, we have a
Desargues’ Configuration with π ∩ π1 as the axis.
PUT NEW FIGURE HERE (3.5.6 FROM CASSE)
In view of the two preceding theorems the following theorem is superfluous,
but we like the idea of seeing the proof using coordinates along with the
synthetic proof.
Theorem 2.2.5. Let V be a vector space of dimension d + 1 over a skewfield
F . Then the Theorem of Desargues holds in P(V ).
Proof. Suppose that the hypotheses of the theorem are fulfilled. Let vi be a
vector such that Ai = 〈vi〉, i = 1, 2, 3. Since A1, A2, A3 are not collinear, the
set {v1, v2, v3} is linearly independent, so form a basis of the 3-dimensional
vector space V ′ := 〈v1, v2, v3〉. There are two cases.
Case 1. The point P lies in the plane spanned by the points A1, A2, A3.
Then there are a1, a2, a3 ∈ F with P = 〈a1v1 + a2v2 + a3v3〉. Since no three
of the points A1, A2, A3, P are collinear, we have a1a2a3 = 0. So replacing vi
with aivi if necessary, we may assume that P = 〈v1 + v2 + v3〉. Since P, Ai
and Bi are collinear there are a1, a2, a3 ∈ F such that
B1 = 〈v + 1 + v2 + v3 + a1v1 + a1v1〉 = 〈(a1 + 1)v1 + v2 + v3〉;
B2 = 〈v1 + (a2 + 1)v2 + v3〉;
B3 = 〈v1 + v2 + (a3 + 1)v3〉.

2.2. THE THEOREMS OF DESARGUES AND PAPPUS 71
Hence Bi = (ai + 1)vi + vi−1 + vi+1, where subscripts are modulo 3. Then
a little computation shows that
Ci = 〈vi−1, vi+1〉 ∩ 〈(ai−1 + 1)vi−1 + vi + vi+1, (ai+1 + 1)vi+1 + vi + vi−1〉
= 〈ai−1vi−1 − ai+1vi+1〉.
This shows that all three points C1, C2, C3 lie on the line
〈a1v1 − a2v2, a2v2 − a3v3〉.
Case 2. The point P = 〈v〉 does not lie in the plane 〈A1, A2, A3〉. The vectors
v, v1, v2, v3 are linearly independent. Therefore we may assume WOLG
that Bi = 〈v + vi〉, i = 1, 2, 3. Hence
Ci = Ai−1Ai+1 ∩ Bi−1Bi+1
= 〈vi−1, vi+1〉 ∩ 〈v + vi−1, v + vi+1〉
= 〈vi−1 − vi+1〉.
Hence the points Ci all lie on the line 〈v1 − v2, v2 − v3〉, proving the
theorem.
Definition 2.2.6. Let P be a projective space. We say that the Theorem of
Pappus holds in P provided the following statement is true:
Let g and h be two lines of P meeting at a point P . Let A1, A2, A3 be three
distinct points of g different from P ; let B1, B2, B3 be three distinct points of
h different from P . For i = 1, 2, 3, put Ci = Ai+1Bi−1 ∩ Ai−1Bi+1. Then C1,
C2 and C3 lie on a line.
Note that the ten points of a Pappus’ configuration always lie in a plane
(spanned by g and h).
Theorem 2.2.7. Let V be a vector space over the skewfield F . Then the
theorem of Pappus holds in P(V ) if and only if F is a field.
Proof. First consider only the points A1, A2, B1, B2 and P = g ∩ h = A1A2 ∩
B1B2. Let u, v, w be vectors with P = 〈u〉, A1 = 〈v〉 and B1 = 〈w〉. Then
A2 = 〈u+av〉 and B2 = 〈u+a ′ v〉 for some a, a ′ ∈ F \{0}. So we may assume
WOLG that
A1 = 〈v〉, A2 = 〈u + v〉, B1 = 〈w〉, and B2 = 〈u + w〉.

72 CHAPTER 2. ANALYTIC GEOMETRY
Let a, b be any elements of F \ {0, 1} (since 0 and 1 commute with each
element of F ). Put
A3 := 〈u + av〉, B3 := 〈u + bw〉.
Since a, b = 0, 1, we have A3 = A1, A2, P and B3 = B1, B2, P .
We claim that ab = ba if and only if the points Ci = Ai+1Bi−1 ∩Ai−1Bi+1,
i = 1, 2, 3, are collinear. Note that in computing a spanning vector of Ci we
do not use commutativity of F .
Similarly,
C3 = A1B2 ∩ A2B1 = 〈v, u + w〉 ∩ 〈u + v, w〉 = 〈u + v + w〉.
C2 = A1B3 ∩ A3B1 = 〈u + av, w〉 ∩ 〈v, u + bw〉 = 〈u + av + bw〉.
Finally, we get
C1 = A2B3 ∩ A3B2 = 〈u + v, u + bw〉 ∩ 〈u + av, u + w〉
= 〈u + av + bw〉.
Clearly this must be a 1-dimensional vector space, and we claim it is
C ′ = 〈(a + (a − 1)(b − 1) −1 ) · u + a · v + (a − 1)(b − 1) −1 · bw〉.
To see this, note first that
C ′ = 〈a · (u + v) + (a − 1)(b − 1) −1 · (u + bw)〉 ∈ 〈u + v, u + bw〉,
and second that
C ′ = (a + (a − 1)(b − 1) −1 + (a − 1)(b − 1) −1 b − (a − 1)(b − 1) −1 b) · u
+a · v + (a − 1)(b − 1) −1 · bw
= (a + (a − 1)(b − 1) −1 (1 − b) + (a − 1)(b − 1) −1 b) · u
+a · v + (a − 1)(b − 1) −1 · bw
= (a − (a − 1) + (a − 1)(b − 1) −1 b) · u + a · v + (a − 1)(b − 1) −1 · bw
= (u + av) + (a − 1)(b − 1) −1 b · (u + w)
∈ 〈u + av, u + w〉.
This proves the claim. We are now ready to analyze just when C1 ∈ C2C3.

2.2. THE THEOREMS OF DESARGUES AND PAPPUS 73
C1 ∈ C2C3
⇐⇒ 〈(a + (a − 1)(b − 1) −1 ) · u + a · v + (a − 1)(b − 1) −1 · bw〉
⊆ 〈u + v + w, u + av + bw〉
⇐⇒ there exist x, y ∈ F with
(a + (a − 1)(b − 1) −1 ) · u + a · v + (a − 1)(b − 1) −1 · bw〉
= xu + xav + xbw + yu + yv + yw
⇐⇒ the following equations hold :
a + (a − 1)(b − 1) −1 = x + y;
a = xa + y;
(a − 1)(b − 1) −1 b = xb + y.
The first equation means y = a + (a − 1)(b − 1) −1 − x, and together with
the second equation gives
hence
x = (a − 1)(b − 1) −1 (1 − a) −1 ;
y = a + (a − 1)(b − 1) −1 (1 − (1 − a) −1 ).
Substituting these formulas for x and y into the third equation we get (with
some computation!)

74 CHAPTER 2. ANALYTIC GEOMETRY
C1 ∈ C2C3
⇐⇒ (a − 1)(b − 1) −1 b
= (a − 1)(b − 1) −1 (1 − a) −1 b + a + (a − 1)(b − 1) −1 (1 − (1 − a) −1 )
⇐⇒ b = (1 − a) −1 b + (b − 1)(a − 1) −1 a + (1 − (1 − a) −1 )
⇐⇒ (1 − a)b = b + (1 − a)(b − 1)(a − 1) −1 a + (1 − a) − 1
⇐⇒ (1 − a)b
= b + (1 − a)(b − 1)(a − 1) −1 a
+(1 − a)(b − 1)(a − 1) −1 − (1 − a)(b − 1)(a − 1) −1 − a
⇐⇒ (1 − a)b
= b + (1 − a)(b − 1)(a − 1) −1 (a − 1) + (1 − a)(b − 1)(a − 1) −1 − a
⇐⇒ (1 − a)b
= b + (1 − a)(b − 1) − a + (1 − a)(b − 1)(a − 1) −1
⇐⇒ 0 = b − (1 − a) − a + (1 − a)(b − 1)(a − 1) −1
⇐⇒ 1 − b = (1 − a)(b − 1)(a − 1) −1
⇐⇒ (1 − b)(a − 1) = (1 − a)(b − 1)
⇐⇒ ba = ab.
Hence the theorem of Pappus holds in projective spaces of the form P(V )
if and only if the coordinatizing skewfield F is actually a field.
Corollary 2.2.8. By Wedderburn’s theorem we see that if F is finite, both
the theorem of Desargues and the theorem of Pappus hold in P(V ).
We are going to prove later that any projective space of dimension 3 or
greater can be viewed as a P(V ) for some vector space over some skewfield

2.2. THE THEOREMS OF DESARGUES AND PAPPUS 75
F . We have seen that in any projective space of dimension 3 or greater the
Theorem of Desargues must hold, and that any projective plane that can be
embedded as a plane in a projective space of dimension 3 or greater must be
Desarguesian. If the space P(V ) is finite, then the skewfield F is actually a
field, so the theorem of Pappus holds also. It will also be shown that any
Desarguesian plane can be coordinatized by a skewfield, which is a field in
the finite case. In the infinite case if we start with a 3-dimensional vector
space V over a skewfield F which is not a field (say, the quaternions), the
corresponding projective plane P(V ) is Desarguesian but not Pappian.
On the other hand, it was proved by Hessenberg that any Pappian plane
must be Desarguesian. We just sketch a proof of this below. There are many
finite projective planes that are not Desarguesian and hence not Pappian,
but we give them only passing mention in this book.
Theorem 2.2.9. Let π be a given Pappian plane (i.e., a projective plane in
which the Theorem of Pappus is valid).
(a) Let ABC and DEF be two triangles in π such that lines AD, BE,
CF meet in a point O and the lines AE, BF , CD meet in a point O ′ . Prove
that AF , BD, CE meet in a point.
(b) Prove that π is Desarguesian.
Proof. (Mostly left as an exercise.) For part (a), use the principle of duality.
For part (b), suppose the triangles ABC and A ′ B ′ C ′ are in perspective from
V . Define the following points:
B ′ C ′ ∩ AC = S, B ′ A ∩ CC ′ = T,
B ′ A ′ ∩ V S = P, BAV S = U,
BC ∩ B ′ C ′ = L, CA ∩ C ′ A ′ = M,
AB ∩ A ′ B ′ = N.
Use the Theorem of Pappus to show that
(i) the points U, T, L are collinear (consider the lines 〈B, B ′ , V 〉 and 〈S, C, A〉).
(ii) the points P, M, T are collinear (consider the lines 〈A, A ′ , V 〉 and 〈C ′ , S, B ′ 〉).
(iii) the points L, M, N are collinear (consider the lines 〈B ′ , A, T 〉 and 〈U, P, S〉).

76 CHAPTER 2. ANALYTIC GEOMETRY
2.3 Coordinates
From now on when we consider a projective space P(V ) we may consider V
to be a finite-dimensional vector space over the finite field F = Fq = GF (q)
for some prime power q. For such a V choose a basis {v0, v1, . . . , vd}. Then
each vector v ∈ V has a unique representation in the form v = d i=0 aivi,
i.e., it is uniquely determined by its coordinates (a0, a1, . . . , ad).
We defiine an equivalence relation ∼ on the set of (d+1)-tuples of elements
of F different from (0, 0, . . . , 0) by
(a0, a1, . . . , ad) ∼ (b0, b1, . . . , bd) : ⇐⇒
there is an a ∈ F \ {0} with (a0, . . . , ad) = a · (b0, . . . , bd).
We say that a point 〈v〉 of P(V ) has homogeneous coordinates (a0, a1, . . . , ad)
provided
d
〈v〉 = 〈 aivi〉.
i=0
Clearly two such (d + 1)-tuples are equivalent if and only if they are
homogeneous coordinates of the same point of P(V ). It is also common to
denote the equivalence class containing (a0, a1, . . . , ad) by (a0 : a1 : · · · : ad)
and then write
P = (a0 : a1 : · · · : ad) if P = 〈
d
aivi〉.
We then also call (a0 : a1 : · · · : ad) the homogeneous coordinates of P .
Homogeneous coordinates are quite convenient to use. For example, WOLG
we may assume that the first (or last) nonzero coordinate of a point is equal
to 1. The line through the points (a0 : a1 : · · · ad) and (b0 : b1 : · · · bd) consists
of the points with the following coordinates:
(a0 : a1 : · · · ad) and (b0 : b1 : · · · bd) + a · (a0 : a1 : · · · ad), a ∈ F.
In particular, the line through (1 : 0 : · · · : 0) and (0 : 1 : 0 : · · · : 0)
consists of the points
i=0
(1 : 0 : · · · : 0) and (a : 1 : 0 : · · · : 0), a ∈ F.

2.3. COORDINATES 77
Now we consider the description of higher-dimensional subspaces by means
of coordinates.
Theorem 2.3.1. Let P1, P2, . . . , Pt be points of P(V ) with homogeneous coordinates
Pi = (ai0 : ai1 : · · · : aid), 1 ≤ i ≤ t.
Then the set {P1, P2, . . . , Pt} is linearly independent provided the matrix
⎛
⎞
⎜
⎝
a10 a11 · · · a1d
.
.
at0 at1 · · · atd
has rank t
In particular, d + 1 points form a basis of V if and only if the matrix
whose rows are the homogeneous coordinates of the points is nonsingular.
Proof. If the above matrix has rank t, then the homogeneous coordinates
of the points span a t-dimensional subspace of V . So by Lemma 2.1.3 the
points P1, . . . , Pt span a (t − 1)-dimensional subspace of P(V ) and form an
independent set.
Conversely, suppose that P1, . . . , Pt form an independent set. If the rank
of the above matrix were smaller than t, then the homogeneous coordinates
of the points would span a subspace of V with dimension smaller than t.
Thus P1, . . . , Pt would span a subspace of P(V ) with dimension smaller than
t − 1, forcing P1, . . . , Pt not to be independent.
Theorem 2.3.2. Let V be a vector space of dimension d + 1 over the field
F , and let P(V ) be the corresponding projective space. Let H be a hyperplane
of P(V ). Then the homogeneous coordinates of the points of H are the
solutions of a homogeneous equation with coefficients iin F . Conversely, any
homogeneous equation that is different from the ’zero equation’ describes a
hyperplane of P(V ).
Proof. Let H ′ be the hyperplane of V that corresponds to H (Lemma 2.1.3(b)).
So P(H ′ ) = H. The coordinates of the vectors in H ′ are solutions of a homogeneous
equation. Conversely, we know from linear algebra that the solutions
of any homogeneous equation different from the zero equation are a subspace
of dimension d. Hence the space of solutions is a hyperplane of V and therefore
a hyperplane of P(V ).
.
⎟
⎠

78 CHAPTER 2. ANALYTIC GEOMETRY
Corollary 2.3.3. Any t-dimensional subspace U of a projective space of dimension
d given by homogeneous coordinates can be described by a homogeneous
system of d − 1 linear equations. More precisely, there exists a
(d − t) × (d + 1) matrix H such that a point P = (a0 : a1 : · · · : ad) is
a point of U if and only if
(a0 : a1 : · · · : ad) · H T = 0.
Proof. By Lemma 1.8.13 U is the intersection of d − 1 hyperplanes.
We can describe any hyperplane by a homogeneous equation. For example,
by the ’hyperplane x0 = 0’ we mean the hyperplane whose points
have homogeneous coordinates (0 : a1 : · · · : ad). We can also describe
a hyperplane by the coefficients of a corresponding homogeneous equation.
For example, the hyperplane x0 = 0 will be represented by homogeneous
coordinates [1 : 0 : · · · : 0].
Theorem 2.3.4. Let P = P(V ) be a projective space whose points are given
by homogeneous coordinates (a0 : · · · : ad) with ai ∈ F , not all zero. Then
any hyperplane of P can be represented by [b0 : b1 : · · · : bd], bi ∈ F not all
zero. Conversely, to any such (d+1)-tuple [b0 : b1 : · · · : bd] there corresponds
a hyperplane. Moreover, we have
(a0 : a1 : · · · : ad) I [b0 : b1 : · · · : bd] ⇐⇒
d
aibi = 0.
The proof follows directly from Theorem 2.3.2, since any homogeneous
linear equation in d + 1 variables can be described by a (d + 1)-tuple [b0 : b1 :
· · · : bd] of homogeneous coordinates.
Now we consider the dual geometry P △ of a projective space P. This
means that we consider P as a rank 2 geometry (two types of objects) whose
points are the hyperplanes of P and whose lines are the points of P.
Corollary 2.3.5. Let P = P(V ) be a coordinatized projective space. Then
P △ ∼ = P. In particular, P △ is coordiinatized as well. Therefore the principle
of duality holds for the class of all coordinatized projective spaces of fixed
dimension d.
At this point we introduce coordinates for affine spaces.
i=0

2.3. COORDINATES 79
Theorem 2.3.6. Let H∞ be the hyperplane of P(V ) with equation x0 = 0.
Then the affine space A = P \ H∞ can be described as follows:
• the points of A are the vectors (a1, . . . , ad) of the d-dimensional vector
space F d .
• the lines of A are the cosets of the 1-dimensional subspaces of F d ; i.e.,
the set u + 〈v〉, where u, v ∈ F d , v = 0.
• incidence is set-theoretical containment.
Proof. Since the homogeneous coordinates of the points on H∞ have as first
entry 0, any point P outside H∞has homogeneous coordinates (a0 : a1 : · · · :
ad) with a0 = 0. Thus it has homogeneous coordinates (1 : a1 : · · · : ad) with
uniquely determined a1, . . . , ad ∈ F . Hence we can identify a point P of A
with the d-tuple (a1, . . . , ad). We say that (a1, . . . , ad) are the inhomogeneous
coordinates of the point P .
For the lines of A, we know that any line g of A has precisely one point
(0 : b1 : · · · : bd) in common with H∞. If (1 : a1 : · · · : ad) is an arbitrary
point of A on g, then the points of A on g have the following homogeneous
coordinates:
(1 : a1 : · · · : ad) + a · (0 : b1 : · · · : bd) with a ∈ F.
It follows that the inhomogeneous coordinates of the points on g are as
follows:
(a1, . . . , ad) + a · (b1, . . . , bd) with a ∈ F.
In other words, the affiine points on g are precisely the points of the coset
(a1, . . . , ad) + 〈(b1, . . . , bd)〉
of the 1-dimensional subspace 〈(b1, . . . , bd)〉 of F d .
Conversely, let (a1, . . . , ad) + 〈(b1, . . . , bd)〉 be a coset of a 1-dimensional
subspace 〈(b1, . . . , bd)〉 of F d . Then, by the above construction, this coset
corresponds to the 2-dimensional subspace 〈((1, a1, . . . , ad), (0, b1, . . . , bd)〉 of
F d+1 . It is consequently a line of P(V ). Since this line intersects H∞ just in
the point (0 : b1 : · · · : bd), the coset in question is a line of A.

80 CHAPTER 2. ANALYTIC GEOMETRY
If H∞ is a hyperplane of P = P G(d, F ), then we denote the affine space
A = P \ H∞ by AG(d, F ) and call it the affine space of dimension d coordinatized
by F . If F = Fq, we also denote A by AG(d, q).
Later, when we study collineations of projective space, we will see that any
two hyperplanes of P G(d, F can be mapped onto each other by a collineation
. Hence AG(d, F ) is independent of the choice of the hyperplane H∞.
Remark: If the theorem of Desargues holds in P, then it also holds in
A = P \ H∞. However, when formulating the theorem of Desargues in affine
spaces one has to observe that two lines in a plane do not necessarily meet.
2.4 Semilinear Maps: A First Glance
In this section and later we often abbreviate the symbol P(V ) to P V .
Recall that σ : Fq → Fq : a ↦→ ap is an automorphism of Fq; σi : Fq →
Fq : a ↦→ api, and that {σ, σ2 , . . . , σe = id} is a complete list of all the
automorphisms of Fq. So Aut(Fq) is a cyclic group of order e generated by
σ : a ↦→ xp . See the Appendix 2 on algebra in finite fields for a great deal
more.
Let A be an invertible (n + 1) × (n + 1) matrix over Fq, α ∈ Aut(F ). If
a = (a0, . . . , an) ∈ V , then LA : V → V : a ↦→ (a0, . . . , an)A is an invertible
linear operator on V . Note that since LA is linear, LA(λa) = λ · LA(a), so
LA actually defines (i.e., induces) an invertible map ¯ LA : P V → P V . These
linear maps on P G(n, q) are called homographies and the set of all of them
forms a group denoted P GL(n + 1, q) and called the projective linear group.
Let α ∈ A = Aut(Fq). Consider the map α : V → V : (a0, . . . , an) ↦→
(aα 0 , . . . , aαn ). Note that α : (λa) ↦→ λα ·a α . So if a and λa represent the same
point in P V , then a α and λα ·a α represent the same point in P V . Hence T =
α ◦ LA : a ↦→ LA(a α ) = a αA is a bijection on V called a projective semilinear
map, and it induces a bijection on P V called a projective semilinear map.
The set of all these maps is a group denoted P ΓL(n + 1, q) and called the
projective semilinear group.
Let θ : P G(n, q) → P G(n, q) : (a0, . . . , an) ↦→ (a0, . . . , an)A. The hyperplane
[x0, . . . , xn] is mapped by θ to A−1 [x0, . . . , xn] T , since
(a0, . . . , an) · [x0, . . . , xn] T = 0 ⇐⇒ (a0, . . . , an)A · A −1 [x0, . . . , xn] T = 0.
Hence when θ is to be given explicitly it is usually helpful to give both the
matrix A and its inverse A −1 .

2.5. SOME COMBINATORICS OF FINITE GEOMETRIES 81
Let θ be a given homography. Later we shall show that here is a hyperplane
π such that θ fixes each point of π if and only if there is a point a such
that θ fixes all hyperplanes through a, in which case π is called the axis of θ
and a is called the center of θ.
The general projective linear (respectively, semilinear) groups and certain
of their subgroups will play an important role throughout this book.
We recall without proof (but will prove it later) a standard theorem from
the theory of projective spaces.
Theorem 2.4.1. Let P0 = (1, 0, . . . , 0), P1 = (0, 1, 0, . . . , 0), . . . ,
Pn = (0, . . . , 0, 1), Pn+1 = (1, 1, . . . , 1). So P0, P1, . . . , Pn+1 is an ordered
set of n + 2 points of P G(n, q), no n + 1 of which lie in a hyperplane . If
Q0, Q1, . . . , Qn+1 is any other (not necessarily distinct) ordered set of n + 1
points with no n + 1 in a hyperplane, then there is a unique homography of
P G(n, q) mapping Qi to Pi, 0 ≤ i ≤ n + 1.
If θ is a permutation of the points of P G(n, q) that maps the points of
each line to the points of some image line, it follows that it maps the points
of each r-dimensional subspace to the points of some r-dimensional subspace
and it is called a collineation of P G(n, q). According to the Fundamental
Theorem of Projective Geometry (to be proved later), each collineation of
P G(n, q) is a projective semilinear map.
Theorem 2.4.2. (Fundamental Theorem of Projective Geometry) Each
collineation of P G(n, q) is a projective semilinear map.
A duality of P G(n, q) is a bijection from the point-set of P G(n, q) to
the set of hyperplanes of P G(n, q) that induces a bijection from the set of
r-dimensional subspaces to the set of subspaces with co-dimension equal to
r. A polarity is a duality that has order 2 as a map. For example, the map
(x0, . . . , xn) ↦→ [x0, . . . , xn] T from points to hyperplanes is a duality that is a
correlation.
2.5 Some Combinatorics of Finite Geometries
The material in this section extends that in Section 1.10. Here we view P V
as a geometry, i.e., not only points and lines are elements of the geometry,
but subspaces of all dimensions.

82 CHAPTER 2. ANALYTIC GEOMETRY
A lattice L is a partially ordered set (poset) with the property that any
finite subset S ⊆ L has a meet (or greatest lower bound), that is, an
element b ∈ L for which
1. b ≤ a for all a ∈ S, and
2. if c ≤ a for all a ∈ S, then c ≤ b.
And dually, there is a join (or least upper bound), i.e., an element
b ∈ L for which
1. ′ a ≤ b for all a ∈ S, and
2. ′ if a ≤ c for all a ∈ S, then b ≤ c.
The meet and join of a two element set S = {x, y} are denoted, respectively,
by x ∧ y and x ∨ y. It is easily seen that ∧ and ∨ are commutative,
associative, idempotent binary operations. Moreover, if all 2-element subsets
have meets and joins, then any finite subset has a meet and a join.
The lattices we will consider have the property that there are no infinite
chains. Such a lattice has a (unique) least element (denoted ˆ0 or 0L), because
the condition that no infinite chains exist allows us to find a minimal element
m, and any minimal element m is a minimum, since if m ≤ a, then m ∧ a
would be less than m. Similarly, there is a unique largest element 1L (or ˆ1).
For elements a and b of a poset, we say a covers b and write a ·> b,
provided a > b but there are no elements c with a > c > b. For example,
when U and W are linear subspaces of a vector space, then U ·> W iff U ⊇ W
and dim(U) = dim(W ) + 1. A point of a lattice with ˆ0 is an element that
covers ˆ0. A copoint of a lattice with ˆ1 is an element covered by ˆ1.
Let Vn(q) denote an n-dimensional vector space over Fq = GF (q). The
term k-subspace will denote a k-dimensional subspace. It is fairly easy to
see that the poset Ln(q) of all subspaces of Vn(q) is a lattice with ˆ0 = {0}
and ˆ1 = Vn(q). We begin with some counting.
Exercise 2.5.0.1. The number of ordered bases for a k-subspace of Vn(q)
is: (q k − 1)(q k − q)(q k − q 2 ) · · · (q k − q k−1 ). How many ordered, linearly
independent subsets of size k are there in Vn(q)?
Solution: Choose a nonzero element v1 of Vn(q) in q n − 1 ways. Then
choose v2 so that {v1, v2} is independent in q n − q ways. Then choose v3 so
that {v1, v2, v3} is independent in q n − q 2 ways. Proceeding in this fashion

2.5. SOME COMBINATORICS OF FINITE GEOMETRIES 83
we see that there are (q n − 1)(q n − q)(q n − q 2 ) · · · (q n − q k−1 ) ordered k-tuples
of vectors of Vn(q) that form linearly independent sets.
To obtain a maximal chain (i.e., a chain of size n + 1 containing one
subspace of each possible dimension) in the poset Ln(q) of all subspaces of
Vn(q), we start with the 0-subspace. After we have chosen an i-subspace Ui,
0 ≤ i < n, we can choose an (i + 1)-subspace Ui+1 that contains Ui, in qn−qi qi+1−qi ways, since we can take the span of Ui and any of the qn − qi vectors not in
Ui. But an (i + 1)-subspace will arise exactly qi+1 − qi times in this manner.
Hence the number of maximal chains of subspaces in Vn(q) is:
This implies that
M(n, q) = (qn − q 0 )
(q 1 − q 0 ) · (qn − q 1 )
(q 2 − q 1 ) · · · (qn − q n−1 )
(q n − q n−1 ) =
= (qn − 1)q(q n−1 − 1)q 2 (q n−2 − 1) · · · q n−1 (q − 1)
(q − 1)q(q − 1)q 2 (q − 1) · · · q n−1 (q − 1)
= (qn − 1)(qn−1 − 1) · · · (q − 1)
(q − 1) n .
M(n, q) = (q n−1 + q n−2 + · · · + q + 1)(q n−2 + · · · q + 1) · · · (q + 1).
We may consider M(n, q) as a polynomial in q for each integer n. When
the indeterminate q is replaced by a prime power, we have the number of
maximal chains in the poset P G(n, q).
Note: When q is replaced by 1, we have M(n, 1) = n!, which is the
number of maximal chains in the poset of subsets of an n-set.
n
The Gaussian number (or Gaussian coefficient) can be de-
k
q
fined as
the number of k-subspaces of Vn(q). This holds for 0 ≤ k ≤ n, where
n
= 1.
0
q
n
To evaluate , count the number N of pairs (U, C) where U is a
k
q
k-subspace and C is a maximal chain that contains U. Since every maximal
chain contains one subspace of dimension k, clearly N = M(n, q). On the
=

84 CHAPTER 2. ANALYTIC GEOMETRY
other hand, we get each maximal chain uniquely by appending to a maximal
chain in the poset of subspaces of U – of which there are M(k, q) – a maximal
chain in the poset of all subspaces of Vn(q) that contain U. There are M(n −
k, q) of these, since the poset {W : U ⊆ W ⊆ V } is isomorphic to the poset
of subspaces of V/U, and dim(V/U) = n − k. Hence
which implies that
M(n, q) =
n
k
q
=
n
k
q
· M(k, q) · M(n − k, q),
M(n, q)
M(k, q)M(n − k, q) =
n
n − k
= (qn−1 + q n−2 + · · · + q + 1)(q n−2 + q n−3 + · · · + q + 1) · · · (q + 1)
(q k−1 + · · · + 1) · · · (q + 1)(q n−k−1 + · · · + q + 1) · · · (q + 1)
= (qn−1 + · · · + 1)(q n−2 + · · · + 1) · · · (q n−k + · · · + 1)
(q k−1 + · · · + 1) · · · (q + 1)
= (qn − 1)(qn−1 − 1) · · · (qn−k+1 − 1)
(qk − 1)(qk−1 .
− 1) · · · (q − 1)
There is a satisfactory way to generalize the notion and the notation
of Gaussian coefficient to the multinomial case. However, for our present
purposes it suffices to consider just the binomial case. Define (0)q = 1, and
for a positive integer j put (j)q = 1 + q + q 2 + · · · + q j−1 . Then put (0)!q = 1
and for a positive integer k, put (k)!q = (1)q(2)q · · · (k)q. So (n)!q = M(n, q).
With this notation we have
n (n)!q
=
. (2.1)
k
q (k)!q(n − k)!q
n
For some purposes it is better to think of as a polynomial in an
k
q
n
indeterminate q rather than as a of function of a prime power q. That
k
q
is a polynomial in q is an easy corollary of the following exercise.
q

2.5. SOME COMBINATORICS OF FINITE GEOMETRIES 85
Exercise 2.5.0.2. Prove the following recurrence:
n
k
n − 1
=
k
+ q n−k
n − 1
k − 1
q
Solution of exercise: Pick a hyperplane H, i.e., an (n − 1)-subspace of
n − 1
Vn(a). There are
k-subspaces contained in H. The others meet
k
q
n − 1
H in a (k−1)-subspace. Each of the
(k−1)-subspaces in H is con-
k − 1
q
n − k + 1
tained in
=
1
qn−k+1
−1
n − k
k-subspaces of V , of which
=
q−1
1
q
qn−k−1 q−1 are contained in H. This leaves qn−k+1−1 q−1 − qn−k−1 q−1
= qn−k+1−q n−k
= q q−1
n−k
not contained in H, from which the desired recurrence follows.
Exercise 2.5.0.3. Prove the following recurrence:
n + 1
k
n
=
k − 1
+ q k
n
k
q
Solution: (Hint: Use the two previous exercises.) Alternatively, the number
of k-dimensional subspaces of Vn(q) is equal to the number of k × n
matrices over F which are in reducedechelon form and have no zero rows.
n
This gives another way to calculate . But now the Gaussian coef-
k
q
ficients can be given a combinatorial interpretation for all positive integer
values of q greater than 1, not just prime powers. For let Q be any set of size
q, containing two distinguished elements called 0 and 1. Then the definition
of a matrix in reduced echelon form over Q makes sense, even through the
algebraic interpretation is lost. The number of k × n matrices in reduced
echelon form with no zero rows is given by a polynomial in q. But, for infinitely
many values (all the prime powers), this polynomial conicides with
the Gaussian coefficient (which is also a polynomial). Hence they are identically
equal. (Explain this.) Using this matrix interpretation we are able
to establish the recurrence relation for the Gaussian coefficients given in this
exercise.
Consider k × (n + 1) matrices in reduced echelon form, with no zero rows.
Divide them into two classes: those for which the leading 1 in the last row
q
q
.
q
.
q
q

86 CHAPTER 2. ANALYTIC GEOMETRY
occurs in the last column; and the others. Those of the first type correspond
to (k − 1) × n matrices in reduced echelon with no zero rows, since the last
row and column are zero apart from the bottom-right entry. Those of the
second type consist of a k × n matrix in reduced echelon with no zero rows,
with a column containing arbitrary elements adjointed on the right. Since
there are q k choices for this column, the recurrence relation follows.
Note that the relation of the previous exercise reduces to the binomial
recurrence when q = 1. However, unlike the binomial recurrence, it is not
‘symmetric’.
n
Exercise 2.5.0.4. Show that =
k
q
l≥0 αlql , where αl is the number of
partitions of l into at most k parts, each of which is at most n − k.
Solution: If A is a k × n matrix in reduce echelon form, suppose that the
leading 1 of row i is in position ci, 1 ≤ i ≤ k. The ith row has its first nonzero
entry equal to 1 in the ci position, and then zeros in k−i positions to the right
of the leading 1. The remaining (n−ci)−(k −i) = n−k −ci +i positions can
each be filled with any of q elements, giving qn−k−ci+i ways to fill in the ith
row. This means that there are ql ways to complete this row reduced echelon
matrix, where l = (n − k − c1 + 1) + (n − k − c2 + 2) + · · · + (n − k − ck + k).
It is easy to check that (n − k − c1 + 1, n − k − c2 + 2, . . . , n − k − ck + k) is a
partition of l into at most k (nonzero) parts, each of which is at most n − k.
Conversely, if (λ1, λ2, . . . , λk) is a partition of l into at most k parts, each of
which is at most n − k, we can put ci = n − k + i − λi to obtain the pattern
of 1’s in a row reduced matrix which can then be completed in ql ways.
n
If we regard a Gaussian coefficient as a function of the real variable
k
q
q (where n and k are fixed integers), then we find that the limit as q goes to
1 of a Gaussian coefficient is a binomial coefficient.
Exercise 2.5.0.5.
limq→1
n
k
q
=
n
.
k
Solution: Using L’Hospital’s Rule we have (for positive a and b):
q
limq→1
a − 1
qb − 1
= limq→1
aqa−1 a
=
bqb−1 b .

2.5. SOME COMBINATORICS OF FINITE GEOMETRIES 87
So
= lim
q → 1
n q − 1
qk
lim
q → 1
− 1
limq→1
n
k
n−1 q − 1
qk−1
· · ·
− 1
= n n − 1 n − k + 1
· · ·
k k − 1 1
q
=
lim
q → 1
n
.
k
n−k+1 q − 1
q − 1
n
So when the indeterminate q is replaced by 1 in we obtain
k
q
n
. k
Because of this reason the Gaussian coefficients are called the “q-analogs” of
the binomial coefficeint.
Exercise 2.5.0.6. (The q-binomial Theorem) Prove that:
(1 + x)(1 + qx) · · · (1 + q n−1 n
x) = q i(i−1)
n
2 x
i
q
i , for n ≥ 1.
Solution: The proof is a straightforward induction. For n = 1, both sides
are 1 + x. Suppose that the result is true for n. Then
n
(1 + q i
n
x) = q k)k−1)/2
n
· (1 + q
k
n x).
i=0
k=0
i=0
x
q
k
The coefficient of xk on the right is
q k(k−1)/2
n
+ q
k
q
(k−1)(k−2)/2
n
k − 1
= q k(k−1)/2
n
+ q
k
q
n−k+1
n
k − 1
= q k(k−1)/2
n + 1
,
k
q
as required.
q
q
n
Letting q → 1, we obtain the usual Binomial Theorem.
q

88 CHAPTER 2. ANALYTIC GEOMETRY
Exercise 2.5.0.7. Prove that:
n + m
k
q
=
k
i=0
n
i
q
m
k − i
Solution: From the previous exercise we have:
n+m
k=0
n + m
k
q
q
k(k−1)
2 x k
q
q
(n−i)(k−i) .
= (1 + x)(1 + qx) · · · (1 + q n−1 x) · (1 + q n x)(1 + q n+1 x) · · · (1 + q n+m−1 x)
= (1 + x)(1 + qx) · · · (1 + q n−1 x) · (1 + (q n x))(1 + q(q n x)) · · · (1 + q m−1 · (q n x))
=
n
i=0
n
i
q
q
i(i−1)
2 x i
·
m
j=0
m
j
q
· q j(j−1)
2 · (q n x) j
From the top line it is clear that the coefficient on xk is
n + m
k
q
q
k(k−1)
2 .
On the other hand, the coefficient on x k is seen to be
k
i=0
n
i
q
q
i(i−1)
2 ·
m
k − i
q
q
(k−i)(k−i−1)
2 q n(k−i) .
After some simplification, we obtain the equality asked for in the exercise.
Define the Gaussian polynomials gn(x) ∈ R[x] as follows: g0(x) =
1; gn(x) = (x − 1)(x − q) · · · (x − q n−1 ) for n > 0. Clearly the Gaussian
polynomials form a basis for R[x] as a vector space over R.
Theorem 2.5.1. The Gaussian coefficients connect the usual monomials to
the Gaussian polynomials, viz.:
(i) xn = n n
k
k=0 (x − 1) ;
k
(ii) xn =
n n
k=0 gk(x).
k
q
.

2.5. SOME COMBINATORICS OF FINITE GEOMETRIES 89
Proof. (i) is a special case of the binomial theorem. And (ii) becomes (i) if
q = 1. To prove (ii), suppose V, W are vector spaces over F = GF (q) with
dim(V ) = n and |W | = r. Here r = q t is any power of q with t ≥ n. Then
|HomF (V, W )| = r n .
Now classify f ∈ HomF (V, W ) according to the kernel subspace f −1 (0) ⊆
V . Given some subspace U ⊆ V , let {u1, . . . , uk} be an ordered basis of
U and extend it to an ordered basis {u1, . . . , uk, uk+1, . . . , un} of V . Then
f −1 (0) = U iff f(ui) = 0, 1 ≤ i ≤ k, and f(uk+1), . . . , f(un) are linearly
independent vectors in W . Now
r n =
(r − 1)(r − q) · · · (r − q n−r(U)−1 )
=
=
n
k=0
n
k=0
U⊆V
n
k
n
k
(Use the fact that
(r − 1)(r − q) · · · (r − q
q
n−k−1 )
=
(r − 1)(r − q) · · · (r − q
q
k−1 )
n
k=0
n
k
n
k
q
=
gk(r).
q
n
n − k
As r can be any power of q as long as r ≥ qn , the polynomials xn
and
n n
k=0 gk(x) agree on infinitely many values of x and hence must be
k
q
identical.
The inverse connection can be obtained from the q-binomial theorem (c.f.
Ex. 2.5.0.6
Exercise 2.5.1.1. Prove that
gn(x) =
n
i=0
n
i
q
q
(n−i 2 ) n−i i
(−1) x .
)
q

90 CHAPTER 2. ANALYTIC GEOMETRY
(Hint: In Ex. 2.5.0.6 first replace x with −x and then replace q with q −1
and simplify.)
If {an} ∞ n=0 is a given sequence of numbers we have considered its ordi-
nary generating function
n≥0 anxn and its exponential generating function
xn
n≥0
an . There is a vast theory of Eulerian generating functions defined
n!
by xn
n≥0
an . The next exercise shows that two specific Eulerian generating
n!q
functions are inverses of each other.
Exercise 2.5.1.2.
(−t)
k≥0
kq (k2) k!q
k≥0
tk
= 1.
k!q
(Hint: Compute the coefficient on t n separately for n = 0 and n ≥ 1.
Then use the q-binomial theorem with x = −1.)
Exercise 2.5.1.3. Gauss inversion: Let {ui} ∞ i=0 and {vi} ∞ i=0 be two sequences
of real numbers. Then
vn =
n
i=0
n
i
q
ui (n ≥ 0) ⇔ un =
n
i=0
(−1) n−i q (n−i
2 )
n
vi (n ≥ 0).
i
(Hint: Use Exercise 2.5.1.2
Solution: Let A(t) =
k≥0 aktk /k!q and B(t) =
k≥0 bktk /k!q. Then
A(t) = B(t)
i≥0 ti /i!q if and only if B(t) = A(t)
i≥0 (−1)iq (i 2) i t /i!q. Now
compute the coefficients on tn /n!q on the two sides of these two equations.
2.6 Some Geometry in P G(2, q)
We have not discussed conics yet, but the reader can check out the appropriate
latter sections if necessary to follow the next historical comment. In 1975
Bruen and Thas [BT75] showed that if F is a set of q + 1 points of P G(2, q)
and C is a nondegenerate conic in P G(2, q) such that each line meeting at
least two points of F misses C, then F must be a line (exterior to C). This
was extended by Segre and Korchmáros [SK77] to include the case q odd.
Ten years later Blokhuis and Wilbrink [BW87] generalized the result to the
following.
q

2.6. SOME GEOMETRY IN P G(2, Q) 91
Theorem 2.6.1. Let A and B be subsets of P G(2, q) such that
(i) |A| = |B| = q + 1;
(ii) A ∩ B = ∅;
(iii) Every line meeting A meets B.
Then B is a line.
Proof. Each line containing a point P of A contains precisely one point of B,
as otherwise there would be a line through P missing B. If B is not a line,
let Q and R be two points of B, so there must be some point R of 〈Q, R〉
through which there is some line ℓ containing no point of B, and hence no
point of A. Hence we may take A and B to be subsets of the affine plane
AG(2, q) = P G(2, q) \ ℓ, which we identify with F q 2. So each point a of
AG(2, q) will be denoted by a = a1 + i · a2, where a1, a2 ∈ Fq and i is and
element of F q 2 for which 1, i form a basis for F q 2 as a vector space over Fq.
Let a = a1 + i · a2 ∈ A and let b = b1 + i · b2, b ′ = b ′ 1 + i · b′ 2 ∈ B with
b = b ′ . Suppose we could have (a − b) q−1 = (a − b ′ ) q−1 , so there exists a
nonzero t ∈ Fq such that a − b = t(a − b ′ ). This implies that
(t − 1)a1 + b1 = tb ′ 1 (2.2)
(t − 1)a2 + b2 = tb ′ 2. (2.3)
Let ℓx + my + n = 0 be the equation of a line L containing the points a
of A and b of B.
Hence
so
From Eqs. 2.2 and 2.3 we find
ℓa1 + ma2 + n = 0 (2.4)
ℓb1 + mb2 + n = 0. (2.5)
(t − 1)(ℓa1 + ma2) + (ℓb1 + mb2) = t(ℓb ′ 1 + mb′ 2 ),
(t − 1)(ℓa1 + ma2 + n) + (ℓb − 1 + mb2 + n) = t(ℓb ′ 1 + mb′ 2 + n).
Then from Eqs. 2.4 and 2.5 it follows that
ℓb ′ 1 + mb′ 2 + n = 0,

92 CHAPTER 2. ANALYTIC GEOMETRY
implying that b ′ is on the line L, contradicting the observation made above
that each line through a point of A lies on a unique point of B. Hence we
have established that if a ∈ A and b, b ′ are distinct points of B, then
Put
(a − b) q−1 = (a − b ′ ) q−1 . (2.6)
f(x) =
(x − b) q−1 .
b∈B
By Eq. 2.6, we see that for each a ∈ A, f(a) is the sum of q + 1 distinct
elements, each of which is a (q + 1) st root of 1. This says that f(a) is the
sum of the distinct roots of x q+1 −1. Hence f(a) = 0. Sof(x) is a polynomial
of degree at most q − 1 for which f(a) = 0 for each of the q + 1 values of
a ∈ A, i.e., f(x) is identically zero. On the other hand the coefficient of x q−1
in f(x) is q + 1 = 1 ∈ Fq. This contradiction leads us to conclude that B
actually must be a line.
2.7 Some Geometry in P G(3, q)
Here we collect some rather elementary results that are needed later.
Lemma 2.7.1. In Σ = P G(3, q) the number of lines
(i) in Σ is (q 2 + q + 1)(q 2 + 1);
(ii) skew to a given line is q 4 ;
(iii) meeting two skew lines is (q + 1) 2 ;
(iv) skew to each of two skew lines is q(q 2 − 1)(q − 1);
(v) meeting three skew lines is q + 1.
Proof. These facts have easy proofs that we leave as exercises.
Lemma 2.7.2. (J. A. Thas [Th73]) If ℓ is a set of q + 1 points in P G(3, q)
which has a non-empty intersection with every every plane, then ℓ is a line
of P G(3, q).
Proof. Let ℓ = {x0, x1, . . . , xq} and suppose that ℓ is not a line. Then the
line ℓ ′ = 〈x1, x2〉 contains a point y with y ∈ ℓ. Let ℓ ′′ be a line such that
y ∈ ℓ ′′ and ℓ ′′ ∩ ℓ = ∅. Since each plane through ℓ ′′ contains at least one
point of ℓ (by hypothesis), and since ℓ ′′ ∩ ℓ = ∅ and |ℓ| = q + 1, each plane
through ℓ ′′ must contain exactly one point of ℓ. As |ℓ ∩ 〈ℓ ′ , ℓ ′′ 〉| ≥ 2, we have
a contradiction. Hence ℓ must be a line.

2.7. SOME GEOMETRY IN P G(3, Q) 93
Start with three mutually skew lines ℓ1, ℓ2, ℓ3, so no plane contains any
two of these lines. If P is any point on ℓ2, the plane π = 〈ℓ1, P 〉 meets ℓ3
in a unique point R. So the line 〈P.R〉 meets all three lines ℓ1, ℓ2, ℓ3. If we
do this for each point P of ℓ2, we get q + 1 lines m1, m2, . . . , mq+1 that are
pairwise skew and are all transversals of the the original triple of lines. Now
take any three of the transversals and repeat the process to get q + 1 lines
ℓ1, ℓ2, . . . , ℓq+1 (including the original triple of lines) that are pairwise skew.
The set {ℓ1, . . . , ℓq+1} of q + 1 pairwise skew lines is called a regulus, and the
set {m1, . . . , mq+1} is its opposite regulus. Using the fact that the group of
homographies is sharply transitive on ordered sets of five points with no four
on a plane, it is a routine exercise to show that the homography group is
transitive on reguli.
Let S = (P, B, I) be a point-line geometry with (nonempty) point set P ,
(nonempty) line set B and incidence relation I. We say that S is a partial
geometry P aG(α, s, t) provided the following hold:
P aG (i) Each line of S has 1 + s points, s ≥ 2.
P aG (ii) Each point of S is on t + 1 lines.
P aG (iii) If x ∈ P , M ∈ B, and x is not incident with M, then there are
α points of M collinear with x.
Often we use the following notation. If x and y are points of an incidence
geometry for which there is a line ℓ incident with both x and y, we write
x ∼ y. If we want to indicate that no such line exists, we write x ∼ y..
Similarly, of ℓ and m are two lines through a common point, we write ℓ ∼ m,
and if the lines ℓ and m are known to have no point in common, we write
ℓ ∼ m.
We now give a ”classical” example. The reader will probably see immediately
how to generalize this example, but we concentrate on this particular
case because of its usefulness in the study of certain generalized quadrangles.
Construction 2.7.3. The partial geometry H3 q . Start with a fixed line
ℓ0 in P G(3, q). Let P be the set of q3 + q2 points of P G(3, q) \ ℓ0. Let B
be the set of q4 lines of P G(3, q) not intersecting ℓ0. Finally, let I be the
incidence inherited from P G(3, q). Then S = (P, B, I) := H 3 q
clearly is an
incidence geometry with q + 1 points on each line and q 2 lines through each
point. Now suppose x ∈ P and ℓ ∈ B with x not incident with ℓ. Let π be
the plane π = 〈x, ℓ〉. Then ℓ0 must meet π in a unique point y. The line
m = 〈x, y〉 of π meets ℓ in a point z. But m is the only line joining x with a

94 CHAPTER 2. ANALYTIC GEOMETRY
point of ℓ that also meets ℓ0. Hence the other q points of ℓ are collinear with
x on a line of S. It follows that S is a partial geometry P aG(q, q, q 2 − 1)
which we denote by H 3 q .
Suppose that ℓ1 and ℓ2 are two distinct lines of S meeting at a point y.
Let m1 and m2 be two distinct lines both meeting ℓ1 and ℓ2 at points different
from y. Since the plane π = 〈ℓ1, ℓ2〉 must contain both lines µ1 and µ2, those
two lines also must meet at a point. This says that S satisfies the Axiom of
Veblen. It turns out to be useful in the study of GQ that a converse result
holds.
Theorem 2.7.4. (J. A. Thas and F. De Clerck [TDC77]) Let S = (P, B, I)
be a partial geometry satisfyinhg P = ∅ = B and
(i) Each line of S has 1 + q points, q ≥ 2.
(ii) Each point of S is on q 2 lines.
(iii) If x ∈ P , M ∈ B, and x is not incident with M, then there are
α = q points of M collinear with x.
(iv) S satisfies the Axiom of Veblen: Let L1, L2 be distinct lines of S
concurrent at a point x. Let M1, M2 be distinct lines of S not incident with
x but both concurrent with both L1 and L2. Then M1 meets M2.
Then S is isomorphic to H 3 q .
Proof. 1. |P | = q 3 + q 2 ; |B| = q 4 .
Proof. Fix L ∈ B. L is incident with 1 + q points. Each point of
P \ L is collinear with q points of L. Each point of L is collinear
with (∗q 2 − 1)q points of P \ L. So the total number of points is
1 + q + (1 + q)(q 2 − 1)q/q = q 2 + q 3 . And (q 2 + q 3 )q 2 = |B|(1 + q),
implying |B| = q 4 .
2. Let L, M be distinct lines concurrent at the point x. Let N and N ′ be
distinct lines, not through x, meeting L and M at four distinct points.
The q lines through x meeting N are also the q lines through x meeting
N ′ .
Proof. By the axiom of Veblen (AV), N and N ′ meet at a point y. Label
z = N ∩ M. Let M ′ be any line through x meeting N, L = M ′ = M.

2.7. SOME GEOMETRY IN P G(3, Q) 95
So N and M are two lines through z, and both M ′ and N ′ meet both
N and M (at four distinct points different from z). Hence by AV it
must be that M ′ meets N ′ .
3. Let y, z be noncollinear points, L1IyIL2, L1 = L2 and M1IzIM2,
M1 = M2. If M1 meets both L1 and L2 and M2 meets L2, then M2
meets L1 also.
Proof. Let x = M1 ∩ L2. Then M2 meets L2 and M1, L1 meets L2 and
M1. So by AV, M2 meets L1.
This last result says, in effect, that if x and y are not collinear, if xIL1,
and if M1, . . . , Mq xL1, . . . , Lq are the lines through x meeting M1, then
Li meets Mj for all i, j, 1 ≤ i, j ≤ q.
4. Now suppose that L, M are distinct lines concurrent at x. We define a
substructure S(L, M) = (P ∗ , B ∗ , I ∗ ) of S. Here B ∗ = B1 ∪ B2, where
B1 is the set of q(q − 1) lines N for which x IN and N meets both L
and M, and B2 is the set of the q lines through x and concurrent with
some line N ∈ B1 (and hence concurrent with all lines of B1 by Step
2. P ∗ is the set of points incident with a line of B ∗ .
Each point on some line of S(L, M) through x is on q lines of S(L, M).
If y is a point of S(L, M) not collinear with x, it follows from Step 3
that y is on q lines of B1. Hence S(L, M) is a partial geometry with
parameters s ∗ = q, t ∗ = q − 1, α ∗ = q. In fact, this just says S(L, M)
is the dualo of an affine plane of order q. Hence any two lines N1, N2
of S(L, M) must concurrent, and S(L, M) = S(N1, N2). We also note
that if x ∈ P ∗ , N ∈ B ∗ , and x is not on N, then there is just one
substructure S(L, M) = (P ∗ , B ∗ , I ∗ ) for which x ∈ P ∗ and N ∈ B ∗ .
Hence this structure is also denoted S(x, N).
Note: S(L, M) has q 2 lines, q + q 2 points, each two lines meet, and
“parallelism” is an equivalence relation on the points.
5. Let S(L, M) = (P ∗ , B ∗ , I ∗ ) be as above. Then each line not in B ∗ has
a unique point in P ∗ .
Proof. Clearly any line with at least two points of P ∗ must lie in B ∗ .
Let N be a line of B \ B ∗ . There is a unique point z of N with x

96 CHAPTER 2. ANALYTIC GEOMETRY
not collinear with z, so each line through z is incident with at most
one point of P ∗ . If L = L1, M = L2, . . . , Lq are the lines of S(L, M)
through x, there are q lines through z meeting each Li, accounting for
q 2 lines through z. But this is all the lines through z, so N is one of
the lines meeting a line of B ∗ 2.
Let x, y ∈ P , x ∼ y. There are q2 /q = q subgeometries S(L, M) which
contain x and y. They are denoted S∗ ∗
i = (Pi , B∗ i , I∗ i ), 1 ≤ i ≤ q. The
“line” xy is defined to be the set xy = ∩iP ∗
i .
6. Each two distinct points of xy are non-collinear in S.
Proof. Suppose z1, z2 ∈ xy, z1 = z2 and z1 ∼ z2, i.e., z1 and z2 are
collinear in S, say on a line L. Then L ∈ B∗ i for 1 ≤ i ≤ q. If x is not
incident with L, then there is only one structure S(L, M) containing x
and L. But x and L are in all the S∗ i , so it must be that xIL. Similarly,
yIL, contradicting the hypothesis that x ∼ y. It is also clear that the
lines xy and z1z2 coincide.
The lines L ∈ B are said to be of the first type, and the lines xy, x ∼ y,
are of the second type. Since xy is a set of pairwise noncollinear points
of the dual affine plane S ∗ 1, we can show that |xy| ≤ q.
Since S(L, M) is a dual affine plane, the relation on P ∗ of being noncollinear
is an equivalence relation on P ∗ , with equivalence classes of
size q. So clearly |xy| ≤ q. One goal is to show that always |xy| = q.
First, note that each poiknt of S is collinear with 1 + q 3 points, and
hence not collinear with (q 2 +q 3 )−(q 3 +1) = q 2 −1 points. And if x, y, z
are three points with x ∼ y ∼ z, then if x ∼ z, y would be not collinear
with two points of xz. Hence “noncollinearity” is an equivalence relation
of the q 2 + q 3 points of S, with equivalence classes of size q 2 . (We
allow a point to be not collinear with itself for this relation.)
7. Let x, y be two noncollinear points of S. Let S ∗ i
∗ = (Pi , B∗ i ), i = 1, 2,
be two subgeometries of the type S(L, M) that contain x and y. Then
|P ∗ 1 ∩ P ∗ 2 | = q.
Proof. If z1 ∈ P ∗ 1 with x ∼ z1 ∼ y, then z1 ∈ P ∗ 2 , so each of the q2
lines through z1 has a unique point of P ∗ 2 , leaving only q points of P ∗ 2

2.7. SOME GEOMETRY IN P G(3, Q) 97
not collinear with z1. Hence we may choose z1 ∈ P ∗ 1 , z2 ∈ P ∗ 2 , with
x ∼ z1 ∼ y, x ∼ z2 ∼ y, z ∼ z2.
Put Li = xzi, i = 1, 2; Mi = yzi, i = 1, 2; N = z1z2; and let u be the
point of N not collinear with y, so also x ∼ u, since being noncollinear
is an equivalence relation.
Let x1 ∈ L1, x = x1 = z1. So u ∼ x1 , and we put U = x1u. Since U
and L2 both meet L1 and N (at points other than z1, their intersection),
by AV, U meets L2 at a point x2. Now x1 is collinear with z1 on M1,
and x1 ∼ y, since x is the unique point of L1 not collinear with y. Let
w1 be the unique point of M1 not collinear with x1, so z1 = w1 = y.
Put V = uw1. Since V and M2 meet M1 and N (at points other than
z1), by AV, V meets M2 at a point w2.
Coonsider a point ri on M1, ri ∈ {z1, w1, y}, 1 ≤ i ≤ q −2. Since u ∼ y,
clearly u ∼ ri, so put Ri = uri. Ri and M2 both meet N and M1 at
four points distiinct from z1, so they have a point si in common.
Since x1 ∼ w1, we know x1 ∼ ri in S ∗ 1. If x2 ∼ si for a particular i,
then U and Ri met at u; x1ri meets Ri at ri and meets U at x1. And
x2si meets Ri at si and meets U at x2. By AV, x1ri meets x2si at a
point ti. As x2 ∈ L2 in S ∗ 2 , si ∈ S ∗ 2 , x2si ∈ B ∗ 2 . So ti ∈ P ∗ 1 ∩ P ∗ 2 .
There are now two cases:
Case 1: w2 happens to be the point of M2 not collinear with x2. In
this case, x2 ∼ si for 1 ≤ i ≤ q −2, giving P ∗ 1 ∩P ∗ 2 = {x, y, t1, . . . , tq−2},
so |P ∗ 1 ∩ P ∗ 2 | = q.
Case 2: x2 ∼ w2. In this case the above construction of the points ti
leads to q − 3 points t1, . . . , tq−3, so even in this case |P ∗ 1 ∩ P ∗ 2 | ≥ q − 1.
But {x, y, t1, . . . , tq−3} is part of a class of q pairwise noncollinear points
of S∗ 2 . Let n be the missing point. So n ∈ P ∗ 2 , and n ∼ x, n ∼ y,
n ∼ ti, 1 ≤ i ≤ q − 3.
Let T be a line of B ∗ 1 containing x. Let S ∗ 3 = (P ∗ 3 , B ∗ 3, I ∗ 3) be the dual
affine plane S(n, T ). So P ∗ 2 ∩P ∗ 3 ⊃ {x, n}, and by the above paragraphs,
we know |P ∗ 2 ∩ P ∗ 3 | ≥ q − 1. And P ∗ 2 ∩ P ∗ 3 is contained in the set
of q mutually noncollinear points of P ∗ 2 containing n. Consequently
|P ∗ 1 ∩ P ∗ 2 ∩ P ∗ 3 | ≥ q − 2 ≥ 2 (use q > 3). Hence {y, t1, . . . , tq−3} ∩ P ∗ 3 =
∅. Let d ∈ {y, t1, . . . , tq−3}. Since P ∗ 1 , respectively P ∗ 3 , is uniquely

98 CHAPTER 2. ANALYTIC GEOMETRY
determined by T and d, it must be that S ∗ 1 = S∗ 3 . So n ∈ P ∗ 1 and
|P ∗ 1 ∩ P ∗ 2 | = q.
8. Let S ∗ i
= (P ∗
i , B∗ i , I∗ ) be the q dual affine planes S(L, M) containing x
and y, 1 ≤ i ≤ q. Put xy = ∩iP ∗
i . Then |xy| = q.
Proof. Since |P ∗
i ∩ Pi| = q, i = 2, 3, . . . , q, the set P ∗
i
points of P ∗ 1 not collinear with x.
contains the q
Let xy and xz be two different lines of the second type. Then y ∼ z,
since noncollinearity is an equivalence relation. Let P∗ denote the set
of points of S not collinear with x, including x. So |P∗| = q 2 . If B∗
is the set of lines of the second type contained in P∗, and if I∗ is the
natural incidence relation. then S∗ = (P∗, B∗, I∗) is an affine plane of
order q.
From now on: Subgeometries (dual affine planes) of the form S(L, M)
are planes of the first type. Affine planes S∗ are planes of the second
type.
9. Each three points which do not belong to a line of the first or second
type are contained in exactly one plane (of the first or second type).
Proof. Almost immediate.
Definition Two lines of the second type are parallel if they coincide
or if they are disjoint subsets of the pointset of a plane of the first or
second type.
10. Parallelism of lines of the second type is an equivalence relation.
Proof. Clearly parallelism is reflexive and symmetric. We show that it
is transitive. Suppose that xy, uv, pq are distinct lines of the second
type, and that xyuv and xypq. Assume that the plane containing
xy and uv (or xy and pq) is of the first type. Consider the planes w =
〈xy, uv〉, w ′ = 〈xy, pq〉. If w = w ′ , then clearly uvpq. So suppose w =
w ′ and that w is of the first type. Since w = w ′ , and w ′ is determined
by xy and any point of pq, clearly pq has no point in common with the
plane w.

2.7. SOME GEOMETRY IN P G(3, Q) 99
There are q + 1 planes through pq, one of them being of the second
type. (Let L be a line of the first type of w. L has one point r not
collinear with the points of pq. And r is in a unique line of the second
type parallel to pq. Each of the other q points of L determines with
pq a unique plane of the first type, which must meet w in a line of the
second type parallel to pq.) But these q + 1 lines of the second type in
which the planes through pq meet w are all the q + 1 lines of w of the
second type. Hence one of them must be uv, implying pquv.
The remaining case is when both w = 〈xy, uv〉 and w ′ = 〈xy, 〉pq〉 are
planes of the second type. Here the points of xy, uv and pq belong to
one equivalence class C of noncollinear points, i.e., w = w ′ = C. And
C is an affine plane in which line parallelism is an equivalence relation.
Any parallel class of lines is a partitiion of the pointset P of S. the
parallel classes are called points of the second type, and the set of these
classes is denoted by P .
Next we define a parallelism on the set of planes of the second type. Let
w be a plane of the second type, and x a point not in w. Clearly each
of the lines of the first type through x has a unique point of w. Let w ′
be the unique plane of the second type containing x. Let L1, . . . , Lq+1
be the lines (of the second type) through x (contained in w ′ ). There
is a unique line xyi ⊆ w ′ lying in the dual affine plane containing Li
and x, and hence parallel to Li. It follows that if w, w ′ are any two
planes of the second type, then any linie of one is parallel to some line
of the second. Hence we say that each two planes of the second type
are parallel, and this parallel class partitions the points of S. The set
consisting of this parallel class is denoted L.
Now we introduce a new incidence structure S = (P , B, I):
P = P ∪ P (points of S, and parallel classes of lines of second type);
B = the set of lines of the first, second and third types.

100 CHAPTER 2. ANALYTIC GEOMETRY
Incidence I requires a few cases:
For x ∈ P, L ∈ B, xIL iff xIL.
For x ∈ P, yz a line of second type, xIyz iff x ∈ yz.
For x ∈ P, x is not Iincident withL.
For [yz] ∈ P (a parallel class of lines of second type), L ∈ B,
[yz] is not I incident with L.
For [yz] ∈ P , uv a line of second tpe,
[yz] is I incident with uv iff uv ∈ [yz]
For [yz] ∈ P , [yz] is Iincident with L.
Note : yz is contained in one of the q + 1 planes
We prove that S ∼ = P G(3, q).
of the second type.
First, it is clear that each two elements of P are I- incident with exactly
one of element of B. For example, if x ∈ P , [yz] ∈ P , then if x ∼ M ∼ y
for M ∈ B, there is a line xr of the second type (in the dual affine plane
containing x and yz). Then xIxr and xrI[yz].
Now we prove that any three distinct elements of P not incident (for
I) with a common element of B generate a projective plane. Let x, y, z
be three such points. they cannot all be in P , since all such points are
I-incident with L. So let x ∈ P .
Case 1 y, z ∈ P , xy, xz, yz all lines of the second type.
Here the points x, y, z belong to an affine plane (plane of second type)
that is coompleted to a projective plane by adjoining L and its points.
Case 2 x, y, z ∈ P with some two on a line of B, say L = yz ∈ B.
So S(x, L) is a dual affine plane, partitioned by parallel lines of second
type, say belonging to the point [pq] ∈ P . Then S(x, L) is completed
to a projective plane by adjoining the point [pq] and the q + 1 lines of
[pq] contained in S(x, L).

2.8. POLARITIES, ETC. 101
Hence by (Veblen and Young [VY38]), S is the design of points and
lines of a projective space. Clearly S ∼ = P G(3, q). Also clearly P
consists of the points not I-incident with L, and B consists of the lines
not concurrent with L. Hence S ∼ = H 3 q .
2.8 Polarities, Etc.
Let Π(n, q) denote the set of hyperplanes of P G(n, q). Think of Π(n, q) as the
dual of P G(n, q) in the sense that Π(n, q) is a projective space of dimension
n. The subspaces of dimension r of Π(n, q) are the subspaces of dimension
n − r − 1 of P G(n, q).
Def. A reciprocity σ of P G(n, q) is a collineation from P G(n, q) onto the
dual space Π(n, q). Let σ be a reciprocity of P G(n, q). Let Sr be a subspace
of dimension r of P G(n, q). Let Sr = 〈P0, P1, . . . , Pr〉, where the points Pi
necessarily are linear independent. We define σ(Sr) = S σ r by
S σ r = ∩{P σ
i : 0 ≤ i ≤ r}.
Note: The dimension of Sσ r is n − r − 1. For example, in P G(3, q) a
reciprocity σ maps points to planes, lines to lines, and planes to points.
It follows from the definition that under a reciprocity σ of P G(n, q), incidence
is preserved. Inother words, if X is a point, and Sn−1 is a hyperplane,
under a reciprocity σ,
X is incident with Sn−1 ⇐⇒ Sn−1 σ is incident with X σ .
Since a reciprocity is a collineation between dual projective spaces, it follows
from the fundamental theorem of projective geometry that a reciprocity
σ of P G(n, q) is of the type:
X ↦→ X σ = ((X) α A) T = A T (X α ) T = A T (X T ) α ,
where A is a non-singular (n + 1) × (n + 1) matrix over Fq, and α ∈ Aut(Fq).
Consider a point X of a hyperplane Sn−1, where X is identified with a
row vector of homogeneous coordinates and Sn−1 is identified with a column
vector of homogeneous coordinates. Under a reciprocity σ we have:

102 CHAPTER 2. ANALYTIC GEOMETRY
Therefore
XSn−1 = 0 ⇐⇒ (XSn−1) αT = 0
⇐⇒ (S T n−1) α (X T ) α = 0
⇐⇒ (S T n−1 )α (A T ) −1 A T (X T ) α = 0
⇐⇒ (S T n−1 )α A −T X σ = 0.
σ : Sn−1 ↦→ (S T n−1 )α A −T .
If we replace A by A T , we have the following:
Lemma 2.8.1. X σ = A(X T ) α ⇐⇒ S σ n−1 = (S T n−1) α A −1 .
Thus we have for a point X of P G(n, q):
σ 2 (X) = σ(A(X T ) α ) = (((A(X T ) α ) T ) α A −1 = X α2
(A T ) α A −1 .
Hence σ 2 is a collineation of P G(n, q) with matrix (A T ) α A −1 and automorphism
α 2 .
Consider the possibility that a reciprocity σ is such that σ 2 acts as the
identity. In that case σ is said to be involutory and is called a polarity. We
have for all points P :
σ 2 (P ) = P ⇐⇒ (A T ) α A −1 = λI for some nonzero λ ∈ Fq, and α 2 = 1,
⇐⇒ (A T ) α = λA, and α 2 = 1,
⇐⇒ A α = λA T , and α 2 = 1,
⇐⇒ A = A α2
= λ α (A T ) α = λ α λA, and α 2 = 1.
It follows that λ α+1 = 1. If α = 1, then λ 2 = 1, so λ = ±1. The case for
q odd is somewhat different from that for q = 2 e .
Def. Let σ be a polarity of P G(n, q):
(a) Let q = p e with p odd.
1. If n is odd, A T = −A and α = 1, then the polarity σ is called a
symplectic polarity or null polarity.

2.8. POLARITIES, ETC. 103
2. If A T = A and α = 1, then σ is said to be an orthogonal polarity.
3. If A α = λA T , α 2 = 1, then σ is called a unitary or hermitian polarity.
(b) Let q = 2 e .
1. If n is odd, A T = A, α = 1, and if all the diagonal elements of A are
zero, then σ is called a symplectic polarity or null polarity.
2. If A T = A and α = 1, and if not all the diagonal elements of A equal
zero, then σ is called a pseudo polarity.
3. If A α = λA T and α 2 = 1 with α = 1, then σ is called a unitary polarity
or hermitian polarity.
Two cases occur depending on whether the associated automorphism α
is the identity or not.
Def. A reciprocity σ whose associated automorphism α is the identity is
called a correlation.
Let σ be an involutory correlation with matrix A. Then for all points P
σ 2 (P ) = P ⇐⇒ A = λ(A T ) = λ 2 A ⇐⇒ A = λ(A T ) and λ 2 = 1.
In any case a point P is called the pole of its polar P σ .
If σ is a polarity, i.e., σ 2 = 1, then P ∈ Q σ ⇐⇒ (P σ ) σ ∈ Q P σ ⇐⇒ P ∈
Q σ .
Similarly, if σ is a null polarity (i.e., symplectic polarity), then each point
P lies on its polar P σ = AP T . This is equivalent to P AP T = 0. If the
characteristic of Fq is not 2, then
P AP T = (P AP T ) T , since P AP T is a 1 × 1 matrix,
Therefore P AP T = 0.
= P A T P T
= −P AP T , since A is skew-symmetric.

104 CHAPTER 2. ANALYTIC GEOMETRY
Now suppose that the characteristic of Fq is 2. Let P = (p0, p1, . . . , pn)
be any point of P G(n, Q). Then
P AP T =
2aijpipj, where A = (Aij) with aii = 0 for all i,
i=j
= 0, since the characteristic of Fq is 2.
Let σ be a null polarity of P G(3, q). If ℓ is a line, and if ℓ σ = ℓ, then ℓ is
said to be a self-polar or totally isotropic line of P G(3, q).
Lemma 2.8.2. The self-polar lines that pass through a given point P are all
the lines through P in the polar plane P σ of P .
Proof. Let Q be a point of P σ . It is clear that Q σ contains both P and Q.
Thus (P Q) σ = P Q. In other words, lines of the pencil in P σ with P as vertex
are self-polar. Conversely, if ℓ is a self-polar line through P and P1 (= P ),
then ℓ = ℓ σ = P σ ∩ p σ 1, forcing ℓ ⊂ P σ .
Note: The case of an involutory reciprocity σ in P G(n, q), q = p h , p a
prime, where the associated automorphism α is not the identity is studied in
more detail in Chapter 21 Such an involutory reciprocity is called a unitary
or hermitian polarity. The pole-polar property still holds. Recall that in
this case we have
A α = λA T ; α 2 = 1, A = λ α λA T .
Here AutFq is a cyclic group of order h. Therefore a hermitian polarity
can occur only if h is even, and consequently q is a square and the automorphism
α is the Frobenius automorphism a ↦→ a √ q . If w is a primitive element
of Fq, then since λ √ q+1 = 1, we have
λ = w i(√ q−1) , 0 ≤ i ≤ q.
Consider the case where λ = w i(√ q−1) , for some i with 0 ≤ i ≤ q. Put
B = w −i A. Then
A α = λA T =⇒
B α = w −i√ q A α = w −i √ q λA T = w −i √ q · w i( √ q−1) A T = w −i A T = B T .
As B = w −i A is defined over Fq, the matrices A and B yield the same hermitian
polarities. Thus later when we study hermitian polarities of P G(n, q)
it will be sufficient to consider only the case A T = A √ q .

2.9. THE PROJECTIVE LINE - CROSS RATIO 105
2.9 The Projective Line - Cross Ratio
Let ℓ be a projective line (embedded in a projective space of higher dimension
or not). So we can view ℓ as a 2-dimensional vector space V over the field
F . Let ∞ be an abstract symbol and extend the algebra of F to that in
˜F = F ∪ {∞} in the following way.
∞ + ∞ = ∞; ∞ × ∞ = ∞; a × ∞ = ∞ ∀a ∈ F ∗ = F \ {0}
a
∞
= 0 and a + ∞ = ∞ ∀a ∈ F ; ∞
∞
and 0 × ∞ are not defined.
Let (P1, P2, P3, P4) be an ordered 4-tuple of points on a line ℓ with P1, P2, P3
distinct. Choose a basis B ′ = (e0, e1) such that the coordinate matrix [Pi]B ′
of Pi with respect to the ordered basis B ′ is given as follows:
0
[P1]B ′ =
1
1
; [P2]B ′ =
0
1
; [P3]B ′ =
1
If [P4]B ′ = [a0, a1] T , the affine coordinate (or non-homogeneous coordinate)
for P4 is the element a1/a0 ∈ ˜ F . So P1 has affine coordinate ∞, P2
has affine coordinate 0, and P3 has affine coordinate 1. The points of ℓ are
in one-to-one correspondence with the elements of ˜ F . A general element θ of
P ΓL(V ) is given by a map (on the coordinate matrices) given by a matrix
A and an automorphism α of F as:
θ : [a0 : a1] T ↦→ [b0 : b1] T = A[a α 0 : a α 1 ] T , where A =
d c
b a
.
, △ = ad−bc = 0.
As a permutation of the affine coordinates of points on ℓ (in case a0 = 0 = b0)
this is easily interpreted to be
θ : x ↦→ axα + b
cxα + d .
In fact with the appropriate interpretation of algebra with ∞ this works out
all the time. We use ∞α a∞+b a
a∞+b
= ∞; = when c = 0; = ∞ when
c∞+d c c∞+d
c = 0; axα +b
cxα +d = ∞ when xα = −d/c if c = 0.
Each homography of ℓ is given by a map on the affine coordinates in the
form
ax + b
θ : x ↦→ , with ad − bc = 0.
cx + d

106 CHAPTER 2. ANALYTIC GEOMETRY
Since for each t ∈ F ∗ , det(tA) = t 2 ·det(A), each element of P SL(V ) (the
group of θ ∈ P ΓL(V ) with α = id and det(A) = 1) has the form
θ : x ↦→
ax + b
, with ad − bc a nonzero square in F.
cx + d
Now let (P1, P2, P3, P4) be an ordered 4-tuple of points on ℓ with P1, P2, P3
distinct and with a basis B ′ as chosen above, but use the notation [x ′ 0 , x′ 1 ]T =
[P4]B ′. Define the cross-ratio (P1, P2; P3, P4) by the following:
If x ∈ ˜ F is the affine coordinate of P4 with respect to the ordered basis
(e0, e1), then (P1, P2; P3, P4) = x.
It follows that
(P1, P2; P3, P1) = ∞; (P1, P2; P3, P2) = 0; (P1, P2; P3, P3) = 1.
Now let B = (f0, f1) be an arbitrary ordered basis of V and put
Let B =
b00 b01
i x0 [Pi]B =
b10 b11
Then for arbitrary P ∈ ℓ,
implying
x i 1
, and Dij =
xi0 xj0
xi 1 x j
1
.
be the invertible matrix such that
(e0, e1) = (f 0, f 1) · B.
P = (e0, e1)[P ]B ′ = (f 0, f 1)[P ]B = (f 0, f 1)B[P ]B ′,
[P ]B =
b00 b01
b10 b11
[P ]B ′. (2.7)
Since the coordinates of a point are determined only up to a nonzero
scalar product, there are nonzero scalars k1, k2, k3 for which the following
hold:

2.9. THE PROJECTIVE LINE - CROSS RATIO 107
P = P1 =⇒ k1
=⇒
P = P2 =⇒ k2
x 1 0
x 1 1
= B
b01 = k1x1 0
b11 = k1x1 1 ;
x 2 0
x 2 1
= B
b00 = k2x 2 0
=⇒
b10 = k2x2 1;
3 x0 P = P3 =⇒ k3 = B
=⇒
x 3 1
0
1
1
0
1
1
b00 + b01 = k3x 3 0
b10 + b11 = k3x 3 1;
=
=
b01
b11
b00
b10
b00 + b01
=
b10 + b11
After making the appropriate substitutions we obtain
1 x0 x2 0
k1
3 x0 = k3
x 1 1 x2 1
Using Cramer’s Rule to solve for k1 and k2 in terms of k3, we find
D32
k1 = k3 ;
D12
D13
k2 = k3 .
D12
Put k3 = D12 to get k1 = D32; k2 = D13. It follows that
2 x0D13 B =
x1 0D32 x2 1D13 x1 1D32
; B −1 =
−x2 1D13 x2 0D13 k2
1
det(B)
Since [P ]B ′ = B−1 [P ]B, with P = P4 we have
=
[P4]B ′ =
1
det(B)
x 3 1
x 1 1D32 −x 1 0D32
4 1 x0 (x
det(B)
1 1D32 − x1 ) x41 D32
x4 0 (−x21 D13) + x2 0x41 D13
−x 2 1 D13 x 2 0 D13
=
x 1 1 D32 −x 1 0 D32
1
det(B)
x 4 0
x 4 1
D32 D41
D13 C24
It follows that the affine coordinate of P4 with respect to the basis B ′ is
D13D24
D14D23
.
.
.

108 CHAPTER 2. ANALYTIC GEOMETRY
If we write xi = x i 1 /xi 0 for the affine coordinate of Pi with respect to B,
then a routine simplification shows that
(P1, P2; P3, P4) = (x3 − x1)(x4 − x2)
. (2.8)
(x4 − x1)(x3 − x2)
In the preceding paragraphs we have defined the cross-ratio only for those
4-tuples with P1, P2, P3 distinct. However, by definition we may allow two of
them to be the same:
(P1, :1; P3, P4) = 1 provided P1, P3, P4 are distinct;
(P1, P2; P1, P4) = 0 provided P1, P2, P4 are distinct;
(P1, P2; P2, P4) = ∞ provided P1, P2, P4 are distinct.
Theorem 2.9.1. Consider projective lines P G(V ) and P G(W ) over respective
fields K and K ′ . Let θ : P G(V ) → P G(W ) be an isomorphism with
associated field isomorphism α : K → K ′ . Then
(p θ 1 , pθ 2 ; pθ 3 , pθ 4 ) = (p1, p2; p3, p4) α
for each choice of points p1, p2, p3, p4 of P G(V ) with at most two of them
equal.
Proof. The cases where two of the points are the same all turn out to be
trivial, so we assume thefour points
are distinct. Choose a basis B = (e0, e1)
0
1
1
for V such that [p1]B = ; [p2]B = and [p3]B = .
1
0
1
Let φ be the nonsingular linear map from V into W induced by θ. Say
e φ
0 = d0 and e φ
1 = d1. So B ′ = (d0, d1) is a basis for W . We have
p θ 1
This implies
= 〈eφ1
〉 = 〈d1〉; p θ 2 = 〈eφ0
〉 = 〈d0〉; p θ 3 = 〈(e0 + e1) φ 〉 = 〈d0 + d1〉.
[p θ
0
1 ]B ′ =
1
; [p θ
1
2 ]B ′ =
0
; [p θ
1
3 ]B ′ =
1
If p4 = 〈e0 + xe1〉, then x = (p1, p2; p3, p4). Therefore
.
p θ 4 = 〈(e0 + xe1) φ 〉 = 〈e φ
0 + xα e φ
1 〉 = 〈d0 + x α d1〉.
So (p θ 1 , pθ 2 , pθ 4 ) = xα = (p1, p2; p3, p4) α .

2.9. THE PROJECTIVE LINE - CROSS RATIO 109
Corollary 2.9.2. State this result with K = K ′ and V = W .
As a kind of converse we have:
Theorem 2.9.3. Consider projective lines P G(V ) and P G(W ) over respective
fields K and K ′ . If θ is a bijection from the pointset of P G(V )
to the pointset of P G(W ) with the property that there is an isomorphism
α : K → K ′ such that (p θ 1, p θ 2; p θ 3, p θ 4) = (p1, p2; p3, p4) α for each choice of four
points p1, p2, p3, p4 of P G(V ) with at most two of them the same, then θ may
be extended to an isomorphism from P G(V ) to P G(W ) with corresponding
field isomorphism α. (Here {0} θ = {0} and V θ = W .)
Proof. Choose bases B = (e0, e1) of V and B ′ = (d0, d1) of W just as in the
preceding proof. Then consider the nonsingular semilinear map φ from V to
x0
W given by the following. If for a general p ∈ P G(V ) we have [p]B = ,
x1
and [pφ
′ x 0
]B ′ = , then φ is defined so that x ′ 0 = aα 0 and x′ 1 = xα1 . With
x ′ 1
respect to these bases we have
φ : e0 ↦→ d0; φ : e1 ↦→ d1; (e0 + e1) φ = d0 + d1,
so that for the induced isomorphism γ : P G(V ) → P G(W ) it holds that:
p γ
1 = p θ 1, p γ
2 = p θ 2 and p γ
3 = p θ 3.
Let p4 be any point of P G(V ) with p4 ∈ {p1, p2, p3}. If [p γ
and [p4]B ′ =
x ′′
0
x ′′
1
(x ′ 1 )(x′ 0 )1 = (p γ
1
, then
, pγ
2
4 ]B ′ =
x ′ 0
; pγ3
, pγ4
) = (p1, p2; p3, p4) α = (p θ 1 , pθ2 ; pθ3 , pθ4 ) = (x′′ 1 )(x′′ 0 )−1 .
It follows that p γ
4 = pθ 4 . The restriction of this isomorphism γ to the pointset
of P G(V ) is the same as θ.
Corollary 2.9.4. A permutation θ of the points of a projective line P G(V )
over the field K is induced by an element of P ΓL(V ) if and only if there is
an automorphism α of K such that (p θ 1 , pθ 2 ; pθ 3 , pθ 4 ) = (p1, p2; p3, p4) α for each
choice of points p1, p2, p3, p4 of P G(V ) with at most two of them equal. A
permutation θ of the points of a projective line P G(V ) over the field K is induced
by an element of P GL(V ) if and only if (p θ 1 , pθ 2 ; pθ 3 , pθ 4 ) = (p1, p2; p3, p4)
for each choice of four points p1, p2, p3, p4 of P G(V ) with at most two of them
the same,
x ′ 1

110 CHAPTER 2. ANALYTIC GEOMETRY
2.10 Theorem of the Complete Quadrilateral
Consider a projective plane P G(V ) over the field K. If {L1, L2, L3, L4} is a
set of lines of P G(V ) with no three concurrent, then we call the set
{L1, L2, L3, L4, L1 ∩ L2 = p1, L1 ∩ L3 = p2, L1 ∩ L4 = p3,
L2 ∩ L3 = p4, L2 ∩ L4 = p5, L3 ∩ L4 = p6}
a complete quadrilateral. The points p1, . . . , p6 are the corner points of the
complete quadrilateral. The corner points p1, p6 (resp., p2, p5; resp., p3, p4)
are opposite cornerpoints. The lines p1p6, p2p5, p3p4 are the diagonals of the
complete quadrilateral. The three intersection points of the pairs of diagonals
are the diagonalpoints of the complete quadrilateral.
Interchanging the roles of the points and lines gives the definition of a
complete quadrangle.
Theorem 2.10.1. Let P G(V ) be a projective plane over a field K. In P G(V )
let a1, a2, a3 be any three distinct collinear points. If T is a complete quadrilateral
with a1 and a2 as opposite corner points and a3 as a diagonal point,
then the second diagonal point a4 on the diagonal line a1a2 is independent of
the choice of T . Moreover, (a1, a2; a3, a4) = −1.
Proof. Assume that T = {L1, L2, L3, L4, L1 ∩ L2 = a1, L1 ∩ L3 = b1, L1 ∩
L4 = c1, L2 ∩ L3 = c2, L2 ∩ L4 = b2, L3 ∩ L4 = a2}. Further, suppose that
a1a2 ∩ b1b2 = a3 and a1a2 ∩ c1c2 = a4. Choose a basis B = (e0, e1, e2) such
that a1 = 〈e1〉, a2 = 〈e0〉, b1 = 〈e2〉, b2 = 〈e0 + e1 + e2〉, from which it follows
that
⎡
[a1]B = ⎣
We also have
0
1
0
⎤
⎡
⎦ ; [a2]B = ⎣
⎡
[a3]B = ⎣
1
1
0
⎤
1
0
0
⎤
⎦ ; [c2]B = ⎣
⎡
⎦ ; [b1]B = ⎣
⎡
1
0
1
⎤
0
0
1
⎤
⎦ ; [c1]B = ⎣
⎡
⎦ ; [b2]B = ⎣
From this we calculate that [a4]B =
1
⎣ −1 ⎦.
0
Hence we have a2 = 〈e0,
a1 = 〈e1, a3 = 〈e0 + e1〉, a4 = 〈e0 − e1〉. It follows that (a1, a2; a3, a4) = −1,
so that in general a4 is uniquely determined by a1, a2 and a3.
⎡
⎤
⎡
0
1
1
⎤
⎦ .
1
1
1
⎤
⎦ .

2.10. THEOREM OF THE COMPLETE QUADRILATERAL 111
Definition 2.10.2. If a1, a2, a3, a4 are four collinear points with (a1, a2; a3, , a4) =
−1, we say that a4 is the harmonic conjugate of a3 with respect to the points
a1 and a2.
Observation 1. If a4 is the harmonic conjugate of a3 w. r. t. a1 and a2,
it follows from the preceding theorem that a3 is the harmonic conjugate of a4
w. r. t. a1 and a2. So we say that a3 and a4 are harmonic conjugates w. r.
t. a1 and a2.
Observation 2. a3 = a4 if and only if the characteristic of K is 2. The
three diagonals of a complete quadrilateral are concurrent if and only if the
characteristic of K is 2. Dually, the three diagonal points of a complete
quadrangle are collinear if and only if the characteristic of K is 2.
Theorem 2.10.3. Let a1, a2, a3, a4 be four distinct points of a projective
line P G(V ) over the field K. Then a3 and a4 are harmonic conjugates w. r.
t. a1 and a2 if and only if there is an element of order 2 in P G(V ) that fixes
a1 and a2 and interchanges a3 and a4.
Proof. Choose in V a basis B = (e0, e1) such that
[a2]B =
1
0
; [a1]B =
0
1
; [a3]B =
Then a4 is the harmonic conjugate of a3 w. r. t. a1 and a2 if and only if
1
(a1, a2; a3, a4) = −1, which holds if and only if [a4]B = .
−1
Now suppose that a3 and a4 are harmonic conjugates w. r. t. a1 and
a2. Since a3 = a4, the characteristic of K must be different from 2. In affine
coordinates an element θ of P GL(V ) is given by
θ : x ↦→ x ′ ax + b
= , with ad − bc = 0.
cx + d
Put a = −1, b = c = 0, d = 1. One can then check that θ : a1 ↦→ a1,
θ : a2 ↦→ a2, θ : a3 ↦→ a4 and θ : a4 ↦→ a3; θ2 = 1; θ = 1.
For the converse, suppose that θ ∈ P GL(V ) satisfies θ = 1 = θ2 , that θ
fixes a1 and a2 and maps a3 to a4. If we write out what it means for θ to fix
a1 and a2 we find that b = c = 0. This says that x ′ = kx for some nonzero
k ∈ K. Since θ = 1 = θ2 in P GL(V ), we see that k = 1 = k2 in K. Thus
k = −1 and the characteristic of Kis different from 2. Thus a4 has affine
1
coordinate -1, so that [a4]B = and a4 is the harmonic conjugate of a3
−1
w. r. t. a1 and a2.
1
1
.

112 CHAPTER 2. ANALYTIC GEOMETRY
2.11 Involutions of the Projective Line
Let P G(V ) be a projective line over the field K. Recall that θ ∈ P ΓL(V ) is
an involution provided θ 2 = 1. Let θ ∈ P ΓL(V ) be given with respect to a
basis B of V by
θ :
x0
x1
′ x 0 ↦→
x ′ 1
=
=
α d c x0 b a xα 1
α dx0 + cxα
1
bx α 0 + ax α 1
, (ad − bc = 0)
Now write out the assumption that θ is an involution. It follows that
there is a nonzero scalar ℓ such that
α
x0 d c dx0 + cx
ℓ =
x1 b a
α 1
bxα 0 + axα α 1
α+1 α α (d + cb )x
=
2
0 + (dcα + caα )xα2
1 ,
obtain
(bd α + ab α )x α2
0 + (bc α + a α+1 )x α2
1
∀ x0, x1 ∈ K with (x0, x1) = (0, 0).
First put x0 = 1, x1 = 0 and then x0 = 0, x1 = 1, to obtain
Hence we have
ℓ = d α+1 + cb α = bc α + a α+1 = 0; (2.9)
0 = bd α + ab a = dc α + ca α . (2.10)
ℓx0
ℓx1
=
(d α+1 + cb α )x α 2
0 ;
(bc α + a α+1 )x α2
1
. (2.11)
Multiply the first row of Eq. 2.11 by x1 and the second row by x0 to
x0x1ℓ = (d α+1 + cb α )x α2
0 x1 = (bc α + a α+1 )x0x α2
1 .
Hence x0x α2
1 = x α2
0 x1. Take x0 = 1 to see that α 2 = 1.
We summarize these computations with the folowing lemma.
Lemma 2.11.1. For θ given by
x0
θ : ↦→
d
b
c
a
x1
x α 0
x α 1
, (ad − bc = 0),

2.11. INVOLUTIONS OF THE PROJECTIVE LINE 113
it is necessary and sufficient that
(i) bd α + ab α = 0;
(ii) dc α + ca α = 0;
(iii) d α+1 + cb α = bc α + a α+1 = 0;
(iv) α 2 = 1.
The case α = id.
In this case we have (a+d)b = 0 = (a+d)c. First suppose that a+d = 0,
so that b = c = 0 and a = ±d. If a = d then θ = 1. So we have a non-identity
involution only when K has characteristic different from 2 and with a = 1
and d = −1. In this case θ has two fixed points if the characteristic of K
is not 2, and θ is the identity map if the characteristic of K equals 2. So
suppose that either b = 0 or c = 0. In this case a + d = 0 and the matrix
looks like −a c
b a
, with b = 0 or c = 0.
The eigenvalues (if they exist!) are λ = ± √ a 2 + bc with associated eigen-
vectors ±λ − a
b
So in odd characteristic there are at most two fixed points of θ, and in
characteristic 2 there is at most one. In terms of affine coordinates we have
θ : x ↦→ x ′ =
.
ax + b
cx − a .
The case α = id.
First we assume that c = 0 and put c α = ℓc. Then
c = c α2
= (ℓc) α = l α · ℓc,
implying ℓ α+1 = 1 and hence ℓ α = ℓ −1 .
From 0 = dc α + ca α = dℓc + ca α = c(dℓ + a α ) we get dℓ + a α = 0. Now
use α 2 = 1 to get d α ℓ α + a = 0 and multiply by ℓ to get d α + aℓ = 0. Now
from dℓ + a α = 0 = d α + aℓ, we have −adℓ = a α+1 = d α+1 . So in this case
we have:
c α = cℓ; b α = bℓ; a α = −dℓ and α 2 = 1 = α.

114 CHAPTER 2. ANALYTIC GEOMETRY
Now suppose that c = 0 = b. Put b α = bℓ. Then b = b α ℓ α which forces
ℓ α = ℓ −1 or ℓ α+1 = 1. Using Eq. 2.10 we see d α + aℓ = 0 and a α+1 = d α+1 ,
α 2 = 1. The bottom line here is that
α.
or
or
c = 0; b = 0; b α = bℓ; a α = −dℓ; α 2 = 1 = α.
The last case is b = c = 0. Here we just have a α+1 = d α+1 with α 2 = 1 =
Collecting the above into two cases we have the following:
(a) (b, c) = (0, 0), b α = bℓ; c α = cℓ; a α = −dℓ,
(b) (b, c) = (0, 0), a α+1 = d α+1 , α 2 = 1.
In affine coordinates we have
(a) x ′ = axα +b
cx α +d with (b, c) = (0, 0), bα = bℓ; c α = cℓ; a α = −dℓ,
(b) x ′ = kx α with k α+1 = 1, α 2 = 1.
Note 2.11.2. : Let θ ∈ P ΓL(V ) be an involution with θ = 1. Choose a basis
(e0, e1) with θ interchanging those two points. Then using affine coordinates
w. r. t. that basis, so e0 and e1 have affine coordinates 0 and ∞, respectively,
we have a = d = 0. If α = 1, then θ is given in affine coordinates by x ′ = k/x.
So the fixed points have affine coordinates X that satisfy X 2 = k. So suppose
that α = 1. Then θ is given in affine coordinates by x ′ = k/x α with k = k α
and α 2 = 1. The fixed points have affine coordinates X that satisfy X α+1 = k.
The finite case. The Galois field GF (p h ) admits a nontrivial involutory
automorphism α if and only if h is even, in which case α : x ↦→ x q . In this
case we usually write x q = x. If α = 1, then the involution θ (with respect
to a certain basis) is given by x ′ = k/x. When the characteristic of Fq 2 is 2,
then k = ℓ 2 for some k ∈ F q 2. If we replace the original basis with a new one
so that the coordinate x is replaced by y = x/ℓ, then the involution θ is given
by y ′ = 1/y. The unique fixed point is the point with affine coordinate equal
to 1. Now suppose that the characteristic of the field is odd. If k = ℓ 2 , with
the same change of basis as above we again have that θ is given by x ′ = 1/x.
The two fixed points are the points with affine coordinates equal to ±1. If
ν is a fixed nonsquare of F q 2 and k = νℓ 2 , then the change of basis yℓ = x

2.12. THE HYPERBOLIC QUADRIC OF P G(3, Q) - A FIRST LOOK115
as before leads to θ being represented by x ′ = ν/x. In this case there are no
fixed points.
Now suppose that α = 1 and that with respect to some fixed basis the
involution θ is given by x ′ = k/x q with k = k q , so k is in the subfield Fq of
F q 2. We showed above that in this case k = ℓ q+1 , ℓ ∈ F q 2. The change of
basis yℓ = x leads to θ being given by y ′ = 1/y q . The fixed points are those
with affine coordinates Y satisfying Y q+1 = 1. Hence there are q + 1 fixed
points.
2.12 The Hyperbolic Quadric of P G(3, q) - A
First Look
Later we will study quadrics in great detail. For the present, we merely
study a specific substructure of P G(3, q) that turns out to be an important
example of a hyperbolic quadric.
In any projective space a set S of subspaces is skew provided no two of
the subspaces in S have a point in common. Also, the subspaces are said
to be skew. Then if S is a set of skew subspaces, a line ℓ is said to be a
transversal of S provided it meets each member of S in exactly one point.
Lemma 2.12.1. Let P be a projective space. Let g1 and g2 be two skew lines,
and let P denote a point not on either g1 or g2. Then there is at most one
transversal of g1 and g2 through P .
Proof. Assume that there are two transversals h1 and h2 through P . Then
each of these transversals meets the lines g1 and g2 in different points. So h1
and h2 span a plane which contains the two skew lines g1 and g2, a contradiction.
Now suppose that P is 3-dimensional. Then the plane 〈P, g1〉 must intersect
the line g2 in some point Q. Hence the line P Q intersects g1 and g2, so
it is a transversal of g1 and g2.
Theorem 2.12.2. (16 point theorem of Dandelin) Let P be a 3-dimensional
projective space over the field K. Let G = {g1, g2, g3} and H = {h1, h2, h3}
be two sets of skew lines with the property that each line gi meets each line
hj. Then the following is true: each transversal g ∈ G of H meets each
transversal h ∈ H of G.

116 CHAPTER 2. ANALYTIC GEOMETRY
Proof. Fix the following notation:
〈v1〉 := g1 ∩ h1; 〈v2〉 := g1 ∩ h2;
〈v3〉 := g2 ∩ h1; 〈v4〉 := g2 ∩ h2.
Then g3 ∩ h1 = 〈av1 + bv3〉 for some a, b ∈ F \ {0}. We may assume that
b = 1. Replacing v1 with v ′ 1 = av1 we get g3 ∩ h1 = 〈v ′ 1 + v3〉. Writing v1 iin
place of v ′ 1 we have g3 ∩ h1 = 〈v1 + v3〉. Multiplying v2 and v4 by suitable
scalars we can arrange to have g1 ∩ h3 = 〈v1 + v2〉 and g2 ∩ h3 = 〈v3 + v4〉.
Finally, we have g3 ∩ h2 = 〈v2 + av4〉. By hypothesis there is a unique point
of intersection of g3 and h3. This forces
g3 ∩ h3 = 〈a1(v1 + v3) + a2(v2 + av4)〉 = 〈b1(v1 + v2) + b2(v3 + v4)〉.
Since the vectors v1, v2, v3, v4 are linear independent we get b1 = a1 =
b2 = a2 and a = 1. Hence g3 ∩ h2 = 〈v2 + v4〉 and
g3 ∩ h3 = 〈v1 + v2 + v3 + v4〉.
Let g ∈ G be a transversal of H and h ∈ H be a transversal of H. Define
the elements a and b of K by
g ∩ h1 := 〈v1 + av3〉;
h ∩ g1 := 〈v1 + bv2〉.
We need to show that the lines g and h intersect each other. By Lemma 2.12.1
we know that there is at most one line through 〈v1 + av3〉 that intersects
both h2 and h3. This must be the line g. But now the points 〈v1 + av3〉,
〈v2 + av4〉 ∈ 〈v2, v4〉 = h2 and 〈v1 + v2 + a(v3 + v4)〉 lie on a line. Therefore
necessarily
g = 〈v1 + av3, v2 + av4〉.
Similarly we get
h = 〈v1 + v2, v3 + bv4〉.
Now we compute the intersection of the subspaces g and h. Since ab = ba it
follows that
〈v1 + av3 + bv2 + bav4〉
is a common point of
g = 〈v1 + av3, v2 + av4〉 and h = 〈v1 + bv2, v3 + bv4〉.

2.12. THE HYPERBOLIC QUADRIC OF P G(3, Q) - A FIRST LOOK117
Definition 2.12.3. Let P = P G(3, q). A nonempty set R of skew lines of
P is called a regulus provided the following hold:
(i) Through each point of each line of R there is a transversal of R.
(ii) Through each point of each transversal of R there is a line of R.
So the set R ′ of all transversals of a regulus R is a regulus called the
opposite regulus of R. It is clear that any regulus in P G(3, q) contains q + 1
lines.
The following theorem is now an easy corollary of the 16 point theorem.
Theorem 2.12.4. Let g1, g2, g3 be three skew lines of P = P G(3, q). Then
there is exactly one regulus containing g1, g2 and g3.
Proof. Exercise.
We now show that the points on the lines of a regulus form a ”quadric.”
This just means that their coordinates are the solutions of a quadratic equation.
Theorem 2.12.5. Let g1, g2, g3 be three skew lines of P = P G(3, q). We
may choose coordinates so that
g1 = 〈(1 : 0 : 0 : 0), (0 : 1 : 0 : 0)〉,
g2 = 〈(0 : 0 : 1 : 0), (0 : 0 : 0 : 1)〉
g3 = 〈(1 : 0 : 1 : 0), (0 : 1 : 0 : 1)〉.
Then the set Q of points on the uniquely determined regulus R through
g1, g2, g3 can be described as
Q = {(a0 : a1 : a2 : a3) : a0a3 = a1a2 with ai ∈ Fq not all zero}.
Hence the coordinates of the points of Q satisfy the quadratic equation
x0x3 − x1x2 = 0.
Proof. Use the notation of the 16 point theorem. If v1 = (1 : 0 : 0 : 0),
v2 = (0 : 1 : 0 : 0), v3 = (0 : 0 : 1 : 0) and v4 = (0 : 0 : 0 : 1), then the points
(a0 : a1 : a2 : a3) on Q satisfy
(a0 : a1 : a2 : a3) = k · v1 + k · bv2 + k · av3 + k · abv4 = (k : kb : ka : kab),

118 CHAPTER 2. ANALYTIC GEOMETRY
where a, b ∈ Fq are arbitrary and k ∈ Fq \ {0}. Hence the coordinates of any
point of Q satisfy the condition
0 = k · kab − ka · kb = a0a3 − a1a2.
Conversely, let P = (a0 : a1 : a2 : a3) denote a point with a0a3 = a1a2. If
a0 = 0, then this point can be written as
P = (a0 : a1 : a2 : a1a2/a0),
and this is a point on Q. If a0 = 0, then a1 = 0 or a2 = 0. Without
loss of generality we may assume that a1 = 0. Then P can be written as
P = (0 : 0 : a2 : a3). Without loss of generality we may assume that a2 = 0.
So P = (0 : 0 : a : ab) with a, b ∈ Fq, and it follows that P ∈ Q.
This type of quadric will be called a ruled quadric in P G(3, q), or sometimes
called a hyperboloid.
Recall Lemma 2.7.1 to see the following.
Lemma 2.12.6. The number of
(i) ruled quadrics containing two skew lines is q(q 2 − 1)(q − 1);
(ii) reguli in P G(3, q) is q 4 (q 2 + q + 1)(q 2 + 1)(q − 1).
(iii) ruled quadrics in P G(3, q) is q 4 (q 2 + q + 1)(q 2 + 1)(q − 1)/2.
2.13 3-Dimensional Projective Spaces Are Desarguesian
The following very important theorem eventually will be used to show that
up to isomorphism the only finite projective geometries of dimension at least
3 are the P G(n, q) with n ≥ 3.
Theorem 2.13.1. Let P be a projective space of dimension n. If n ≥ 3 then
the theorem of Desargues holds in P.
Proof. Let A1, A2, A3, B1, B2, B3, P12, P23, P31 and C be points of P that satisfy
the hypothesis of the theorem of Desargues. We must show that the
points P12, P23, P31 are collinear.

2.13. 3-DIMENSIONAL PROJECTIVE SPACES ARE DESARGUESIAN119
Case 1. The planes π = 〈A1, A2, A3〉 and π ′ = 〈B1, B2, B3〉 are different.
(FIGURE would be nice - 2.7 p. 79 of Beut)
Since Ai and Bi are collinear with C, we have Bi ∈ U = 〈C, A1, A2, A3〉,
i = 1, 2, 3. Therefore all points and lines mentioned above are in the 3dimensional
subspace U of P. The points P12, P23, P31 all lie in π ∩ π ′ . Since
any two planes of U intersect in a line, the points P12, P23, P31 are collinear.
Case 2. The points A1, A2, A3, B1, B2, B3, P12, P23, P31 and C lie in a
common plane π.
(FIGURE 2.8 of Beut)
The idea is to reduce this case to Case 1. To do this we construct three
noncollinear points D1, D2, D3 not in π and two points C ′ , C ′′ such that for
{D1, D2, D3, A1, A2, A3, C ′ } and {D1, D2, D3, B1, B2, B3, C ′′ the hypotheses
of Case 1 are satisfied.
Let C ′ , C ′′ be two distinct points outside π such that the line C ′ C ′′
intersects the plane π in the point C. Since C ∈ C ′ C ′′ ∩A1B1, the lines C ′ C ′′
and A1B1 span a plane. Therefore the lines C ′ A1 and C ′′ B1 intersect a point
D1 which is outside π.
Similarly, there exist points D2 and D3 outside π with
D2 = C ′ A2 ∩ C ′′ B2 and D3 = C ′ A3 ∩ C ′′ B3.
We now show that {D1, D2, D3, A1, A2, A3, C ′ } satisfy the hypotheses of the
theorem of Desargues.
If three of the points C ′ , D1, D2, D3 are collinear, then the dimension of
〈D1, D2, D3, C ′ 〉 is at most 2. Since by construction Ai is on C ′ Di, i = 1, 2, 3,
we have that the points A1, A2, A3 lie in π∩〈D1, D2, D3, C ′ 〉. Since the dimension
of this intersection is at most 1, it is a point or a line. Hence A1, A2, A3
would be collinear, a contradiction. It follows that {D1, D2, D3, A1, A2, A3, C ′ }
satisfies the hypotheses of the theorem of Desargues.
By the construction of C ′ it is clear that no three of the ponts C ′ , A1, A2, A3
are collinear. Similarly it follows that {D1, D2, D3, B1, B2, B3, C ′′ } also satisfy
the hypothesis of the theorem of Desargues.
Define π ′ = 〈D1, D2, D3〉. By Case 1 the lines DiDi+1 and AiAi+1 intersect
in points of the line g := π ∩ π ′ , i = 1, 2, 2, where subscripts are 1, 2 or 3
modulo 3.

120 CHAPTER 2. ANALYTIC GEOMETRY
Similarly the lines DiDi+1 and BiBi+1, i = 1, 2, 3, intersect in points of
the line g. In particular, AiAj ∩ BiBj is also a point of g. If, for instance,
X := D1D2 ∩ A1A2 and Y := D1D2 ∩ B1B2, then X = D1D2 ∩ g = Y , and
this point is also X = Y = A1A2 ∩ B1B2 = P12. It follows that the ponts
P12, P23 and P31 are all incident with the line g.
Corollary 2.13.2. Let P be a projective plane. If P can be embedded as
a plane in a projective space of dimension at least 3, then the theorem of
Desargues holds in P.
Exercise 2.13.2.1. (FIGURE P. 89) (Sometimes called the theorem of Fano)
Let the four points P1, P2, P3, P4 form a quadrangle in a projective plane embedded
in P(V ). Then the points
Q1 := P1P4 ∩ P2P3; Q2 := P2P4 ∩ P1P3 and Q3 := P3P4 ∩ P1P2
are on a common line if and only if the underlying field has characteristic 2.
Exercise 2.13.2.2. (FIGURE P. 90) (Theorem of Menelaus) Let P = 〈u〉,
Q = 〈v〉, R = 〈w〉 be three noncollinear points of a projective plane P(V ).
Let P ′ = 〈v + aw〉 be a point on QR; Q ′ = 〈w + bu〉 be a point on RP ; and
R ′ = 〈u + cv〉 be a point on P Q. Prove that the points P ′ , Q ′ , R ′ are collinear
if and only if abc = −1.
Exercise 2.13.2.3. (FIGURE P. 91) (Theorem of Ceva) Let P = 〈u〉, Q =
〈v〉, R = 〈w〉 be three noncollinear points of a projective plane P(V ). Let
P ′ = 〈v + aw〉 be a point on QR, Q ′ = 〈w + bu〉 be a point on RP , and
R ′ = 〈u + cv〉 be a point on P Q. Prove that the lines P P ′ , QQ ′ and RR ′
pass through a common point if and only if abc = 1.

Chapter 3
The Fundamental Theorem for
P G(n, q)
In this chapter we study representations and collineations of projective spaces.
The main goal is to show that any projective space of finite dimension at least
3 must be isomorphic to some P(V ) where V is a finite dimensional vector
space over a skewfield K. Of course our main interest is in finite projective
and affine spaces so our skewfields are finite fields.
3.1 Central Collineations
In this chapter P will always denote a projective space of dimension d ≥ 2.
The main aim is to construct, using the theorem of Desargues, a vector space
V such that P = P(V ). For this we have to construct a division ring K. It
will be composed of collineations (automorphisms) of P. Hence our first aim
must be to use the theorem of Desargues to construct many collineations of
P.
Recall that a collineation of P is a pair of bijections, one from the point set
of P to itself and one from the line to itself such that incidence is preserved.
If θ is such a collineation we let θ denote the bijection on points and also the
bijection on lines. Hence we have that for two points X and Y it must be
true that
θ(XY ) = θ(X)θ(Y ). (3.1)
Clearly the set of all collineations of P forms a group with respect to
composition of maps.
121

122 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
Definition 3.1.1. A collineation θ of P is called a central collineation if
there is a hyperplane H (the axis of θ) and a point C (the center of θ) with
the following properties:
• Every point X of H is a fixed point of θ;
• Every line x on C is a fixed line of θ.
Note that the image of each pont on a fixed line g lies on g, but this does
not mean that every point on g is a fixed point.
Lemma 3.1.2. Let H be a hyperplane and C a point of P . Then the set
of central collineations with axis H and center C is a group with respect to
composition of maps.
Proof. Easy exercise.
Lemma 3.1.3. Let θ be a central collineation of P with axis H and center
C. Let P = C be a point not on H, and let P ′ = θ(P ) be the image of P .
Then θ is uniquely determined. In particular, the image of each point X that
is neither on H nor on P P ′ (= P C) satisfies
(FIGURE P. 97)
θ(X) = CX ∩ F P ′ where F = P X ∩ H.
Proof. By the definition of central collineation, the image X ′ = θ(X) of a
point X (not on H nor on P C is determined by the following restrictions:
On the one hand the line CX is mapped onto itself (as is any line through
C); hence X ′ must be on θ(CX) = CX.
On the other hand, consider the point P := P X ∩ H. Being a point of
the axis H it is fixed by θ. Hence
X ′ = θ(X) I θ(P X) = θ(F P ) = θ(F )θ(P ) = F P ′ .
Since X is not on P P ′ , F is not on CX. Hence X ′ is the intersection of the
two distinct lines CX and F P ′ , so is uniquely determined.
It now follows that the image of each point Y on CP is also uniquely
determined: for just replace (P, P ”) with a pair (F, R ′ ) with R ′ = R and
R ∈ CP . Then it follows that Y ′ = CY ∩ F ∗ R ′ (with F ∗ = RY ∩ H) is
uniquely determined.

3.1. CENTRAL COLLINEATIONS 123
Corollary 3.1.4. (Uniqueness of central collineations) Let θ be a central
collineation of P with axis H and center C that is not the identity. Then:
(a) If P is a point different from C and not on H, then P is not fixed by
θ.
(b) The central collineation θ is uniquely determined by one pair (P, θ(P ))
with P = θ(P ).
Proof. For (a), assume that P is fixed by θ. The goal is to show that each
point X is fixed. First, let X not be on the line CP . Then using the notation
of Lemma 3.1.3 we have that θ(X) = CX ∩ F P ′ = CX ∩ F P = X since X
is on F P .
Using a (fixed) point X0 outside P C we see that each point on P C is also
fixed by θ. Hence θ is the identity map, contradicting the hypothesis. This
proves (a), and (b) also follows immediately from Lemma 3.1.3.
A central collineation whose center lies on its axis is called a translation
or an elation - see below.
The previous corollary is useful in the following situation. If we have two
translations θ and φ whose centers lie on their common axis H, in order to
show that θ = φ it suffices to show that they agree on just one point not on
H.
Corollary 3.1.5. The axis and center of a central collineation θ = id of P
are uniquely determined.
Proof. If θ had two axes H and H ′ , then any of the (at least two) points of
H ′ \ H would be fixed by θ, contradicting the previous corollary. Suppose
that θ had two centers C and C ′ . Let P be a point outside the axis H and
line CC ′ . Then P ′ = θ(P ) satisfies
P ′ ∈ P C and P ′ ∈ P C ′ ,
forcing P ′ = P . So again by Cor. 3.1.4 we have a contradiction.
Definition 3.1.6. Let θ = id be a central collineation of a projective P. We
call θ an elation if its center is incident with its axis and a homology if its
center and axis are not incident. The identity is considered to be both an
elation and an homology. A set of elations all having the same axis H will
be called translations of the affine space A = P \ H.

124 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
Lemma 3.1.7. Let θ = id be a central collineation of the projective space P
with center C and axis H. If U is a subspace of P with C ∈ U but C ∈ H,
then the restriction of θ to U is a central collineation which is not the identity.
Proof. Since C ∈ U and each line through C is fixed, clearly U is invariant
under θ. So the restriction θ ′ of θ to U is a collineation of U with C as center
and U ∩ H as axis. Hence θ ′ is a central collineation. Since U ⊆ H, θ ′ is not
the identity.
In the situation of Lemma 3.1.7 we say that θ induces a central collineation
of U. Later we shall see that each central collineation of a subspace is induced
by a central collineation of the whole projective space.
The next lemma is a general one on the extendibility of collineatons.
Lemma 3.1.8. Let P be a projective space of dimension of at least 2, and let
g0 be a line of P. Let P ′ be the point-line geometry consisting of the points
P that not on g0 and the lines of P that are different from g0. Let θ ′ be a
collineation of P ′ . Then θ can be uniquely extended to a collineation θ ∗ of
P. This collineation θ ∗ fixes the line g0.
Proof. The main step is to prove the following:
Claim: Let g and h be two lines that intersect g0 in the same point P .
Then the lines g ′ = θ(g) and h ′ = θ(h) also intersect each other in a point
P ′ of g0.
Since θ preserves the incidence of P ′ , the lines g ′ and h ′ have no point of
P ′ in common. We have to distinguish two cases.
Case 1. The order of P is greater than 2.
Then there are a point Q in P ′ and two lines m and n through Q that
intersect each of the line g ′ and h ′ in distinct points. The collineation θ maps
m and n onto two lines m ′ and n ′ that pass through a common pont Q ′ of
P ′ and intersect each of the line g ′ and h ′ in distinct points. So g ′ and h ′ lie
in a common plane of P, i.e., the plane spanned by m ′ and n ′ . Hence these
two lines must intersect at a point of P. Since they have no point of P ′ in
common, they must intersect in a point of g0.
Case 2. The order of P equals 2.
Here, if g, h and g0 do not lie in a common plane, then there is a point
Q as above and the proof can be completed just as above. However, if g,

3.1. CENTRAL COLLINEATIONS 125
h and g0 all lie in a common plane, then these lines cover all points of that
plane (since they are concurrent) and we cannot find a point Q as above.
Again we have to consider two cases. If P is only a plane (the seven-point
Fano plane), then it is an easy exercise to complete the proof. If P is not
only a plane, then there exists a line ℓ through the point P = g ∩ h ∩ g0
that is not contained in the plane spanned by g and h. Then we apply our
results obtained so far to the pairs (ℓ, g) and (ℓ, h). Since ℓ, g and g0 are not
in a common plane, the lines ℓ ′ and g ′ pass through a common point of g0.
Similarly ℓ ′ and h ′ contain a common point of g0. Therefore g ′ and h ′ also
pass through the same point of g0, proving the claim.
From this we obtain the following: For a point P on g0 let GP be the set
of line P ′ through P . Then there is a point P ′ on g0 such that all images of
the lines of GP pass through P ′ . In other words: θ(GP ) = GP ′.
Now we define θ ∗ in such a way that the restriction of θ ∗ to P ′ is θ; that
θ ∗ (g0) = g0; and that for each point P on g) we have θ ∗ (P ) = P ′ . Then we
see that not only is θ ∗ the only possible extensioin of θ to P, but also that
θ ∗ is in fact a collineation.
Lemma 3.1.9. Let θ be a collineation of P such that there is a hyperplane
H with the property that each of point of H is fixed by θ. Then there exists a
point C of P such that each line through C is fixed by θ. Briefly: each axial
collineation is central. Dually, each central collineation is axial, but we leave
the proof of that as an exercise.
Proof. If there is a point C ∈ H with θ(C) = C, then clearly C is a center.
So consider the case where no point outside H is fixed by θ. Let P be
an arbitrary point outside H. Then the line P θ(P ) must be fixed since its
intersection C with H is fixed. In fact, we claim the point C is a center for
θ. Suppose that g is a line through C. Clearly each line in H is fixed, so we
may assume that g does not lie in H.
Claim: For each point Q ∈ H, Q ∈ P θ(P ), the line Qθ(Q) passes
through the point C = P θ(P ) ∩ H. For let S = P Q ∩ H. Then
S = θ(S) ∈ θ(P Q) = θ(P )θ(Q).
Hence the points S, P , Q, θ(P ), θ(Q) are contained in a common plane
π. There the lines Qθ(Q) and P θ(P ) intersect in some X of π. Since the
lines Qθ(Q) and P θ(P ) are fixed by θ, the point X satisfies
θ(X) = θ(P θ(P )) ∩ θ(Qθ(Q)) = P θ(P ) ∩ Qθ(Q) = X.

126 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
Therefore X lies in H and hence must coincide with the point P θ(P )∩H = C.
Hence all lines of the form Qθ(Q) pass through C, and each line through
C is fixed by θ.
Now we are ready to prove an important result which in the general
form given here was first given by R. Baer in 1942, although in essence it
was already known long before. Basically it says that in a Desarguesian
projective space all possible central collineations exist.
Theorem 3.1.10. (existence of central collineations) If in the projective
space P the theorem of Desargues is valid we have: if H is a hyperplane and
C, P and P ′ are distinct collinear points of P with P , P ′ ∈ H, then there is
precisely one central collineation of P with axis H and center C mapping P
to P ′ .
Proof. The uniqueness of the central collineation has already been proved
above. So what we have to do now is to prove the existence of the central
collineation.
The basic idea is to start with the (rank 2) point-line geometry P ′ whose
points are the points of P not on CP and the lines of P different from CP .
We shall define a map θ and show that it is a collineation of P ′ . Then the
previous lemma implies that θ can be extended to a collineation of P.
Definition of the map θ: Each point of H is fixed by θ. For a point X
outside H define X ′ = θ(X) := CX ∩ F P ′ , where F = XP ∩ H.
The heart of the proof consists in showing that this θ is a collineation of
P ′ .First we show that θ is a bijection of the pointset of P ′ onto itself.
Suppose that X1 and X2 are two distinct points. To show that they have
different images we may certainly suppose that they are not in H. If X1, X2
and C are collinear, then the ponts F1 = X1P ∩ H and F2 = X2P ∩ H are
distinct. Hence θ(X1) and θ(X2) are two distinct points on the line through
C, X1 and X2. If X1, X2 and C are not collineat then θ(X1) and θ(X2) are
distinct since θ(CX1) is on X1 and θ(X2) is on CX2. Hence θ is injective.
The map θ is surjective since the point Y0 := CY ∩ F P , where F =
Y P ′ ∩ H, is a preimage of Y .
We now show that any three collinear points are mapped onto collinear
points. This is enough to show that θ is a collinetion. For this we first need
to prove the following
Claim: For any two points X and Y of P ′ , the lines XY and X ′ Y ′
intersect in a point of the axis H. Clearly we may suppose that X, Y ∈ H.

3.1. CENTRAL COLLINEATIONS 127
Recall the construction of X ′ and Y ′ . First deal with the “trivial” case that
P , X and Y are collinear. (FIGURE 3.3 P. 101) Then X ′ and Y ′ are also on
the line through F = P X ∩ H and P ′ . Hence P ′ , X ′ and Y ′ are collinear. In
particular, XY = P X and X ′ Y ′ = X ′ P ′ intersect in the point F of H.
Now study the general case where X, Y and P are not collinear (FIGURE
P. 102). Then the points F1 = P X ∩ H and F2 = P Y ∩ H are distinct.
Since P ′ is not a fixed point it does not lie in H. Thus the lines F1P ′ and
F2P ′ are distinct. So X ′ , Y ′ and P ′ are also noncollinear, and therefore the
triangle △(X, Y, P and △(X ′ , Y ′ , P ′ ) satisfy the hypothesis of the theorem
of Desargues (with center C).
The theorem of Desargues, applied to these triangles, says that the points
F1, F2 and XY ∩ X ′ Y ′ (the point we seek) are collinear. In particular,
XY ∩ X ′ Y ′ lies on H.
It now follows that θ is a collineation of P ′ : Let X, Y, W be three points
on a line g, and let X ′ , Y ′ and W ′ be their images under θ. In view of the
above claim, Q = XY ∩ X ′ Y ′ = g ∩ X ′ Y ′ is also a point of H, and this must
be the intersection of g and H, hence it is the pont Q. This means that
X ′ W ′ = X ′ Q = g ′ . So W ′ also lies on QX ′ = g ′ . Therefore all three images
lie on a common line, showing that θ is a collineation of P ′ . By Lemma 3.1.8
we can extend a collineation θ ofP ′ to a collineation θ ∗ of P. It remains to
show that θ ∗ is a central collineation. By construction, H is an axis of θ ∗ . As
a consequence of Lemma 3.1.9 we see that θ ∗ must also have a center, which
is in fact the point C.
A simple application of the theorem of Baer is the converse of Lemma 3.1.7.
Theorem 3.1.11. Let P be a Desarguesian projective space of dimension
d ≥ 2. Then each central collineation γ ∗ of a subspace U of P is induced by
a central collineation of P. More precisely, for each point P ∈ U there is a
central collineation of P whose axis passes through P and induces θ ∗ .
Proof. Let C be the center and H ∗ the axis of θ ∗ . Let H be a hyperplane
of P through P such that H ∩ U = H ∗ . Consider an arbitrary point Q of U
with Q = C and Q ∈ H ∗ . Let Q ′ = θ ∗ (Q). By the theorem of Baer there is
a central collineation θ of P with center C and axis H that maps Q onto Q ′ .
Let θ ′ be the central collineation of U that is induced by θ. Since θ ′ and
θ ∗ share the same center and the same axis and map Q onto Q ′ , it follows
that θ ′ = θ ∗ . Hence θ ∗ is induced by the central collineation θ of P.

128 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
3.2 The Group of Translations
Our present basic goal is to construct a vector space that will be used to
coordinatize the projective space P. A main idea is first to coordinatize an
affine space A = P \ H for some hyperplane H of P. The next theorem says
that this approach should be a good one.
Theorem 3.2.1. Let θ be a collineation of the affine space A = P \H. If the
order of P is greater than 2 then there is precisely one collineation of P such
that its restriction to A is θ. Briefly, θ has a unique projective extension.
Proof. First we show that θ has at most one extension to a collineation of
P. Let θ ∗ and θ + be extensions of θ. Then β := θ + θ ∗−1 is a collineation of
P which fixes each point and each line of A. Since each line of A and the
hyperplane H are fixed, β also fixes each point at infinity. Hence β = id and
θ ∗ = θ + .
The existence is more difficult. We start with the following
Claim: If g and h are parallel lines, then θ(g) and θ(h) are parallel.
Clearly we may assume that g = h. Then the images of g and h also
have no point in common, but they could be skew! Since g and h are parallel
they are contained in some plane π. The immediate goal is to show that the
images of g and h also lie in a common plane.
Since the order of P is greater than 2 there is a point P in π that is
neither on g nor on h. Let ℓ and m be two lines in π through P that are not
parallel to g. Then the lines ℓ and m both intersect both g and h. Then θ(ℓ)
and θ(m) intersect each other and also both intersect both θ(g) and θ(h)..
Thus θ(g) and θ(h) lie in the plane 〈θ(ℓ), θ(m)〉 and are therefore parallel,
proving the Claim.
Now define an extension θ ∗ of θ as follows. For P ∈ A = P \ H put
θ ∗ (P ) = θ(P ). For P ∈ H let g be a line through P that is not in H. Put
θ ∗ (P ) = θ(g) ∩ H. Since g is an affine line, θ(g) is well-defined.
We have to show that θ ∗ is well-defined. So if g and h are two lines
through P that are not in H, then they are parallel as lines of A. By the
Claim, θ(g) and θ(h) are also parallel, hence they intersect each other in a
point of H, namely in θ ∗ (P ). Thus θ ∗ is well-defined.
The map θ ∗ is bijective on the points of H since it is bijective on the line
set of A.
In order to show that θ ∗ is a collineation it is sufficient to show that θ ∗
maps collinear points to collinear points. This is clear for points of A, and

3.2. THE GROUP OF TRANSLATIONS 129
also if only one of the points lies in H. Let P1, P2, P3 be three collinear
points of H and let X be any point of A. Then π = 〈X, P1, P2, P3〉 is a plane.
By definition of θ ∗ , θ ∗ (π) is a plane that intersects H in a line containing the
points θ ∗ (P1), θ ∗ (P2) and θ ∗ (P3). Hence these points are collinear.
The case where the order of P is 2 is truly an exceptional case. The
existence part of the result does not remain true. The problem comes in
because any permutation of the points of A preserves collinearity of points
and might not preserve planes. However, if a collineation θ preserves the
planes of A it can be shown that θ can be extended to a colliineation of P.
Note 3.2.2. For the remainder of this chapter we suppose that P is Desarguesian.
Furthermore, we fix a hyperplane H. Then the theorem of Desargues
also holds in the affine space A = P \ H.
We denote the set of all elations with axis H and center on H by T (H).
(In the affine space A = P \ H the elements of T (H) are called translations
of A.)
Theorem 3.2.3. For each hyperplane H of P, T (H) is an abelian group
acting regularly on the points of A = P \ H.
Proof. Let P and Q be any points of A. WOLG we may assume P = Q. If
P is mapped by a translation onto Q, then the center of this translation is
C = P Q∩H. By the Theorem of Baer there is exactly one central collineation
with axis H and center C mapping P onto Q, the set T (H) acts even sharply
transitively on the points of A.
Now we show that T (H) is a group. Since the set of all collineations of
P is a group with respect to composition of maps, we need only show that
T (H) is a subgroup of this group. Since T (H) is nonempty, we need only
show that for θ, φ ∈ T (H) both θ −1 and θφ are in T (H). So let θ, φ be
arbitrary element of T (H). Then each point of H is fixed by θ −1 and by θφ.
So by Lemma 3.1.9 θ −1 and θφ are central collineations. WOLG we may
suppose that θ = id. Then θ fixes no point of A, implying that θ −1 fixes no
point of A. In particular the center of θ −1 is in H, so θ −1 ∈ T (H). Now
we may suppose that θφ = id and we must show that the center of θφ is on
H. Assume that some point P ∈ H is fixed by θφ. Then φ(P ) = θ −1 (P ).
By Corollary 3.1.4 (b) we have θ −1 = φ, so θφ must be the identity. This
completes a proof that T (H) is a group.

130 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
Finally, we must show that T (H) is abelian. Let θ1 and θ2 be arbitrary
elements of T (H). It suffices to show that for even one point X ∈ H we have
θ1θ2(X) = θ2θ1(X). Suppose that θi has center Ci (and neither elation is the
identity) i = 1, 2.
Case 1. C1 = C2.
Let X be an arbitrary point of A. Consider the points X, θ1(X), θ2(X)
and θ2(θ1(X)). Since the point θ2(θ1(X)) is the image of θ1(X) under θ2, the
point θ2(θ1(X)) lies on the line C2θ1(X). Moreover, we have
C1 = θ2(C1) is on θ2(Xθ1(X)) = θ2(X)θ2(θ1(X)).
Thus the point θ2(θ1(X)) is also incident with C1θ2(X). Similarly, we have
that the point θ(θ2(X)) is incident with C2θ1(X) as well as with C1θ2(X).
Therefore θ1θ2(X) = θ2θ1(X). Hence θ1θ2 = θ2θ1.
Case 2. C1 = C2. Consider a point C3 = C1 and a θνT (H), θ3 = id,
with center C3. Since the translations with center C1 form a group, θ1θ3 also
has a center different form C1. By Case 1 we get
θ2θ3 = θ3θ2 and θ2(θ1θ3) = (θ1θ3)θ2.
From these results together it follows that
(θ1θ2)θ3 = θ1(θ2θ3) = θ1(θ3θ2) = (θ1θ3)θ2 = θ2(θ1θ3) = (θ2θ1)θ3,
hence θ1θ2 = θ2θ1.
Now distinguish an arbitrary (but fixed) point O of A. Since T (H) acts
regularly on the points of A, each point P ∈ A may be identified with the
unique translation τP ∈ T (H) that maps O to P . In particular, τO = id and
τP (O) = P . This identification allows us to define an addition ‘+ ′ on the
points of H.
Define:
P + Q := τP (Q) = τP (τQ(O)) = τQ(τP (O)).
Theorem 3.2.4. Let P ∗ denote the set of points of A. Then (P ∗ , +) is a
group isomorphic to (T (H), ◦).

3.2. THE GROUP OF TRANSLATIONS 131
Proof. Define a map
f : P ∗ → T (H) : P ↦→ τP .
Clearly f is a bijection, so what we must show is that f(P + Q) =
f(P ) ◦ f(Q) for all points P, Q ∈ P ∗ . By the definition of “ + ” we have
f(P + Q) = θP +Q = ττP (Q).
So f(P + Q) is the translation that maps O onto the point τP (Q). Now
consider the image of O under f(P ) ◦ f(Q) = τP τQ. Here τQ maps O onto Q,
and then this point is mapped onto τP (Q). Thus the translatons f(P + Q)
and f(P ) ◦ f(Q) map O onto the same point. Hence these translations are
equal.
In particular we have
τP +Q = τP ◦ τQ. (3.2)
Given a point P ∈ A different from O we note that its “negative” −P
is the point τ −1
P
(O). Since τ −1
P must also be a translation with center P ∗ =
OP ∩ H, the point −P lies on the line OP ∗ = OP . Clearly −O = O.
It is sometimes useful to see how to obtain the sum of two points in a
geometric way. First, suppose that O = P ∈ P ∗ with P ∗ = P O ∩H as above.
Then if P , Q and O are noncollinear, we claim
P + Q = P ∗ Q ∩ P Q ∗ . (3.3)
To see this, note that P ∗ is the center of τP , hence P + Q = τP (Q) l ies
on the line QP ∗ . On the other hand, τP maps the line OQ ∗ onto P Q ∗ , since
Q ∗ is fixed by τP . Hence τP (Q) is also incident with P Q ∗ .
From an affine point of view, the lines OP and Q(P + Q), as well as the
lines OQ and P (P + Q) are parallel. Thus the points O, P, Q, P + Q determine
a parallelogram. So we can express the above geometric observation as:
P + Q is the fourth point that completes O, P, Q to the four corner points
of a parallelogram in A.
Keep the above points and lines in mind and suppose that the order of
P is greater than 2. Then let P ′ be any point of the line OP different from
O, P and P ∗ . To see where P + P ′ is, note that the line P ′ Q meets H in a
point R ∗ . Since τP : Q ↦→ P + Q and τP fixes R ∗ , it maps the line P ′ Q to
the line 〈P + Q, R ∗ . So
P + P ′ = τP (P ′ ) = 〈P + Q, R ∗ 〉 ∩ OP.

132 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
Of course this must be independent of the point Q, essentially because the
theorem of Desargues is assume to hold.
Definition 3.2.5. For a point P ∈ H let T (P, H) be the set of all central
collineations with axis H and center P .
Clearly for each P ∈ H the set T (P, H) is an (abelian) subgroup of T (H).
Moreover, we may identify the elements of T (P, H) with the points of A on
the line OP . So the lines through O are in a one-to-one correspondence with
the subgroups T (P, H) of T (H). With the help of T (H) we already have a
fairly nice description of the affine space A. However, we want T (H) not
merely to be an abelian group, we want it to be a vector space in order to
obtain the usual description of an affine space. The following theorem is a
step in this direction.
Theorem 3.2.6. The points of A form an abelian group, which we denote
by T . The lines through O are certain (necessarily normal) subgroups of T .
The other lines are the cosets with respect to these subgroups on O.
Proof. The first statement was established in the previous theorem. For the
remainder we introduce some helpful notation. Let g be a line of A and
define
T (g) = {τP : P ∈ g}.
We need to show the following:
(a) If g is a line through O then T (g) = T (C, H), where C is the point
at infinity of g. Conversely, for any subgroup of T (H) of the form T (C, H)
there is a line g through O such that T (g) = T (C, H). (This is clear since
the translations that fix g are exactly the translations with center C.)
(b) If g is a line not through O, then
T (g) = τP ◦ T (C, H),
where P is an arbitrary point of A on g and C is the point at infinity of g.
Conversely, for each point P and each point C at infinity there is a line g
such that
T (g) = τP ◦ T (C, H).
The proof of (b) is a little more involved. Let g be a line not through O,
let P be a point of A on g, and let C = g ∩ H.
Claim: g = {P + X : X is an affine point of OC} =: P + OC.

3.3. THE DIVISION RING 133
To see this, think of C as X ∗ = OX ∩ H for X ∈ OC and P ∗ = OP ∩ H
(as in the previous notation), so that
P + X = P ∗ X ∩ P X ∗ = P ∗ X ∩ P C = P ∗ X ∩ g ∈ g.
On the other hand let Q be a point of A on g. Then the point X :=
OC ∩ P ∗ Q is a preimage of Q under τP . This establishes the claim, and by
(a) we have:
T (g) = {τP +X : X ∈ OC} = {τP ◦ τX : X ∈ OC} = τP ◦ T (C, H).
Conversely, suppose that P is a point of A and C a point at infinity, then
for g = P C we have T (g) = τP ◦ T (C, H).
If g is a line of A meeting H at the point C, we often will write g = P +OC.
3.3 The Division Ring
As before O is a point of P not in the hyperplane H. Let DO denote the set of
all central collineations of P with center O and axis H. The elements of DO
are also called dilatations with center O and axis H. It follows directly from
the theorem of Baer (Theorem 3.1.10) that D0 is a group which acts sharply
transitively on the affine points different from O on each line of A = P \ H
through O. Also, consider the image of a line P + OC of A under σ ∈ DO.
Since O and C are fixed by σ and the image must contain σ(P ), we obtain
Lemma 3.3.1. Let
σ(g) = σ(P ) + OC.
µ : P ∗ → P ∗ : X → −X.
This means that µ maps an arbitrary affine point X to the point −X =
(τX) −1 (O). Then µ (actually it projective extension) is an element of DO.
Proof. Clearly µ is a bijective map fixing the point O. A routine computation
shows that µ maps the line g = P + OC to the line −P + OC. This is
because for each point X on OC the point −X is also on OX = OC, forcing
µ(P + X) = −(P + X) = −P + (−X) is a point of −P + OC. Hence µ is
a collineation of A. So µ has a projective extension (which we also denote

134 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
by µ). (If the order of P is 2, then µ is the identity.) Since along with the
point P , the point −P is also on OC, µ maps any line throught O onto itself.
Since any central collineation must also be axial, µ is a central collineation
with center O and some hyperplane as axis. Since µ has no affine fixed point
except for O”, its axis must be H. This says µ ∈ DO.
Note: Since X + −X = τX(−X) = τX(τ −1
X (O)) = 0, we see that every
point X satisfies
X + −X = O.
The elements of DO act on the points of P ∗ , so they also act on the
elements of the group P ∗ , +). We need to see how they act on this group.
Lemma 3.3.2. Each element of DO is an automorphism of (P ∗ , +). Moreover,
σ ◦ µ = µ ◦ σ, i.e., σ(−X) = −σ(X) for all σ ∈ DO and for all points
X.
Proof. Let σ be an arbitrary element of DO. Since it is clear that σ acts
bijectively on P ∗ , it suffices to show that for arbitrary points X and Y we
have
σ(X + Y ) = σ(X) + σ(Y ).
(FIGURE P. 112) For this we may assume that X, Y = O. First suppose that
X, Y and O are not collinear. Since X +Y = XY ∗ ∩X ∗ Y and σ(X)+σ(Y ) =
σ(X)σ(Y ) ∗ ∩ σ(X) ∗ σ(Y ), we must show that
σ(XY ∗ ∩ X ∗ Y ) = σ(X)σ(Y ) ∗ ∩ σ(X) ∗ σ(Y ).
Since σ is a collineation, this just means
σ(X)σ(Y ∗ ) ∩ σ(X ∗ )σ(Y ) = σ(X)σ(Y ) ∗ ∩ σ(X) ∗ σ(Y ).
We know that σ(X ∗ ) = X ∗ since X ∗ is a point of H. Similarly we get
that σ(Y ∗ ) = Y ∗ . Morever, σ(X) ∗ = X ∗ , since σ moves the point X on
the line OX ∗ . Hence OX and Oσ(X) have the same point at infinity, i.e.,
X ∗ = σ(X) ∗ . Since Y ∗ = σ(Y ) ∗ follows similarly, the assertion holds in the
case that X, Y, and O are not collinear.
Next we show that σ(−X) = −σ(X) for each point X = O. For this, let
P be a point not on the line OX. Since O, X and −X are collinear, P + X

3.3. THE DIVISION RING 135
is also not on the line O(−X). Using what we have already proved we have
σ(P ) = σ(P + X + (−X))
= σ(P + X) + σ(−X)
= σ(P ) + σ(X) + σ(−X)
=⇒
0 = σ(X) + σ(−X),
forcing σ(−X) = −σ(X). Finally we are abler to show that σ(X + Y ) =
σ(X) + σ(Y ) holds for points X, Y with O ∈ XY . If Y = −X then this
is the statement just proved. So assume that Y = −X. Consider a point
P ∈ XY , so also −P ∈ XY . Moreover, since Y = X we have that O, X + P
and (Y − P are not collinear. (EXERCISE: prove this.) Using the first two
parts of the proof it now follows that
σ(X + Y ) = σ(X + P − P + Y )
= σ(X + P ) + σ(Y − P )
= σ(X) + σ(P ) + σ(Y ) + σ(−P )
= σ(X) + σ(P ) + σ(Y ) − σ(P )
= σ(X) + σ(Y ).
The next step is to define addition in DO. The main work consists in
showing that this operation is closed and is contained in the next Lemma.
Lemma 3.3.3. Let σ1, σ2 ∈ DO. Define the map
σ1 + σ2 : P ∗ → P ∗ : X ↦→ σ1(X) + σ2(X).
If σ1 + σ2 is not the zero map (sending each point of A to O), then the
projective extension of σ1 + σ2 is an element of DO.
Proof. Define σ := σ1 + σ2.
Step 1. We have σ(O) = O and σ(X + Y ) = σ(X) + σ(Y ) for all
X, Y ∈ P ∗ . To see this,

136 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
σ(O) = (σ1 + σ2)(O) = σ1(O) + σ2(O) = O + O = O. By the preceding
lemma we see that σi(X + Y ) = σi(X) + σi(Y ) for i = 1, 2; hence
σ(X + Y ) = (σ1 + σ2)(X + Y ) = σ1(S + Y ) + σ2(X + Y )
= σ1(X) + σ1(Y ) + σ2(X) + σ2(Y )
= (σ1 + σ2)(X) + (σ1 + σ2)(Y )
= σ(X) + σ(Y ).
Step 2. If g = P + OC denotes a line with C a point on H, then
σ(g) ⊆ σ(P ) + OC.
To see this, first suppose that P ∈ OC. Hence each point of OC is mapped
onto a point of OC. Now let g = P + OC be an arbitrary line where P is
not on OC. Then by Step 1,
σ(g) = {σ(P + X) : X ∈ OC}
= {σ(P ) + σ(X) : X ∈ OC}
⊆ {(σ(P ) + Y : Y ∈ OC}.
Step 3. If σ is not injective, then σ is the zero map. Suppose there are
distinct points X and Y for which σ(X) = σ(Y ). This means that
so
(σ1 + σ2)(X) = (σ1 + σ2)(Y ),
σ1(X) + σ2(X) = σ1(Y ) + σ2(Y ).
Since µ ∈ DO and since all elements of DO are automorphisms of (P ∗ , +) it
follows that
σ1(X − Y ) = σ1(X) − σ1(Y ) = σ2(Y ) − σ2(X) = σ2(Y − X)
= σ2 ◦ µ(X − Y ) = µ ◦ σ2(X − Y ),
since by Lemma 3.3.2 the automorphism µ commutes with σ2. Since X −U =
O, this implies σ1 = µ ◦ σ2, hence
where 0 denotes the zero map.
σ1 + σ2 = µ ◦ σ2 + σ2 = 0,

3.3. THE DIVISION RING 137
Step 4. If σ is injective then σ ∈ DO. Since σ is injective we have
σ(X) = σ(O) for each point X = O. Consider an arbitrary point X0 with
X0 = O. Since σ(X0) = O, Step 2 implies that σ(X0) ∈ OX0. Hence by the
theorem of Baer there is a σ ′ ∈ DO such that σ ′ (X0) = σ(X0). The goal is to
prove that σ = σ ′ so that σ ∈ DO. To this end consider an arbitrary point
Y ∈ P ∗ , where we first suppose that Y ∈ OX0. Then
Y = OY ∩ [X0 + (Y − X0)O],
since Y − X0 ∈ (Y − X0)O. Therefore we have that
σ(Y ) ∈ σ(OY ) ∩ σ(X0 + (Y − X0)O) ⊆ OY ∩ [σ(X0) + (Y − X0)O].
On the other hand, since σ ′ (X0) = σ(X0), it follows that
σ ′ (Y ) = OY ∩ [σ ′ (X0) + (Y − X0)O] = OY ∩ [σ(X0) + (Y − X0)O],
forcing σ(Y ) = σ ′ (Y ).
From this we also obtain the assertion for points on OX0, by letting a
point Y0 outside OX) play the role of X0. This completes a proof of the
lemma.
Theorem 3.3.4. Let 0 be the zero map on P ∗ . On the set F = DO ∪ {0} we
define (as above) an addition by
(σ1 + σ2)(X) = σ1(X) + σ2(X)
and a multiplication as follows:
σ1 ◦ σ2, if σ1, σ2 ∈ DO,
σ1 · σ2 :=
0, if σ1 = 0 or σ2 = 0.
Then (F, +, ·) is a division ring.
Proof. The fact that F is closed under addition follows from Lemma 3.3.3.
The associativity and commutativity of (F, +) can be reduced to the associativity
and commutativity of (P ∗ , +). The neutral element (or zero) is 0,
and µ ◦ σ is the additive inverse (or negative) of σ. We shall usually just
write −σ instead of µ ◦ σ. This takes care of addition.
For the multiplication, since (Do, ◦) is a group, (F \{0}, ·) is also a group.
So it remains only to show that the distributive laws hold. Let σ1, σ2, σ3 ∈ F .
Then for each X ∈ P ∗ we have

138 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
and
σ1(σ2 + σ3)(X) = σ1((σ2 + σ3)(X)) = σ1(σ2(X) + σ3(X))
= σ1σ2(X) + σ1σ3(X) = (σ1σ2 + σ1σ3)(X),
(σ1 + σ2)(σ3(X) = (σ1 + σ2)(σ3(X)) − σ1(σ3(X)) + σ2(σ3(X))
= σ1σ3(x) + σ2σ3(X) = (σ1σ3 + σ2σ3)(X).
Corollary 3.3.5. Define a “scalar multiplication” on P ∗ by
σ · X := σ(X) for σ ∈ F, X ∈ P ∗ .
Then P ∗ is a vector space over the division ring F .
Proof. We already know that (P ∗ , +) is a commutative group. Moreover, F
is a division ring, and σ(O) = O; σ(X + Y ) = σ(X) + σ(Y ) for all σ ∈ DO
and all X, Y ∈ P ∗ . The equation (σ1 +σ2)(X) = σ1(X)+σ2(X) follows from
the definition of addition in DO. Finally, if σ is the identity map, i.e., the
multiplicative identity, then σ · X = σ(X) = X.
At this point we have made a great deal of progress in our push to describe
each Desarguesian spaces as a P (V ) for some vector space over a division ring.
We know that the set DO of all dilatations with center O, together with the
zero map, is a division ring F , and the set P ∗ of points of A is a vector space
over F .
3.4 The Representation Theorems
We start with a representation theorem for affine spaces. This will quickly
lead to a representation theorem for projective spaces.
Theorem 3.4.1. Let A = P \ H be an affine space in which the theorem of
Desargues is valid. Then there are a division ring F and a vector space V ∗
over F such that

3.4. THE REPRESENTATION THEOREMS 139
• the elements of V ∗ are the points of A;
• the cosets of the 1-dimensional subspaces of V ∗ are the lines of A.
Proof. Obviously we let F be as in Theorem 3.3.4 and put V ∗ := P ∗ .
Each line g through O is of the form OP . Since, by the theorem of Baer,
DO acts transitively on the points of g that are different from O and g ∩ H,
the multiplicative group DO of the division ring F acts sharply transitively
on the set of points of g different from O (and also different from γ ∩ H).
Hence g is the 1-dimensional subspace of V ∗ spanned by P .
Each line g not through O is of the form g = P + OX. Since OX is the
1-dimensional subspace 〈X〉, the line g is the coset P + OX.
Conversely, let 〈X〉 be a 1-dimensional subspace of V ∗ . Then we have
〈X〉 = {σ(X) : σ ∈ F } = OX. So the coset P + 〈X〉 is the line P + OX.
We are now ready for our major representation of projective spaces.
Theorem 3.4.2. Let P be a projective space of dimension at least 2. If P
is Desarguesian, then there is a vector space V over a division ring F such
that P = P(V ).
This theorem says that geometries of the form P(V ) are precisely those
that are Desarguesian.
Proof. Fix an arbitrary hyperplane H of P. Let F and V ∗ be as in Theorem
3.4.1. Define
V := F × V ∗ .
Then V is a vector space over F with elements (a, v) ∈ F × V ∗ . We now
define a map α that assigns to each point X of P a 1-dimensional subspace
of V :
• If X is a point of P \ H, then by Theorem 3.4.1 X is a vector of V ∗ .
In this case we define
α(X) = 〈(1, X)〉.
• If X is a point of H, consider the line OX of P \ H. By Theorem 3.4.1
this line is a 1-dimensional subspace 〈v〉 of V ∗ . In this case define
α(X) = 〈(0, v)〉.

140 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
Claim 1: α is bijective. Clearly α is injective. Consider an arbitrary
1-dimensional subspace 〈(a, v)〉 of V . If a = 0, the vector v/a of V ∗ (or the
corresponding point of P \ H) is the preimage of 〈(a, v)〉. If a = 0, then the
point of H on the line 〈v〉 of P \ H is a preimage of 〈(a, v)〉.
Claim 2: The map α sends lines of P onto lines of P(V ). For, if g is a
line of P \ H, then g = u + 〈v〉 with u, v ∈ V ∗ , and it follows that
α(g) = 〈(1, u), (0, v)〉.
On the other hand suppose that g is a line of H. Let C1, C2 be two points
on g. Let v1, v2 be the two vectors of V ∗ for which
OC1 = 〈v1〉; OC2 = 〈v2〉.
Then 〈(0, v1), (0, v2)〉 is the line α(g). One must now show that α is bijective
on lines, but this is fairly routiine, so the proof is essentially complete.
Since every projective space with dimension at least 3 is Desarguesian,
we have the following importnt corollary.
Corollary 3.4.3. If dim(P) ≥ 3, then P = P(V ).
3.5 The Fundamental Theorem
In this section we shall determine all collineations of Desarguesian affine
and projective spaces. The collineations will be described in terms of the
underlying vector spaces, and as in the preceding section we begin with affine
spaces.
Let A = P \H be an affine space in which the theorem of Desargues holds.
Fix a point O of A. By T = T (H) we denote the group of all translations of
A, and by Γ the group of all collineations of A. These are the collineations
of P that fix the hyperplane H as a set. Put
ΓO = {θ ∈ Γ : θ(O) = O}.
The group ΓO is the subgroup of Γ fixing O and will contain more than
merely the elements of DO.
As we saw in the previous section, there are a division ring F and a vector
space V ∗ such that the points of A are the elements of V ∗ , and the lines of
A are the cosets of the 1-dimensional subspaces of V ∗ .

3.5. THE FUNDAMENTAL THEOREM 141
Lemma 3.5.1. Using the above notation we have:
(a) Γ is a group (with respect to composition of maps).
(b) ΓO is a subgroup of Γ.
(c) T is a normal subgroup of Γ.
(d) Each θ ∈ Γ can be uniquely written as
θ = τσ with τ ∈ T, σ ∈ ΓO.
Proof. Parts (a) and (b) are more or less obvious, so we begin with part (c).
Let τ ∈ T and θ ∈ Γ be arbitrary elements. We have to show that θτθ −1 ∈ T .
Since τ fixes all points of H, for each point P of H we have
so
τθ −1 (P ) = θ −1 (P ),
θτθ −1 (P ) = θθ −1 (P ) = P.
Thus the collineation θτθ −1 has axis H. If θτθ −1 fixes a point Q ∈ H, then
θτθ −1 (Q) = Q and applying θ −1 gives τθ −1 (Q) = Q,
so τ (and hence θτθ −1 ) is the the identity and trivially contained in T . If
θτθ −1 fixes no point outside H, then its center is in H and it belongs to T .
For part (d), given a θ put τ = τθ(O) be the translation moving O to θ(O).
Put σ = τ −1 θ. Then θ = τσ where τ ∈ T and σ(O) = τ −1 (θ(O) = O, so
that σ ∈ ΓO. Suppose also that θ = τ ′ σ ′ where τ ′ ∈ T and σ ′ ∈ GO. Then
Hence τ = τ ′ and σ = σ ′ .
τ ′ τ −1 = σ ′ σ −1 ∈ T ∩ GO = {id}.
This result means that the problem of describing all collineations of A is
reduced to describing all elements of T and all elements of GO.
Lemma 3.5.2. Let τ ∈ T be an arbitrary translation. Choose an arbitrary
point P of A and put P ′ = τ(P ). Consider the points P and P ′ as vectors
of V ∗ . Then we may describe τ as
τ(X) = X + P ′ − P for all points X of A.

142 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
Proof. In view of Theorem 3.2.3 we know that the map τ ′ defined by
τ ′ (X) := X + P ′ − P
is a translatiion. Thus τ and τ ′ are translations which both map the point
P onto P ′ . This implies that τ ′ = τ. This implies that τ ′ = τ, proving the
Lemma.
We now show that the elements of ΓO are ‘semilinear’ maps of the vector
space V ∗ .
Definition 3.5.3. Let V be a vector space over the division ring F , and let
α be an automorphism of F . A map θ of V into V is called a semilinear map
with associated field automorphism α if for all v, w ∈ V and all a ∈ F we
have
θ(v + w) = θ(v) + θ(w),
and
θ(a · v) = α(a)θ(v).
It is clear that those semilinear maps whose associated field automorphism
is the identity are precisely the linear maps. Let {v1, . . . , vs} be a basis for
V and let α be an automorphism of F . Define a map θα : V → V by
θα :
d
i=1
aivi ↦→
d
α(ai) · (vi).
It is almost immediately clear that θα is a semilinear transformation of
A with associated field automorphism being α.
We now wish to provide a description of the elements of Γ0. Since the
elements of V ∗ are the points of A, the collineations of A also act on V ∗ .
For the remainder of this section we will assume that the order of V ∗ is
at least 3.
Lemma 3.5.4. For σ ∈ ΓO and u, v ∈ V ∗ we have that
i=1
σ(v + w) = σ(v) + σ(w).
Proof. Without loss of generality we may assume that v, w = 0

3.5. THE FUNDAMENTAL THEOREM 143
Case 1: 〈v〉 = 〈w〉. Recall from earlier that v + w = vw ∗ + v ∗ w. Since σ
may be extended to a collineation of the projective space P we have
Similarly,
σ(v + w) = σ(v)σ(w ∗ ) ∩ σ(v ∗ )σ(w).
σ(v) + σ(w) = σ(v)σ(w) ∗ ∩ σ(v) ∗ σ(w).
Since σ is a collineation that fixes O and H we conclude that
and similarly
σ(v) ∗ = Oσ(v) ∩ H = σ(Ov ∩ H) = σ(Ov) ∩ σ(H) = σ(v ∗ ),
Putting these together we get
Since 〈v − w〉 = 〈w〉 we also get
and hence
On the other hand we also have
σ(x) ∗ = σ(w ∗ ).
σ(v + w) = σ(v)σ(w ∗ ) ∩ σ(v ∗ )σ(w)
= σ(v)σ(w) ∗ ∩ σ(v) ∗ σ(w)
= σ(v) + σ(w).
σ(v) = σ(v − w + w) = σ(v − w) + σ(w),
σ(v − w) = σ(v) − σ(w).
σ(v − w) = σ(v) + σ(−w),
so in particular σ(−w) = −σ(w). This is true for each vector w ∈ V ∗ since
there is always a vector v ∈ V ∗ with 〈v〉 = 〈w〉.
Case 2: 〈v〉 = 〈w〉.
If v + w = 0, then
σ(v + w) = σ(0) = 0 = σ(v) − σ(v) + σ(v) + σ(−v) = σ(v) + σ(w).
Thus we may assume that v + w = 0. Since dim(P) ≥ 2 there is a u ∈ V ∗
such that 〈u〉 = 〈v〉. By the first case we now get

144 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
σ(v + w) = σ((v + w + u) − u)
= σ(v + w + u) − σ(u) (since 〈u〉 = 〈v + w + u〉)
= σ(v) + σ(w + u) − σ(u) (since 〈v〉 = 〈w + u〉)
= σ(v) + σ(w) + σ(u) − σ(u) (since 〈w〉 = 〈u〉)
= σ(v) + σ(w).
The proof of the next theorem is the main challenge of this section.
Theorem 3.5.5. Each collineation σ ∈ ΓO is a semilinear map of the vector
space V ∗ .
Proof. We have just shown that each element of ΓO is additive. What remains
to be shown is that for each σ ∈ ΓO there is a field automorphism α such
that for all a ∈ F and all X ∈ V ∗ it is true that
σ(a · X) = α(a) · σ(X).
So let σ ∈ ΓO. For a ∈ ˜ F = F \ {0} and a point X = O in P \ H = A, the
points O, X and a · X are collinear. Since is a collineation with σ(O) = O,
the points O, σ(X) and σ(a · X) are also collinear. So σ(a · X) is a scalar
multiple of σ(X). Let αX(a) be the corresponding element of F . This means
that
σ(a · X) = αX(a) · σ(X) ∀a ∈ ˜ F , X = O.
Claim 1. For all a ∈ ˜ F and all X, Y = O we have
αX(a) = αY (a).
The proof is fairly long and we have two cases.
Case 1. The points O, X and Y are not collinear.
By the definition of α, we have on the one hand
σ(a · (X + Y )) = αX+Y (a) · σ(X + Y ) = αX+Y (a) · (σ(X) + σ(Y ))
On the other hand, we have
= αX+Y (a) · σ(X) + αX+Y (a) · σ(Y ).
σ(a · (X + Y )) = σ(a · X + a · Y ) = σ(a · X) + σ(a · Y )
= αX(a) · σ(X) + αY (a) · σ(Y ).

3.5. THE FUNDAMENTAL THEOREM 145
Since σ is a collineation, the points O = σ(O), σ(X) and σ(Y ) are also
noncollinear. Hence σ(X) and σ(Y ) are - considered as vectors - linearly
independent. Thus we have
αX(a) = αX+Y (a) = αY (a).
Case 2. The points O, X and Y are collinear.
Let Z be a point of A not on OX. Then by Case 1 we have that
This completes a proof of Claim 1.
αX(a) = αZ(a) = αY (a).
Define αO(a) := 0. Then for each X = O we have a map α from F into
itself defined by
α(a) := αX(a)
which satisfies
σ(a) · (X) = α(a) · σ(X) ∀α ∈ F and all points X ∈ A.
Obviously α is our candidate for the associated field automorphism.
Claim 2. The map α is an automorphism of F .
First we show that α preserves addition and multiplication. For a, b ∈ F
and X ∈ A we have
α(a + b+) · σ(X) = σ((a + b) · X) = σ(a · X + b · X)
which implies
Also we have
= σ(a · X) + σ(b · X) = α(a) · (X) + α(b) · σ(X)
= (α(a) + α(b)) · σ(X),
α(a + b) = α(a) + α(b).
α(ab) · σ(X) = α(ab · X) = α(a) · σ(b · X) = α(a)α(b) · σ(X),
which implies that
α(ab) = α(a)α(b).

146 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
To show that α is one-to-one, suppose that α(a) = α(b). Recall that σ
acts bijectively on V ∗ , so
σ(a · X) = α(a) · σ(X) = α(b) · σ(X) = σ(b · X), so a · X = b · X,
which implies that a = b and α is injective.
In truth we are mostly interested in the case that F is finite, so F is a
field and an injection α is automatically surjective. But we have gone so far
with a proof in the general case of F being a division ring that we include a
proof that α is surjective.
Suppose a ∈ F and X = O is a point of A. Since a · σ(X) is a point of
the line through O and σ(X), the preimage Y of a · σ(X) must be a point of
the line through O and X. Thus there is a b such that Y = b · X. Therefore
This implies
σ(b · X) = σ(Y ) = a · σ(X).
a · σ(X) = σ(b · X) = α(b) · σ(X), and hence α(b) = a.
This shows that α is an automorphism.
Theorem 3.5.6. Let A = P \H be a Desarguesian affine space of dimension
d ≥ 2 that is represented by a vector space V ∗ over the division ring F . Then
the following assertions are true:
(a) If σ is an invertible semilinear map of V ∗ , then σ is a collineation of
A.
(b) If τ is a translation of A and σ is a semilinear map of V ∗ , then τ ◦ σ
is a collineation of A.
(c) Each collineation θ of A can be represented uniquely in the form
θ = τ ◦ σ,
where τ is a translation of A and σ is a semilinear map on V ∗ .
Proof. Once part (a) is established, parts (b) and (c) follow immediately from
what has been established before. We leave part (a) as an easy exercise.
It is now time to consider the projective version.
Theorem 3.5.7. Let V be a vector space over the division ring F . If θ is a
bijective semilinear map of V then θ induces a collineation of P(V ).

3.5. THE FUNDAMENTAL THEOREM 147
Proof. Let α be the field automorphism associated with θ. We define
Then φ is well-defined since
φ(〈v〉) := 〈θ(v)〉.
φ(〈a · v〉) = 〈θ(a · v)〉 = 〈α(a) · θ(v)〉
for a = 0. Moreover, φ maps lines onto lines:
φ(〈v, w〉) = φ({〈a · v + b · w〉 : a, b ∈ F })
= {〈θ(a · v + b · w)〉 : a, b ∈ F }
= {〈α(a) · θ(v) + α(b) · θ(w)〉 : a, b ∈ F }
= 〈θ(v), θ(w)〉.
Since θ is bijective, φ is als bijective, and the proof is complete.
The next theorem essentially completes the fundamental theorem of projective
geometry.
Theorem 3.5.8. Let P be a Desarguesian projective space of dimension
d ≥ 2, and let V be a vector space such that P = P(V ). Then for each
collineation θ of P there is a bijective semilinear map φ of V that induces θ.
Proof. Let H, O, V ∗ and V be as in the proof of Theorem 3.4.2. Then the
point O is represented by the subspace 〈(1, 0)〉 of V . Let B ∗ = {v1, . . . , vd}
be a basis of V ∗ . Then
is a basis of V and
BV = {ui := (0, vi) : 1 ≤ i ≤ d} ∪ {u0 := (1, 0)}
B = {〈(0, vi)〉 : 1 ≤ i ≤ d} ∪ {〈(1, 0)〉}
is a basis of P. Hence B \ {O} = (B \ {(1, 0)〉}) is a set of d independent
points of H. Since θ is a collineation, θ(B) is also a basis of P.
Let w0, w1, . . . , wd be vectors of V such that
and
〈wi〉 := θ(〈(0, vi)〉), (1 ≤ i ≤ d),
〈w0〉 := θ(〈(1, 0)〉).

148 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
Then {w0, w1, . . . , wd} is a basis of V . We define the map φ : V → V so that
φ maps vectors as follows:
φ : x =
d
i=0
kiui ↦→
d
kiwi.
Then φ is a bijective linear map of V onto itself, so φ induces a collineation
ϕ of P. It follows that
σ := ϕ −1 θ
is a collineation of P which fixes all points of B. In particular, σ fixes the
pont O and a basis of H, hence it also fixes H as a set. Thus we may also
consider σ as a collineation of A = P \ H. Therefore, the ‘affine part’ of
σ, i.e., the restriction of σ to A, is a semilinear map of V ∗ into itself with
associated field automorphism α.
Hence the map ρ of V into itself that is defined by
i=0
ρ(a, v) := (α(a), σ(v)) (a ∈ F, v ∈ V ∗ )
is a semilinear map of V into itself with associated automorphism α.
Since σ conicides with the collineation of P \ H induced by ρ, and since
the extensions of these collineations to P are unique, ρ is the collineation
induced by σ on P.
Thus θ ′ := θρ is the desired semilinear map.
3.6 Projective Collineations
Definition 3.6.1. A collineation of a projective space P(V ) is called projective
if it is induced by a linear map of V .
Clearly the set of projective collineations of P(V ) forms a group. We
want to see how big this group is and find a way to describe the elements of
this group in a useful way.
Theorem 3.6.2. Each central collineation of P(V ) is a projective collineation.
Proof. It is sufficient to prove that if H is an arbitrary hyperplane of P =
P(V ), then any central collineation with axis H is projective.

3.6. PROJECTIVE COLLINEATIONS 149
Let P be coordinatized in such a way that O and V have the same meaning
as in the proof of the last theorem of the preceding section. It is sufficient to
show that all translations with axis H and all elements of DO are projective,
since each central collineation with axis H is a product of a translation with
axis H and an element of DO.
First we show that each element of DO is projective. So let g be the line
through the point O = 〈(1, O)〉 and a point C = 〈(0, u1)〉 on H. Then the
points different form O and C on g have the form 〈(a, u1)〉 with a = 0. Let
P = 〈(a, u1)〉 and Q = 〈(b, u1)〉 be two distinct points of g different from O
and C. The idea is to show that there is a projective collineation θ in DO
mvoing P to Q. So then we know that θ is THE element in DO moving P
to Q. It follows that all members of DO are projective.
To this end we extend 〈(0, u1)〉 to a basis
{〈(0, u1)〉, 〈(0, u2)〉, . . . , 〈(0, ud)〉}
of H. Then the linear map φ that is defined by
φ(1, O) := (b/a, O) and φ(0, ui) = (0, ui), 1 ≤ i ≤ d
induces a collineation θ. Since φ fixes each (0, ui), it fixes each vector in the
span of these vectors. So φ fixes each point of H. So clearly φ has axis H
and center O. Finally, since P = 〈a · (1, O) + (0, u1)〉 is mapped by φ to the
point 〈b · (1, O) + (0, u1)〉 = Q, θ is the projective collineation of DO mapping
P to Q.
Now let τ be a translation with axis H mapping the point O to a point P =
O. In the language of P(V ) this means the following: Let {〈(0, u1)〉, . . . , 〈(0, ud)〉}
be a basis of H. Then τ maps each of the points 〈(0, ui)〉 to itself and each
point of the form 〈(1, X)〉 to 〈(1, X + P )〉. Then the linear map φ that is
defined by
φ : (0, ui) ↦→ (0, ui) and φ : (1, O) ↦→ (1, P )
induces the translation τ. (The collineation induced by φ has H as axis and
moves O to P = O.)
Definition 3.6.3. A set of d + 2 points in general position is called a frame.
In other words, a frame is a set R of d + 2 points of P such that for each
point P of R, the set R \ {P } is a basis of P. An ordered frame is a list
(P0, P1, . . . , Pd+1) of d + 2 points of P such that the set {P0, P1, . . . , Pd+1} is
a frame.

150 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
In a projective plane, a frame is a set of four points no three of which are
collinear. A frame in a 3-dimensional space is a set of five points, no four
of which are contained in a plane. Let {〈v0〉, 〈v1〉, . . . , 〈vd+1〉} be a frame of
P = P(V ). Then without loss of generality we may assume that
vd1 = v0 + v1 + · · · + vd
by replacing each vi with some scalar multiple of itself, 0 ≤ i ≤ d.
We can use frames to put a bound on the number of projective collineations.
Theorem 3.6.4. If a projective collineation of P = P(V ) fixes each point of
a frame of P, then it must be the identity.
Proof. Let θ be a projective collineation that fixes each point of the frame
R = {〈v0〉, . . . , 〈vd+1〉}.
Since θ is projective, there is a linear map φ of V that induces θ. Since each
point of R is fixed, it follows that
and
φ(vi) = ai · vi, (ai ∈ ˜ F )
φ(v0 + · · · vd) = a · (v0 + · · · + vd) (a ∈ ˜ F ).
Since φ is linear and the vi are linearly independent, it follows easily that
a = a0 = · · · ad. Hence φ maps each vector in V to a · v, so it acts as the
identity on the set of subspaces of V . This says it is the identity on P.
Corollary 3.6.5. The group of projective collineations of P(V ) acts sharply
transitively on the the set of ordered frames of P(V ).
Proof. Let R = {P0, . . . , Pd+1} be a frame of P(V ) with Pi = 〈vi〉, 0 ≤
i ≤ d and vd+1 = v0 + · · · + vd. Suppose R ′ = {P ′ 0, . . . , P ′ d+1 is a second
frame with the appropriate notation. Let φ be the linear map defined by
φ(vi) = v ′ i . Then φ induces a projective collineation that maps Pi to P ′
i ,
0 ≤ i ≤ d + 1. The uniqueness of this collineation follows immediately from
Theorem 3.6.4.
An important theorem in projective geometry says that the projective
collineations are precisely the products of central collineations. Our major
goal for the remainder of tis section is to develop a proof of this fact.

3.6. PROJECTIVE COLLINEATIONS 151
Lemma 3.6.6. Let {P0, P1, . . . , Pd} and {Q0, Q1, . . . , Qd} be bases of P.
Then there is a product φ of at most d + 1 central collineations such that
φ(Pi) = Qi for 0 ≤ i ≤ d.
Proof. We start by showing that for each s ∈ {0, 1, . . . , d} there is a product
φs of at most s + 1 central collineations such that
φs(Pi) = Qi for 0 ≤ i ≤ s.
For s = 0, put φ0 equal to any central collineation mapping P0 to Q0. As an
induction hypothesis, now suppose that φs−1 is a central collineation that is
the product of at most s central collineations and maps Pi to Qi for 1 ≤ i ≤
s − 1, and put φs−1(Ps) = P ′ s.
We now need to construct a central collineation θs that fixes Q0, . . . , Qs−1
and maps P ′ s to Qs. Since s − 1 ≤ d − 1, we have that dim〈Q0, . . . , Qs−1i〉 =
s − 1 ≤ d − 1. Since {Q0, . . . Qs−1, Qs} and
{Q0, . . . , Qs−1, P ′ s } = {φs−1(P0), φs−1(P1), . . . , φs−1(Ps−1), φs−1(Ps)}
are independent sets, neither Qs nor P ′ s is a point of 〈Q0, Q1, . . . Qs−1〉.
If we can show that there is a hyperplane H through 〈Q0, . . . , Qs−1〉 that
contains neither Qs nor P ′ s, then there is a central collineation θs with axis
H that maps P ′ s onto Qs. It follows that φs = θs ◦ φs−1 maps Ps to Qs and
is a product of at most s + 1 central collineations.
So what remains is to prove that the hyperplane H with the desired
properties exists.
Claim: There is a hyperplane through 〈Q0, Q1, . . . , Qs−1〉 containing neither
Qs nor P ′ s .
If Qs = P ′ s, then
〈Q0, . . . , Qs−1, Qs+1, . . . , Qd〉
is the desired hyperplane.
So we may assume that Qs = P ′ s . Let P be an arbitrary point on the
line QsP ′ s that is different from Qs and P ′ s. Then there is a subset B of
{Q0, Q1, . . . , Qd} such that P ∈ 〈B〉, but there is no proper subset B ′ of B
such that P ∈ 〈B ′ 〉. By the construction (choice?) of P there is an element
Qi = Qs in B. In view of the exchange lemma, [{Q0, . . . , Qd} \ {Qi}] ∪ {P }
is a basis of P. Hence [〈{Q0, . . . , Qd} \ {Qi, Qs}] ∪ {P }〉 is a hyperplane that
contains P but not Qs. and hence it also does not contain P ′ s .

152 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
Lemma 3.6.7. Let {P0, P1, . . . , Pd, P } and {P0, P1, . . . , Pd, Q} be frames of
P. Then there is a product φ of at most d central collineations such that
and
φ(Pi) = Pi for 0 ≤ i ≤ d
φ(P ) = Q.
Proof. In fact, using induction on d we prove the following slightly stronger
statement: Let Pj be an arbitrary element of {P0, P1, . . . , Pd}. Then there is
a product φ of at most d central collineations with φ(Pi) = Pi for 0 ≤ i ≤ d
and φ(P ) = Q, where Pj is contained in the axis of each of these cental
collineations.
WOLG we may suppose that Pj = P0. First suppose that d = 2. There
is a central collineation θ2 with center P2 and axis P0P1 that maps the point
P to the point P ′ = P P2 ∩ P1Q. Then there is a central collineation θ1 with
center P1 and axis P0P2 that maps P ′ onto Q. Since θ1 and θ2 fix the points
P0, P1 and P2, the same thing holds for φ = θ1θ2 and φ maps P ′ to Q.
Suppose now that d > 2 and assume that the main claim is true for d − 1.
Consider the hyperplane H = 〈P1, . . . Pd−1, P 〉. We first construct two frames
for H to which we may apply the induction hypothesis. Put P ′′ = P0Pd ∩ H
and P ′ = QPd ∩ H.
Claim 1: {P1, . . . , Pd−1, P ′′ , P } is a frame of H.
Since these points are all in H, what we have to show is that any d
of them form an independent set. Since {P1, . . . , Pd−1, P } is a subset of a
frame, it certainly is an independent set. Assume that P ′′ is dependent on
a set B of d − 1 points of {P1, . . . , Pd−1, P }. Then P ′′ ∈ 〈B〉 and therefore
P0 ∈ P ′′ Pd ⊆ 〈B, Pd〉, which is a contradiction since {P0, . . . , Pd, P } is a
frame.
Claim 2: {P1, . . . , Pd−1, P ′′ , P ′ } is a frame of H.
By the first claim, {P1, . . . , Pd−1, P ′′ } is an independent set. Assume that
P ′ is is dependent on {P1, . . . , Pd−1}. Then Q ∈ P ′ P d ⊆ {P1, . . . , Pd}, a
contradiction since {P0, P1, . . . , Pd, Q} is a frame. Assume that P ′ is dependent
on {P1, . . . , Pd−1, P ′′ } \ {Pi} (1 ≤ i ≤ d − 1). WLOG i = 1. It follows
that Q ∈ P ′ Pd ⊆ 〈P2, . . . , Pd, P ′′ 〉 = 〈P2, . . . , Pd, P0〉, a contradiction since
{P0, . . . , Pd, Q} is a frame.
So we have now established that both

3.6. PROJECTIVE COLLINEATIONS 153
{P1, . . . , Pd−1, P ′′ , P } and {P1, . . . , Pd−1, P ′′ , P ′ } are frames of H.
By the induction hypothesis there is a product φ∗ d−1 of d − 1 (at most) d − 1
central collineations φ1∗, . . . , φ∗ d−1 of H such that
φ ∗ d−1(Pi) = Pi for 1 ≤ i ≤ d − 1,
φ ∗ d−1 (P ) = P ′ ,
and with the property that P ′′ is contained in the axis of φ∗ i (1 ≤ i ≤ d − 1).
By Theorem 3.1.11 each central collineation φ∗ i of H is induced by a central
collineation φi of P whose axis passes through P0. Hence the line P ′′ P0, and
therefore also the point Pd ∈ P ′′ P0, is contained in the axis of φi (1 ≤ i ≤
d − 1). In particular, for θd−1 = Φ1 ◦ φ2 ◦ · · · ◦ φd−1 that
φd−1(Pd) = Pd.
We now have to find a central collineation φd that fixes all the points
P0, . . . , Pd, has P0 in its axis, and maps P ′ to Q.
Define φd to be the central collineation with axis 〈P0, P1, . . . , Pd−1〉 and
center Pd that maps P ′ to Q. This is possible if P ′ and Q are not in the axis of
φd. Clearly Q is not in the axis since {P0, . . . , Pd, Q} is a frame. Suppose that
P ′ is contained in the axis of φd. Recall that P ′ = QPd ∩ 〈P1, . . . , Pd−1, P 〉.
Then
P ′ ∈ 〈P0, . . . , Pd−1〉 ∩ 〈P1, . . . , Pd−1, P 〉 = 〈P1, . . . , Pd−1〉,
and therefore
Q ∈ P ′ Pd ⊆ 〈P1, . . . , Pd−1, Pd〉.
But this is impossible since {P0, . . . , Pd, Q} is a frame.
Thus the collineation
has the following properties:
θd = φd ◦ φd−1 ◦ · · · ◦ φ1
θd(Pi) = Pi for 0 ≤ i ≤ d,
θd(P ) = φd ◦ θd−1(P ) = · · · = φd(Q ′ ) = Q.
Hence P0 is contained in the axis of φi (1 ≤ i ≤ d), completing the proof.

154 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)
An easy corollary of the two preceding results is that at most 2d + 1
central collineations are needed to map one ordered frame to another. This
is not too good a result, however, since it is known that at most d + 2 central
collineations are needed.
We are now in a good position to see the following:
Theorem 3.6.8. The projective collineations of P(V ) are precisely the products
of central collineations.
Proof. Since each central collineation is projective, and the product of projective
collineations is projective, one direction easily follows. For the other
direction, let φ be an arbitrary projective collineation. Then φ maps a frame
R = {P0, . . . , Pd+1} to another frame R ′ = {Q0, . . . , Qd+1}. But the previous
Lemma says that there is product θ of central collineations such that
θ(Pi) = Qi, 0 ≤ i ≤ d + 1. Then θ(Pi) = φ(Pi) for 0 ≤ i ≤ d + 1, hence we
have θ = φ. This means φ is a product a central collineations.
Corollary 3.6.9. Let Σ be the set of collineations of P(V ) that fix each
point of a frame. Then Σ is a group which is isomorphic to the group of
automorphisms of F .
Proof. Let
R = {〈v0〉, 〈v1〉, . . . , 〈vd〉, 〈v0 + . . . vd〉}
be a frame that is fixed pointwise by each element of Σ. Let σ ∈ Σ and let θ
be the semilinear map of V with associated automorphism α that induces σ.
As before, σ maps each vi onto avi. WOLG we assume that a = 1. Therefore
we have
θ : (a0v0 + · · · + advd) ↦→ α(a0)v0 + · · · + α)ad)vd.
We denote this map more accurately by θα, and note that the linear part
of θ is just the identity map. Consider the map Σ → Aut(F ) : σ ↦→ α. It
is routine to check that this map is a group homomorphism. If α = α ′ , the
collineations θα and θα ′ are distinct. So the proof is essentially complete.
3.7 A Collection of Collineations
Let π = P G(2, q) and consider the projective collineations θA : (x, y, z) ↦→
(x, y, z)A for a nonsingular matrix A over Fq. Recall that such collineations
are called homographies.

3.7. A COLLECTION OF COLLINEATIONS 155
Obs. 3.7.1. θA is an elation with axis [1, 0, 0] T and center (0, b, c) if and
only if A has the form ⎛
1 bx
⎞
cx
A = ⎝ 0 1 0 ⎠
0 0 1
with b, c, x ∈ Fq, (b, c) = (0, 0).
Obs. 3.7.2. The group of all homographies θ of P G(3, q) which are
(a) elations with axis [0, 0, 1, 0] and center (a, b, 0, d) consists of those
homographies with matrix
⎛
1
⎜
A = ⎜ 0
⎝ ax
0
1
bx
0
0
1
⎞
0
0 ⎟
dx ⎠ , x ∈ Fq.
0 0 0 1
(b) homologies with axis [0, 0, 1, 0] and center (0, 0, 1, 0) consist of those
homographies with matrix
⎛
1
⎜
A = ⎜ 0
⎝ 0
0
1
0
0
0
z
⎞
0
0 ⎟
0 ⎠ , 0 = z ∈ Fq.
0 0 0 1
Obs. 3.7.3. In P G(3, q) there are q4−1 q−1 · q4−q q−1 · q4−a2 q−1 · q4−q3 ordered 4-tuples
q−1
of points in general position. Using the principle of inclusion-exclusion we
can determine that the number of 5th points completing an ordered frame is
(q − 1) 3 . Hence
There are (q 4 − 1)(q 4 − q)(q 4 − q 2 )q 3 ordered frames in P G(3, q).
This with Corollary 3.6.5 shows that the order of the projective group P GL(3, q)
of homographies is (q 4 − 1)(q 4 − q)(q 4 − q 2 )q 3 .

156 CHAPTER 3. THE FUNDAMENTAL THEOREM FOR P G(N, Q)

Chapter 4
The Ubiquitous Oval
4.1 Arcs and Ovals
Let π be a projective plane of order n. (Of course, our main interest is in
the case π ∼ = P G(2, q).) A k-arc in π is a set K of k points of π with no
three collinear. It follows that for any line l of π, either l is an exterior line
(meeting K in 0 points), a tangent line (meeting K in a unique point), or a
secant line (meeting K in 2 points).
Lemma 4.1.1. If K is a k-arc in a plane π of order n, then k ≤ n + 2.
Proof. If K = {P1, P2, . . . , Pk}, then the lines P1P2, P1P3, . . . , P1Pk are k −1
distinct lines through P1, implying that k − 1 ≤ n + 1.
Def. A (0,2)-set in π is a set Ω of points of π such that each line of π
meets Ω in 0 or 2 points.
Lemma 4.1.2. A nonempty set Ω of points of π is a (0, 2)-set iff Ω is a
k-arc with k = n + 2.
Proof. Let Ω be a (0,2)-set with k points, k ≥ 1, and suppose that P ∈ Ω.
The n + 1 lines through P each contain a unique second point of Ω, so
k ≥ n + 2. But as Ω is a k-arc, by Lemma 4.1.1 we have k ≤ n + 2, i.e.,
k = n + 2. Conversely, it is easy to see that if Ω is an (n + 2)-arc, it is also
a (0,2)-set.
Lemma 4.1.3. If n is odd, any k-arc of π has k ≤ n + 1.
157

158 CHAPTER 4. THE UBIQUITOUS OVAL
Proof. Suppose π actually has an (n + 2)-arc Ω. Then Ω must be a (0,2)-set.
So if P ∈ π \ Ω, the points of Ω lie in pairs on those lines through P meeting
Ω in at least one point. Hence n + 2 is even.
Def. An oval of π is a k-arc Ω with k = n + 1. Note that through each
point of an oval there is a unique tangent line.
Let Ω be an oval of π. For 0 ≤ i ≤ n + 1, let ei be the number of points
of π \ Ω on exactly i tangent lines.
n+1
ei = n 2 = |π \ Ω|. (4.1)
i=0
n+1
iei = n(n + 1) = (4.2)
i=0
= |{(P, l) : P ∈ π \ Ω; l is tangent to Ω, and P is incident with l}|.
n+1
i(i − 1)ei = n(n + 1) = (4.3)
i=0
= the number of ordered pairs of intersecting tangent lines.
Lemma 4.1.4. Let n be odd. Each point of π \ Ω is on 0 or 2 tangents,
i.e., is an interior point or an exterior point of Ω. There are n(n−1)
interior
2
points and n(n+1)
exterior points.
2
Proof. Fix a point X ∈ π \ Ω. The secants through X pick off points of Ω in
pairs, leaving an even number of tangents through X. So ej = 0 if j is odd.
Hence we may rewrite equations Eq. 4.1, 4.2, 4.3 as follows:
(4.1)’
(4.2)’
(4.3)’
n+1
2
j=0 e2j = n 2 ;
n+1
2
j=0 2je2j = n(n + 1);
n+1
2
j=0 2j(2j − 1)e2j = n(n + 1).

4.1. ARCS AND OVALS 159
n+1
2
j=0 2j(2j−2)e2j = 0 =
n+1
2
j=0 4j(j−
Subtract the second from the third:
1)e2j. Since e2j ≥ 0, and 4j(j − 1) > 0 for j > 1, it must be that e2j = 0
for j ≥ 2. Hence e0 and e2 are the only ej’s that might be nonzero. From
the middle equation we see e2 = n(n+1)
is the number of exterior points.
2
Then from the first equation, e0 = n2 − e2 = n(n−1)
is the number of interior
2
points.
Lemma 4.1.5. Let n be even. There is a unique point N of π \ Ω that is
on q + 1 tangents to Ω. (N is called the nucleus or knot of Ω.) Each other
point of π \ Ω is on a unique tangent.
Proof. Since n + 1 is odd and the secants through a point X ∈ π \ Ω pick off
the points of Ω in pairs, X must lie on an odd number of tangents. So we
may rewrite Eqs. 4.1, 4.2, 4.3 as follows:
(4.1)
(4.2)
(4.3)
n
′′ 2
j=0 e2j+1 = n2 ;
n
′′ 2
j=0 (2j + 1)e2j+1 = n(n + 1);
n
′′ 2
j=0 (2j + 1)2je2j+1 = n(n + 1).
Now compute (n + 1)(3.2) ′′ − (n + 1)(3.1) ′′ − (3.3) ′′ :
n/2
j=0 [(n + 1)(2j + 1) − (n + 1) − (2j + 1)2j]e2j+1 = (n + 1)n(n + 1) −
(n + 1)n 2 − n(n + 1) simplifies to: n/2
j=0 2j(n − 2j)e2j+1 = 0. So the only
e2j+1 that could be nonzero are for j = 0 and j = n
2 , i.e., e1 and en+1. Using
(4.3) ′′ , we see en+1 = 1. Then from (4.1) ′′ , e1 = n 2 − 1.
This result shows that when n is even and Ω is an oval of the projective
plane π of order n, if N is the nucleus of Ω, then Ω + = Ω ∪ {N} is an
(n + 2)-arc (i.e., a (0,2)-set).
Def. An (n+2)-arc in a projective plane π of order n is called a hyperoval.
Note that if X is any point of a hyperoval Ω + , Ω + \ {X} is an oval with
nucleus X.

160 CHAPTER 4. THE UBIQUITOUS OVAL
4.2 Conics: The Classical Examples
Put
C = {(x0, x1, x2) ∈ P G(2, q) : x 2 1 = x0x2} =
= {(1, t, t 2 ) : t ∈ Fq} ∪ {(0, 0, 1)}.
It is easy to check that each three points of C are linearly independent,
i.e., noncollinear in P G(2, q). Hence C is an example of an oval in P G(2, q),
and when q = 2 e , the nucleus of C is N = (0, 1, 0).
A quadric Q in π = P G(2, q) is the set of points (x, y, z) of π satisfying
a0x 2 + a1xy + a2y 2 + a3xz + a4yz + a5z 2 = 0, (4.4)
for some scalars a0, . . . , a5, not all zero.
Given five points Pi = (xi, yi, zi) on Q, we see that five linear equations
in the six unknowns a0, . . . , a5 arise, so a nontrivial solution exists. Hence
any five points lie on at least one quadric. We have seen that the quadric
given by C : x 2 1 = x0x2 is a (q + 1)-arc, i.e., an oval. In this case C is called a
conic, and the quadric is nonsingular (irreducible, nondegenerate). The other
possible point-sets consisting of a quadric in P G(2, q) are:
• a line counted twice – e.g., x 2 0 ;
• a pair of distinct lines – e.g., x0x1;
• a point, i.e., a pair of conjugate lines in P G(2, q 2 ) with just their intersection
in P G(2, q)) – e.g., x 2 0 + ax0x1 + bx 2 1 irreducible).
In these cases the quadric is singular. See [Hi98] for a much more thorough
analysis of the situation. For our purposes it suffices to prove the following
result.
Lemma 4.2.1. Let Q be⎛a nondegenerate ⎞ quadric in P G(2, q). If the quadric
x
is Q(x, y, z) = (x, y, z)A ⎝ y ⎠ with
z
⎛
a b c
⎞
A = ⎝ 0 d e ⎠ ,
0 0 f

4.2. CONICS: THE CLASSICAL EXAMPLES 161
put B = A + A T . Then the tangent line at the point P of the quadric Q is
BP T . Moreover,
(a) A 5-arc in P G(2, q) lies on a unique conic which is an oval.
(b) A 4-arc in P G(2, q) with a specified tangent line at one of them lie in
a unique conic which is an oval.
(c) A 3-arc in P G(2, q) with tangent lines specified at two of them lie in
a unique conic.
Proof. Suppose that P is a point of Q, i.e., P AP T = 0, so in particular,
P BP T = 0. Let L = BP T . Clearly P is on L. Suppose R is a point on
L ∩ Q. It is easy to check that each point of the line 〈P, R〉 is on Q. But a
nondegenerate quadric in P G(2, q) must contain no line. Hence BP T must
be tangent to Q at P . If we first just fix a 3-arc in Q, since P GL(3, q)
is (sharply) transitive on ordered 4-arcs, in particular we may assume that
P1(1, 0, 0), P2(0, 1, 0), and P3(0, 0, 1) are three points of Q. Putting these
into ⎛ the equation ⎞ for Q (as in Eq. 4.4) yields a = d = f = 0. Then B =
0 b c
⎝ b 0 e ⎠. Using the fact that P GL(3, q) is transitive on 4-arcs we may
c e 0
assume without loss in gvenerality ⎛ ⎞ that ⎡ the ⎤ tangent L to Q at P (1, 0, 0) is the
1 0
line L = [0, 1, 1] = λ·B ⎝ 0 ⎠ = λ ⎣ b ⎦ (for some nonzero λ), which forces
0 c
b = c. The tangent M to Q at P2(0, 1, 0) must have the form M = [1, 0, m] T
for some nonzero m ∈ Fq. Also, M = BP T 2 = [b, 0, e]T , i.e., e = bm with
b = 0. Hence we may assume without loss of generality that b = 1, so
⎛
A = ⎝
0 1 1
0 0 m
0 0 0
This proves part (c) of the theorem. Now suppose Q contains the 4-arc
P1(1, 0, 0), P2(0, 1, 0), P3(0, 0, 1), P4(1, 1, 1) and the tangent [0, 1, m] T at P1,
m = 0, 1. It follows that b = 0 so we may assume that b ⎛=
1. Then c = m and ⎞
0 1 m
from P4 being on Q we have e = −(1+m). Hence A = ⎝ 0 0 −(1 + m) ⎠.
0
Hence part (b) follows.
0 0
⎞
⎠ .

162 CHAPTER 4. THE UBIQUITOUS OVAL
Let K be a 5-arc in P G(2, q). As above, we may assume that K contains
the points (1, 0, 0), (0, 1, 0, (0, 0, 1) and (1, 1, 1). Putting these four points
into Eq. 4.4 reduces the quadratic form to
a1xy + a3xz − (a1 + a3)yz = 0. (4.5)
The six lines joining the four points in pairs are
l1 : y = 0; l2 : y = 0; l3 : z = 0;
l4 : x = y; l5 : y = z; l6 : x = z.
So the fifth point of K is of the form P = (1, y0, z0) with y0z0(y0−z0)(y0−
1)(z0 − 1) = 0. Putting this point in Eq. 4.5 yields
a1 = −a3z0(1 − y0)
.
y0(1 − z0)
This determines the quadratic form up to a scalar multiple, i.e., as a set of
points it is uniquely determined and is equivalent to a form
axy + bxz − (a + b)yz = 0,
ay
with ab = 0, and having points (1, y, (a+b)y−b ), b
a+b = y ∈ Fq, together with
(0, 1, 0) and (0, 0, 1). It is now a routine exercise to show that C is an oval.
Our goal for the remainder of this section is to show that
Theorem 4.2.2. Every homogeneous quadratic equation in three variables
over Fq has a nontrivial solution.
where
To start with we recall that every such quadratic equation has the form
ax 2 + bxy + cxz + dy 2 + eyz + fz 2 ⎛ ⎞
x
= (x, y, z)A ⎝ y ⎠ = 0,
z
⎛
A = ⎝
a b c
0 d e
0 0 f
⎞
⎠ .

4.2. CONICS: THE CLASSICAL EXAMPLES 163
It is clear that if adf = 0 there is a non-trivial solution. Hence we may
assume that adf = 0.
First take care of the case that q = 2e . Then if c = 0, put y = 0 and
the equation becomes ax2 + fz 2 = 0. Put z = √ a, x = √ f to obtain a nontrivial
solution. So suppose c = 0 and put x = e
y. The equation becomes
c
a ′ y2 + fz 2 = 0 for some element a ′ ∈ Fq, which clearly has a non-trivial
solution. This takes care of the case where q is a power of 2. So now assume
that q is odd.
The proof of the theorem depends on the following lemmas.
Lemma 4.2.3. If 0 = γ ∈ Fq, then γ = x 2 + y 2 has a solution in Fq.
Proof. If γ is a square, there is nothing to prove. Hence we may suppose that
q is odd and γ is a non-square in Fq. Then let β1, . . . , β q−1 be the non-zero
2
squares of Fq and put
S = {γ − βi : 1 ≤ i ≤
q − 1
2 }.
If S contained no squares, S would contain precisely all non-squares. But
γ ∈ S implies that S must contain some square, say γ − βi = x 2 , βi = y 2 ,
x, y ∈ Fq. Hence γ = x 2 + y 2 .
Lemma 4.2.4. The equation ax2 0 + dx21 + fx22 nontrivial solution.
= 0 with adf = 0 has a
Proof. At least one of a/d, d/f, f/a is a square, since their product is a
square. Since the problem is symmetric in the three variables, we may suppose
that a/d = g 2 with g ∈ Fq. Divide the equation by d and put x2 = 1.
The equation then becomes
which has a solution by Lemma 4.2.3.
(gx0) 2 + x 2 1 = − f
d ,
Since q is odd, we may replace A with A ′ = 1
2 (A + AT ), i.e., we may
assume that A is symmetric. Then put B(x, z) = q(x + z) − q(x) − q(z) =
x(A + AT )zT = 2q(x + z). The set x⊥ = {z ∈ F 3 q : B(x, z) = 0} is a subspace
of dimension 2 or 3.

164 CHAPTER 4. THE UBIQUITOUS OVAL
Lemma 4.2.5. When q is odd, the vector space F 3 q has a basis {x0, x1, x3}
such that B(xi, xj) = 0 if i = j.
Proof. Since q is odd, q(x) = 0 iff B(x, x) = 0. Hence if q(x) = 0 for all
x ∈ F 3 q there is nothing to prove. So suppose q(x) = 0, so x ∈ x⊥ . Put
x = x0. Then F 3 q = 〈x〉 ⊕ x⊥ . If q(y) = 0 for all y ∈ x⊥ , let x1, x2 be any
basis of x⊥ . Otherwise, let x1 be an element of x⊥ for which q(x1) = 0. Then
x⊥ 1 ∩x⊥0 is a 1-dimensional space of x⊥ with a basis x2. Then B = {x0, x1, x2}
is a basis of F 3 q of the type desired.
The quadratic form q(x) = xAx T , when expressed in terms of the basis
B has a diagonal matrix, i.e., if [x] is the coordinate matrix of the vector x
with respect to the basis B, then q(x) = [x]A ′ [x] T , where A ′ is a diagonal
matrix. By Lemma 4.2.4 such a quadratic form has a nontrivial solution.
This completes a proof of Theorem 4.2.2.
Exercise 4.2.5.1. Let P1, P2, P3, P4 be four points of P G(2, q) with no three
on a line. Show that there are three degenerate conics containing these four
points, each consisting of two lines so that each of the points Pi is on just
one of the two lines. There are q − 2 irreducible conics partitioning the
(q − 2)(q − 3) ⎛points
not on any ⎞ of these six lines.
0 a b
Let A = ⎝ 0 0 −a − b ⎠. Then q(x) = xAx
0 0 0
T is degenerate if and
only if ab(a + b) = 0. Suppose a = 1, b ∈ {0, −1}. Then if q = 2e , the point
(1 + b, b, 1) is the nucleus. Such a point lies on the line [1, 1, 1]. The quadric
Q : q(x) = 0 contains the fundamental quadrangle.
Exercise 4.2.5.2. Show that in P G(2, q) there are
(i) q 4 − q 2 conics on a fixed point P ;
(ii) q 5 − q 2 conics;
(iii) q 5 − q 4 conics not on a fixed point P .
Now suppose that q = 2 e and let N be a fixed point of P G(2, q). Show
that there are
(iv) q 3 − q 2 conics with nucleus N.
(v) q 2 (q − 1)(q 2 − 1) conics not on a fixed point P and having nucleus
different from P .
(vi) q(q − 1) 2 conics with nucleus N and not containing a fixed point P
different from N.

4.3. THE GROUP OF THE CONIC 165
4.3 The Group of the Conic
Let
C = {(x0, x1, x2) ∈ P G(2, q) : x 2 1 = x0x2}
= {(1, t, t 2 ) : t ∈ Fq} ∪ {(0, 0, 1)}.
The subgroup of P ΓL(3, q) leaving C invariant is denoted P Γ0(3, q). Note
that the field automorphisms induce collineations of P G(2, q) that leave C
invariant, so we need only to determine the group P GO(3, q) of homographies
leaving C invariant. This is given by the following theorem.
Theorem 4.3.1. The group P GO(3, q) of homographies leaving C invariant
consist of those homographies θ with matrices of the form
⎛
d
[θ] = ⎝
2 −bd b2 −2cd ad + bc −2ab
c2 −ac a2 ⎞
⎠ , ∆ = ad − bc = 0.
The determinant of this [θ] is ∆3 , hence we may multiply the matrix by ∆−1 to obtain a matrix with determinant 1. The inverse matrix (which acts on
the planes of P G(2, q)) is then given by
[θ] −1 = ∆ −1
⎛
⎝
a 2 ab b 2
2ac ad + bc 2bd
c 2 cd d 2
Further, when q is odd we have the following:
(i) P GO(3, q) is sharply triply transitive on the points of C; |P GO(3, q)| =
q 3 − q;
(ii) P GO(3, q) is transitive on the interior points of C;
(iii) P GO(3, q) is transitive on the exterior points of C;
(iv) P GO(3, q) is transitive on the tangent lines to C;
(v) P GO(3, q) is transitive on the secant lines to C;
(vi) P GO(3, q) is transitive on the point-line pairs (P, ℓ), where P is an
exterior point on the exterior line ℓ. The subgroup fixing such a pair has
order 4.
Proof. It is an easy exercise to show that the homographies given above really
preserve C and act sharply triply transitively on the points of C.
⎞
⎠ .

166 CHAPTER 4. THE UBIQUITOUS OVAL
The matrix of C is
⎛
0 0
⎞
1
AC = ⎝ 0
0
−1
0
0 ⎠ ;
0
BC = AC + A T ⎛
C = ⎝
0 0 1
0 −2 0
1 0 0
The tangent to C at (0, 0, 1) is the line [1, 0, 0], whose exterior points are
those of the form (0, 1, m), m ∈ Fq. Put d = 1, b = c = 0 in [θ] to map
(0, 1, m) to (0, 1, ma), a ∈ Fq. So the exterior points (0, 1, m) are all in the
same orbit of P GO(3, q). Each point not on the tangent [1, 0, 0] is of the
form (1, t, t 2 + k), t, k ∈ Fq. The tangent at (1, s, s 2 ) is [s 2 , −2s, 1]. Then
(1, t, t 2 + k) is on some tangent [s 2 , −2s, 1] if and only if ((s − t) 2 = −k.
Hence
Interior points of C are (1, t, t 2 + k), t ∈ Fq, −k = ❆ ∈ Fq;
Exterior points of C are (1, t, t 2 + k), 0 = −k = ∈ Fq; (0, 1, m); m ∈ Fq
For arbitrary t ∈ Fq, in [θ] put b = at, 0 = a, d = −1, c = 0 to map
(1, t, t2 + k) to
(1, t, t 2 ⎛
1 at a
+ k) ⎝
2t2 0 −a −2a2t 0 0 a2 ⎞
⎠ = (1, 0, a 2 k).
From this it is clear that P GO(3, q) is transitive on the interior points of
C, and it is transitive on the exterior points of the form (1, t, t 2 +k), −k = .
Since k = −1 gives −k = 1 = , (1, 1, 0) is one of these exterior points. Then
suppose that −k = s 2 = 0. In [θ] put a = 0, c = d = 1, b = s to get the
following:
⎛
θ : (1, 1, 0) ↦→ (1, 1, 0) ⎝
1 −s −k
−2 s 0
1 0 0
⎞
⎞
⎠ .
⎠ = (−1, 0, −k) ≡ (1, 0, k).
Hence P GO(3, q) is transitive on the set of all exterior points.
We notice that there are q(q−1)
interior points and 2
q(q+1)
exterior points.
2
So the order of the stabilizer of an interior (resp., exterior) point is much

4.3. THE GROUP OF THE CONIC 167
smaller than the size of the set of the other interior (resp., exterior) points.
Hence P GO(3, q) is far from doubly transitive on the set of interior (resp.,
exterior) points.
The only part of the theorem still awaiting proof is part (vi). For this it
suffices to consider the set of lines passing through an exterior point of our
choice. So consider the lines through P = (1, 1, 0). The two tangent lines are
[0, 0, 1] and [1, −1, 1/4]. There are q−1
exterior (resp., secant) lines through
2
P . The line [1, −1, w] with 0 = w = 1/4 may be secant or exterior, and will
be an exterior line if and only if 1 − 4w = ❆ . First compute the stabilizer of
(1, 1, 0).
The subgroup G of P GO(3, q) that fixes the point (1, 1, 0) consists of the
2(q − 1) homographies with matrices of the following types:
1
(1 − 2g)
⎛
⎝
1 0 0
−2g 1 − 2g 0
g2 −g(1 − 2g) (1 − 2g) 2
⎛
⎝
1 2 4
2g 2g − 1 −4
g 2 −g 1
⎞
⎠ , g = 1
⎞
;
2
(4.6)
⎠ , g = − 1
.
2
(4.7)
The homographies of the first type form a subgroup G0 of order q − 1.
The inverse of such a matrix is
⎛
(1 − 2g)
1
⎝
(1 − 2g)
2 2g(1 − 2g)
g
0
1 − 2g
0
0
2 g
⎞
⎠ .
1
Let G1 denote the other coset of G0 in G. Each member of G1 is an
involution.
With this we can see that an element of G0 maps [1, −1, w] to ℓg =
[1, −1, w+g2−g (1−2g) 2 ]. So [1, −1, w] is fixed if and only if (g2 − g)(1 − 4w) = 0. The
two tangents [0, 0, 1] and [4, −4, 1] through (1, 1, 0) are fixed by each element
of G0, and interchanged by each element of G1. The collineation with g = 1 is
an homology with center (1, 1, 0) and axis [0, −2, 1], the perp of (1, 1, 0) with
respect to the conic, so it fixes all lines through (1, 1, 0). But for g = 0, 1,
the two tangents are the only lines through (1, 1, 0) that are fixed.
Now consider the other coset G1. Given a line [1, −1, w], it is stabilized
by the two homographies in G1 with g = −2w or g = 2w − 1. But these two
homographies, with matrices

168 CHAPTER 4. THE UBIQUITOUS OVAL
⎛
⎝
1 2 4
−4w −4w − 1 −4
4w 2 2w 1
⎞
⎠ , and
⎛
⎝
1 2 4
2(2w − 1) 4w − 3 −4
(2w − 1) 2 1 − 2w 1
⎞
⎠ , respectively,
also fix exactly one other line [1, −1, (1 − 2w)/2]. Now 1 − 4
1−2w = −1 +
2
4w = −(1 − 4w). If q ≡ 1 (mod 4), the two lines [1, −1, w] and [1, −1, (1 −
2w)/2] are of the same type, but when q ≡ −1 (mod 4) they are of opposite
types. In any case we see that each element of G1 fixes two lines through
(1, 1, 0). By the orbit counting lemma we have that the number of orbits of
G on the set of q − 1 non-tangent lines through (1, 1, 0) is equal to
1
[(q − 1) + (q − 1) + (q − 1)2] = 2,
2(q − 1)
i.e., the two orbits are the secants and the exterior lines through (1, 1, 0).
Now consider the case where (1, 0, k) is an internal point, i.e., −k = ❆ .
This point is fixed by the group of homographies of order 2(q+1) with matrics
of the following types:
⎛
⎝
1 0 0
0 ±1 0
0 0 1
⎞
⎠ , and
⎛
a
⎝
2 a 1
± 2a
k ± 1
k − a2 ∓2a
a2 ⎞
⎠ . (4.8)
1
k 2
The nonidentity homography of the first type is an homology with center
(0, 1, 0) and axis [0, 1, 0]. Here [0, 1, 0] is a secant through (1, 0, k) (on (1, 0, 0)
and (0, 0, 1)). The other fixed line through (1, 0, k) is [−k, 0, 1], which is a
secant if and only if q ≡ 3 (mod 4).
The other type of homography fixing (1, 0, k) has inverse (up to a scalar
multiple) equal to ⎛
a
⎝
2 ±a 1
2a ± ⎞
1
k
1
k 2
∓ a
k
−a
k
k − a2 −2a
a2 ⎠ .
The fixed lines through (1, 0, k) when we use the top sign (of ±) and have
a = 0 are the two lines [−k, v, 1] with v = 2ak or v = −2.
With the bottom
a
sign and a = 0 there is no solution. When a = 0 we find no solution with the
top sign. With the bottom sign the collineation is the homology with center
(1, 0, k) and axis [k, 0, 1], the perp of (1, 0, k) with respect to the conic, so

4.3. THE GROUP OF THE CONIC 169
there are q + 1 lines through (1, 0, k) fixed. By the orbit counting lemma we
see that the number of orbits on the lines through (1, 0, k) of the group fixing
(1, 0, k) is
1
[(q + 1) + 2 + 2(q − 1) + 0(q − 1) + 2 + (q + 1)] = 2.
2(q + 1)
Hence the subgroup of P GO(3, q) fixing (1, 0, k) acts transitively on the secants
through (1, 0, k) and transitively on the exterior lines through (1, 0, k).
In the case of using the top sign with a = 0, the fixed line [−k, 2ak, 1] is
secant if and only if (a2k + 1)k = , while the other fixed line [−k, −2,
1] is a
a secant if and only if (a2k + 1) = . So the two lines through (1, 0, k) fixed
by such a collineation are of the same type if and only if q ≡ 3 (mod 4).
In any case, the group fixing (1, 0, k) has order 2(q + 1); the subgroup
fixing a line has order 4.
Recap: P GO(3, q) is transitive on the point-line pairs (P, ℓ) where P is
an exterior point and ℓ is an exterior (resp., secant ) line through P . The
subgroup G fixing the conic and the exterior point P has order 2(q − 1). It is
transitive on the (q − 1)/2 exterior (resp., secant) lines through P , and the
subgroup of G fixing a given exterior (resp. secant) line through P has order
4. The identity and one nonidentity involution (an homology with center P
and axis P ⊥ ) fix all the lines through P (and obviously permute the q − 1
points of C not on the two tangents through P in pairs). The other two
involutions that fix a given exterior line ℓ through P also fix the same second
line ℓ ′ . When q ≡ 1 (mod 4), ℓ ′ is also an exterior line. When q ≡ 3 (mod 4),
ℓ ′ is a secant line.
P GO(3, q) is transitive on the point-line pairs (P, ℓ) where P is an internal
point and ℓ is an exterior (resp., secant) line through P . The subgroup G
fixing the conic and the point P has order 2(q +1), has a subgroup of order 4
fixing each line through P and acts transitively on the secant (resp., exterior)
lines through P . Again there is a unique homology in G with center P and
axis P ⊥ . In those cases in which the given element of G fixes exactly two lines
through P , the two lines are of the same type if and only if q ≡ 3 (mod 4).
We now consider certain collections of conics in P G(2, q).

170 CHAPTER 4. THE UBIQUITOUS OVAL
The conics C ′ of π = P G(2, q) that have ℓ = [1, 0, 0] as tangent line at
∞ = (0, 1, 0) ∈ C ′ are those
with upper triangular matrix
C ′ = {(1, as 2 + bs + c, s) : s ∈ Fq} ∪ {(0, 1, 0)}
AC ′ =
⎛
⎝
c −1 b
0 0 0
0 0 a
⎞
⎠ . (4.9)
Our favorite conic is C with matrix
⎛
0 −1
⎞
0
AC = ⎝ 0 0 0 ⎠ . (4.10)
0 0 1
The homography φ with matrix
⎛
[φ] = ⎝
1 b c
0 1 0
0 0 1
is the elation with center (0, b, c) and axis [1, 0, 0].
⎛
⎝
1 λe 0
0 1 0
0 e 1
⎛
⎝
⎞
1 a 0
0 1 0
0 b 1
⎛
⎝
⎞
⎠ (4.11)
⎠ : e ∈ Fq gives all elations with center (0, 1, 0) and axis [λ, 0, 1].
⎞
1 c 0
0 a 0
0 b 1
(4.12)
⎠ : a, b ∈ Fq gives all elations with center (0, 1, 0). (4.13)
⎞
homographies with center (0, 1, 0).
⎠ : a, b, c ∈ Fq gives all q 2 (q − 1) central (4.14)

4.3. THE GROUP OF THE CONIC 171
The general homography with center (0, 1, 0) (as given in Eq. 4.14) maps
C to
C ′ = {(1, as 2 + bs + c, s) : s ∈ Fq} ∪ {(0, 1, 0)} (4.15)
which is the general conic in P G(2, q) with [1, 0, 0] as tangent at (0, 1, 0).
Hence the conics of P G(2, q) with [1, 0, 0] as tangent at (0, 1, 0) which are images
of C under elations with center (0, 1, 0) are precisely those C ′ in Eq. 4.15
with a = 1.
For f(s) a quadratic polynomial in s, let C(f) denote the conic whose
general point appears as (1, f(s), s), s ∈ Fq, together with ∞. Then we want
to investigate the intersection of the conics, keeping in mind that they always
share the point ∞ = (0, 1, 0).
C (s 2 ) ∩ C (as 2 +bs+c) contains (1, t 2 , t) iff (a − 1)t 2 + bt + c = 0. (4.16)
There are precisely the following possibilities:
(i) Suppose a = 1, so C (as 2 +bs+c) is not the image of C (s 2 ) under an elation
with center (0, 1, 0). Then there is a point (1, t 2 , t) ∈ C (s 2 ) ∩ C (as 2 +bs+c) if and
only if
(a) q is odd and b2 − 4(a − 1)c = ∈ Fq;
(a−1)c
(b) q is even and tr = 0.
b 2
(ii) Suppose a = 1, so C(s 2 +bs+c) is the image of C(s 2 ) under an elation with
center (0, 1, 0).
In this case C (s 2 ) and C (s 2 +bs+c) have the point (b 2 , c 2 , −bc) in common. If
b = 0, this is a point in addition to ∞. If b = 0, (so c = 0 if the two conics
are really distinct), then the two conics have only the point ∞ in common.
We have the following conclusion:
Lemma 4.3.2. C (s 2 ) ∩ C (as 2 +bs+c) = {(0, 1, 0)} if and only if either
(i) a = 1 (and b = 0), and
(a) q is odd and b2 − 4(a − 1)c = ❆ ∈ Fq, or
(a−1)c
(b) q is even and tr = 1;
b 2

172 CHAPTER 4. THE UBIQUITOUS OVAL
(In this case C (as 2 +bs+c) is not the image of C (s 2 ) under an elation of
P G(2, q) with center ∞.)
or
(ii) a = 1 and b = 0. (In this case C (as 2 +bs+c) is the image of C (s 2 ) under
an elation of P G(2, q) with center ∞.)
Then
Let φ be the homography of P G(2, q) with matrix
⎛
[φ] = ⎝
[φ] −1 AC[φ] −T =
⎛
= ⎝
⎛
⎝
1 c 0
0 a 0
0 b 1
⎞
1 −c/a 0
0 1/a 0
0 −b/a 1
c/a −1/a b/a
0 0 0
0 0 1
⎛
⎠ , and [φ] −1 = ⎝
⎞ ⎛
⎠ ⎝
⎞
0 −1 0
0 0 0
0 0 1
⎛
⎠ ≡ ⎝
1 −c/a 0
0 1/a 0
0 −b/a 1
⎞ ⎛
⎠ ⎝
c −1 b
0 0 0
0 0 a
⎞
⎠ .
1 0 0
−c/a 1/a −b/a
0 0 1
⎞
⎠ = AC ′.
Now we want to embed P G(2, q) into π = [0, 0, 0, 1] ⊂ P G(3, q) by
(x, y, z) ↦→ (x, y, z, 0), and then lift φ to a homography Φ of P G(3, q) that
acts like φ on π and fixes the point (1, u, v, 1). A simple computation shows
that for each 0 = w ∈ Fq the following homography Φ will do this:
[Φ] =
⎛
⎜
⎝
1 c 0 0
0 a 0 0
0 b 1 0
w − 1 (w − a)u − bv − c v(w − 1) w
⎞
⎟
⎠ .
The conics C ′ of π = P G(2, q) that have ℓ = [1, 0, 0] as tangent line at
∞ = (0, 1, 0) ∈ C ′ are those
with upper triangular matrix
C ′ = {(1, as 2 + bs + c, s) : s ∈ Fq} ∪ {(0, 1, 0)}
⎞
⎠

4.3. THE GROUP OF THE CONIC 173
AC ′ =
⎛
⎝
c −1 b
0 0 0
0 0 a
⎞
⎠ . (4.17)
Our favorite conic is C with matrix
⎛
0 −1
⎞
0
AC = ⎝ 0 0 0 ⎠ . (4.18)
0 0 1
The homography φ with matrix
⎛
[φ] = ⎝
1 b c
0 1 0
0 0 1
is the elation with center (0, b, c) and axis [1, 0, 0].
⎛
⎝
1 λe 0
0 1 0
0 e 1
⎛
⎝
⎞
1 a 0
0 1 0
0 b 1
⎛
⎝
⎞
⎠ (4.19)
⎠ : e ∈ Fq gives all elations with center (0, 1, 0) and axis [λ, 0, 1].
⎞
1 c 0
0 a 0
0 b 1
(4.20)
⎠ : a, b ∈ Fq gives all elations with center (0, 1, 0). (4.21)
⎞
⎠ : a, b, c ∈ Fq gives all q 2 (q − 1) central (4.22)
homographies with center (0, 1, 0).
The general homography with center (0, 1, 0) (as given in Eq. 4.22) maps
C to
C ′ = {(1, as 2 + bs + c, s) : s ∈ Fq} ∪ {(0, 1, 0)} (4.23)
which is the general conic in P G(2, q) with [1, 0, 0] as tangent at (0, 1, 0).
Hence the conics of P G(2, q) with [1, 0, 0] as tangent at (0, 1, 0) which are images
of C under elations with center (0, 1, 0) are precisely those C ′ in Eq. 4.23
with a = 1.

174 CHAPTER 4. THE UBIQUITOUS OVAL
Then
Let φ be the homography of P G(2, q) with matrix
⎛
1 c
⎞
0
[φ] = ⎝ 0 a 0 ⎠ , and [φ]
0 b 1
−1 ⎛
1 −c/a 0
= ⎝ 0 1/a 0
0 −b/a 1
[φ] −1 AC[φ] −T =
⎛
= ⎝
⎛
⎝
1 −c/a 0
0 1/a 0
0 −b/a 1
c/a −1/a b/a
0 0 0
0 0 1
⎞ ⎛
⎠ ⎝
⎞
0 −1 0
0 0 0
0 0 1
⎛
⎠ ≡ ⎝
⎞ ⎛
⎠ ⎝
c −1 b
0 0 0
0 0 a
⎞
⎠ .
1 0 0
−c/a 1/a −b/a
0 0 1
⎞
⎠ = AC ′.
Now we want to embed P G(2, q) into π = [0, 0, 0, 1] ⊂ P G(3, q) by
(x, y, z) ↦→ (x, y, z, 0), and then lift φ to a homography Φ of P G(3, q) that
acts like φ on π and fixes the point (1, u, v, 1). A simple computation shows
that for each 0 = w ∈ Fq the following homography Φ will do this:
[Φ] =
⎛
⎜
⎝
1 c 0 0
0 a 0 0
0 b 1 0
w − 1 (w − a)u − bv − c v(w − 1) w
Recap: ∞ = (0, 1, 0, 0); π∞ = [1, 0, 0, 0]; π = [0, 0, 0, 1].
C = {(1, s 2 , s, 0) : s ∈ Fq} ∪ {∞} is our favorite conic in π.
C ′ = {(1, as 2 + bs + c, s, 0) : s ∈ Fq} ∪ {∞} gives all conics in π with
[1, 0, 0, 0]∩[0, 0, 0, 1] as the line tangent at the point ∞. Here a, b, c ∈ Fq with
a = 0. These are also all the conics that are images of C under homographies
of π with center ∞. These are homologies if a = 1 and are elations if a = 1.
4.4 Segre’s Theorem for q Odd
In this section we prove the celebrated theorem of B. Segre [Se55] which says
that each oval in P G(2, q) with q odd must be a conic.
Let Ω be an oval of P G(2, q), q odd. If a triangle has its vertices on Ω,
then we say it is an inscribed triangle of Ω; if its sides are tangents to Ω
⎞
⎟
⎠ .
⎞
⎠

4.4. SEGRE’S THEOREM FOR Q ODD 175
we say it is circumscribed. Each three points (respectively, tangents) of Ω
determine a unique pair of triangles, one inscribed and one circumscribed.
We call these two triangles mates.
Theorem 4.4.1. (Lemma of tangents) Every inscribed triangle of Ω and its
mate are in perspective.
Proof. The collineation group of P G(2, q) is transitive on 3-arcs. Hence
without loss of generality we may assume that the inscribed triangle has as
its points the following basic points:
A1 = (1, 0, 0); A2 = (0, 1, 0); A3 = (0, 0, 1).
If a1, a2, a3 are the tangents to Ω at A1, A2, A3, respectively, then the
equations of a1, a2, a3 are of the form:
a1 : x2 = k1x3; a2 : x3 = k2x1; a3 : x1 = k3x2,
where k1, k2, k3 are nonzero elements of Fq. If B = (b1, b2, b3) is any one of
the q − 2 points of Ω distinct from A1, A2, A3, then b1b2b3 = 0 and the lines
A1B, A2B, A3B have respective equations x2 = h1x3, x3 = h2x1, x1 = h3x2,
where h1 = b2b −1
3 ,h2 = b3b −1
1 , h3 = b1b −1
2 . Clearly the hi satisfy
h1h2h3 = 1. (4.24)
Conversely, if h1 is any of the q − 2 elements of Fq distinct from zero
and from k1, the line x2 = h1x3 meets Ω at A1 and at a second point, B
say, distinct from A1, A2, A3. Hence the coefficients h2, h3 in the equations
x3 = h2x1, x1 = h3x2 of the lines A2B, A3B are functions of h1, connected
by Eq. 4.24, which take once each of the nonzero values of Fq distinct from
k2, k3, respectively. On multiplying together the q − 2 equations for h1h2h3
thus obtained, we have
T 3 = k1k2k3,
where T denotes the product of the q−1 nonzero elements of Fq. But T = −1
(easy exercise), so k1k2k3 = −1.
The points a2a3 = (k3, 1, k2k3), a3a1 = (k3k1, k1, 1), a1a2 = (1, k1k2, k2)
are joined to A1, A2, A3, respectively, by the lines:
x3 = k2k3x2, x1 = k3k1x3, x2 = k1k2x1.

176 CHAPTER 4. THE UBIQUITOUS OVAL
But these lines are concurrent at K = (1, k1k2, −k2), and this proves the
lemma.
(1, k1k2, k2)
a2 : x3 = k2x1
A2(0, 1, 0) (k3, 1, k2k3)
❧▲ ☞
▲
❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧
x2 = k1k2x1
x3 = k2k3x2 ☞
▲
☞
▲
☞
▲
☞
▲
☞
▲
☞
▲
☞
▲
☞
a1 : x2 = k1x3 ▲
☞ a3 : x1 = k3x2
▲
☞
▲
K = (1, k1k2, ☞−k2)
▲
☞
▲
☞
▲
☞
▲
☞
▲ ☞ ☞
A1(1, 0, 0) ▲☞
☞ A3(0, 0, 1)
▲
☞
▲
☞
▲
☞
▲ x1 = k3k1x3 ☞
▲
☞
▲
☞
▲
☞
▲
☞
▲
☞
▲
☞
▲
☞
▲ ☞
▲ ☞
▲ ☞
▲ ☞
▲ ☞
▲ ☞
▲☞
(k3k1, k1, 1)
☞☞☞☞☞☞
☞ ☞☞☞☞☞☞☞☞☞☞▲ . ✱
▲ ✱
✱
▲▲▲▲▲▲▲▲▲▲▲▲▲▲▲▲
✱
✱
✱
✱
✱
✱
✱
✱
✱
✱
✱
✱
✱
✱
✱
✱
✱
✱
✱
Figure 4.1: Figure for the Lemma of Tangents

4.4. SEGRE’S THEOREM FOR Q ODD 177
In order to prove the theorem of Segre, we need to observe that the
group of linear transformations (which induce homograqphies of the plane
π = P G(2, q)) which fix each of the points A1, A2, A3 is transitive on the set
of points of K for which the four fundamental points mentioned earlier form a
quadrangle. But this group is the set of invertible diagonal matrices, and it is
transitive on the set of points (a, b, c) with abc = 0. Hence in the proof of the
previous theorem we may assume that K = (1, 1, 1), i.e., k1 = k2 = k3 = −1
Theorem 4.4.2. (B. Segre [Se55]) Every oval in P G(2, q), q odd, is a conic.
Proof. We continue to use the notation of the preceding proof. If B =
(c1, c2, c3) is any of the q − 2 points of Ω distinct from A1, A2, A3, we denote
by b : x1b1 + x2b2 + x3b3 = 0 the tangent to Ω at B. Since B is on b we have
3
i=1 cibi = 0. None of A1, A2, A3 is on b, so each bi is nonzero. And since
no point is on as many as three tangents, each of the points aiaj is not on b.
If subscripts are taken modulo three, ai−1ai+1 has −1 in positions i − 1 and
i + 1 and +1 in position i. So if we put d1 = b1 − b2 − b3, d2 = −b1 + b2 − b3,
d3 = −b1 − b2 + b3, then each di is nonzero.
By the lemma of tangents, the triangles BA2A3 and ba2a3 are in perspective.
By straightforward calculations this implies b2(c1 + c2) = b3(c1 + c3). A
similar consideration for the triangles BA3A1, BA1A2 and their mates gives
b3(c2 + c3) = b1(c2 + c1) and b1(c3 + c1) = b2(c + 3 + c2). But these last three
equations imply b1 : b2 : b3 = (c2 + c3) : (c3 + c1) : (c1 + c2). Substitution for
b1, b2, b3 in 3
i=1 cibi = 0 gives
which, since q is odd, gives
c1(c2 + c3) + c2(c3 + c1) + c3(c1 + c2) = 0,
c1c2 + c2c3 + c3c1 = 0.
In fact this last equation shows that each of the q − 2 points of Ω distinct
from A1, A2, AQ3 lies on the conic with equation
x2x3 + x3x1 + x1x2 = 0.
This conic clearly also contains A1, A2, A3. Thus since the conic also
has exactly q + 1 points, the points of Ω are the points of the conic with the
equation displayed above.

178 CHAPTER 4. THE UBIQUITOUS OVAL
(b2, b3 − b1, −b2) a2 : x3 = −x1
A2(0, 1, 0) (−1, 1, 1)
▲❧
☞
▲
❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧
☞
⎡ ⎤
⎡ ⎤
▲
☞
b1 − b3
c2 − c3
▲
☞
▲ ⎣ b2 ⎦
⎣ −c1 − c3 ⎦ ☞
▲ 0
c1 +
☞
c2
▲
☞
▲
☞
▲
☞
xibi = 0 ▲
☞ a3 : x1 = −x2
▲
☞
=⇒
▲
P
☞
cibi = 0 ▲
☞
▲
☞
▲
☞
▲
☞
▲ ☞
B(c1, c2, c3) ▲☞
A3(0, 0, 1)
▲ ⎡ ⎤
☞☞☞☞☞☞☞☞☞☞☞☞☞☞☞☞☞▲ . ✱
▲ ✱
▲▲▲▲▲▲▲▲▲▲
✱
✱
✱
✱
✱
✱
✱
✱
✱
✱
✱
✱
✱
✱
▲
✱
▲▲▲▲▲
✱
✱
✱
✱
✱
☞
☞
☞
☞
☞
☞
▲
b1 − b2
▲
▲ ⎣ 0 ⎦
▲
b3
☞
▲
☞
▲
☞
▲
☞
▲
☞
▲
☞
▲
☞
▲ ☞
▲ ☞
▲ ☞
▲ ☞
▲ ☞
▲ ☞
▲☞
(b3, −b3, b2 − b1)
P = (b2b3, b3(b3 − b1, b2(b2 − b1)
Figure 4.2: Figure for Segre’s Theorem

4.5. O- POLYNOMIALS & THE CRITERION OF GLYNN 179
4.5 o- Polynomials & the Criterion of Glynn
In this section q = 2 e , so each oval Ω of P G(2, q) is contained in a unique
hyperoval, i.e., (0, 2)-set. An oval is called regular if it is a conic. A hyperoval
is regular provided it is a conic plus its nucleus. And a regular hyperoval
minus a point different from its natural nucleus is called a pointed conic.
Theorem 4.5.1. If q = 2, then every hyperoval is regular.
Proof. In P G(2, 2) a hyperoval consists of four points, any three of which
form a conic.
By the transitivity of the collineation group P ΓL(3, q) of P G(2, q) on
quadrangles, every hyperoval can be mapped by an element of P ΓL(3, q)
(even by an element of P GL(3, q)) to one containing the fundamental quadrangle
(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1). A hyperoval which is the image
under an element of P ΓL(3, q) of a hyperoval H is equivalent to H.
From now on we restrict our attention to P G(2, q) with q > 2 and for
the most part consider only hyperovals which contain at least three points of
the fundamental quadrangle. (Sometimes we do not require that (1, 1, 1) be
on the hyperoval.) We now show that these hyperovals are related to certain
permutation polynomials.
Theorem 4.5.2. Let H be a hyperoval containing the points (0, 1, 0) and
(0, 0, 1). Then H = D(f), where
D(f) = {(1, t, f(t)) : t ∈ Fq} ∪ {(0, 1, 0), (0, 0, 1)},
where f is a permutation polynomial of degree at most q − 2. Further, for
each s ∈ Fq, the polynomial fs(x) is a permutation polynomial, where
fs(x) =
f(x + s) + f(s)
, fs(0) = 0.
x
Moreover, without loss of generality we may assume that H contains the
point (1, 0, 0) so that f(0) = 0. Also, we may assume that H contains the
point (1, 1, 1), in which case f(1) = 1.
Conversely, if f is a permutation polynomial satisfying the above condition
on fs(x), then D(f) is a hyperoval.

180 CHAPTER 4. THE UBIQUITOUS OVAL
Proof. Start with the assumption that H is an oval containing the two points
(0, 1, 0) and (0, 0, 1). The line [1, 0, 0] contains the points (0, y, z) ∈ P G(2, q),
of which (0, 1, 0) and (0, 0, 1) are the only points of H. So each other point
of H is of the form (1, y, z).
For each t ∈ Fq, the line [t, 1, 0] through (0, 0, 1) contains a unique point
of the form (1, t, z), so we write z = f(t) for some function f. For 0 = b ∈
Fq, the line [1, 0, b −1 ] through (0, 1, 0) must contain a unique point of the
form (1, y, b) on H, so y = f −1 (b) and f must be a permutation. So by
Theorem 20.3.2, f must be given by a permutation polynomial of degree at
most q − 2.
Clearly (1, 0, 0) is on H∩[0, 1, 0] if and only if f(0) = 0. Similarly, (1, 1, 1)
is on H ∩ [1, 1, 0] if and only if f(1) = 1.
At this point we know that f(x) is a permutation polynomial of degree
at most q − 2, and H = D(f). Since no three points of H are collinear, if s,
t, u are distinct elements of Fq, it must be that
if and only if
0 =
1 s f(s)
1 t f(t)
1 u f(u)
=
1 s f(s)
0 t − s f(t) − f(s)
0 u − s f(u) − f(s)
= (t − s)f(u) − f(s)) − (u − s)(f(t) − f(s))
f(t) − f(s)
t − s
f(s + a) + f(s)
a
=
=
f(u) − f(s)
.
u − s
f(s + b) + f(s)
.
b
=
Hence a ↦→ fs(a) = f(s+a)+f(s)
permutes the nonzero elements. As fs(x)
a
has degree at most q − 3 < q − 2, by Lemma 20.3.3 fs(x) is a permutation
polynomial with fs(0) = 0. The converse is now easy.
Def. Permutation polynomials f(x) satisfying all the conditions of Theorem
4.5.2 are called o-polynomials. If f(x) satisfies all the conditions except
for f(1) = 1, f is called an o-permutation. If f(x) = x k is an o-polynomial,
it is a monomial o-polynomial.

4.5. O- POLYNOMIALS & THE CRITERION OF GLYNN 181
The homography
⎛
(x, y, z) ↦→ (x, y, z) ⎝
1 0 0
0 1 a
0 0 1
⎞
⎠ = (x, y, ay + z)
is an involution that interchanges N = (0, 1, 0) and (0, 1, a), maps Q =
(0, 0, 1) to itself, and maps the general point (1, x, f(x)) of Of to the point
(1, x, ax + f(x)). It is now an easy exercise to prove the following Lemma.
Lemma 4.5.3. Those ovals that contain the point Q = (0, 0, 1) and have a
nucleus on the line L are precisely those of the form {(1, x, ax + f(x)) : x ∈
Fq}∪{Q} for some o-permutation f(x), and the nucleus is the point (0, 1, a).
If we start with an oval {(1, x, ax + f(x)) : x ∈ Fq} ∪ {Q} having nucleus
(0, 1, a), consider the homography
⎛
1 0
⎞
f(0)/A
(x, y, z) ↦→ (x, y, z) ⎝ 0 1 a/A ⎠ ,
0 0 1/A
where A = f(0) + f(1). Then the image oval is {(1, x, f(0)+f(x)
) : x ∈
A
Fq} ∪ {Q}, and the o-permutation x ↦→ f(0)+f(x)
A
gives an oval that together
with its nucleus is a hyperoval that contains the fundamental quadrangle
P1(1, 0, 0), P2(0, 2, 0), P3(0, 0, 1), P4(1, 1, 1).
Theorem 4.5.4. Let D(f) be a hyperoval containing (1, 0, 0), (0, 1, 0), (0, 0, 1)
and with associated polynomial (i.e., o-permutation) f. Since f is a bijection,
f(x) = q−1
i=0 aix i with aq−1 = 0 by Lemma 20.3.2. Since 0 = f(0), also
a0 = 0. Then since q > 2, ai = 0 if i is odd, i.e.,
f(x) = a2x 2 + a4x 4 + · · · + aq−2x q−2 .
Proof. Since f is a permutation polynomial of degree at most q − 2 with
f(0) = 0, b ↦→ f(a)−f(b)
is a bijection from Fq \ {a} to F a−b
∗ q . Put b = T + a.
f(a)−f(T +a)
Then T ↦→ g(T ) = is a bijection of F −T
∗ q . Using a Taylor expansion
for f(T +a) we have f(T +a) = f(a)+f ′ (a)T +o(T 2 ). So g(T ) = f ′ (a)+o(T ).
But since g is a bijection of F ∗ q and is a polynomial function of degree at most
q − 2 (actually ≤ q − 3) it must be that g(0) = 0 by Lemma 20.3.3 Hence
f ′ (a) = 0. But this holds for all a ∈ Fq. Hence f ′ has no nonzero terms.
This means all nonzero terms of f must have even degree, as claimed.

182 CHAPTER 4. THE UBIQUITOUS OVAL
Corollary 4.5.5. Each oval in P G(2, 4) is a conic.
Proof. Let Ω be an oval of P G(2, 4). Without loss of generality we may
assume that Ω is given by an o-polynomial f(t) as in Theorem 4.5.2 with
f(t) clearly not the identity permutation. Since F4 = {0, 1, w, w 2 } where
w = 1 = w 3 , and we may assume that f(0) = 0 and f(1) = 1, it must be
that f(w) = w 2 and f(w 2 ) = w, i.e., f(t) = t 2 .
Theorem 4.5.6. Let f(t) be a polynomial over Fq with q = 2 e > 4 and with
f(0) = 0, f(1) = 1.
(i) If deg f = 2, then D(f) is a hyperoval iff D(f) = D(2).
(ii) If deg f = 4, then D(f) is a hyperoval iff e is odd and D(f) = D(4).
(iii) If deg f = 6, then D(f) is a hyperoval iff e is odd and
f(t) = (t 6 + at 4 + a 2 t 2 )/(1 + a + a 2 ) for some a ∈ Fq.
In this case D(f) ∼ D(6).
Proof. (i) Using the preceding results with f(t) = 1 we have immediately
that f(t) = t 2 .
(ii) Here f(t) = at 2 + bt 4 with b = 0. Since f(t) = 0 implies t = 0, clearly
a = 0. Then f(1) = 1 implies f(t) = t 4 . As 4 = 2 2 with (2, e) = 1 iff D(f) is
a hyperoval, we see this holds iff e is odd.
(iii) Here D(f) is a hyperoval iff we may take f(t) = at 2 + bt 4 + ct 6 with
c = 0. The homography (x0, x1, x2) ↦→ (x0, x1, c −1 x0) maps D(f) to D(f/c),
so without loss of generality we may assume c = 1. Now just consider when
this polynomial is a permutation polynomial. Since t ↦→ √ t is a permutation,
we see t ↦→ f(t) is a permutation iff t ↦→ at + bt 2 + t 3 is a permutation.
Similarly, since t ↦→ t + b is a permutation, t ↦→ at + bt 2 + t 3 is a permutation
iff t ↦→ a(t + b) + b(t 2 + b 2 ) + t 3 + t 2 b + tb 2 + b 3 = ab + t(a + b 2 ) + t 3 is a
permutation iff t ↦→ (a + b 2 )t + t 3 is a permutation, If this latter really is a
permutation, then a = b 2 , so that it becomes t ↦→ t 3 , which is a permutation
if and only if e is odd. So if we divide f(t) by f(1) to get a polynomial whose
value at 1 is 1, we see that we may assume that e is odd and
If the projectivity
f(t) = (b 2 t 2 + bt 4 + t 6 )/(b 2 + b + 1).
(x0, x1, x2) ↦→ (x0, x1 + √ bx0, (1 + b + b 2 )x2 + b 3 x0)

4.5. O- POLYNOMIALS & THE CRITERION OF GLYNN 183
is applied, then
D(f) = {(1, t, t6 + bt4 + b2t2 1 + b + b2 ) : t ∈ Fq} ∪ {(0, 0, 1), (0, 1, 0)}
∼ {(1, t + √ b, (t + √ b) 6 ) : t ∈ Fq} ∪ {(0, 0, 1), (0, 1, 0)}
= {(1, t, t 6 ) : t ∈ Fq} ∪ {(0, 0, 1), (0, 1, 0)} (4.25)
= D(6).
We have shown above that for D(6) to be a hyperoval e must be odd.
That it is a hyperoval when e is odd we have separated into its own theorem
that follows the next result.
Let k be any positive integer. The map a ↦→ a k for a ∈ Fq is a bijection
if and only if (k, q − 1) = 1. Put
D(k) = {(1, t, t k ) : t ∈ Fq} ∪ {(0, 0, 1), (0, 1, 0)}.
Theorem 4.5.7. D(k) is a hyperoval if and only if
b ↦→ fk(b) = (1 + b)k + 1
b
is a permutation, and
(k, q − 1) = (k − 1, q − 1) = 1. By Theorem 4.5.4 k must be even.
Proof. We use the proof of Theorem 4.5.3 to check that D(k) is an oval if
and only if (k, q − 1) = 1 and b ↦→ ak−bk is a permutation for each a ∈ Fq. If
a−b
a = 0, b ↦→ ak−bk a−b = bk−1 will be a permutation if and only if (k −1, q −1) = 1.
For a = 0, and for b = a = c, ak−bk a−b = ak−ck b
1−(
iff a)
a−c k
1− b =
a
1−( c
a) k
1− c .
a
So w ↦→ ak −w k
a−w
1, q − 1) = 1 and w ↦→ 1−wk
1−w
w = v+1 ↦→ 1+(v+1)k
v
for w = a, is an injection for all a if and only if (k −
is a permutation. This latter is equivalent to
being a permutation. So we have the desired result.
Theorem 4.5.8. (B. Segre [Se62] and B. Segre and U. Bartocci [SB71]) If
q = 2 e with e odd, then D(6) is a hyperoval. (If q ≥ 32 it is irregular and
not equivalent to one of the translation hyperovals to be introduced later.).

184 CHAPTER 4. THE UBIQUITOUS OVAL
Proof. By Theorem 4.5.7, to show that D(6) is a hyperoval it suffices to show
that the map x ↦→ 1+(1+x)6
= x + x x
3 + x5 permutes the elements of F ∗ q . So
suppose that for x = y we have
0 = (x + x 3 + x 5 ) + (y + y 3 + y 5 ) = (4.26)
= (x + y)((x 2 + y 2 + 1) 2 + (x 2 + y 2 + 1)(xy + 1) + (xy + 1) 2 ).
Since e is odd, z 2 + zw + w 2 = 0 has no non-trivial solution over Fq, from
which it follows rather easily that Eq. 4.26 has no solution.
Lemma 4.5.9. Let f(x) be a polynomial of degree at most q − 2 over Fq.
Put
D(f) = {(1, t, f(t) : t ∈ Fq} ∪ {(0, 1, 0), (0, 0, 1)}.
Then D(f) is a hyperoval if and only if the q 2 −q lines of P G(2, q) passing
neither through (0, 1, 0) nor through (0, 0, 1) always intersect D(f) in an even
number of points.
Proof. If D(f) is a hyperoval, then all lines of P G(2, q) intersect D(f) in 0
or 2 points. So assume the converse. Clearly (1, t, f(t), (0, 1, 0) and (0, 0, 1)
can never be collinear. Let P be one of the q points of D(f) different from
(0, 1, 0) and (0, 0, 1). The q − 1 lines which pass through P but not through
(0, 1, 0) or (0, 0, 1) all contain at least one further point of D(f). Since there
are only q − 1 further points of D(f), these q − 1 lines each contain exactly
two points of H(f). And the two lines through P containing (0, 1, 0) and
(0, 0, 1), respectively, each contain exactly two points of D(f). As P ranges
over all q − 1 points of D(f) different from (0, 1, 0) and (0, 0, 1), we see that
D(f) must be a hyperoval.
Lemma 4.5.10. Let g(t) be a polynomial over Fq. Then g(t) = u has an
even number of solutions t ∈ Fq for all u ∈ Fq if and only if the following
holds:
g(a) r = 0 for all r = 1, 2, . . . , q − 1.
a∈Fq
Proof. The non-trivial part is to show that the above algebraic condition
implies that g(t) = u always has an even number of solutions. Let A =
{u : g(t) = u has an odd number of solutions}. Then the sum given in the

4.5. O- POLYNOMIALS & THE CRITERION OF GLYNN 185
above condition can be reduced to summing over a ∈ A. But Vandermonde’s
determinant formula (20.24) implies that the vectors (1, u, u 2 , . . . , u q−1 ) with
u varying over all elements of Fq are linearly independent. However, the
above condition implies that the sum of the vectors with u ∈ A is zero. Thus
A = ∅.
Lemma 4.5.11. Each of the q 2 − q lines of P G(2, q) passing neither through
(0, 1, 0) nor through (0, 0, 1) always intersects D(f) in an even number of
points if and only if the following condition holds:
(f(a) + ax) r = 0 for 1 ≤ r ≤ q − 1 and for all nonzero x ∈ Fq. (4.27)
a∈Fq
Proof. A line not passing through (0, 1, 0) or (0, 0, 1) has coordinates [u, x, 1],
x = 0. The condition that this line contain a point (1, t, f(t)) of D(f) is
f(t) + tx = u. Hence, for all x ∈ Fq \ {0},
(f(a) + ax) r = 0 for 1 ≤ r ≤ q − 1
a∈Fq
if and only if for all u ∈ Fq each line [u, x, 1] intersects D(f) in an even
number of points.
Remark It follows that if f is a polynomial with degree at most q − 2,
then D(f) is a hyperoval if and only if
(f(a) + ax) r = 0 for 1 ≤ r ≤ q − 1, 0 = x ∈ Fq.
a∈Fq
When x is zero and f(t) has degree at most q −2, by the Hermite-Dickson
criterion f(t) is a permutation polynomial if and only if f(a) = 0 has a unique
solution a ∈ Fq and the coefficient on t q−1 in [f(t)] r (modulo t q −t) is zero for
all 1 ≤ r ≤ q − 2 and r odd. And this is the case if D(f) is an oval, because
the lines [a, 0, 1] (different from [1, 0, 0] through (0, 1, 0) and (0, 0, 1)) passing
through (0, 1, 0) each intersect D(f) in a unique additional point (1, t, f(t))
with a = f(t). By Theorem 20.3.1 the coefficient on t q−1 in [f(t)] r (modulo
t q − t) equals −
a∈Fq [f(a)]r . Hence if D(f) is an oval, it is also true that

186 CHAPTER 4. THE UBIQUITOUS OVAL
the equation of Lemma 4.5.11 holds when x = 0 as long as 1 ≤ r < q − 1
and r is odd. But if r = k2i with k odd and 1 ≤ k2i ≤ q − 2, then
a∈Fq [f(a)]r = a∈Fq [f(a)]k
2i = 0. Hence the equation of Lemma 4.5.11
holds when x = 0 as long as 1 ≤ r < q − 1.
Def. The partial ordering ≪ (with companion dual ≫) on the set Nq =
{n : 0 ≤ n ≤ q − 1} is defined as follows: For a, b ∈ Nq, write a = e−1 i
i=0 ai2
and b = e−1 i=0 bi2i , with ai, bi ∈ {0, 1} for all i. Then a ≪ b iff ai ≤ bi for all
i.
Lemma 4.5.12. If k is a positive integer, then the expansion of (1+t) k (mod tq− t) over Fq may be calculated as follows.
For k ∈ Z, (1 + t) k ≡
t c (mod t q − t). (4.28)
c≪k
c∈Nq
Proof. First, if k = 0, then (1 + t) k = 1. If k = 0, then first reduce k modulo
q − 1 so that 1 ≤ k ≤ q − 1. Write k = e−1
i=o ki2 i , with ki ∈ {0, 1}. Then
(1 + t) k = (1 + t) P ki2i e−1
= (1 + t 2i
) ki
= t c .
i=0
c≪k
c∈Nq
For example, (1 + t) 2i = 1 + t2i for 1 ≤ i ∈ Z. Since q − 1 = e−1 i=0 2i ,
(1+t) q−1 =
c∈Nq tc . And for 0 < i ∈ Z, (1+t) (q−1)i = (1+t) (q−1)(i−1)+q−1 =
(1 + t) q(i−1)+q−i ≡ (1 + tq ) i−1 (1 + t) q−i ≡ (1 + t) q−1 =
c∈Nq tc .
Of course this is immediately generalized to
Lemma 4.5.13. If k is a positive integer, then the expansion of (a+bt) k (mod tq− t) over Fq may be calculated as follows:
For k ∈ Z, (a + bt) k ≡
a k−c b c t c (mod t q − t). (4.29)
.
Hence
c≪k
c∈Nq
(f(λ) + λx) r =
λ∈Fq
0≤s≪r
⎛
⎝
f(λ) r−s λ s
⎞
⎠ x s . (4.30)
λ∈Fq

4.5. O- POLYNOMIALS & THE CRITERION OF GLYNN 187
As a corollary we have
Corollary 4.5.14. For k ≤ q − 1, we have
fk(t) = (1 + t)k + 1
t
=
0

188 CHAPTER 4. THE UBIQUITOUS OVAL
Note: Glynn also shows that d ≪ kd for all d with 1 ≤ d ≤ q − 2 implies
that (k, q − 1) = (k − 1, q − 1) = 1. Hence the last theorem does not need
this as part of its hypothesis.
In [Gl89] the preceding theorem was generalized to the following:
Theorem 4.5.16. Let f ∈ Fq[t] be a polynomial with degree ≤ q − 2 and
f(a) = 0 if and only if a = 0. D(f) is a hyperoval if and only if the following
condition holds: The coefficient of t a in [f(t)] b (mod t q − t) is zero, for all
pairs of integers (a, b) with 1 ≤ b ≪ a ≤ q − 1, b = q − 1.
Before beginning the proof of this theorem we need a lemma.
Lemma 4.5.17. For 0 ≤ i ≤ r ≤ q − 1 we have i ≪ r if and only if
r − i ≪ q − 1 − i.
Proof. Note that the binary representation of i is the bitwise complement of
the binary representation of q −1−i, i.e., there is a 1 in the binary expansion
of i in position k if and only if there is a 0 there in the expansion of q − 1 − i.
First assume that i ≪ r. Then the complement of i in r is contained in the
complement of i in q − 1, so r − i ≪ q − 1 − i. For the converse, suppose that
0 ≤ i ≤ r ≤ q − 1 and suppose that i ≪ r, so in fact i < r. First suppose
that r − i and i have no common 1’s. So where i has a 1, r − i has a 0 and
hence q − 1 − (r − i) has a 1, so i ≪ q − 1 − (r − i) = q − 1 − r + i. Hence
(q − 1 − r + i) − i = q − 1 − r has no common 1’s with i. Where i has a 1,
q − 1 − r has a 0, so r has a 1. This says i ≪ r, a contradiction. Hence i ≪ r
implies that r − i and i have a common 1. This says that q − 1 − i has a 0
in some spot where r − i has a 1, implying r − i ≪ q − 1 − i, completing a
proof of the lemma.
We now start the proof of the theorem.
Proof. First note that when b = 0, if a > 0, then automatically the coefficient
of t a in [f(t)] b is zero. So the condition of the theorem may be restated as
1. The coefficient of t a in [f(t)] b (mod t q − t) is zero for all 0 ≤ b ≪ a ≤
q − 1 with (b, a) not equal to (q − 1, q − 1) or to (0, 0).
Put b = r − i and a = q − 1 − i. Then rewrite 1. as:
2. The coefficient of t q−1−i ≡ t −i in [f(t)] r−i is 0 for all 0 ≤ r − i ≪
q − 1 − i ≤ q − 1 with (r − i, q − 1 − i) not equal to (q − 1, q − 1) or (0, 0).
Using the lemma we can restate this as

4.5. O- POLYNOMIALS & THE CRITERION OF GLYNN 189
3. The coefficient of t −i in [f(t)] r−i is 0 for all 0 ≤ i ≪ r ≤ q − 1 with
(i, r) not equal to (0, q − 1) or (q − 1, q − 1).
4. Let h(t) = f(t) r−i . If 0 < q − 1 − i < q − 1 (i.e., 0 < i < q − 1) and
i ≪ r, then the coefficient aq−1−i on tq−1−i = t−i in h(t) = f(t) r−i reduced
modulo tq − q is
aq−1−i =
f(λ) r−i λ i .
λ∈Fq
Also, with i = 0 and 1 ≤ r ≤ q − 2, so i ≪ r, the coefficient of t−i in f(t) r−i
is equal to
aq−1 =
f(λ) r .
λ∈Fq
We are now ready to put these conditions together. First suppose that
f(t) is a polynomial of degree at most q−2, f(0) = 0 and D)f) is a hyperoval.
So f is a permutation polynomial and by the Remark after Lemma 4.5.11
we know
(f(a) + ax) r = 0 for 1 ≤ r ≤ q − 1 and x ∈ Fq, (4.33)
a∈Fq
except for (x, r) = (0, q − 1).
The idea is to use the condition appearing in Lemma 4.5.11 with r < q−1,
thinking of (f(λ) + λx) r as a polynomial in x. This condition becomes
f(λ) r−s λ s x s
= 0 for all r = 1, 2, . . . , q − 2.
λ∈Fq
0≤s≪r
Since this polynomial in x has degree at most q − 2 and this equality holds
for all x ∈ Fq, the coefficient of x i is zero for all 0 ≤ i ≤ q − 2. Thus
f(λ) r−i λ i = 0 for all 0 ≤ i ≪ r ≤ q − 2.
By the first part of 4., this says the coefficient of t−i in f(t) r−i is 0 for
0 ≤ i ≪ r ≤ q − 2. Now suppose r = q − 1 and 1 ≤ i ≤ q − 2, so of course
i ≪ r. By Eq. 4.33
(f(λ) + λx) q−1 = 0 for 0 = x ∈ Fq.
λ∈Fq

190 CHAPTER 4. THE UBIQUITOUS OVAL
Hence
λ∈Fq (f(λ) + λx)q−1 is the polynomial
g(x) = g(0)(x q−1 − 1) =
λ∈Fq
0≤s≪q−1
f(λ) q−1−s λ s x s
The coefficient on xq−1 is
λ∈Fq f(λ)0λq−1 = 1, which implies that g(x) =
xq−1 − 1. Hence the coefficient on xi
is 0 for 1 ≤ i ≤ q − 2, i.e.,
λ∈Fq f(λ)q−1−iλ i = 0. This means that we now know that the condition
in 3. is satisfied, so that the condition in 1. is satisfied. This shows that if
D(f) is a hyperoval, Glynn’s condition is satisfied.
Now we consider the converse. Suppose that Glynn’s condition is satisfied
for a polynomial f(t) with degree ≤ q − 2 and f(0) = 0. What we need to
show is that
If 1 ≤ r ≤ q − 1 and 0 = x ∈ Fq, then (4.34)
⎛
⎝
f(λ) r−s λ s
⎞
⎠ x s = 0.
0≤s≪r
λ∈Fq
What we do know by 4. is that Glynn’s condition implies that
aq−1−i =
f(λ) r−i λ i = 0 for 0 < i < q − 1 and i ≪ r ≤ q − 1. (4.35)
λ∈Fq
Also, with i = 0 and 1 ≤ r ≤ q − 2, so 0 = i ≪ r ≤ q − 2, the coefficient of
tq−1 is
aq−1 =
f(λ) r = 0. (4.36)
λ∈Fq
At this stage we know the following:
f(λ) r−i λ i = 0 for 0 ≤ i ≤ q − 1 and i ≪ r ≤ q − 1,
λ∈Fq
except for (i, r) = (0, 0) or (q − 1, q − 1), where when i = 0, t −i is to be
interpreted as t q−1 . Hence the needed condition in Eq. 4.34 simplifies to
f(λ) q−1 +
λ q−1 x q−1 = 0 if x = 0. (4.37)
λ∈Fq
λ∈Fq
.

4.6. ARCS IN OVALS 191
This is just the condition that
λ∈Fq
f(λ) q−1 = 1,
which would be true if f(t) were a permutation, which in turn is true by the
Hermite-Dickson Criterion since f(a) = 0 only when a = 0.
Historical Note: Frequently in the statement of Glynn’s Theorem (our
Theorem 4.5.16) that part of the hypothesis that says “f(a) = 0 if and only
if a = 0” is replaced with something weaker, such as merely “f(0) = 0.” But
the constant function f(a) = 0 for all a ∈ Fq then is a trivial counterexample.
So some additional hypothesis is needed. On the other hand, when Glynn’s
Criterion is used to check whether or not some function f is an o-permutation,
the function f is usually already known to be a permutation mapping 0 to 0,
so the theorem is not likely to have been applied incorrectly.
4.6 Arcs in Ovals
When q = 2 the results we want to develop are easily given and often this
case provides exceptions to some general rule. So throughout this section we
assume that q = 2 e > 2. Also, we use the following notation:
P0 = (1, 0, 0); P1 = (0, 1, 0); P2 = (0, 0, 1).
The following lemma has an easy proof that we leave to the reader.
Lemma 4.6.1. Let li be a line through Pi but not through Pj for j ∈ {0, 1, 2}\
{i}. Then
l0 = [0, 1, e] T
for some nonzero e ∈ Fq.
l1 = [f, 0, 1] T for some nonzero f ∈ Fq. (4.38)
l2 = [1, g, 0] T for some nonzero g ∈ Fq.
Moreover, the three lines l0, l1, l2 are concurrent if and only if efg = 1.
Lemma 4.6.2. Let P = (d0, d1, d2) ∈ K \ {P0, P1, P2}. Since K is an arc,
d0d1d2 = 0. The secants to K through P and Pi, 0 ≤ i ≤ 2, respectively, are:

192 CHAPTER 4. THE UBIQUITOUS OVAL
Here clearly efg = 1.
(i) [0, 1, d1/d2] T = [0, 1, e] T ; e = 0.
(ii) [d2/d0, 01] T = [f, 0, 1] T ; f = 0. (4.39)
(iii)[1, d0/d1, 0] T = [1, g, 0] T ; g = 0.
Lemma 4.6.3. Let m0, m1, m2 be any three tangents to K at P0, P1, P2,
respectively. Here
(i) m0 = [0, 1, b0] T
(ii) m1 = [b1, 0, 1] T
(iii) m2 = [1, b2, 0] T
for some b0 = 0;
for some b1 = 0; (4.40)
for some b2 = 0.
Then m0, m1, m2 are concurrent at a point Q (not on K) iff b0b1b2 = 1.
Now suppose that K = {P0, P1, P2, P3, . . . , Pk−1}. For each i, 0 ≤ i ≤ 2,
there are k − 3 secants through Pi but not through Pj with j ∈ {0, 1, 2} \ {i},
and there are t = q − k + 2 tangents to K at Pi.
Lemma 4.6.4. Choose labels so that
(i) The secant through P0 and Pj is [0, 1, ej] T , 3 ≤ j ≤ k − 1.
(ii) The secant through P1 and Pj is [fj, 0, 1] T , 3 ≤ j ≤ k − 1.
(iii) The secant through P2 and Pj is [1, gj, 0] T , 3 ≤ j ≤ k − 1.
Then by Lemma 4.6.2 ejfjgj = 1 for 3 ≤ j ≤ k − 1.
Choose labels on the t tangents at Pi, 0 ≤ i ≤ 2, so that
(i) The tangents to K at P0 are [0, 1, ai] T , 1 ≤ i ≤ t, ai = 0.
(ii) The tangents to K at P1 are [bi, 0, 1] T , 1 ≤ i ≤ t, bi = 0.
(iii) The tangents to K at P2 are [1, ci, 0] T , 1 ≤ i ≤ t, ci = 0.
Then
1 =
a∈F ∗ q
a = aiej = bifj = cigj,
where the product in each case is over all i and j with 3 ≤ j ≤ k − 1 and
1 ≤ i ≤ t. This implies that
This proves
1 = (aibici) (ejfjgj) = aibici.

4.6. ARCS IN OVALS 193
Lemma 4.6.5. (Lemma of Tangents)
t
aibici = 1.
i=1
Lemma 4.6.6. If hyperovals K and K ′ have 1
(q + 2) + n points in common,
2
with n > 0, then K = K ′ .
Proof. If K = K ′ , fix P ′ ∈ K ′ \ K. Then for each P ∈ K ∩ K ′ , the line P P ′
meets K again in a point Q of K \ K ′ . So for every point of K ∩ K ′ there is
a point of K \ K ′ . Hence K has 1
2 (q + 2) + n = q + 2 + 2n points, forcing
n = 0, i.e., K = K ′ .
Now let K be a q-arc. Each point of K is on q − 1 secants and and on
two tangents.
Lemma 4.6.7. Let P be any point of K and l a tangent to K at P . Then
there is a point Q on l \ {P } for which Q is on at least three tangents to K.
Proof. The 2(q − 1) tangents at points of K \ {P } meet l at points different
from P . If they all meet l at distinct points, then q ≥ 2(q − 1), forcing q ≤ 2,
contradicting our basic hypothesis. This shows that some point Q of l \ {P }
is on at least three tangents to K.
The next theorem is due to G. Tallini [Ta57], but the proof we give is due
to J. A. Thas.
Theorem 4.6.8. In P G(2, q), q = 2 e > 2, a q-arc K is contained in a unique
oval.
Proof. Let K be a q-arc and let Q be a point off K through which 3 tangents
l0, l1, l2 pass. Choose a coordinate system so that Li ∩ K = Pi, 0 ≤ i ≤ 2.
Here we may put
l0 = [0, 1, a0] T ; l1 = [a1, 0, 1] T ; l2 = [1, a2, 0] T , with a0a1a2 = 1.
Now let m0, m1, m2 be the other tangents to K at P0, P1, P2, respectively.
Here
m0 = [0, 1, b0] T ; m1 = [b1, 0, 1] T ; m2 = [1, b2, 0] T , with b0b1b2 = 0.

194 CHAPTER 4. THE UBIQUITOUS OVAL
By the Lemma of Tangents 2
i=0 aibi = 1. Hence 2
i=0 bi = 1, implying
that m0, m1, and m2 are concurrent at a point Q ′ not on K.
CLAIM: K ′ = K ∪ {Q, Q ′ } is a hyperoval.
To establish this claim, let P be an arbitrary point of K \ {P0, P1, P2}.
Let l, m be the two tangents to K at P . We claim that {l, m} = {P Q, P Q ′ }.
If this is true for all P ∈ K \ {P0, P1, P2}, then P Q and P Q ′ are tangents
to K, implying that K ′ is a hyperoval. Without loss of generality we may
assume that coordinates are chosen so that P = (1, 1, 1). Then we have the
following:
l = [1, c, 1+c] T ; m = [1, d, 1+d] T ; P Q = [1, c ′ , 1+c ′ ] T ; P Q ′ = [1, d ′ , 1+d ′ ] T .
We now want to apply the Lemma of Tangents to the triangles △(P0P1P )
and △(P0, P2P ). To do this, first apply the homography T : (x0, x1, x2) ↦→
(x0 + x2, x1 + x2, x2). The map T fixes P0 and P1 and interchanges P2 ⎛ ⎞
and
1 0 0
P = (1, 1, 1). If A = [T ], then A = ⎝ 0 1 0 ⎠. Clearly A = A
1 1 1
−1 . So T
maps the plane [a, b, c] T to the plane [a, b, a + b + c] T . Then

4.6. ARCS IN OVALS 195
⎡
l0 = ⎣
⎡
l1 = ⎣
⎡
l = ⎣
⎡
P Q = ⎣
0
1
a0
a1
0
1
1
c
1 + c
1
c ′
1 + c ′
⎡
m0 = ⎣
m1 = ⎣
⎡
m = ⎣
⎡
P Q ′ = ⎣
⎡
0
1
b0
b1
0
1
1
d
d + 1
1
d ′
d ′ + 1
⎤
⎦ T
↦→
⎤
⎦ T
↦→
⎤
⎦ T
↦→
⎤
⎦ T
↦→
⎤
⎦ T
↦→
⎤
⎦ T
↦→
⎤
⎦ T
↦→
⎤
⎦ T
↦→
⎡
⎣
⎡
⎣
⎡
⎣
⎡
⎣
⎡
⎣
⎡
⎣
⎡
⎣
0
1
a0 + 1
a1
0
a1 + 1
1
c
0
1
c ′
0
⎤
⎦
⎤
⎤
⎡
⎦ = ⎣
⎤
⎡
⎦ = ⎣
0
1
t0
t1
0
1
⎤
⎦ ;
⎤
⎦ ;
⎦ . (4.41)
0
1
1 + b0
b1
0
b1 + 1
1
d
0
⎤
⎦ .
⎤
⎡
⎦ = ⎣
⎤
⎡
⎦ = ⎣
If we apply the Lemma of Tangents to the triangle △(P0P1P ), we see
that t0t1c · t ′ 0 t′ 1 d = 1. As l0, l1, P Q are concurrent at Q and m0, m1, P Q ′
are concurrent at Q ′ , so their images are also concurrent at some points,
⎡
⎣
1
d ′
0
⎤
⎦ .
implying t0t1c ′ = 1 and t ′ 0t ′ 1d ′ = 1. Hence t0t1 = 1
c
c ′ · d
d ′ = 1, i.e.,
0
1
t ′ 0
t ′ 1
0
1
⎤
⎦ .
⎤
⎦ .
c ′ and t ′ 0t ′ 1 = 1
d ′ , implying
cd = c ′ d ′ . (4.42)
For triangle △(P0P2P ) use (x0, x1, x2) S
↦→ (x0 + x1, x1, x1 + x2). Here S
fixes P0 and P2 and interchanges P1 and P , and S = S −1 , so [a, b, c] T ↦→
[a, a + b + c, c] T .

196 CHAPTER 4. THE UBIQUITOUS OVAL
First consider the tangents at vertices of △(P0P2P ) meeting at Q, then
the line P Q, then the corresponding lines through Q ′ :
⎡
l0 = ⎣
⎡
l2 = ⎣
⎡
l = ⎣
⎡
P Q = ⎣
0
1
a0
1
a2
0
1
c
1 + c
1
c ′
1 + c ′
⎡
m0 = ⎣
m2 = ⎣
⎡
m = ⎣
⎡
P Q ′ = ⎣
⎡
0
1
b0
1
b2
0
1
d
1 + d
1
d ′
1 + d ′
⎤
⎦ S
↦→
⎤
⎦ S
↦→
⎤
⎦ S
↦→
⎤
⎦ S
↦→
⎤
⎦ S
↦→
⎤
⎦ S
↦→
⎤
⎦ S
↦→
⎤
⎦ S
↦→
⎡
0
⎤ ⎡
⎣ 1 + a0 ⎦ = ⎣
⎡
⎣
⎡
⎣
⎡
⎣
⎡
⎣
a0
1
1 + a2
0
1
0
1 + c
1
0
1 + c ′
0
1 + b0
b0
⎤
⎡
⎦ = ⎣
⎤
⎡
⎦ = ⎣
⎤
⎡
⎦ = ⎣
⎤
⎡
⎦ = ⎣
0
1
x0
1
x2
0
1
1+c
0
1
1
1+c ′
0
1
0
1
x ′ 0
1
x ′ 2
0
⎡
1
⎤ ⎡
⎣ 1 + b2 ⎦ = ⎣
⎡
⎣
0
1
0
⎤ ⎡
⎦ = ⎣
1
1+d
0
1 + d
⎡
1
⎣ 0
1
1 + d ′
⎤ ⎡
⎦ = ⎣
1
1+d ′
0
1
⎤
⎦ ;
⎤
⎦ ;
⎤
⎦ ;
⎤
⎤
⎦ ; (4.43)
So from the Lemma of Tangents applied to the image under S of △(P0P2P )
we have
1
x0x2
1 + c x′ 0x′ 1
2 = 1. (4.44)
1 + d
Since l0, l2, and P Q meet at Q, their images under S meet at a point,
and the same for the images of m0, m2 and P Q ′ :
1
x0x2
1 + c ′ = 1 = x′ 0x′ 2
1
1 + d ′
⎦ ;
⎤
⎦ ;
⎤
⎦ ;
⎤
⎦ .
(4.45)

4.7. WHAT IS A TRANSLATION OVAL? 197
Now using cd = c ′ d ′ from above, we quickly see
c + d = c ′ + d ′ . (4.46)
Eliminating c from the two equations Eq. 4.42 and Eq. 4.46, we find (c ′ +
d)(d ′ + d) = 0. But this implies either d = d ′ (and c = c ′ ) or d = c ′ (and
c = d ′ ). Hence {l, m} = {P Q, P Q ′ }, completing the proof.
4.7 What is a translation oval?
If an oval O of P G(2, q) has a tangent line which is the axis of each element
of a group of order q of elations that stabilize the oval, then O is called a
translation oval, and the tangent line is called an axis of the oval. We may
choose coordinates so that the tangent line, i.e., the axis is the line [1, 0, 0],
containing the oval point (0, 0, 1) and the nucleus (0, 1, 0).
Lemma 4.7.1. Let α be an o-polynomial, i.e.,
Ω(α) = {(1, t, t α ) : t ∈ Fq} ∪ {(0, 0, 1)} is an oval with nucleus (0, 1, 0).
In particular we know that α is a permutation of the elements of Fq fixing 0
and 1. Suppose that there is a group of order q of elations that stabilize the
oval and have [1, 0, 0] as axis. Then α is an additive map for which x ↦→ x α /x
is a permutation of the nonzero elements of Fq.
Proof. The elations with axis [1, 0, 0] are of the form (x, y, z) ↦→ (x, y, z)A,
where
⎛
1 b
⎞
c
A = ⎝ 0 1 0 ⎠ .
0 0 1
Such an elation maps the point (1, t, tα ) to the point (1, t + b, tα + c),
which must be of the form (1, x, xα ). Hence (t + b) α = tα + c for all t ∈ Fq.
With t = 0 we find c = bα . Then it is clear that α must be additive. Since
xα +yα = x+y xα +zα for distinct x, y, z, and α is additive, it follows immediately
x+z
that x ↦→ xα−1 must permute the nonzero elements of Fq.
In 1957 B. Segre proved the following.

198 CHAPTER 4. THE UBIQUITOUS OVAL
Theorem 4.7.2. Let α be a generator of Gal(Fq), q even. Then
is a hyperoval.
D(α) = {(1, t, t α )} ∪ {(0, 1, 0), (0, 0, 1)}
Proof. If α : x ↦→ x2i for some i with 1 ≤ i ≤ e − 1, D(α) is a hyperoval
if and only if for each fixed t ∈ Fq the map ft : x ↦→ tα−xα t−x = (t + x)α−1 is
a bijection from the elements of Fq \ {t} to the set of nonzero elements of
Fq. This holds if and only if gcd(2i − 1, 2e − 1) = 1, which is if and only if
gcd(i, e) = 1. This proves that D(α) is a hyperoval if and only if α generates
Gal(Fq).
In 1971 S. E. Payne proved that each translation oval is projectively
equivalent to one of those given by Segre. Before we can prove this we need
to study linearized polynomials.
4.8 Linear Maps
Until further notice F = GF (q n ) and K = GF (q), where q = p e is any power
of a prime and n is any positive integer. We may view F as a vector space
over K with dimension n. Recall that if V and W are vector spaces over
a field K with dimensions m, n, respectively, then HomK(V, W ) is a vector
space over K with dimension mn. Moreover, if K is a field with q elements
and V is a vector space over K with dimension n, then |V | = q n . Then
V = HomK(F, F ), the vector space of all K-linear maps from F to F , is a
vector space over K with dimension n2 , so that |HomK(F, F )| = qn2. On the
other hand, we can construct elements of V = HomK(F, F ) in the following
way. Let σ : x ↦→ xq , for x ∈ F . So σ generates the Galois group Gal(F/K),
i.e., σi : x ↦→ xqi, i = 0, 1, 2, . . . , n − 1, gives the distinct automorphisms of
F that are the identity on K. For each choice of ai ∈ F , 0 ≤ i ≤ n − 1,
consider the map f : x ↦→ f(x) = n−1 i=0 aiσi (x) = n−1 qi
i=0 aix . Clearly
f ∈ HomK(F, F ). Because each such polynomial f(x) has degree at most
qn−1 , no two of them can represent the same function. Hence there are (qn ) n
such functions, showing that each K-linear function from F to F must be of
this form.
This proves the following theorem:

4.8. LINEAR MAPS 199
Theorem 4.8.1. Each f ∈ HomK(F, F ) is given by a linearized q-polynomial,
i.e.,a polynomial of the form
n−1
f(x) =
i=0
aix qi
, ai ∈ F. (4.47)
Consequently, if f(x) ∈ F [x] has degree at most q n − 1 and has the property
that f(a + b) = f(a) + f(b) for all a, b ∈ F , and f(ca) = cf(a) for all c ∈ K
and all a ∈ F , then f(x) is a linearized q-polynomial.
For ai ∈ F , 0 ≤ i ≤ n − 1, write
¯α = (a0, a1, . . . , an−1), and [¯α] =
a qi
[j−i] . (4.48)
(0≤i,j≤n−1)
Here we are using the following convention. The symbol [k −j] as a subscript
means that k − j is to be reduced modulo n to one of 0, 1, . . . , n − 1. Hence
the top row of [¯α] is just the vector ¯α. Then each row is obtained by cycling
the row above one place to the right and replacing each element by its image
under the automorphism x ↦→ x q .
For x ∈ F , put
[x] = (x, x q , x q2
, . . . , x qn−1
) T .
Then ¯α uniquely determines an element α of HomK(F, F ) by
It is easy to check that for each j,
It follows that
α : x ↦→ x α n−1
= aix qi
= ¯α · [x].
(x α ) qj
n−1
=
i=0
i=0
a qj
i xqi+j
n−1
=
k=0
a qj
[k−j] xqk.
[x α ] = [¯α · [x]] = [¯α][x]. (4.49)
Lemma 4.8.2. Let x1, x2, . . . , xr be elements of F . Then {x1, . . . , xr} is linearly
independent over K if and only if {[x1], . . . , [xr]} is linearly independent
over F .

200 CHAPTER 4. THE UBIQUITOUS OVAL
Proof.
First suppose that {x1, . . . , xr} is linearly dependent over K, say
r
i=1 cixi = 0, with ci ∈ K and at least one of the ci not zero, 1 ≤ i ≤ r. Then
0 = r i = r qj
cixi , from which it follows that (0, 0, . . . , 0)T =
r
i=1 cqj i xqj
i=1
i=1 ci[xi], so that {[x1], . . . , [xr]} is linearly dependent over K, and hence
over F .
Now suppose that {[x1], . . . , [x]r} is linearly dependent over F , say
(0, 0, . . . , 0) T = r
i=1 ci[xi] with ci ∈ F and some ci = 0. The goal is to show
that each ci ∈ K, so that the top row of the equation of the hypothesis says
that 0 = r
i=1 cixi with ci ∈ K. The proof proceeds by induction on r. First
note that for r = 1 the result is clear, and suppose that r = 2. Here we may
assume that [x1] = c[x2] with c ∈ F . Then the first row says that x1 = cx2
and its image under x ↦→ x q says that x1 q = c q x2 q . The second row says that
x1 q = cx2 q . Hence c = x1/x2 = x1 q /x2 q = c q , which forces c to be in K,
i.e., {x1, x2} is linearly dependent over K. This shows that the desired result
holds when r = 2.
Suppose that for 1 ≤ r − 1 whenever {x1, . . . , xr−1} is linearly independent
over K, then {[x1], [x2], . . . , [xr−1]} is linearly independent over F , and
suppose that {x1, . . . , xr−1, xr} is linearly independent over K. In particular
we know that {[x1], [x2], . . . , [xr−1]} is linearly independent over F , so that if
some vector in F n is an F -linear combination of the [xi], 1 ≤ i ≤ r − 1, the
coefficients in the linear combination are unique.
Suppose r i=1 ci[xi] = [0]. If cr = 0, then all the ci’s equal zero. Without
loss of generality we may assume that cr = 1, and that [xr] = r−1 i=1 di[xi],
di ∈ F . Looking at row j of this equality we see xr qj = r−1 qj
i=1
dixi . But
taking the q-th power of row j we find xr qj+1 = r−1 i=1 di q xi qj+1.
By letting j
vary, this says
r−1
r−1
[xr] = di[xi] =
i=1
i=1
d q
i [xi].
Since the coefficients are unique when writing a vector in F n as an F -
linear combination of {[x1], . . . , [xr−1]}, it must be that di = di q , i.e., di ∈ K,
for 1 ≤ i ≤ r − 1. Hence the first row says that x1, . . . , xr are linearly
dependent over K.
Theorem 4.8.3. Let ¯α = (a0, . . . , an−1) ∈ F n be given. The kernel

4.8. LINEAR MAPS 201
U = x ∈ F = GF (q n ) : x α n−1
= aix qi
= 0; ai ∈ F
is a subspace over K = GF (q). If U has dimension r (i.e., is a projective
(r − 1)-dimensional subspace of F ), then the F -rank of the matrix [¯α] =
(a qi
[j−i] ) is equal to n − r.
Proof. It is easy to see that the set U of zeros of n−1 i=0
and any element of U is also a zero of
n−1
j=0
a qi
j xqi+j
n−1
=
j=0
i=0
a qi
[j−i] xqi,
qi aix forms a subspace
where the indices are taken modulo n. For all x ∈ U, the vector [x] in
V (n, q n ) is in the right null space of [¯α]. There are r elements of U linearly
independent over K, and therefore by Lemma 4.7.2 there are r vectors of the
form [x] linearly independent over F and in the right null space of [¯α]. Hence
the nullspace of [¯α] is a subspace over F of dimension at least r, implying
that the F -rank of [¯α] is at most n − r.
As before, r is the K-dimension of the kernel of α, so there are n − r
elements x1 α , . . . , xn−r α in the image of α that are K-independent, so that
by Lemma 4.8.2
[x1 α ] = [¯α][x1], . . . , [xn−r α ] = [¯α][xn−r]
are F -linearly independent. This says the F -rank of [¯α] is at least n − r.
Hence the F -rank of [¯α] must be exactly n − r.
Corollary 4.8.4. The additive map α is one-to-one and onto if and only if
[¯α] is nonsingular, and in general det[¯α] ∈ K.
Proof. The first part of the Corollary is contained in the preceding theorem.
Now put ∆ = det[¯α]. Then x ↦→ x q is an automorphism of F fixing K
pointwise, so ∆ q = det([¯α] q ) = det(B), where B is obtained from [¯α] by
shifting rows one place up and columns one place left. Hence detB = det[¯α],
i.e., ∆ = ∆ q , implying that ∆ ∈ K. This completes the proof.
Note:
[α ◦ µ][x] = [x α◦µ ] = [(x α ) µ ] = [¯µ · [x α ]] = [¯µ · [¯α][x]] = [¯µ][¯α][x].

202 CHAPTER 4. THE UBIQUITOUS OVAL
This says ([α ◦ µ] − [¯µ][¯α])[x] = 0 for all x ∈ Fq. Let ¯ρ = (r0, . . . , re−1)
be the ith row of [α ◦ µ] − [¯µ][¯α]. So e−1 pi
i=0 rix = 0 for all x, i.e., the map
ρ : x ↦→ ¯ρ[x] is trivial. This implies ¯ρ = ¯0. Hence it follows that
4.9 Additive Maps on Fq
[α ◦ µ] = [¯µ][¯α]. (4.50)
Theorem 4.9.1. Let α and β be permutations of Fq such that 0 α = 0 = 0 β
and such that x ↦→ x α /x β permutes the nonzero elements of Fq. Then p = 2.
Proof. Let ζ be a generator of the multiplicative group F ∗ q , i.e., a primitive
i α
root for Fq. Then α and β are given by ζ ↦→ ζ i′ i β
, ζ ↦→ ζ i′′ , where i ↦→ i ′ and
i ↦→ i ′′ are permutations of the integers 1, 2, . . . , pe − 1. If pe − 1 = |F ∗ q | = 2k,
then 1 + 2 + · · ·+ 2k = k(2k + 1) ≡ k (mod 2k). Thus 0 =
1≤i≤2k (i′ − i ′′
) ≡
i ≡ k (mod 2k), an impossibility.
1≤i≤2k
Corollary 4.9.2. Let α and β be additive permutations of the elements of
Fq for which x ↦→ x α /x β permutes the nonzero elements of Fq. Then p = 2.
For the remainder of this section Fq = GF (2 e ).
Theorem 4.9.3. Let α and β be additive permutations of the elements of
Fq = GF (2 e ) that fix 1 and for which x ↦→ x α /x β permutes the nonzero
elements of Fq. Then α −1 β is an automorphism of Fq of maximal order e.
Proof. Let α and β satisfy the hypotheses of the theorem. Then there are
scalars ai, bi in Fq, 0 ≤ i ≤ e − 1, for which
e−1
α : x ↦→
i=0
aix 2i
; β : x ↦→
e−1
i=0
bix 2i
.
Let A = [¯α] = (aij), so aij = a 2i−1
[j−i] , and B = [ ¯ β] = (bij), so bij = b 2i−1
[j−i] ,
1 ≤ i, j ≤ e.
It is convenient to make the following observations. Let P be the permutation
matrix with a 1 in the (i, j) position if and only if j ≡ i + 1 (mod e).
So (d1, . . . , de)P = (de, d1, . . . , de−1). If D (2) denotes the matrix obtained
from the matrix D by each element dij by its square (dij) 2 , then

4.9. ADDITIVE MAPS ON FQ 203
(P −1 A (2) P )ij =
1≤r,k≤e
(P T )ir(A (2) )rkPkj
= (A (2) )i−1,j−1 = Aij, implying P −1 A (2) P = A.
Then continuing with the proof, since α and β are permutations, by
Cor. 4.8.4 both A and B are nonsingular. Since x ↦→ xα /xβ is a permutation
of the elements of F ∗ q , for each c ∈ F ∗ q there must a be a (unique) nonzero
solution x to xα−cx β = 0. Hence e−1 2i
i=0 (ai−λbi)x = 0 has a unique nonzero
solution x for each λ ∈ F ∗ q . By Cor. 4.8.4, the matrix
Cλ = ((a[j−i] − λb[j−i]) 2i−1
), 1 ≤ i, j ≤ e,
has zero determinant for each λ ∈ F ∗ q . And 0 = detA · detB, so detA =
detB = 1. So det Cλ is a polynomial in λ of degree 2e − 1 with constant term
1, leading term 1, and having each nonzero element of Fq as a simple root.
This implies
det(Cλ) = λ 2e −1 + 1. (4.51)
For 1 ≤ t ≤ 2e − 2, we now calculate the coefficient of λt in det(Cλ) and
set it equal to 0.
Let ti1, ti2, . . . , tir be the nonzero coefficients in the binary expansion of
t = e−1 i=0 ti2i . Then the coefficient of λt in det(Cλ) is easily seen to be the
determinant of the matrix obtained by replacing rows i1, . . . , ir of A with rows
i1, . . . , ir of B Hence we know the following: the rows of A are independent,
the rows of B are independent, and any set of rows formed by taking some
r rows of B and the complementary e − r rows of A is a linearly dependent
set, 1 ≤ r ≤ e − 1. In particular, the ith row of B is a linear combination
of rows 1, . . . , i − 1, i + 1, . . . , e of A. Let βi be the ith row of B, 1 ≤ i ≤ e.
Then
βi = (b 2i−1
[1−i] , . . . , b2i−1 [e−i] ).
Then there are scalars d1, . . . , di−1, di+1, . . . , de such that
βi = (d1, . . . , di−1, 0, di+1, . . . , de)A. (4.52)
Apply the automorphism x ↦→ x 2 to Eq. 4.52

204 CHAPTER 4. THE UBIQUITOUS OVAL
(b 2i
[1−i] , . . . , b2i [e−i] ) = (d21 , . . . , d2i−1 , 0, d2 i+1 , . . . d2e )
a 2k
[j−k] . (4.53)
1≤k,j≤e
In Eq. ?? first permute the columns cyclically, moving column j (on the
left-hand side) to position j + 1 (mod e), then effect the same permutation
of columns on the right. The result is an equation that says:
that is,
βi+1 = β (2)
i P = (d2 1, . . . , d 2 i−1, 0, d 2 i+1, . . . d 2 e)P · P −1 A (2) P,
βi+1 = (d 2 e , d21 , . . . , d2i−1 , 0, d2i+1 , . . . , d2 e−1 )A. (4.54)
Repeating this k times we obtain
βi+k = (d 2k
[1−k] , d2k [2−k] , . . . , d2k [e−k] )A, (4.55)
where the d ′ js are unique and di = 0. By permuting rows cyclically we may
adopt a new notation:
β1 = (d1, . . . , de)A with d1 = 0, so β1+i = (d 2i
1−i, . . . , d 2i
e−i)A. (4.56)
Let αi denote the i th row of A, 1 ≤ i ≤ e. For some λ1, λ2 in Fq, not
both zero, we have
λ1β1 + λ2β2 =
e
fjαj = (λ2d 2 e, λ1d2, λ1d3 + λ2d 2 2, . . . , λ1dj + λ2d 2 j−1, . . .)A.
j=3
(4.57)
Hence as the rows of A are independent, λ2d 2 e = 0 = λ1d2. If λ1 = 0, then
d2 = 0. If λ2 = 0, then de = 0. Since d1 = 0, we have either both de and d1
are zero, or both d1 and d2 are zero, i.e., there is a string of length two of
consecutive d ′ is (including d1) equal to zero. By changing notation we may
suppose d1 = d2 = 0 and complete the proof by a finite induction.
Suppose there is a string of t zeros, say d1 = · · · = dt = 0, 1 ≤ t ≤ e − 2.
We then wish to show that there is a string of t + 1 zeros. This yields the
following.

4.9. ADDITIVE MAPS ON FQ 205
β1 = (0, . . . , 0, dt+1, . . . , de)A;
β2 = (d 2 e , 0, . . . , 0, . . . , d2 e−1 )A;
.
βt = (d2t−1 e+2−t , . . . , d2t−1 e
, 0, . . . , 0, d 2t−1
↑
t
↑
2t − 1
t+1
βt+1 = (d 2t
e+1−t, . . . , d 2t
e , 0, . . . , 0, . . . , d 2t
↑
t + 1
↑
2t
, . . . , d2t−1
e+1−t )A;
e−t)A.
(4.58)
Moreover, there are scalars λ1, . . . , λt+1, not all zero, for which t+1 i=1 λiβi
is some linear combination of αt+2, . . . , αe, since 1 < t+1 < e. Use Eq. 4.52 to
calculate the coefficients on α1, . . . , αt+1 (which must be zero) in t+1 i=1 λiβi,
starting with column t + 1 of Eq. 4.52 and working backwards to column 1.
0 = λ1dt+1
0 = λt+1d 2t
e
0 = λtd 2t−1
e
0 = λt−1d 2t−2
e
2t + λt+1d
e−1
+ λtd 2t−1
e−1 + λt+1d 2t
e−2
.
0 = λ2d 2 e + λ3d 22
e−1 + · · · + λt+1d 2t
e−(t−1)
(4.59)
If λt+1 = 0, then de = 0, giving a string of zeros of length t + 1.
If λt+1 = 0, λt = 0, then again de = 0.
If λt+1 = λt = 0, λt−1 = 0, then again de = 0.
Continue in this fashion. . . .
If λt+1 = λt = · · · = λ3 = 0, λ2 = 0, then again de = 0.
If λt+1 = · · · = λ2 = 0, λ1 = 0, then dt+1 = 0, giving a string of zeros of
length t + 1.
It follows that only one di can be nonzero, say d = dt = 0, t = 1. This
says:
b[j] = da 2t−1
[j−(t−1)] , 0 ≤ j ≤ e − 1. (4.60)
Our assumption that 1 = 1 α = 1 β implies that d = 1. Put u = t − 1, so
Eq. 4.60 becomes:
b[j] = a 2u
[j−u] . (4.61)

206 CHAPTER 4. THE UBIQUITOUS OVAL
If x ↦→ x2u is an automorphism of F of order n < e, it is routine to find
n rows of B which together with the complementary e − n rows of A form
a linearly independent set. Hence it must be that gcd(u, e) = 1. Moreover,
from Eq. 4.55 we obtain
x β e−1
= bjx 2j
e−1
=
j=0
j=0
Hence Eq. 4.55 is equivalent to
Since x α /x β = x α /x α◦2u
(u, e) = 1, we are done.
a 2u
j−u x2j
x β = (x α ) 2u
=
= (x α−β ) 1−2u
e−1
aj−ux 2j−u
2u j=0
= (x α ) 2u
.
⇒ β = α · 2 u . (4.62)
is a permutation if and only if
Corollary 4.9.4. Let β be an additive permutation of the elements of Fq for
which x ↦→ x/xβ permutes the nonzero elements of Fq. Then β has the form
xβ = dx2u for fixed d ∈ F ∗ q , (u, e) = 1.
4.10 A Characterization of Translation Ovals
Theorem 4.10.1. Let O be an oval of P G(2, q), q = 2 e , and let ℓ be a
tangent line to O. Then the following are equivalent.
(1) The oval O is a translation oval with axis ℓ.
(2) There is a collineation of P G(2, q) which maps O to Oα = {(1, t, t α ) :
t ∈ Fq}∪{(0, 0, 1)} with nucleus (0, 1, 0) and which maps ℓ to the line [1, 0, 0],
for some generator α of Aut(Fq).
(3) There is a collineation of P G(2, q) induced by a matrix in GL(3, q)
which maps O to Oα and ℓ to [1, 0, 0] for some generator α of Aut(Fq).
(4) If P1, P2, P3, P4 are four points of O not on ℓ, and if P1P2 ∩ P3P4 ∈ ℓ,
then P1P3 ∩ P2P4 ∈ ℓ.
Proof. Basically we have already seen that (1), (2) and (3) are equivalent.
We now show that (1) and (4) are equivalent.
So suppose that (1) holds. Let P1, P2, P3, P4 be four points of O not on
ℓ, and suppose that P1P2 ∩ P3P4 ∈ ℓ. By (1) there must be an elation θ with
axis ℓ mapping P1 to P2 and leaving the oval O invariant. If P1P2 ∩ ℓ = R,
then R must be the center of the elation, so the line P3P4 meets ℓ in the point

4.11. STABILIZERS OF ARCS 207
R and must be fixed by θ. Let Q = P1P3 ∩ ℓ. The elation θ maps QP1P3 to
QP2P ′ 3, where P ′ 3 must be a point of O on the line QP2. But P ′ 3 must also
be on the fixed line P3R, which contains only the additional point P4 of O.
Hence P ′ 3 = P4, and P1P3 ∩ P2P4 = Q ∈ ℓ, showing that (4) holds.
Now suppose that (4) holds. Let θ be the elation of P G(2, q) with axis ℓ
and moving P1 to P2 and having center R = P1P2 ∩ ℓ. We want to show that
θ preserves the oval O. So let P3 be any point of O different from P1, P2,
and not on ℓ. Let Q be the point Q = P1P3 ∩ ℓ. The line QP2 must be a
secant to O, hence must contain an additional point P4. Since P1P3 meets
P2P4 at a point Q of ℓ, it must be by (4) that P1P2 meets P3P4 at a point
of ℓ, in fact at the center R of θ. It follows that θ maps P3 = QP1 ∩ P3R to
P4 = QP2 ∩ P3R, which is a point of O.
It seems likely that Part (4) of the theorem is of interest only in that it
provides a synthetic definition of a translation oval. On the other hand the
automorphism α is extremely important. It is uniquely determined by O and
is calle the automorphism associated with O.
4.11 Stabilizers of Arcs
In this section we recall without proof a few well-known results from the theory
of collineation groups of P G(2, q). First, each collineation θ of P G(2, q)
is of the form
θ : (x, y, z) ↦→ (x σ , y σ , z σ )A, (4.63)
where A is an invertible matrix unique only up to a scalar multiple and σ is
an automorphism of Fq.
All invertible matrices A and automorphisms σ combine this way to give
such a collineation. If σ = id, the map θ is called an homography. Let θ be
a collineation. Then θ fixes all the points on some line (called the axis of θ)
if and only if θ fixes all the lines through some point (called the center of θ).
In this case θ is called a central collineation. If θ is a nonidentity central
collineation with center P and axis ℓ, then θ is called an elation provided P
is incident with ℓ, and an homology otherwise.
The following lemma has a routine proof:

208 CHAPTER 4. THE UBIQUITOUS OVAL
Lemma 4.11.1. The homographies that are central collineations with axis
[1, 0, 0] T are those with matrix
⎛
A = ⎝
a b c
0 1 0
0 0 1
If a = 1, the point (a − 1, b, c) is the center and clearly not incident with the
axis. So the elations with [1, 0, 0] T as axis are those with
⎛
A = ⎝
1 b c
0 1 0
0 0 1
Moreover, it is clear that each elation is an involution (since q = 2 e ), and
the set of elations with common axis [1, 0, 0] T is a group with order q 2 .
Let x ↦→ x α be a function such that
D(α) = {(1, t, t α ) : t ∈ Fq} ∪ {(0, 0, 1), (0, 1, 0)} is a hyperoval.
Put Ω(α) = D(α) \ {(0, 1, 0)}, so Ω(α) is an oval with nucleus (0, 1, 0). Put
Ω− (α) = Ω(α) \ {(0, 0, 1)}, and Ω0(α) = {(1, t, tα ) : 0 = t ∈ Fq}. In this
section we consider some stabilizers of large arcs contained in an oval.
If an oval O of P G(2, q) has a tangent line which is the axis of each
element of a group of order q of elations that stabilize the oval, then O is
called a translation oval, and the tangent line is called an axis of the oval.
All translation ovals were determined in the previous section.
At this point we know that if α = 2i with (i, e) = 1, then D(α) is a
hyperoval with [1, 0, 0] T as an axis. Moreover, we also know that for each
σ ∈ Aut(Fq) and for a, b ∈ Fq with b = 0, the map
(x, y, z) ↦→ (x σ , y σ , z σ ) ⎝
is a collineation of P G(2, q) stabilizing D(α).
⎛
⎞
⎠ .
⎞
⎠ .
1 a a α
0 b 0
0 0 b α
⎞
⎠ (4.64)
Theorem 4.11.2. If α is additive and Ω(α) has two axes with q = 2 e ≥ 8,
then α = 2 and the oval has q + 1 axes.

4.11. STABILIZERS OF ARCS 209
Proof. We may assume that α = 2i with (i, e) = 1. Since Ω(α) admits a group
of elations transitive on the points (1, t, tα ), t ∈ Fq, we may assume that the
second axis is the line [0, 0, 1] T through (0, 1, 0) and (1, ⎛0,
0). The ⎞elations
1 0 0
with this line as axis are of the form (x, y, z) ↦→ (x, y, z) ⎝ 0 1 0 ⎠. The
c b 1
point (0, 0, 1) maps to (c, b, 1), which is on the oval iff c = b = 0 or b = 0 and
c = bα∗ = b α
α−1 . Then (1, t, tα ) maps to (1 + tαbα∗, t + btα , tα ). Expressing the
fact that this point is on the oval (yα = xα−1z) leads to bαtα2 = bα∗t2α for all
t ∈ Fq. This quickly forces α = 2. In this case each tangent is an axis.
Corollary 4.11.3. If α is additive with D(α) an irregular hyperoval , then
the group of collineations stabilizing the oval Ω(α) fixes the point (0, 0, 1) and
acts transitively on the points of the form (1, t, t α ), t ∈ Fq.
Note that D( 1
) has q + 1 axes all passing through the point (0, 0, 1), but
2
Ω( 1
2 ) has only one axis [1, 0, 0]T .
At this point we can complete the determination of the collineation stabilizer
of D(2). Let G be the full collineation stabilizer fixing the point (0, 1, 0).
It is clear that for each σ ∈ Aut(Fq) the map (x, y, z) ↦→ (xσ , yσ , zσ ) is in
G. Then the group of homographies preserving the hyperoval, and fixing the
nucleus (0, 1, 0), consists of the maps (x, y, z) ↦→ (x, y, z)A where
⎛
A = ⎝
a √ ab b
0 1 0
c √ cd d
⎞
⎠ , where det(A) = 1. (4.65)
Note that with a = d = 0 and b = c = 1 there is a collineation preserving
the hyperoval and interchanging the points (1, 0, 0) and (0, 0, 1). Hence the
group is triply transitive on the q + 1 points of the conic. If a projective
collineation stabilizing the oval fixes three points of the oval as well as the
nucleus, clearly it must be the identity map. So the group is sharply triply
transitive on the conic itself. This means that each tangent to the conic Ω(2)
is the axis of a group of q elations with that tangent as axis.
Suppose that a collineation (which may be assumed to be an homography)
moves the point (0, 1, 0). Then the full homography collineation group
is 4-ply transitive on the points of the hyperoval, and the homography interchanging
points (0, 1, 0) and (0, 0, 1) and interchanging (1, 0, 0) and (1, 1, 1)
must stabilize the hyperoval. This collineation has matrix

210 CHAPTER 4. THE UBIQUITOUS OVAL
⎛
A = ⎝
1 1 1
0 0 1
0 1 0
Clearly this homography maps (1, t, t 2 ) to the point (1, 1 + t 2 , 1 + t), which
is on the hyperoval if and only if 1 + t = (1 + t 2 ) 2 = 1 + t 4 . This occurs if
and only if q ≤ 4.
So now consider a secant line [0, 1, 0] T through the points (0, 0, 1) and
(1, 0, 0) of the conic. The group of elations with this line as axis are of the
form (x, y, z) ↦→ (x + ay, y, z + cy). But now it is routine to verify that this
map stabilizes the conic if and only if a = c = 0.
This completes a proof of the following theorem.
Theorem 4.11.4. If q ≥ 8, the complete stabilizer of the regular hyperoval
D(2) fixes the point (0, 1, 0). The homographies are sharply triply transitive
on the points of Ω(2) = D(2) \ {0, 1, 0)}. Each tangent to Ω(2) is the axis of
a group of order q of elations stabilizing the conic. No other secant to D(2)
is the axis of such a group of elations stabilizing the hyperoval.
At this point we can also determine the hyperovals with a transitive
collineation group. The proof given here is taken from a 1994 paper by
O’Keefe and Penttila, but the result was originally due to G. Korchmáros in
1978.
Theorem 4.11.5. Let K be a hyperoval in P G(2, q) with a transitive collineation
group G. Then q ∈ {2, 4, 16}. For q = 2 and q = 4 it is easy to check that the
only hyperovals are regular with transitive collineation groups. For q = 16,
the regular hyperoval D(2) does not have a transitive collineation group by
the previous theorem. However, there is a second hyperoval that does.
Proof. Let G be a group of collineations stabilizing K and acting transitively
on it. Since G has an orbit of length q + 2 on K, q + 2 must divide |G|,
and hence must divide |P ΓL(3, q)| = (q 2 + q + 1)(q − 1) 2 q 3 (q + 1)e. But
q ≡ −1 (mod q + 2), so |P ΓL(3, q)| ≡ 3 · 9 · 8(−1)e ≡ 0 (mod q + 2), which
implies that 2(2 e−1 + 1) divides 3 3 · 2 3 e. This implies 2 e−1 + 1 divides 3 3 e
if e > 1. It follows fairly easily that e = 1, 2, or 4. The cases for q = 2
and 4 are easily checked. When q = 16, by Theorem 4.11.5 D(2) does not
have a transitive collineation group, however the Lunelli-Sce hyperoval does.
This was first established by M. Hall, Jr., in 1975 using a computer. We may
return to this example later.
⎞
⎠ .

4.12. MONOMIAL HYPEROVALS 211
Theorem 4.11.6. Let α be additive with D(α) an irregular hyperoval. Then
the complete homography stabilizer of D(α) has order q(q − 1) and is given
by Eq. 4.64 with σ = id.
Proof. Let α be additive and let G be the complete homography stabilizer
of D(α). We know that the points of the form (1, t, t α ) for t ∈ Fq are in one
orbit and we may assume that q > 16. Hence there is no orbit of length q +2.
Suppose there is an orbit of length q + 1. Then the map (x, y, z) ↦→ (x, z, y)
interchanges (0, 1, 0) and (0, 0, 1), and interchanges D(α) and D(α −1 ). As
α −1 is also additive, we may assume that {(0, 1, 0)} is the orbit of length 1.
By Theorem 4.11.2 we have a contradiction. Hence G must have an orbit
of length q. If G stabilizes both (0, 1, 0) and (0, 0, 1), then it is easy to see
that G consists of exactly the homographies given in Eq. 4.64. Otherwise,
we may suppose that there is a collineation θ : (0, 1, 0) ↔ (0, 0, 1). Since the
group fixing each of (0, 1, 0) and (0, 0, 1) is transitive on the points of the
form (1, t, t α ), for t ∈ Fq, we may suppose that θ fixes (1, 0, 0). This means
that we may write the matrix [θ] in the form
⎛
[θ] = ⎝
1 0 0
0 0 b
0 c 0
Then θ : (1, t, t α ) ↦→ (1, ct α , bt). Hence (ct α ) α = bt, from which it quickly
follows that α 2 = id. This forces e = 1 or 2, completing the proof.
4.12 Monomial Hyperovals
Even though our main interest will be in translation ovals, in general a hyperoval
that contains a translation oval will also contain other ovals that are
not translation ovals but are still monomial ovals. Therefore it is reasonable
for us to spend some time on this more general class of oval.
Recall that a conic in P G(2, q) is an oval, and a conic plus its nucleus
is a hyperoval called a regular hyperoval. Moreover, an oval which is a
regular hyperoval minus the natural nucleus is called a pointed conic. We
have already noted that all hyperovals of P G(2, q) are regular for q = 2 and
q = 4. For q > 4 we have also seen that a conic is not equivalent to a pointed
conic. At this point we have done most the work needed to give a complete
determination of all ovals of P G(2, 8).
⎞
⎠ .

212 CHAPTER 4. THE UBIQUITOUS OVAL
Theorem 4.12.1. In P G(2, 8) all hyperovals are regular. Each oval is either
a conic or a pointed conic.
Proof. Using Theorems 4.5.4 and 4.5.6 we see that each hyperoval of P G(2, 8)
is equivalent to one of D(2), D(4) or D(6), each of which must be a hyperoval
since q = 23 and 3 is odd. But modulo q − 1 = 7, 4 = 1
, so we know D(4)
2
is regular. Note that 6 ≡ 1 − 2 (mod 7), and the homography (x, y, z) ↦→
(y, x, z) interchanges D(2) and D(6). Hence in P G(2, 8), D(6) is regular. At
the same time we know that Ω(4) and Ω(6) are pointed conics and are not
equivalent to the conic Ω(2).
At this point we have seen that for q ≤ 8 all ovals in P G(2, q) are monomial.
In this section we want to consider further the general case of monomial
ovals.
For any integer k for which D(k) is a hyperoval in P G(2, q), if σ is an
automorphism of Fq, then (x k ) σ = (x σ ) k , so that
(x, y, z) ↦→ (x σ , y σ , z σ ) (4.66)
is a collineation of P G(2, q) that preserves the hyperoval and fixes the points
(1, 0, 0), (0, 1, 0), (0, 0, 1).
Similarly, the map
(x, y, z) ↦→ (x, ya, (za) k ) (4.67)
preserves the hyperoval D(k), fixes the points (1, 0, 0), (0, 1, 0) and (0, 0, 1),
and maps (1, t, t k ) ↦→ (1, at, (at) k ) for t = 0.
Let A = (0, 1, 0), B = (0, 0, 1), C = (1, 0, 0), D = (1, 1, 1). Let O
denote the set of permutations (i.e., permutation polynomials) associated
with hyperovals that contain the four points A, B, C, and D, i.e., O is the
set of permutations α of Fq such that 0 α = 0, 1 α = 1, and aα −b α
whenever a, b, c are distinct elements of Fq.
Then for each α ∈ O
a−b = aα −c α
a−c ,
D(α) = {A, B} ∪ {(1, t, t α ) : t ∈ Fq} is a hyperoval containing A, B, C, D.
(4.68)
Conversely, every hyperoval in P G(2, q) is projectively equivalent to one of
these, and
Ω(α) = {(x, y, z) ∈ P G(2, q) : y α = x α−1 z} (4.69)
is the oval contained in D(α) that has A as nucleus.

4.12. MONOMIAL HYPEROVALS 213
Now put
M = {k : gcd(k − 1, q − 1) = 1 = gcd(k, q − 1)} (4.70)
Here the symbol k denotes both an integer (mod q − 1) and a map from
Fq to itself. So we think of M as the set of candidates for being monomial
hyperovals. Now put
Also put
I = M ∩ O = the set of monomial hyperovals (4.71)
A = {α ∈ O : x α + y α = (x + y) α ∀x, y ∈ Fq} (4.72)
So α ∈ A is an additive map for which α − 1 is also a permutation. By
Cor. 4.9.4 we know that
Theorem 4.12.2. α ∈ A if and only if α : x ↦→ x2i gcd(i, e) = 1, 1 ≤ i ≤ e.
for some i with
Consider the projective collineation (x, y, z) ↦→ (x, z, y). For each k ∈ M,
this maps D(k) to D(k −1 ). So we have the following:
Theorem 4.12.3. k ∈ I iff k −1 ∈ I, and D(k) and D(k −1 ) are projectively
equivalent.
The projective collineation (x, y, z) ↦→ (z, y, x) fixes the point A and for
each k ∈ M maps D(k) to D(k ∗ ), where k ∗ = k/(k − 1). Hence it follows
that also
Theorem 4.12.4. k ∈ I iff k ∗ = k/(k − 1) ∈ I, and in this case Ω(k) is
projectively equivalent to Ω(k ∗ ).
For k ∈ M, put
T (k) = {k, k ∗ = k
k − 1 , (k∗ ) −1 k − 1
=
k ,
(1 − k) −1 = 1
1 − k ,
k − 1
k
∗
∗ 1
= k
1 − k
−1 }.
= 1 − k, (4.73)
It follows that for k ∈ M, the six permutations of Fq contained in T (k) are
all in I or none are in I.

214 CHAPTER 4. THE UBIQUITOUS OVAL
For k ∈ I and fixed d ∈ Fq, D(k) \ {(1, d, dk )} is projectively equivalent
to Ω(kd), where
kd : x ↦→ x[dk + (d + x−1 ) k ]
dk + (1 + d) k
. (4.74)
In particular, k0 : x ↦→ x 1−k .
Before determining isomorphisms between monomial ovals we give a general
lemma that turns out to be useful.
Lemma 4.12.5. Suppose a, b, c, d ∈ Fq with ac = 0, and recall
M = {i : 1 ≤ i ≤ q − 1 with (i, q − 1) = (i − 1, q − 1) = 1}.
Finally, suppose i, j, k ∈ M and that (at i + b) j = ct k + d for all t ∈ Fq.
Then, reducing all exponents modulo q − 1, we have j is additive (and hence
an automorphism of maximal order e), ij ≡ k (mod q−1), b j = d and a j = c.
Proof. First note that since (i, q − 1) = 1, ir ≡ is (mod q − 1) iff r ≡
s (mod q − 1). Then use the binomial theorem to write
(at i + b) j =
j
r=0
j
a
r
r t ir b j−r = ct k + d for all t ∈ Fq.
Since the exponents are all reduced mod q−1 and no two of the exponents on
the left side of the equation are the same (mod q −1), the various coefficients
on both side of the equality must be the same. Clearly for r = 0 and r = j
the corresponding two coefficients on the left are nonzero, the exponents 0
and ij on the left must be the same as the exponents k and 0 on the right.
Also, for 0 < r < j, the coefficients on t ir on the left must be zero. This shows
that j must be additive, and the rest of the lemma follows immediately.
Now suppose that α ∈ A. It is clear that α−1 ∈ A also. But an easy
check shows that α∗ ∈ A if and only if α = 2. Similarly, 1 − α ∈ A iff α = 1
2 .
It now follows immediately that if α ∈ A and if D(α) is irregular, then the
two elements α and α−1 of T (α) are in A, and the remaining elements of
T (α) are not in A. We want to know when T (α) has six distinct elements.
The following Lemma has a straightforward proof.

4.12. MONOMIAL HYPEROVALS 215
Lemma 4.12.6. (i) α = α∗ iff α = 2;
(ii) α = (α∗ ) −1 iff α2 − α + 1 = 0;
(iii) α = ((α∗ ) −1 ) ∗ = ((α−1 ) ∗ ) −1 = 1 − α iff α = 1
2 ;
(iv) α = (α−1 ) ∗ = 1
1−α iff α2 − α + 1 = 0.
(v) α = α−1 iff e = 1 or 2.
When T (α) contains an element of A, i.e., D(α) is a translation hyperoval,
we know what the complete collineation group of D(α) is. So now we
suppose that α ∈ I but T (α) contains no element of A. Since each field automorphism
clearly induces a collineation of D(α), it suffices just to consider
the homography stabilizer of D(α). So let G be the complete homography
stabilizer of D(α). We may assume that G does not have an orbit of length
q + 2 (since there is only one possible case here with q = 16, and this case
will be dealt with later). So suppose first that G has an orbit of length q + 1.
Without loss of generality we may assume (by replacing α with some other
member of T (α) if necessary) that the single point orbit is {(0, 1, 0)}.
First suppose that every homography in G that moves (1, 0, 0) also moves
(0, 0, 1), and vice versa. Since the homographies (x, y, z) ↦→ (x, ay, a α z) for
0 = a ∈ Fq fix (1, 0, 0) and (0, 0, 1) and move the q − 1 other points in a
single orbit, clearly each of the points (1, 0, 0), (0, 0, 1) uniquely determines
the other. Then the points of Ω(α) must be divided into pairs which are sets
of imprimivity. This is impossible, since q + 1 is odd. By interchanging α
and α ∗ if necessary we may assume that there is a θ ∈ G that fixes (0, 1, 0)
and (0, 0, 1) and moves (1, 0, 0) to (1, 1, 1). Hence the matrix representing θ
is of the form
⎛
A = ⎝
1 1 1
0 b 0
0 0 c
⎞
⎠ , 0 = b, c ∈ Fq.
Hence θ : (1, t, t α ) ↦→ (1, 1 + bt, 1 + ct α ), forcing (1 + bt) α = 1 + ct α . By
Lemma 4.12.5 α must be additive and hence an automorphism. In this case
we know the complete stabilizer of D(α), and it does not have an orbit of
length q + 1 if it is not regular.
Now suppose that G has an orbit of length q. By interchanging α with
some member of T (α) if necessary, we suppose that the orbit of length q
consists of the points of the form (1, t, t α ), t ∈ Fq. If the points (0, 1, 0) and
(0, 0, 1 each form singleton orbits, then the argument above shows that α
must be additive. So suppose the θ ∈ G that moves (1, 0, 0) to (1, 1, 1) also

216 CHAPTER 4. THE UBIQUITOUS OVAL
interchanges (0, 1, 0) and (0, 0, 1). In this case the matrix of θ has the form
⎛
A = ⎝
1 1 1
0 0 b
0 c 0
⎞
⎠ , 0 = b, c ∈ Fq.
Hence θ : (1, t, t α ) ↦→ (1, 1 + ct α , 1 + bt), forcing (1 + ct α ) α = 1 + bt. So again
α must be additive and this time e ≤ 2. We have now finished a proof of the
following:
Theorem 4.12.7. If α ∈ I but no member of T (α) is in A, then the complete
homography group G of D(α) has an orbit of length q−1 which we may assume
to be {(1, t, t α ) : 0 = t ∈ Fq}.
It now follows that G has one orbit of length q − 1 and the remaining
orbits are subsets of S = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. If G has an orbit of
length 2 in its action on S, it is easy to show that some member of T (α) is
additive. If G has an orbit of length 3 on S, we may assume there is a θ ∈ G
for which θ : (1, 0, 0) ↦→ (0, 0, 1) ↦→ (0, 1, 0) ↦→ (1, 0, 0). In this case it is easy
to show that α2 ⎛ − α ⎞+
1 = 0. In this case we see that the homography with
0 0 1
matrix ⎝ 1 0 0 ⎠ really is in G. This completes a proof of the following:
0 1 0
Theorem 4.12.8. If α ∈ I but no element of T (α) is in A, then the complete
homography stabilizer of D(α) has order q − 1 with one orbit of length q − 1
and three orbits of length 1, unless α 2 −α+1 = 0, in which case |G| = 3(q−1)
and there is also one orbit of length 3.
Only two examples of monomial o-polynomials x α are known to exist with
α 2 − α + 1 = 0. The first is α = 6 with q = 32, which we have already seen
does give an oval. The other is α = 20 with q = 128. In this case, if we use
the notation of Theorem 4.12.9, σ = 16 and γ = 4, so σ + γ = 20. It seems
likely that there are no other examples.
So far we we have given only one family of monomial hyperovals other
than the translation hyperovals, i.e., those containing the Segre oval D(6)
with q = 2 e . At this stage the remaining known ovals given by monomial opolynomials
are not so easy to establish. They were both given by D. Glynn
in 1983 and are as follows.

4.12. MONOMIAL HYPEROVALS 217
Theorem 4.12.9. (The Glynn Hyperovals) Let q = 22e+1 , so q is not a
square and there are automorphisms σ and γ of Fq for which σ2 ≡ γ4 ≡
2 (mod q − 1). Here σ : x ↦→ x2e+1, and γ : x ↦→ x2(e+1)/2 if e is odd
and x ↦→ x2(3e+2)/2 when e is even. Then xσ+γ and x3σ+4 polynomials.
are monomial o-
Proof. After we have developed Cherowitzo’s theory of α-flocks we give a
proof by Cherowitzo that these monomials of Glynn really give ovals.
Using a computer, D. Glynn in 1989 showed that the only hyperovals
with a monomial o-polynomial in the plane P G(2, 2 e ) with e ≤ 30 are those
given by the constructions we have already seen.
The following conjectures are also due to Glynn and are still wide open.
Conjecture 4.12.10. For P G(2, q), q = 2 2k , the only hyperovals with a
monomial o-polynomial are the translation hyperovals.
Conjecture 4.12.11. There are no more hyperovals with monomial o-polynomials.
To complete this section we need to consider just when two monomial
ovals are projectively equivalent.
Theorem 4.12.12. So let α and β be two monomial maps in O and suppose
that q = 2e ≥ 8 and that Ω(α) ∼ Ω(β).
(i) If D(α) is regular, then so is D(β). In this case α = β ∈ {2, 1,
−1}. 2
(ii) If D(α) is an irregular translation hyperoval, then so is D(β). By
replacing α with some member of T (α) and similarly for β, we may assume
that the unique axis in each case is the line [1, 0, 0] T with points (0, 1, 0)
and (0, 0, 1). Then by replacing α with α−1 if necessary, and the same for
β, we may assume that (0, 1, 0) in D(α) is mapped to (0, 1, 0) in D(β). It
follows readily that α = β or α = β∗ . In general, using a knowledge of the
collineation stabilizer it is easy to check that if Ω(α) is a translation oval and
β ∈ T (α) with Ω(α) ∼ Ω(β), then β ∈ {α, α∗ }. (iii) So let D(α) and D(β)
contain no translation oval. Clearly the orbit of length q−1 in D(α) is mapped
to the orbit of length q − 1 in D(β) by any collineation taking the one oval to
the other. Using our knowledge of the complete collineation stabilizer of these
hyperovals and Lemma 4.12.6, it is now routine to check that β ∈ T (α), and
then Ω(α) ∼ Ω(β) if and only if β = α∗ , unles α2 − α + 1 = 0, in which case
α = (α∗ ) −1 = (α−1 ) ∗ , and the other three members of T (α) are also equal to
each other. Moreover, Ω(α) and Ω(α−1 ) are not projectively equivalent (since
q ≥ 8).

218 CHAPTER 4. THE UBIQUITOUS OVAL
4.13 The Trace Map
In this section we just begin to collect important computational facts about
the absolute trace function.
For q = 2 e we define the (absolute) trace map from Fq to F2 by
e−1
tr(x) = x 2i
=
i=0
σ∈Aut(Fq)
x σ , ∀ x ∈ Fq.
Theorem 4.13.1. These are the most basic results concerning the trace function.
1. The trace map is additive, i.e., tr(x+y) = tr(x)+tr(y) for all x, y ∈ Fq.
2. The trace map is invariant under automorphisms of Fq, i.e., tr(x σ ) =
tr(x) for all x ∈ Fq and for all σ ∈ Aut(Fq).
3. The kernel K of the trace map is a subgroup of (Fq, +) of index 2, and
tr is onto F2.
4. K σ = K for all σ ∈ Aut(Fq).
5. There are q − 1 subgroups of (Fq, +) of index 2, namely, aK for each
nonzero a ∈ Fq. In particular, if aK = bK, then a = b.
6. The only subgroup of Fq of index 2 which is invariant under each automorphism
of Fq is K.
7. For α ∈ Aut(Fq), the set {x + x α : x ∈ Fq} ⊆ K, with equality if and
only if α generates Aut(Fq).
8. For a, b ∈ Fq \ {0}, and α a generator of Aut(Fq), we have
{at α + bt : t ∈ Fq} =
b α
α−1
a 1
α−1
9. For a ∈ Fq, x 2 + x + a is irreducible over Fq if and only if tr(a) = 1.
K.

4.13. THE TRACE MAP 219
10 If α is a generator of Aut(Fq) and a, b, c are nonzero elements of Fq,
then
ax α + bx + c = 0 has zero or two solutions according as
tr
a 1
α−1 c
b α
= 1 or 0.
α−1
Proof. Set A = Aut(Fq). Then for x ∈ Fq, tr(x) =
α∈A xα .
(1) Since the trace is the sum of additive maps, the trace is itself additive.
(2) Since for α ∈ A we have A α = A, so for all x ∈ Fq, we have
tr(x α ) =
x αβ =
x β = tr(x).
β∈A
(3) Since tr(x), considered as an element of F2[x] has degree less than q,
it follows that tr: Fq → F2 is not the zero map. Thus, as trace is additive,
(3) follows.
(4) If x ∈ K, then by (2), x α ∈ K for α ∈ A.
(5) The subgroups of (Fq, +) of index 2 can be identified with the hyperplanes
of the projective geometry P G(e − 1, 2), and the multiplicative group
F ∗ q of Fq acts regularly on the points and hence also on the hyperplanes of
P G(e − 1, 2). Thus there are q − 1 such subgroups, each of which is of the
form aK for some a ∈ F ∗ q , and if aK = bK, then a = b. More generally,
β∈A
aK + c = bK + d ⇐⇒ a = b and d − c ∈ K.
(6) For α ∈ A and a ∈ F ∗ q , it follows that (aK)α = a α K α = a α K, so
(aK) α = aK if and only if a α = a, by (5). Hence (6) follows.
(7) For α ∈ A and x ∈ Fq it follows from (1) and (2) that tr(x +
x α ) =tr(x)+tr(x α ) = 0, so that {x+x α : x ∈ Fq} ⊆ K. The map x ↦→ x+x α
is additive with kernel equal to the fixed field of α, so the image of this map,
{x + x α : x ∈ Fq},
has size |K| = q/2 if and only if the fixed field of α is F2, that is, if and only
if α generates A.

220 CHAPTER 4. THE UBIQUITOUS OVAL
(8) Let α be a generator of A. Then we have
{at α + bt : t ∈ Fq} = a{t α + (a −1 b)t : t ∈ Fq
= a{((a −1 b) 1/(α−1) s) α + (a −1 b)(a −1 b) 1/(α−1) s : s ∈ Fq}
= a −1/(α−1) b α/(α−1) {s α + s : s ∈ Fq}
= a −1/(α−1) b α/(α−1) K by Part (7).
(9) Note that if a = b = 1 in (8), or use (7), since 2 is a generator of A, we
have
K = {t α + t : t ∈ Fq}.
Then for a ∈ Fq, x 2 + x + a is irreducible over Fq if and only if x 2 + x + a = 0
has no solutions in Fq if and only if a ∈ {x 2 + x : x ∈ Fq} = K if and only if
tr(a) = 1.
(10) ax α + bx + c = 0 ⇐⇒ x α + b c
x + = 0
a
a
α
x
⇐⇒
+
⇐⇒ tr
(b/a) 1
α−1
a 1
α−1 c
b α
α−1
= 0.
x
(b/a) 1
α−1
+
c
a · b
a
α
α−1
Lemma 4.13.2. Suppose that tr( q−1
i=1 aix i ) = 0 for all x ∈ Fq. Then for all
i with 1 ≤ i ≤ q − 1, we have
α∈A
a α
iα−1 = 0,
where the subscripts are considered mod q −1 and automorphisms are written
as powers of 2.
Proof. Now
tr
q−1
aix i
i=1
=
≡
q−1
tr(aix i q−1
) =
i=1
q−1
j=1
α∈A
a α
jα −1
i=1 α∈A
a α i x iα
x j (mod x q + x).
= 0

4.13. THE TRACE MAP 221
.
So tr( q−1
i=1 aix i ) = 0 for all x ∈ Fq if and only if
for all j with 1 ≤ j ≤ q − 1.
α∈A
a α
jα−1 = 0,
Keep in mind that each positive integer less than q = 2e has a unique
binary expansion
0≤i

222 CHAPTER 4. THE UBIQUITOUS OVAL
proving part (b). Finally, to verify part (c), consider
x α+1 y + xy α+1
(x + y) α+2
= xαy + xyα +
(x + y) α+1
x α/2 y + xy α/2
(x + y) (α+2)/2
By part (a), both terms on the right-hand side have trace 0.
2
.
4.14 Translation Ovals and the External Lines
Lemma
Throughout this section α is a generator of Aut(Fq), q = 2 e . Let D(α) be
the hyperoval
D(α) = {(1, t, t α ) : t ∈ Fq} ∪ {(0, 0, 1), (0, 1, 0)}.
Let G be the stabilizer of D(α) in P GL(3, q).
subgroups
We have seen that G has
⎧⎛
⎨ 1
D = ⎝ 0
⎩
0
0
a
0
0
0
aα ⎞
⎠ : a ∈ F ∗ ⎫ ⎧⎛
⎬ ⎨ 1
q , E = ⎝
⎭ ⎩
b bα 0
0
1
0
⎞ ⎫
⎬
0 ⎠ : b ∈ Fq
⎭
1
.
We saw that E is an elation group of order q with axis ℓ = [1, 0, 0]. The
following lemma has a routine proof that we leave as an exercise.
Lemma 4.14.1. The orbits of the group 〈D, E〉 on the points of P G(2, q)
are the following:
(i) ∆1 = {(0, 1, 0)} (the nucleus of D(α));
(ii) ∆2 = {(0, 0, 1)} (the point of D(α) on the axis ℓ);
(iii) ∆3 = {(0, 1, t) : t ∈ F ∗ q
(the other q − 1 points of ℓ);
(iv) ∆4 = {(1, t, t α ) : t ∈ Fq} (the other q points of D(α));
(v) ∆5 the remaining q 2 − q points.
We now examine the images of translation ovals under collineations of
P G(2, q).

4.14. TRANSLATION OVALS AND THE EXTERNAL LINES LEMMA223
Theorem 4.14.2. Suppose that O is a translation oval of P G(2, q) (q = 2 e )
with associated automorphism α and containing the point (0, 0, 1). Let ℓ be
the line ℓ = [1, 0, 0].
(i) Suppose that O has nucleus (0, 1, 0). Then O = Of for some opolynomial
f : Fq → Fq. If ℓ is an axis of O, then
f(t) = a + bt α
for some a, b ∈ Fq with b = 0; if ℓ is not an axis of O, then
for some a, b, c ∈ Fq with b = 0.
f(t) = a + b(t + c) α
α−1
(ii) Suppose that O has nucleus (0, 1, a) for some a ∈ F ∗ q , and that ℓ is an
axis of O. Then O = Of where
for some b, c ∈ Fq with b = 0.
f(t) = c + at + bt α
Proof. By Theorem 4.10.1 there is a matrix M ∈ GL(3, q) for which Oα·M =
O. Since M maps the nucleus (0, 1, 0) of Oα to the nucleus of O, we have
(0, 1, 0) · M = (0, e, 0) (the second row of M)
for some nonzero e in Fq. First, suppose that ℓ = [1, 0, 0] is an axis of O. If
O is a conic the stabilizer of O is transitive on tangent lines to O, so we may
assume that M fixes ℓ. On the other hand, if O is not a conic, then ℓ is the
unique axis of both Oα and of O, and again M fixes ℓ. Thus M fixes ℓ, and
since (0, 0, 1) is the pint of intersection of ℓ and Oα, and also of ℓ and O, it
follows that
(0, 0, 1)M = (0, 0, g), g = 0 (the third row of M).
Since M is invertible, the (1, 1) entry of M must be nonzero, so we may
assume that it is 1. This shows that for some d, h ∈ Fq we must have
⎛
1 d
⎞
h
M = ⎝ 0 e 0 ⎠ .
0 0 g

224 CHAPTER 4. THE UBIQUITOUS OVAL
Then M maps the point (1, t, t α ) to the point (1, d+et, h+gt α ). Set s = d+et,
so that t = e −1 (s + d) and h + gt α = h + g(e −1 d) α + ge −α s α . Then setting
a = h + g(e −1 d) α and b = ge −α we obtain
Oα · M = {(1, s, a + bs α ) : P s ∈ Fq} ∪ {(0, 0, 1)}.
Second, suppose that ℓ is not an axis of O.
(1, t, t
Then M maps the point
α ) of Oα to (0, 0, 1) for some t ∈ Fq. Set
⎛
t
L = ⎝
α t
⎞
1
0 1 0 ⎠ , so L
1 0 0
−1 ⎛
0 0 1
= ⎝ 0 1 0
1 t tα ⎞
⎠ .
So L−1 · M fixes both points (0,1,0) and (0,0,1). Hence, on multiplying L by
a scalar if necessary, we have
L −1 ⎛
1 c
⎞
a
· M = ⎝ 0 e 0 ⎠ ,
0 0 g
for some a, c, e, g ∈ Fq with e and g nonzero.
So
Oα · M = OαL(L −1 M)
= ({(s α + t α , s + t, 1) : s ∈ Fq} ∪ {(1, 0, 0)}) · L −1 M
= ({(u α , u, 1) : u ∈ Fq} ∪ {(1, 0, 0)}) · L −1 M
= ({(1, v, v α
α−1 ) : v ∈ Fq} ∪ {(0, 0, 1)} (on setting v = u 1−α )
= {(1, c + ev, a + gv α
α−1 ) : v ∈ Fq} ∪ {(0, 0, 1)}
= {(1, w, a + g((e −1 (w + c)) α
α−1 )) : w ∈ Fq} ∪ {(0, 0, 1)}.
Thus setting b = ge−α/(α−1) we have Oα · M = Of with
f(w) = a + b(w + c) α
α−1 . This completes the proof of part ⎛(i).
1 0
⎞
0
Part (ii) follows from part (i) on applying the matrix ⎝ 0 1 a ⎠.
0 0 1
⎛
1 0
⎞
b
We now review the results of Theorem 4.14.2. Put Ma,b = ⎝ 0 1 0 ⎠,
0
a = 0. Then
0 a

4.14. TRANSLATION OVALS AND THE EXTERNAL LINES LEMMA225
Ma,b : (1, t, t α ) ↦→ (1, t, at α + b);
(0, 0, 1) ↦→ (0, 0, a) ≡ (0, 0, 1);
(0, 1, 0) ↦→ (0, 1, 0).
With a natural extension of the usual notation we see that Ma,b maps Oα
to Oat α +b. This gives all translation ovals equivalent to Oα with axis [1, 0, 0]
containing (0, 0, 1) and having nucleus (0, 1, 0).
Corollary 4.14.3. The line [c, v, 1] through (0, 1, v), v = 0, is a secant to
Oatα +b if and only if c ∈ vα∗ 1
· a ( α) ∗
· K + b.
Proof. To see this recall that in general atα + bt + c = 0 has a solution t if
and only if tα +
b c t + a a = 0 has a solution (now recall that α∗ = α/(α − 1)
and divide by
b α∗ 1
) if and only if c ∈ a ( α) ∗
· bα∗ · K. Then [c, v, 1] has a
a
point (1, t, atα + b) if and only if c + vt + atα + b = 0 has a solution, from
which the desired result follows immediately.
This immediately gives us the following also.
Corollary 4.14.4. By fixing v and a, and varying b ∈ Fq we get a family of
q hyperovals in two classes. In one class (of size q/2) all the ovals have the
same secants through (0, 1, v). Each point of the line [c, v, 1] is on a unique
oval in one of the classes, with each oval in that class meeting [c, v, 1] in two
points. If O1 and O2 are in opposite classes, they are compatible at (0, 1, v).
Of O1 and O2 are compatible at (0, 1, v) and O1 and O3 are compatible at
(0, 1, v), then O2 and O3 have the same secants through (0, 1, v) and the same
external lines.
Now for a, b, v ∈ Fq with a = 0 put
⎛
1 0 b
T(a,b,v) = ⎝ 0 1 v
0 0 a
takes Of with nucleus (0, 1, 0) and containing (0, 0, 1) to an equivalent oval
Oaf(t)+vt+b with nucleus (0, 1, v) and containing (0, 0, 1).
Start with: [c, w, 1] on (0, 1, w), w = 0, is secant to Oα if and only if
c ∈ wα∗ · K. Map by the homography with matrix T(a,b,v) to obtain
⎞
⎠

226 CHAPTER 4. THE UBIQUITOUS OVAL
[ac + b, aw + v, 1] on (0, 1, aw + v) is secant to Oatα +vt+b iff c ∈ w α∗
· K.
Put x = ac + b, y = aw + v. Then we have (for y = v) that
[x, y, 1] on (0, 1, y) is secant to Oatα +vt+b iff x ∈ (y +v) α∗
1
a ( α) ∗
K +b. (4.75)
Here Oat α +vt+b (for a, v, b ∈ Fq with a = 0) gives all ovals equivalent to
Oα containing (0, 0, 1), having nucleus on [1, 0, 0] and having [1, 0, 0] as axis.
We generalize this one step further. For a = 0, v = w, put
⎛
1 b
⎞
bw
M(a,b,v,w) = ⎝ 0 1 v ⎠ .
0 a aw
Then the homography with matrix M(a,b,v,w) maps points as follows:
M(a,b,v,w) : (1, t, t α ) ↦→ 1, at α + t + b, w(at α + b) + vt);
(0, 1, 0) ↦→ (0, 1, v); (4.76)
(0, 0, 1) ↦→ (0, 1, w).
This gives all ovals equivalent to Oα with [1, 0, 0] as axis, nucleus (0, 1, v),
and containing (0, 1, w).
Finally, for a, b, c ∈ Fq with a = 0 put
⎛
0
M = ⎝ 0
1
0
1
b
⎞
a
0 ⎠ ,
c
so M −1 = 1
⎛
⎝
a
c ab a
0 a 0
1 0 0
Corollary 4.14.5. Then the homography with matrix M maps
(1, t, t α ) ↦→ (t α , bt α + t, ct α + a).
If t = 0, the image point is (1, s, a(s + b) α∗ + c). If t = 0 this says (1, 0, 0) ↦→
(0, 0, 1). The line [1, 0, 0] maps to the new axis [c, 0, 1] and (0, 1, 0) to the same
nucleus (0, 1, 0). This gives all ovals equivalent to Oα containing (0, 0, 1),
having nucleus (0, 1, 0) and having [c.0, 1] as axis.
⎞
⎠ .

4.14. TRANSLATION OVALS AND THE EXTERNAL LINES LEMMA227
We emphasize the following:
Theorem 4.14.6. [c, v, 1] on (0, 1, v) is secant to Otα +b if and only if
c ∈ vα∗ · K + b. Also, Otα and Otα +b are compatible at (0, 1, v) if and only if
b/vα∗ ∈ K if and only if tr b/vα∗ = 1. This implies there are q/2 points of
compatibility on [1, 0, 0] and where they are incompatible they share secants
completely.
Suppose we are given a point N of P G(2, q) with q even as usual, and a
set S of q/2 lines on N, and a point M distinct from N and not on any of
the lines of S. We would like to know which ovals have M as nucleus and S
as their set of external lines on N. As we shall see later, a good answer to
this question would be helpful in determining all ovoids in P G(3, q). For the
present we give a partial answer for translation ovals. This result is known
as The External Lines Lemma and is a key to later results characterizing
ovoids having plane sections that are translation ovals.
Theorem 4.14.7. (The External Lines Lemma of Penttila and Praeger) Let
O be a translation oval of P G(2, q), q = 2 e , with nucleus N, and let ℓ be
an axis of O. Let P be the point of O on ℓ and let Q be another point of ℓ
distinct from P and N. Suppose that O ′ is a translation oval containing P
such that its nucleus N ′ is a point on ℓ distinct from Q. If every line on Q
external to O is also external to O ′ , then ℓ is an axis of O ′ .
Proof. We may assume that O is Oα, that the nucleus N is (0, 1, 0), that ℓ
is [1, 0, 0], and hence that P = (0, 0, 1). Similarly, there must be a matrix
M taking O ′ to Oβ, where α and β are generators of the Galois group of Fq.
Also, since by Lemma 4.14.1 the q − 1 points of ℓ distinct from P and N
form one orbit of the stabilizer in GL(3, q) of Oα, we may assume that Q is
(0, 1, 1). If the image QM of Q lies on ℓ, then as M maps N ′ (the nucleus
of O ′ ) to N = (0, 1, 0) (the nucleus of Oβ), it follows that M maps the line
ℓ = 〈N ′ , Q〉 to the line 〈N, QM〉 = ℓ, i.e., ℓM = ℓ. Thus if QM lies on ℓ,
then as ℓ is an axis of Oβ, ℓ must be an axis of O ′ .
Hence we may assume that QM is not on ℓ. Since by Lemma 4.14.1 the
set of all points not on ℓ or on Oβ is one orbit of the stabilizer of Oβ in
GL(3, q), we may assume that QM is (1, 1, 0).
A line on Q = (0, 1, 1) different from [1, 0, 0] is of the form ℓa = [a, 1, 1].
Note that ℓa contains a point (1, t, t α ) of Oα if and only if a = t + t α . So ℓa
is an external line to Oα if and only if a ∈ K, where K is the kernel of the

228 CHAPTER 4. THE UBIQUITOUS OVAL
trace map. Similarly, a line on QM = (1, 1, 0) is of the form mb = [1, 1, b].
This line contains a point (1, t, t β of Oβ (where clearly t = 0) if and only if
1 + t + bt β = 0, hence a line on QM is external to Oβ if and only if it is is of
the form
mb = [1, 1, b], with b ∈ X = {t −β + t 1−β : t ∈ Fq}.
By hypothesis, the lines on Q external to O are external to O ′ . Hence
{ℓa : a ∈ K} · M = {mb : b ∈ X}.
As N ′ is a point on ℓ distinct from P and Q, we have N ′ = (0, 1, c) for
some c = 1. Moreover, M maps N ′ to N = (0, 1, 0), maps Q = (0, 1, 1) to
QM = (1, 1, 0) and maps P = (0, 0, 1) = ℓ ∩ O ′ to ℓM ∩ Oβ = (1, 0, 0) (since
ℓM is the line on N = N ′ M and QM). Since m31 = 0 we may assume that
m31 = 1. Then (0, 1, c)M = λ(0, 1, 0), and all of this forces M to look like
⎛
M = ⎝
m11 m12 m13
c m22 0
1 0 0
⎞
⎠ , and M −1 ⎛
⎜
= ⎜
⎝
0 0 1
0
1
m13
1
m22
m12
m13m22
c
m22
m11m22+cm12
m13m22
for some mij ∈ Fq with m22 = c+1. If we replace m22 with c+1 and simplify
we see that M maps ℓa to mb where b = a+m11+m12 . Set d = m13
1 and and
m13
e = (m11 + m12)/m13, noting that m13 is nonzero since M is nonsingular.
Then M maps ℓa to mb where b = da + e. Hence X = dK + e. Since the
line m0 contains P and is therefore not external, it must be that 0 ∈ X
and d−1e ∈ K. Then X = dK + e says that X is a coset of an additive
group, and since X contains 0, it must be that X = dK. By definition
X = {t−β + t1−β : t ∈ Fq} = {sβ∗ + s : s ∈ Fq}, which is clearly invariant
under Aut(Fq), and hence by Theorem 4.13.1 we have X = K. We now
want to use Lemma 4.13.2 to see that β/(β − 1) is an automorphism, so
that from Lemma 4.13.3 it must be that β = 2. Thus O ′ is a conic and
all tangent lines are axes. In particular ℓ is an axis of O ′ . So suppose that
tr(xk + x) = 0 for all s ∈ Fq. Since xk + x = q−1 i=1 aixi with ai = 1 for i = k
and i = 1. By Lemma 4.13.2
jα−1 =
α 1jα = 0. Put j = 1 to
α∈Aut(Fq) 1α
get
α 1α = 0, forcing k to be an automorphism. By Lemma 4.13.3 β∗ = k
is an automorphism if and only if β = 2.
⎞
⎟
⎠ ,

4.15. OVAL DERIVATION 229
4.15 Oval Derivation
The main idea of this section is to start with one Desarguesian affine plane of
order q and to replace most of its lines (as sets of points) with certain q-arcs
to obtain what appears to be a new affine plane but which is isomorphic to
the original.
Let α : Fq → Fq be an arbitrary permutation of the elements of Fq with
0α = 0. Let P = {(1, y, z) : y, z ∈ Fq}, so P is a set of q2 elements that
are going to play the role of points in the planes we are about to construct.
Given the permutation α, define a point-line geometry Aα as follows. The
points of Aα are just the elements of P. The lines are to be described as sets
of q points each. They are of two classes:
(a) For arbitrary c, m ∈ Fq let [c, m, 1]α be the set of points (1, y, z) for
which
⎡
(1, y α , z) ⎣
c
m
1
⎤
⎦ = c + my α + z = 0.
So [c, m, 1]α contains the points (1, y, c + my α ), for all y ∈ Fq.
(b) For arbitrary c ∈ Fq, let [c, 1, 0]α be the set of points (1, y, z) for which
(1, y α ⎡ ⎤
c
, z) ⎣ 1 ⎦ = c + y
0
α = 0.
The line [c, 1, 0]α contains the points (1, cα−1, z), or in other words, [cα , 1, 0]α
contains the points (1, c, z) for all z ∈ Fq. It is now a straight forward exercise
to show that with these points and lines Aα is an affine plane of order q. For
fixed m ∈ Fq, the lines of the form [c, m, 1]α constitute a parallel class, and
the lines of the form [cα , 1, 0]α form a parallel class. Now let α and β both be
arbitrary permutations of the elements of Fq with 0α = 0β = 0, so we have
two planes Aα and Aβ.
Define a map φα,β on the points of P by
φα,β : (1, y, z) ↦→ (1, y α◦β−1
, z) for all y, z ∈ Fq. (4.77)
It is clear that φα,β is a permutation of the points of Aα, and the typical
point (1, y, c + myα ) of [c, m, 1]α is mapped to the point (1, yα◦β−1, c + myα )
of [c, m, 1]β. Similarly, the typical point (1, cα−1, z) of [c, 1, 0]α is mapped to

230 CHAPTER 4. THE UBIQUITOUS OVAL
the point (1, cβ−1, z) of [c, 1, 0]β. It follows that φα,β is a permutation of the
points of P mapping the lines of Aα to the lines of Aβ. This means it is an
isomorphism from Aα to Aβ. If α = 1 is the identity mapping on Fq, then A1
is just the usual coordinatization of the classical Desarguesian affine plane
AG(2, q) whose projective completion to the projective plane P G(2, q) has
the usual line [1, 0, 0] as line at infinity. This means that all the planes just
constructed are isomorphic to the usual Desarguesian plane AG(2, q).
We now consider what the lines of Aβ look like as sets of points in Aα for
distinct permutations α and β with 0α = 0β = 0.
First, for fixed c ∈ Fq, the points (1, c, z), z ∈ Fq, constitute the points of
the line [c α , 1, 0]α of the plane Aα (which may be considered to pass through
the infinite point (0, 0, 1) of the projective completion of Aα). But they also
constitute the points of the line [c β , 1, 0]β of Aβ (through the point (0, 0, 1)
of the projective completion of Aβ). Similarly, if m = 0 and c ∈ Fq is fixed,
the points (1, y, c), y ∈ Fq, of the line [c, 0, 1]α are also the points of the line
[c, 0, 1]β, all of which may be considered to pass through the infinite point
(0, 1, 0). So the two parallel classes of lines just considered are both parallel
classes of lines in each of the affine planes Aα and Aβ. Moreover, we have
the following for arbitrary c, c1, m ∈ Fq with m = 0:
[c α 1 , 1, 0]α ∩ [c, m, 1]β = {(1, c1, c + mc β
1 )}; (4.78)
[c1, 0, 1]α ∩ [c, m, 1]β = {(1,
β−1 c + c1
, c1)}. (4.79)
m
This shows that the β-lines [c, m, 1]β with m = 0 meet the α-lines through
the infinite points (0, 1, 0) and (0, 0, 1) in exactly one point. We now want
to determine under what conditions the β-lines [c2, m2, 1]β meet the α-lines
[c1, m1, 1]α in at most two points. So suppose that y1, y2 and y3 are distinct
elements of Fq and that m1 and m2 are nonzero elements of Fq. Then
(1, yi, zi) ∈ [c1, m1, 1]α ∩ [c2, m2, 1]β holds if and only if
zi = c1 + m1y α i = c2 + m2y β
i
for i = 1, 2, 3.
A little rearranging shows that this holds if and only if
m2
m1
= yα 1 + yα 2
y β
1 + y β =
2
yα 1 + yα 3
y β
1 + y β
3
.

4.15. OVAL DERIVATION 231
From this it is clear that each β-line [c2, m2, 1]β is a q-arc in Aα, and dually,
each α-line [c1, m1, 1]α is a q-arc in Aβ if and only if
β −1 ◦ α is an o-permutation if and only if α −1 ◦ β is an o-permutation.
Moreover, we see that the affine plane Aβ is derived from the plane Aα
by the following procedure. The lines [c, 1, 0]α through the infinite point
(0, 0, 1) (as sets of points) remain as lines but are re-coordinatized as [c, 1, 0]β.
Similarly, the lines [c, 0, 1]α through the infinite point (0, 1, 0) (as sets of
points) remain as lines but are re-coordinatized as [c, 0, 1]β. Finally, for
m = 0, the line [c, m, 1]α is replaced (as a point set) by the q-arc [c, m, 1]β.
Moreover, the q-arc [c, m, 1]β together with the infinite point (0, 0, 1) is an
oval in the projective completion of Aα having nucleus (0, 1, 0). The function
φα,β maps the line [c, m, 1]α to the oval [c, m, 1]β (as a point set) which is a
β-line. It is clear that the roles of α and β may be interchanged.
In the plane Aα we now give the point (1, y, z) coordinates P (1, y α , z).
Then P (1, y α , z) is on the q-arc [c, m, 1]β if and only if (1, y, z) ∈ [c, m, 1]β
if and only if z = c + my β = c + m(y α ) α−1 ◦β , so in the usual coordinates
the q-arc [c, m, 1]β is the q-arc {(1, t, c + mt α−1 ◦β ) : t ∈ Fq}, which completes
to a unique hyperoval which may be viewed as containing the infinite points
(0, 0, 1) and (0, 1, 0).
This last observation is worth stating as a theorem.
Theorem 4.15.1. Let α and β be permutations of the elements of Fq with
0 α = 0 β = 0. Suppose that α −1 ◦ β (and hence also β −1 ◦ α) is an opermutation
of Fq. Then the q-arc [c, m, 1]β of Aα completes to an oval
projectively equivalent to O α −1 ◦β in the projective completion of Aα. Similarly,
the q-arc [c, m, 1]α of Aβ completes to an oval projectively equivalent
to Oβ −1 ◦α in the projective completion of Aβ. In particular, if α = 1, so Aα
is just the usual representation of the Desarguesian plane A(2, q), the q-arc
[c, m, 1]β of A(2, q) completes to an oval projectively equivalent to Oβ, and
the q-arc [c, m, 1]1 of Aβ completes to an oval isomorphic to O β −1.
We now review, from a slightly different point of view, the process used
to replace the plane Aα with the plane Aβ. View Aα as the classical affine
plane A(2, q) by giving the point (1, y, z) coordinates P (1, y α , z) so that it
is naturally completed to the projective plane P G(2, q). We start with the
o-permutation γ = α −1 ◦ β and the oval {(1, t, t γ ) : t ∈ Fq} ∪ {(0, 0, 1)} with

232 CHAPTER 4. THE UBIQUITOUS OVAL
nucleus N = (0, 1, 0). This is the oval Oγ : {(x, y, z) : y γ = x γ−1 z. For each
pair (c, m) ∈ Fq × Fq with m = 0 the homography
⎛
(x, y, z) ↦→ (x, y, z) ⎝
1 0 c
0 1 0
0 0 m
maps the oval Oγ : yγ = xγ−1z of P G(2, q) to the oval Oγ(c, m) : yγ =
xγ−1 ( 1 c z + m mx). These ovals are all isomorphic to Oγ, all contain the point
Q = (0, 0, 1) and all have nucleus N = (0, 1, 0). Let the line L = 〈Q, N〉 be
the line at infinity to obtain the affine plane A(2, q). If we delete the point
Q from each of these ovals, we obtain point sets which are q(q − 1) lines of
Aβ, with the other two parallel classes of lines being the lines different from
L through the two points Q and N. The construction of the plane Aβ from
Aα is called α −1 ◦ β-derivation with respect to (Q, N). The lines of Aβ form
the α −1 ◦ β-derivation net of Aα with respect to (Q, N). Since the parallel
class of lines of Aα with point at infinity Q is also a parallel class of lines
of Aβ with point at infinity Q, the notation Q will be used to denote both
parallel classes. Similarly, the parallel classes with point at infinity N will
both be denoted N .
It is clear that if we start with the plane Aβ and perform β −1 ◦α-derivation
we obtain the plane Aα back again. Note that if α = 1, Aα is just the usual
classical affine plane A(2, q).
It is important for us to review the preceding material under the assumption
that β = 1. In this case the affine plane Aβ = A1 is just the usual
classical affine plane A(2, q). The map
φα,β = φα : (1, y, z) ↦→ (1, y α , z) for all y, z, ∈ Fq
maps the typical point (1, y, c + my α ) of [c, m, 1]α (which completes to an
oval equivalent to Oα) to the point (1, y α , c + my α ) of the (ordinary line)
[c, m, 1]1. Similarly the typical point (1, cα−1, z) of [c, 1, 0]α is mapped to the
point (1, c, z) of the line [c, 1, 0]1. So φα maps the lines of Aα (ovals of A(2, q))
to lines of A(2, q).
The q-arc [c, m, 1]α of A(2, q) completes to the oval Oα in the projective
completion of A(2, q). The construction of the affine plane A(2, q) from
Aα is α−1-derivation and the construction of Aα from A(2, q) using ovals
isomorphic to Oα is α-derivation.
⎞
⎠

4.16. TRANSLATION OVALS AGAIN 233
4.16 Translation Ovals Again
As usual in these notes, q = 2e , Fq is the Galois field with q elements and α
is a generator of the Galois group of Fq. Say α : a ↦→ a2i for a fixed integer i
with 0 ≤ i ≤ e − 1 and (i, e) = 1. Let Oα be the translation oval defined by
Oα = {(1, t, t α ) ∈ P G(2, q) : t ∈ Fq} ∪ {(0, 0, 1) (4.80)
= {(x0, x1, x2) ∈ P G(2, q) : x1 α = x0 α−1 x2}.
This oval has nucleus N = (0, 1, 0), and we let Dα denote the corresponding
hyperoval
Dα = Oα ∪ {(0, 1, 0)}.
If α ∈ {2, 2 −1 }, then the complete oval stabilizer in this case is given by
T = {T(σ,a,b) : σ ∈ Aut(Fq); a, b ∈ Fq, b = 0}, where
⎛
T(σ,a,b) : (x, y, z) ↦→ (x σ , y σ , z σ ) ⎝
1 a a α
0 b 0
0 0 b α
⎞
⎠ . (4.81)
When b = 1 we get a subgroup of T consisting of elations of P G(2, q) fixing
the oval and having axis [1, 0, 0] and center (0, a1−α , 1) (when a = 0) on the
lines [c, aα−1 , 1]. The homography with matrix
⎛
⎝
m 1
α−1 0 cm 1
α−1
0 1 0
0 0 1
maps the oval Oα to the oval
⎞
⎠ with inverse
⎛
⎝
m 1
1−α 0 c
0 1 0
0 0 1
[c, m, 1]α = {(1, t, c + mt α ) : t ∈ Fq} ∪ {(0, 0, 1)} (4.82)
with nucleus N = (0, 1, 0). This oval is also a translation oval with axis
[1, 0, 0] and with homography stabilizer (complete when α ∈ {2, 2 −1 }) given
by the collineations with matrices
⎛
⎝
1 am 1
1−α c + cb α + m 1
1−α a α
0 b 0
0 0 b α
⎞
⎞
⎠
⎠ , with a, b ∈ Fq, b = 0. (4.83)

234 CHAPTER 4. THE UBIQUITOUS OVAL
Again putting b = 1 gives the elations with [1, 0, 0] as axis. Note that if
α = 2 the translation axis is unique. Also, using the new notation for these
translation ovals, we have
Oα = [0, 1, 1]α.
The collineations given above still exist when α = 2, but in this case there
are additional collineations stabilizing the oval Oα.
Define the map φα on the points of the affine plane AG(2, q) consisting
of the points (1, y, z) not on the line [1, 0, 0] by
φα : (1, y, z) ↦→ (1, y α , z). (4.84)
It is clear that φα is a bijection mapping the point (1, y, c + my α ) of
[c, m, 1]α, (m = 0), to the point (1, y α , c + my α ) of the line [c, m, 1]. It
also maps the points (1, c, z) of [c, 1, 0] to the points (1, c α , z) of [c α , 1, 0] =
[c α , 1, 0]α and the points (1, y, c) of [c, 0, 1] to the points (1, y α , c)] also on
[c, 0, 1].
Theorem 4.16.1. Let A = A(2, q) be the affine plane whose projective completion
is P G(2, q) with L as line at infinity. Define another geometry as
follows. Let α be a generator of the Galois group Aut(Fq). Let (Q, N) be an
ordered pair of distinct points of L. Define Aα to be the point-line incidence
structure whose points are just the points of A, and whose lines are of the
following three types:
(a) translation ovals isomorphic to Oα containing the point Q, with nucleus
N, and with the line 〈Q, N〉 as translation axis;
(b) the lines of A with point at infinity Q;
(c) the lines of A with point at infinity N.
Incidence is defined to be just that induced by incidence in A. Then Aα
is isomorphic to A, i.e., it is an affine plane isomorphic to A(2, q).
Proof. Without loss of generality we may choose coordinates so that the line
at infinity is L = [1, 0, 0]. As the group of homographies of P G(2, q) is
doubly transitive on the points of L, we may asume that Q = (0, 0, 1) and
N = (0, 1, 0). Then clearly the map φα defined above is an isomorphism from
Aα to A.

4.16. TRANSLATION OVALS AGAIN 235
The construction of the plane Aα from AG(2, q) is called α-derivation
with respect to (Q, N). The lines of Aα form the α-derivation net of A with
respect to (Q, N). Since the parallel class of lines of A with point at infinity
Q is also a parallel class of lines of Aα with point at infinity Q, the notation
Q will be used to denote both parallel classes. Similarly, the parallel classes
with point at infinity N will both be denoted N .
We note in passing that if we start with the plane Aα as a canonical
plane isomorphic to AG(2, q) and perform α −1 -derivation we obtain the plane
AG(2, q) back again. All this is clearly just a special case of the general
treatment in the preceding section.

236 CHAPTER 4. THE UBIQUITOUS OVAL

Chapter 5
Quadratic Forms
This chapter is devoted to the study of quadratic forms and their associated
symmetric bilinear polar forms. In an appendix (Chapter 21) the more general
theory of sesquilinear forms is reviewed, and several of the results given
here on quadratic forms are included there within the general setting. However,
many results on quadratic forms in this chapter have no counterpart
in the appendix. For much of this book the reader familiar with the present
chapter will rarely need to consult the appendix, but it is available.
5.1 Basic Definitions and Results
Let K be a field and let Vn+1 be the (n + 1)-dimensional vector space of
(n + 1)-tuples (row vectors), whose elements will be represented either as
x = (x0, x1, . . . , xn) or as x = (xi), xi ∈ K. Two nonzero vectors (xi),
(yi) are said to be equivalent provided there is a nonzero λ in K such that
λyi = xi, i = 0, . . . , n. The projective n-space Pn(K) = P G(n, K) consists
of the equivalence classes containing the nonzero vectors of Vn+1. Usually,
it is clear from the context whether (xi) is representing a vector in Vn+1 or
a point in P G(n, K), so we use the same notation for both. For that rare
occasion on which we need separate symbols for the two concepts we will let
the reader know what notation we are using when the time arrives.
A t-dimensional subspace of Pn(K) is to consist of all the classes of points
containing vectors in some (t + 1)-dimensional subspace of Vn+1. In terms of
the vector space structure of Vn+1, a point of Pn(K) can be identified with
a 1-dimensional subspace of Vn+1. Lines correspond to 2-dimensional vector
237

238 CHAPTER 5. QUADRATIC FORMS
subspaces, and hyperplanes of Pn(K) correspond to n-dimensional subspaces
of Vn+1. An m-dimensional subspace of Pn(K) is a subspace isomorphic to
P G(m, q). Such a subspace is the set of points x = (xi) satisfying Ax T = 0,
where x T is the transpose of x and A is an (n − m) × (n + 1) matrix over
K of rank n − m. In particular, a hyperplane is the locus of points whose
coordinates satisfy an equation of the form
a0x0 + a1x1 + · · · anxn = 0,
where at least one ai is not zero.
An algebraic hypersurface in Pn(K) is the locus in Pn(K) of a homogeneous
polynomial equation f(x0, . . . , xn) = 0 with coefficients in K. It seems
consistent with this definition that (0, 0, . . . , 0) not be considered a point of
Pn(K): such a point would be on all hypersurfaces. Concerning the homogeneity
of the equation, we note that the coordinates of a point in Pn(K)
are determined only up to a proportionality factor, and it would be desirable
for all coordinates of a given point on the surface to satisfy the equation.
This actually forces the homogeneity of the equation provided that the field
is large enough.
A quadric (or hyperquadric) in Pn(K) is defined in various ways. Since
our main interest in projective spaces will be the combinatorial properties of
sets of points defined in terms of quadric surfaces, we begin with a study of
the equivalence of three alternative definitions. It should cause no confusion
if we let x = (xi) denote a vector or a point as is convenient.
Defn. 1 A quadric (or quadric surface) in Pn(K) is an algebraic hypersurface
given by an homogeneous polynomial equation of degree 2.
Defn. 2 A quadric (or quadric surface) in Pn(K) is the locus of points
x satisfying xAx T = 0, where A is a square upper triangular matrix over K
with n + 1 rows.
Defn. 3 Let q be a mapping from Vn+1 into K with the properties that:
(i) q(ax) = a 2 q(x), a ∈ K, x ∈ Vn+1;
(ii) The mapping (x, y) ↦→ B(x, y) = q(x+y)−q(x)−q(y) is a (symmetric)
bilinear form B on Vn+1 × Vn+1.
Then q is called a quadratic form on Vn+1, and the set of points x in
Pn(K) such that q(x) = 0 is called a quadric Qn (or Qn(K)) or hyperquadric.
And the bilinear form B is the associated polar form of q or Qn. Sometimes
we write fq for the polar form of the quadratic form q.
Theorem 5.1.1. The three definitions of quadric surface are equivalent.

5.1. BASIC DEFINITIONS AND RESULTS 239
Proof. Let Qn be a quadric according to Defn.1 given by the polynomial
equation
f(x0, . . . xn) =
aijxixj = 0.
0≤i,j≤n
i≤j
Then let A be the upper triangular matrix A = (aij), where aij is as given
in f(xo, . . . , xn) = xAx T , and the equivalence of the first two definitions is
easily established.
If A is an upper triangular matrix, then defining q(x) = xAx T yields a
quadratic form with
B(x, y) = q(x + y) − q(x) − q(y) = x(A + A T )y T
serving as the associated bilinear form. This is easy to check. For the converse,
let q be a quadratic form as in Defn.3, with B(x, y) as defined there.
For any vectors α1, . . . , αr, we have
q(α1 + · · · + αr) = q(α1 + · · · + αr−1) + q(αr) + B(α1 + · · · αr−1, αr)
r−1
= q(α1 + · · · + αr−1) + q(αr) + B(αi, αr).
By a suitable induction step this becomes
q(
r
αi) =
i=1
r
q(αi) +
i=1
1≤i

240 CHAPTER 5. QUADRATIC FORMS
So suppose that A is an upper triangular matrix corresponding to the
quadratic form q and the quadric surface Q. Also B will denote either the
symmetric bilinear form associated with q or the symmetric matrix B =
A + A T , since B(x, y) = xBy T for vectors x, y. If K has characteristic
2, then B(x, x) = xAx T + xA T x T = 2xAx T = 0 for each x. If K has
characteristic different from 2, then xBx T = 0 if and only if 2xAx T = 0, i.e.,
iff q(x) = 0 iff B(x, x) = 0. In this case xBx T = 0 could have been taken as
the defining equation of a quadric surface Q, and ˜q(x) = B(x, x) = xBx T as
the quadratic form associated with the symmetric matrix B. This is probably
the customary procedure when K is not allowed to have characteristic 2.
However, we will consistently use the definitions given earlier.
In algebraic geometry a point x is said to be a singular point of Q provided
Bx T = 0 and x is on Q. If K has characteristic different from 2, then Bx T = 0
implies xAx T = 0 and x is on Q automatically. If x is a point of Q such that
Bx T = 0, then x is called a simple point of Q. We retain this meaning for
simple and singular point, but also any point x of Q such that Bx T = 0 will
be called a nonsimple point of Q. On the other hand, a source of confusion
is the fact that a vector v with q(v) = 0 is called a singular vector.
Points x and y are said to be conjugate with respect to Q (or q or B)
provided xBy T = 0. So if K has characteristic different from 2, Q is just
the set of self-conjugate points. However, if K has characteristic 2, all points
are self-conjugate. The polar (conjugate, or orthogonal) space τ(x) of a point
x is the set of all points conjugate to x. So the polar of x is the subspace
determined by the equation x(A+A T )y T = 0. More generally, the polar space
τ(Σp) of a p-dimensional subspace Σp is the set of all points that are conjugate
to each point of Σp. Two subspaces are said to be mutually conjugate provided
each point of one is conjugate to each point of the other. If x is a point of Q,
the polar space τ(x) of x is called the tangent space of Q at the point x. The
dimension of the tangent space depends on the rank of (the matrix) B and on
the point x, in particular on whether or not Bx T = 0. If B is a nonsingular
matrix, the tangent space of Q at any point of Q is a hyperplane. If B is
singular, the tangent space of Q at a point x of Q is a hyperplane if x is a
simple point of Q, and is the whole space otherwise.
Q is said to be degenerate if it has a nonsimple (i.e., singular) point.
Otherwise it is nondegenerate. Thus if K has characteristic not equal to 2,
Q is nondegenerate iff B is a nonsingular matrix.

5.1. BASIC DEFINITIONS AND RESULTS 241
Exercise 5.1.1.1. Show that if K has characteristic equal to 2, then Q can
be nondegenerate even if B is a singular matrix.
Suppose that Q is given by f(x0, . . . , xn) = ¯xAx T = 0 where
f(x0, . . . , xn) =
i≤j
aijxixj.
Lemma 5.1.2. (i) Points y and z are conjugate (w.r.t. Q) iff
n
∂f
yi = 0.
∂xi
i=0
(ii) z is a nonsimple point of Q iff
z
∂f
∂xi
z
= 0 for each i (and z ∈ Q if
K has characteristic 2).
(iii) If g is a homogeneous polynomial of degree m in t variables x1, . . . , xt,
then mg = t i=1
Proof.
∂g
∂xi xi.
∂f
∂xi
= 2aiixi +
aijxj +
ij
ajixj.
n
∂f
yi = 0.
∂xi
Clearly the set of points conjugate to z forms a hyperplane iff
for some i. This proves (i) and (ii). To prove (iii), note that since the partial
derivative operator is additive, it suffices to establish the result in the special
case that g is a monomial of the form ax ei 1
1 xei 2
2 · · · x ei t
t for some scalar a and
nonnegative integer exponents ei that sum to m. This is straightforward.
A set of points in Pn(K) will be called independent (respectively, dependent)
provided it corresponds to some set of linearly independent (respectively,
dependent) vectors in Vn+1. Define analogously the subspace spanned
(or determined) by a set of points of Pn(K).
z
∂f
∂xi
z
= 0

242 CHAPTER 5. QUADRATIC FORMS
Lemma 5.1.3. (i) If a point x is conjugate to each point in some set of
points of Pn(K), then x is in the polar space of the subspace spanned by
those points.
(ii) The polar space of a p-dimensional subspace Σp is the intersection of
the polar spaces of each point in a set of p + 1 independent points of Σp.
(iii) Let α0, . . . , αp be independent points on a quadric Q. Then the pdimensional
subspace Σp determined by these points is contained in Q iff
these points are pairwise conjugate.
(iv) If three distinct points of Q lie on a line, then each point of the line
lies on Q and any two points of the line are conjugate.
(v) Let x be any nonsimple point of Q and y any other point of Q. Then
the line containing x and y lies in Q.
(vi) If x ∈ Q, y ∈ Q, L = 〈x, y〉 ⊆ y ⊥ , then L ∩ Q = {x}.
Proof. A routine exercise, but the reader should do all the parts.
A quadric Q in Pn(K) with a unique nonsimple point x is called a cone
. For example, if K has characteristic not 2, it is easy to see that Q is a
cone iff B has rank n. By part (v) of the preceding lemma this terminology
should seem reasonable.
Let V1, V2 be vector spaces over the field K equipped with quadratic
forms q1, q2, respectively. Let T : V1 → V2 be an invertible semilinear map
with companion automorphism σ. Then T is a semisimilarity from (V1, q1)
to (V2, q2) provided there is a nonzero c ∈ K such that
q2(T (v)) = c · (q1(v)) σ ∀ v ∈ V1.
T is a similarity iff σ = 1;
T is an isometry iff σ = 1 and c = 1. .
NOTE: If g : V1 → V2 is a semisimilarity/similarity/isometry then g
induces a map ¯g : P V1 → P V2 which is called a projective semisimilarity/similarity/isometry.
Let S = (V, q) be an orthogonal space with q a quadratic form (with polar
form fq).
ΓS = group of semisimilarities of V ;
GS = group of similarities of V ;
S = group of isometries of V.

5.2. STRUCTURE OF QUADRICS 243
P ΓS, P GS, P S are the corresponding projective groups.
Let Z(S) be the group of form-preserving scalar maps. Then
P ΓS ∼ = Γ/Z(S);
P GS ∼ = GS/Z(S);
P S ∼ = S/Z(S).
Let f be a symmetric bilinear form on V (for example, the polar form of
a quadratic form).
v ∈ V is isotropic iff f(v, v) = 0.
W ≤ V is totally isotropic iff f(v, w) = 0 ∀ (v, w) ∈ W × W.
5.2 Structure of Quadrics
Throughout most of this book we have used q to denote the order of a finite
field. This leads to confusion in this chapter since we so often want q to
denote a quadratic form. Hence throughout this chapter we use t to
denote the size of the finite field (i.e., t = p e for some prime p), and
q will denote a quadratic form.
Lemma 5.2.1. Let E be a line spanned by vectors x and y, where x is on
a nondegenerate quadric surface Q in E. Then there is a vector z on E ∩ Q
such that B(x, z) = 1.
Proof. Since q(x) = 0, B(x, x) = 0, and the polar τ(x) is spanned by x. Thus
B(x, y) = 0, and we may solve 0 = q(ax + y) = q(y) + aB(x, y) for a. Then
solve B(x, b(ax + y)) = bB(x, y) = 1 for b and put z = b(ax + y).
Such a line is called a hyperbolic line (w.r.t q or Q or B). Thus there
is no ambiguity in defining a hyperbolic space to be the orthogonal direct
sum of hyperbolic lines, where this means the hyperbolic lines are mutually
conjugate with respect to a quadratic form on the entire space which when
restricted to the individual lines gives their quadratic forms.
Note: This Lemma implies that if E is a line with a nondegenerate
quadratic form q defining a quadratic surface Q, and if E contains one point
.

244 CHAPTER 5. QUADRATIC FORMS
of Q, then it contains precisely two points of Q. So suppose that E is
anisotropic, i.e., no point of E is on Q. If the matrix A giving the quadratic
a b
form is A = , we are assuming that ax
0 c
2 + bxy + cy2 = 0 has no
solutions. If the characteristic p is odd, having Q be nondegenerate is the
same as having B = A + A T be nonsingular. So suppose that t = 2 e . Then
0 b
B = . So B is nonsingular if and only if b = 0. And if b = 0, then
b 0
clearly E is not anisotropic. Hence if E is an anisotropic line, with t odd
or even, for each point x on E we have x⊥ is a single point of E. This is
equivalent in this case to saying E ∩ E⊥ = ∅.
Lemma 5.2.2. Any two orthogonal hyperbolic lines are isometric.
Proof. Let (V, Q) be an orthogonal hyperbolic line. So V = 〈u, v〉, Q(u) =
Q(v) = 0 and fQ(u, v) = fQ(v, u) = 1. Similarly, V ′ = 〈u ′ , v ′ 〉, Q ′ (u ′ ) = 0,
. . . , etc. Define T : V → V ′ by T (u) = u ′ , T (v) = v ′ . Then
Q(au + bv) = Q(au) + Q(bv) + fQ(au, bv)
= a 2 Q(u) + b 2 Q(v) + ab · fQ(u, v)
= 0 + 0 + ab.
Similarly, Q ′ (T (au + bv)) = Q ′ (au ′ + bv ′ ) = ab = Q(au + bv), implying that
T is an isometry.
Lemma 5.2.1 is the first step in the proof of the following theorem.
Theorem 5.2.3. Let Q be a nondegenerate quadric in Vn+1. Let N be an
r-dimensional subspace lying in Q, r ≥ 1. Then there is an r-dimensional
subspace P of Q such that N and P are disjoint and such that N ⊕ P is
disjoint from its polar space. If x1, . . . , xr form a basis for N and if P has the
properties just described, there is a basis y 1, . . . , y r of P such that B(xi, y j) =
δij, 1 ≤ i, j ≤ r. If N is a maximal subspace of Q, then q(x) = 0 for all
nonzero x in the polar space of N ⊕ P .
Proof. The proof proceeds by a finite induction. We first construct y1, then
y2, etc.
Since Q is nondegenerate and (x1, . . . , xr) is a basis for N ⊂ Q, we know
that (x1B, . . . , xrB) is linearly independent. Then N ′ = span(x2B, . . . , xrB)
has dimension r−1, so its polar space has dimension (n+1)−(r−1) = n−r+2.

5.2. STRUCTURE OF QUADRICS 245
The polar space of N has dimension n − r + 1. Hence there is a vector
y ∈ τ(N ′ ) \ τ(N), i.e., B(xi, y) = 0 for 2 ≤ i ≤ r and B(x1, y) = 0. We may
multiply y by a scalar so that without loss of generality we may assume that
y has the properties
(i) B(xi, y) = 0 for 2 ≤ i ≤ r, and
(ii) B(x1, y) = 1.
At this stage we do not know whether or not y ∈ Q. If we apply Lemma 5.2.1
to E = 〈x1, y〉 we can find y 1 ∈ E with y 1 ∈ Q and B(x1, y 1) = 1. (It is a
simple exercise to check that the restriction of the quadratic form q to E is
nonsingular.) Clearly B(xi, y 1) = 0 for 2 ≤ i ≤ r.
If r = 1, we are done. So suppose 1 ≤ p < r and that we have obtained
y 1, . . . , y p such that
(i) B(xi, y j) = δij for 1 ≤ i ≤ r, 1 ≤ j ≤ p, and
(ii) (y 1, . . . , y p) is independent and spans a subspace of Q.
As above we know that (x1B, . . . , xrB) is independent. Let
N ′ = span(x1, . . . , xp, xp+2, . . . , xr),
so the polar space τ(N ′ ) has dimension (n+1)−(r−1) = n−r+2 while τ(N)
has dimension only n − r + 1. Thus there is a vector y such that B(y, xi) = 0
for i = p + 1, B(y, xp+1) = 0, and we may assume that B(y, xp+1) = 1.
Furthermore, since N is contained in its own polar space, if y ′ = y + x
with x ∈ N, then B(xi, y ′ ) = B(xi, y + x) = 0 + 0 if i = p + 1. And
B(xp+1, y ′ ) = B(xp+1, y) + B(xp+1, x) = 1 + 0 = 1.
Put
y ′ p
= y − B(y, yi)xi. So y ′ has the properties just mentioned, and for 1 ≤ j ≤ p,
B(y ′ p
, yj) = B(y − B(y, yi)xi, yj) = B(y, yj) − B(y, yj) = 0.
i=1
i=1
Now put y p+1 = y ′ − q(y ′ )xp+1. Then y p+1 has the properties just mentioned
and
q(y p+1) = q(y ′ ) + [q(y ′ )] 2 q(xp+1) + B(y ′ , −q(y ′ )xp+1
= q(y ′ ) + 0 − q(y ′ ) = 0.

246 CHAPTER 5. QUADRATIC FORMS
Since B(xp+1, y p+1) = 1 and B(xp+1, y j) = 0 for 1 ≤ j ≤ p, it is easy to
show that (y 1, . . . , y p+1) is independent. So we have vectors y 1, . . . , y p+1 that
satisfy the following:
(i) B(xi, y j) = δij for 1 ≤ i ≤ r, 1 ≤ j ≤ p + 1, and
(ii) (y 1, . . . , y p+1) is independent and spans a subspace of Q.
Proceeding in this fashion we complete the selection of vectors y 1, . . . , y r
satisfying
(i) B(xi, y j) = δij for 1 ≤ i, j ≤ r, and
(ii) (y 1, . . . , y r) is independent and spans a subspace P of Q.
Let R be the polar of N ⊕P . Then it is clear that R and N ⊕P are disjoint,
so by a dimension argument, Vn+1 = R ⊕ (N ⊕ P ). If z ∈ R, and if q(z) =
0, z = 0, then B(z, x) = 0 for x ∈ N, so q(x +z) = q(x)+q(z)+B(x, z) = 0.
Thus N ⊕ Kz is a subspace of Q properly containing N. Hence if N is a
maximal subspace lying in Q there is no such point z, i.e., q(z) = 0 for all
nonzero z ∈ R. This completes a proof of the theorem.
If r is the maximum dimension of any subspace lying in a nondegenerate
quadric Q in Vn+1, we say that Q has Witt index r. Also, a subspace is called
anisotropic provided it contains no singular vector.
If we were being strictly logical here the next Corollary would not appear
until after we have proved Witt’s theorem which appears in the book as
Theorem 5.6.2. However, it is convenient to have it here, and we are careful
not to use this corollary in the proof of Witt’s theorem.
Corollary 5.2.4. Any non-degenerate quadric of Witt index r is an orthogonal
direct sum of r hyperbolic lines and an anisotropic space A, where A is
determined up to isometry. A is called the germ of the space.
Proof. Let Q be a nondegenerate quadric in Vn+1 with Witt index r. Let
N and P be r-dimensional subspaces of Q as described in Theorem 5.2.3
with bases x1, . . . , xr and y 1, . . . , y r satisfying B(xi, y j) = δij, 1 ≤ i, j ≤ r.
Put A = τ(N ⊕ P ). Then A is anisotropic by Theorem 5.2.3. If we put
Hi = 〈xi, yi〉, then Hi is a hyperbolic line, and Vn+1 is the orthogonal direct
sum H1 ⊥ H2 ⊥ . . . ⊥ Hr ⊥ A.

5.2. STRUCTURE OF QUADRICS 247
To show that A is determined up to isometry, suppose V = H1 ⊥ . . . ⊥
Hr ⊥ A = H ′ 1 ⊥ . . . ⊥ H ′ r ⊥ A ′ . By Lemma 5.2.2 any two hyperbolic lines
are isometric, so clearly U = H1 ⊥ . . . ⊥ Hr and W = H ′ 1 ⊥ . . . ⊥ H′ r are
isometric. By Witt’s Theorem 5.6.2 there exists an isometry g of V with
g(H1 ⊥ . . . ⊥ Hr) = H ′ 1 ⊥ . . . ⊥ H ′ r. But A = τ(H1 ⊥ . . . ⊥ Hr) and
A ′ = τ(H ′ 1 ⊥ . . . ⊥ H′ r ). Then
g(A) = g({v ∈ V : B(u, v) = 0 ∀ u ∈ U})
= {gv ∈ V : B(gu, gv) = 0 ∀ gu ∈ W }
= τ(W ) = A ′ .
Lemma 5.2.5. Let Q be a nondegenerate quadric in Vn+1. We want to know
how large the germ of Q could be.
(i) The (vector space) dimension of an anisotropic orthogonal space over
Ft is at most 2.
(ii) The 2-dimensional anisotropic orthogonal spaces over Ft are all isometric.
(iii) When t is even, the 1-dimensional anisotropic orthogonal spaces over
Ft are all isometric.
(iv) When t is odd, the 1-dimensional anisotropic orthogonal spaces over
Ft are all similar, and they fall into two isometry classes.
Proof. The proof of (i) follows from the fact that any quadratic polynomial
in three (or more) variables has a nontrivial zero. This fact is actually proved
within the proof of Theorem 5.4.2, where the number of points on each type
of quadric is determined. It also follows easily from Theorem 20.23.1.
For the proof of (ii), let L be an anisotropic line, so L∩Q = {0} = L∩L ⊥ .
For each x ∈ L, x ⊥ ∩ L is one point, say {y}. When t is odd, x = y. When
t = 2 e , x = y. But in either case B(x, y) = 0.
Let t be odd. Then q(ax + by) = a 2 q(x) + b 2 q(y) = 1 has a solution
(a, b) ∈ Ft × Ft. So we may assume that L = 〈x, y〉 with q(x) = 1 and
B(x, y) = 0. Then B(x, y + λx) = λB(x, x) = 1 if and only if λ = B(x, x) −1 ,
which is possible since we know B(x, x) = 0. For this λ replace y with y +λx
so that we have
L = 〈x, y〉 with q(x) = 1, B(x, y) = 1.

248 CHAPTER 5. QUADRATIC FORMS
Next let q = 2 e . So B(x, x) = 0 for all x ∈ L and x ⊥ = {x}. Say L = 〈x, y〉
with q(x) = λ = u 2 = 0. Then q(x/u) = 1. So we may suppose q(x) = 1.
Then B(x, λy) = λB(x, y) = 1 if and only if λ = B(x, y) −1 , which is welldefined
since we know B(x, y) = 0. So again we have
L = 〈x, y〉 with q(x) = 1, B(x, y) = 1.
Suppose that q(y) = a = 0 since L is anisotropic. For the same reason
0 = q(λx + y) = λ 2 + λ + a for each λ ∈ Ft.
If t is odd this implies that 1 − 4a = ❆ ∈ Ft. If t = 2 e , then tr(a) = 1.
Now let
L ′ = 〈z, w〉, with q ′ (z) = 1, B ′ (z, w) = 1, q ′ (w) = b,
where 0 = q(λx + y) = λ 2 + λ + b for each λ ∈ Ft.
For t odd this implies that (1−4a)(1−4b) = ∈ Ft. For t = 2 e it implies
that tr(a + b) = 0. It follows that for any characteristic we may choose an
s ∈ Ft for which
s 2 (1 − 4a) + s(1 − 4a) + b − a = 0, and then put t = 2s + 1.
Note that: s 2 + s + b = a(4s 2 + 4s + 1) = at 2 .
Define a linear map T by
T : (L, q) → (L ′ , q ′ ) : λx + µy ↦→ (λ + s µ
µ)z +
t t w.
Then q(λx + µy) = λ2 + λµ + µ 2a, and
q ′ (T (λx + µy)) = q ′
(λ + s µ
µ)z +
t t w
= (λ + s
t µ)2 + (λ + s
t µ)µ
t + µ2b t2 = λ 2 + 2λµ s
t + s2 µ 2 λµ
+
t2 t + µ2s t2 + µ2b t2 = λ 2
2s + 1
+ λµ + µ
t
2
2 s + s + b
t2
= λ 2 + λµ + µ 2 a.

5.3. CHANGE OF BASIS 249
It follows that q ′ (T (v)) = q(v) for each v ∈ L. This completes the proof
of part (ii).
For parts (iii) and (iv) let E = 〈x〉 with a given quadratic form q for
which q(x) = a ∈ Ft. Then q is nondegenerate if and only if a = 0, in which
case E is anisotropic. If T : E → E is the linear map T : x ↦→ bx with
b = 0, then q(T (v)) = b 2 q(v) for each v ∈ E. If t = 2 e , we can put b = a −1/2
to obtain q(T (x)) = 1, and all anisotropic points are isometric. If t is odd,
then all anisotropic points are similar, but two of them, say q1(x) = a and
q2(x) = b, are isometric if and only if a/b = ∈ Ft.
Corollary 5.2.6. Any non-degenerate orthogonal space of Witt index n over
Ft is isometric to
(i) The orthogonal direct sum of n hyperbolic lines (sometimes denoted
O + (2n, t) or by Q + (2n − 1, t)),
or
(ii) The orthogonal direct sum of n hyperbolic lines and a 1-dimensional
anisotropic space of discriminant a square/non-square (denoted O(2n + 1, t)
or Q(2n, t)). These spaces fall into two isometry classes and one similarity
class for t odd, and one isometry class for t even),
or
(iii) The orthogonal direct sum of n hyperbolic lines and a 2-dimensional
anisotropic space (denoted O − (2n + 2, t) or Q − (2n + 1, t)).
5.3 Change of Basis
Let B1 = {v0, . . . , vn} be any ordered basis for Vn+1. If w ∈ Vn+1 satisfies
w = n j=0 cjvj, cj ∈ K, we write [w]B1 = (c1, c2, . . . , cn) and call [w]B1 the
coordinate matrix of w with respect to the ordered basis B1. Moreover, we
often write B1 as a column ‘vector’ B1 = [v0, . . . , vn] T and write
w = [w]B · B. (5.1)
If B2 = [u0, . . . , un] T is a second ordered basis for Vn+1 with ui = n j=1 pijvj,
for 1 ≤ i ≤ n, we put P = (pij) and write B2 = P B1. If T : Vn+1 → Vn+1
is a linear operator, let [T ]Bi
denote the n × n matrix for which [T (w)]Bi =
[w]Bi · [T ]Bi for all w ∈ Vn+1, i = 1, 2. Putting w equal to the kth member of
the basis Bi, we see that the kth row of [T ]Bi is the coordinate matrix of the

250 CHAPTER 5. QUADRATIC FORMS
image under T of the kth member of Bi. Using this notation we immediately
obtain the following theorem.
Theorem 5.3.1. The following are equivalent:
(i) B2 = P B1;
(ii) [w]B2 = [w]B1 · P −1 ;
(iii) [T ]B2 = P [T ]B1P −1 .
Now let q be a quadratic form on Vn+1, and let B(x, y) = q(x+y)−q(x)−
q(y) be its associated polar form. Given a basis B1 = {v0, . . . , vn} of Vn+1,
define the Gram matrix B = (Bij) of the form B by Bij = B(vi, vj). For
vectors x = [x]B1·B1 and y = [y]B1·B1, we have B(x, y) = B( civi,
djvj) =
i,j ciBijdj = [x]B1B[y] T B1 (where [x]B1 = (c0, . . . , cn) and [y]B1 = (d0, . . . , dn)).
If B2 is a second ordered basis as above with B2 = P · B1, then B(x, y) =
[x]B1 · B · [y] T B1 = [x]B1 · P −1 · P BP T · (P T ) −1 [y] T B1 = [x]B2 · P BP T · [y] T B2
follows that P BP T is the matrix that is the Gram matrix of B with respect
to the basis B2.
Similarly, q(x) = [x]B1A[x] ′ B1 for the matrix A = (Aij) defined by
⎧
⎨
aij =
⎩
B(vi, vj), i < j;
q(vj), i = j;
0, i > j.
Here the matrix B is the sum B = A + AT .
Now suppose that B2 is a second basis as above, so that [x]B1 = [x]B2 · P .
Then q(x) = [x]B1A[x] T B1 = [x]B2 · P · A · P T [x] T B2 for all x ∈ Vn+1. Hence the
upper triangular matrix A is replaced by P AP T when B1 is replaced by B2 =
P · B1. However, the matrix P AP T may not be upper triangular. Suppose
P AP T = A∗ and P BP T = B∗ where B = A + AT and B∗ = A ∗ +(A∗) T .
However, B∗ is symmetric and B∗ii = (P AP T + P AT P T )ii = 2(P AP T )ii.
So we can write B∗ = Ā + ĀT where Ā is upper triangular and has diagonal
equal to that of P AP T . Now Āij = (P AP T )ij + (P A T P T )ij for i < j,
Āii = (P AP T )ii, and Āij = 0 for i > j. If we use the standard basis so
we can identify a row vector with its coordinate matrix, it follows that for
any row vector y, y ĀyT = yP AP T y T . Hence xAx T = 0 iff yP AP T y T = 0
iff yP = x where xAx T = 0. The mapping x ↦→ xP −1 permutes the points
of Pn(K) in a one-to-one fashion preserving containment relations among
subspaces. Furthermore, we have just seen that it takes points of the quadric
Q given by xAx T = 0 to the point of the quadric ¯ Q given by y ĀyT = 0 where
. It

5.3. CHANGE OF BASIS 251
Ā is derived in the natural fashion from B∗ = P BP T . The points x and y
are conjugate w.r.t. Q iff xP −1 and yP −1 are conjugate w.r.t. ¯ Q. All of this
can be expressed for a general basis.
Exercise 5.3.1.1. Using the notation of the preceding paragraph, show that
x is a singular point of Q iff xP −1 is a singular point of ¯ Q. Hence Q is
nondegenerate iff ¯ Q is.
Two matrices A and B are congruent provided there is an invertible matrix
P for which B = P AP T . One consequence of the above discussion is that
two symmetric matrices B1 and B2 (with zeros on the main diagonal in case
K has characteristic 2) are congruent iff they represent the same quadratic
forms w.r.t. some pair of bases. Hence given a quadratic form on Vn+1, we
need to specify a basis of Vn+1 before the matrices B and A are determined.
Until now we just always used the standard basis.
Theorem 5.3.2. If K has characteristic different from 2, then Vn+1 has a
basis α0, . . . , αn such that B(αi, αj) = 0 for i = j.
Proof. Use induction on n. If n = 0, there is nothing to prove. Assume the
theorem is true for n = m − 1. If q(x) = 0 for all x there is nothing to
prove. So assume x is a vector such that q(x) = 0. Then the polar space
of x has dimension n or n + 1. Since the characteristic of K is not 2 and
q(x) = 0, it must be that B(x, x) = 0, so x ∈ τ(x). Thus Vn+1 = Kx ⊕ τ(x),
and n = dim (τ(x)). By the induction hypothesis there is a suitable basis
x0, . . . , xn−1 for τ(x). Putting x = xn gives the desired basis.
As a consequence of this theorem we see that if K does not have characteristic
2 there is a basis for which the matrix A associated with q is
diag(a00, a11, . . . , ann), and the polynomial f(x) has the form f(x) = a00x 2 0 +
a11x 2 1 + · · · + annx 2 n.
We now give an alternative characterization of degeneracy. Let q be a
quadratic form on Vn+1 with Q and B as usual. Let A be a matrix assoiated
with Q by a basis B1.
Theorem 5.3.3. The following are equivalent:
(i) q (or Q, or B) is degenerate, i.e., Q has a singular point.
(ii) There is a basis B2 of Vn+1 such that if A ∗ is the upper triangular
matrix of q with respect to B2, then the right-hand column of A ∗ contains
only zeros.

252 CHAPTER 5. QUADRATIC FORMS
(iii) The polynomial of q associated with A can be transformed by a nonsingular
matrix D into one with fewer variables.
(iv) There is a nonsingular matrix D mapping the points of Q into those
of a quadric ¯ Q which is degenerate by the original definition.
Proof. If we show that (ii) is equivalent to the original definition, the equivalence
of the other characterizations should be clear.
If the bottom row and right hand column of B∗ = A∗ +A∗T contains only
zeros, and if the characteristic of K is not 2, then the right hand column of
A∗ contains only zeros. On the other hand, if K has characteristic 2, then
all we can say is that the right hand column of A∗ contains only zeros off
the main diagonal. Thus the characterization (ii) of degeneracy must be in
terms of A∗ , not just of A or of B∗ .
Suppose that [x]B2 = (0, . . . , 0, 1) and that q is degenerate in the sense
of (ii). Then q(x) = [x]B2A∗ [x] T B2 = 0, and [x]B2B ∗ = [x]B2(A∗ + (AT ) ∗ ) = 0.
On the other hand, if C is the nonsingular matrix transforming B2 into the
standard basis, then 0 = [x]B2B ∗ = xCT B∗ = xCT B∗C, where CT B∗C is the
symmetric matrix of the bilinear form of q w.r.t. the standard basis. Thus
x is a nonsimple (i.e., singular) point of Q. Let B1 be the standard basis,
so 0 = xB = x(A + AT ) and 0 = xAxT . (Note that 0 represents a zero
matrix of whatever size is needed.) Let D be any nonsingular matrix with
bottom row x = [x]B1. Let B∗ = DBDT , and let A∗ be the upper triangular
matrix such that A∗ + (A∗ ) T = B∗ . for i = n, the (i, n) entry of A∗ is
(DADT )in + (DAT DT )in =
j,k (DijAjkD T kn + DijAT jkDT
kn ) = jk (Dij(Ajk +
AT jk )xk) =
j Dij((A+A T )xT )jn = 0. The (n, n) entry of A∗ is (DADT )nn =
xAxT = 0.
Theorem 5.3.4. The intersection ¯ Q of a nondegenerate quadric Q in Pn(K)
with some hyperplane Pn−1(K) is either a cone or a nondegenerate quadric.
Proof. It is more convenient to work in Vn+1. So let Q be a nondegenerate
quadric in Vn+1, with B = {α0, . . . , αn} a basis of Vn+1 such that ¯ B =
{α0, . . . , αn−1} is a basis of the space Vn corresponding to Pn−1(K). Let A and
B = A + AT be the matrices for Q relative to B. Finally, suppose that x and
y are linearly independent vectors representing singular points of ¯ Q = Q∩Vn.
If Ā is the matrix obtained by deleting the bottom row and right hand column
of A, and ¯ B = Ā + ĀT , then Ā and ¯ B are the matrices of ¯ Q associated with
the basis ¯ B. Let [x]B = (x0, . . . , xn−1, 0) and [y]B = (y0, . . . , yn−1, 0). Then for

5.4. CANONICAL FORMS OVER FINITE FIELDS 253
T Ā z
d1 and d2 in K not both zero, 0 = d1x + d2y. In fact, if A =
and
0 c
if we write (abusing the notation!) ¯x = (x0, . . . , xn−1), ¯y = (y0, . . . , yn−1),
then
(d1x + d2y) Ā(d1x + d2y) T = d 2 1¯x Ā¯xT + d 2 2¯y Ā¯yT +
d1d2¯x Ā¯yT + d1d2¯y Ā¯xT = 0,
so d1x+d2y ∈ Q∩Vn = ¯ Q. Furthermore, B(d1x+d2y) T
¯B T z
=
z 2c
(d1x+
d2y) = (0, . . . , 0, d1z¯x T +d2z¯y T ) T by the hypothesis that x and y are singular
points of ¯ Q. Thus any choice of d1, d2 in K, not both zero, such that d1z¯x T +
d2z¯y T = 0 determines a singular point of Q, contradicting the hypothesis that
Q is nondegenerate. Since ¯ Q does not have distinct singular points (points
of ¯ Q conjugate to all points of ¯ Q), it must be a cone or nondegenerate.
Exercise 5.3.4.1. Show that both possibilities mentioned in the previous
theorem actually occur:
(i) If Q is a nondegenerate quadric in P G(n, t) and x ∈ Q, then Q ∩ x ⊥
is a cone over a nondegenerate quadric in P G(n − 2, t) of the same type as
Q.
(ii) If Q is a nondegenerate quadric in P G(n, t) and x ∈ Q, then Q ∩ x ⊥
is nondegenerate in P G(n − 1, t). (Show that all three possibilities do occur.)
5.4 Canonical Forms over Finite Fields
From now on we assume that the field K has the property that any
quadric surface in a vector space of dimension at least 3 over K
has a nonzero vector. This property certainly holds for any algebraically
closed field K, and later in this section we establish the fact that it holds for
all finite fields.
Theorem 5.4.1. Let Q be a nondegenerate quadric surface (as a set of vectors
only) in Vn+1 with n ≥ 2. Then the matrix A determining Q with respect
to some basis B is determined up to a scalar factor, so that conjugacy with
respect to Q is completely determined.
Proof. Let x0 ∈ Q. Since τ(x0) is some n-dimensional subspace of Vn+1 and
x0 is clearly a nonsinple vector of Q ∩ τ(x0), that intersection is a cone with

254 CHAPTER 5. QUADRATIC FORMS
vertex x0. In particular this says that if x0 and y0 are linearly independent
vectors of Q then τ(x0) ∩ Q = τ(y0) ∩ Q. Furthermore, the points of the
quadric determine which lines lie in the quadric and thus which points in
the quadric are conjugate to each other. If A1 and A2 are any two upper
triangular matrices determining Q relative to the same basis, we may assume
that A1 and A2 have a specific form (by changing basis if necessary). For we
know there are subspaces N and P of maximum possible dimension r in Q
with bases {x1, . . . , xr} and {y1, . . . , yr}, respectively, such that xi and yj are
conjugate iff i = j. We claim that the polar τ(N ⊕P ) is uniquely determined.
This would have to be the intersection R of the polars τ(xi), τ(yi), 1 ≤ i ≤ r,
if we knew that each of these polars was uniquely determined. But this follows
from the following two easily checked results:
1. If 0 = x ∈ Q, y ∈ Q, then the line containing x and y has no points
different from x on Q if x and y are conjugate.
2. If 0 = x ∈ Q, y ∈ Q, then the line containing x and y has precisely
one point different from x and lying on Q if x and y are not conjugate.
Since each of xi, yi, 1 ≤ i ≤ r, lies on Q, by checking the tangents and
secants to Q through each of these point their polars may be determined.
Basically, the core of the proof is divided into three cases corresponding
to whether R = τ(N ⊕P ) has dimension 0, 1 or 2. For all three cases we have
arranged it so that B1(xi, yj) = δijci, B1(xi, xj) = B1(yi, yj) = B2(xi, xj) =
B2(yi, yj) = 0, B2(xi, yj) = δijdi, for some scalars ci, di = 0.
Case (i) R = {0}. Then the two associated polynomials are
f1(x) =
cixiyi; f2(x) =
dixiyi.
i
The assumption is that these two polynomials have the same zeros and
that r i=1 cidi = 0. Necessarily r ≥ 2 in this case, since n + 1 ≥ 3. If we
set xi = yi = 0 for i > 2, then the assumption that c1x1y1 + c2x2y2 = 0 and
d1x1y1 + d2x2y2 = 0 have the same zeros implies that c1 c2 = . By similar
d1 d2
computations with x1, y1 and xj, yj, for j = 3, . . . , r, we see that ci is a di
constant for i = 1, . . . , r. Thus A1 is some nonzero multiple of A2.
Case(ii) Let R be spanned by the single nonzero vector z1. Here f1(x) =
c1x1y1+· · ·+crxryr+cz 2 1 , f2(x) = d1x1y1+· · ·+drxryr+dz 2 1 and cd r 1 cidi =
0.
For each i, 1 ≤ i ≤ r, our basic assumption inplies that cixiyi + cz2 1 = 0
= 0 have the same solutions. Since these are quadratic
and dixiyi + dz 2 1
i

5.4. CANONICAL FORMS OVER FINITE FIELDS 255
forms on vector spaces of dimension 3, there will be nontrivial solutions by
hypothesis. In fact z1 = 1, xi = 1 and yi = − c
is a solution to the first.
ci
Since it must be a solution of the second, we have c d = ci di . Again A1 is some
nonzero scalar multiple of A2.
Case (III) Let R be spanned by the independent vectors z1, z2. Since no
point of R is in Q,
f1(x) =
f2(x) =
r
1
r
1
cixiyi + cr+1z 2 1 + cr+2z 2 2 + cz1z2;
dixiyi + dr+1z 2 1 + dr+2z 2 2
+ dz1z2;
where cr+1, cr+2, dr+1, dr+2, at least, are nonzero. Considering in turn
solutions with z1 = z2 = 0, then z1 = 0, then z2 = 0, we have from the
preceding cases that ci
di is a constant for 1 ≤ i ≤ r + 2. Then put xi = yi = 0
for 2 ≤ i ≤ r, and z1 = z2 = 1 = x1. Then y1 = −(cr+1+cr+2+c)
gives a solution
x1
of the first equation, which being a solution of the second, implies d1c = c1d.
So c = d = 0 or c1
d1
= c
d .
As observed earlier, any algebraically closed field K has the property that
any quadric surface in a vector space of dimension at least 3 over K has a
nonzero vector, and the next theorem establishes the property for finite fields.
From now on suppose that K = GF (t), t = p e where p is a prime. Since
any degenerate quadric has nonzero vectors by definition, we only consider
nondegenerate ones.
Theorem 5.4.2. A nondegenerate quadric surface Q in Pn(K) has exactly
t + 1 points if n = 2, and Q is nonempty if n > 2, in case K = GF (t), p
prime and t = p e .
Proof. Case (i) p > 2. By Theorem 5.3.2 there is a basis of Vn+1 for which
the associated polynomial has the form
f(x) = a00x 2 0 + · · · + annx 2 n.
It is clear that any quadric which has a polynomial in this form will be
nondegenerate if and only if n i=0 aii = 0 (odd characteristic or not). If
n > 2, put xj = 0 for j > 2. Then we need to show that there are t + 1
distinct points [xi], 0 ≤ i ≤ 2, satisfying a00x2 0 + a11x2 1 + a22x2 2 = 0. At

256 CHAPTER 5. QUADRATIC FORMS
least one of a00 a00 a11
a11
, , is a square in K. For example, suppose is a
a22 a11 a22 a22
square. Then the above equation may be replaced by an equation of the
form ax2 0 + x21 + x22 = 0. And the existence of at least one point on Q follows
from the fact that if 0 = γ ∈ K, then there are elements α and β of K such
that α2 + β2 = γ. (For p = 2 this is trivial. Also, we may of course assume
that γ = ❆ .)
This combinatorial lemma is actually a special case of Theorem 20.23.1.
However, this special case has a cute combinatorial proof which we give now.
Define the multiplicative homomorphism h on the nonzero elements of K
by h : a ↦→ a2 . Since the kernel of h is {+1, −1}, precisely half of the
nonzero elements of K are squares, say β1, . . . , βr, r = t−1
2
are the squares.
Let S = {γ − βi : 1 ≤ i ≤ r}. If S contained no squares, then S would
contain precisely all the nonsquares of K, especially γ. But clearly γ ∈ S, so
γ = βi + βj for some i and j. This completes the proof that Q is nonempty.
Now suppose that x is a point of Q and n = 2. By Theorem 5.2.2 there
is a y ∈ Q which is not conjugate to x. Thus x and y span a hyperbolic line
H. If z is in the conjugate space R of H, then z ∈ Q. Since the conjugate
space τ(x) has dimension 2, it is clear that τ(x) = Kx ⊕ τ(x). For a vector
z in τ(H), q(x + z) = q(x) + q(z) + B(x, z) = q(z) = 0. Thus τ(x) has only
t − 1 nonzero vectors α such that q(α) = 0: viz, the nonzero multiples of x.
So we need to count the points of Q not conjugate to x. There are t3−t2 = t2
points y not conjugate to the point x. For each such point y, xy is a line
= t
(which must be hyperbolic by Lemma 5.2.1. Each such line will give t2 −t
t−1
points not conjugate to x. In this way there are t2
t
t−1
= t distinct hyperbolic
lines determined containing x. If H = xy, x, y ∈ Q is a hyperbolic line with
B(x, y) = 0, it is clear that the only points of Q in H are x and y. Thus
there will be x in Q plus the t other points on the hyperbolic lines through
x also on Q. So Q has exactly t + 1 points.
Case (ii) p = 2. In this case each element of K is a square. For n = 2
f(x) has the form f(x) = a00x2 0 +a11x 2 1 +a22x 2 2 +a01x0x1 +a02x0x2 +a12x1x2.
If a02 = 0, put x1 = 0 and solve a00
a22 =
2
x2
a12
. Otherwise, put x0 = x1,
x0
a02
a00a
so f(x) becomes f(x) =
2 12
a2
x
02
2 1 + a11x2 1 + a22x2
a01a12
2 + x a02
2 1 + a12x1x2 +
a12x1x2 = cx 2 1 + a22x 2 2. From this it is easy to find at least one nontrivial
solution to f(x) = 0, so Q has a least one point. That it has t + 1 points
follows from the argument given in case (i).

5.4. CANONICAL FORMS OVER FINITE FIELDS 257
It follows from Theorems 5.2.2 and 5.4.2 that if n is even, any nondegenerate
quadric in Pn(K) contains a subspace N of dimension n
and no subspace
2
of higher dimension. If n is odd, there are two possibilities: a subspace of Q
of maximum dimension has dimension 1
1 (n + 1) or (n − 1). Furthermore, Q
2 2
will have a subspace of dimension 1
2 (n + 1) iff Vn+1 is an orthogonal direct
sum of hyperbolic lines. The space Vn+1 (or Pn(K)) endowed with a particular
nondegenerate quadric surface (or quadratic form, unique up to a scalar
factor) will be called a hyperbolic or elliptic space, and the quadric surface Q
will be called hyperbolic or elliptic according as Q has subspaces of dimension
1(n
+ 1) or not.
2
When n is odd, any two hyperbolic quadric surfaces have the same matrix
A but relative to distinct bases. Thus a linear transformation mapping one
of these bases into the other (in an appropriate manner) maps one of the
quadrics into the other. This implies that any two hyperbolic quadrics are
essentially (i.e., projectively) equivalent.
If n is even, any nondegenerate quadric has a polynomial of the form
f(x) = r i=1 xiyi + cz2 for some 0 = c ∈ K and where r = n.
Multiplying
2
each xi by c gives the form c ( r i=1 xiyi + z2 ). So when n is even, any two
nondegenerate quadric surfaces are essentially the same. In this case, the
vector space endowed with a nondegenerate quadratic form will be called a
parabolic space.
If n is odd, any elliptic quadric has the form
r
i=1
xiyi + c1z 2 1 + c2z 2 2 + cz1z2, where r =
n − 1
.
2
Replacing each xi by c1xi and then dividing the entire form by c1, we see
that we may suppose the quadric has the form
r
i=1
xiyi + z 2 1 + c2z 2 2 + cz1z2.
Replacing z1 by z ′ 1 = z1 + c
2 z2 yields the form
f(x) =
r
xiyi + z 2 1 + cz 2 2, where 0 = c.
i=1
Put ¯z1 = (0, . . . , 0, 1, 0) and ¯z2 = (0, . . . , 0, 1). The fact that no point of
the line R spanned by ¯z1 and ¯z2 is on Q is equivalent to the assumption that

258 CHAPTER 5. QUADRATIC FORMS
−c is not a square in K. Clearly, by adjusting z2 by some factor, −c may be
assumed to be any particular nonsquare. Thus K has essentially only one
elliptic quadric if K has odd characteristic. If K has characteristic 2, the
transformation x1 = x ′ c
1− 2 x2 is undefined. But each element of K is a square.
So replace zi by zi
√ci , i = 1, 2, to get f of the form xiyi + z 2 1 + z 2 2 + cz1z2,
where z 2 1 + z2 2 + cz1z2 = 0 has no solution in K. Replacing z2 by c −1 z2 yields
a polynomial of the form xiyi + z2 1 + z1z2 + δz2 2 where δ is an element of
K = GF (2e ) such that 0 = z2 1 + z1z2 + δz2 2 has no solution in K (other than
the trivial solution, of courwse). This is equivalent to z2 1 + z1 + δ = 0 having
no solution in K. If z ′ 0 = z0 + λz1, z ′ 1 = z1, then the former equality in two
variables becomes 0 = z ′ 2 ′
1 + z 1z ′ 2 + δ′ z ′ 2 ′ 2
2 , where δ = δ + λ + λ . Clearly
z 2 + z + δ = 0 has a root in K iff (z + λ) 2 + (z + λ) + δ = z 2 + z + δ ′ = 0 has
a solution. We claim that in the finite case if δ ′ + z + z 2 = 0 has no solution,
then there is some λ ∈ K = GF (2 e ) such that δ ′ = δ + λ + λ 2 . This would
imply that any two elliptic quadrics in Pn(K) are equivalent. To prove this
we need some additional information about the elements of GF (2 e ). (We
include this material here for the reader who has not digested the material
in Section 4.13.)
First, suppose ζ 2 = δ + ζ for two elements δ and ζ of K. Then ζ 4 =
δ 2 + ζ 2 = ζ + δ + δ 2 . By an induction argument we find
implying that
ζ 2e
= ζ + δ + δ 2 + δ 22
+ δ 23
+ · · · + ζ 2e
,
δ + δ 2 + δ 22
+ δ 23
+ · · · + δ 2e−1
= 0.
This shows that if δ is an element of K for which z 2 + z + δ = 0 has a
solution z ∈ K, then the absolute trace of δ equals 0. We know that the
absolute trace map
tr : GF (2 e ) → GF (2) : δ ↦→ δ + δ 2 + δ 22
+ δ 23
+ · · · + δ 2e−1
is an additive function which is onto GF (2), i.e., it has kernel of size 2 e−1 .
Let θ be the map: θ : λ ↦→ λ + λ 2 . Then θ is an additive homomorphism
of the additive group of K. θ(λ1) = θ(λ2) iff λ1 = λ2 or λ1 = 1+λ2. Thus the
kernel of θ has cardinality 2, which implies that the range of θ has cardinality
2 e−1 . Moreover, each element δ of the range C0 of θ clearly has the property
that δ + z + z 2 has a solution z ∈ K. It follows that the range C0 of θ is

5.4. CANONICAL FORMS OVER FINITE FIELDS 259
exactly the kernel C0 of the trace map tr, and that C0 is precisely the set of
elements δ ∈ K for which δ + z + z 2 = 0 has a solution. If δ0 is any element
of K for which δ0 + z + z 2 does not have a solution z ∈ K, the other coset
C1 = C0 + δ0 of C0 is exactly the set of elements of K for which δ0 + z + z 2
does not have a solution z ∈ K. It follows that δ1 and δ2 are two elements
in the same coset of C0 iff there is some λ ∈ K for which δ1 = δ2 + λ + λ 2 ,
which proves our claim.
We record some of this for easy reference.
Lemma 5.4.3. If K = GF (2 e ), and if tr : K → GF (2) is the absolute trace
function, then for a given δ ∈ K,
δ + z + z 2 = 0 has 2m solutions in K provided tr(δ) = m, m = 0 or 1.
Now suppose that Vn+1 is an orthogonal direct sum Vn+1 = W1 ⊥ W2
relative to some nondegenerate quadratic form. If n + 1 is odd, then one
of W1, W2 has odd dimension, the other has even dimension. Since the
odd dimensional (vector) spaces are essentially of one kind, the type of the
even dimensional subspace makes no difference on the type of the whole
space. However, suppose that n + 1 is even. Clearly the orthogonal sum of
two hyperbolic spaces is hyperbolic, and the orthogonal sum of a hyperbolic
and an elliptic space must be elliptic (consider the form of the associated
polynomials). So consider the orthogonal direct sum of two elliptic spaces.
Case (i) Odd characteristic. Clearly we may suppose that the associated
polynomial has the form
f =
r
xiyi + z 2 1 + cz 2 2 + w 2 1 + cw 2 2, where − c is a nonsquare.
i=1
We know we can choose scalars α, β such that α 2 + β 2 = c −1 . Replacing z2
by z ′ 2 = αz2 − βw2 and w2 by w ′ 2 = βz2 + αw2 amounts to changing bases so
that the polynomial becomes
r
i=1
xiyi + z 2 1 + z2 2 + w2 1 + w2 2 .
Thus the whole space will be hyperbolic if we show that a form 4
i=1 x2 i
is hyperbolic. Choose α, β ∈ K such that α 2 + β 2 = −1. Then the points

260 CHAPTER 5. QUADRATIC FORMS
(α, β, 1, 0) and (β, −α, 0, 1) generate a line in Q, showing that the space is
hyperbolic.
Case (ii) Even characteristic. In this case we may assume that the associated
polynomial has the form f = r i=1 xiyi+z 2 1 +z1z2+δz 2 2 +w2 1 +w1w2+δw 2 2 ,
where δ is some element of K such that z2 + z + δ = 0 has no solution. So
here the whole space will be hyperbolic if we show that z2 1 +z1z2 +δz 2 2 +w2 1 +
w1w2 + δw2 2 = 0 is hyperbolic. But the points (z1 = w1 = 1; z2 = w2 = 0)
and (z1 = w1 = 0; z2 = w2 = 1) are linearly independent conjugate points on
the quadric.
So what about the orthogonal direct sum of two parabolic spaces? Such a
quadratic form would have a polynomial of the form f = r i=1 xiyi +w 2 +z 2 .
If t = 2e , this is a cone over a parabolic quadric in a hyperplane (with vertex
(0, 0, . . . , 0, 1, 1)). If t is odd, it is a hyperbolic space when −1 = ∈ Ft and
is an elliptic quadric when −1 = ❆ ∈ Ft.
We summarize the results of the preceding paragraphs in the following
theorem.
Theorem 5.4.4. Let q be a nondegenerate quadratic form on a vector space
Vn+1 over K = GF (t).
(i) If n+1 is odd, there is essentially only one kind of quadric in the sense
that with respect to some basis its polynomial has the form f = r
i=1 xiyi+z 2 .
(ii) If n + 1 is even and q is hyperbolic, then with respect to some basis
its polynomial has the form
i xiyi.
(iii) If n + 1 is even and q is not hyperbolic, then it must be elliptic.
Furthermore, with respect to some basis its polynomial has the form:
(a) Odd characteristic: f = r
i=1 xiyi + z 2 1 + cz 2 2; −c any nonsquare.
(b) Characteristic 2: f = r
i=1 xiyi + z 2 1 + z1z2 + δz 2 2 ; with z2 + z + δ
irreducible over K.
Furthermore, any two elliptic quadrics on Vn+1 are equivalent.
(iv) If Vn+1 is the orthogonal direct sum of two subspaces W1, W2, each
of even dimension, then Vn+1 is hyperbolic or elliptic according as W1 and
W2 are of the same kind or of opposite kinds.
(v) The form 4
i=1 x2 i
is hyperbolic (actually for any characteristic not
zero).
(vi) The orthogonal sum of two parabolic spaces is a cone over a parabolic
space in a hyperplane when t = 2 e . When t is odd, it is either elliptic or
hyperbolic, depending on whether −1 is a nonsquare or a square.

5.4. CANONICAL FORMS OVER FINITE FIELDS 261
Remark: The preceding theorem really contains a great deal of information.
We mention one type of application of part (iv) that is very useful.
Suppose Vn+1 is hyperbolic and W is a subspace of even dimension on which
the restriction of the nondegenerate quadratic form q is also nondegenerate.
Then Vn+1 is the orthogonal direct sum of W and W ⊥ . This means that W
and W ⊥ must be of the same type, i.e., either both elliptic or both hyperbolic.
Theorem 5.4.5. Let Q be a nondegenerate quadric in Pn(K) with K =
GF (t). Then the number of points on Q is:
, if n is even;
(i) tn −1
t−1
(ii) ( (tk+1 +1)(t k −1)
t−1
(iii) (tk+1 −1)(t k +1)
t−1
, if Q is elliptic, n = 2k + 1 ≥ 3.
, if Q is hyperbolic, n = 2k + 1 ≥ 3.
Proof. Case (i) For n = 2 we have already established the result in Theorem
5.4.2 The polar τ(x0) of any point x0 ∈ Q intersects A in a cone Q ′ in
τ(x0) ∼ = Pn−1(K) whose equation may be expressed in terms of n − 1 variables.
Hence the number of lines through x0 lying in Q ′ is the same as the
number of points lying on a nondegenerate quadric in Pn−2(K), i.e., tn−2 −1
t−1 by
an induction hypothesis. Since each point on Q ′ is on one of the lines through
x0, there are t(tn−2−1) t−1 + 1 = tn−1−1 t−1 points on Q′ . There are tn+1−1 t−1 − tn−1 = tn
t−1
points y not conjugate to x0. For each such y the line x0y is a hyperbolic line
(a secant to Q) containing t points different from x0. Thus there are tn−1 hyperbolic lines containing x0, each containing one other point of Q not in
τ(x0). Hence there are tn−1−1 t−1 + tn−1 = tn−1 points on Q, as claimed.
t−1
For (ii) and (iii), let n = 3 and x0 ∈ Q. By a now familiar calculation
there are t2 lines through x0 not lying in τ(x0), each containing precisely one
point of Q different from x0. If Q is elliptic, no line lies in Q, so τ(x0) ∩ Q is
just the point x0. Thus there are exactly 1+t2 points on Q. If Q is hyperbolic,
Q ∩ τ(x0) is a cone in P2(K) with 1 + tr points, where r is the number of
points on a quadric in some P1(K) with points. But a nondegenerate quadric
in P1(K) with a point x must have a unique other point y, so r = 2. Thus
Q has t2 + (1 + 2t) = (1 + t) 2 points as claimed in the theorem.
Proceeding by induction (on n), let Q be of either type, x0 ∈ Q, y ′ ∈
τ(x0). Then there is a unique point y0 ∈ Q on the line x0y ′ such that
y0 ∈ τ(x0). And x0y0 IS a hyperbolic line whose polar space τ(x0) ∩ τ(y0)
is a Pn−2(K). But Q ∩ (τ(x0) ∩ τ(y0)) is a quadric of the same type as Q
(Theorem 5.4.4, part (iv)). By induction Q ∩ (τ(x0) ∩ τ(y0)) contains φn−2
points where φn−2 is the number of points on a nondegenerate quadric in

262 CHAPTER 5. QUADRATIC FORMS
Pn−2(K) of the appropriate type. Let y = cx0 + z ∈ Kx0 ⊥ (τ(x0) ∩ τ(y0)) =
τ(x0), with z ∈ τ(x0) ∩ τ(y0). Clearly y ∈ Q iff z ∈ Q. Thus τ(x0) contains
t · φn−2 points of Q different from x0. Just as in the proof of part (i), there
are t n−1 hyperbolic lines through x0, each containing precisely one point of
Q other than x0. Thus φn = 1 + t n−1 + t · φn−2. Our induction hypothesis is
that φn−2 =
n−1
(t 2 ±1)(t n−3
2 ∓1)
t−1
recurrence leads easily to φn =
show.
. This together with the immediately preceding
n+1
(t 2 ±1)(t n−1
2 ∓1)
t−1
, which is what we wanted to
5.5 The Discriminant in Odd Characteristic
Now let K = GF (t), t = p e , p an odd prime. In this case the quadratic
form q is nonsingular and its polar form B (with Gram matrix also denoted
B) is nondegenerate provided |B| = det(B) = 0. If this is the case, then
|P BP T | = |P | 2 · |B|, so that |B| and |P BP T | are either both squares in K
or are both nonsquares in K. Hence if B is nondegenerate we may define the
discriminant of q (or of B) to be disc(B) = |B| modulo the group of squares
in K ∗ = K \ {0}.
Let V and W be two n-dimensional vector spaces over K. Let q, q ′ be two
nonsingular quadratic forms on V , W , respectively. A linear transformation
T : V ↦→ W is an isometry from q to q ′ provided q(v) = q ′ (T (v)) for all v ∈ V ,
in which case T is invertible. Suppose this is the case and let B, B ′ be ordered
bases of V , W , respectively. Let A be the matrix representing T with respect
to the pair B, B ′ . This means that [T (v)]B ′ = [v]B · A for all v ∈ V . Let B,
B ′ be the matrices representing the polar forms B, B ′ of q, q ′ , with respect
to the bases B, B ′ , respectively. Then clearly B(v, w) = B ′ (T (v), T (w)) for
all v, w ∈ V , so that
[v]B · B · [w] T B = [v]B · A(A −1 BA −T )A T [w] T B
= [T (v)]B ′(A−1 BA −T )[T (w)]B ′
= [T (v)]B ′B′ [T (w)]B ′
for all v, w ∈ V . Hence B ′ = A −1 BA −T , and disc(B ′ ) = disc(B).
As above, let q : V ↦→ F be a nonsingular quadratic form on V with polar
form B. Since q(v) = 0 if and only if q(cv) = c 2 q(v) = 0 for all c ∈ K ∗ , the

5.5. THE DISCRIMINANT IN ODD CHARACTERISTIC 263
quadric Q(n − 1, t) = {v ∈ P (V ) : q(v) = 0} is a well defined set of points of
the projective space P (V ) associated with V . If W is a subspace of V , define
W ⊥ = {v ∈ V : B(v, w) = 0 for all w ∈ W }. If W ∩ W ⊥ = {0}, it is possible
to show that V = W ⊕ W ⊥ , and we write V = W ⊥ W ⊥ . If B1 is a basis of
W and B2 is a basis of W ⊥ , then B1 ∪ B2 is a basis of V , and the matrix B
that is the Gram matrix of q with respect to the basis B1 ∪ B2 is the direct
sum of the matrices B1, B2 that are the Gram matrices of q restricted to W ,
W ⊥ , respectively. Hence
disc(q) = disc(q|W) · disc(q| W ⊥). (5.2)
Now suppose W =< x, y > is a line, i.e., a 2-dimensional subspace of
V . Put B(x, x) = a, B(x, y) = B(y, x) = b, B(y, y) = c. Then disc(q|W
) =
a b
b c = ac − b2 . And q(dx + ey) = 0 if and only if 0 = B(dx + ey, dx +
ey) = d2a + 2deb + e2c, which has no solutions (d, e) = (0, 0) if and only if
4b2 − 4ac = −4(ac − b2 ) = −disc(q|W ) = ❆ ∈ K. This proves the following
theorem.
Theorem 5.5.1. The line W is anisotropic (or elliptic), i.e., contains no
nonzero vector v with q(v) = 0 if and only if −disc(q|W ) = ❆ .
Given a non-degenerate even-dimensional orthogonal space over Ft with
t odd, the discriminant of the quadratic form tells us whether the space is of
type (i) or (iii) in Cor. 5.2.6. Hence the discriminant is a complete invariant.
Since any O + (2n, t) space can be written as the orthogonal direct sum of
n hyperbolic lines, the matrix of the polar form (with respect to a basis
of hyperbolic pairs from these lines) has determinant (−1) n . Thus, a nondegenerate
orthogonal space (V2n, Q ′ ) of even rank (vector space dimension)
with t odd is an O + (2n, t) space if and only if det(B) = (−1) n modulo the
group of squares, where B is the matrix of the polar form associated to Q ′ .
The O + , O and O − spaces will be referred to as hyperbolic, parabolic and
elliptic spaces, respectively.
The set of singular vectors of an O(3, t) space will be called a conic, and
the set of singular vectors of an O − (4, t) space is an elliptic quadric.
We now give an application of this section that will turn out to be quite
useful when we study generalized quadrangles constructed from BLT-sets.

264 CHAPTER 5. QUADRATIC FORMS
Let t be odd, and let P1, P2, P3 be three pairwise non-orthogonal points of
Q(4, t). We may take V = F 5
t with quadratic form Q(x) = x0x1 − x2 2 + x3x4
and associated polar form B(x, y) = x0y1 + x1y0 − 2x2y2 + x3y4 + x4y3. So
we are assuming that B(Pi, Pj) = 0 if and only if i = j, 1 ≤ i, j ≤ 3. So
W = 〈P1, P2, P3〉 is a plane with polar line W ⊥ . We showed above that W ⊥
is anisotropic if and only if disc(Q| W ⊥) = −❆ . Suppose that B(P1, P2) = a,
B(P2, P3) = b, B(P3, P1) = c. So the gram matrix for Q|W is
⎛
BW = ⎝
0 a c
a 0 b
c b 0
implying that disc(Q|W ) = 2abc. It is also easy to check that disc(Q)) = −2.
Since disc(Q) = disc(Q|W ) · disc(Q| W ⊥), we have W ⊥ is anisotropic iff −2 ≡
2abc · (−❆ ) (modulo the nonzero squares in F ∗
t ). Hence we have proved the
following
Lemma 5.5.2. Let t be odd. Then let Q be a nondegenerate (parabolic)
quadratic form on V = F 5
t with associated polar form B. Let P1, P2, P3 be
three linearly independent vectors in V with B(Pi, Bj) = 0 if and only if
i = j, 1 ≤ i ≤ j ≤ 3. Then W ⊥ is an anisotropic line (i.e., there is no
nonzero R ∈ V with Q(R) = 0 and B(Pi, R) = 0 for 1 ≤ i ≤ 3) if and only
if B(P1, P2) · B(P2, P3) · B(P3, P1) = ❆ .
Corollary 5.5.3. With Q and B as in the preceding lemma, let P0, P1, P2, P3
be four points of V such that any three of them form a linearly independent
set with B(Pi, Pj) = 0 iff i = j, 0 ≤ i, j ≤ 3. Define Wk = 〈Pi, Pj, Pl〉,
where {i, j, l, k} = {0, 1, 2, 3}. Suppose that all three lines W ⊥ k , 1 ≤ k ≤ 3
are anisotropic. Then W ⊥ 0 is also anisotropic.
Proof. By the lemma we have
⎞
⎠ ,
B(P0, P1)B(P1, P2)B(P0, P2) = ❆
B(P0, P1)B(P1, P3)B(P0, P3) = ❆
B(P0, P2)B(P2, P3)B(P0, P3) = ❆
Take the product of these three lines. Since the product of three nonsquares
is a nonsquare, we see that B(P1, P2)B(P2, P3)B(P3, P1) = ❆ .

5.6. ACTION OF THE ORTHOGONAL GROUP 265
5.6 Action of the Orthogonal Group
Let M and N be finite dimensional vector spaces over the field K. A bilinear
form B on M × N is a mapping of M × N into K with the property that for
each x ∈ M and y ∈ N the functions λx : y ↦→ B(x, y) and µy : x ↦→ B(x, y)
are linear on N and M, respectively. Let P and R be subspaces of M
and N, respectively. Put P ′ = {y ∈ N : B(x, y) = 0 for all x ∈ P } and
R ′ = {x ∈ M : B(x, y) = 0 for all y ∈ R}. Then P ′ (resp., R ′ ) is a subspace
of M (resp., N)called the right (resp., left) conjugate space of P (resp., R).
The following relations are simple consequences of the definitions:
(i) P ⊆ (P ′ ) ′ for any subspace P of M;
(ii) R ⊆ (R ′ ) ′ for any subspace R of N;
(iii) (P1 + P2) ′ = P ′ 1 ∩ P ′ 2 , if P ′ 1 , P ′ (iv) (R1 + R2)
2 are sub spaces of M;
′ = R ′ 1 ∩ R ′ 2, if R1, R2 are subspaces of N.
(5.3)
We say B is nondegenerate if both M ′ = 0 and N ′ = 0, i.e., iff both
λ : x ↦→ λx and µ : y ↦→ µy are isomorphisms of M onto N ∗ and N onto M ∗ ,
respectively. Here M ∗ represents the dual space of M; similarly for N ∗ .
Theorem 5.6.1. Let B be a nondegenerate bilinear form on the product of
two m-dimjensional vector spaces M and N over K. If P is a p-dimensional
subspace of M, then its conjugate P ′ is of dimension m − p, and (P ′ ) ′ = P .
If Q is a q-dimensional subspace of N, then Q ′ is of dimension m − q and
(Q ′ ) ′ = Q.
Proof. From elementary linear algebra.
Until further notice M will denote an m-dimensional vector space over
K, q a quadratic form on M, and B the bilinear polar form on M × M associated
with q. Furthermore, we assume B is nondegenerate. Here we mean
something slightly stronger than the concept of nondegeneracy employed in
the previous section, viz., even if K has characteristic 2 we assume that any
associated matrix B is nonsingular.
Exercise 5.6.1.1. Show that our assumptions imply either K has characteristic
different from 2 or m is even.

266 CHAPTER 5. QUADRATIC FORMS
A linear mapping s of M into itself is called orthogonal (relative to q)
provided q(s · x) = q(x) for all x ∈ M. Clearly if s is orthogonal, then
B(s · x, s · y) = B(x, y) for all x, y ∈ M. In particular, if s · x = 0, then
B(x, y) = 0 for all y ∈ M if s is orthogonal, implying x = 0. Hence any
orthogonal map is invertible. Under composition the orthogonal mappings
form a group called the orthogonal group G of q.
A vector space isomorphism s of a subspace N of M onto a subspace P
of M is called a q-isomorphism provided q(s · x) = q(x) for all x ∈ N. Then
B(s · x, s · y) = B(x, y) for all x, y ∈ N. Note that if x ∈ M, we write Kx
for the linear span of x.
The following result is a famous one known as Witt’s theorem.
Theorem 5.6.2. The assumptions and notations being as above, any
q-isomorphism s of a subspace N of M onto a subspace P may be extended
to an element of G.
Proof. We proceed by induction on the dimension n of N, the theorem being
obviously true if n = 0. So assume n > 0 and that the statement of the
theorem is true for subspaces of dimension n − 1. Let U be an (n − 1)dimensional
subspace of N. Then the inductive hypothesis says that the
restriction of s to U may be extended to an element s0 of G. Define s ′
by s ′ (x) = s −1
0 s(x) for all x ∈ N. Then s ′ is a q-isomorphism of N (onto
some subspace of M) which leaves the elements of U fixed. If s ′ extends to an
element s ′ 0 of G, then s0s ′ 0 is an element of G which extends s. Hence without
loss of generality we may assume that the original map s leaves elements of
U fixed.
Let M be the set of subspaces V of M with the property that s may be
extended to a q-isomorphism of V +N leaving the elements of V fixed. Since
M has finite dimension, M has a maximal element V1 with a corresponding qisomorphism
s1 of N1 = V1 +N which extends s and leaves fixed the elements
of V1. Let P1 = s1(N1), U1 = V1 +U. If U1 = N1, i.e., V1 +U = V1 +N = N1,
then s1 is the identity on N1. But clearly the identity map on N1 can be
extended to the identity map on all of M, implying that N1 = M. So suppose
otherwise, i.e., U1 is a hyperplane of N1, and let x1 ∈ N1 \ U1, y1 = s1 · x1.
Then N1 = U1 + Kx1, P1 = U1 + Ky1, and q(x1) = q(y1). Clearly we may
also assume that s1 · x1 = y1 = x1.
The remainder of the proof is divided into two parts. The first part
consists of finding elements z, z ′ ∈ M with special properties, and the second
part consists of using z and z ′ to complete the proof.

5.6. ACTION OF THE ORTHOGONAL GROUP 267
Claim: There are elements z, z ′ ∈ M with the following properties:
(i) z ∈ N1;
(ii) z ′ ∈ P1;
(iii) z ′ − z ∈ U ′ 1, where U ′ 1 is the conjugate of U1;
(iv) B(z ′ , y1) = B(z, x1);
(v) q(z) = q(z ′ ).
Let H be the conjugate of the space K(x1 − y1). We claim H = N1 = P1.
First suppose H contains an element z which is in neither N1 nor P1, and
extend s1 to an isomorphism s ′ of N1 + Kz onto P1 + Kz which maps z onto
itself. Then we claim s ′ would be a q-isomorphism, for suppose u + cx1 ∈ N1,
u ∈ U1. Then (note that z ∈ (K(x1 − y1)) ′ implies that B(z, x1) = B(z, y1))
q(s ′ · (u + cx1 + dz)) = q(s ′ · u) + c 2 q(s ′ · x1) + d 2 q(z)
+B(s ′ · u, s ′ · cx1) + B(s ′ · u, s ′ · dz) + B(s ′ · cx1, s ′ · dz)
= q(s1 · u) + c 2 q(y1) + d 2 q(z) + B(s1 · u, cy1) + B(s1 · u, dz) + B(s1 · cx1, dz)
= q(u) + c 2 q(x1) + d 2 q(z) + B(u, cx1) + B(u, dz) + B(cx1, dz)
= q(u + cx1 + dz).
But this says V1 + Kz is an element of M larger than V1, contradicting
the choice of V1. Hence each element of H is in either N1 or P1, so clearly
H ⊆ N1 or H ⊆ P1. If N1 = M there is nothing to prove. So suppose
N1 = M. Then dim (H) = m − 1, dim (N1) = dim (P1) < dim (M) = m,
and H ⊆ N1 or H ⊆ P1. So H = N1 or H = P1. But H is the conjugate
space of K(x1 − y1), and x1 ∈ N1, y1 ∈ P1. So x1 − y1 is conjugate to one
of x1, y1. But B(x1, x1) = B(y1, y1), so B(x1, x1 − y1) = B(y1 − x1, y1) = 0,
implying both x1 and y1 are in H. From N1 = U1 + Kx1, P1 = U1 + Ky1, it
follows that N1 = P1 = H. In particular, B(x1, x1 −y1) = B(y1, x1 −y1) = 0.
Now let z ∈ M \H. Thus B(z, x1 −y1) = 0, and M = H +Kz = N1+Kz.
For this z we can find a z ′ such that the pair z, z ′ have the desired properties.
Since x1 ∈ N1 \ U1, clearly y1 ∈ U1. Thus U ′ 1 contains a vector not
conjugate to y1, and indeed there is a vector u ∈ U ′ 1 such that B(u, y1) =
B(z, x1 − y1) = 0. Since B(x1 − y1, y1) = 0, u ∈ K(x1 − y1), i.e., u ∈ H ′ .
But u ∈ U ′ 1 and H = N1 = U1 + Kx1, so B(u, x1) = 0. B(z + u, x1 − y1) =
B(z, x1 −y1)+B(u, x1)−B(u, y1) = B(u, x1) = 0. Thus z+u ∈ P1 = H. But
note that x1 − y1 ∈ H ∩ H ′ , and U1 ⊆ H, so H ′ ⊆ U ′ 1 , implying x1 − y1 ∈ U ′ 1 .
So for c ∈ K, we have z + u + c(x1 − y1) ∈ (z + U ′ 1 ) \ P1. Since q(x1 − y1) =

268 CHAPTER 5. QUADRATIC FORMS
q(x1)+q(y1)−B(x1, y1) = 2q(x1)−B(x1, y1) = B(x1, x1)−B(x1, y1) = 0, we
have q(z+u+c(x1−y1)) = q(z+u)+cB(z+u, x1−y1). Since B(z+u, x1−y1) =
0, we can choose c such that q(z + u + c(x1 − y1)) = q(z). For this c, set
z ′ = z + u + c(x1 − y1). Then z and z ′ do have the desired properties. In
fact, we have already observed that all the properties hold except perhaps
B(z ′ , y1) = B(z, x1). But B(z ′ , y1) = B(z+u+c(x1−y1), y1) = B(z+u, y1) =
B(z, y1) + B(z, x1 − y1) = B(z, x1).
Using z and z ′ we can now finish the proof. For we may extend s1 to an
isomorphism s2 of N1 + Kz onto P1 + Kz ′ such that s2 · z = z ′ . It remains
only to show that s2 is a q-isormphism, since M = N1 + Kz = P1 + Kz ′ .
If x ∈ N1, then x = u + ax1, with u ∈ U1, a ∈ K, and s1 · x = u + ay1.
Since B(z ′ − z, u) = 0, and B(z, x1) = B(z ′ , y1), it follows that B(z, x) =
B(z, u + ax1) = B(z ′ , u + ay1) = B(z ′ , s1 · x). Also, q(x) = q(s1 · x) since s1 is
assumed to be a q-isomorphism of N1. So for b ∈ K, q(bz+x) = b 2 q(z)+q(x)+
bB(z, x) = b 2 q(z ′ ) + q(s1 · x) + bB(z ′ , s1 · x) = q(bz ′ + s1 · x) = q(s1 · (bz + x)),
proving s2 is a q-isomorphism of M.
Corollary 5.6.3. Let N be a subspace of M lying entirely in the quadric
surface Q determined by q. Then each automorphism (i.e., K-linear operator)
on N may be extended to an element of G.
Corollary 5.6.4. All maximal subspaces of Q have the same dimension and
are permuted transitively among themselves by the elements of G.
Proof. If N and P are subspaces of Q having the same dimension, any isomorphism
between them must be a q-isomorphism, and thus must be extendable
to an element s of G. If P ′ is a subspace of Q containing P , then s −1 (P ′ ) is
a subspae of Q containing N. Thus if N is a subspace of Q not properly contained
in any subspace of Q, any other maximal subspace of Q is isomorphic
to N, and by an element of G.
Of course we knew some of this much earlier. The common dimension of
all the maximal subspaces of Q is called the index of q (and of Q and of B).
Let r be the index of q. By Theorem 5.4.4 we know 2r ≤ m.
Theorem 5.6.5. Let s be an element of G and define the linear operator µ
on M by µ : x ↦→ s · x − x. Then the image µ(M) of µ is the conjugate of
the kernel L of µ.

5.6. ACTION OF THE ORTHOGONAL GROUP 269
Proof. If x ∈ M, y ∈ L (so s·y = y), then B(y, s·x) = B(s·y, s·x) = B(y, x),
so B(y, s · x − x) = 0, implying µ(M) ⊆ L ′ . But µ(M) and L ′ have the same
dimension, so they are equal.
Corollary 5.6.6. Suppose q has index r = m and let N be a maximal
2
subspace of Q. Let s be an element of G leaving points of N fixed. Then
s · x − x ∈ N for all x ∈ M.
Proof. By Theorem 5.6.5 s · x − x is in the conjugate N ′ of N, but of course
N ′ = N.
Before proceeding with the next theorem we recall that if Γ is a bilinear
form on M × M with the property that Γ(x, x) = 0 for all x ∈ M, then Γ
is called alternating. Furthermore, there is an integer ρ such that M has a
basis y1, . . . , ym for which
⎧
⎨ 1, i = 2k − 1, j = 2k, 1 ≤ k ≤ ρ;
Γ(yi, yj) = −1 i = 2k, j = 2k − 1, 1 ≤ k ≤ ρ;
⎩
0, otherwise.
The integer 2ρ is called the rank of Γ.
Theorem 5.6.7. Let Q have index r = m,
m = the dimension of M. Let N
2
and P be maximal subspaces of Q such that M = N ⊕ P , and let H be the
subgroup of G fixing N pointwise. For each s ∈ H define a mapping Γs on
P × P by Γs(y, y ′ ) = B(y, s · y ′ ). Then the following are true:
(i) Γs is an alternating bilinear form on P × P and s → Γs is an isomorphism
of H with the additive group of all alternating bilinear forms on
P × P .
(ii) The rank of Γs is the dimension of µs(M) where µs : x ↦→ s · x − x
for all x ∈ M.
(iii) If s, s ′ ∈ H are such that Γs and Γs ′ have the same rank, then s and
s ′ are conjugate to each other in G.
Proof. Clearly Γs is bilinear. For y ∈ P , 0 = q(y) = q(s·y) = q(y+s·y−y) =
q(y)+q(s·y−y)+B(y, s·y−y) = 0+0+B(y, s·y)−0 = B(y, s·y) = Γs(y, y),
since s · y − y ∈ N by Corollary 5.6.6 and B(y, −y) = −2q(y) = 0. Thus
Γs is alternating. Furthermore, for s, s ′ ∈ H, s · (s ′ · y − y) = s ′ · y − y, so
s · s ′ · y − y = s · (s ′ · y − y) + s · y − y = (s ′ · y − y) + (s · y − y). Thus
Γs·s ′(y′ , y) = B(y ′ , ss ′ · y) = B(y ′ , ss ′ · y − y + y) = B(y ′ , s ′ · y) + B(y ′ , s ·

270 CHAPTER 5. QUADRATIC FORMS
y) − B(y ′ , y) = Γs(y ′ , y) + Γs(y ′ , y). This shows that Γiss ′ = Γ ′ s + Γs, and
s ↦→ Γs is a group homomorphism. If Γs = 0, then for y ∈ P we know
s · y − y ∈ N ∩ P ′ = N ∩ P = {0}. Thus s fixes all elements of N and of P ,
so s is the identity of H, proving that s ↦→ Γs s one-to-one.
To complete the proof of (i) we now show that s ↦→ Γs is onto. So let Γ
be any alternating bilinear form on P × P . For x ∈ N, let λx be the linear
form y ↦→ B(x, y) on P . Then x ↦→ λx is a linear mapping of N into the dual
P ∗ of P which is actually an isomorphism onto since B is nondegenerate.
On the other hand, if y ′ ∈ P , the mapping y ↦→ Γ(y, y ′ ) is an element of
P ∗ . So for each y ′ ∈ P there is a unique µ(y ′ ) ∈ N such that λµ(y ′ ) is the map
y ↦→ Γ(y, y ′ ), i.e., B(y, µ(y ′ )) = Γ(y, y ′ ) for all y ∈ P . It is easy to see that
µ : P → N is linear. Then define a map s on M by s(x + y) = x + y + µ(y)),
x ∈ N, y ∈ P . Clearly s is linear with zero kernel, so s is an automorphism
of M. Also, for x ∈ N, y ∈ P , q(x + y + µ(y)) = B(y, µ(y)) + B(x, y) =
B(x, y) = q(x + y) since 0 = Γ(y, y) = B(y, µ(y)). Then s ∈ G, and in fact
s ∈ H with Γs = Γ. This shows that s ↦→ Γs is onto and completes the proof
of i).
Now suppose Γ = Γs is an alternating bilinear form on P × P with rank
2ρ. Then there is a basis y1, . . . , yr of P satisfying
⎧
⎨ 1, i = 2k − 1, j = 2k, 1 ≤ k ≤ ρ;
Γ(yi, yj) = −1,
⎩
0,
i = 2k, j = 2k − 1, 1 ≤ k ≤ ρ;
otherwise.
Using the same notation as above, put x2k−1 = µ(y2k), x2k = −µ(y2k−1),
if 1 ≤ k ≤ ρ. Then B(xi, yj) = δij, if 1 ≤ i, j ≤ 2ρ. For example,
B(x2k−1, yj) = B(yj, µ(y2k)) = Γ(yj, y2k) = δj,2k−1. Also, B(xi, yj) =
B(µ(yi ′), yj) = Γ(yi ′, yj) = 0 if 1 ≤ i ≤ 2ρ, 2ρ < j ≤ r. Since x ↦→ λx is an
isomorphism of N with P ∗ , there must be an xj ∈ N, 2ρ < j ≤ r, such that
λxj (yi) = δij, i.e., B(xj, yi) = δij for 2ρ < j ≤ r, 1 ≤ i ≤ r. Then x1, . . . , xr
form a basis for N since λx1, . . . , λxr is the basis of P ∗ dual to y1, . . . , yr.
Then
s · y2k−1 = y2k−1 + µ(y2k−1) = y2k−1 − x2k
s · y2k = y2k + µ(y2k) = y2k + x2k−1
⎫
⎬
, for k ≤ ρ.
⎭
For i > 2ρ, Γ(y, yi) = 0 for all y ∈ P , so µ(yi) = 0 ∈ N because
B(y, x) = 0 for all y ∈ P iff x = 0. Thus for i > 2ρ, s · yi = yi. For this s

5.6. ACTION OF THE ORTHOGONAL GROUP 271
(such that Γ = Γs) we consider the effect of the mapping x ↦→ s · x − x on
the basis {x1, . . . , xr, y1, . . . , yr} of M. (Recall that for s ∈ H, s · x = x, for
all x ∈ N.)
s · y2k−1 − y2k−1 = x2k, k ≤ ρ;
s · y2k − y2k = x2k−1, k ≤ ρ;
s · yi − yi = 0, i > 2ρ;
s · xi − xi = 0, 1 ≤ i ≤ r.
Thus a basis for the range of x ↦→ s · x − x is {x1, . . . , x2ρ}, implying that
the rank of Γs is indeed the rank of the linear map x ↦→ s · x − x.
Finally, to prove (iii) let s be as above and suppose that s ′ ∈ H also
has the property that the rank of Γs ′ is 2ρ. Suppose also that the basis
{x ′ 1, . . . , x ′ r, y ′ 1, . . . , y ′ r} has been determined from Γs ′ in the manner used for
Γs. Since q(xi) = q(x ′ i ) = q(yi) = q(y ′ i ) = 0, and since B(xi, yj) = B(x ′ i , y′ j ),
the automorphism τ of M defined by τ · xi = x ′ i , τ · yi = y ′ i (for 1 ≤ i ≤ r) is
in G, and τsτ −1 = s ′ .
Exercise: Check this last statement on a basis.
This completes a proof of the theorem.
Suppose that H is a hyperplane of M whose conjugate contains a vector
z such that q(z) = a = 0. For x ∈ M put s · x = x − a −1 · B(x, z)z. Then we
have the following:
Theorem 5.6.8. The linear map s is the unique element of G different from
the identity which leaves fixed the points of H.
Proof. Clearly s is linear, and q(s · x) = q(x − a −1 B(x, z)z) = q(x) +
q(−a −1 B(x, z)z)+B(x, −a −1 B(x, z)z) = q(x)+a −2 B 2 (x, z)q(z)−a −1 B 2 (x, z) =
q(x). So s ∈ G. And s is not affected if z is replaced by cz, 0 = c ∈ K.
Thus s depends only on H and is called the symmetry with respect to H .
Then s · x = x iff 0 = B(x, z), so H is the space of fixed points of s. And
s · z = z − a −1 2q(z)z = −z. Conversely, suppose s ′ ∈ G is not the identity on
M but is on H. Then if x ∈ H, s ′ ·x−x is a nonzero element in the conjugate
of H (by Theorem 5.6.5, whence s ′ · x = x + cz. Since q(s ′ · x) = q(x) by
hypothesis, q(x) + c 2 q(x) + cB(x, z) = q(x), implying c · a + B(x, z) = 0, i.e.,
c = −a −1 B(x, z). Since c cannot be zero, s ′ and s agree on H and and on
Kx (since s ′ · x = x − a −1 B(x, z)z = s · x). Hence s ′ = s.

272 CHAPTER 5. QUADRATIC FORMS
Exercise: Show that if τ ∈ G, then τsτ −1 is the symmetry with respect
to the hyperplane τ(H).
5.7 The Cartan-Dieudonné Theorem
This section is devoted to the proof of the so-called Cartan-Dieudonné Theorem,
one of the more famous results dealing with the orthogonal group. We
begin by stating the theorem itself, but the proof depends on a fairly long
sequence of lemmas.
Theorem 5.7.1. Except in the case where K has only two elements, M is
of dimension 4, and Q is of index 2, every element of G belongs to the group
G ′ generated by symmetries with respect to the hyperplanes whose conjugates
contain vectors not in Q.
We use the following notation:
For s ∈ G, L(s) is the set of fixed points of s;
ν(s) is the dimension of L(s);
µs is the linear transformation µs : x ↦→ s · x − x;
µs(M) is the conjugate space of L(s).
Lemma 5.7.2. Let s ∈ G, z ∈ µs(M) such that q(z) = 0, and let τ be the
symmetry with respect to the conjugate hyperplane H of z. Then ν(τs) > ν(s)
(i.e., dim µτs(M) < dim µs(M)). Furthermore, there is a τ ∈ G ′ such that
ν(τs) is maximal, and in that case µτs(M) ⊆ Q.
Proof. Let s ∈ G, z = s · y − y ∈ µs(M) such that q(z) = 0. Let τ be the
symmetry with respect to the conjugate hyperplane H of z. Thus τ · z = −z
and τ · h = h for all h ∈ H. And q(z) = q(s · y − y) = q(s · y) + q(−y) −
B(s · y, y) = 2 · q(y) − B(s · y, y) = B(y, y) − B(s · y, y) = −B(z, y). From the
original definition of τ we have τ · y = y − q(z) −1 · B(y, z)z = y + z = s · y.
Thus (τs) · y = τ · (z + y) = −z + s · y = y, implying y ∈ L(τs). On the
other hand, since µs(M) is the conjugate space of L(s) and z ∈ µx(M), the
conjugate space H of z must contain L(s). Thus any vector in L(s) is fixed
under τ, so L(s) + Ky ⊂ L(τs). Since clearly y ∈ L(s), ν(τs) > ν(s). Since
the dimension of M is finite, the rest of the lemma is clearly true.
We require a definition for the next lemma.

5.7. THE CARTAN-DIEUDONNÉ THEOREM 273
Defn. For s ∈ G, if µs(M) ⊂ Q, we say s is singular . If s is singular,
the index of s is defined to be the dimension ρ(s) = m − ν(s) of µs(M).
Lemma 5.7.3. Any two maximal subspaces of Q may be transformed into
each other by elements of G ′ .
Proof. Let N and N ′ be maximal subspaces of Q. It is clearly sufficient to
prove that if N ∩ N ′ has dimension l < dimN, then there is a hyperplane
H whose conjugate contains a vector z ∈ Q such that the symmetry τ with
respect to H transforms N ′ into a space τ(N ′ ) such that dim(N ∩τ(N ′ )) > l.
Since N + N ′ has dimension greater than dimN and N is a maximal
subspace of Q, N + N ′ must contain a vector z = x + x ′ such that q(z) = 0,
x ∈ N, x ′ ∈ N ′ . Then q(z) = q(x) + q(x ′ ) + B(x, x ′ ) = B(x, x ′ ) = 0, so that
x ∈ N ′ , x ′ ∈ N. Let H be the conjugate space of z and τ the associated
symmetry. Then τ(x ′ ) = x ′ −q(z) −1 B(x ′ , z)z = x ′ −B(x, x ′ ) −1 B(x ′ , x+x ′ )z =
x ′ −z = x. From τ(x ′ ) = x it follows that x ∈ N ∩τ(N ′ ). Also, if x ′′ ∈ N ∩N ′ ,
then B(x, x ′′ ) = B(x ′ , x ′′ ) = 0, so B(z, x ′′ ) = 0. Whence x ′′ ∈ H, τ · x ′′ = x ′′ ,
and x ′′ ∈ τ(N ′ ) ∩ N. Thus (N ∩ N ′ ) + Kx ⊆ N ∩ τ(N ′ ), which completes
the proof of the lemma.
We note that if s0 is any singular element of G, then µs0(M) is contained
in a maximal subspace N of Q. Since the conjugate of N is then contained
in the conjugate of µs0(M), it follows that the conjugate of N is contained
in L(s0). For a fixed N we define HN by
HN = {s ∈ G : s fixes each vector in the conjugate of N.
Let P be a maximal subspace of Q such that N ⊕ P is disjoint from its
conjugate R. So M = R ⊕ N ⊕ P . If r = dimN, then the conjugate of N,
which has dimension m − r, contains N + R, also of dimension m − r. So
N + R is the conjugate of N. Thus
HN = {s ∈ G : s fixes each element of N + R}.
Consider the subspace N ⊕ P with restricted Q ′ , B ′ , G 0 . Since any Qisomorphism
of N ⊕ P (i.e., any Q ′ -isomorphism) of N ⊕ P can be extended
to an element of G, G 0 is the full orthogonal group on N + P with respect
to Q ′ . Note that N is its own conjugate in N + P and define H by
H = {s ′ ∈ G 0 : s ′ · x = x for all x ∈ N}.

274 CHAPTER 5. QUADRATIC FORMS
We are now ready to state the next lemma using this notation.
Lemma 5.7.4. Any two elements of HN are singular and are conjugate in
G if they have the same f . HN is abelian and isomorphic to H, the subgroup
of G fixing N pointwise.
Proof. If s ′ ∈ H, then N ⊆ L(s ′ ), so µs ′(N + P ) ⊆ N. Thus µs ′(N + P ) ⊆
Q, and H contains only singular elements of G0 . By Theorem 5.6.7 H is
isomorphic with the additive abelian group of all alternating bilinear forms
on P × P under an isomorphism s ′ ↦→ Γs ′ where the rank of Γs ′ is the index
of s ′ , i.e., dimi µs ′(N + P ), and two elements of H are conjugate in G0 if
they have the same index.
Any Q-automorphism of N + P has a unique extension to a linear endomorphism
of M which leaves R fixed pointwise and which is easily seen to
be an element of G. Thus HN = {s ′ ∈ G : s ′ fixes N + R pointwise} may
be identified with H = {s ′ ∈ G0 : s ′ fixes N pointwise} by identifying each
s ′ ∈ H with its unique extension s ∈ HN. The index of s ′ is dim µs ′(N + P ),
and the index of s is dim µs(N +P +R) = dim µs(N +P ) = dim µx ′(N +P ),
since µs(R) = 0. Thus two elements of HN with the same index correspond
to elements of H of the same index. Two elements s ′ , s ′′ of H are conjugate
in G0 if they have the same index, say s ′ = τ −1s ′′ τ, τ ∈ G0 , s ′ , s ′′ ∈ H. By
extending s ′ , s ′′ , τ to elements of G fixing R pointwise, we obtain that the
extensions of s ′ and s ′′ in HN are conjugate in G.
Lemma 5.7.5. Each singular element of G may be transformed by some
element of G ′ into an element of HN.
Proof. Let s be a singular element of G, N ′ a maximal subspace of Q containing
µs(M), and let τ ∈ G ′ be such that τ(N ′ ) = N. Then µ(s s (M) ⊆ N ′ implies
that the conjugate of N ′ is contained in L(s), whence s fixes each point
of the conugate of N ′ and τ takes the conjugate of N ′ into the conjugate of N.
Thus τsτ −1 fixes the conjugate of N pointwise, implying τsτ −1 ∈ HN.
If s is an arbitrary element of G and τ is the symmetry with respect
to some hyperplane H ′ , we noted earlier that sτs −1 is the symmetry with
respect to the hyperplane s(H ′ ). Thus G ′ is normal in G.
Lemma 5.7.6. G = HNG ′ , and G/G ′ is abelian.

5.7. THE CARTAN-DIEUDONNÉ THEOREM 275
Proof. For s ∈ G, we know there is a τ ∈ G ′ such that τs is singular. Then
there is a τ1 ∈ G ′ such that τ1τsτ −1
1
= h ∈ HN, and s = τ −1 τ −1
1 hτ1 ∈
G ′ HNG ′ = HNG ′ , since G ′ ⊳ G. Thus G = HNG ′ . Since HN is abelian,
and G/G ′ = HNG ′ /G ′ ∼ = HN/(HN ∩ G ′ ) is a homomorphic image of HN, it
follows that G/G ′ is abelian.
Lemma 5.7.7. If there are singular elements s, s ′ ∈ G such that s, s ′ , ss ′
have the same index, then s, s ′ ∈ G ′ .
Proof. Let s, s ′ ∈ G be singular with the same index. Then they are conjugate
to h, h ′ ∈ HN. For any τ ∈ G, it is easily checked that L(τsτ −1 ) = τL(s),
so dim L(s) = dim L(τsτ −1 ). Thus s and h (similarly, s ′ and h ′ ) have the
same index. Whence h and h ′ have the same index and are conjugate in G,
implying that s and s ′ are conjugate in G. Let τsτ −1 = s ′ for some τ ∈ G. So
τsτ −1 G ′ = s ′ G ′ , implying s ′ G ′ = τG ′ sG ′ τ −1 G ′ = sG ′ , since G/G ′ is abelian,
i.e., s ′ G ′ = sG ′ . Thus singular elements of G with the same index belong to
the same coset of G ′ .
Now suppose that s, s ′ and ss ′ are all singular with index r. We have
just seen that sG ′ = s ′ G ′ = ss ′ G ′ = sG ′ · s ′ G ′ , so sG ′ = s ′ G ′ = G ′ = the
identity of G/G ′ . Thus s, s ′ ∈ G ′ .
We are now finally ready for the proof of Theorem 5.7.1
Proof. We know that G = HNG ′ , so the aim is to show that, with the
exception stated in the theorem, HN ⊆ G ′ . In the proof of Lemma 5.7.4 it
was noted that H ∼ = HN is isomorphic to the group of all alternating bilinear
forms on P × P under an isomorphism s ′ ↦→ Γs ′, where the rank of Γs ′ is the
index of s ′ , which is dim µs ′(N + P + R) = dim µs ′(N + P ).
Suppose that K has more than two elements. Let a ∈ K, 0 = a = −1,
and let s ∈ HN. Clearly aΓs is an alternating bilinear form on P × P with
the same rank as that of Γs. Thus aΓs = Γs ′ for some s′ ∈ HN. Similarly
(1 + a)Γs = Γss ′ has the same rank as does Γs. So s, s ′ , ss ′ are elements of
HN (thus all singular) with the same rank, implying by the last lemma that
s ∈ G ′ .
Suppose finally, that K has exactly two elements, and let r be the index
of Q. If r = 0 or r = 1, HN contains only the identity (the rank of any
alternating bilinear form being even). Then clearly HN ⊆ G ′ . Let r > 2, so
m > 4. Every alternating form is representable as a sum of forms of rank 2.

276 CHAPTER 5. QUADRATIC FORMS
Thus each s ∈ HN is a product of elements in HN of index 2. So it suffices
to show that s ∈ G ′ whenever s ∈ HN has index 2. So let s ∈ HN have index
2.
Suppose that x1, x2 form a basis for µs(M) ⊆ Q. Since L(s) ⊇ N +R (by
the definition of HN), taking conjugates we have µs(M) ⊆ N. So s1, x2 ∈ N.
Let y1 and y2 be elements of P such that µs(y1) = −x2, µs(y2) = x1. Then
the subspace X0 spanned by x1, x2, y1, y2 is disjoint from its conjugate R0
and has dimension 4. By hypothesis R0 has positive dimension (m > 4).
Since µs(M) ⊆ X0, L(s) ⊇ R0, so s fixes the elements of R0. Furthermore
R0 contains some vector z such that q(z) = 0. Let τ1, τ2, τ3, τ4 be the
symmetries with respect to the conjugates of z, z + x1 + x2, z + x2, z + x1,
and let τ = τ1τ2τ3τ4. Clearly the conjugates of z, z + x1 + x2, z + x2, z + x1
all contain {x1, x2}, so τi fixes x1 and x2. A straight forward calculation
using the assumption that the characteristic of K is 2 shows that for y ∈ R0,
τ2τ3τ4(y) = τ1(y). Thus τ(y) = τ1(τ1(y)) = y. So τ fixes R0 pointwise. Then
since M = {x1, x2} ⊕ {y1, y2} ⊕ R0, we will have τ = s if we can show that
τ · yi = s · yi, i = 1, 2.
Now τ(yi) = yi+B(yi, x2)x1+B(yi, x1)x2, after a few steps. Since {x1, x2}
is not in the conjugate space of yi, and for i = i ′ B(yi, xi ′) = B(yi, s·yi−yi) =
0 by the first part of the proof of Theorem 5.6.7, it must be that B(yi, xi) = 0.
But K has only two elements, so B(yi, xj) = δij. Hence τ · yi = s · yi.
The theorem has now been proved except in the case where K has 2
elements, dim M = 4, and Q has index 2. In that case one can check that
G ′ actually has index 2 in G.

Chapter 6
Quadrics in P G(3, q)
By Chapter 5 we know that there are two types of nonsingular quadrics in
P G(3, q), viz., hyperbolic and elliptic. These are of special interest to us
along with the degenerate quadratic cones. We start with a study of the
hyperbolic quadrics.
6.1 Hyperbolic quadrics in P G(3, q)
Let H3 be a hyperbolic quadric in P G(3, q), q any prime power. It follows
that H3 has lines but no planes, and H3 defines a polarity denoted by ”⊥”.
If ℓ is a line disjoint from ℓ ⊥ , then ℓ ∩ H3 and ℓ ⊥ ∩ H3 must be the same type
since H3 = ℓ ⊥ ℓ ⊥ is hyperbolic. Also, it follows easily from Theorem 5.4.4,
part (ii), that with respect to some basis it has the homogeneous form
f(x0, x1, x2, x3) = x0x1 + x2x3. (6.1)
From Theorem 5.4.5 we see that H3 has (t + 1) 2 points. In fact, it
follows fairly easily that the lines of H3 form a regulus together with its
opposite regulus. Put ℓ∞ = 〈(1, 0, 0, 0), (0, 0, −1, 0)〉, and for b ∈ Fq put
ℓb = 〈(b, 0, 0, 1), (0, 1, −b, 0)〉. Then put m∞ = 〈(1, 0, 0, 0), (0, 0, 0, 1)〉, and
for a ∈ Fq put ma = 〈(a, 0, −1, 0), (0, 1, 0, a)〉. Then L = {ℓb : b ∈ ˜ Fq} is
a regulus with opposite regulus M = {ma : a ∈ ˜ Fq}, and their points are
precisely the points of H3.
Let G0 be the group of homographies determined by the matrices
277

278 CHAPTER 6. QUADRICS IN P G(3, Q)
⎧⎛
⎪⎨ ⎜
G0 = ⎜
⎝
⎪⎩
⎞ ⎫
1 0 b 0
⎪⎬
0 1 0 0 ⎟
0 0 1 0 ⎠ , b ∈ Fq .
⎪⎭
0 −b 0 1
Then G0 fixes each point of m0, is sharply transitive on the lines ma : a ∈
˜F \ {0}, and fixes each line ℓa of the opposite regulus.
Let G∞ be the group of homographies determined by the matrices
⎧⎛
⎪⎨ ⎜
G∞ = ⎜
⎝
⎪⎩
1 0 0 0
0 1 0 −a
a 0 1 0
0 0 0 1
⎞ ⎫
⎪⎬
⎟
⎠ , a ∈ Fq.
⎪⎭
Then G∞ fixes each point of m∞, is sharply transitive on the lines ma :
a ∈ F , and fixes each line ℓa of the opposite regulus.
So G∞ and G0 generate a group that is doubly transitive on the lines
ma : a ∈ ˜ F and fixes each line of the opposite regulus. If G0,∞ is the group
with matrices
⎧⎛
⎪⎨ ⎜
G0,∞ = ⎜
⎝
⎪⎩
1 0 0 0
0 c 0 0
0 0 c 0
0 0 0 1
⎞
⎟
⎠
: 0 = c ∈ Fq
⎫
⎪⎬
,
⎪⎭
then G0,∞ is transitive on the lines ma with 0 = a ∈ Fq and fixes m0, m∞,
and all the lines of the opposite regulus L.
Now suppose that g is some homography with matrix [g] = (aij) that
fixes each point of both m∞ and m0 and fixes (1,0,1,0). It is easy to check
that this forces g to be the identity map.
Hence the group 〈G0, G∞, G0,∞〉 acts sharply triply transitively on the
lines of the regulus M and fixes each line of the opposite regulus L. Finally,
let θ be the homography with matrix
[θ] =
⎛
⎜
⎝
1 0 0 0
0 1 0 0
0 0 0 1
0 0 1 0
⎞
⎟
⎠ .

6.1. HYPERBOLIC QUADRICS IN P G(3, Q) 279
Then θ leaves invariant the hyperbolic quadric but interchanges the two reguli
M and L.
This gives us a group of homographies of order at least 2(q 3 − q) 2 leaving
the hyperbolic quadric H3 invariant. In fact, this is the order of the complete
homography stabilizer of H3, which is |P GL(3, q)| divided by the number of
ruled quadrics in P G(3, q). So the result follows from Lemma 2.12.6 and
Obs. 3.7.3. We record this as
Theorem 6.1.1. The group of homographies of P G(3, q) leaving invariant
a ruled quadric H3 has order 2(q 3 − q) 2 .
Note: If H3 contains the lines ℓ∞, ℓ0, m∞, m0 described above, it is easy
to check that the quadratic form looks like
fa(x0, x1, x2, x3) = x0x1 + ax2x3, a = 0.
The q − 1 ruled quadrics given by fa, a = 0, are the only such quadrics
containing the four lines and pairwise intersect in precisely those four lines.
THIS GIVES A SET OF q − 1 HYPERBOLIC QUADRICS MEETING
PAIRWISE IN THE FOUR LINES OF AN ORDINARY QUADRANGLE
NOT CONTAINED IN A PLANE.
By Theorem 5.3.4 it must be that if π is any plane of P G(3, q), then
π ∩ H3 is one of two possible types:
(i) π ∩ H3 is an irreducible quadric in π, i.e., a conic;
(ii) π ∩ H3 is a degenerate quadric, i.e., a pair of lines. This occurs if π
contains a point and the two lines of H3 through that point. Note that if ℓ
is a line of H3, then each plane through ℓ contains a unique line m of the
opposite “ruling” and is the tangent plane at the point ℓ ∩ m.
The conics lying in H3 are called circles of H3. A flock of H3 is a set F
of q + 1 mutually disjoint circles of H3. If ℓ is a line of P G(3, q) which has
no point in common with H3, then the circles π ∩ H3, where π is a plane
containing ℓ, constitute a linear flock of H3.
Theorem 6.1.2. (J. A. Thas [Th75]) Each flock of the non-singular ruled
quadric H of P G(3, q), q even, is linear.

280 CHAPTER 6. QUADRICS IN P G(3, Q)
Proof. Suppose that F = {C1, C2, · · · , Cq+1} is a flock of H. The nucleus of
the circle Ci is denoted by ni. We shall prove that ℓ = {n1, · · · , nq+1} is a
line. By Lemma 2.7.2 it suffices to show that each plane of P G(3, q) has at
least one point in common with ℓ.
(a) Let π be the tangent plane to H at the point x ∈ H. Through x
passes some circle Ci of the flock. The tangent line of Ci at x is contained in
π, so ni ∈ π.
(b) Let π be a plane for which π ∩ H is a conic Ci belonging to the flock
F. Then ni must belong to π.
(c) Finally, suppose that π is a plane for which C = π ∩ H is a circle of H
but not belonging to F. For each i, 1 ≤ i ≤ q+1, |Ci∩C| ∈ {0, 1, 2}. As q+1
is odd, there must be at least one Cj for which |Cj ∩ C| = 1. Hence Cj and
C have a common tangent line T at their common point. So nj ∈ T ⊂ π, as
claimed. Hence each plane of P G(3, q) contains a point of ℓ, so that ℓ must
be a line of P G(3, q).
Let ν be the symplectic polarity determined by H (so derived from
Eq. 6.1). The polar planes of n1, . . . , nq+1 with respect to ν are the planes
of the circles C1, . . . , Cq+1. As ℓ is a line, these q + 1 planes all pass through
the polar line of ℓ with respect to ν. Hence the flock F is linear.
Note that the above proof shows that if ℓ is a line disjoint from H, each
of the q + 1 planes through ℓ meets H in a conic whose nucleus is a point of
ℓ, and each point of ℓ is the nucleus of just one of the conics.
The hypothesis that q be even really is necessary. Using ideas developed
in the Ph.D. thesis of W. F. Orr [Or73] in the study of elliptic quadrics, J.
A. Thas [Th75] gave examples of nonlinear flocks of H in the case that q is
odd. We have developed this material (adapting the approach given by J. C.
Fisher and J. A. Thas [FT79]) in Section 6.7 of the present chapter. In that
section there is a construction of a non-linear flock of H first given in [Th75].
There is an interesting corollary due to A. A. Bruen and J. A. Thas [BT75]:
Theorem 6.1.3. Let C be an irreducible conic of P G(2, q), q even. If F =
{x1, . . . , xq+1} is a set of q + 1 points such that any line xixj, i = j, is an
exterior line of C, then F is an exterior line of C.
Proof. Let C be embedded in a hyperbolic quadric H of P G(3, q), q even.
Since ℓij = 〈xi, xj〉 is an exterior line to C, with respect to H it is elliptic, so
ℓ ⊥ ij must also be elliptic, i.e., disjoint from H. This says that (x⊥ i ∩H)∩(x⊥ j ∩

6.1. HYPERBOLIC QUADRICS IN P G(3, Q) 281
H) = ∅. Hence the polar planes π1, . . . , πq+1 of x1, . . . , xq+1 with respect
to H intersect H in q + 1 mutually disjoint circles (since each line xixj is
disjoint from H). These circles constitute a flock F ∗ of H. As q is even, by
Theorem 6.2.1 F ∗ is linear, so all the planes πi pass through one exterior line
ℓ of H. Consequently their poles xi all lie on one exterior line of H. This
implies that F is an exterior line of C.
This result was first generalized by Segre and Korchmaros [SK77] to include
the case q odd, then it was generalized even further by Blokhuis and
Wilbrink [BW87] to Theorem 2.6.1.
Theorem 6.1.4. (See Theorem 2.6.1) Let A and B be subsets of P G(2, q)
such that
(i) |A| = |B| = q + 1;
(ii) A ∩ B = ∅;
(iii) Every line meeting A meets B.
Then B is a line.
Let P be a fixed point of H = H3, and let L0, L1, . . . , Lq be the lines
through P in the plane πP tangent to H at P , with L0 and L1 the two lines
of πP in H. Note that each plane that meets H in a circle through P meets
πP in one of the lines L2, . . . , Lq. Also note that there are exactly q(q − 1)
circles through P . Fix a plane π through L2, π = πP and let C = π ∩H. The
q − 1 planes through L2 different from π and πp meet H in circles tangent to
C at P and even pairwise tangent at P . This leaves q(q − 1) − q = q(q − 2)
other circles through P , each in a plane meeting πP in one of the lines Lj,
3 ≤ j ≤ q. On the other hand, there are q points in C \ {P }, each lying
on q − 2 circles different from C but containing P . This accounts for all the
circles through P and establishes the following.
Theorem 6.1.5. Let π and π ′ be two planes that meet the hyperbolic quadric
H in the two circles C and C ′ that contain a common point P . Then C and
C ′ are tangent at P if and only if their planes π and π ′ meet the tangent
plane πP at P in the same line. Otherwise they meet in two points. Hence
tangency at P is an equivalence relation on the set of circles in H through
P . It follows that there is an affine plane AP defined as follows. Let L0 and
L1 be the two lines of H through P . The points of AP are the q 2 points of
H \ (L0 ∪ L1). The lines of AP are the circles of H through P and the lines
of H \ {L0, L1}.

282 CHAPTER 6. QUADRICS IN P G(3, Q)
We want to investigate the hyperbolic (i.e., ruled) quadrics in P G(3, q)
that contain a given conic. Let Q be the general upper triangular matrix
giving a nondegenerate quadric which is denoted Q:
Let C be the conic:
Q =
⎛
⎜
⎝
a11 a12 a13 a14
0 a22 a23 a24
0 0 a33 a34
0 0 0 a44
⎞
⎟
⎠ .
C = {(x0, x1, x2, x3) ∈ P G(3, q) : x2 = x3 and x0x1 + x 2 2 = 0}.
It follows that
C = {(−1, t 2 , t, t) : t ∈ Fq} ∪ {(0, 1, 0, 0)}.
If we assume that (0, 1, 0, 0) ∈ Q, clearly a22 = 0.
Next assume that (−1, t 2 , t, t) ∈ Q for all t ∈ Fq. This means that
0 = (−1, t 2 , t, t)Q(−1, t 2 , t, t) T
= (a23 + a24)t 3 + (−a12 + a33a34 + a44)t 2 + (−a13 − a14)t + a11 ∀t ∈ Fq.
It then follows that: a11 = 0, a14 = −a13, a24 = −a23, a12 = a33 + a34 + a44.
At this point we have proved the following:
Lemma 6.1.6. Q contains the conic C : x0x1 + x2 2 = 0; x2 = x3 if and only
if Q has the form
⎛
⎞
Q =
⎜
⎝
0 a33 + a34 + a44 a13 −a13
0 0 a23 −a23
0 0 a33 a34
0 0 0 a44
⎟
⎠ .
Now we want to investigate the special case of the quadric (a33 = a44 =
0; a13 = a; a23 = b; a34 = 1)
Qa,b =
⎛
⎜
⎝
0 1 a −a
0 0 b −b
0 0 0 1
0 0 0 0
⎞
⎟
⎠ .

6.1. HYPERBOLIC QUADRICS IN P G(3, Q) 283
So the quadratic form is
Qa,b : x0x1 + x2x3 + ax0(x2 − x3) + bx1(x2 − x3)
= x0x1 + x2x3 + (ax0 + bx1)(x2 − x3). (6.2)
If we compute the determinant of Q + Q T we find that it is δ = 1 + 4ab.
Hence Q is degenerate if and only if q is odd and δ = 1 + 4ab = 0. So from
now on we assume that 1 + 4ab = 0, and we want to determine whether Q is
hyperbolic or elliptic. In either case we know it contains the conic C.
Clearly the point (0, 1, 0, 0) is on Q. The polar of (0, 1, 0, 0) is the plane
[1, 0, b, −b] T . Q will be hyperbolic precisely when this plane contains two
lines of points in Q. The general point (b(x3 − x2), x1, x2, x3) of [1, 0, b, −b] T
is on Q if and only if
0 = b(x3 − x2)x1 + x2x3 + ab(x3 − x2)(x2 − x3) + bx1(x2 − x3)
= x2x3 − ab(x2 − x3) 2 .
Note that if there are values for x2 and x3 (not both zero) that satisfy this
equation, then any value of x1 will give a point on the intersection of the
polar and the quadric, i.e., the quadric is hyperbolic if and only if there is a
(nontrivial) solution to
if and only if there is a solution to
abx 2 2 − (1 + 2ab)x2x3 + abx 2 3 = 0,
abT 2 − (1 + 2ab)T + ab = 0. (6.3)
From this it is easy to see that the following holds:
Lemma 6.1.7. Assume that 1 + 4ab = 0. Then Qa,b is nondegenerate.
(i) If q is odd, Qa,b is hyperbolic if and only if 1 + 4ab is a (nonzero)
square.
(ii) If q is even, Qa,b is hyperbolic if and only if tr(ab) = 0.
Now suppose that Qa,b and Qa ′ ,b ′ are hyperbolic quadrics with (a, b) =
(a ′ , b ′ ), q odd or even. We want to consider the intersection of the two
quadrics. Let P = (x0, x1, x2, x3) be a point on both quadrics. Then the
coordinates of P must also satisfy
((a − a ′ )x0 + (b − b ′ )x1)(x2 − x3) = 0.

284 CHAPTER 6. QUADRICS IN P G(3, Q)
From this we see that a point in both quadrics either lies in the plane x2 = x3,
which meets both quadrics in the conic C, or it lies in the plane
π : (a − a ′ )x0 + (b − b ′ )x1 = 0.
At this point we know that either a−a ′ = 0 or b−b ′ = 0. By interchanging
the roles of x0 and x1, without loss of generality we may assume that a−a ′ =
0, so that a point in the plane has x0 = b ′ −b
a−a ′
x1 and then lies on the quadric
Qa,b if and only if
0 =
b ′ − b
a − a ′
x 2 1 + x2x3 +
′ ′ ab − a b
a − a ′
x1(x2 − x3).
A corresponding matrix for the quadric in the plane π is
⎛
b
¯Q = ⎝
′ − b ab ′ − a ′ b −(ab ′ − a ′ 0 0
b)
a − a ′
⎞
⎠ .
0 0 0
Put △ = ab ′ − a ′ b. The determinant of ( ¯ Q + ¯ Q T ) equals
| ¯ Q + ¯ Q T | = −2(a − a ′ ) △ 2 − (a − a ′ )(b − b ′ ) .
We consider some special cases. First consider Q1,0 ∩ Q0,1. So △ = 1,
a − a ′ = 1, b − b ′ = −1, and △ 2 − (a − a ′ )(b − b ′ ) = 2.
Q1 = Q1,0 : x0x1 + x2x3 + x0(x2 − x3);
Q2 = Q0,1 : x0x1 + x2x3 + x1(x2 − x3).
If (x0, x1, x2, x3) ∈ Q1 ∩ Q2, then (x0 − x1)(x2 − x3) = 0.
In x2 = x3 both intersections are x0x1 + x2 2 = 0, which we recognize as a
conic. In the plane x0 = x1 both intersections are x2 1 + x2x3 + x1(x2 − x3),
which is also easily checked to be a conic when q is odd and is degenerate
(x1 + x2)(x1 + x3) when q = 2e . So for q odd, Q1 ∩ Q2 is the union of 2
conics, C and C ′ . Then C ∩ C ′ is the set of points of the form (x, x, y, y) where
x2 + y2 = 0. It follows that if q is odd and −1 is a square, there are two
such points. If −1 is not a square, the intersection is empty. If q = 2e , the
intersection is a unique point (1, 1, 1) and the intersection of Q1 and Q2 is a
conic together with two lines through some point of the conic.

6.1. HYPERBOLIC QUADRICS IN P G(3, Q) 285
Second, suppose that q is odd and consider Q1,0 ∩ Q0,−1. In this case
△ = −1, a − a ′ = 1, b − b ′ = 1, and △ 2 − (a − a ′ )(b − b ′ ) = 0. In this
case Q1,0 ∩ Q0,−1 is a conic in the plane x2 = x3 union with a pair of lines
in π : x0 = −x1 meeting at a point not on the line of intersection of the two
planes and meeting x2 = x3 in the points (1, −1, 1, 1) and (−1, 1, 1, 1) of the
conic C.
For b = b ′ , Q0,b ∩ Q0,b ′ is the conic C in x2 = x3 together with the two
lines 〈(1, 0, 0, 0), (0, 0, 1, 0)〉 and 〈(1, 0, 0, 0), (0, 0, 0, 1)〉 meeting at the point
(1, 0, 0, 0) of C.
Suppose that the hyperbolic quadric Q contains the conic C and some
additional point P not on the plane x2 = x3 of C. There must be two lines of
Q through P , each containing some point of C. The group of homographies
of P G(2, q) leaving a conic invariant is sharply triply transitive on the points
of that conic, so we may assume that P = (0, 0, 0, 1) and the lines through
P lying on Q are the lines ℓ = 〈(1, 0, 0, 0), (0, 0, 0, 1) containing the point
(1, 0, 0, b) for each b ∈ Fq and m = 〈(−1, 1, 1, 1), (0, 0, 0, 1)〉 with points
(−1, 1, 1, b), b ∈ Fq. Starting with Lemma 6.1.5 it is easy to see that Q is
given by a matrix of the form
Q =
⎛
⎜
⎝
0 a33 + a34 0 0
0 0 a34 −a34
0 0 a33 a34
0 0 0 0
Each point of ℓ lies on a second line of Q and it must go through a point
of C, so for some b ∈ Fq, b = 0, the second line n of Q through (1, 0, 0, b)
passes through the point (0, 1, 0, 0) of C. Hence (1, 1, 0, b) is on Q, forcing
Qb =
⎛
⎜
⎝
0 ba34 0 0
0 0 a34 −a34
0 0 (b − 1)a34 a34
0 0 0 0
⎞
⎟
⎠ =
⎛
⎜
⎝
⎞
⎟
⎠ .
0 b 0 0
0 0 1 −1
0 0 (b − 1) 1
0 0 0 0
⎞
⎟
⎠ ,
for each b ∈ Fq \ {0, 1}.
Since |Qb + Q T b | = b2 = 0, Qb is nondegenerate, so it must be hyperbolic
since it contains lines. Putting b = 1 we obtain the following.

286 CHAPTER 6. QUADRICS IN P G(3, Q)
Lemma 6.1.8. The unique nondegenerate quadric containing the conic C :
x0x1 + x 2 2 = 0; x2 = x3 and containing the lines
and
ℓ = 〈(1, 0, 0, 0), (0, 0, 0, 1)〉, m = 〈(0, 0, 0, 1), (−1, 1, 1, 1)〉,
n = 〈(1, 0, 0, 1), (0, 1, 0, 0)〉,
must have the matrix Q (putting a34 = 1) given by
Q =
⎛
⎜
⎝
0 1 0 0
0 0 1 −1
0 0 0 1
0 0 0 0
⎞
⎟
⎠ .
It is a routine exercise to show that the two quadrics given by b = 1
and 0 = b = 1 meet in exactly the points of C and the points of the lines
ℓ and n. (Start with points (0, 1, 0, s) and (−1, t 2 , t, s) on x0x1 + x 2 2 = 0,
the “difference” of the two quadratic forms, and put them back in the two
original forms to obtain the result.)
For the remainder of this section we assume that q = 2 e .
Using Theorem 4.3.1 (and interchanging coordinates x1 and x2) we can
establish the following:
Lemma 6.1.9. The homographies of P G(3, q) that leave invariant the conic
x0x1 + x 2 2 = 0 in the plane x2 = x3 are given by matrices of the form:
A =
⎛
⎜
⎝
a b √ ab √ ab
c d √ cd √ cd
0 0 e f
0 0 e + 1 f + 1
⎞
⎟
⎠ , ad + bc = 1; e = f.
Recall that the order of the group of homographies fixing a conic in a
plane is q 3 − q. This is the number of ordered 4-tuples (a, b, c, d) such that
ad + bc = 1. Then there are q(q − 1) ordered pairs (e, f) with e = f. This
gives
Corollary 6.1.10. The group of homographies of P G(3, q) leaving invariant
the plane x2 = x3 and the conic x0x1 = x 2 2 has order (q + 1)q 2 (q − 1) 2 .

6.1. HYPERBOLIC QUADRICS IN P G(3, Q) 287
Lemma 6.1.11. If A is the matrix in Lemma 6.1.9, then
A −1 ⎛
d b d
⎜
= ⎜
⎝
√ ab + b √ cd d √ ab + b √ cd
c a c √ ab + a √ cd c √ ab + a √ ⎞
cd
⎟
f ⎟
0 0
⎠ .
We know that Q =
0 0
f+1
e+f
e+1
e+f
⎛
⎜
⎝
0 1 0 0
0 0 0 0
0 0 0 1
0 0 0 0
⎞
e+f
e
e+f
⎟
⎠ is the matrix giving the hyperbolic
quadric x0x1 + x2x3 = 0.
The homography (x0, . . . , x3) ↦→ (x0, . . . , x3)A with A as in Lemma 6.1.9,
replaces Q with A −1 QA −T , which is equivalent to the upper triangular matrix
⎛
0 1 d√ab+b √ cd
e+f
0 0 c√ab+a √ cd
Q ′ ⎜
= ⎜
⎝ 0 0
0 0 0
e+f
f(f+1)
(e+f) 2
d √ ab+b √ cd
e+f
c √ ab+a √ cd
e+f
1
e+f
(e+1)e
(e+f) 2
⎞
⎟
⎠ =
⎛
0
⎜
0
⎜
⎝ 0
1
0
0
√
bd
e+f √
ac
e+f
f(f+1)
(e+f) 2
0 0 0
√
bd
e+f √
ac
e+f
1
e+f
(e+1)e
(e+f) 2
⎞
⎟
⎠ .
So these are the hyperbolic quadrics that contain the conic C : x0x1+x 2 2 =
0 in the plane x2 = x3.
We ask: How many of these quadrics are of the form
A little work yields the following:
Hb : x0x1 + x2x3 + b(x 2 2 + x 2 3) = 0?
Lemma 6.1.12. There are 2q(q − 1) homographies leaving C invariant and
sending H0 to a hyperbolic quadric of the form Hb. They have matrices of
the form
⎛
⎜
⎝
a 0 0 0
0 a −1 0 0
0 0 e e + 1
0 0 e + 1 e
⎞
⎟
⎠ , a = 0, e ∈ Fq;

288 CHAPTER 6. QUADRICS IN P G(3, Q)
or ⎛
⎜
⎝
0 b 0 0
b −1 0 0 0
0 0 e e + 1
0 0 e + 1 e
⎞
⎟
⎠ , b = 0, e ∈ Fq.
The 4(q − 1) homographies fixing H0 and C are those with e ∈ {0, 1} and
either a = 0, or b = 0, respectively.
The hyperbolic quadric H with matrix Q determines a null polarity ν :
x ↦→ (Q + Q T )x T . If x ∈ H, then x ν is the tangent plane to H at x which
contains one line in each ruling of H. If x ∈ H, then x is the nucleus of the
conic x ν ∩ H. The nucleus of the conic C in Hb is (0, 0, 1, 1), and the ruled
quadrics of the form Hb have an interesting property with respect to the lines
through N = (0, 0, 1, 1).
Lemma 6.1.13. If a line through N not in the plane x2 = x3 has one point
in common with some hyperbolic quadric of the form Hb, then it meets each
of the q/2 quadrics Hb in exactly two points. As each of the quadrics has
q(q + 1) points not in the plane x2 = x3, it takes q(q + 1)/2 lines through
N to partition the points (not in the given plane) in the union of the hyperbolic
quadrics, leaving exactly q(q−1)
lines through N covering precisely all the
2
points of P G(3, q) not on the plane x2 = x2 and not on any of the Hb. Each
line contained in one of the Hb meets the plane x2 = x3 in a unique point
of the conic C. If we pick one ruling of each of the Hb along with the lines
through N not meeting any Hb we have a partition of the points of P G(3, q)
not in the plane x2 = x3.
Proof. The matrix Q ′ on the previous page gives the quadratic form
HB : x0x1 + B(x 2 2 + x2 3 )
if and only if √ bd = √ ac = 0 and e + f = 1. It is then of the form
⎛
0
⎜ 0
⎝ 0
1
0
0
0
0
e(e + 1)
0
0
1
⎞
⎟
⎠ , with B = e(e + 1).
0 0 0 e(e + 1)
As e ranges over the q elements of Fq, B assumes each field element with
trace 0 exactly twice. Also, e = 0 and e = 1 give H0.

6.2. QUADRATIC CONES IN P G(3, Q) 289
Suppose that x = (x0, x1, x2, x3) ∈ H0 \ (x2 = x3), so x0x1 + x2x3 = 0
with x2 = x3. N = (0, 0, 1, 1) is the nucleus of C. So the line ℓ = 〈x, N〉
contains x ∈ H0 \ (x2 = x3). The point P = x + λN (an arbitrary point of ℓ
not in (x2 = x2)) is on HB (with B = e 2 + e) if and only if
satisfies
if and only if
if and only if
if and only if
(x0, x1, x2, x3) + λ(0, 0, 1, 1) = (x0, x1, x2 + λ, x3 + λ)
x0x1 + (x2 + λ)(x3 + λ) + B((x2 + λ) 2 (x3 + λ) 2 ) = 0
λ 2 + λ(x2 + x3) + (e 2 + e)(x2 + x3) 2 = 0
λ
x2 + x3
∈ {e, e + 1}
λ ∈ {e(x2 + x3), (e + 1)(x2 + x3)}.
Hence each point of ℓ different from N is in exactly one of the Hb. There
are q(q +1) points of H0 not on the plane x2 = x3. Hence q(q+1)
lines through
2
N exactly cover the points of all the Hb not in the plane x2 = x3, so that
there are q2 − q(q+1) q(q−1)
= lines through N not containing any point of
2
2
any Hb and whose union minus the point N exactly covers the points of
P G(3, q)not on an Hb and not in the plane x2 = x3.
6.2 Quadratic cones in P G(3, q)
Let π = P G(2, q) be the hyperplane [0, 0, 0, 1] in Σ = P G(3, q), q odd or
even. Let O be an oval in π and let V be the point V = (0, 0, 0, 1) of Σ \ π.
The cone K with base oval O and vertex V is the set of points on the lines
(called generators) joining V to the points of O. If O is a conic (for example,
O = C = {(x, y, z, 0) ∈ π : y 2 = xz}, then K is called a quadratic cone, and
K = {(x, y, z, w) ∈ Σ : y 2 = xz}.
When q = 2 e , O has a nucleus N, and the line 〈N, V 〉 is called the nuclear
generator of the cone. Although quadratic cones play a central role in the
theory of interest to us, and indeed are the only cones we consider when q is

290 CHAPTER 6. QUADRICS IN P G(3, Q)
odd, in the case q = 2 e we also study cones over more general ovals, especially
over monomial ovals. Since much of the general theory of quadratic cones
generalizes to translation oval cones and some of it to monomial oval cones,
we post pone some of our development of cones in characteristic 2 to the
chapter on monomial oval cones.
For the remainder of this section we suppose that O is the conic C given
above.
Probably one of the first things we should examine is the group of homographies
of Σ = P G(3, q) leaving the cone K invariant. First note that each
field automorphism induces a collineation of Σ that leaves the cone invariant,
so we really only need to look at homographies. Each plane that does not
contain the vertex V meets each generator in a unique point. Let (1, 0, 0, r),
(1, 1, 1, s) and (0, 0, 1, t) be three arbitrary points on the respective generators
containing them. They determine the plane π = [−r, r + t − s, −t, 1].
Consider the homography θ which is the elation with center V and axis
[−r, r + t − s, −t, 0]. The matrix of θ is
[θ] = A =
⎛
⎜
⎝
1 0 0 −r
0 1 0 r + t − s
0 0 1 −t
0 0 0 1
The map θ leaves K invariant and maps the plane π to the plane π3 =
[0, 0, 0, 1]. Hence we now need only to determine the homographies that fix
the plane π3 and leave the conic C invariant. At this point we recall that
the complete group of homographies of π3 leaving C invariant consists of all
θ with matrices
⎛
[θ] = ⎝
d 2 −bd b 2
−2cd ad + bc −2ab
c 2 −ac a 2
⎞
⎞
⎟
⎠ .
⎠ , ∆ = ad − bc = 0. (6.4)
The determinant of this matrix is ∆3 , so we assume the matrix has been
replaced by [θ]/∆, so it has determinant 1. Then the inverse matrix (which
acts on the planes of Σ) is
⎛
[θ] −1 = ⎝
a 2 ab b 2
2ac ad + bc 2bd
c 2 cd d 2
⎞
⎠ .

6.2. QUADRATIC CONES IN P G(3, Q) 291
We have essentially proved the following.
Theorem 6.2.1. The complete group of homographies of P G(3, q) that leave
invariant the cone K = {(x, y, z, w) ∈ P G(3, q) : xz = y2 } consists of those
θ given by all matrices of the form
[θ] =
⎛
⎜
⎝
a 2 −bd b 2 x
−2cd ad + bc −2ab y
c 2 −ac a 2 z
0 0 0 w
⎞
⎟
⎠ , ∆ = ad − bc = 1; w = 0.
A collection of q conics of the cone K that partition the points of K \ {V }
is called a flock of the quadratic cone K. . However, if C is one of the conics
of a flock F, there is a unique plane π for which C = π ∩ K. It is often more
convenient to consider a flock of a cone to be the set of these q planes whose
intersections with K give the actual flock, hence we shall often do this.
Obs. 6.2.2. Let K be a quadratic cone in P G(3, q) with vertex V . Let
P ∈ K \ {V }, ℓ = 〈P, V 〉 and let πℓ be the plane meeting K in ℓ. Let π be any
plane containing neither V nor P . Project the points of K \ {V }from P onto
π. Then there is a one-to-one correspondence between the points of K \ ℓ and
the points of π \(πℓ ∩π), while the points of ℓ\{P, V } project onto P ′ = ℓ∩π.
The q3 − q2 plane sections of K containing neither P nor V project onto the
q3 − q2 conics of π containing P ′ and with tangent line π ∩ πℓ. If C1 and C2
are two planar sectioins of K both containing the point Q ∈ ℓ \ {P, V }, then
their respective projections C ′ 1 , C′ 2 in π have the property that C′ 2 is one of the
q2 images of C ′ 1 under an elation of π with center P ′ . The q2 planar sections
of K containing P project onto the q2 lines of π not incident with P ′ . Hence
a flock {C1, . . . , Cq} of K projects to a set {C ′ 1 , . . . , C′ q−1 , m} where C′ 1 , . . . , C′ q−1
are conics of π with common point P ′ , common tangent π ∩ πℓ, m is a line of
π not incident with P ′ , and C ′ 1 , . . . C′ q−1 , m partition the points of π \ (π ∩ πℓ).
Further, no C ′ i is the image of a C ′ j, i, j ∈ {1, . . . , q − 1}, i = j, under an
elation of π with center P ′ . Conversely, any such set {C ′ 1 , . . . , C′ q−1 , m} with
these properties corresponds to a flock of K.
Proof. We make the above comments explicit, starting with the cone K as
given above. The homography φ with matrix
⎛
⎞
1 0 0 −u
⎜
[φ] = ⎜ 0 1 0 −v ⎟
⎝ 0 0 1 −w ⎠
0 0 0 1

292 CHAPTER 6. QUADRICS IN P G(3, Q)
maps the plane [u, v, w, 1] to the plane [0, 0, 0, 1] without moving any generators
of the cone. So we may assume that π = [0, 0, 0, 1]. Let ℓ be the
generator through P = (0, 0, 1, 1) ∈ K and the vertex V = (0, 0, 0, 1), and
let π∞ = [1, 0, 0, 0] be the plane meeting K in ℓ. ℓ meets π in the point
P ′ = (0, 0, 1, 0). The homographies θ of P G(3, q) fixing the vertex V and
inducing elations on π with center P ′ have matrices
[θa,b] =
⎛
⎜
⎝
The conic C = K ∩ [0, 0, 0, 1] is
The image of C under θa,b is
1 0 a 0
0 1 b 0
0 0 1 0
0 0 0 1
⎞
⎟
⎠ , a, b ∈ Fq.
C = {(1, x, x 2 , 0) : x ∈ Fq} ∪ {(0, 0, 1, 0)}.
Ca,b = {(1, x, a + bx + x 2 , 0) : x ∈ Fq} ∪ {(0, 0, 1, 0)}.
The planes through P not through V are those of the form [u, v, −1, 1].
The conic which is the intersection of such a plane with the cone K projects
from P to the line {(1, x, vx + u, 0) : x ∈ Fq} ∪ {(0, 1, v, 0)}. Now let Q =
(0, 0, 1, λ) with 1 = λ ∈ Fq. The general plane through Q but not V is the
plane [a, b, −λ, 1]. It meets the cone in the conic
Ca,b,λ = {(0, 0, 1, λ)} ∪ {(1, x, x 2 , λx 2 − bx − a) : x ∈ Fq} ∪ {(0, 0, 1, λ)}.
Then Ca,b,λ projects to the conic
C ′ a,b,λ = {(0, 0, 1, 0)} ∪ {(1, x, (1 − λ)x2 + bx + a, 0) : x ∈ Fq}, when λ = 1.
It is now clear that for a fixed λ = 1, the conics C ′ a,b,λ
form one orbit under
the group of elations with center P ′ , as do the lines of π not through the
point P ′ .
For the remainder of this section we broaden our point of view. Let π be
a plane of P G(3, q) and V a point of P G(3, q) \ {V }. Let O be an oval of
π and let K be the cone with vertex V and generators the line joining V to
the points of O. Then each plane section of K by a plane not containing V
is an oval projectively equivalent to O. When q is even let N be the nucleus

6.2. QUADRATIC CONES IN P G(3, Q) 293
of O and let n be the line joining V with N. Each plane through n must be
tangent to the cone. Clearly each oval O ′ that is the intersection of K and
some plane π ′ has a nucleus N ′ on the line n. We define a flock of K to be a
set F of q planes π1, . . . , πq cutting K in pairwise disjoint ovals O1, . . . , Oq.
Of course, if q is odd, each oval is a conic and there is no nuclear generator.
Lemma 6.2.3. Let F = {π1, . . . , πq} be a flock the oval cone K meeting
K in the ovals O1, . . . , Oq, respectively. If Ni is the nucleus of Oi, then
V, N1, . . . , Nq are the q + 1 distinct points of the nuclear generator n.
Proof. Suppose that Ni = Nj for some i = j. Then πi ∩ πj is a line through
Ni = Nj tangent to Oi and to Oj. Hence πi ∩ πj must meet the hyperoval
Oi ∪ {Ni} in a second point and also must meet the hyperoval Oj ∪ {Nj} in
a second point. These two points must be the same point, forcing Oi and Oj
not to be disjoint, contradicting the assumption that F is a flock.
Remark: It follows that F is a flock of the hyperoval cone obtained by
adjoining the nuclear generator n to the cone K.
In the next theorem we show that a partial flock of deficiency 1, i.e., q −1
planes cutting the cone in disjoint ovals, uniquely extends to a complete flock.
This result first appeared in [PT91] with separate proofs for q odd or even.
The result can be strengthened considerably (see [ST95], but we give a proof
only for partial flocks of deficiency 1. The first proof is the proof for q even
from [PT91]. The second proof is a simple one discovered by Peter Sziklai
that works for all conical cones.
Theorem 6.2.4. ([PT91]) Let K be an oval cone in P G(3, q) with vertex V .
For 2 ≤ i ≤ q, let πi be a plane not through V cutting the cone in an oval Oi
so that Oi ∩ Oj = ∅ if i = j. So on each generator there is a unique point
different from V and not belonging to any of O2, . . . , Oq. Let these points
be Z0, . . . , Zq. Then there is a plane π meeting K in exactly these points
Z0, . . . , Zq. In other words, a partial flock of deficiency 1 can be completed
to a flock.
Proof. Case 1. q even. In this case we let n be the nuclear generator of the
oval cone K. For 1 ≤ i ≤ q − 1 let Ni be the nucleus of Oi on n. The same
proof as for the preceding lemma shows that V, N1, . . . , Nq−1 are distinct
points of n. Let Oq be the set of q + 1 points (one on each generator of K)
not lying on any Oi for 1 ≤ i ≤ q − 1, and let Nq be the remaining point of
n. Finally, put O + q = Oq ∪ {Nq}.

294 CHAPTER 6. QUADRICS IN P G(3, Q)
First we show that each plane π of P G(3, q) must meet O + q in an even
number of points. If π is some πi, 1 ≤ i ≤ q − 1, then |O + q ∩ π| = 0. If π
contains the vertex V , then π contains 0 or 2 generators of K and hence 0 or
2 points of O + q . If V ∈ π, |(K ∪n)∩π| = q +2 is even, and π meets each πi in
a line with 0 or 2 points of K ∪ n, implying |π ∩ O + q | is even. Project from Nq
onto a plane not containing Nq. Then Oq projects onto q + 1 distinct points,
since the generators of K together with n form a set of q + 2 lines with no
three in a plane. All planes through Nq project onto all lines of π, with each
such plane meeting Oq in an odd number of points, so in at least one point.
So each plane through Nq projects to a line that contains one of the points to
which Oq projects. Say the points Z0, . . . , Zq of Oq project to W0, . . . , Wq of
π. Put m = W0W1, and suppose that Z is a point of m not in the projection
of Oq. The lines ZW0 = ZW1 = m, ZW2, . . . , ZWq are at most q in number,
leaving at least one line ℓ through Z missing all of W0, . . . , Wq. Hence the
plane 〈ℓ, Nq〉 must miss Oq. This contradiction forces m to contain all of
W0, . . . , Wq. Hence mNq is the desired plane πq for which πq ∩ K = Oq.
Case 2. q odd or even in the quadratic cone case.
Let F = {πi = [ai, bi, ci, 1] : 2 ≤ i ≤ q} be a partial flock of deficiency 1
of the standard cone K : x0x2 = x 2 1. For fixed t ∈ Fq the map i ↦→ fi(t) =
ai + bit + cit 2 , 2 ≤ i ≤ q, is an injection into Fq, and the point of πi on
the generator through the point (1, t, t 2 , 0) is (1, t, t 2 , −fi(t)). Similarly, the
point of πi on the generator through (0, 0, 1, 0) is (0, 0, 1, −ci). Also, when
q > 2 the sum of the elements of Fq is equal to 0. Hence the missing point
on the generator 〈(0, 0, 0, 1), (1, t, t 2 , 0) (not on any of the πi) is
(1, t, t 2 ,
q
(ai + bit + cit 2 ).
i=2
Similarly, the missing point on the generator 〈(0, 0, 0, 1), (0, 0, 1, 0)〉 is
(0, 0, 1,
q
ci).
All these points lie on the plane [− ai, − bi, − ci, 1]. Hence F extends
to a flock.
i=2

6.3. ELLIPTIC QUADRICS IN P G(3, Q) 295
6.3 Elliptic quadrics in P G(3, q)
Let Ω be an elliptic quadric in P G(3, q). By Theorem 5.4.5 Ω has 1 + q 2
points. Since Ω contains points but no lines, it must be that no line contains
three points of Ω, i.e., Ω is a k-cap of P G(3, q) with k = q 2 + 1. Hence by
looking ahead to Theorem 7.2.6 we see that Ω is an ovoid. In the chapter
on ovoids we will show that each plane of P G(3, q) is either tangent to Ω or
meets Ω in a conic. Recall that Ω defines a polarity that here we denote by
⊥.
If ℓ is a line of P G(3, q) that has two points of Ω, it is hyperbolic, so
ℓ ⊥ must be elliptic (i.e., anisotropic). Hence the perp of a secant line is an
exterior line, and vice versa, while the perp of a tangent line is a tangent
line.
A flock of Ω is a set F of mutually disjoint conics of Ω such that, with
the exception of precisely two points x and y, each point of Ω is on a unique
conic of F. The points x and y are called the carriers of the flock.
Let ℓ be a line of P G(3, q) which has no point in common with Ω. The
conics π ∩ Ω, where π varies over the planes through ℓ with |π ∩ Ω| > 1, form
a flock called a linear flock. When q = 2 e this is the only kind of flock.
Theorem 6.3.1. (J. A. Thas [Th73]) Let q = 2 e > 2 and let F = {C1, . . . , Cq−1}
be a flock of the elliptic quadric Ω. The planes of the q − 1 circles Ci all pass
through the same line ℓ. This line ℓ is the intersection of the tangent planes
to Ω at the carriers x, y of the flock.
Proof. First note that Ω = C1 ∪ C2 ∪ · · · ∪ Cq−1 ∪ {x, y}. The nucleus of the
circle Ci is denoted by ni. We now show that ℓ = {x, y, n1, n2, · · · , nq−1} is a
line. For that purpose we show that each plane of P G(3, q) has at least one
point in common with ℓ. Then by Lemma 2.7.2 it must be that ℓ is a line.
(a) Every plane through x or y certainly has a point in common with ℓ.
(b) Let π be a plane tangent to Ω at the point p ∈ Ω, x = p = y. Through
p there passes a circle of the flock, say Ci. The tangent line of Ci at p is
contained in π, forcing ni to be in π.
(c) Let π be a plane for which π ∩ Ω ∈ {C1, C2, . . . , Cq−1}. If Ci = π ∩ Ω,
then ni ∈ π.
(d) Finally, let π be a plane for which |π ∩ Ω| > 1 and x ∈ π, y ∈ π,
π ∩ Ω ∈ {C1, . . . , Cq−1}. If π ∩ Ω = C, then |Ci ∩ C| ∈ {0, 1, 2}, and
π ∩ Ω = ∪{Ci : 1 ≤ i ≤ q − 1}. As q + 1 is odd there exists a Cj such that

296 CHAPTER 6. QUADRICS IN P G(3, Q)
|Cj ∩C| = 1. It follows that Cj and C have a common tangent line T at their
common point. Hence nj ∈ T ⊂ π, completing the proof that ℓ is a line.
Next we note that the polar planes of x, y, n1, . . . , nq−1, with respect to
the symplectic polarity ν defined by Ω, are the tangent planes of Ω at x and
y, respectively, and the the planes of the conics Ci. As ℓ is a line, these q + 1
planes all pass through the polar line of ℓ with respect to ν. So we conclude
that the planes of the q − 1 circles Ci all pass through the intersection of the
tangent planes of Ω at the carriers x, y of the flock F.
This theorem says that when q = 2 e each flock of an elliptic quadric in
P G(3, q) must be linear. That this theorem is also true for q odd was proved
by W. F. Orr. The proof is a bit long for inclusion right here, but it does
appear later as Theorem 6.7.10.
Theorem 6.3.2. (W. F. Orr [Or73]) When q is odd, each flock of an elliptic
quadric in P G(3, q) must be linear.
There are two corollaries we mention here.
Corollary 6.3.3. Let C be an oval of P G(2, q), q > 2, which can be embedded
in an ovoid of P G(3, q) (for example, an irreducible conic). If F =
{x1, x2, . . . , xq−1} is a set of q − 1 points of P G(2, q) \ C, such that any line
xixj, i = j, is a secant of C, then the points of F all lie on one secant of C.
Proof. Let C be embedded in an ovoid Ω of P G(3, q). Since each xi is exterior
to Ω, its polar plane πi with respect to Ω meets Ω in an oval. Since 〈xi, xj〉
is a secant line, its perp, which is πi ∩ πj, must be an exterior line. Hence
the polar planes π1, π2, . . . , πq−1 of x1, . . . , xq−1 with respect to Ω, intersect
Ω in q − 1 mutually disjoint circles. These circles constitute a flock F ∗ of
Ω. As F ∗ must be linear by Theorems 6.3.1 and 6.3.2, the planes πi all pass
through one exterior line ℓ of Ω. Consequently their poles xi all lie on one
exterior line of Ω. We conclude that F is an exterior line of C.
Corollary 6.3.4. Let Ω be an elliptic quadric of P G(3, q), q even. Let F be
a set of q + 1 circles, any two of which have at least two points in common,
If the planes of these circles all meet in one point p ∈ Ω, then F is a bundle
of circles (i.e., the q + 1 conics all meet in two fixed points on a secant line
to the quadric).

6.4. THE (ELLIPTIC) ORTHOGONAL GROUP 297
Proof. The poles (with respect to Ω) of the q + 1 planes containing the
elements of F, all lie in the polar plane π of p. The line joining any two
of these poles is an exterior line of the irreducible conic π ∩ Ω. Since q is
even, the set F ∗ of these poles is an exterior line of 〈π, Ω〉 by Theorem 6.1.2.
Consequently the planes of the q + 1 conics of F all contain one fixed secant
of Ω. We conclude that F is a bundle of conics of the elliptic quadric Ω.
6.4 The (Elliptic) Orthogonal Group
In Chapter 5 we determined a canonical form for elliptic quadrics in case q is
even or odd. In this section we want to give explicit examples and determine
the corresponding orthogonal groups. First suppose that q is odd and let η
be a non-square in Fq. Put
Then
A =
⎛
⎜
⎝
0 1 0 0
1 0 0 0
0 0 −2 0
0 0 0 2η
⎞
⎟
⎠ .
O = {x = (x0, x1, x2, x3) : xAx T = 2x0x1 − 2x 2 2 + 2ηx23 = 0}
is a typical elliptic quadric in P G(3, q). Alternatively,
O = {(1, s 2 − ηt 2 , s, t) ∈ P G(3, q) : s, t ∈ Fq} ∪ {(0, 1, 0, 0)}.
For any a, b ∈ Fq, define the homography τa,b by the matrix
[τa,b] =
⎛
⎜
⎝
1 a 2 − ηb 2 a b
0 1 0 0
0 2a 1 0
0 −2ηb 0 1
⎞
⎟
⎠ .
It is routine to check that τa,b preserves the elliptic quadric O, fixes the point
(0, 1, 0, 0) and maps (1, 0, 0, 0) ↦→ (1, a 2 − ηb 2 , a, b), i.e., it is transitive on the
q 2 points of O different from (0, 1, 0, 0). Next, consider the homography φ

298 CHAPTER 6. QUADRICS IN P G(3, Q)
given by the matrix
[φ] =
⎛
⎜
⎝
0 1 0 0
1 0 0 0
0 0 1 0
0 0 0 1
It is easy to check that φ leaves O invariant and interchanges the points
(0, 1, 0, 0) and (1, 0, 0, 0). This shows that the orthogonal group G = P GO − (4, q)
is transitive on the points of O, and in fact is doubly transitive.
Now let (0, 0, ) = (c, d) ∈ Fq × Fq and let ϕc,d be the homography given
by the matrix
[ϕc,d] =
⎛
⎜
⎝
⎞
⎟
⎠ .
1 0 0 0
0 −c 2 − ηd 2 0 0
0 0 c d
0 0 ηd c
Again it is routine to check that ϕc,d ∈ G, it fixes (0, 1, 0, 0) and (1, 0, 0, 0) and
is transitive on the remaining q2 − 1 points of O, hence G is triply transitive
on the points of O. At this point it is easy to verify that the subgroup of
G fixing the three points (1, 0, 0, 0), (0, 1, 0, 0), (1, 1, 1, 0) has order 2 with
homographies given by the matrices
A =
⎛
⎜
⎝
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 ±1
⎞
⎟
⎠ .
⎞
⎟
⎠ .
Now let q = 2 e and let ρ ∈ Fq with tr(ρ) = 1. Then a typical elliptic
quadric is given by
O = {x = (x0, x1, x2, x3) ∈ P G(3, q) : xAx T = x0x1+x 2 2 +x2x3+ρx 2 3 = 0} where
⎛
0
⎜
A = ⎜ 0
⎝ 0
1
0
0
0
0
1
⎞
0
0 ⎟
1 ⎠
0 0 0 ρ
.
Alternatively, if f(s, t) = s 2 + st + ρt 2 , then
O = {(1, f(s, t), s, t) : s, t ∈ Fq} ∪ {(0, 1, 0, 0)}.

6.5. ELLIPTIC QUADRICS CONTAINING A CONIC 299
For a, b ∈ Fq, let τa,b be the homography with matrix
⎛
1
⎜
[τa,b] = ⎜ 0
⎝ 0
f(a, b)
1
b
a
0
1
⎞
b
0 ⎟
0 ⎠
0 a 0 1
.
Then τa,b ∈ G = P GO−1 , fixes (0, 1, 0, 0) and maps (1, 0, 0, 0) to (1, f(a, b), a, b).
The homography φ with matrix
⎛
0
⎜
[φ] = ⎜ 1
⎝ 0
1
0
0
0
0
1
⎞
0
0 ⎟
0 ⎠
0 0 0 1
preserves O and interchanges the points (1, 0, 0, 0) and (0, 1, 0, 0), showing
that G is transitive on the points of O, and in fact doubly transitive. Now
let (0, 0) = (c, d) ∈ Fq × Fq, put λ = f(c, d), which is not zero, and let ϕc,d
be the homography with matrix
[ϕc,d] =
⎛
⎜
⎝
1 0 0 0
0 λ 0 0
0 0 c d
0 0 ρd c + d
Then ϕc,d is in G, fixes (1, 0, 0, 0) and (0, 1, 0, 0) and maps (1, 1, 1, 0) to
(1, f(c, d), c, d). This shows that G is triply transitive on the points of O.
Finally, we want to determine the elements of G that fix the three points
(1, 0, 0, 0), (0, 1, 0, 0) and (1, 1, 1, 0). A little work shows that this group is of
order 2 and is generated by the homography with matrix
P =
⎛
⎜
⎝
1 0 0 0
0 1 0 0
0 0 1 0
0 0 1 1
⎞
⎟
⎠ .
⎞
⎟
⎠ .
6.5 Elliptic Quadrics Containing a Conic
Let O be a quadric containing the point ∞ = (0, 1, 0, 0) and having the plane
π∞ = [1, 0, 0, 0] as its tangent plane at ∞.

300 CHAPTER 6. QUADRICS IN P G(3, Q)
This is enough to force O to be given by an upper triangular matrix of
the form
[O] = A =
⎛
⎜
⎝
a 1 c d
0 0 0 0
0 0 h i
0 0 0 j
⎞
⎟
⎠ .
We want O to be elliptic, so the plane π∞ should contain no point of O
other than ∞.
Compute:
π∞ ∩ O = {(0, y, z, w) : hz 2 + izw + jw 2 = 0}.
If π∞ is tangent to O at ∞, it must be that the quadratic equation
hz 2 + izw + jw 2 = 0
has no nontrivial solutions (z, w). This is true if and only if
If q is odd, then i 2 − 4hj = ❆ ; (6.5)
hj
If q is even, then tr
i2
= 1. (6.6)
In particular, both h and j are nonzero, and when q is even also i is
nonzero.
Later on we are going to want the following.
Let q be even and let O be an elliptic quadric in P G(3, q) with π∞ =
[1, 0, 0, 0] as its tangent plane at the point ∞ = (0, 1, 0, 0). This means it has
a matrix
AO = [O] =
⎛
⎜
⎝
a 1 c d
0 0 0 0
0 0 h i
0 0 0 j
where hz 2 + izw + jw 2 = 0 has no nontrivial solution (z, w). In particular
h, i, j are all nonzero. The bilinear form is a polarity, so [1, 0, 0, 0] has
⎞
⎟
⎠

6.5. ELLIPTIC QUADRICS CONTAINING A CONIC 301
(0, 1, 0, 0) as its pole and (0, 1, 0, 0) is a point of [1, 0, 0, 0]. Here
A + A T ⎛
0
⎜
= ⎜ 1
⎝ c
1
0
0
c
0
0
⎞
d
0 ⎟
i ⎠
d 0 i 0
and
(A + A T ) −1 =
⎛
⎜
⎝
0 1 c d
1 0 d/i c/i
0 d/i 0 1/i
0 c/i 1/i 0
Let O ′ be a second such elliptic quadric with parameters a ′ , c ′ , d ′ , h ′ , i ′ , j ′ .
Then there is a plane π (not π∞) through the point ∞ having the same
pole N with respect to both symplectic polarities. In fact, there are planes
sharing a pole with the original elliptic quadric, containing ∞, and covering
all the points not on π∞. We now determine just when [a, 0, b, e] is such a
plane.
(A + A T ) −1
⎡
⎢
⎣
a
0
b
e
⎤
⎞
⎟
⎠
⎥
⎦ = [0, a + bd/i + ce/i, e/i, b/i]T .
Replacing A by A ′ , then computing the pole of [a, 0, b, e] T and setting it equal
to λ times the plane displayed above, and using the fact that at least one of
b, e must be nonzero, yields the system of equations
a(i ′ − i) + b(d ′ − d) + e(c ′ − c) = 0.
In addition to wanting the plane [a, 0, b, e] T to have the same pole with
respect to both polarities, we want to find one containing an arbitrary point
(1, x, y, z) of P G(3, q) \ π∞. So we are looking for nontrivial solutions to the
system
i ′ − i d ′ − d c ′ − c
1 y z
⎡
⎣
a
b
e
⎤
⎡
⎦ = ⎣
This system has rank 1 or 2, forcing at least a one-dimensional space of
solutions, as desired. So we have proved the lemma:
0
0
0
⎤
⎦ .

302 CHAPTER 6. QUADRICS IN P G(3, Q)
Lemma 6.5.1. Let q be even. Let O and O ′ be any two elliptic quadrics
having (∞, π∞) as pole-polar pair. Then each point off π∞ belongs to some
plane π through ∞ for which π has the same pole N with respect to both
polarities defined by O and O ′ .
Let π = [0, 0, 0, 1], so π is a secant plane through ∞.
π ∩ O = {(x, y, z, 0) : ax 2 + xy + cxz + hz 2 = 0}.
If O = {(1, f(s, t), s, t) : s, t ∈ Fq} ∪ {(0, 1, 0, 0)}, then
⎛
a
⎜
(1, f(s, t), s, t) ⎜ 0
⎝ 0
1
0
0
c
0
h
⎞ ⎛ ⎞
d 1
0 ⎟ ⎜
⎟ ⎜ f(s, t) ⎟
i ⎠ ⎝ s ⎠
0 0 0 j t
implies
= a + f(s, t) + cs + dt + hs 2 + ist + jt 2 = 0
f(s, t) = −hs 2 − ist − jt 2 − cs − dt − a. (6.7)
We can choose coordinates in π so that
C = π ∩ O = {(1, s 2 , s, 0) : s ∈ Fq} ∪ {∞}.
This means that h = −1 and a = c = 0, so
⎛
0
⎜
A = ⎜ 0
⎝ 0
1
0
0
0
0
−1
⎞
k
0 ⎟
i ⎠
0 0 0 j
,
where we have renamed d as k.
This completes a proof of the following theorem.
Theorem 6.5.2. Let π∞ = [1, 0, 0, 0], π = [0, 0, 0, 1], so π∩π∞ = {(0, y, z, 0) ∈
P G(3, q) : y, z ∈ Fq}. Let C be the conic in π given by
C = {(1, s 2 , s, 0) : s ∈ Fq} ∪ {∞}.
Then the elliptic quadric O contains C and has π∞ as tangent plane at ∞ if
and only if O is given by an upper triangular matrix of the form

6.5. ELLIPTIC QUADRICS CONTAINING A CONIC 303
where
A =
⎛
⎜
⎝
0 1 0 k
0 0 0 0
0 0 −1 i
0 0 0 j
⎞
⎟
⎠ ,
−z 2 + izw + jw 2 = 0 (6.8)
has no nontrivial solution.
This means that when q is odd, i2 + 4j = ❆ ∈ Fq, and when q is even,
tr( j
i2 ) = 1.
So O = {(1, s2 −ist−jt 2 −kt, s, t) : s, t ∈ Fq}∪{(0, 1, 0, 0)}. By choosing
coordinates appropriately we can even normalize O a bit more.
1. If q is odd and η = ❆ ∈ Fq, i = k = 0, j = η, we have
⎛
0
⎜
A = ⎜ 0
⎝ 0
1
0
0
0
0
−1
⎞
0
0 ⎟
0 ⎠
0 0 0 η
,
and O = {(1, s 2 − ηt 2 , s, t) : s, t ∈ Fq} ∪ {∞}.
2. If q is even and tr(ρ) = 1, i = 1, j = ρ, k = 0, we have
⎛
0
⎜
A = ⎜ 0
⎝ 0
1
0
0
0
0
−1
⎞
0
0 ⎟
1 ⎠
0 0 0 ρ
,
and O = {(1, s 2 + st + ρt 2 , s, t) : s, t ∈ Fq} ∪ {∞}.
Now let O ′ be any other elliptic quadric containing C and having π∞ as
tangent plane at ∞. There must be an homography φ mapping O to O ′ . The
group of homographies of Σ = P G(3, q) leaving O invariant is triply transitive
on the points of O. Hnce we may assume φ fixes (0, 1, 0, 0), (1, 0, 0, 0), and
(1, 1, 1, 0). Of course φ must also leave π∞ invariant. It is a routine exercise
to show that this is enough to force the matrix of φ to look like
[φ] =
⎛
⎜
⎝
1 0 0 0
0 a 0 0
0 1 − a 1 0
0 b c d
⎞
⎟
⎠ , a, d = 0.

304 CHAPTER 6. QUADRICS IN P G(3, Q)
It is also clear that φ must fix the conic C. Since φ : (1, s 2 , s, 0) ↦→
((1, as 2 + s(1 − a), s, 0) ∈ C, it follows that a = 1, so
[φ] =
⎛
⎜
⎝
1 0 0 0
0 1 0 0
0 0 1 0
0 b c d
⎞
⎟
⎠ , d = 0.
It follows that φ is a central collineation with axis the plane π and center
the point (0, b, c, d − 1), which is in π if and only if d = 1.
and
Starting with
[φ] =
A =
⎛
⎜
⎝
⎛
⎜
⎝
0 1 0 k
0 0 0 0
0 0 −1 i
0 0 0 j
1 0 0 0
0 1 0 0
0 0 1 0
0 b c d
⎞
⎞
⎟
⎠ ,
⎟
⎠ , d = 0,
φ transforms the quadratic form associated with A to that given by
A ′ = [φ] −1 A[φ] ≡
⎛
⎜
⎝
0 1 0 k−b
d
0 0 0 0
0 0 −1 i+2c
d
0 0 0
j−ic−c 2
d 2
Since d = 0, it first appears that there are q 2 (q−1) matrices A ′ . However,
with fixed i, j, k, it might be that as b, c, d vary the matrices A ′ are not
all distinct. Suppose that i, j, k are fixed with −z 2 + izw + jw 2 = 0 having
no nontrivial solution (z, w), and that
A ′ :=
⎛
⎜
⎝
0 1 0 k−b
d
0 0 0 0
0 0 −1 i+2c
d
0 0 0
j−ic−c 2
d 2
⎞
⎛
⎟
⎠ = A′′ ⎜
:= ⎜
⎝
⎞
⎟
⎠ .
0 1 0 k−b ′
d ′
0 0 0 0
0 0 −1 d ′
0 0 0
i+2c ′
j−ic ′ −c ′2
d ′2
⎞
⎟
⎠ .

6.5. ELLIPTIC QUADRICS CONTAINING A CONIC 305
First suppose that d = d ′ . Then b = b ′ . If q is odd, also c = c ′ , so no
duplication occurs. Suppose that q = 2 e and d = d ′ . By Eq.4, when q is
even i must be nonzero, so there are two choices for φ with A ′ = A ′′ ; viz.
(b, c, d) = (b ′ , c ′ , c ′ ) = (b, c + i, d).
Now suppose that d = d ′ and A ′ = A ′′ . From the third row of A ′ it follows
that d = d ′ if q is even. Hence q is odd and it is easy to see that
b ′ = k(d − d′ ) + bd ′
d
and c ′ = 2cd′ + i(d ′ − d)
.
2d
Substitute this value for c ′ into the equality j−ic−c2
d2 plify. In a finite number of steps it appears that
(4j + i 2 )(d ′2 − d 2 ) = 0.
= j−ic′ −c ′2
d ′2
and sim-
When q is odd we have seen that 4j + j 2 = ❆ ∈ Fq, so it must be that
d ′2 = d 2 , i.e., d ′ = −d. This completes a proof of the following theorem.
Theorem 6.5.3. Let O be an elliptic quadric containing the conic C =
{(1, s2 , s, 0) : s ∈ Fq} ∪ {∞} in the plane π = [0, 0, 0, 1] and having π∞ =
[1, 0, 0, 0] as its tangent plane at the point ∞ = (0, 1, 0, 0).
We have seen that this means that the upper triangular matrix giving O
has the form given in Theorem 1. If O is a fixed one of these quadrics, then
any other one of them can be obtained from O by applying a collineation φ
with matrix
[φ] = [φb,c,d] =
⎛
⎜
⎝
1 0 0 0
0 1 0 0
0 0 1 0
0 b c d
⎞
⎟
⎠ , d = 0,
which transforms the quadratic form associated with A to that given by
⎛
⎞
A ′ = [φ] −1 A[φ] −T ≡
⎜
⎝
0 1 0 k−b
d
0 0 0 0
0 0 −1 d
0 0 0
j−ic−c2 d2 1. Suppose that q is odd. Then φb,c,d transforms O to the same quadric
as does φb ′ ,c ′ ,d ′ if and only if either (b′ , c ′ , d ′ ) = (b, c, d) or (b ′ , c ′ , d ′ ) = (2k −
b, −i − c, −d). Since b and c can be any elements of Fq and d can be any
i+2c
⎟
⎠ .

306 CHAPTER 6. QUADRICS IN P G(3, Q)
nonzero element of Fq, there must be q 2 (q − 1)/2 distinct elliptic quadrics
containing C and having π∞ = [1, 0, 0, 0] as tangent plane at the point ∞ =
(0, 1, 0, 0).
2. Suppose that q = 2 e . In this case we know that i = 0. Then
φb,c,d transforms O to the same quadric as does φb ′ ,c ′ ,d ′ if and only if either
(b ′ , c ′ , d ′ ) = (b, c, d) or (b ′ , c ′ , d ′ ) = (b, c + i, d). Again, there must be
q2 (q − 1)/2 distinct quadrics containing C and having π∞
tangent plane at the point ∞ = (0, 1, 0, 0).
= [1, 0, 0, 0] as
Note: In both the q odd and the q even cases the two φ that send O to
the same quadric have the same formula: (b ′ , c ′ , d ′ ) = (2k − b, −i − c, −d).
Moreover, the unique nonidentity homography leaving O and π∞ fixed is
φ2k,−i,−1.
Now we want to consider the intersection of two elliptic quadrics that
contain C and have π∞ as the tangent plane at ∞.
Suppose O is given by an upper triangular matrix with given parameters
(i, j, k) and is mapped by φb,c,d to the elliptic quadric O ′ with parameters
(i ′ , j ′ , k ′ ) = ( i+2d
d
, j−ic−c2
d 2
, k−b
d ).
The functions for O and O ′ , respectively, are
f(s, t) = s 2 − ist − jt 2 − kt, s, t); (6.9)
g(s, t) = s 2 2 i + 2c c + ic − j
− st +
d
d2
t 2 b − k
+ t. (6.10)
d
The two ovoids have points in common for values of s and t for which
f(s, t) = g(s, t). When t = 0 exactly the points of C turn up. So we may
assume that t = 0 and ask: When does f(s, t) = g(s, t)? This turns out to
be if and only if
s[i(d 2 − d) − 2cd] + t[c 2 + ic + (d 2 − 1)j] + k(d 2 − d) + bd = 0. (6.11)
The Case q is odd
Suppose q is odd and let η be an arbitrary nonsquare in Fq. Coordinates
may be chosen so that our special ovoid O is
O = {(1, s 2 − ηt 2 , s, t) : s, t ∈ Fq}.

6.5. ELLIPTIC QUADRICS CONTAINING A CONIC 307
Then Eq. 6.7 becomes
It follows that
s[−2cd] + t[c 2 + (d 2 − 1)j] + bd = 0. (6.12)
(O ∩ O ′ ) \ π = (O ∩ [bd, 0, −2cd, c 2 + (d 2 − 1)η]) \ {∞}. (6.13)
If we write πb,c,d = [bd, 0, −2cd, c2 + (d2 − 1)η], then ∞ ∈ πb,c,d, and
πb,c,d = π∞ if and only if c = 0 and d2 = 1. (in which case it must be that
b = 0).
Note: If c = 0, then πb,c,d is a secant plane to O through the point ∞, and
O ∩ O ′ is the union of two conics that share the points ∞ and (1, b2
4c2 , b , 0). 2c
Lemma 6.5.4. Suppose that q is odd and
O = {(1, s 2 − ηt 2 , s, t) : s, t ∈ Fq} ∪ {∞},
where η = ❆ ∈ Fq. Let φb,c,1 be a nontrivial elation with center (0, b, c, 0)
and axis π = [0, 0, 0, 1], so φb,c,1 maps O to
O ′ = {(1, s 2 − 2cst + (c 2 − η)t 2 + bt, s, t) : s, t ∈ Fq} ∪ {∞}.
Then O ∩ O ′ = C if and only if c = 0. Moreover, if c = 0, then each of the
q − 1 elliptic quadrics O ′ that together with O form an orbit of size q of the
group of elations with center ∞ = (0, 1, 0, 0) and axis π∞meets O in precisely
C, and each two of these quadrics meet pairwise in precisely C.
Proof. From Eq. 6.10 we see that O ∩ O ′ has a point outside C if and only
if there is a nonzero t and associated s for which −2cs + c 2 t + b = 0. When
c = 0 there clearly is such a solution. If c = 0, and if b = 0, there is no such
solution. If P is a point of O different from ∞, then φb,0,1 maps P to a point
on the line through ∞ and P , which has no further point of O.
Since φb,c,d maps O to the same quadric as does φ−b,−c,−d, if d 2 = 1 in
Eq. 6.10 we may assume that d = 1. So now suppose that d ∈ Fq \{0, 1, −1}.
We want to determine which quadrics arise which meet O in exactly C. So
basically the answer is that they are those arising from φb,c,d with Eq. 6.10
having no solution. Clearly if c = 0 there will be additional points in the
intersection. So suppose that c = 0 in Eq. 6.10: t(d 2 − 1)η + bd = 0. But
here if b = 0, there will always be a nonzero solution for t. Hence we must
have b = 0. Then since φ0,0,d = φ0,0,−d, we have the following.

308 CHAPTER 6. QUADRICS IN P G(3, Q)
Lemma 6.5.5. The only elliptic quadrics containing C, having π∞ as tangent
plane at ∞, meeting O in precisely C and NOT included in the preceding
Lemma lie in the orbit containing O of size (q − 1)/2 under the group of
homologies with axis π and center (0, 0, 0, 1). These quadrics pairwise meet
in precisely C.
Theorem 6.5.6. Let q be odd. The complete list of all elliptic quadrics that
meet O in precisely C and have π∞ as tangent plane at ∞ is as follows:
(i) The q − 1 images of O under the q − 1 nontrivial elations with axis
π = [0, 0, 0, 1] and center ∞ = (0, 1, 0, 0). These pairwise intersect in C and
have ovoidal functions
g(s, t) = s 2 − ηt 2 + bt. (6.14)
(ii) The (q −3)/2 images of O under the q −3 nontrivial homologies (with
d = 0, 1, −1) having axis π and center (0, 0, 0, 1). These pairwise intersect
precisely in C and have ovoidal functions of the form
h(s, t) = s 2 − ηd 2 t 2 (where we have replaced d with its inverse). (6.15)
(iii) Each elliptic quadric (not O) of the type in (i) meets each elliptic
quadric (not O) of the type in (ii) in a at least one point not in C.
(iv) Any set of more than (q − 1)/2 elliptic quadrics containing O and
whose elements meet pairwise in precisely the conic C must consist entirely
of conics of type (i).
Now we consider the case that q is even.
We let q = 2 e and put f(s, t) = s 2 + st + ρt 2 where ρ is a fixed element
of Fq with trace(ρ) = 1. It follows that the image of O under φb,c,d has ovoid
function
g(s, t) = s 2 2 1 c + c − ρ
− st +
d
d2
Eq. 6.8 becomes
t 2 + b
t. (6.16)
d
s(d 2 − d) + t(c 2 + c + (d 2 − 1)j) + bd = 0. (6.17)
If d = 1 there will be solutions of Eq. 6.17 for each t = 0. Hence if the
intersection with O is to be precisely C, we must have d = 1. Then the
equation becomes
t(c 2 + c) + b = 0.
This will fail to have solutions with t = 0 if and only if c 2 + c = 0 but b = 0
or c 2 + c = 0 but b = 0.

6.5. ELLIPTIC QUADRICS CONTAINING A CONIC 309
Theorem 6.5.7. Let q = 2 e . The complete list of all elliptic quadrics that
meet O in precisely C and have π∞ as tangent plane at ∞ is as follows:
(i) The q − 1 images of O under the q − 1 nontrivial elations with axis
π and center ∞. These pairwise intersect in C and have ovoidal functions
g(s, t) = s 2 + st + ρt 2 + bt, b = 0.
(ii) The (q − 2)/2 images of O (different from O) under the group of
elations with axis π and center (0, 0, 1, 0) . (Remember φ0,c,1 and φ0,c+1,1 give
the same ovoid O ′ .) These pairwise meet in precisely C and have ovoidal
functions h(s, t) = s 2 + st + (c 2 + c + ρ)t 2 .
(iii) Each elliptic quadric (not O) of the type in (i) meets each elliptic
quadric of type (ii) in point(s) not on C.
(iv) Any set of more than q/2 elliptic quadrics containing O and whose
elements meet pairwise in precisely the conic C must consist entirely of conics
of type (i).
Proof. It is routine to show that g(s, t) = h(s, t) has a solution s for each
nonzero t.
The following theorem is an immediate consequence of the preceding results.
Theorem 6.5.8. For q odd or even let Ω be a collection of q elliptic quadrics
meeting pairwise precisely in the conic C of the plane π and having tangent
plane π∞ = π at the point ∞ ∈ C. Then Ω must be an orbit of the group of
q elations with center ∞ and axis π.
The group of homographies of Σ leaving O invariant is transitive on the
secant planes to O containing ∞. Hence if π is any such secant plane, and if
C is the conic C = O ∩ π, any set of exactly q − 1 other ovoids of Σ meeting
O in precisely C must be those in the orbit containing O of the group of q
elations with center ∞ and axis π. Moreover, we will show that the group
G1 generated by these elations (as π varies over the secant planes containing
(∞) is the group of order q3 containing all the elations of Σ with center ∞
(including those with axis π∞).
Let φa,b,c be the elation with
[φa,b,c] =
⎛
⎜
⎝
1 a 0 0
0 1 0 0
0 b 1 0
0 c 0 1
⎞
⎟
⎠ , which has axis [a, 0, b, c].

310 CHAPTER 6. QUADRICS IN P G(3, Q)
Then G1 = {φa,b,c : a, b, c ∈ Fq} and it is easy to see that [φa,b,c]·[φa ′ ,b ′ ,c ′] =
[φa+a ′ ,b+b ′ ,c+c ′]. So by putting a′ = 0, b ′ = −b and c ′ = −c where (b, c) =
(0, 0), we get [φa,b,c] · [φ0,−b,−c] = [φa,0,0]. Hence the group of all elations with
center ∞ and axis π∞ = [1, 0, 0, 0] is contained in the group G1. Let Ω be
the set of ovoids in the orbit of G1 containing O. These q3 ovoids come in
q2 + q sets of size q that contain the ovoids meeting pairwise exactly in some
conic of O containing ∞ along with a set of q ovoids including O that are
the orbit containing O of the group of elations with center ∞ and axis π∞.
Moreover, the elations with center ∞ and axis π∞ map O in an orbit of size
q consisting of ovoids that form a rosette of ovoids with base point ∞ and
base plane π∞. (A rosette of ovoids is a set of q ovoids meeting pairwise in a
fixed point P (called the base point) and all having the same tangent plane
at P (called the base plane) This proves the following theorem.
Theorem 6.5.9. The G1-orbit Ω containing O consists of q 3 ovoids. There
are q of these including O that form a rosette with base point ∞ and base
plane π∞. The remaining q 3 − q elliptic quadrics in Ω share a common conic
with O. Hence two elliptic quadrics in Ω either intersect in the point ∞ or
in a conic containing ∞. We also note that the q 3 elliptic quadrics in Ω are
partitioned into q 2 disjoint rosettes with base point ∞, each rosette being an
orbit of the group of elations with center ∞ and axis π∞.
With O = {(1, f(s, t), s, t) : s, t ∈ Fq} ∪ {∞}, we want to see what the
elliptic quadrics in its G1-orbit look like.
φa,b,c : O ↦→ {(1, f(s, t) + bs + ct + a, s, t) : s, t ∈ Fq} ∪ {∞}.
Let G2 be the group of elations of P G(3, q) with axis π∞ = [1, 0, 0, 0]. If
θa,b,c has matrix
⎛
1
⎜
[θa,b,c] = ⎜ 0
⎝ 0
a
1
0
b
0
1
⎞
c
0 ⎟
0 ⎠
0 0 0 1
,
then G2 = {θa,b,c : a, b, c ∈ Fq}. Moreover,
θa,b,c : O ↦→ O ′ = {(1, f(s, t) + a, b + s, c + t) : s, t ∈ Fq} ∪ {∞}.

6.6. A CONDITION THAT DETERMINES A QUADRIC 311
It is then easy to see that
O ′ = {(1, f(s − b, t − c) + a, s, t) : s, t ∈ Fq} ∪ {∞}.
When q is odd and f(s, t) = s 2 − ηt 2 ,
a + f(s − b, t − c) = f(s, t) + −2bs + 2ηct + f(b, c) + a.
When q is even and f(s, t) = s 2 + st + ρt 2 , then
f(s − b, t − c) + a = f(s, t) + sc + bt + f(b, c) + a.
This makes it clear that G2 leaves Ω invariant. Now put G = 〈G1, G2〉. Since
φ −1
a,b,c ◦ θd,e,f ◦ φa,b,c = θd+be+cf,e,f, G = G1 · G2. Clearly |G1 ∩ G2| = q, so
|G| = q5 .
Hence we have proved the following:
Lemma 6.5.10. The group G of order q 5 generated by the elations with
center ∞ and the elations with axis π∞ leaves invariant the G1-orbit Ω containing
O.
6.6 A Condition that Determines a Quadric
This section is devoted to a proof of the following theorem:
Theorem 6.6.1. Let ∞ and R be points of P G(3, q) and let π, π∞, πR be
planes of P G(3, q) such that ∞ ∈ π∞ ∩ π, R ∈ πR \ (π ∪ π∞) and the three
planes π, π∞, πR meet in a point. If C is any conic of π such that ∞ ∈ C,
π ∩ π∞ is the tangent to C at ∞ and π ∩ πR is external to C, then there exists
a unique elliptic quadric O such that C ∪ {R} ⊂ O, π∞ is the tangent plane
to O at ∞ and πR is the tangent plane to O at R.
Further, suppose that C ′ is a second conic of π containing ∞, with tangent
line π∩π∞ at ∞, external line π∩πR and that O ′ is the unique elliptic quadric
containing C ′ ∪ {R} and such that π∞ is the tangent plane to O ′ at ∞ and
πR is the tangent plane to O ′ at R. Then O ∩ O ′ = {∞, R} if and only if
C ∩ C ′ = {∞} and C ′ is not the image of C under an elation of π with center
∞.

312 CHAPTER 6. QUADRICS IN P G(3, Q)
Proof. By choosing coordinates appropriately we may assume that ∞ =
(0, 1, 0, 0), π∞ = [1, 0, 0, 0], π = [0, 0, 0, 1], and C = {(1, s 2 , s, 0) : s ∈ Fq} ∪
{∞}.
Let φ be the homography with matrix
[φ] =
⎛
⎜
⎝
t 2 0 0 0
0 1 0 0
0 0 t 0
0 0 0 1
⎞
⎟
⎠ .
Then φ fixes C, the point ∞, the plane π∞. It maps (0, λ, 1, 0) to (0, λ/t, 1, 0).
The plane πR will meet the line π ∩ π∞ at a point of the form (0, λ, 1, 0), and
we have just seen that we may assume
λ ∈ {0, 1}.
Now suppose that the quadratic form is given by the matrix
A =
⎛
⎜
⎝
a b c d
0 e f g
0 0 h i
0 0 0 j
⎞
⎟
⎠ ,
and let Q denote the associated quadric. Use the fact that ∞ ∈ Q to see
e = 0. Then (1, s 2 , s, 0) ∈ Q for all s ∈ Fq implies a = c = f = 0 and h = −b.
If we compute ∞ ⊥ = [b, 0, 0, g], which must be [1, 0, 0, 0], we see that g = 0
and we may put b = 1. At this point
A =
⎛
⎜
⎝
0 1 0 d
0 0 0 0
0 0 −1 i
0 0 0 j
⎞
⎟
⎠ , so A + AT =
⎛
⎜
⎝
0 1 0 d
1 0 0 0
0 0 −2 i
d 0 i 2j
⎞
⎟
⎠ .
It is now convenient to split the argument into two cases: q odd or even.
Case 1 q odd. We want R to be a point on the quadric but not in π or
π∞. So we may write R = (1, u, v, w) with w = 0. πR is supposed to be a
plane containing R for which R ⊥ = πR and πR is to contain no other point of
the quadric. πR must meet the line π ∩ π∞ in a point of the form (0, λ, 1, 0).

6.6. A CONDITION THAT DETERMINES A QUADRIC 313
Let φ be the homography with matrix
⎛
1
⎜
[φ] = ⎜
⎝
a2 0
0
1
2a
a
0
1
⎞
0
0 ⎟
0 ⎠
0 0 0 1
.
This φ leaves C invariant and fixes both ∞ and π∞. It maps (0, λ, 1, 0) to
(0, λ + 2a, 1, 0), so by picking a appropriately we may assume that π ∩ πR ∩
π∞ = (0, 0, 1, 0). Now apply the homography with matrix
⎛
⎜
⎝
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 r
⎞
⎟
⎠ , r = 0.
This homography leaves fixed each point of π and the plane π∞, while it
moves R = (1, u, v, w) to (1, u, v, rw). Hence we may assume R = (1, u, v, 1)
This tells us that
πR = R ⊥ = [u + d, 1, −2v + i, d + iv + 2j].
Since (0, 0, 1, 0) ∈ πR we know i = 2v. Then R = (1, u, v, 1) ∈ πR =
[u+d, 1, 0, d+2v2 +2j] says u+d+u+0+d+2v 2 +2j = 0, so 2(u+d+v2 +j) = 0.
This says that d = −u − j − v2 . Hence πR = [−(j + v2 ), 1, 0, j + v2 − u]. Now
we have
A =
⎛
⎜
⎝
0 1 0 −u − j − v 2
0 0 0 0
0 0 −1 2v
0 0 0 j
The assumption that (1, s2 , s, 0) is not in πR for all s ∈ Fq says that
s2 = j + v2 has no solution s. Let η be a fixed non-square of Fq and put
j = η − v2 . Then
A =
⎛
⎜
⎝
0 1 0 −u − η
0 0 0 0
0 0 −1 2v
0 0 0 η − v 2
So πR = R ⊥ = [−η, 1, 0, η − u]. We claim that the resulting quadric is
elliptic. This will be the case if πR has no point of the quadric other than
⎞
⎞
⎟
⎠ .
⎟
⎠ .

314 CHAPTER 6. QUADRICS IN P G(3, Q)
R. So suppose (x0, x1, x2, x3) ∈ πR. This implies x1 = ηx0 + (u − η)x3. If we
suppose (x0, ηx0 + (u − η)x3, x2, x3) is in the quadric, we get η(x0 − x3) 2 =
(x2 − vx3) 2 . Since η is a nonsquare, it must be that x0 = x3 and x2 = vx3,
and x1 = ux3, from which we see that (x0, x1, x2, x3) = R.
This concludes the case q odd, since R = (1, u, v, 1) and πR = [−η, 1, 0, η−
u] being fixed completely determines A.
Case 2. q = 2e . If we now pick up where we were before assuming that
q is odd, we find in characteristic 2 that
A =
⎛
⎜
⎝
0 1 0 d
0 0 0 0
0 0 1 i
0 0 0 j
In this case πR cannot meet π∩π∞ in the point (0, 0, 1, 0), since this point
is the nucleus of of the conic C, so any line in π through (0, 0, 1, 0) must have
a point of C. So π ∩ π∞ ∩ πR must be a point of the form (0, λ, 1, 0) for λ = 0.
But since the homography with matrix diag(1, t2 , t, 1) maps (0, λ, 1, 0) to
(0, tλ, 1, 0), we may suppose that (0, 1, 1, 0) is the intersection of the three
planes.
Then with R = (1, u, v, 1), R⊥ = [u + d, 1, i, d + iv] and with (0, 1, 1, 0) ∈
πR we have i = 1. Then R on the the quadric implies d = j + u + v + v2 . At
this point
A =
⎛
⎜
⎝
⎞
⎟
⎠ .
0 1 0 j + u + v + v 2
0 0 0 0
0 0 1 1
0 0 0 j
So πR = R ⊥ = [j + v + v 2 , 1, 1, j + u + v 2 ]. This finishes the case q even,
since if R = (1, u, v, 1) is fixed and πR = [j + v + v 2 , 1, 1, j + u + v 2 ] is fixed,
then clearly A is determined.
Moreover, πR fails to contain a point of C \ {∞} if and only if (s + v) 2 +
(s + v) + j = 0 has no solution, i.e., if and only if tr(j) = 1.
This completes a proof of the first paragraph of the theorem.
To begin on a proof of the second paragraph of the theorem we start
by recalling certain facts about conics, homographies, etc., in P G(2, q) that
were developed in Section 4.3. For the convenience of the reader we mention
only the following:
⎞
⎟
⎠ .

6.6. A CONDITION THAT DETERMINES A QUADRIC 315
Recap: ∞ = (0, 1, 0, 0); π∞ = [1, 0, 0, 0]; π = [0, 0, 0, 1].
C = {(1, s 2 , s, 0) : s ∈ Fq} ∪ {∞} is our favorite conic in π.
C ′ = {(1, as2 + bs + c, s, 0) : s ∈ Fq} ∪ {∞} gives all conics in π with
[1, 0, 0, 0]∩[0, 0, 0, 1] as the line tangent at the point ∞. Here a, b, c ∈ Fq with
a = 0. These are also all the conics that are images of C under homographies
of π with center ∞. These are homologies if a = 1 and are elations if a = 1.
If O is an elliptic quadric of P G(3, q) having π∞ as tangent plane at ∞,
and meeting π in the conic C, then the upper triangular matrix AO giving O
is
AO =
⎛
⎜
⎝
0 −1 0 k
0 0 0 0
0 0 1 i
0 0 0 j
⎞
⎟
⎠ .
The point (0, y, z, w) of π∞ is on O if and only if z 2 + izw + jw 2 = 0.
So O is elliptic if and only if z 2 + izw + jw 2 = 0 has no nontrivial solution
(z, w).
From this we can calculate that
O = {(1, f(s, t), s, t) : s, t ∈ Fq} ∪ {∞}, where
f(s, t) = s 2 + (u − v 2 − iv − j)t + ist + jt 2 . (6.18)
If we take a point R not in π or π∞ we saw above that we could assume
that it is R = (1, u, v, 1) for some u, v ∈ Fq.
Asserting that R be on O gives k = u − v 2 − iv − j. The general point
(1, f(s, t), s, t) of O is on
πR = R(A + A T ) = [−v 2 − iv − j, −1, 2v + i, u − v 2 + j]
if and only if (after some routine computation)
(s − v) 2 + i(s − v)(t − 1) + j(t − 1) 2 = 0. (6.19)
When t = 1 equality holds if and only if s = v; when s = v equality holds
if and only if t = 1. If t = 1 and s = v, then we want Eq. 6.19 to have no
solution. This can be interpreted as follows:
Lemma 6.6.2. It holds that πR meets O in precisely {R, ∞} if and only if:
(i) When q is odd, i 2 − 4j = ❆ ∈ Fq;

316 CHAPTER 6. QUADRICS IN P G(3, Q)
and
(ii) when q is even, tr j
i2
= 1.
Not surprisingly, this is the same criterion for O to be elliptic discovered
above.
Now we redo the above calculations (as much as possible) with C replaced
by an arbitrary conic that is the image of C under a central homography of
π with center ∞:
C ′ = {(1, as 2 + bs + c, s, 0) : s ∈ Fq} ∪ {∞}, a, b, c ∈ Fq, a = 0.
Suppose that O ′ is an elliptic quadric in P G(3, q) that has the following
properties:
(i) π∞ = [1, 0, 0, 0] is the tangent plane to O at ∞ = (0, 1, 0, 0).
(ii) π ∩ O ′ = [0, 0, 0, 1] ∩ O ′ = C ′ .
(iii) R = (1, u, v, 1) ∈ O ′ , and
(iv) the plane πR as above is tangent to O ′ at R.
We leave a proof of the following to the reader.
Lemma 6.6.3. Since O ′ has π∞ as tangent plane at ∞ and has O ′ ∩ π = C ′
(as above), the upper triangular matrix giving O ′ is
⎛
c
⎜
AO ′ = ⎜ 0
⎝ 0
−1
0
0
b
0
a
⎞
a14
0 ⎟
a34 ⎠
0 0 0 a44
.
If we insist that (1, u, v, 1) be on O ′ , we get
⎛
c
⎜
AO ′ = ⎜
⎝
−1 b u − c − (b + a34)v − av2 0
0
0
0
0
a
− a44
0
a34
0 0 0 a44
⎞
⎟
⎠ .
We want O ′ to be elliptic, so π∞ = [1, 0, 0, 0] should meet O ′ in precisely
the point ∞. This is iff (0, y, z, w) ∈ O ′ iff
az 2 + a34zw + a44w 2 = 0 has no nontrivial solution (z, w). (6.20)
Lemma 6.6.4.
πR = [(1, u, v, 1) AO ′ + ATO ′
] =
= [c − a34v − av 2 − a44, −1, b + 2av + a34, −c + u − bv − av 2 + a44].

6.6. A CONDITION THAT DETERMINES A QUADRIC 317
If we now requuire that πR be be the same plane given above, we find
a34 = 2v + i − b − 2av; a44 = c + (a − 1)v 2 + bv + j.
This completely determines O ′ with
⎛
c
⎜
AO ′ = ⎜
⎝
−1 b u − 2c − v2 0
0
0
0
0
a
− iv − bv − j
0
(1 − a)2v + i − b
0 0 0 c + (a − 1)v2 + bv + j
⎞
⎟
⎠ . (6.21)
This allows us to compute the function g(s, t) for which (1, g(s, t), s, t) is
the general point of O ′ .
g(s, t) =
as 2 +bs+c+(u−2c−v 2 −iv−bv−j)t+((1−a)2v+i−b)st+(c+(a−1)v 2 +bv+j)t 2 .
(6.22)
At this point we break up the situation into cases for q odd and q even.
Case 1 q odd. Then as we saw above, i = −2v and j = v 2 − η, so
f(s, t) = s 2 + (v 2 − η)t 2 + (u + η)t − 2vst where η is a fixed but arbitrary
nonsquare of Fq
AO =
⎛
⎜
⎝
0 −1 0 u + η
0 0 0 0
0 0 1 −2v
0 0 0 v 2 − η
⎞
⎟
⎠ .
In this case πR = [η, −1, 0, u − η].
Making the appropriate substitutions for a34 and a44 in AO ′ we have
⎛
c
⎜
AO ′ = ⎜ 0
⎝ 0
−1
0
0
b
0
a
u − 2c − bv + η
0
−b − 2av
0 0 0 av2 ⎞
⎟
⎠
+ bv + c − η
.
This allows us to compute
g(s, t) = as 2 + bs + c + (u − 2c − bv + η)t − (b + 2av)st + (av 2 + bv + c − η)t 2 .
The point (1, g(s, t), s, t) is on O ′ ∩O if and only if f(s, t) = g(s, t). When
t = 0 we get the two conics C and C ′ which meet in precisely ∞ if and only

318 CHAPTER 6. QUADRICS IN P G(3, Q)
if either b 2 − 4(a − 1)c = ❆ ∈ Fq, or a = 1 and b = 0 . But for t = 0 we get
f(s, t) = g(s, t) if and only if (after some computation)
(a − 1)(s − vt) 2 + b(1 − t)(s − vt) + c(1 − t) 2 = 0. (6.23)
If a = 1 and b = 0, so c = 0 if the two elliptic quadrics are really different,
then any value of s gives (1, f(s, t), s, t) = (1, g(s, t), s, t). In this case O and
O ′ also contain the conic C ′′ = {((1, s 2 − 2vs + v 2 + u, s, 1) : s ∈ Fq} ∪ {∞}
lying in the plane [1, 0, 0, −1]. Here C ∩ C ′′ = {∞}, so |O ∩ O ′ | = 1 + 2q. So
now we suppose that it is not the case that a = 1 and b = 0.
Of course t = 1 and s = vt = v gives a solution, so the point R =
(1, u, v, 1) is on both elliptic quadrics. Suppose there is a solution with s = vt.
Then c(1 − t) 2 = 0. If we want the two elliptic quadrics to meet in precisely
{∞, R}, then b 2 − 4(a − 1)c = ❆ , so c = 0, forcing t = 1. With s = vt and
t = 1 we just get the point R. So s = vt. Similarly, if there is a solution
with t = 1, then 0 = (a − 1)(s − v) 2 . But a = 1 if b 2 − 4(a − 1)c = ❆ , so
s = v and the point R is the only point (1, f(s, t), s, t) = (1, g(s, t), s, t). So
we have the following lemma.
Lemma 6.6.5. When q is odd and the ovoid O is normalized as above,
O ∩ O ′ = {∞, R} if and only if
b 2 − 4(a − 1)c = ❆ ∈ Fq.
Basically we have that O ∩ O ′ = {∞, R} if and only if C ∩ C ′ = {∞} and
a = 1, i.e., C ′ is not the image of C under an elation of π with center ∞.
Case 2 q = 2e . By the computations in the first part of this section we
see that we may assume that
⎛
0
⎜
AO = ⎜
⎝
1 0 v2 0
0
0
0
0
1
⎞
+ v + u + j
0 ⎟
1 ⎠ , where tr(j) = 0.
0 0 0 j
and
In the notation used above, i = 1 and we have
f(s, t) = s 2 + (v 2 + v + u + j)t + st + jt 2 ,
πR = [v 2 + v + j, 1, 1, v 2 + u + j].

6.6. A CONDITION THAT DETERMINES A QUADRIC 319
We also have that (1, f(s, t), s, t) ∈ πR if and only if
(s − v) 2 + (s − v)(t − 1) + j(t − 1) 2 = 0. (6.24)
When equality holds, s = v if and only if t = 1, in which case the only
point obtained is R = (1, u, v, 1). So suppose that (s − v)(t − 1) = 0. Then
equality holds if and only if tr(j) = 1.
From Eq. 6.20 we have
Also
AO ′ =
⎛
⎜
⎝
c 1 b v 2 + v + u + bv + j
0 0 0 0
0 0 a b + 1
0 0 0 c + (a + 1)v 2 + bv + j
g(s, t) =
⎞
⎟
⎠ .
as 2 + bs + c + (v 2 + v + u + bv + j)t + (c + (a + 1)v 2 + bv + j)t 2 + (b + 1)st.
Recall that C ∩C ′
(a+1)c
= {∞} if and only if either (i) a = 1 and tr b2
= 1,
or (ii) a = 1 and b = 0. Note that c = 0 in both cases if the two elliptic
quadrics are really different.
Now check to see when (1, f(s, t), s, t) = (1, g(s, t), s, t). This turns out
to be the case if and only if
(a + 1)(s + vt) 2 + b(s + vt)(1 + t) + c(1 + t) 2 = 0.
If s = vt, then equality holds if and only if c(1 + t) 2 = 0, i.e., t = 1. This
gives the point R. Similarly, if t = 1, then (a + 1)(s + v) 2 = 0. Here if a = 1
then clearly R is the only point of the form (1, f(s, t), s, t) obtained in O∩O ′ .
But if a = 1 then O ∩O ′ contains the conic C ′′ = {(1, s 2 +s+v 2 +v +u, s, 1) :
s ∈ Fq} ∪ {∞} in the plane [1, 0, 0, 1]. So we have the following.
Lemma 6.6.6. When q = 2e and the ovoid O is normalized as above, then
O ∩ O ′
(a+1)c
= {∞, R} if and only if tr = 1. This says that O ∩ O ′ =
{∞, R} if and only if C ∩ C ′ = {∞} and a = 1, i.e., C ′ is not the image of C
under an elation of π with center ∞.
.
This completes our proof of Theorem 6.6.1
b 2

320 CHAPTER 6. QUADRICS IN P G(3, Q)
6.7 Nonsingular Quadrics in PG(3,q) with q
odd
Throughout this section q is odd, and let α be a nonsquare in Fq. Let Q be
the quadric with equation
x 2 − αy 2 + βz 2 − w 2 = 0.
⎛
1 0 0 0
⎜
The associated matrix is A = ⎜ 0 −α 0 0
⎝ 0 0 β 0
⎞
⎟
⎠
(6.25)
0 0 0 −1
.
Since α = ❆ , it follows that x2 − αy2 = 0 is an elliptic (anisotropic) line.
Also, βz2 − w2 = 0 is elliptic if β = α and hyperbolic if β = 1. It follows
that
hyperbolic if β = α;
Q is
(6.26)
elliptic if β = 1.
Let P = (p0, p1, p2, p3) be a point of P G(3, q) with polar plane
πP = P ⊥ = [p0, −αp1, bp2, −p3] T .
If P ∈ Q, then CP = πP ∩ Q is a circle. For Y = (y0, y1, y2, y3) and Z =
(z0, z1, z2, z3) ∈ P G(3, q), put
Y · Z = Y AZ T = y0z0 − αy1z1 + βy2z2 − y3z3. (6.27)
Y = Y · Y ; Y = 0 ⇐⇒ Y ∈ Q. (6.28)
Y × Z = (Y · Z) 2 =
− Y Z
(6.29)
y0
2 2
y3 y0 y1
z0 z3 + α
z0 z1 −
y1
z1
2
y3
z3
2
y0 y2
−β
z0 z2
−
y2
2
y3
z2 z3
+αβ
y1
2
y2
.
z1 z2

6.7. NONSINGULAR QUADRICS IN PG(3,Q) WITH Q ODD 321
Consider two distinct circles, so in particular Y = 0 = Z. The
common points of the line ℓ = 〈Y, Z〉 and the quadric Q are of the form
Y + hZ where
(y0 + hz0) 2 − α(y1 + hz1) 2 + β(y2 + hz2) 2 − (y3 + hz3) 2 = 0.
This easily works out to be equivalent to
Zh 2 + 2(Y · Z)h + Y = 0. (6.30)
The discriminant of this quadratic equation in h is △ = 4(Y ·Z) 2 −4ZY :=
4 · (Y × Z). Put ℓ = 〈Y, Z〉 and note that ℓ ⊥ ∩ Q = CY ∩ CZ. Moreover, if Q
is hyperbolic, then ℓ and ℓ ⊥ have the same type. If Q is elliptic, then ℓ and
ℓ ⊥ have opposite types (in the first and third cases).
Q Y × Z |Y Z ∩ Q| |CY ∩ CZ|
hyperbolic 2 2
hyperbolic 0 1 1
hyperbolic ❆ 0 0
elliptic 2 0
elliptic 0 1 1
elliptic ❆ 0 2
Write CY ∼ CZ if and only if circles CY and CZ have a common tangent
circle. Otherwise write CY ∼ CZ.
Lemma 6.7.1. The relation ∼ is an equivalence relation on circles. In fact,
CY ∼ CZ ⇐⇒ Y Z = ∈ Fq. It follows that there are two classes of
circles, with CP in one class if and only if P = (= 0), and in the other
class iff P = ❆ .
Proof. Clearly ∼ is reflexive and symmetric, so we just need to show that it
is transitive. This will be clear from the following computations.
Suppose CY ∼ CZ and let CU be a common tangent circle. So Y × U =
Z × U = 0, i.e., (Y · U) 2 = Y U and (Z · U) 2 = ZU. This implies
Y ZU 2 = (Y · U) 2 (Z · U) 2 ,
so that Y Z is a (nonzero) square in Fq.

322 CHAPTER 6. QUADRICS IN P G(3, Q)
Conversely, suppose that CY and CZ are circles with Y Z a (necessarily
nonzero) square. If CY = CZ, they have a common tangent circle. So
suppose CY = CZ. Let V = (v0, v1, v2, v2) ∈ CY \ CZ. Then V · Y = 0,
V · Z = 0, V · V = 0. Let T ∈ 〈Y, V 〉 \ {Y, V }, i.e., T = Y + hV with h = 0.
First note that T = h 2 V + 2h(Y · V ) + Y = Y . Then
T × Y = (T · Y ) 2 − T Y
= [(Y + hV ) · Y ] 2 − Y 2
= Y 2 − Y 2 = 0.
This says that |〈T, Y 〉 ∩ Q| = 1, which implies that |CT ∩ CY | = 1. It
follows that for each nonzero h, CT ∩ CY = {V }. Also,
T × Z = [(Y + hV ) · Z] 2 − Y Z
= (V · Z) 2 h 2 + 2(Y · Z)(V · Z)h + (Y · Z) 2 − Y Z.
So the equation T × Z = 0 is a quadratic in h with discriminant equal to
4(V · Z) 2 Y Z = = 0.
This says that there is at least one T for which T ×Z = 0, i.e., |CT ∩CZ| = 1,
so that CY ∼ CZ. At this stage we have shown that
CY ∼ CZ ⇐⇒ Y Z = ∈ Fq. (6.31)
If Y Z = and ZW = , then Y W = , so ∼ is transitive
and hence an equivalence relation.
(We abuse notation by letting a point 〈C〉 denote the circle 〈C〉 ⊥ ∩ Q.)
Now let 〈C〉 be a circle and let P be any point of 〈C〉. Put E = C + xP ,
x ∈ Fq. This much tells us that
C = 0;
P = 0;
(P · C) = 0;
(C · E) = C · C + xC · P = C = 0;
E = (C + xP ) · (C + xP ) = C = 0.

6.7. NONSINGULAR QUADRICS IN PG(3,Q) WITH Q ODD 323
So E is any point of 〈C, P 〉 different from P , and 〈E〉 is a circle for each
x ∈ Fq.
C × E = (C · E) 2 − CE
= C 2 − C 2 = 0,
so 〈C〉 is tangent to the circle corresponding to any point of 〈C, P 〉 different
from 〈P 〉 and from 〈C〉 itself. Since P · E = P · C + xP · P = 0, 〈P 〉 is a
point on each circle 〈E〉. It also follows easily that (C + xP ) × (C + yP ) = 0,
so the circles {〈C + xP 〉 : x ∈ Fq} form a pencil.
Now suppose that 〈D〉 is any circle distinct from 〈C〉, and suppose that
〈P 〉 ∈ 〈C〉 \ 〈D〉. So P · D = 0.
D × E = C × (C + xP )
= (C · D + xP · D) 2 − C + xP D
= x 2 (P · D) 2 + 2(C · D)(P · D)x + C × D.
So D × E = 0 is a quadratic equation in x with discriminant equal to 4(P ·
D) 2 CD.
Lemma 6.7.2. If 〈C〉 and 〈D〉 are distinct circles tangent at a point P they
have a pencil of common tangent circles at 〈P 〉 and one common tangent at
each point of 〈C〉\〈D〉 (resp., 〈D〉\〈C〉). If 〈C〉 and 〈D〉 are not tangent but
C ∼ D, they have two common tangents at each point of 〈C〉 \ 〈D〉 (resp.,
〈D〉 \ 〈C〉).
Proof. Case 1: C ∼ D (so CD = = 0) with 〈C〉 and 〈D〉 not
tangent.
In this case there are two solutions x to the quadratic equation D×E = 0,
so there are two circles 〈E〉 = 〈C + xP 〉 that are tangent to both C (at 〈P 〉)
and E. This says that if C and D are equivalent but not tangent (so meet in
0 or 2 points), there are two circles tangent to 〈C〉 at each point of 〈C〉 \ 〈D〉
which are also tangent to 〈D〉.
Case 2: C ∼ D (so CD = = 0) with 〈C〉 and 〈D〉 tangent, so
C × D = 0. In this case D × E = 0 reduces to
x[x(P · D) + 2(C · D)] = 0.

324 CHAPTER 6. QUADRICS IN P G(3, Q)
One of the solutions is x = 0, giving E = C, so 〈C〉 and 〈D〉 have a
pencil of common tangent circles through their point of tangency. The other
is x = −2(C.D)
, giving a second circle tangent to both. Hence 〈C〉 and 〈D〉
P.D
have a unique common tangent circle through each point of 〈C〉 \ 〈D〉 (resp.,
〈D〉 \ 〈C〉).
Lemma 6.7.3. Let CY and CZ be distinct circles, so Y Z = 0. For
a, b, c, d ∈ Fq with (ad − bc) = 0, let E = aY + bZ and F = cY + dZ. So 〈E〉
and 〈F 〉 are distinct points on the line ℓ = 〈Y, Z〉. Then
E × F = (aY + bZ) × (cY + dZ) = (ad − bc) 2 (Y × Z). (6.32)
Proof. A straightorward computation.
It follows that |CE ∩ CF | = |CY ∩ CZ| for any two circles CE and CF with
〈E〉 and 〈F 〉 on ℓ.
Lemma 6.7.4. If ℓ is a non-tangent line, then half the points of ℓ not on Q
lie in each of the two equivalence classes. Clearly all points on a tangent line
are in the same class.
Proof. We start by recalling Theorem 20.23.1 in the following form:
Theorem: Let a and k be nonzero elements of Fq.
The number of solutions (A, B) to A 2 − aB 2 = k is
q − 1, if a = ;
q + 1, if a = ❆ .
Let ℓ = 〈Y, Z〉 be a line meeting Q in 0 or 2 points, so Y ×Z = 0. Clearly
we may assume that Z ∈ Q, so Z = 0. Put a = Z, b = 2Y · Z, c = Y ,
and let E = Y + xZ, for x ∈ Fq. Then
E · E = ax 2 + bx + c (= 0 since b 2 − 4ac = 4(Y × Z) = 0). (6.33)
Basically we need to know for how many x ∈ Fq we have ax 2 +bx+c = y 2

6.7. NONSINGULAR QUADRICS IN PG(3,Q) WITH Q ODD 325
for some square y 2 = 0.
ax 2 + bx + c = y 2
4a 2 x 2 + 4abx + b 2 + 4ac − b 2
(2ax + b) 2 − a(2y) 2
A 2 − aB 2
Here we put A = 2ax + b and B = 2y.
⇐⇒
= 4ay 2
⇐⇒
= b 2 − 4ac := k (= 0)
⇐⇒
= k.
If a = (= 0), there are q − 1 solutions (A, B) to A2 − aB2 = k. If
k = , two of these have B = 0 (i.e., y = 0, giving E ∈ Q). But q − 3 of the
solutions (A, B) have B = 0. Since (A, B) and A, −B) give the same value
of x, we have q−3
2 values of x that have ax2 + bx + c = = 0. This gives q−3
2
points E of 〈Y, Z〉 plus the point Z in one equivalence class. Since 〈Y, Z〉 \ Q
has q − 1 points in this case, half of them are in each class.
If k = ❆ = Z (still with a = ), clearly B = 0 and 〈Y, Z〉 ∩ Q = ∅.
There are are q−1
solutions x to ax 2 2 + bx + c = . This gives q−1
(plus Z, so
points E
2 q+1
) points D on 〈Y, Z〉 with D = 2
and that many in the other.
, i.e., in one of the classes
Now assume that a = ❆ = Z. There are q + 1 solutions (A, B) to
A2 − aB2 = k. If k = , two solutions have B = 0, accounting for two
values of x giving points E on Q, and leaving q − 1 solutions with B = 0. In
this case there are q−1
2 values of x with ax2 + bx + c = = 0, i.e., (q − 1)/w
points D on 〈Y, Z〉 \ Q in one class, leaving that many in the other class.
If k = ❆ , so 〈Y, Z〉 ∩ Q = ∅, B = 0, implying that there are q+1
values of
2
x for which ax2 + bx + c = = 0, giving q+1
points of 〈Y, Z〉 in the ”square
2
class” and that many in the other one.
Lemma 6.7.5. When Q is hyperbolic and q ≡ 1(mod 4), orthogonal circles
intersect if and only they are in the same class. When Q is hyperbolic and
q ≡ −1(mod 4), orthogonal circles intersect if and only if they are in different
classes. The conditions are reversed when Q is elliptic.
Proof. The circles CY and CZ are orthogonal if and only if Y · Z = 0, in
which case Y ×Z = (Y ·Z) 2 −Y Z = −Y Z. The result follows from
the fact that −1 = iff q ≡ 1 (mod 4).

326 CHAPTER 6. QUADRICS IN P G(3, Q)
Theorem 6.7.6. When Q is hyperbolic (and q is odd), there is a nonlinear
flock.
Proof. Let Q be given by x2 − αy2 + αz2 − w2 = 0. We define two sets of
planes (or of circles) as follows.
First put Y = (1, 0, 0, 0) and Z = (0, 1, 0, 0), so Y · Z = 0 and Y × Z =
−Y Z = α = ❆ . So the circles CV and CW for V, W ∈ ℓ1 = 〈Y, Z〉 are
pairwise disjoint. Put S1 = {P = (k0, k1, 0, 0) : P = k2 0 − αk2 1 = (= 0)}.
So the points P of S1 are the points on the line ℓ1 : x2 = x3 = 0 in the
”square” class and they determine circles that are pairwise disjoint. Put
X = (0, 0, 1, 0) and W = (0, 0, 0, 1) and ℓ2 = 〈X, W 〉. Then X · W = 0 and
X × W = −XW = α = ❆ . So CU and CV for U, V ∈ ℓ2 are pairwise
disjoint. If Q = (0, 0, k2, k3), so Q = αk2 2 −k2 3 , put S2 = {Q = (0, 0, k2, k3) :
−Q = −αk2 2 + k2 3 = ❆ }. So if P ∈ S1 and Q ∈ S2, then P · Q = 0. When
q ≡ 1 (mod 4), −1 = so Q = ❆ , implying that P ∈ S1 and Q ∈ S2 are
in different classes, which is precisely when CP and CQ are disjoint. When
q ≡ −1 (mod 4), then −1 = ❆ , so Q = , implying P ∈ S1 and Q ∈ S2
are in the same class, which is precisely when CP and CQ are disjoint. It
follows that F = {CR : R ∈ S1 ∪ S2} is a nonlinear flock of Q.
We now turn our attention to the elliptic case. Suppose that Q has
quadratic form
q(x0, x1, x2, x3) = x 2 0 − αx2 1 + x2 2 − x2 3 .
Denote by L the set of q 2 + q circles through the ”North Pole” N =
(0, 0, 1, 1). Let the point S = (0, 0, −1, 1) be called the ”South Pole” and
note that both N and S belong to Q. So of course N · S = −2 = 0. L
has exactly q + 1 circles containing S. Let C1, C2, . . . , Cq 2 −1 be the circles
in L not containing the South Pole. Let D = {p0, p1, . . . , pq} be a circle not
containing N. Let Li be the line in the plane of D tangent to D at pi. Then
if πi = 〈Li, N〉, πi ∩ Q is the unique circle tangent to D at pi and containing
N.
Lemma 6.7.7. Let F be a flock of Q with carriers N and S. So F is a set
of q − 1 pairwise disjoint circles whose union is Q \ {N, S}. Then a circle of
L either passes through S and is tangent to no circles of F, or it is tangent
to precisely one circle of F.

6.7. NONSINGULAR QUADRICS IN PG(3,Q) WITH Q ODD 327
Proof. Say F = {D1, D2, . . . , Dq−1}. Let T = {(Ci, Dj) : 1 ≤ i ≤ q2 − 1, 1 ≤
j ≤ q − 1}. By the comments just preceding the statement of the Lemma
we know that each Dj is tangent to a unique member of L at each of its
q + 1 points, but that member of L could conceivably contain S. So at least
we have |T | ≤ (q − 1)(q + 1). Each Ci has q points different from N (and
S), each of which must be in some member of F, and each Ci meets each
circle of F in 0, 1 or 2 points. Since q is odd, Ci must be tangent to at least
one Dj. Say Ci is tangent to θi circles of F, where we know θi ≥ 1. Then
|T | = q2−1 i=1 θi ≥ q2 − 1. Hence |T | = q2 − 1. This means that each θi = 1
and no circle through both N and S can be tangent to any member of F.
Lemma 6.7.8. Let D1 ∈ F, a flock with carriers N and S. The circles of
the pencil intersecting in N and S meet D1 in 0 or 2 points by Lemma 6.7.7.
Those that contain a point of D1 are all in the same equivalence class.
Proof. Retain the notation used above: F = {D1, D2, . . . , Dq−1} and fix a
point P in D1. Consider the q + 1 circles through P and N. One of them,
denoted C∞, passes through the South Pole S and is tangent to no plane
of F (so contains a point different from P in D1. Let C1, C2, . . . , Cq be
the other circles through P and N, with C1 that circle through N which is
tangent to D1 at P . Fix attention on a circle Dj, 2 ≤ j ≤ q − 1. The line
〈N, P 〉 meets the plane πj of Dj in a point Yj which must not be a point of
Q. So Yj is either an interior or an exterior point of πj, and hence on either 0
or 2 tangents to Dj. If a circle through P and N is tangent to Dj at a point
Q, its plane must contain the tangent line to Q in πj, which must contain the
point Yj. Of course, then Yj is an exterior point and is also on the tangent
line to a second point Q ′ of Dj. So in this case the two circles through P and
N containing Q and Q ′ , respectively, are exactly the circles through P and
N tangent to Dj. If Yj is an interior point of Dj, then no circle through P
and N can be tangent to Dj. In this way we see that the circles C2, . . . , Cq
come in pairs, with two in a given pair tangent to the same Dj. Recall that
tangent circles belong to the same class. So the circles through N and P
and not through S come in pairs belonging to the same class, so an even
number belong to each class. From Lemma 6.7.4 exactly half of the circles
intersecting in N and P are in each class.
So C2, . . . , Cq come in pairs tangent to q−1
2 of the circles D2, . . . , Dq−1.
For notational convenience suppose that C2i and C2i+1 are tangent to Di+1,
for 1 ≤ i ≤ (q − 1)/2. So C2i and C2i+1 are in the same class. First suppose

328 CHAPTER 6. QUADRICS IN P G(3, Q)
that q ≡ 3 (mod 4). In this case it must be that q−3
2 of the pairs C2i, C2i+1
are in one class and q−1
2
of them are in the other class. It then must be the
case that C∞ and C1 are in the first class. Since C1 is in the class of D1,
it must be that C∞ is in the class of D1. As P varies over the points of
D1 it follows that in each case the circle containing N, S, and P belong to
the class of D1. Second, suppose that q ≡ 1 (mod 4). In this case q−1
4 of
the pairs C2i, C2i+1 fall in each of the two classes. This means that one of
C∞, C1 falls in one class and the other in the other class. Since C1 is in the
class of D1, it must be that C∞ in not in the class of D1. As P varies over
the points of D1, it follows that in each case the circle containing N, S, and
P belong to the other class than the class of D1. So in all cases the circles
through N and S that contain a point of D1 fall into the same class, proving
the lemma.
Now let C be the linear flock with carriers N and S. So C is the linear
flock given by planes through N ⊥ ∩ S ⊥ = 〈N, S〉 ⊥ .
Lemma 6.7.9. If D is any circle of F for which all of the circles of the
pencil intersecting in N and S that contain a point of D are in the same
class, then D ∈ C.
Proof. Suppose that D1 ∈ C. All circles of C meeting D1 are in the same
class. But the circles of C cover all the points of D1, at most 2 at a time. This
requires at least q+1
circles of C to be in the same class. By Lemma 6.7.4
2
only half of the circles on the planes through 〈N, S〉 ⊥ are in one class. This
contradiction forces D1 ∈ C.
As D1 is an arbitrary circle of F, it must be that F = C, and we have
proved the following theorem.
Theorem 6.7.10. (Theorem of Orr; Proof by J. A. Thas) Any flock of an
elliptic quadric in P G(3, q) with q odd must be linear.

Chapter 7
Ovoids in 3-Dimensional Space
7.1 Lines and Planes related to an Ovoid
A set O of points of P G(3, q) is called an ovoid provided it satisfies the
following:
(i) Each line meets O in at most two points.
(ii) Each point of O lies on exactly q + 1 tangent lines, all of which lie in
a plane.
Suppose that O is a given ovoid in P G(3, q). Lines and planes are given
labels that reflect their relationship to O.
Lines A line l of P G(3, q) is a tangent line to O provided |l ∩ O| = 1. If
l ∩ O = {p}, we say l is tangent to O at p. If |l ∩ O| = 2, l is called a secant
line. If |l ∩ O| = 0, l is an exterior line.
Planes A plane π of P G(3, q) is a tangent plane (TP) if |π ∩ O| = 1. π
is called a secant plane (SP) or cutting plane provided |π ∩ O| > 1.
Theorem 7.1.1. Each ovoid has q 2 + 1 points.
Proof. Let P be a point of O, so P is on q + 1 tangent lines. Hence the remaining
q 2 lines on P must each have a unique second point of O, accounting
for all points of O. So |O| = 1 + q 2 .
We leave the routine proof of the following theorem to the reader.
Theorem 7.1.2. To each ovoid O there are:
(i) (q2 + 1)(q + 1) tangent lines;
(ii) q2 +1 (q
= 2
2 +1)q2 secant lines; and
2
(iii) (1 + q2 )(1 + q + q2 ) − (q2 + 1)(q + 1) − (q2 +1)q2 329
2
= q2 +1
exterior lines.
2

330 CHAPTER 7. OVOIDS IN 3-DIMENSIONAL SPACE
Lemma 7.1.3. Each plane α with |α ∩ O| > 1 meets O in an oval.
Proof. Let P ∈ O ∩ α with |O ∩ α| > 1. Clearly α is not a tangent plane,
so it meets the tangent plane at P in a line, which must be tangent to O at
P . Each other line of α through P (not being in the plane of tangent lines)
must be a secant to O, giving q other points of α ∩ O. So |O ∩ α| = q + 1,
and hence O ∩ α is an oval in α.
Theorem 7.1.4. There are q 2 +1 tangent planes, and q(q 2 +1) secant planes,
giving all planes of P G(3, q).
Proof. Clearly there are q2 + 1 tangent planes, one at each point of O. And
there are q2 +1 q+1 2 2 2 / = q(q + 1) secant planes. But (q + 1) + q(q + 1) =
3 3
(q + 1)(q2 + 1) is the total number of planes.
Theorem 7.1.5. For a line l of a given type, we count the planes of each
type through it.
type of line TP SP
tangent 1 q
secant 0 q + 1
exterior 2 q − 1
Proof. For tangent and secant lines the count is easy. So let l be an exterior
line. There are q +1 planes through l, say a of them are tangent planes and b
of them are secant planes. And each point of O is on a unique plane through
l. So:
a + b = q + 1 is the number of planes on l.
a + b(q + 1) = q 2 + 1 is the number of points on O.
=⇒ bq = q 2 − q =⇒ b = q − 1 and a = 2, giving the third line of the
table.
7.2 Ovoids as Caps and with many Conics
In P G(3, q) a set K of k points, no three on a line, is a k-cap. A k-cap is
complete if it is not contained in a (k + 1)-cap.
Lemma 7.2.1. If q is odd, a k-cap K in P G(3, q) satisfies k ≤ q 2 + 1.

7.2. OVOIDS AS CAPS AND WITH MANY CONICS 331
Proof. If P1, P2 ∈ K, each plane through P1P2 meets K in a k ′ -arc, so
k ′ ≤ q + 1. So k ≤ 2 + (q + 1)(q − 1) = q 2 + 1.
Now let K be a k-cap in Π3 = P G(3, q), q = 2 e . For P ∈ K, let t be the
number of tangents to K through P .
Lemma 7.2.2. t + k = q 2 + q + 2, so k ≤ q 2 + q + 2.
Proof. There are t tangents through P and k − 1 secants, accounting for all
lines through P . Hence t + k − 1 = q 2 + q + 1.
Theorem 7.2.3. A k-cap K in Π3, q even, has no tangents if and only if
q = 2, k = 8, and K is the complement of a plane.
Proof. Suppose K is a k-cap with no tangents. So t = 0 and k = q2 + q + 2.
So the number of secants is k 1 = 2 2 (q2 + q + 2)(q2 + q + 1), which is less than
(q2 + 1)(q2 + q + 1), the total number of lines. Hence there is an exterior
line l. Suppose π is a plane through l having a point P of K. Then π ∩ K
is a set of points of π such that each line of π meets π ∩ K in 0 or 2 points,
i.e., π ∩ K is a (0, 2)-set, so |π ∩ K| = q + 2. So each plane through l meets
K in 0 or q + 2 points. This implies (q + 2)|(q2 + q + 2), which implies
q2 + q + 2 = (q − 1)(q + 2) + 4 ≡ 4 ≡ 0 (mod q + 2). then q + 2|4, implying
q = 2, so k = 4 + 2 + 2 = 8.
Now Π3 = P G(3, 2) has q3 + q2 + q + 1 = 15 points. And Π3 \ K has 7
points. As each line meets K in 0 or 2 points, each line meets Π3 \ K in 1 or
3 points. As each line meeting Π3 \ K in 2 points must lie in Π3 \ K, Π3 \ K
is a subspace, i.e., a plane.
For the converse, note that K = Π3\ (plane) is an 8-cap with no tangents.
The following lemma is often useful and holds for q odd or even.
Lemma 7.2.4. Let K be a k-cap in Π3, q odd or even. Let l be a line
tangent to K at P . Let π be any plane through l. Then π ∩ K is a k ′ -arc
with k ′ ≤ q + 1.
Proof. The q + 1 lines of π through P meet K in at most one other point of
K and l is one of these lines.
Lemma 7.2.5. If K is a complete k-cap in P G(3, q), q = 2 e , and q > 2,
then k ≤ q 2 + 1.

332 CHAPTER 7. OVOIDS IN 3-DIMENSIONAL SPACE
Proof. To obtain a contradiction we suppose q 2 + 1 < k. Theorem 7.2.3
(with the help of Lemma 7.2.2) implies that k < q 2 + q + 2. Let l be a
tangent to K at P . Then by Lemma 7.2.2 any plane through l meets K in
at most q + 1 points. If all planes through l meet K in at most q points,
then k ≤ 1 + (q + 1)(q − 1) = q 2 . Hence there is some plane π through l
for which π ∩ K is a (q + 1)-arc with a nucleus Q. If every line through Q
were external or tangent to K, then K ∪ {Q} would be a (k + 1)-cap. So
there must be a secant b of K through Q, and b cannot lie in π. Each plane
through b meets π in a line through Q, a tangent in π through Q. So no
plane through b meets K in a (q + 2)-arc by Lemma 7.2.4. Count the points
of K on planes through b: k ≤ 2 + (q + 1)(q − 1) = q 2 + 1.
Let q be odd or even.
Theorem 7.2.6. Let K be a (q2 + 1)-cap in P G(3, q).
(i) At each point P of K there is a unique tangent plane πP , so πP ∩ K =
{P }.
(ii) If π is a plane but not one of the q2 + 1 tangent planes, then π meets
K in a (q + 1)-arc.
(iii) The lines through P ∈ K consist of q2 bisecants and q + 1 tangents,
the latter all lying in the tangent plane at P . (Hence K is an ovoid of
P G(3, q).)
(iv) Through a point R off K, there are 1
1
q(q − 1) secants, q(q + 1)
2 2
external lines, and q + 1 tangents. When q is even, the q + 1 tangents are
coplanar. When q is odd, no three of the tangents are coplanar. But in any
case the q + 1 points of K on tangents through R form an oval (for which R
is the nucleus when q is even).
Proof. First let q be odd. Any plane through 2 points of K meets K in a
(q + 1)-arc, since q 2 + 1 = 2 + (q + 1)(q − 1). For a point P ∈ K, count the
planes α through P meeting K in a (q +1)-arc (which must be a conic). This
number is: q 2 (q + 1)/q = q 2 + q, leaving only one plane π tangent to K at P .
The q + 1 lines of π through P are tangents to K at P and are necessarily
all the tangents through P .
Let R be a point of Π3 \ K. If R lies on one tangent l to K at P , then
each plane w through l other than the tangent plane meets K in a conic.
So this plane w contains exactly one other tangent through R. Hence there
are q + 1 tangents through R. No three of these tangents can lie in a plane,
because if they did lie in a plane γ, then γ ∩ K would be an oval in γ and R

7.2. OVOIDS AS CAPS AND WITH MANY CONICS 333
would be a point collinear with three points of the oval. This cannot happen
when q is odd. So each point R of Π3 \ K is on 0 or q + 1 tangent lines. Say
there are m points off K on q + 1 tangents and count pairs (t, R), t a tangent
through R. This number is (q 2 + 1)(q + 1)q = m(q + 1), implying m = q 3 + q.
But |K| = q 2 + 1 and (q 3 + q) + (q 2 + 1) is the total number of points of Π3.
So each point off K must be on exactly q + 1 tangents, and no three of them
are coplanar.
What remains to be proved in the case q is odd is that if R is a point
off K, then the q + 1 points of K on the tangent lines through R all lie in a
plane. Since no three of the tangents through R lie in a plane, the points of
these lines form a cone C1. Each plane not containing R must intersect C1
in an oval which must be a conic. Let P1, P2, P3 be the points of K on three
distinct tangents through R. Then these three points determine a plane α
which intersects C1 in an oval O1, and intersects the ovoid K in an oval O2
(which might not contain all the other points of K on the remaining tangents
through R). But the oval O2 determines a second cone C2 with vertex R. In
particular the plane α meets C1 in an oval O1 and meets C2 in an oval O2.
These ovals have the points P1, P2 and P3 in common. Let πi be the plane
tangent to K at Pi, i = 1, 2, 3. So R ∈ πi. Then πi ∩ α must be the line
of α tangent to both C1 and C2. So C1 and C2 share three points and the
tangents at them. Hence we may conclude that C1 = C2. This says that the
points of K lying on the tangents through R must all lie in the plane α.
Now suppose q is even, K a (q 2 + 1)-cap. If P ∈ K, there are q 2 secant
lines through P leaving q + 1 tangents through P . Let r be a tangent to K
at P . By Lemma 7.2.4, no plane through r has q + 2 points of K. If each
of the q + 1 planes through r had at most q points of K, this would account
for only (q + 1)(q − 1) + 1 = q 2 points of K. Hence there is some plane α on
r that has q + 1 points of K. So α ∩ K is an oval Γ with a nucleus Q. The
lines joining Q to points of K cannot all be tangents, since then K ∪ {Q}
would be a (q 2 + 2)-cap. So let m be a line through Q meeting K in two
points A and B. Clearly m does not lie in α. If π is any plane through m,
then π ∩ α is a line of α through Q, i.e., a tangent to K. Hence π contains
at most q + 1 points of K by Lemma 7.2.4. So the q + 1 planes through m
account for at most 2 + (q + 1)(q − 1) = q 2 + 1 points of K, implying that
each plane through m meets K in an oval.
Now suppose l is any line through Q different from m and not lying in
α. The plane γ containing l and m meets α in a line s through Q. From
the preceding paragraph we know that γ meets K in an oval Γ ′ . The lines

334 CHAPTER 7. OVOIDS IN 3-DIMENSIONAL SPACE
l, m, s lie in γ and meet at Q. But m is a secant to Γ ′ . Hence Q is not the
nucleus of Γ ′ , so l cannot be a tangent, i.e., the only lines through Q that
are tangents to K are the q + 1 lines through Q in α.
Let π be any plane different from α containing the tangent line r. Since
r is a tangent to Γ, Q belongs to r, so Q ∈ π. If π has a second point P ′
of K (P = P ′ ), then QP ′ must be a secant line. Hence by the preceding
paragraph the plane π through QP ′ must meet K in an oval. So q of the
planes through r account for 1 + q2 points of K, leaving one tangent plane.
The q + 1 lines through P in this plane must then be all q + 1 tangent lines
through P . This proves that at each point P of K there is a unique tangent
plane πP containing all the tangent lines through P . Every nontangent plane
through P meets πP in a tangent line r, and meets K in an oval Γ with a
nucleus Q not in K. And all tangents through Q are tangent to Γ.
There are q2 + 1 tangent planes, and every plane meeting K in at least
two points meets K in an oval whose nucleus lies off K. This accounts for
q+1
/
q 2 + 1 + q 2 +1
3
3
= q 2 + 1 + q 3 + q planes, i.e., all planes. Also, the q 3 + q
cutting planes account for q 3 +q points off K, the nuclei of their intersections
with K. Hence each point off K is the nucleus of a unique oval. It lies on
q +1 tangents and (q 2 −q)/2 secants, leaving q 2 +q +1−(q +1)−(q 2 −q)/2 =
(q 2 + q)/2 exterior lines through it.
Now let Q be any point off K and let α be the plane meeting K in the
oval Γ for which Q is the nucleus. If π is any plane through Q, π = α, then π
meets α in a line l through Q having a point P of Γ. The line l is a tangent
through P , so lies in the unique tangent plane πP at P . If πP = π = α, then
π is a cutting plane. Hence Q lies on q + 1 tangent planes, one through each
line of α through Q.
The classification of ovoids in P G(3, q) with q odd was obtained early
in the study of ovoids. It depends on the classification by Segre of ovals
in P G(2, q) and was foound independently in 1955 by Barlotti([Ba55]) and
Panella ([Pa55]). Barlotti showed that if all oval sections of an ovoid in
P G(3, q), q odd, are conics, then the ovoid is an elliptic quadric. He then
noticed that this characterization was also valid for q even, resulting in the
following theorem.
Theorem 7.2.7. Let Ω be an ovoid of P G(3, q) with q > 2. If every secant
plane section of Ω is a conic, then Ω is an elliptic quadric.
Proof. We consider separately the case where q = 3, so |Ω| = 10. Let ℓ be a

7.2. OVOIDS AS CAPS AND WITH MANY CONICS 335
line of P G(3, 3) exterior to Ω. Then of the four planes on ℓ two π1 and π2
meet Ω in a conic and two, π3 and π4 are tangent to Ω. Let πi ∩ Ω = Ci,
i = 1, 2. Also, let πj be tangent at Pj, j = 3, 4. The nine points C1∪C2∪{P3}
define an elliptic quadric E. We need to show that P4 ∈ E so that Ω = E.
Suppose the Ω = E, so that P4 ∈ E and E \ Ω = {X} for some point X.
Since E and Ω are maximal sized sets of points no three collinear, it must be
that the line 〈P4, X〉 contains a further point Y of E ∩ Ω, i.e., m is a secant
to both Ω and E. Each plane on m meets both E and Ω in a conic, and for
a given plane its intersection with E and its intersection with Ω have three
(out of four) points in common. So let π be a plane on m, with π ∩ E = C,
π ∩ Ω = C ′ and C ∩ C ′ = {Y, A, B}. The line 〈A, B〉 cannot contain P4, Y or
X, so must meet m in the remaining point of M, say Q. It follows that Q
lies on no tangents of E or Ω, which is impossible. Hence Ω = E is an elliptic
quadric.
We now assume that q ≥ 4.
Let π1 be a plane meeting Ω in a conic C1 and let P1, . . . , P5 be five points
on the conic C1. Take another plane π2 through P1 and P2, and let Q1, Q2,
Q3 be three other points of the Conic C2 = π2 ∩ Ω. The tangent plane at Pi,
i = 1, 2, to Ω meets π2 in a line which is tangent to C2. Take a third plane
π3 through P1 and P2 containing a point R of Ω not in π1 or π2. Likewise π3
meets the tangent plane at Pi, i = 1, 2, in a line tangent to C3 = π3 ∩ Ω.
The quadric Q through the nine points P1, . . . , P5, Q1, Q2, Q3, and R contains
five points of each of the conics C1 and C2, so contains C1 and C2
themselves. The tangent lines to C1 and C2 at P1 are therefore tangents to
Q at P1. Hence the tangent plane to Ω at P1 is also tangent to Q. Hence the
conic section C = Q ∩ π3 will have as its tangent at P1 the tangent which
is the intersection of the tangent plane (to both Ω and Q) at P1 with π3.
Similarly, the tangent to C at P2 is the intersection of the tangent plane at
P2 with π3. So C3 and C have three points in common as well as tangents
at two of them, and hence C3 = C. (See Lemma 4.2.1)
Now take a point P , (P = P1, P2), P on Ω, and π a plane through P P1
which contains neither P2 nor any of the tangents at P1 to C1, C2 or C3.
This plane meets C1, C2, C3 in three distinct points apart from P1. Hence
π ∩ Q is a conic that contains these four points. Moreover, the tangent plane
to Ω at P1 meets π in a line tangent to π ∩ Q at P1. Since π ∩ Q and π ∩ Ω
share four points and a tangent at one of them, they must be the same conic,
forcing P ∈ Q. Hence Ω ⊆ Q. As Q is then a quadric containing q 2 + 1

336 CHAPTER 7. OVOIDS IN 3-DIMENSIONAL SPACE
points with no three collinear, Q can only be an elliptic quadric with q 2 + 1
points. Hence Ω = Q.
7.3 Null polarities and skew-symmetric matrices
In this section we deal only with P G(3, q), but there are natural generalizations
for P G(n, q) with n odd. A polarity of P G(3, q) is a map ρ that is an
isomorphism from P G(3, q) to its point-hyperplane dual P G(3, q) ˆ with the
property that its square is the identity map. So for each point P , P ρ is a
plane and for each plane π, π ρ is a point. Then for points P and Q we have
Q ∈ P ρ if and only if P ∈ Q ρ . The polarity ρ is a null polarity provided
P ∈ P ρ for all points P ∈ P G(3, q). Given any polarity, a point and its
associated plane are respectively called pole and polar.
Let A be an invertible, 4 × 4 skew-symmetric matrix over Fq with zeros
on the diagonal. Define ρA as follows:
ρA : point (¯x) ↦→ plane A(¯x) T ; plane [ū] T ↦→ (ū)A −1 .
Then ρA is a typical null polarity.
Theorem 7.3.1. Let ρ be a null polarity of P G(3, q). Then there is a 4 × 4
nonsingular, skew-symmetric matrix over Fq with zero diagonal such that
ρ = ρA. Moreover, any two null polarities are projectively equivalent.
Proof. Let ρ be a null polarity. Since ρ is a correlation, i.e., an isomorphism
from P G(3, q) to its point-hyperplane dual, there must be a 4×4 nonsingular,
skew-symmetric matrix over Fq and a field automorphism σ such that
ρ : (¯x) ↦→ A[¯x σ ] T ; [ū] T ↦→ λ(ū σ )A −1 , (7.1)
for some nonzero scalar λ.
Writing out the fact that ρ 2 = id we find
ρ 2 : (¯x) ↦→ λ · (¯x σ2
)(A σ ) T A −1 = µ · (¯x),
for all points (¯x). By choosing appropriate values for the point (¯x) we easily
show first that A = c · (A σ ) T for some nonzero scalar c, and then that
σ 2 = id. (7.2)

7.4. OVOIDS YIELD NULL POLARITIES IN CHARACTERISTIC 2 337
Now use the fact that (¯x)A(¯x σ ) T = 0 for all points (¯x). First put (¯x) =
λ · (ēi), (the vector with a λ in position i and zeros elsewhere), and then
(¯x) = (ēi) + (ēj), to show that
A is skew-symmetric with zero diagonal. (7.3)
Now use (¯x)A(¯x σ ) T = 0 for all (¯x) of the form (¯x) = λ · (ēi) + µ · (ēj) to
discover that σ = id. At this point we know that
It follows that
ρ : (¯x) ↦→ A[¯x] T ; [ū] ↦→ (ū)A −1 .
(ēi)A(ēj) T = Aij = −Aji = −(ēj)A(ēi) T ,
and f((¯x), (¯y)) = (¯x)A(¯y) T is a nonsingular alternating form. Now let (ū)1 be
any nonzero vector representing a point of P G(3, q), and choose a nonzero
vector (¯v)1 such that (¯v)1 ∈ (ū) ρ
1 . By multiplying (¯v)1 by an appropriate
scalar we may suppose that (ū)1A(¯v) T 1 = 1 and (¯v)1A(ū) T 1 = −1.
Let ℓ be the line ℓ = 〈(ū)1, (¯v)1〉, so ℓρ = (ū) ρ
1 ∩ (¯v) ρ
1. Then ℓ = ℓρ , so
(ℓρ ) ρ = ℓ = ℓρ , which implies ℓρ must have two points with representative
vectors (ū)2, (¯v)2 with (ū)2A(¯v)2 = −1. If we now let B be the ordered
basis of the underlying vector space B = {(ū)1, (¯v)1, (ū)2, (¯v)2} , then if the
coordinate matrices of (¯x) and (¯y) are (¯c) and ( ¯ d), respectively, we have
f((¯x), (¯y)) = (¯c)C( ¯ d) T ⎛
0
⎜
, where C = ⎜ −1
⎝ 0
1
0
0
0
0
0
⎞
0
0 ⎟
1 ⎠
0 0 −1 0
.
Replacing the standard ordered basis with the ordered basis B amounts
to replacing the matrix A with the matrix C.
7.4 Ovoids Yield Null Polarities in Characteristic
2
Theorem 7.4.1. Let Ω be an ovoid in P G(3, q), q even q > 4. Associate
to each tangent plane its point of contact. Any other plane π meets Ω in
a (q + 1)-arc with nucleus Q. Associate π to Q. Then this point ⇐⇒
hyperplane correspondence is a null polarity.

338 CHAPTER 7. OVOIDS IN 3-DIMENSIONAL SPACE
Proof. Suppose the point P corresponds to the plane πP . It must be shown
that the poles of the planes through P are all the points of πP . Let P ρ = πP
and π ρ = P where pi = πP . For ρ to be a polarity means Q ∈ P ρ iff P ∈ Q ρ .
For the polarity ρ to be a null polarity means that P ∈ P ρ for every point P
of P G(3, q).
First suppose P ∈ Ω, and suppose Q ∈ P ρ = πP . Then Q is the nucleus
of the plane Q ρ containing the tangents through Q, including the line QP .
So P ∈ Q ρ . Conversely, if P belongs to the plane π = πP , then π is a cutting
plane and π = Q ρ for a unique point Q off Ω. But QP is a tangent to π ∩ Ω,
so QP lies in πP .
Second, suppose P ∈ Ω. From the above, if Q ∈ Ω, then P ∈ Q ρ if and
only if Q ∈ P ρ . So suppose Q ∈ P ρ and Q ∈ Ω. Since Q ∈ P ρ , P is the
nucleus of P ρ ∩Ω, so QP is tangent to P ρ ∩Ω. But the only tangents through
Q lie in Q ρ , so P ∈ Q ρ .
Corollary 7.4.2. (i) Let ℓ be a tangent to Ω at P . Then ℓ = ∩Q∈lQ ρ . (ℓ is
its own polar.)
(ii) Let ℓ be a secant to Ω at points A, B. Then ∩Q∈ℓQ ρ = A ρ ∩B ρ , which
is an exterior line.
(iii) The polar of an exterior line m is the line AB where m = A ρ ∩ B ρ ,
and A, B ∈ Ω. Hence secant lines are interchanged with exterior lines.
Proof. Let ℓ be tangent to Ω at P , so ℓ ⊆ P ρ . Let Q ∈ ℓ, Q = P . So Q ∈ P ρ
implies P ∈ Q ρ . Then P, Q ∈ Q ρ , so ℓ ⊆ Q ρ , implying ℓ = ∩Q∈ℓQ ρ . This
proves (i).
(ii) Let ℓ ∩ Ω = {A, B}. Put ℓ ρ = A ρ ∩ B ρ . Clearly B ∈ A ρ and A ∈ B ρ ,
since each line through A in A ρ must be a tangent. Let Q ∈ ℓ, A = Q = B.
We need to show that A ρ ∩ B ρ ⊆ Q ρ . So let R ∈ A ρ ∩ B ρ . R ∈ A ρ ∩ B ρ
implies A, B ∈ R ρ , implying ℓ ⊆ R ρ . So Q ∈ ℓ implies Q ∈ R ρ , which implies
R ∈ Q ρ . Hence A ρ ∩ B ρ ⊆ Q ρ , implying A ρ ∩ B ρ ⊆ ∩Q∈ℓQ ρ ⊆ A ρ ∩ B ρ .
For (iii), let m be an exterior line and suppose it is on a tangent planes
and on b cutting planes. Then a + b = q + 1 and 1 · a + (q + 1) · b = q 2 + 1,
hence a = 2 and b = q −1. Hence the polar of an exterior line m is the secant
line AB where m = A ρ ∩ B ρ , and A, B ∈ Ω.
Corollary 7.4.3. For distinct points P1 and P2 we have that P1 is incident
with P ρ
2 if and only if the line P1P2 is tangent to O.
Proof. If P2 lies on O, then P ρ
2 is the tangent plane to O at P2, so it is clear
that if P1 is on P ρ
2 then P1P2 is a tangent line to O at P2. So suppose P2 is

7.5. SYMPLECTIC GEOMETRY AND NORMALIZED COORDINATES339
not on O. Then P2 is the nucleus of the oval P ρ
2
∩ O. Thus every line on P2
in the plane P ρ
2 is tangent to P ρ
2 ∩ O and hence a tangent to O. Conversely,
if P1P2 is tangent to O, say at Q, then we have just shown that P2 is incident
with Qρ , implying Q is incident with P ρ
2 . Thus the line P1P2 = P2Q lies in
P ρ
2 so P1 is incident with P ρ
2 .
7.5 Symplectic Geometry and Normalized Coordinates
Let ρ be a null polarity of P G(3, q). We are going to construct a point-line
geometry W (q). The points of W (q) are all the absolute points of ρ (i.e.,
the points (¯x) for which (¯x) is incident with the plane (¯x) ρ ), which in fact
includes all the points of P G(3, q). The lines of W (q) are the lines of P G(3, q)
which are absolute lines of ρ, i.e., the lines ℓ for which ℓ ρ = ℓ. Incidence is
the one inherited from P G(3, q). It is an easy exercise to show that W (q) is
a GQ of order q. (See Section 4 of the chapter on GQ.)
Corollary 7.4.2 makes it clear that if q = 2 e and ρ is the null polarity
arising from an ovoid Ω, then the absolute lines of ρ are the lines tangent to
Ω.
At this point we want to explain
how to normalize the coordinates for
0 1
P 0
the ovoid Ω. Let P = and let C be the matrix C = , so
1 0
0 P
C is skew-symmetric (with 0’s on the diagonal). For ¯x = (x0, x1, x2, x3) and
¯y = (y0, y1, y2, y3) in F 4 q , put
¯x ∗ ¯y = ¯xC ¯y T .
Then (¯x, ¯y) ↦→ ¯x ∗ ¯y is a nonsingular, bilinear alternating form. If we put
¯x ⊥ = {¯y : ¯x ∗ ¯y = 0}, then for nonzero ¯x, ¯x ⊥ is a plane containing ¯x, and
the map ν : ¯x ↦→ ¯x ⊥ is a symplectic polarity. Recall that any two symplectic
polarities are equivalent. So if Ω ′ is an ovoid with a plane section being
an oval projectively equivalent to some Of, then there is an homography of
P G(3, q) mapping Ω ′ to an ovoid Ω whose symplectic polarity is exactly ν,
and Ω will also contain a plane section that is an oval O equivalent to Of.
If W (q) is the symplectic geometry (i.e., generalized quadrangle) defined by
Ω, then the group of homographies of P G(3, q) leaving W (q) invariant is
known as the symplectic group Sp(4, q) and is transitive on the points of

340 CHAPTER 7. OVOIDS IN 3-DIMENSIONAL SPACE
W (q). Hence there is an homography preserving Ω and mapping the nucleus
of O to the point N = (0, 0, 1, 0). Then by our assumption that ν is the
polarity defined by Ω, we know that the plane containing O is [0, 0, 0, 1],
i.e., O = Ω ∩ [0, 0, 0, 1]. Furthermore, the subgroup of Sp(3, q) fixing N is
transitive on the points collinear with N in W (q), and hence also transitive
on the q + 1 lines of W (q) through N. So if O were a translation oval, for
example, we could without loss of generality assume that the line x0 = x3 = 0
is a translation axis of O.
Let N be the point of P G(3, q) given above, with ν : N ↔ N ⊥ . The
subgroup of Sp(4, q) leaving N ⊥ invariant induces a group of collineations
of N ⊥ containing the group of elations of N ⊥ with center N. (To see this,
consider a line ℓ of N ⊥ and a point Q different from N on ℓ. The group
of elations with center Q and axis Q ⊥ induces in N ⊥ a group containing all
elations of N ⊥ with center Q and axis ℓ. Letting Q range over all points
collinear in W (q) with N, the group generated contains all elations of N ⊥
with center N.) For points in N ⊥ = [0, 0, 0, 1] we may temporarily drop the
final coordinate, which is always 0. Then for the record, the homography
with matrix
⎛
⎝
1 0 c
0 1 bc
0 0 1
for b, c ∈ Fq with c = 0, is an elation of [0, 0, 0, 1] with center N = (0, 0, 1)
and axis [1, b, 0]. The homography with matrix
⎛
⎝
1 0 0
0 1 b
0 0 1
for b = 0, is an elation of [0, 0, 0, 1] with center N = (0, 0, 1) and axis [0, 1, 0].)
Now the group of elations of N ⊥ with center N fixes each of the q + 1 lines of
W (q) through N and is transitive on the q points different from N on each of
the lines. Hence we may assume that O contains the point Q∞ = (0, 1, 0, 0),
that the point N is the nucleus (and when O is a translation oval that the
line L∞ = 〈Q∞, N〉 is a translation axis of O). This means that there is an
o-permutation f such that
O = {(1, f(t), t, 0) : t ∈ Fq} ∪ {(0, 1, 0, 0)} and has nucleus N = (0, 0, 1, 0).
⎞
⎠ ,
⎞
⎠

7.5. SYMPLECTIC GEOMETRY AND NORMALIZED COORDINATES341
Now consider the homography of P G(3, q) with matrix
⎛
⎞
1 f(0)/A 0 0
⎜
H = ⎜ 0 1/A 0 0 ⎟
⎝ 0 0 1 0 ⎠
0 0 0 1/A
where A = f(0) + f(1). It may be checked that HCH T =
1 C, so that
A
the homography commutes with the symplectic polarity. The nucleus N is
left invariant, so of course the plane N ⊥ = [0, 0, 0, 1] is left invariant. The
oval point Q∞ is fixed and the general oval point (1, f(t), t, 0) is mapped to
(1, f(0)+f(t)
, t, 0). In particular, the points (1, 0, 0, 0) and (1, 1, 1, 0) belong
A
to Ω. But if the oval is really a translation oval with translation axis L∞,
then since the function t ↦→ f(0)+f(t)
maps 0 to 0 and 1 to 1, this must
A
be the oval Oα for α a generator of the automorphism group of Fq, where
Oα = {(1, tα , t, 0) : t ∈ Fq} ∪ {(0, 1, 0, 0)} with nucleus N = (0, 0, 1, 0).
We state this as a theorem.
Theorem 7.5.1. Let Ω be an ovoid of P G(3, q) containing an oval section
O equivalent to the oval Oα = {(1, t, t α ) : t ∈ Fq} ∪ {(0, 0, 1)} with nucleus
(0, 1, 0), where α is a generator of Aut(Fq). Then coordinates of P G(3, q)
may be chosen so that the symplectic polarity of P G(3, q) defined by Ω has the
alternating form x0y1+x1y0+x2y3+x3y2 and Ω contains the oval {(1, t α , t, 0) :
t ∈ Fq} ∪ {(0, 1, 0, 0)} with nucleus (0, 0, 1, 0).
7.5.2 Coordinates for the known examples
In this subsection we assume as usual that q = 2 e . The only known ovoids of
P G(3, q) are the elliptic quadrics and the Tits ovoids. We study both these
examples elsewhere, but here we recall standard coordinates for them. But
first we make a general observation about the shape of an ovoid Ω. We have
seen above that we may assume that Q∞ = (0, 1, 0, 0) ∈ Ω along with having
the standard alternating bilinear form associated with Ω. This means that
the tangent plane to Ω at (0, 1, 0, 0) is [1, 0, 0, 0]. Hence all other points of Ω
are not on this plane and hence have coordinates of the form (1, a, b, c). Also,
if there were two points of the form (1, a, b, c) and (1, a ′ , b, c) for a = a ′ , then
the line through these two points of Ω contains the third point (0, 1, 0, 0), an
impossibility. Hence there must be a function f(s, t) of two variables such
that

342 CHAPTER 7. OVOIDS IN 3-DIMENSIONAL SPACE
Ω = {(1, f(s, t), s, t) : s, t ∈ Fq} ∪ {(0, 1, 0, 0)}.
Then with this new function f of two variables, N ⊥ = [0, 0, 0, 1] ∩ Ω =
{(1, f(s, 0), s, 0) : s ∈ Fq} ∪ {(0, 1, 0, 0)}.
Writing the known ovoids in this form we have the following:
Elliptic Quadrics:
O = {(1, s 2 + st + at 2 , s, t) : s, t ∈ Fq} ∪ {(0, 1, 0, 0)},
where x 2 + x + a is irreducible over Fq.
Tits Ovoids (σ ∈ Aut(Fq) with σ 2 = 2):
O = {(1, s σ + st + t σ+2 , s, t)} ∪ {(0, 1, 0, 0)}.
In fact, we can generalize this description of an ovoid and prove a major
theorem of Penttila and Praeger.
For α ∈ AutFq and a function d : Fq → Fq, let O(α, d) denote the set
O(α, d) = {(1, t α + ts + d(s), t, s) : s, t ∈ Fq} ∪ {(0, 1, 0, 0)}.
It is easy to see that with d(s) = as 2 for some a ∈ Fq with tr(a) = 1,
then O(2, d) is an elliptic quadric. And if σ ∈ A with σ 2 = 2, O(σ, σ + 2) is
the Tits Ovoid.
Theorem 7.5.3. Suppose that Ω is an ovoid in P G(3, q), where q = 2 e > 2,
and that π is a secant plane such that π ∩ Ω is a translation oval with associated
automorphism α. Let ℓ be an axis of π ∩ Ω. Suppose that each secant
plane to Ω on ℓ meets Ω in a translation oval. Then there is a collineation
of P G(3, q mapping Ω to O(α, d) where d : Fq → Fq satisfies d(0) = 0 and
d(a) + d(b) ∈ (a + b) α/(α−1) K (7.4)
for all a, b ∈ Fq with a = b, and where K is the kernel of the trace map.
Conversely, if α is a generator of Aut(Fq) and d : Fq → Fq has d(0) = 0 and
satisfies Eq. 7.4 for all a = b, then O(α, d) is an ovoid.

7.5. SYMPLECTIC GEOMETRY AND NORMALIZED COORDINATES343
Proof. By Theorem 7.5.1 we may assume that π = [0, 0, 0, 1], that ℓ =
[1, 0, 0, 0]∩[0, 0, 0, 1], that the nucleus of π ∩Ω is N = (0, 0, 1, 0), that ℓ∩Ω is
the point Q = (0, 1, 0, 0), and that the other points of π ∩ Ω are those of the
form (1, tα , t, 0), t ∈ Fq. For a ∈ Fq, let πa be the plane πa = [a, 0, 0, 1]. With
this notation, π0 = π. The tangent plane to Ω at Q is (0, 1, 0, 0) ⊥ = [1, 0, 0, 0],
so the q planes πa, a ∈ Fq, constitute all the secant planes to Ω through ℓ.
By hypothesis each πa ∩ Ω is a translation oval with some associated field
automorphism βa. The nucleus of πa is π⊥ a = (0, a, 1, 0).
The next step is to use the External Lines Lemma (See Theorem 4.14.7) to
show that ℓ is an axis of πa∩Ω for 0 = a ∈ Fq. For t ∈ Fq, set Pt = (1, t α , t, 0),
a point on Ω ∩ π0 but not in πa. Since Pt ∈ Ω, P ⊥
t ∩ Ω = {Pt}, so P ⊥
t ∩ πa is
a line on N = (0, 0, 1, 0) external to πa ∩ Ω. What is P ⊥
t ∩ πa? It is
[t α , 1, 0, t] ∩ [a, 0, 0, 1] = {(x0, (t α + at)x0, x2, ax0) ∈ P G(3, q) : x0, x2 ∈ Fq}
= {(x0, dx0, x2, ax0) ∈ P G(3, q) : x0, x2 ∈ Fq}, where d ∈ a α/(α−1) K
by part (8) of Theorem 4.12.1. Since K has size q/2 and there are q/2 lines
on N in πa external to πa ∩ Ω, it follows that all such lines are of the form
P ⊥
t ∩ πa (where t automatically satisfies tα + at ∈ aα/(α−1) K).
Now consider the following oval in πa:
O ′ = {(1, t α + at + c, t, a) : t ∈ Fq} ∪ {(0, 1, 0, 0)},
where c ∈ aα/(α−1) K. This is a translation oval, as it is the image of Oα ∼ =
π ∩ Ω under the homography with matrix
⎛
1
⎜ 0
⎝ 0
c
a
1
0
1
0
⎞
a
0 ⎟
0 ⎠
0 0 0 1
.
(Or see Theorem 4.13.2.)
Now O ′ has axis ℓ, nucleus (0, a, 1, 0) on ℓ, and ℓ meets both πa ∩ Ω and
O ′ in the point Q = (0, 1, 0, 0). (In Theorem 4.13.2 interchange the second
and third coordinates.)
Put σd = [d, 1, 0, 0], so that πa ∩ σd = {(x0, dx0, x2, ax0) ∈ P G(3, q) :
x0, x2 ∈ Fq}. So πa ∩ σd is a line of πa through N = (0, 0, 1, 0) not through
Q = (0, 1, 0, 0). The line πa ∩ σd contains a point (1, t α + at + c, t, a) of
O ′ if and only if it is of the form (1, d, t, a) = (1, t α + at + c, t, a). Since

344 CHAPTER 7. OVOIDS IN 3-DIMENSIONAL SPACE
{t α + at : t ∈ Fq} = a α/(α−1) K (of size q/2) and c ∈ a α/(α−1) K, so the
q/2 lines of πa through N external to O ′ are those of the form πa ∩ σd for
d ∈ a α/(α−1) K, that is, the set of line on N in πa external to O ′ and the set
external to πa ∩ O are the same. Hence by the External Lines Lemma, ℓ is
an axis of O. It now follows from Theorem 4.13.2 that
πa ∩ Ω = {(1, c(a)t βa + at + d(a), t, a) : t ∈ Fq},
for some c(a), d(a) ∈ Fq with c(a) = 0. The points (1, s α , s, 0) and (c(a)t βa +
at+d(a), t, a) of Ω are not perpendicular with respect to the form determined
by Ω, so s α + c(a)t βa + at + d(a) + as = 0 for all s and t in Fq. Hence d(a)
does not lie in the set
X = {s α + c(a)t β a + a(s + t) : s, t ∈ Fa}.
But X is a proper subset of Fq which is closed under addition, and X
contains both
{s α + as : s ∈ Fq} = a α/(α−1) K
and
{c(a)t βa + at : t ∈ Fq} = c(a) −1/βa−1)(βa/(β−1)) K
by Theorem 4.12.1, part (8). By part 5 of the same result it follows that
which yields
a α/(α−1) = c(a) −1/(βa−1) a βa/(βa−1)
c(a) = a (α−βa)/(α−1)
for all non-zero a in Fq. Again, the points (1, c(a)s βa + as + d(a), s, a) and
(1, c(b)t βb + bt + d(b), t, b) are not perpendicular. Hence
c(a)s βa + as + d(a) + c(b)t βb + bt + d(b) + at + bs = 0 ∀s, t ∈ Fq.
This says that d(a) + d(b) does not lie in the set
X = {c(a)s βa + (a + b)s + c(b)t βb + (a + b)t : s, t ∈ Fq}.
So X is a proper subset of Fq closed under addition and containing both
A and B where
A = {c(a)s βa + (a + b)s : s ∈ Fq} and B = {c(b)t βb + (a + b)t : t ∈ Fq}.

7.5. SYMPLECTIC GEOMETRY AND NORMALIZED COORDINATES345
and
Here
A = c(a) −1
β b −1 · (a + b) βa
βa−1 K with c(a) = a α−βa
1 ,
B = c(b) −1
]ba−1 · (a + b) β b
β b −1 K with c(b) = b α−β b
α−1 .
It follows that A = B, so
a α−βa
βa−1
After some simplification we find
1
1−α
· b βb−α 1
1−α
βb−1 = (a + b) β b
β b −1 .
(a + b) (1−α)(βb−βa) = a (α−βa)(1−βb) · b (α−βb)(βa−1) . (7.5)
If βa = βb, we find that 1 = (ab −1 ) α−βa , which forces ab −1 to be in the
fixed field of βaα −1 .
If βa = α, then (a + b) (1−α)(βb−α) = b (α−βb)(α−1) , from which we get
1 = (a + b)b −1 βb−α ,
implying that (a + b)b −1 is in the fixed field of βbα −1 , so that again ab −1 is
in the fixed field of βbα −1 .
Suppose there is some b for which βb = α. then the fixed field of βbα −1 can
have at most √ q elements. Since there are only φ(e) generators of Aut(Fq),
and βa can be any one of these for at most √ q values of a, it must be that
φ(e) √ q ≥ q. In particular e ≥ √ q. With q = 2 e this is not possible. Hence
βa must equal α for all nonzero a in Fq. This forces c(a) = 1 for all nonzero
a. If we set d(0) = 0, then Ω = ∪{πa ∩ Ω : a ∈ Fq}, so
where
Ω = O(α, d) = {(1, t α + at + d(a), t, a) : a, t ∈ Fq} ∪ {(0, 1, 0, 0)},
d(a) + d(b) ∈ {(s + t) α + (a + b)(s + t) : s, t ∈ Fq} = (a + b) α
α−1 K. (7.6)
Conversely, O(α, d) is an ovoid corresponding to the alternating form
x0y1 + x1y0 + x2y3 + x3y2 if d satisfies Eq. 7.6, since this is just the condition
that no two points of the set O(α, d) are perpendicular with respect to the
given form.

346 CHAPTER 7. OVOIDS IN 3-DIMENSIONAL SPACE
Theorem 7.5.3 is a major step in the proof that if each secant plane in a
pencil of an ovoid Ω meet Ω in a translation oval, then Ω is either an elliptic
quadric or a Tits ovoid. However, at the present stage in our development
the best we can do is to prove the following.
Theorem 7.5.4. Suppose that Ω is an ovoid in P G(3, q), q = 2 e > 2, and
that π is a secant plane such that π ∩ Ω is a translation oval. Let ℓ be an axis
of π ∩ Ω. If each secant plane to Ω on ℓ meets Ω in in a translation oval,
then there is a groupM of q homographies that fixes each plane πs, and in
πs induces a group of elations all having axis ℓ and acting transitively on the
points of πs ∩ Ω different from (0, 1, 0, 0). (This will be enough later to show
that O is either an elliptic quadric or a Tits ovoid. First we need some very
technical results of D. Glynn [Gl84] that are given in Chapter 8.)
Proof. By the preceding theorem there is a generator α of Aut(Fq) and a
function d such that Ω = O(α, d) and ℓ = [1, 0, 0, 0] ∩ [0, 0, 0, 1]. For u ∈ Fq
let
⎛
1
⎜
Mu = ⎜
⎝
uα 0
0
1
0
u
0
1
⎞
0
0 ⎟
0 ⎠
0 u 0 1
.
Then x ↦→ xMu is an homography (also denoted Mu) of P G(3, q) that maps
the point (1, t α +st+d(s), t, s) to the point (1, (t+u) α +s(t+u)+d(s), t+u, s).
So Mu leaves Ω fixed and induces in each πs an elation with axis [1, 0, 0, 0] ∩
[0, 0, 0, 1] with center (0, u α−1 + s, 1, 0). The group M = {Mu : u ∈ Fq} is
transitive on the points different from (0, 1, 0, 0) in πs ∩ Ω.
7.6 The plane representation theorem
We are now prepared to prove the “plane representation theorem” found
independently by D. Glynn and T. Penttila.
Let ℓ be a line of P G(3, q) tangent to the ovoid Ω at a point Q. So ℓ is
on the unique plane tangent to Ω at Q and on q secant planes meeting Ω
in q different ovals whose nuclei are the q points of ℓ different from Q. This
set of q secant planes on a tangent line ℓ is called a pencil of Ω and ℓ is the
carrier line of the pencil.
Let O1 and O2 be two ovals of P G(2, q) (always q = 2 e ) having a point
Q in common and the same nucleus N. Let ℓ be the line ℓ = 〈Q, N〉 and let

7.6. THE PLANE REPRESENTATION THEOREM 347
P be any point of ℓ different from Q and N. We say that O1 and O2 are
compatible at P provided every secant line to O1 through P is external to
O2, which is equivalent to having each tangent line to O1 through P be a
secant to O2.
Let {Os : s ∈ Fq} be a set of ovals in P G(2, q) all having nucleus N =
(0, 1, 0) and satisfying Os ∩ Ot = {Q = (0, 0, 1)} for all s, t ∈ Fq with s = t
and such that Os and Ot are compatible at (0, 1, s + t). Any image under a
collineation of P G(2, q) of such a set of ovals is called a fan of ovals and the
corresponding image of the line [1, 0, 0] is called the common tangent of the
fan. The Plane Equivalent Theorem (also known as the Plane Representation
Theorem) establishes a correspondence between pencils of ovoids of P G(3, q)
and fans of ovals of P G(2, q).
Theorem 7.6.1. (Glynn and Penttila) There is a one-to-one correspondence
between orbits of P ΓL(4, q) on (ovoid, tangent line)-pairs in P G(3, q) and
orbits of P ΓL(3, q) on fans of ovals of P G(2, q). Moreover, if the fan is
{Os : s ∈ Fq},
then for each plane π of the pencil there is a parameterization πs, s ∈ Fq of
the planes of the pencil with π0 = π and such that there are homographies
Ms taking πs to P G(2, q) with the image of the intersection of the ovoid with
πs under Ms being Os for each s ∈ Fq. Also, the image of the carrier line of
the pencil under each Ms is the common tangent of the fan.
Proof. Let Ω be an ovoid of P G(3, q) that defines the symplectic polarity
ν : (x0, x1, x2, x3) ↔ [x1, x0, x3, x2].
For t ∈ Fq, put Nt = (0, 1, t, 0) and Q∞ = (0, 0, 1, 0). We may suppose
that Q∞ ∈ Ω and that L∞ = 〈Q, N0〉 is a tangent line to Ω. Consider the
pencil with carrier L∞. Since ν : Q∞ ↦→ π∞ = [0, 0, 0, 1], this plane is the
tangent plane to Ω at Q∞. So the secant planes of the pencil with carrier
L∞ are those of the form πa = [1, 0, 0, a], a ∈ Fq. Then ν : πa ↦→ Na. Since
Oa = Ω ∩ πa contains the point Q∞ = (0, 0, 1, 0) and has nucleus Na, there
must be an o-polynomial fa for which
Oa = {(a, t, fa(t) + at, 1) : t ∈ Fq} ∪ {Q∞} with nucleus Na = (0, 1, a, 0).

348 CHAPTER 7. OVOIDS IN 3-DIMENSIONAL SPACE
Let I be the 2 × 2 identity matrix and Ds the matrix Ds =
Let µs be the homography with matrix
I Ds
[µs] = Ms =
Ds I
i.e., µs : (w, y, z, x) ↦→ (w, y, z, x) · Ms.
,
Each of the following facts has a very simple proof.
(i) MsMt = Ms+t and M0 = I4, so µs is an involution.
(ii) MsCM T s = C, so µs commutes with the polarity ν.
(iii) µs : Nt ↦→ Ns+t, hence µs : πt ↦→ πs+t.
(iv) µs : Q∞ ↦→ Q∞; µs : π∞ ↦→ π∞; µs : L∞ ↦→ L∞.
0 0
s 0
Since µs : πs ↦→ π0, µs : Ns ↦→ N0, and µs : Q∞ ↦→ Q∞, clearly µs :
Os ↦→ Ōs = {(0, t, fa(t), 1) ∈ π0 : t ∈ Fq} ∪ {Q∞} with nucleus N = N0. Put
F = { Ōs : s ∈ Fq}. We will show that F is a fan of ovals in π0.
Let Qs,a = (s, 0, a, 1) and recall Nt = (0, 1, t, 0). Put L = 〈Qs,a, Nt〉.
As a varies over elements of Fq, L varies over the lines of πs through the
point Nt. Check that µs+t : Qs,a ↦→ Qt,a and µs+t : Nt ↦→ Ns. Check that
〈Qt,a, Ns〉 ν = Q ν t,a ∩ N ν s = [0, t, 1, a] ∩ [1, 0, 0, s] = L. Hence 〈Qt,a, Ns〉 = L ν ,
i.e., µs+t : L ↦→ L ν . Hence µs+t maps the lines of πs through Nt to the lines
of πt through Ns in such a way that each such line gets mapped to its polar
image. If L is a secant (respectively, external) line to Ω, its polar image is
an external (respectively, secant) line. So µs+t maps the secant lines to Os
through Nt to the external lines to Ot through Ns. But of course µs+t maps
secant lines to Os to secant lines to (Os) µs+t through Ns. Hence (Os) µs+t is
compatible with Ot at Ns, and it follows that
(Os) µs+t◦µt = (Os) µs = Ōs is compatible with (Ot) µt = Ōt at N µt
s
= Ns+t.
Thus { Ōs : s ∈ Fq} is a fan of ovals of π0 with common point Q∞ = (0, 0, 1, 0)
and common nucleus N0 = (0, 1, 0, 0).
Conversely, suppose we have a fan of ovals of P G(2, q). Take the image of
the fan under a collineation so that the common nucleus is (1, 0, 0) and the
.

7.6. THE PLANE REPRESENTATION THEOREM 349
common point is (0, 1, 0), and let the resulting fan be denoted {Os : s ∈ Fq}.
Each oval has the form {(t, fs(t), 1) : t ∈ Fq}∪{(0, 1, 0)} with nucleus (1, 0, 0).
Embed P G(2, q) in P G(3, q) by ι : (y, z, x) ↦→ (0, y, z, x). Then {ι(Os) :=
Ōs : s ∈ Fq} is a fan of ovals of π0 = [1, 0, 0, 0]. Put Ω = {( Ōs) µs : s ∈ Fq}.
We claim that Ω is an ovoid of P G(3, q). The proof amounts to showing that
no two points of Ω can be perpendicular with respect to the form defined by
the polarity ν. Clearly (0, t, fs(t), 1) ∗ (0, r, fs(r), 1) = fs(t) + fs(r) = 0 for
r = t, and (0, t, fs(t), 1) ∗ (0, 0, 1, 0) = 1 = 0. Hence no two points of the
same ( Ōs) µs are perpendicular.
Suppose X and Y are two points of Ω. Suppose that X ∗ Y = 0, and that
X ∈ ( Ōs) µs and Y ∈ ( Ōr) µr . This says that
Then
X = (s, x, sx + fs(x), 1); Y = (r, y, ry + fr(y), 1), for some x, y ∈ Fq.
X ∗ Y = sy + xr + sx + fs(x) + ry + fr(y) = fs(x) + fr(y) + (x + y)(r + s) = 0.
Consider the line in π0 through (0, x, fs(x), 1) ∈ Ōs and Nr+s = (0, 1, r +
s, 0). It is a secant to Ōs through the point Nr+s at which Ōs and Ōr
are compatible. So Ōr should have no point of this line. But a point on
the line is (0, x, fs(x), 1) + (y + x)(0, 1, r + s, 0) = (0, y, (y + x)(r + s) +
fs(x), 1) = (0, y, fr(y), 1) ∈ Ōr, a contradiction. Hence no two points of Ω
are perpendicular with respect to the form defined by ν. Hence Ω is an ovoid
of the GQ W (q) defined by this form. Then by a result of J. A. Thas, Ω must
be an ovoid of P G(3, q). We include a proof of this as a separate lemma.
Lemma 7.6.2. Let Ω be any ovoid of W (q). Then Ω is an ovoid of P G(3, q).
Proof. Let Ω = {Q0, Q1, . . . , Q q 2}. A line of W (q) will be called a W -line,
and a line of P G(3, 1) not in W (q) will be called a W ′ -line. Through Q0 (for
example), there are q + 1 W-lines and q 2 W ′ -lines. Let Mj = 〈Q0, Qj〉, for
1 ≤ j ≤ q 2 . If the lines M1, . . . , Mj are not all distinct, i.e., if some three
points Q0, Qi, Qj of Ω are on a W ′ -line, then there must be a W ′ -line ℓ through
Q0 that contains no other point of Ω. Suppose that ℓ = {Q0, P1, . . . , Pq}.
Each of the W -lines through each Pi must contain a unique point of Ω (clearly
not Q0). Hence some two of these points of ℓ (say Pu, Pv) must be W -collinear
with the same Qj. Since the two lines 〈Pu, Qj〉 and 〈Pv, Qj〉 are both W -lines,
the plane they define must consist of the points that are W -collinear with

350 CHAPTER 7. OVOIDS IN 3-DIMENSIONAL SPACE
Qj. But this says that Q0 is W -collinear with Qj, a contradiction. Hence no
line through Q0 can pass through two other points of Ω. This says that Ω is
an ovoid of P G(3, q).
This can be immediately reworded to have a rather different appearance
although it says the same thing.
Corollary 7.6.3. A set of q 2 + 1 points of P G(3, q) with q = 2 e , q > 2,
such that no two are perpendicular with respect to a fixed non-degenerate
alternating form is an ovoid of P G(3, q).
7.7 Semi-ovals and Semi-ovoids
This section contains some results of J. A. Thas [Th74b]. Let K be a
nonempty pointset of P G(n, q), n ≥ 2. A tangent to K at a point P of
K is a line ℓ such that ℓ ∩ K = {P }. Then K is called a semi-oval (resp.,
semi-ovoid), provided n = 2 (resp., n > 2) and for each P ∈ K the union
KP of all tangents to K at P is a hyperplane of P G(n, q). The semi-oval K
(resp., semi-ovoid K) is an oval (resp., ovoid) of P G(n, q) if and only if each
line not tangent to K intersects K in just two points or in none at all. Recall
that each oval contains q+1 points and each ovoid of P G(3, q) contains q 2 +1
points.
If σ is a polarity of a finite projective plane Π of odd order, then the set
of absolute points of σ is a semi-oval. A nonsingular hermitian polarity of
P G(2, q) with q odd has 1 + q √ q points. If O is a semi oval of the plane Π
and p ∈ O is a point such that all non-tangent lines through p intersect O
in more than two points, then O \ {p} is still a semi oval of P . A quadrangle
in P G(2, 3) together with two of its diagonal points gives a semi oval with
6 points. So for planes, the concept of semi oval is definitely more general
than that of oval. On the other hand, F. Buekenhout [Bu76] conjectured
that every finite semi ovoid in dimension three is an ovoid and that there are
no finite semi ovoids in dimension n > 3. The next theorem by J. A. Thas
shows that the conjecture by Buekenhout is true.
Theorem 7.7.1. (a) If O is a semi-oval of the finite projective plane π of
order q, then q + 1 ≤ |O| ≤ q √ q + 1.
(b) The only semi-ovoids of P G(3, q) are the ovoids.
(c) In P G(n, q) with n > 3 there are no semi-ovoids.

7.7. SEMI-OVALS AND SEMI-OVOIDS 351
Proof. Let O be a semi-oval of the finite projective plane π of order q. If
P ∈ O and if ℓ is a line through P not tangent to O, then |O ∩ ℓ| > 1.
Consequently
|O| ≥ q + 1. (7.7)
Next let O be a semi-ovoid in P G(n, q), n ≥ 3. Let ℓ be a line tangent to
O at the point P of O. If π is a plane through ℓ different from the tangent
hyperplane OP , (so π is not a tangent plane at any point of O) and if Q is
any point of (π ∩ O), then the intersection π ∩ O meets the tangent plane
OQ in a unique line and it follows that π ∩ O is a semi-oval of π. Since there
are q n−2 planes through ℓ not belonging to the tangent hyperplane , we have
|O| = 1 +
(|π ∩ O| − 1) ≥ 1 + q · q n−2 = q n−1 + 1. (7.8)
π
Suppose that n ≥ 2. Let α be the number of points on O and v the
number of points not on O, say {x1, . . . , xv} = P G(n, q) \ O and O =
{P1, . . . , Pα}. Put ti equal to the number of tangent lines on xi, 1 ≤ i ≤ v.
Note that there are n−2 i=0 qi tangent lines on each Pj ∈ O, and q points xi
on each such tangent line but not on O. Count in two ways the number of
ordered pairs (xi, Pj) such that xiPj is a tangent line of O to obtain:
v n−2
ti = αq q i . (7.9)
i=1
Now count in two ways the number of ordered triples (xi, Pj, Pk) where
xi ∈ P G(n, q) \ O, Pj, Pk ∈ O, Pj = Pk, and both xiPj and xiPk are tangent
lines.
Since two tangent hyperplanes meet in a P G(n − 2, q) disjoint from O,
this number is
i=0
v
n−2
ti(ti − 1) = α(α − 1) q i . (7.10)
i=1
Now put Q ′ = n−2
i=0 qi and add Eqs. 7.9 and 7.10 to obtain
t 2 i = α(α + q − 1)Q ′ . (7.11)
Put t = 1
ti. Then
v
0 ≤
(ti − t) 2 =
i
i
t 2 i
i=0
− 2t ti + v · t 2
i

352 CHAPTER 7. OVOIDS IN 3-DIMENSIONAL SPACE
= α(α + q − 1)Q ′ − 1
2
ti .
v
i
Put in the fact that v = n i=0 qi − α to get
n
0 ≤ q i
− α α(α + q − 1)Q ′ − α 2 q 2 Q ′2 .
i=0
Divide through by αQ ′ :
n
0 ≤ q i
− α (α + q − 1) − α
i=0
Now put Q = n
i=0 qi , so we have
n
q i .
i=2
0 ≤ (Q − α)(α + q − 1) − α(Q − q − 1).
After some simplification this becomes
α 2 − 2α − (q − 1)Q ≤ 0, i.e., 1 − q n+1 ≤ α ≤ 1 + q n+1 . (7.12)
It is now clear that part (a) of the theorem is proved, and from Eqs. 7.8
and 7.12 we have the following:
If n ≥ 3, then 1 + q n−1 ≤ α ≤ 1 + q n+1 . But q n−1 ≤ q n+1 implies
that n ≤ 3 and if n = 3 then α = 1 + q 2 .

Chapter 8
Trace Equations
The first characterization theorem for elliptic quadrics in P G(3, q) was the
theorem of Barlotti [Ba55] and (independently) Panella [Pa55] that we gave
in Chapter 7. This theorem says that an ovoid of P G(3, q) for which each
plane section is a conic must be an elliptic quadric. D. Glynn [Gl84] improved
this to say that an ovoid with a pencil of conics is an elliptic quadric. Glynn’s
paper was not quite self-contained. In several papers T. Penttila with various
coauthors strengthened the trace equations of Glynn so that in this chapter
we can give a self-contained version due essentially to Penttila. In view of
the fact that in Chapter 11 we give the result by M. Brown that characterizes
elliptic quadrics as the ovoids with even one conic, and his proof really
depends only on the Barlotti-Panella result, one may question the need for
the present chapter. However, the full story is not in yet. In this chapter we
give several results that were used by Penttila (and coauthors) to give more
general characterizations. These results appear later in this book. It may be
that this material will prove useful in yet further improvements.
In this chapter we collect a variety of rather specialized results concerning
the trace function. These are all used to characterize ovoids with certain
kinds of plane sections and also flock generalized quadrangles to be introduced
later. This chapter is rather dry reading and might better be referred
to from time to time later as certain of the results are needed.
353

354 CHAPTER 8. TRACE EQUATIONS
8.1 Cyclotomic Cosets
In this section q = 2 e for some positive integer e. For each integer x put
[x] = {2 i x (mod q − 1) : i ∈ Z}.
This set [x] of integers modulo q−1 is called the cyclotomic coset of x modulo
q − 1. For 2 i x, 2 j x in [x], define the product
2 i x ◦ 2 j x ≡ 2 i+j x (mod q − 1).
With this product it is easy to show that [x] is a cyclic group with identity
2 0 x = x and generator 2x. Since 2 e x ≡ x (mod q − 1), the order of the group
[x] must divide e. Let ℓx be the order of the group [x]. Here we may think
of x as being a residue modulo q − 1 since if x ≡ y (mod q − 1) then [x] and
[y] are identical. So another way to view ℓx is that it is the smallest positive
integer for which (2 ℓx − 1)x ≡ 0 (mod q − 1). This means that it is the order
of 2 modulo (q − 1)/gcd(x, q − 1). In particular, ℓ0 = 1 for all q − 1, and
ℓ1 = e for q − 1 = 2 e − 1.
Lemma 8.1.1. The residues modulo q − 1 = 2 e − 1 are partitioned into
cyclotomic cosets.
Proof. We need to show that two cyclotomic cosets are disjoint or identical.
So suppose that x and y are integers that are distinct modulo q − 1. Suppose
that there are integers i and j for which 2 i x ≡ 2 j y (mod q − 1). Since any
power of 2 is relatively prime to q − 1, this implies y ≡ 2 i−j x (mod q − 1),
i.e., y ∈ [x], which easily implies [y] ⊆ [x]. Similarly, [x] ⊆ [y]. So Zq−1 is
partitioned into distinct cyclotomic cosets.
Def. A reduced representative set (RRS) (modulo q − 1) is a set of distinct
representatives for the distinct cyclotomic cosets modulo q − 1. The
definitions and Lemma 8.1.1 make it clear that the following corollary holds.
Corollary 8.1.2. Let R be an RRS modulo q − 1. Then each residue (mod
q − 1) is uniquely represented in the form 2 i d for some d ∈ R and 0 ≤ i ≤
ℓd − 1.
For 1 ≤ j ≤ q − 2 there is a unique nonzero dj ∈ R such that j ∈ [dj].
Let ij be the smallest nonnegative integer such that 2 ij · dj ≡ j (modq − 1).

8.2. MORE BASIC TRACE EQUATIONS 355
Then j
:= 2 dj
ij 0 . For j = 0, we have 2 · 0 ≡ 0(modq − 1). Here ℓ0 = 1
and d0 = 0. By convention we put 0 = 1. With this notation we have the
d0
following lemma.
Lemma 8.1.3. For 0 ≤ j ≤ q − 2, we have j ≡ dj · 2 ij (mod q − 1) with
dj ∈ R uniquely determined and ij the smallest nonnegative integer for which
the congruence holds. Then
j ≡ d · 2 i (mod q − 1) with d ∈ R and 0 ≤ i ≤ e − 1 ⇐⇒
d = dj ∈ R and i ≡ ij (mod ℓdj ).
Lemma 8.1.4. Suppose that j ≡ dj ·2 ij with dj ∈ R as above. Then ℓj = ℓdj .
Proof. ℓ ≡ 0 (mod ℓj) if and only if (2 ℓ −1)j ≡ (2 ℓ −1)dj ·2 ij ≡ 0 (mod q −1)
if and only if (2 ℓ − 1)dj ≡ 0 (mod q − 1) (since (2 ij , q − 1) = 1) if and only
if ℓ ≡ 0 (mod ℓdj ). This shows that ℓj = ℓdj .
8.2 More Basic Trace Equations
Lemma 8.2.1. Let α be a generator of Aut(Fq) = A, q = 2 e . Then
tr((x + 1) α/(α−1) + x α/(α−1) + 1) = 0 (8.1)
for all x ∈ Fq if and only if α = 2 or α 2 = 2.
Proof. Suppose that Eq. 8.1 holds for all x ∈ Fq. Since α generates Aut(Fq)
there is a smallest positive integer i such that α i = 2. Hence (α − 1)(1 +
α + · · · + α i−1 ) = (α i − 1) = 2 − 1 = 1 and it follows that α/(α − 1) =
α + α 2 + · · · + α i . Hence
(x + 1) α/(α−1) + x α/(α−1) + 1
= (x + 1) α+α2 +···+αi + x α+α2 +···+αi + 1
= (x + 1) α (x + 1) α2
· · · (x + 1) αi
+ x α+α2 +α 3 +···+α i
= (x α + 1)(x α2
+ 1) · · · (x αi
+ 1) + x α+α2 +···+αi + 1
= x n ,
+ 1

356 CHAPTER 8. TRACE EQUATIONS
where the sum is over all n which are sums of a non-empy proper subset of
{α, α2 , . . . , αi }. Suppose that i > 1. We claim that the only powers of 2
(mod q − 1) occurring as exponents in this summation are α, α2 , . . . , αi .
Before proving this claim we note a couple of things. First, each of the terms
α, α2 , . . . , αi does occur. Second, if the claim is true, then Lemma 4.12.2 is
saying that the number of terms of the form x αj
that appear must be even,
since each would appear with a coefficient a α j = 1 and they must add up to
0.
Then, by hypothesis α = 2 a for some a with 1 ≤ a ≤ e − 1 and (a, e) = 1.
Suppose that for some X = {k1, . . . , kd} ⊆ {1, 2, . . . , i} it is true that
k∈X 2ak ≡ 2 j (mod q − 1) for some 0 ≤ j < e. There are distinct integers
l1, . . . , ld such that, for each m, 1 ≤ m ≤ d, we have 0 < lm < e and
lm ≡ akm (mod e). Then 2 lm = 2 j . It follows that X = {k1} and 2 j =
2 l1 ≡ 2 ak1 ≡ α k1 (mod q − 1). Thus there are exactly i powers of 2 occurring
as exponents in the summation. By the previous lemma this number i must
be even.
Now suppose that i > 2. We claim that the only sums of two powers of
2 of the form 2 j + 2 j+a , where 0 ≤ j < e and α = 2 a , occurring as exponents
in the sumation are α + α 2 , α 2 + α 3 , . . . , α i−1 + α i . With the notation of
the previous paragraph suppose that
2 lm ≡ 2 akm ≡ 2 j + 2 j+a ≡ 2 j + 2 l (mod q − 1),
where j + a ≡ l (mod e) for 0 ≤ l < e. Hence 2 ak1 + 2 ak2 = 2 j + 2 j+a ,
and we deduce as above that X = {k1, k2 = k1 + 2} with 2 j = α k1 and
2 l ≡ α k2 = α k1+1 , proving our claim. By Lemma 4.12.2 the number i − 1 of
terms of the form α j + α j+1 must be even. This is a contradiction. Hence
i ≤ 2, that is α = 2 or α 2 = 2.
Conversely, if α = 2 then α/(α − 1) = 2, so (x + 1) α/(α−1) + x α/(α−1) + 1 =
(x + 1) 2 + x 2 + 1 = 0 and hence tr((x + 1) α/(α−1) + x α/(α−1) + 1) = 0 for all
x ∈ Fq. Similarly if α 2 = 2 then α/(α − 1) = α + α 2 = α + 2, so
(x + 1) α/(α−1) + x α/(α−1) + 1 =
= (x + 1) α (x + 1) 2 + x α+2 + 1
= (x α + 1)(x 2 + 1) + x α+2 + 1
= x α + x 2 .
For all x ∈ Fq we have tr(x α + x 2 ) = tr(x α )+tr(x 2 ) = 0.

8.2. MORE BASIC TRACE EQUATIONS 357
Lemma 8.2.2. Let q = 2 e and let n be a positive integer. If for some a ∈ F ∗ q
either
(1) tr(ax n ) = 0 for all x ∈ Fq, or
(2) tr(ax n ) = 1 for all x ∈ F ∗ q ,
then for some integer i with 0 < i < e it follows that n(2 i −1) ≡ 0 (mod q−1).
Proof. From Lemma 4.12.2 we see that there must be (at least one) α ∈ A,
α = id, such that nα ≡ n (mod q − 1), say α = 2 i . This proves (1). For
(2), if tr(b) = 1, then tr(ax n + bx q−1 ) = 0 for all x ∈ Fq. Since nα ≡
q − 1 (mod q − 1), by Lemma 4.12.2 there must be some α ∈ A with α = id
for which nα ≡ n (mod q − 1).
Lemma 8.2.3. Let α and β be generators of A. Suppose that either
(1) (α/(α − 1) − β/(β − 1))(2 i − 1) ≡ 0 (mod q − 1), or
(2) (α/(α − 1) − β)(2 i − 1) ≡ 0 (mod q − 1)
for some i with 0 < i < e. Then α = β, and in Case (2) α = 2.
Proof. By hypothesis, α = 2 j with 0 < j < e and (j, e) = 1. So (α−1, q−1) =
1. Similarly, both β −1 and (α−1)(β −1) are units in Zq−1. Suppose that (1)
holds. Then multiplying by (α − 1)(β − 1) we have (α(β − 1) − β(α − 1))(2 i −
1) = (β −α)(2 i −1) ≡ 0 )mod q −1), implying β2 i +α ≡ α2 i +β (mod q −1).
By the uniqueness of the binary expansion of a number, {β2 i , α} = {α2 i , β},
(interpreting β2 i and α2 i as powers of 2 less than q). Since α = α2 i , it follows
that α = β.
Now suppose that (2) holds. Then multiplying by α − 1 gives (α + β −
αβ)(2 i −1) ≡ 0 (mod q−1). Hence α2 i +β2 i +αβ ≡ α+β+αβ2 i (mod q−1),
and we interpret each side of this congruence as a sum of three positive powers
of 2, each less than q. If, on each side, the three terms were distinct powers
of 2, then we would have {α2 i , β2 i , αβ} = {α, β, αβ2 i }, but this is not the
case since αβ does not lie in the second set. Thus on at least one side of
the congruence two of the terms are equal. It follows that either α = β or
one of α and β lies in {2 i , 2 e−i }. In the latter case, as α and β are both
generators of A, the integer i must be prime to e, so the hypothesis of (2)
says (α/(α − 1) ≡ β (mod q − 1). By Lemma 4.12.5, α = β = 2. In the
former case, the congruence becomes α2 i+1 + α 2 ≡ 2α + α 2 2 i (mod q − 1), so
{α2 i+1 , α 2 } = {2α, α 2 2 i } and hence α = 2 (since 2 i ≡ 1 (mod q − 1)).
Lemma 8.2.4. Let α, γ be generators of A with α = 2. If
trace((x α−1 + 1) 1/(γ−1) + x) = 1 ∀x ∈ Fq, (8.2)

358 CHAPTER 8. TRACE EQUATIONS
then q is not a square and α = γ = σ, where σ ∈ A is such that σ 2 = 2.
Proof. Putting x = 0 in Eq. 8.2 shows that trace(1) = 1, hence q is not a
square and
trace((x α−1 + 1) 1/(γ−1) + x) = 1 ∀x ∈ Fq
⇐⇒ trace((x α−1 + 1) 1/(γ−1) + x + 1) = 0 ∀x ∈ Fq
⇐⇒ trace((y + 1) 1/(γ−1) + y 1/(α−1) + 1) = 0
⎛
∀y ∈ Fq (8.3)
⎜
⇐⇒ trace ⎝ y k + y 1/(α−1) ⎞
⎟
+ 1⎠
= 0 ∀y ∈ Fq.
k 1
γ−1
Since α and γ are generators of A there exist positive integers n, m such
that α n = 2 and γ m = 2. It follows that 1/(α − 1) = 1 + α + · · · + α n−1 and
1/(γ − 1) = 1 + γ + · · · + γ m−1 . Hence we obtain
⎛
trace ⎝
k1+γ+···+γ m−1
y k + y 1+α+···+αn−1
⎞
+ 1⎠
= 0 ∀y ∈ Fq.
Let bi be the coefficient of y i in the sum and let ai be the coefficient of y i
in the whole expression. Put z = 1 + α + · · · + α n−1 . Then a0 = b0 + 1,
az = bz + 1 and ai = bi for each i = 0, z. We shall apply Lemma 4.12.2 with
i = 1 + γ + · · · + γ m−1 . For β ∈ A we have a iβ −1 = 0 if and only if either
iβ −1 i and iβ −1 = z, or iβ −1 = z and z i.
Suppose that iβ −1 = z for each β ∈ A. then a iβ −1 = 0 for β ∈ A if and
only if iβ −1 i, that is, if and only if β = 1. By Lemma 4.12.2 we have
ai = 1 = 0, an impossibility. Hence there exists β ∈ A such that iβ −1 = z.
This says
(1 + γ + · · · + γ m−1 ) = β(1 + α + · · · + α n−1 ).
It follows that α − 1 = β(γ − 1). Suppose that α = 2 r , β = 2 s , γ = 2 t ,
where (r, e) = (t, e) = 1. Then a little manipulation gives α + β = βγ + 1, or
2 r + 2 s ≡ 2 s+t + 1 (mod q − 1). It follows that 2 s = 1, i.e., s = 0 and r = t,
i.e., α = γ. Now apply the automorphism α to the first line of Eq. 8.3 with
α = γ and x = y 1
α−1 , and use Lemma 8.2.1 to see that α2 = 2 as desired.

8.3. TRACE EQUATIONS BY GLYNN 359
8.3 Trace Equations by Glynn
The first result is Lemma 4.7 of Glynn [Gl84].
Let q = 2e and let R be a reduced representative system mod q − 1, so
R contains one element of each cyclotomic coset mod q − 1. For each d ∈ Z
(but especially for each d ∈ R), ℓd is the smallest positive integer for which
(2ℓd q−1
− 1)d ≡ 0 (mod q − 1), i.e., ℓd is the order of 2 modulo . Put
gcd(d,q−1)
σd = 2ℓd, so σk e
d , 1 ≤ k ≤ ℓd , are the distinct powers of σd. If 0 ≤ j ≤ q − 2,
write j ≡ dj · 2ij for a uniquely determined dj ∈ R and a unique smallest
nonnegative integer ij. (Review Section 8.1 and recall that ℓj = ℓdj .)
For each d ∈ R, let bd be an element of Fq. Put
⎛ „ «
q−2 e/ℓj j
(σd ) dj j
f(t) = ⎝ b
k
⎞
⎠ t dj
j ⎛
q−2 e/ℓj
= ⎝ b 2i ⎞
j +kℓj⎠
t dj
j .
j=0
k=1
Lemma 8.3.1. 1 = tr
d∈R bdt d = e−1
i=0
j=0
k=1
d∈R bdt d 2 i
and only if e/ℓd
k=1 bσk d
d = 0 for all d ∈ R \ {0}, and tr(b0) = 1.
Proof. The key observation here is that
e−1
bdt d
⎛
2i q−2 e/ℓj
= ⎝ b
i=0
d∈R
j=0
k=1
„ «
j
(σd ) dj j k
dj
⎞
⎠ t j = f(t).
for all t ∈ F ∗ q if
If this expression equals 1 for all t ∈ F ∗ q , since deg(f(t)) ≤ q − 2 < q − 1,
then if f(0) = ℓ, f(t) must be the unique polynomial of degree at most q − 1
1, t = 0,
with f(t) =
l, t = 0, Hence f(t) = (1 − ℓ)tq−1 + ℓ. Since f(t) actually
has degree at most q −2, it must be that ℓ = 1 and f(t) = 1. So the constant
= tr(b0). Also, if 1 ≤ j ≤ q − 2, then
term is 1 = f(0) = e/ℓ0
k=1 b1·(2ℓ 0 ) k
d0
implying
e/ℓj
b
k=1
„ «
j
(2 dj ℓj ) k
dj
⎛
e/ℓj
= ⎝
e/ℓj
k=1
b σk d j
dj
k=1
b 2kℓ d j
dj
= 0.
⎞
⎠
„ «
j
dj = 0,

360 CHAPTER 8. TRACE EQUATIONS
This completes the proof.
Our next result generalizes Lemma 4.8 of Glynn [Gl84].
Let m be an integer (mod q −1) such that (m, q −1) = 1. Let g : Fq → Fq
be a function given by
q−2
g(x) = aix i .
We want to investigate just when g(x) can have the property that
g(a) + g(b)
(a + b) m
It is clear that this holds if and only if
Put
g(s + t) + g(s)
t m
K(s, t) =
i=0
has trace 1 for all a, b ∈ Fa, a = b.
has trace 1 for all s, t ∈ Fq, t = 0.
g(s + t) + g(s)
t m
= t −m
q−2
i=0
ai((s + t) i + s i ).
So we want to know when tr(K(s, t)) = 1 ∀ s, t ∈ Fq, t = 0.
Let R be an RRS (mod q − 1). Recall that each residue mod q − 1
is uniquely represented in the form p2 i mod q − 1 where p ∈ R and 0 ≤
i ≤ lp − 1. So if m is a fixed residue mod q − 1, each i mod q − 1 has
a unique representation of the form i ≡ p · 2 k + m (mod q − 1) for some
p ∈ R and 0 ≤ k ≤ lp − 1. Let p · 2 k + m be the unique integer such that
0 ≤ p · 2 k + m ≤ q − 2 and p · 2 k + m ≡ p · 2 k + m (mod q − 1). Then it is
clear that
Hence
{i : 0 ≤ i ≤ q − 2} = {p · 2 k + m : p ∈ R & 0 ≤ k ≤ lp − 1}.

8.3. TRACE EQUATIONS BY GLYNN 361
=
tr ap·2k (s + t) +m
p∈R k=0
p·2k +m p·2
+ s k
+m
t −m
lp−1
2−k
tr ap·2k (s + t) +m
p·2k +m p·2
+ s k
+m
t −m
2−k tr(K(s, t)) = lp−1
= tr
p∈R k=0
lp−1
p∈R
Put αpk =
k=0
2−k ap·2k +m
2−k ap·2k +m
s
p·2k +m
+ 1 +
t
s
p·2k +m
2−k t
t p
.
and z = s.
Then tr(K(s, t)) = 1 for all s, t ∈ Fq
t
with t = 0 if and only if tr(K ′ (z, t) = 1 for all z, t ∈ Fq with t = 0, where
tr{K ′ lp−1
(z, t)} = tr αpk((1 + z) p·2k +m p·2
+ z k +m 2
) −k
· t p
.
p∈R
k=0
Consider K ′ (z, t) as a polynomial in t and apply Lemma 8.1 to get the
following result:
Lemma 8.3.2.
(i) e/lp
j=1
g(s) + g(t)
tr
(s + t) m
= 1 ∀ s = t ∈ Fq ⇐⇒
lp−1
k=0 αpk
(1 + z) p·2k +m p·2 + z k 2−k +m
2jlp = 0 ∀ z ∈ Fq, p ∈
R \ {0},
and
(ii) (p = 0) 1 = tr {α00 [(1 + z) m + z m ]}, for all z ∈ Fq and α00 =
am = am is the coefficient of x m in g(x). In particular tr(am) = 1.
Note: We emphasize that if tr
g(s)+g(t)
(s+t) m
= 1 ∀ s = t ∈ Fq, then if am is
the coefficient of x m in g(x), it must be that 1 = tr {am [(1 + z) m + z m ]} for
all z ∈ Fq. In particular, tr(am) = 1. Going back through the substitutions
made above we see that this is equivalent to
m m a(t + s )
tr
(t + s) m
= 1 ∀ t, s ∈ Fq, s = t. (8.4)

362 CHAPTER 8. TRACE EQUATIONS
Lemma 8.3.3. (Lemma 2.6 of [GOPP96]) Suppose that for some a ∈ Fq
and some integer r,
Then
tr {a[(x + 1) r + x r ]} = 1 ∀ x ∈ Fq.
tr{(x + 1) r + x r + 1} = 0 ∀ x ∈ Fq.
Proof. Put x = 0 to see that tr(a) = 1. Then put T = {(1+x) r +xr +1 : x ∈
Fq}. Since tr{a(x + 1) r + axr + a} = 0, clearly aT ⊆ K. Also T α = T and
aαT = (aT ) α ⊆ Kα = K. Hence t ∈ T implies aαt ∈ T =⇒
α∈A aαt ∈
K =⇒ tr(a)t ∈ K. Since tr(a) = 1, this implies that if t ∈ T , then t ∈ K,
i.e., T ⊂ K.
Recall the relation “ ” on the set {0, 1, . . . , q − 2} of integers mod q − 1,
and write i ≺ n to mean that i n but i = n. For x, y ∈ Fq, we have
(x + y) n =
x i y n−i .
0in
Lemma 8.3.4. (Lemma 2.7 of [GPPP96]) Suppose tr((x + 1) r + x r + 1) = 0
for all x ∈ Fq and for some integer r with (r, q −1) = 1 (so x ↦→ x r permutes
the elements of Fq). Then r = α or r = α + β where α, β ∈ A are distinct.
Conversely, if r = α or r = α+β with α, β ∈ A, then tr((x+1) r +x r +1) = 0
for all x ∈ Fq.
Proof. Clearly, if r = α or r = α + β, then since xα and xβ have the same
trace, tr((x + 1) r + xr + 1) = 0 for all x ∈ Fq. So assume this latter condition
holds. WOLG we may assume that r ∈ A, so r is not a power of 2. Then
tr{(x + 1) r + x r + 1} = tr{
x k } =
x kη =
x kη .
0≺k≺r
η∈A 0≺k≺r
0≺k≺r η∈A
This expression is 0 for all x ∈ Fq if and only if the coefficient on each term
x i is zero. It turns out that it is sufficient to consider the coefficients on those
x i with 0 ≺ i ≺ r. Note that kη = i iff k = iη −1 . So the coefficient on x i for
0 ≺ i ≺ r in
0≺k≺r
η∈A xkη is |{γ ∈ A : iγ ≺ r}| (mod 2). So
tr((x + 1) r + x r + 1) = 0 ∀ x ∈ Fq ⇐⇒ |{γ ∈ A : iγ ≺ r}| ≡ 0 (mod 2)
whenever 0 ≺ i ≺ r.

8.3. TRACE EQUATIONS BY GLYNN 363
If i = r, γ ≺ r would imply that rγ would have fewer summands than r,
impossible.
First consider i = 2j for fixed j with 2j r (of course r not a power of
2 implies 2jγ = r for any γ ∈ A). Put c = |{γ ∈ A : 2jγ r}|, so c must be
even. This says r is the sum of an even number c of distinct powers of 2.
Next, put i = r − 2j r. Here |{γ ∈ A : (r − 2j )γ ≺ r}| must be even.
As 1 is in this set, there must be some nonidentity element δ ∈ A for which
(r − 2j )δ r. Put i = 1 + δ and note that |{γ ∈ A : (1 + d)γ ≺ r}| must be
even. But (1 + δ)γ r implies γ + γδ r =⇒ γδ r (since γ = γδ ∈ A).
There are c elements γ ∈ A with γ r, since r is a sum of c powers of 2.
Suppose that c > 2. If γ ∈ A satisfies γ r and γ = 2j , then γ appears in
r − 2j so that γδ r. Hence (1 + δ)γ = γ + γδ r, but (1 + δ)γ = r since
(1 + δ)γ is the sum of only 2 powers of 2.
On the other hand, if 2ℓ r, then (1 + δ)2ℓ = 2ℓ + 2ℓδ r implies there
are exactly c − 1 automorphisms γ = 2j with (1 + δ)γ r (i.e., those γ ∈ A
with γ r and γ = 2j ). Since c − 1 is odd, and there are an even number of
γ with (1 + δ)γ r, we must also have γ = 2j satisfy (1 + δ)2j r, so that
2jδ r.
Hence γ ≤ r iff (1 + δ)γ r (and γδ r). It follows that
rδ = γ δ =
γδ r.
γr
Since r is a sum of c distinct powers of 2, so is rδ, implying rδ = r, so
r(δ−1) ≡ 0 (mod q−1). But (r, q−1) = 1 implies that δ = 1, a contradiction.
Hence c ≤ 2, completing the proof.
Lemma 8.3.5. (Lemma 4.10 of Glynn generalized) For q = 2 e and a fixed
integer p with 1 ≤ p ≤ q − 2, we suppose that there are αpk ∈ Fq, 0 ≤ k ≤
lp − 1, such that
e/ℓp ℓp−1
αpk[(1 + z) p+2r−k
n=1
k=0
γr
+ z p+2r−k
]
2 nℓp
= 0 ∀ z ∈ Fq. (8.5)
Then αpk = 0 for all k = 0, 1, . . . , ℓp − 1. (Keep in mind that p = 0.)
Proof. Put z = 1 to get e/ℓp
n=1
[(1 + z) p+2r−k
ℓp−1
k=0 αpk
2 nℓp
= 0. Note that
+ z p+2r−k
] = (1 + z) p + [(1 + z) p + z p ]z 2r−k
.

364 CHAPTER 8. TRACE EQUATIONS
Also, recall that (1 + z) p·2nℓp
k and n, Eq. 8.7 becomes
e/ℓp ℓp−1
αpk[(1 + z) p + z p ]z 2r−k
n=1
This is the same as
k=0
e/ℓp
[(1 + z) p + z p ] 2nℓ + p
n=1
= (1 + z) p . Since (1 + z) p is independent of both
ℓp−1
k=0
αpkz 2r−k
2 nℓp
2 nℓp
By definition of ℓp, (2 ℓp − 1)p ≡ 0 (mod q − 1), so
It follows that
(1 + z) p·2nℓp
+ z p·2nℓp
[(1 + z) p z p e/ℓp
] ·
= 0 ∀ z ∈ Fq. (8.6)
= 0 ∀ z ∈ Fq.
= (1 + z) p + z p for n = 1, 2, . . . , e/ℓp.
n=1
ℓp−1
k=0
αpkz 2r−k
2 nℓp
for all z ∈ Fq.
Suppose 0 ≤ k, k ′ ≤ ℓp − 1 and 1 ≤ n, n ′ ≤ e/ℓp. Then
2 r−k+nℓp ≡ 2 r−k′ +n ′ ℓp (mod q − 1) iff
−k + nℓp ≡ −k ′ + n ′ ℓp (mod e) iff
k ′ − k ≡ (n ′ − n)ℓp (mod e).
= 0 (8.7)
This condition implies that k ′ − k ≡ 0 (mod ℓp), so k = k ′ , and then n ′ − n ≡
0 (mod e/ℓp), forcing n = n ′ . Thus [(1 + z) p + zp ] · e−1 2i
i=0 γiz = 0 for all
z ∈ Fq, where for each i with 0 ≤ i ≤ e − 1, γi is just some αpk raised to
some power of 2.
We now need a result from elementary number theory: each cyclotomic
coset different from {0} must contain some element between 1 and q/2 − 1,
inclusive. So for the given p we are using, 1 ≤ p ≤ q − 2, choose an integer
j such that 2jp ≡ x (mod q − 1) where 1 ≤ x < q/2. Raise Eq. 8.6 to the
power 2j to obtain:
[(1 + z) x + z x e−1
] · γi 2j
i=0
· z 2i+j
= 0 ∀ z ∈ Fq.

8.3. TRACE EQUATIONS BY GLYNN 365
Let z a be the largest nonzero term of (1+z) x +z x and let z b be the largest
power of z with nonzero coefficient in e−1 2j
i=0
γi ·z2i+j (with exponents reduced
so they lie between 20 and 2e−1 ).
Then a+b < (2e−1−1)+2e−1 = q−1, so a+b ≤ q−2. Hence za+b is left over
(with nonzero coefficient) after any cancellation of terms of the expression.
Hence the coefficient on zb must have been zero after all, since a polynomial
of degree ≤ q − 2 with more than q − 2 roots must be the zero polynomial.
It follows that γi = 0 for all i, so αpk = 0 for all 0 ≤ k ≤ ℓp − 1.
Recapitulation: Suppose m is an integer with 1 ≤ m ≤ q − 2 and
(m, q − 1) = 1. Then suppose g : Fq → Fq satisfies
g(a) + g(b)
tr
(a + b) m
= 1 ∀ a, b ∈ Fq, a = b. (8.8)
It is clear that g must be a permutation, so we can write
We saw that
q−2
g(x) = aix i .
i=0
{i : 0 ≤ i ≤ q − 2} = {p · 2 k + m : p ∈ R and 0 ≤ k ≤ lp − 1},
where R is an RRS mod q − 1. Put αpk =
g(s) = lp−1
p∈R k=0
For 0 = p ∈ R, Lemma 8.3.2 says that
e/lp lp−1
j=1
k=0
αpk
2−k ap·2k . Then
+m
αpkx p·2k +m .
(1 + z) p·2k +m p·2
+ z k 2
2−k +m
jlp = 0 ∀ z ∈ Fq, p ∈ R \ {0}.
Then Lemma 8.3.5 says that if m = α ∈ A, then αpk = 0 if 0 = p ∈ R,
but this means a p·2 k +m = 0 if p = 0. But the only term with p = 0 is am, so
that

366 CHAPTER 8. TRACE EQUATIONS
g(x) = ax α with tr(a) = 1.
Also, Lemma 8.3.2 says that tr{am[(1 + z) m + z m ]} = 1 for all z ∈ Fq.
Then by Lemma 8.3.3 we have tr{(x + 1) m + x m + 1} = 0 for all x ∈ Fq, so
that by Lemma 8.3.4 m = α or m = α + β for distinct α, β ∈ A. This means
that Eq. 8.7 can be rewritten. First, write a in place of am, for some a with
tr(a) = 1. If m = α ∈ A, then g(s) = as α with tr(a) = 1 satisfies Eq. 8.7.
So suppose m = α + β for distinct α, β ∈ A. Then the equation appears as
α+β α+β a(s + t )
tr
(s + t) α+β
= 1 ∀ s, t ∈ Fq, s = t. (8.9)
If we make the substitution t = s + z, Eq. 8.8 becomes
α β β α α+β a(s z + s z + z )
tr
zα+β
= 1 ∀s, z ∈ Fq, z = 0. (8.10)
Now make the substitution y = s+z , i.e., s = z + yz, to obtain
z
tr(a[y α + y β + 1]) = tr(a[(y + 1) α+β + y α+β ]) = 1 ∀y ∈ Fq, tr(a) = 1, (8.11)
which is equivalent to (put η = α/β ∈ A):
tr(a[(y + 1) η+1 + y η+1 ]) = tr(a[y η + y + 1]) = 1 ∀ y ∈ Fq,
and hence, since we also have tr(a) = 1, to
tr(at η + at) = 0 ∀ t ∈ Fq (for some a ∈ Fq with tr(a) = 1. (8.12)
Now apply Lemma 8.3.1 to get a + aη−1 = 0, i.e., a ∈ Fix(η). Conversely,
if tr(a) = 1 and α, β ∈ A is such that aα/β = a, then Eq. 8.7 holds.
Since tr(a) = 1, and a ∈ Fix(α/β) = Fq0, it must be that q = qh 0 where h
is odd. This completes a proof of the following.
Theorem 8.3.6. Suppose that m is an integer with 1 ≤ m ≤ q − 2 and
(m, q − 1) = 1. Then suppose that g : Fq → Fq satisfies
g(a) + g(b)
tr
(a + b) m
= 1 ∀ a, b ∈ Fq, a = b.

8.4. OVOIDS WITH A PENCIL OF CONICS 367
Then either m = α ∈ A and g(s) = asα for some a ∈ Fq with tr(a) = 1,
or there are distinct α, β ∈ A for which m = α + β and if a is the coefficient
on sα+β
a(sα+β +tα+β )
in g(s), then tr (s+t) α+β
= 1 whenever s = t, for some a ∈ Fq
with tr(a) = 1 and a is in the subfield Fq0 fixed by the automorphism η = α/β,
and q = qh 0 where h is odd. Moreover, the converse holds.
Corollary 8.3.7. If g : Fq → Fq satisfies
g(s) + g(t)
tr
(s + t) 2
= 1 ∀ s, t ∈ Fq with s = t,
then there is some constant a ∈ Fq with tr(a) = 1, and g(s) = as 2 .
We are now in a position to give a proof of a theorem of D. Glynn .
Since it is rather important we give it a separate section of its own. The
proof as given, while based on the computations in Glynn [Gl84], is modified
somewhat as suggested by the article of Penttila and Praeger [PP97].
8.4 Ovoids with a pencil of conics
If ℓ is a tangent line at a point P to an ovoid Ω of P G(3, q), the set of q + 1
plane sections of Ω by the planes on ℓ is called a pencil of conics of Ω at the
point P . The line ℓ is the carrier of the pencil.
Theorem 8.4.1. If an ovoid Ω of P G(3, q) has a pencil of conics, then it is
an elliptic quadric.
Proof. By the theorem of Barlotti and Panella we may assume that q is
even and even greater than 4. Moreover, we have seen that we may choose
coordinates of P G(3, q) so that the symplectic form determined by Ω is:
x ∗ y = x0y1 + x1y0 + x2y3 + x3y2,
and the tangent line ℓ carrying the pencil is: x0 = x3 = 0, i.e., [1, 0, 0, 0] ∩
[0, 0, 0, 1]. Also, Q = ℓ ∩ Ω = (0, 1, 0, 0). The pencil is then {Os = πs ∩ Ω :
s ∈ Fq}, where πs = [1, 0, 0, s], so the nucleus of Oa is π ⊥ s = Ns = (0, 1, s, 0).
This means that the conic Os has the form
Os = {(1, f(s)t 2 + ts + g(s), t, s) : t ∈ Fq} ∪ {(0, 1, 0, 0)}.

368 CHAPTER 8. TRACE EQUATIONS
For fixed but distinct s1, s2, we have for all t1, t2 ∈ Fq,
(1, f(s1)t 2 1 + t1s1 + g(s1), t1, s1) ∗ (1, f(s2)t 2 2 + t2s2 + g(s2), t2, s2) = 0.
Writing out this condition leads to
f(s1)t 2 1 + (s1 + s2)t1 + g(s1) = f(s2)t 2 2 + (s1 + s2)t2 + g(s2).
Since {dt2 + et + f : t ∈ Fq} = e2 K + f (by Theorem 8.3), so we have
d
that
(s1 + s2) 2
f(s1) K + g(s1) is disjoint from (s1 + s2) 2
K + g(s2).
f(s2)
(s1+s2) 2
Hence f(s1) = f(s2) = a, say, and g(s1) + g(s2) ∈ K, which means
a
a(g(s1) + g(s2))
tr
(s1 + s2) 2
= 1 ∀ s1, s2 ∈ Fq, s1 = s2.
Hence g(x) = bs 2 with tr(ab) = 1, so
Ω = {(1, at 2 + st + bs 2 , t, s) : t, s ∈ Fq} ∪ {(0, 1, 0, 0)},
which is an elliptic quadric since ax 2 + x + b is irreducible.
8.5 More Trace Equations
The next result is from O’Keefe and Penttila (Characterizations of Flock
GQ) and is a generalization of Lemma 4.11 of Glynn [Gl84].
Lemma 8.5.1. Let q = 2 e , e odd and e ≥ 3. For a fixed p ∈ Z suppose
1 ≤ p ≤ q − 2. Let γ and δ be elements of A such that γ/δ is a generator of
A. Finally, suppose that
e/lp ℓp−1
αpk[(1 + z) p+(γ+δ)2−k
n=1
k=0
+ z p+(γ+δ)2−k
]
2 nℓp
for all z ∈ Fq. Then αpk = 0 for all k with 0 ≤ k < ℓp, or there is an r ∈ Z
with 2 r p ≡ q − 2 (mod q − 1 (so WLOG p = q − 2, since p and q − 2 would
then be in the same cyclotomic coset), and αpk = 0 for (p, k) = (q − 2, r) and
(q − 2, s), with all other coefficients equal to zero.
= 0

8.5. MORE TRACE EQUATIONS 369
Proof. Recall that ℓp was defined to be the smallest positive integer such that
(2ℓp − 1)p ≡ 0 (mod q − 1). It follows by induction on n that for any positive
integer n it is true that zp·2nℓp = zp for all z ∈ Fq. The expression in the
Lemma is zero for all z ∈ Fq if and only if it reduces to zero modulo (zq − z)
when written as a sum of powers of z. If we equate with zero the coefficient
on each power of z we get a set of q equations in the ℓp variables αpk. To
solve these equations we first put (αpk) 2nℓp
= xn k , 0 ≤ k < ℓp, 1 ≤ n ≤ e/ℓp
and solve the resulting system L of q linear equations
e/ℓp
ℓp−1
n=1 k=0
x n k
(1 + z) p+(γ+δ)2−k
+ z p+(γ+δ)2−k 2 nℓp
= 0, ∀ z ∈ Fq.
Note that in this expression, x n k is not a power of xk but one of e variables.
Now make two simplifications. Bring in the power 2 nℓp and replace p · 2 nℓp
with p. Second, relabel the variables x n k = ym = y nℓp−k so that c ≡ c (mod e)
with 1 ≤ c ≤ e. Then notice that the system of linear equations in the ym,
1 ≤ m ≤ e has coefficients in F2. So if we can solve the original system of
equations, we can now solve
e
ym((1 + z) p+(γ+δ)2m
m=1
+ z p+(γ+δ)2m
) = 0, ∀z ∈ Fq, (8.13)
with ym ∈ F2, 1 ≤ m ≤ e.
Put H = {0, 1, . . . , e−1}. For x ∈ Z, define 〈x〉 by x ≡
i∈〈x〉 2i (mod q−
1). Here, if x ≡ 0 (mod q − 1), put 〈x〉 = ∅ when x = 0 and 〈x〉 = H
otherwise. Then the binomial theorem (mod 2) says that
(1 + z) a = z b , where the sum is over all b such that
〈b〉 ⊆ 〈a〉, 0 ≤ b ≤ q − 1.
Suppose that Eq. 8.13 has a non-trivial solution with ym ∈ {0, 1}, 1 ≤
m ≤ e. Let Y = {〈p + (γ + δ)2m 〉 : ym = 1}. If zi appears with a coefficient
of 1 in one of the expressions ((1 + z) p+(γ+δ)2m + zp+(γ+δ)2m ), then clearly
〈i〉 must be a proper subset of the corresponding 〈p + (γ + δ)2m 〉, and every
subset of H must be strictly contained in an even number of the sets in Y .
The exponent 0 on z corresponds to the empty set of H. As it must appear
in an even number of sets in Y and it appears in each set of Y , it must be
that |Y | is even. Since our solution {ym} is nontrivial, Y = ∅, so |Y | ≥ 2.

370 CHAPTER 8. TRACE EQUATIONS
Let M be the size of the largest set in Y , and let A be one such set of
size M. Every subset of size M − 1 in A is strictly contained in at least two
sets of Y , so at least one set of Y other than A. Such a set necessarily has
size M.
Let A = 〈p + (γ + δ)2 a 〉. It has M subsets of size M − 1. Let Bi =
〈p + (γ + δ)2 bi 〉, 1 ≤ i ≤ M, such that A ∩ B1, A ∩ B2, . . . , A ∩ BM are the
distinct subsets of A of size M − 1.
Suppose γ = 2 r = δ = 2 s , so our hypothesis includes the assumption that
(r − s, e) = 1. Then (p + (γ + δ)2 a ) − (p + (γ + δ)2 bi ) ≡ (2 r + 2 s )(2 a − 2 bi ) ≡
2 xi − 2 yi (mod q − 1) for some xi = yi ∈ H.
We now consider this last congruence:
2 r+a + 2 s+a + 2 yi ≡ x r+bi + 2 s+bi + 2 xi (mod q − 1). (8.14)
We first consider the case where the three powers on the left are distinct,
implying that the three powers on the right are also distinct. Hence, modulo
e, we have that r ≡ s; a ≡ bi; xi ≡ yi, and {r+a, s+a, yi} = {r+bi, s+bi, xi}
are sets of size 3 modulo e. One of the two possibilities is that yi ≡ r + bi,
which forces s + a ≡ xi and hence r + a ≡ s + bi. From this it follows easily
that
a − bi ≡ s − r (mod e).
The other possibility is that yi ≡ s + bi, which forces s + a ≡ r + bi and
r + a ≡ xi. From this it follows that
Hence
a − bi ≡ r − s(mod e).
a − bi ≡ ±(r − s) (mod e) (8.15)
Since there are at most two possible values for bi in this case, M ≤ 2.
The next case to consider is where the three powers on the left in Eq. 8.15
yield two distinct powers of 2 (and the same on the right). If 2 i + 2 j + 2 k
simplifies to a sum of two distinct powers of 2, it must be that some two of
the three exponents are the same, say i ≡ j, and that the third exponent
k satisfies k ≡ i + 1 (mod e). In our situation we cannot have all three
exponents the same, since r ≡ s (mod e). So in the present case there are
two possibilities:

8.5. MORE TRACE EQUATIONS 371
The first is: yi ≡ r+a ≡ s+bi ≡ r+bi+1 and xi ≡ r+bi ≡ s+a ≡ r+a+1}.
Here we see a−bi ≡ s−r ≡ r−s, so 2(r−s) ≡ 0 (mod e). But by hypothesis,
gcd(r − s, e) = 1, implying that e ≤ 2, a possibility we do not allow.
The second possibility is: yi ≡ s+a ≡ r +bi ≡ s+bi +1 and xi ≡ s+bi ≡
r + a ≡ s + a + 1. Here we see a − bi ≡ r − s ≡ s − r, so again e ≤ 2, a
possibility we do not allow. Hence this case does not arise.
The third and final case to be considered is where the three powers on the
left in Eq. 8.14 yield just one power of 2 (and the same on the right). Here
there are two possibilities. The first is that yi ≡ r + a ≡ s + bi ≡ r + bi + 1
and s + a ≡ r + a + 1 ≡ xi ≡ r + bi. It follows that a − bi ≡ s − r ≡ r − s,
and s ≡ r + 1, implying that ±2 ≡ 0, or e ≤ 2. The other possibility is that
yi ≡ s + a ≡ r + bi ≡ s + bi + 1 and xi ≡ s + bi ≡ r + a ≡ s + a + 1. Here
a − bi ≡ r − s ≡ s − r and r ≡ s + 1. So again e ≤ 2. Hence this third case
does not arise for q ≥ 8.
We have seen that in each case there are at most two possible values for
bi. So M ≤ 2. So first suppose that M = 2. Hence we have the following
three sets in Y :
A : 〈p + 2 r+a + 2 s+a = 2 u + 2 v 〉
B1 : 〈 p + 2 r+b1 + 2 s+b1 = 2 u + 2 w 〉
B2 : 〈 p + 2 r+b2 + 2 s+b2 = 2 v + 2 t 〉
By considering A − B1 and A − B2 we quickly find
{r + a, s + a, w} = {r + b1, s + b1, v} must be three distinct residues mod e;
{r + a, s + a, t} = {r + b2, s + b2, u} must be three distinct residues mod e.
Here a, b1, b2 are distinct modulo e; u, v, w are distinct modulo e; and
u, v, t are distinct modulo e. At this point we see that v must be congruent
to r + a or to s + a. The roles of r and s are interchangeable. So WLOG we
assume v ≡ r + a (mod e). It quickly follows that
Consider B1 now:
v ≡ r + a, s + a ≡ r + b1, s + b1 ≡ w,
s + a ≡ u, r + a ≡ s + b2, t ≡ r + b2.

372 CHAPTER 8. TRACE EQUATIONS
p + 2 r+b1 + 2 s+b1 = 2 u + 2 w = 2 s+a + 2 s+b1 = 2 r+b1 + 2 s+b1 ,
which implies that p = 0. As this contradicts our hypothesis, we must
conclude that M = 2, i.e., M = 1.
So with M = 1 we have
p + (2 r + 2 s )2 a = 2 x , and
p + (2 r + 2 s )2 b = 2 y ,
where a ≡ b and x ≡ y, modulo e. Moreover, we saw above that we may
assume that a ≡ b + s − r. Hence
(2 r + 2 s )(2 a − 2 b ) ≡ (2 r + 2 s )(2 s−r − 1)2 b ≡ 2 2s−r+b − 2 r+b ≡ 2 x − 2 y .
It follows that y ≡ r + b, x ≡ 2s − r + b, so that x ≡ y + 2(s − r). Then
p = 2 x − (2 r + 2 s )2 a = 2 y+2(s−r) − (2 b+s + 2 b+2s−r ) ≡ −2 y−r+s .
This implies that
2 r−s−y p ≡ −1 ≡ q − 2 (mod q − 1).
This means that p and −1 are in the same cyclotomic coset, and we could
use −1 in place of p when choosing an RRS. So suppose p ≡ −1 ≡ q−2. Then
r−s−y ≡ 0, so y ≡ r−s; b ≡ y −r ≡ −s ≡ e−s; a ≡ b+s−r ≡ −r ≡ e−r.
At this point we have established that |〈−1+(γ +δ)2 m 〉| = 1 implies that
m ≡ e − r or m ≡ e − s. Solutions to the system L have p ≡ q − 2 ≡ −1 and
ye−r = 0 = ye−s, and ym = 0 otherwise. With p = −1, we have ℓp = e, since
p(2 ℓ − 1) ≡ − (mod q − 1) if and only if 2 ℓ − 1 ≡ 0. Then with p = −1 we
have ye−r = ye·ℓp−r = x e r = αq−2,r. The same thing holds with r replaced by
s, and we have
αq−2,r = 0 = αq−2,s; all other αp,k = 0 for p ∈ R \ {0}.
Lemma 8.5.2. Let f : Fq → Fq be a function (which WLOG may be assumed
to send 0 to 0) and let γ, δ ∈ A be such that γ/δ is a generator of A (so
γ = δ). The equation
f(x) + f(y)
tr
(x + y) γ+δ
= 1
holds for all x, y ∈ Fq with x = y if and only if f(x) = x γ+δ + b δ x γ + b γ x δ for
some b ∈ Fq. Further, in this case q is not a square.

8.5. MORE TRACE EQUATIONS 373
Proof. By Lemma 8.3.2 we have the following:
f(s) + f(t)
tr
(s + t) γ+δ
= 1 ∀ s = t ∈ Fq ⇐⇒
(i) e/ℓp
j=1
ℓp−1
k=0 αpk
(1 + z) p·2k +γ+δ p·2 + z k 2−k +γ+δ
2jℓp = 0 ∀ z ∈ Fq, p ∈
R \ {0},
and
(ii) (p = 0) 1 = tr
γ+δ γ+δ
α00 (1 + z) + z , for all z ∈ Fq and α00 =
aγ+δ = aγ+δ is the coefficient of xγ+δ in g(x). In particular tr(aγ+δ) = 1. It
follows by Lemma 8.5.1 that f(x) = axγ+σ +bxγ +cxδ for some a with tr(a) =
1 and b = 0 = c. Moreover, by Theorem 8.3.6 we have a ∈ F ix(γ/δ) = F2,
so a = 1. Then tr(1) = 1 implies q is not a square.
So at this point we have:
γ+δ γ+δ (x + y )
1 = tr
(x + y) γ+δ
γ δ
b(x + y) + c(x + y)
+ tr
(x + y) γ+δ
,
where the first term has trace equal to 1, so that
tr(b(t) −δ + ct −γ ) = 0 ∀ t ∈ Fq. (8.16)
By Lemma 4.12.2, if we index b = a−δ and c = a−γ, then 0 = a−δ +a δ/γ
−γ =
b + c δ/γ , i.e., b γ = c δ . So put b = d δ and c = d γ for some nonzero d. Then
f(x) appears as f(x) = x γ+δ + d δ x γ + d γ x δ , as desired.

374 CHAPTER 8. TRACE EQUATIONS

Chapter 9
Finite Generalized Quadrangles
9.1 Introduction
During the time since generalized quadrangles (GQ) were first introduced
by J. Tits [Ti59] a rather large body of theory has grown up around them.
In 1984 it was possible to include in [PT84] essentially all that was known
about finite GQ. In the two decades following the appearance of [PT84]
(which we often denote by FGQ) the theory has developed at such a pace
that it is no longer possible to include most of what is known in a single
book. There have been a few surveys that have attempted to bring the
reader up to date. We mention two in particular: Chapters 7 and 9 by
J. A. Thas in the Handbook of Incidence Geometry [Bu95], and the survey
[JP97]. More recently there appeared two other books devoted to finite GQ,
the volume [KT04] by K. Thas and the volume [TTVM06] by J. A. Thas,
K. Thas and H. Van Maldeghem. But these works are not really intended
as first introductions to GQ. In the present book our goal is to provide
an essentially self-contained introduction to the theory of finite generalized
quadrangles, with emphasis on those examples and theorems that are related
to ovals and ovoids in P G(n, q), for n ≤ 5. It is not our intention to replace
[PT84], but rather to provide an alternative introduction that treats those
aspects of the theory most useful in the present setting.
Before we give the definition of a GQ we present the smallest nontrivial
example and a sequence of exercises dealing with it. This section can be
omitted on a first reading, but it does offer a rather detailed look at the
smallest member of an infinite family (the symplectic GQ denoted W (q))
375

376 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
that will play a significant role in our studies.
Start with the set N6 = {1, 2, 3, 4, 5, 6}. A duad is any one of the fifteen
subsets of N6 of size 2. A syntheme is any one of the fifteen sets of three
pairwise disjoint duads. If we think of duads as points and synthemes as lines,
with incidence being containment, the resulting point-line geometry has the
following properties. Two duads are contained in a common syntheme if and
only if they are disjoint, and then that syntheme is uniquely determined.
Hence it is easy to show that each point is incident with the same number
(three) of lines, each line is incident with the same number (three) of points,
no two lines are incident with the same two points, and if x is a point not
incident with a line L, then there is a unique point y incident with L for which
x and y are incident with a common line, i.e., are collinear. For example,
consider the duad {1, 3} that is not in the syntheme {{1, 2}, {3, 4}, {5, 6}}.
Here {5, 6} is the uniqe duad in the given syntheme “collinear with” (i.e.,
disjoint from) {1, 3} via the syntheme {{1, 3}, {5, 6}, {2, 4}}. This synthemeduad
geometry is apparently due to J. J. Sylvester (1844). In general when
this example is recalled later on, the duad {i, j} will be denoted simply as
ij.
Sylvester’s syntheme-duad geometry is the smallest nontrivial example of
a generalized quadrangle. It can be embedded in a larger example as follows.
Start with the syntheme-duad model just given and add twelve additional
points: 1, 2, 3, 4, 5, 6, ˆ1, ˆ2, ˆ3, ˆ4, ˆ5, ˆ6. Then in addition to the fiften synthemes,
there are thirty new lines of the form {i, ij, ˆj}, for 1 ≤ i, j ≤ 6; i = j. This
gives a point-line geometry with 27 points and 45 lines. Here each line is
incident with three points as before, but now each point is incident with five
lines. Moreover, it is easy enough to check that there are no “triangles”. By
counting the number of lines reachable from a given point by a path of length
at most 3 it is easy to check that this example also has the property that if x
is a point not incident with a line L, then there is a unique point y incident
with L for which x and y are collinear.
The following exercises are all concerned with Sylvester’s syntheme-duad
geometry, which will be denoted W (2) (for reasons to be dealt with later).
Exercise 9.1.0.1. In W (2) show that if K is a set of k points, no two of
which are collinear, then k ≤ 5. Such a set K is called a k-arc.
Exercise 9.1.0.2. For each integer i, 1 ≤ i ≤ 6, the set Ki of five duads
ij (j = i) is a 5-arc. Moreover, each line of W (2) is incident with a unique
point of Ki. (Such a k-arc is called an ovoid.)

9.2. AXIOMS AND PRELIMINARY DEFINITIONS 377
Exercise 9.1.0.3. The 5-arcs Ki, 1 ≤ i ≤ 6 are the only ovoids of W (2)
and any two have a unique point in common.
Exercise 9.1.0.4. The symmetric group S6 acts naturally as a group of
collineations (permutations of the points that preserve lines) of W (2), and is
in fact the complete group of collineations of W (2).
Exercise 9.1.0.5. There is a bijection from the points to the lines of W (2)
that maps lines to points and preserves incidence. We say W (2) is self-dual.
9.2 Axioms and Preliminary Definitions
fgq2
finite generalized quadrangle (GQ) A triple S = (P, B, I) in which P
and B are disjoint sets of objects called points and lines, respectively,
and for which I is a symmetric point-line incidence relation (so “x is
incident with L” is denoted xIL or LIx), is called a finite generalized
quadrangle provided it satisfies the following four axioms:
Axiom 1. No two points of S are incident with two lines in common.
Axiom 2. If x is a point not on a line L (xI|L), then there is a unique pair
(y, M) ∈ (P × B) for which xIMIyIL.
Given two points x, y we write x ∼ y and say that x and y are collinear
provided there is some line L ∈ B for which xILIy. Otherwise, x ∼ y
denotes that x and y are not collinear. Dually, for lines L and M, we write
L ∼ M or L ∼ M according as L and M are concurrent or not concurrent,
respectively.
Axiom 3. There are points x, y ∈ P for which x ∼ y; there are lines L, M ∈
B for which L ∼ M.
Axiom 4. P B is a finite set.
It is immediate that there is a point-line duality with respect to which
each of the axioms is self-dual. Hence the point-line dual of a GQ is a GQ.
It follows that any definition or theorem has a point-line dual. Often we take
for granted that such a point-line dual has been given even if it has not been
mentioned explicitly. Moreover, elementary arguments yield the following
basic results which we leave as exercises.

378 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
Theorem 9.2.1. A GQ has some nondegeneracy.
Let S = (P, B, I) be a GQ.
1. There are four distinct points xi and four distinct lines Li with xiILi,
i = 1, . . . , 4; xiILi+1, i = 1, 2, 3; and x4IL1.
2. Each point is incident with at least two lines; each line is incident with
at least two points.
3. If two points are not collinear, they are incident with the same number
of lines; if two lines are not concurrent they are incident with the same
number of points.
A proof of the following theorem is left as an exercise.
Theorem 9.2.2. A GQ is a grid, or a dual grid, or has parameters (s, t),
s, t ≥ 1.
In somewhat greater detail, let S = (P, B, I) be a GQ. Then at least one
of the following must occur:
1. Each point is incident with exactly two lines; the set B of lines is
partitioned into two nonempty disjoint sets B1, B2 such that each line
in B1 is concurrent with each line in B2 and such that any two lines
of Bi are not concurrent, i = 1, 2; each line in Bi is incident with |Bj|
points, {i, j} = {1, 2}.
2. Each line is incident with exactly two points; the set P of points is
partitioned into two nonempty disjoint sets P1, P2 such that each point
in P1 is collinear with each point in P2 and such that any two distinct
point of Pi are noncollinear, i = 1, 2; each point in Pi is incident with
|Pj| points, where {i, j} = {1, 2}.
3. There are positive integers s and t such that each line is incident with
1 + s points and each point is incident with 1 + t lines.
If case 1. of Theorem 9.2.2 holds, S is said to be a grid; if case 2. of
Theorem 9.2.2 holds, S is a dual grid; and if case 3. of Theorem 9.2.2 holds
S is said to have parameters (s, t) or that it has order (s, t). If s = t,
we usually just say that S has order s. Note that grids and dual grids not
having an order were eliminated by the axioms adopted in [PT84] and in

9.2. AXIOMS AND PRELIMINARY DEFINITIONS 379
many articles dealing with GQ. Also note that the point-line dual of a GQ
of order (s, t) is a GQ of order (t, s).
For the following definitions let S = (P, B, I) be a GQ.
perp For x in P define x perp by x ⊥ = {y ∈ P : x ∼ y}. Note that x ∈ x ⊥ .
The perp (or trace) of a pair {x, y} of distinct points is defined to be
the set x ⊥ y ⊥ and is denoted {x, y} ⊥ . More generally, if A ⊆ P , A
perp is defined by A ⊥ = {x ⊥ : x ∈ A}.
span For distinct points x and y, the span (or perp perp) of the pair {x, y}
is {x, y} ⊥⊥ = {u ∈ P : u ∈ z ⊥ ∀z ∈ {x, y} ⊥ }. If x ∼ y, then {x, y} ⊥⊥
is also called the hyperbolic line defined by x and y.
regular If S has order (s, t) and if x and y are noncollinear points, then
|{x, y} ⊥ | = t + 1 and |{x, y} ⊥⊥ | ≤ t + 1. If equality holds, the pair
{x, y} is said to be regular. If x and y are distinct collinear points we
also say that the pair {x, y} is regular. If x is a point for which {x, y}
is regular for all points y different from x, then the point x is called
regular.
triad A triad (of points) is an unordered triple of pairwise noncollinear
points. Given a triad T = {x, y, z}, a center of T is just a point of T ⊥ .
We say T is acentric, centric, or unicentric according as |T ⊥ | is zero,
positive, or equal to 1.
k-arc A k-arc (in a GQ) is a set of k points, no two of which are collinear.
Exercise 9.2.2.1. Prove Theorem 9.2.1
Exercise 9.2.2.2. Prove Theorem 9.2.2
Exercise 9.2.2.3. Let S = (P, B, I) be a GQ with order (s, t) and let (x, y)
be a pair of noncollinear points. Show that {x, y} is a regular pair if and only
if each triad {x, y, z} with at least two centers actually has t + 1 centers.
Exercise 9.2.2.4. With the same hypothesis as in Exercise 9.2.2.3 show that
{x, y} is a regular pair if and only if some pair {z, w} of distinct points in
{x, y} ⊥ is regular if and only if some pair {u, v} of distinct points in {x, y} ⊥⊥
is regular.

380 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
Exercise 9.2.2.5. Let S = (P, B, I) be a GQ with order 2. Show that up
to isomorphism, S is unique. (Hint: First show that S must have 15 points
and 15 lines. Then show that S contains an ordered pentagon with vertices
labeled pi, 1 ≤ i ≤ 5, so that pi ⊥ pi+1, for 1 ≤ i ≤ 5, with subscripts reduced
modulo 5 to one of 1, . . . , 5. Hence the side opposite a vertex pi has one point
qi other than p±2, and it must be that pi ⊥ qi. Now show that the third line
through qi must meet the two diagonals pi+1qi+1 and pi−1qi−1. It is possible
to see that this figure can be completed to a GQ(2, 2) in at most one way.)
The resulting figure is called a doily.
Exercise 9.2.2.6. Show that W (2) is isomorphic to its point-line dual. In
fact, W (2) is self-polar, i.e., there is a duality δ mapping points to lines
and lines to points with the property that δ 2 is the identity mapping.
Exercise 9.2.2.7. This exercise gives an ad hoc description of a GQ of order
3 which turns out to be isomorphic to the FGQ with order 3 later denoted
by W (3). Start with the set N5 = {1, 2, 3, 4, 5}. There are two types of
points: i) the twenty ordered pairs ij, i, j ∈ N5; i = j; ii) the twenty 3cycles
(i, j, k) = ijk with i, j, k distinct elements of N5. So, for example,
ijk = jki = ikj. Then there are three types of lines: a) Li; Mi; i ∈ N5.
b) Dij = Dji; i, j ∈ N5; i = j. c) Tij(= Tji); i, j ∈ N5; i = j. Incidence is
described as follows. Li is incident with the four points ij, i = j ∈ N5. Mj is
incident with the four points ij, j = i ∈ N5. Dij = Dji is incident with the
four points ij, ji, klm, kml, where {i, j, k, l, m} = {1, 2, 3, 4, 5}. Finally, Tij
is incident with the four points ij, ijk, ijl, ijm, where again {i, j, k, l, m} =
{1, 2, 3, 4, 5}.
9.3 Basic Combinatorics of GQ
Let S = (P, B, I) be a GQ of order (s, t), and put |P | = v, |B| = b. Most
the following results are given for points. The corresponding results for lines
that arise from point-line duality will also be considered as given. Proofs of
the first six results are simple and will be left as exercises.
Theorem 9.3.1. S has (1 + s)(1 + st) points, etc.
1. v = (1 + s)(1 + st); b = (1 + t)(1 + st).

9.3. BASIC COMBINATORICS OF GQ 381
2. For distinct x, y ∈ P , |{x, y} ⊥ | = s + 1 or t + 1 according as x ∼ y or
x ∼ y.
3. For each point x, |x ⊥ | = 1 + s + st.
4. For each x ∈ P , |P \ x ⊥ | = ts 2 .
5. If x ∼ y = x, |P \ (x ⊥ y ⊥ )| = ts 2 − st.
6. If x ∼ y, |P \ (x ⊥ y ⊥ )| = ts 2 − st − s + t.
For the remainder of this section we assume that both s > 1 and t > 1.
Let x, y be fixed, noncollinear points with perp given by:
{x, y} ⊥ = {z0, . . . , zt}
And let V = {w1, . . . , wd} be the set of points w for which {x, y, w} is a triad.
By Part 6 of the preceding theorem we know that d = ts 2 − st − s + t. For
1 ≤ i ≤ d let ti be the number of zj collinear with wi. And for 0 ≤ i ≤ t + 1
let ni be the number of triads {x, y, w} with exactly i centers. Counting the
total number of triads, we have
Lemma 9.3.2. t+1
i=0 ni = ts 2 − st − s + t.
Count in different ways the number of ordered pairs (w, z), with w ∈ V ,
z ∈ {x, y} ⊥ , and w and z collinear, to obtain:
Lemma 9.3.3. d
i=1 ti = t+1
i=1 ini = (t 2 − 1)s.
Next count in different ways the number of ordered triples (w, z, z ′ ), where
w is in V , where z, z ′ are distinct members of {x, y} ⊥ , and w is collinear
with both z and z ′ , to obtain:
Lemma 9.3.4. d
i=1 ti(ti − 1) = t+1
i=2 i(i − 1)ni = (t 2 − 1)t.
Lemma 9.3.5. nt = 0.
Proof. Suppose u is a point not collinear with x or y and u ∼ zi for 1 ≤ i ≤ t,
but u ∼ z0. Let Li be the line incident with z0 and concurrent with uzi,
1 ≤ i ≤ t. Then L1, . . . , Lt, xz0, yz0 must be t + 2 distinct lines incident with
z0, a contradiction.

382 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
Theorem 9.3.6. The inequality of D. G. Higman says s ≤ t 2 if t > 1.
Dually, t ≤ s 2 if s > 1. Moreover, the following are equivalent:
1. t = s 2 .
2. For some pair (x, y) of noncollinear points, each triad (x, y, z) has a
constant number of centers, in which case this constant is s + 1.
3. Every triad of points has a constant number of centers, in which case
this constant is s + 1.
Proof. From Lemmas 9.3.3 and 9.3.4 we have d i=1 t2i = (t2−1)(s+t). Define
m by: dm =
i ti. Then 0 ≤ (m − ti) 2 implies, after some simplification,
that t(s − 1)(s2 − t) is nonnegative, with equality holding if and only if each
ti = m.
Corollary 9.3.7. (t + 1)n0 = (s − t)t(s − 1)(t + 1) + t i=2 (i − 1)(t + 1 − i)ni.
Proof. Compute:−(t+1)A+(t+1)B−C, where A, B and C are the equalities
from Lemmas 9.3.2, 9.3.3 and 9.3.4, respectively.
Corollary 9.3.8. n1 = 1+t
i=2 ((si2 − i(s + t))/t)ni.
Proof. Divide B by s and C by t. Subtract one from the other and simplify.
Corollary 9.3.9. n0 = 0 iff each triad {x, y, w} is centric iff
t−1
(t − s)t(s − 1)(t + 1) = (i − 1)(t + 1 − i)ni
So n0 = 0 implies t ≥ s, with equality iff ni = 0 for 2 ≤ i ≤ t.
Note: This corollary says that when s = t, the pair (x, y) of noncollinear
points is regular if and only if each triad (x, y, w) containing (x, y) is centric.
Corollary 9.3.10. Computing C − B we see that n1 = 0 iff
i=2
(t − s)(t 2 t+1
− 1) = i(i − 2)ni
in which case t ≥ s. If t = s, then ni = 0 for i = 0, 2.
i=3

9.3. BASIC COMBINATORICS OF GQ 383
The next two results are now easy exercises.
Corollary 9.3.11. If x and y are distinct collinear points, then there are β
ordered pairs (z, w) with no two of x, y, z, w being collinear except for x and
y, where β = (s − 1)st(ts 2 − 2st + 2t − s).
Corollary 9.3.12. If x ∼ y, then there are α ordered pairs (z, w) with no
two of x, y, z, w being collinear, where α = s 4 t 2 − 3s 3 t 2 − 3s 3 t + 6s 2 t 2 +
5s 2 t − 6st 2 − 6st + 3t 2 + t + 2s 2 .
Continuing with the notation of this section, let Li, 1 ≤ i ≤ t − 1 be the
lines through z0 not through x or y, and let L∗ i be the set of points of Li
different from z0. Put W = {L∗ i }. So |W |= s(t − 1). If ai is the number of
triads of the form {x, y, w}, for w ∈ W , having exactly i centers from among
z1, . . . zt, 0 ≤ i ≤ t, then ai = (t − 1)s. And counting pairs (w, zj), with
w ∈ W and w ∼ zj, 1 ≤ j ≤ t, we have iai = (t−1)t. Subtract the second
equation from t/s times the first, and rearrange so all terms that appear are
nonnegative:
((t/s) − i)ai =
(i − t/s)ai
(9.1)
it/s
It is clear that ai = 0 for all i < t/s iff ai = 0 for all i > t/s iff s divides t
and each triad {x, y, w} has 1 + t/s centers.
We state this as a Corollary:
Corollary 9.3.13. Every triad T = {x, y, w} with center z0 (see the notation
above) has at most 1+t/s centers iff each such triad has at least 1+t/s centers
iff each such triad has exactly 1 + t/s centers.
Exercise 9.3.13.1. Prove Corollary 9.3.11.
Exercise 9.3.13.2. Prove Corollary 9.3.12.
Exercise 9.3.13.3. Let S = (P, B, I) be a GQ with parameters (2, t), where
2 ≤ t. Show that t ∈ {2, 4}.
(Hint: In view of Higman’s inequality, the problem is to show that
there is no GQ with parameters (2, 3). Let S be such a GQ. Starting with
Lemma 9.3.5 it is possible to show that no triad of points can have more
than 2 centers. Then use Cor. 9.3.13. It is also possible to prove that any
GQ with three points on each line must be finite.)

384 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
9.4 The Symplectic GQ W (q)
Before continuing with the abstract theory we want to introduce a classical
family of GQ and try out some of our definitions.
Let F = GF (q) (although for most of this section F could be any
field). Let V = F 4 ,
the 4-dimensional vector space
of 4-tuples over F .
0 1
Put P =
, C = P ˙+P
P 0
=
. So C is 4 × 4 over F
−1 0
0 P
and is skew-symmetric and nonsingular (with zeros on its main diagonal).
For ¯x = (x0, x1, x2, x3), ¯y = (y0, y1, y2, y3) ∈ V , define ¯x ◦ ¯y = ¯xC ¯y T =
x0y1 − x1y0 + x2y3 − x3y2. Define ¯x ⊥ ¯y if and only if ¯x ◦ ¯y = 0. Here
(¯x, ¯y) ↦→ ¯x ◦ ¯y is a nonsingular, bilinear alternating form. We collect the
main consequences of this in a sequence of observations that the reader unfamiliar
with this material should consider to be a series of exercises.
Obs. 9.4.1. ¯x ⊥ ¯x for all ¯x ∈ V , and ¯x ⊥ ¯y if and only if ¯y ⊥ ¯x.
Obs. 9.4.2. ¯x ⊥ ¯y if and only if ¯z ⊥ ¯w for all ¯z, ¯w ∈< ¯x, ¯y >.
Obs. 9.4.3. If D is any 4 × 4 matrix over F , then ¯xC ¯y T = 0 if and only if
¯xD¯y T = 0 for all ¯x, ¯y ∈ V provided D = λC for some λ ∈ F ∗ = F \ {0}.
Obs. 9.4.4. Define ¯x ⊥ := {¯y ∈ V : ¯x ⊥ ¯y} for all ¯x ∈ V . Then for nonzero
¯x and ¯y, ¯x ⊥ = ¯y ⊥ if and only if ¯x = λ¯y for some λ ∈ F ∗ .
Obs. 9.4.5. Associated with V is the 3-dimensional projective space P G(3, q).
A nonzero vector ¯x ∈ V corresponds to the point [¯x] = {λ¯x : λ ∈ F ∗ }. However,
for a nonzero vector ¯x, we usually just write ¯x in place of [¯x] to denote
a point of P G(3, q) and let the context inform the reader whether ¯x is a vector
in V or a point in P G(3, q). Other notations are naturally carried over from
V to P G(3, q). For example, for ¯x ∈ P G(3, q), ¯x ⊥ = {¯y ∈ P G(3, q) : ¯x ⊥ ¯y}.
Then the correspondence ρ : ¯x ↔ ¯x ⊥ between points and planes of P G(3, q)
is called a null polarity. It is a polarity because it maps points to planes,
planes to points, preserves incidence, and as a duality of P G(3, q) (i.e., an
isomorphism from P G(3, q) to its point-hyperplane dual) it has order 2. It
is specifically a null polarity because each point or plane ¯x is incident with
its image under ρ. Here ρ : L =< ¯x, ¯y >↦→ L ρ = ¯x ⊥ ∩ ¯y ⊥ for each line
L =< ¯x, ¯y >, and L = L ⊥ if and only if ¯x ⊥ ¯y.

9.4. THE SYMPLECTIC GQ W (Q) 385
The point-line incidence geometry W (q) = (P, B, I) is defined as follows.
P = {¯x ∈ P G(3, q)}; B = {L : L is a line of P G(3, q) with L = L ⊥ }; I is
the incidence inherited from P G(3, q). Given two points ¯x, ¯y ∈ P G(3, q),
¯x ⊥ ¯y if and only if ¯y ⊥ ¯x if and only if ¯y ∈ ¯x ⊥ if and only if ¯x ∈ ¯y ⊥ . Clearly
the lines of B through ¯x are exactly the q + 1 lines through ¯x lying in the
plane ¯x ⊥ . But also, if ¯x, ¯y, ¯z are three distinct points of ¯x ⊥ , then ¯y ⊥ ¯z if
and only if ¯x, ¯y and ¯z lie on a line. To see this,first suppose that ¯x, ¯y and ¯z
do not lie on a line. Observe that if ¯y ⊥ ¯x ⊥ ¯z ⊥ ¯y, then ¯y ∈< ¯x, ¯y, ¯z > ⊥ , so
¯y ⊥ =< ¯x, ¯y, ¯z >= ¯x ⊥ , which readily implies that ¯x = ¯y as points of P G(3, q).
On the other hand, if ¯x, ¯y and ¯z lie on a line, then ¯y ⊥ ¯w for each point on
〈¯x, ¯y〉, so in particular ¯y ⊥ ¯z. It follows that W (q) is a generalized quadrangle
with parameters (q, q). Moreover, we can say a bit more.
Theorem 9.4.6. A GQ of order s (s > 1) is isomorphic to W (q) if and only
if all its points are regular.
Historically, this was surely the earliest combinatorial characterization of
a class of generalized quadrangles and no doubt was discovered several times.
See [PT84] for some references.
Proof. W (q) is a GQ with order q. If ¯x and ¯y are two distinct points of
W (q), then either ¯x ⊥ ¯y and {¯x, ¯y} ⊥ = {¯x, ¯y} ⊥⊥ is the line of W (q) through
¯x and ¯y, or {¯x, ¯y} ⊥ = L is a line of P G(3, q) and its points in P G(3, q) are
the points of {¯x, ¯y} ⊥ in W (q) considered as a GQ. Then L ρ = L ⊥ = {¯x, ¯y} ⊥⊥
is the span of {¯x, ¯y}, which shows that each pair of points of the GQ W (q)
is regular.
Conversely, suppose that S = (P, B, I) is a GQ of order s with all points
regular, s > 1. Now introduce a new point-line incidence structure S ′ =
(P ′ , B ′ , I ′ ), with P ′ = P and B ′ equal to the set of all spans {x, y} ⊥⊥ of pairs
(x, y) of distinct points of S. Incidence I ′ is the natural incidence induced
by containment. So S is (isomorphic to) the substructure of S ′ consisting of
all points and all spans of pairs of distinct points collinear in S. If x is any
point of P = P ′ , the points of x ⊥ (with respect to I) and all the lines of S ′
containing at least two points of x ⊥ form a projective plane. If {x, y, z} are
three points not on a line of S ′ , then because all points are regular, whether
{x, y, z} is a triad or not, by Cor. 9.3.9 there will always be some point
w ∈ {x, y, z} ⊥ , so that {x, y, z} will generate a projective plane in S ′ . The
number of points in S ′ is the same as that in P G(3, q) and it follows easily
that S ′ is isomorphic to P G(3, q).

386 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
Since each line of P G(3, q) is either a line of W (q) or the span of two
points of W (q), and each plane contains exactly the set of points collinear
in W (q) with a unique point of W (q), it is clear that any collineation of
W (q) determines and is determined uniquely by a collineation of P G(3, q)
that preserves the set of lines of W (q).
First, note that if σ is any automorphism of F , then σ induces a unique
collineation of W (q) by: σ : ¯x = (x0, . . . , x3) ↦→ ¯x σ = (x σ 0, . . . , x σ 3). This
holds since ¯x ◦ ¯y = 0 if and only if (¯x ◦ ¯y) σ = ¯x σ ◦ ¯y σ = 0.
Second, note that for any A ∈ GL(4, q), ¯x ↦→ ¯xA yields a collineation
of P G(3, q) which is a collineation of W (q) if and only if it preserves “⊥”,
which is if and only if ACA T = λC for some λ ∈ F ∗ . The symplectic group
Sp(4, q) is defined by
Sp(4, q) = {A ∈ GL(4, q) : ACA T = λC for some λ ∈ F ∗ }
The quotient group Sp(4, q) modulo its center, the group of nonzero
scalars times the identity matrix, is the projective symplectic group P Sp(4, q)
which acts faithfully as a group of (linear) collineations of W (q). Along with
the group of collineations of W (q) induced by the field automorphisms we
now have the complete group of collineations of W (q).
Write A ∈ GL(4, q) in block form
A1 A2
A =
.
A3 A4
Using Ex. 9.4.9.5 it is easy to show that
ACA T
(det(A1) + det(A2))P A1P A
=
T 3 + A2P AT 4
(det(A3) + det(A4))P
A3P A T 1 + A4P A T 2
Using this equation it is clear that for A ∈ GL(4, q) we have
if and only if
(i) 0 = det(A1) + det(A2) = det(A3) + det(A4),
(ii) A1P A T 3 + A2P A T 4 = 0.
. (9.2)
A ∈ Sp(4, q) (9.3)
Let ¯p be the point ¯p = (1, 0, 0, 0), so that the point ¯y = (y0, y1, y2, y3) is
collinear with ¯p if and only if y1 = 0.

9.4. THE SYMPLECTIC GQ W (Q) 387
Obs. 9.4.7. The stabilizer P Sp(4, q)¯p of ¯p is transitive on the set of points
not collinear with ¯p.
Proof. The points of W (q) not collinear with ¯p are those with coordinates of
the form (x, 1, z, w). If we put
A =
⎛
⎜
⎝
1 0 0 0
x 1 z w
−w 0 1 0
z 0 0 1
⎞
⎟
⎠ ,
then clearly A ∈ GL(4, q), and with the block notation from above, det(A1) =
det(A4) = 1, det(A2) = det(A3) = 0. It is now straightforward to calculate
that both conditions of Eq. 9.3 are satisfied, so that A ∈ Sp(4, q). Also
¯p = ¯pA, so the projective collineation induced by multiplication by A is in
P Sp(4, q)¯p. Now (0, 1, 0, 0) ↦→ (0, 1, 0, 0)A = (x, 1, z, w).
Obs. 9.4.8. P Sp(4, q) is transitive on the points of W (q).
P 0
Proof. Put ¯q = (0, 1, 0, 0) and A =
∈ Sp(4, q). Then multiplica-
0 P
tion by A interchanges ¯p and ¯q. Hence ¯p and all the points not collinear with
¯p are in th same orbit under P Sp(4, q), and thus all points not collinear with
some point not collinear with ¯p are in the same orbit. It follows immediately
that P Sp(4, q) is transitive on the points of W (q).
Our next step is to determine the subgroup of of P Sp(4, q) fixing both ¯p
and ¯q.
¯pA = ((1, 0)A1, (1, 0)A2) = (λ, 0, 0, 0) implies that (A1)12 = 0 = (A2)11 =
(A2)12. Without loss of generality we may assume (A1)11 = A11 = 1.
Similarly, ¯qA = ((0, 1)A1, (0, 1)A2) = (0, µ, 0, 0) implies that (A1)21 = 0 =
0 0
1 0
(A2)21 = (A2)22, which now implies that A2 = and A1 =
0 0
0 b
for some nonzero b ∈ F .
At this stage we know that
λC = ACA T ⎛
= ⎝ bP
1 0
P A
0 b
T ⎞
3 + 0
⎠ .
(det(A4) + det(A3))P
A3P A T 1

388 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
Since
1 0
0 b
P is invertible, AT 3 = 0, so
ACA T
bP 0
=
0 det(A4)P
.
So we must have λ = b = det(A4). This proves the following.
Obs. 9.4.9. P Sp(4, q)¯p,¯q is the group of collineations induced by multiplication
by matrices of the form
A =
⎛
⎜
⎝
for arbitrary A4 ∈ GL(2, q).
1 0
0 det(A4)
0 0
0 0
0 0
0 0
A4
⎞
⎟
⎠ ,
We leave as an exercise the determination of the stabilizer of the point ¯p.
The points of {¯p, ¯q} ⊥ are those with coordinates of the form (0, 0, z, w).
It is also now an easy exercise to show that P Sp(4, q)¯p,¯q is triply transitive
on {¯p, ¯q} ⊥ .
Exercise 9.4.9.1. Let ¯x, ¯y, ¯z be three points of W (q), no two of which are
on a line of W (q). Show that there is a unique point ¯w of W (q) collinear
with each of ¯x, ¯y, ¯z, or that there are 1 + q such points.
Exercise 9.4.9.2. The stabilizer P Sp(4, q)¯p of the point ¯p is the set of
collineations induced by multiplication by matrices of the form
A =
⎛
⎜
⎝
|A4| −1 A4
1 0
a
|A4|
−d
0
c 0
0 0
c d
A4
⎞
⎟
⎠ : a, c, d ∈ F, A4 ∈ GL(2, q).
Exercise 9.4.9.3. The stabilizer P Sp(4, q)¯p,¯q is triply transitive on {¯p, ¯q} ⊥ .
Moreover, any θ ∈ P Sp(4, q)¯p,¯q that fixes three lines of W (q) through ¯p must
fix all lines through ¯p and must be given by multiplication by a matrix of the
form

9.4. THE SYMPLECTIC GQ W (Q) 389
A =
⎛
⎜
⎝
1 0 0 0
0 d 2 0 0
0 0 d 0
0 0 0 d
⎞
⎟
⎠ , 0 = d ∈ F.
It is now easy to see that this last set of matrices gives a group of collineations
isomorphic to the multiplicative group of the field F .
Exercise 9.4.9.4. Determine whether or not the lines of W (q) are regular.
(Answer: when q is even, all lines are regular. When q is odd, there is not
even any 3 × 3 grid.)
Exercise 9.4.9.5. If A is 2 × 2 over F , then AP A T = det(A) · P .
At this point we want to study to orbits of P Sp(4, q) on conics contained
in planes of P G(3, q). First, since P Sp(4, q) is transitive on points of W (q)
(i.e., points of P G(3, q)), it must be transitive on planes of P G(3, q). So if
C were a conic of some plane, we could use P Sp(4, q) to map it to a conic in
the plane p ⊥ = [0, 1, 0, 0] where p = (1, 0, 0, 0). So what are the orbits of the
stabilizer P Sp(4, q)p of p on the points of [0, 1, 0, 0]? Put A4 =
where △ = a1a4 + a2a3 = 0, and then for arbitrary a, c, d ∈ Fq put
A =
⎛
⎜
⎝
1 0 0 0
a △ c d
a1d+a2c
△ 0 a1 a2
a3d+ca2
△ 0 a3 a4
⎞
⎟
⎠ .
a1 a2
a3 a4
The homography given by this matrix belongs to P Sp(4, q)p and maps r =
(0, 0, 0, 1) to the point (a3d + a4c, 0, a3△, a4△). By varying c, d, a3 and a4
(and choosing a1 and a2 so that △ = 0 we can map r to every point of
[0, 1, 0, 0] different from p. An homography in P Sp(4, q) that fixes both p
and r has a matrix of the form
A =
⎛
⎜
⎝
1 0 0 0
a a1a4 0 d
d/a4 0 a1 a2
0 0 0 a4
⎞
⎟
⎠ .
These homographies have 〈p, r〉 \ {p, r} as one orbit of size q − 1 and the
q 2 points of the form (x, 0, 1, w) as an orbit of size q 2 . One of the points in
,

390 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
this latter orbit is s = (0, 0, 1, 0). So consider the stabilizer P Sp(4, q)p,r,s. An
homography in this group has matrix of the form
⎛
1
⎜
A = ⎜ a
⎝ 0
0
a1a4
0
0
0
a1
0
0
0
⎞
⎟
⎠
0 0 0 a4
.
This group has an orbit of size q −1 on the line 〈r, s〉 and an orbit of size (q −
1) 2 consisting of the points not on the sides of the triangle with vertices p, r, s.
Let t = (1, 0, 1, 1) be such a point. We have just shown that P Sp(4, q)p is
transitive on the triples of points on distinct lines through p in p ⊥ , and
(r, s, t) is such a triple.
Now let C be a conic in the plane p ⊥ . We have the following three cases.
Case 1. p is the nucleus of C. Using P Sp(4, q) we may assume that r, s
and t all belong to C. This completely determines C.
C = {(t, 0, t 2 , 1) : t ∈ Fq} ∪ {(0, 0, 1, 0)} with nucleus p = (1, 0, 0, 0).
Note that in this case no two points of C are collinear in W (q) (since the
lines of W (q) in p ⊥ are precisely the lines through p).
Case 2. p ∈ C. In this case we may assume that r is the nucleus of C
and that s and t belong to C. Again using P Sp(4, q)p we may assume that
C is any such C, say
C = {((1, 0, t 2 , t) : t ∈ Fq} ∪ {(0, 0, 1, 0)} with nucleus r = (0, 0, 0, 1).
In this case the only collinearities (in W (q))between points of C are those
between p and the other points.
Case 3. p ∈ C and p is not the nucleus of C. In this case we may assume
that the nucleus is r = (0, 0, 0, 1) and that both s and t belong to C. In this
case one can show that for each choice of a ∈ Fq \ {0, 1} there is a conic C
not containing p and given by
C = {(1, 0, t 2 (a 2 +1)+1, t) : t ∈ Fq}∪{(0, 0, 1, 0)} with nucleus r = (0, 0, 0, 1).
We collect part of the above discussion in the following theorem.

9.4. THE SYMPLECTIC GQ W (Q) 391
Theorem 9.4.10. P Sp(4, q) has three orbits in its action on conics of P G(3, q).
Orbit 1 A conic C of P G(3, q) is in Orbit 1 if and only if it lies in the
plane p ⊥ where p is the nucleus of C, which is if and only if it has no two
points collinear in W (q). Orbit 1 contains (1 + q)(1 + q 2 )(q 3 − q 2 ) conics.
(See Exercise 4.2.5.2.)
Orbit 2 A conic C of P G(3, q) is in Orbit 2 if and only if the pole p of the
plane containing C is contained in C. Orbit 2 contains (1+q)(1+q 2 )(q 4 −q 2 )
conics. In the hyperoval containing C the only collinearities in W (q) are
between p and the other points of the hyperoval.
Orbit 3 A conic C is in Orbit 3 if and only if the pole p of the plane
containing C is not contained in the hyperoval containing C. There are
(1 + q)(1 + q 2 )q 2 (q − 1)(q 2 − 1) conics in Orbit 3. If C + is the hyperoval
containing C, then the points of C + come in collinear pairs (by lines of
W (q)).
Exercise 9.4.10.1. Let R = {L0, L1, . . . , Lq} be a regulus consisting of lines
of W (q) (as given above). So R has 1 + q transversals. Show that if q = 2 e
either 1 or q + 1 of the transversals belong to W (q).
Solution: By Exercise 9.4.9.4 we know that all lines of W (q) are regular,
so that each triad of lines of W (q) (three lines of W (q) that are pairwise
skew) has 0, 1 or q + 1 transversals of lines of W (q). Using this it follows
readily that a regulus of lines of W (q) has 0, 1 or q + 1 transversals that are
lines of W (q). So we just need to show that every such regulus has at least
one transversal that is a line of W (q).
So suppose that R = {L0, . . . , Lq} is a regulus of lines of W (q) with
transversals R opp = {M0, . . . , Mq}. We may suppose that M0 and M1 do
not belong to W (q). This means that there are vectors p1, p2, p3, p4 not
on a plane satisfying the following: First, L0 = 〈p1, p2〉; L1 = 〈p3, p4〉; M0 =
〈p1, p3〉; M1 = 〈p2, p4〉. And
If 1 ≤ i ≤ j ≤ 4, then pi ◦ pj = 0 if and only if
i = j, or i = j and {i, j} = {1, 2} or {3, 4}.

392 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
If pi ◦ pj = aij = 0, then
replace p3 with a −1
13 p3, so p1 ◦ p3 = 1
replace p4 with a −1
14 p4, so p1 ◦ p4 = 1
replace p2 with a −1
23 p2, so p2 ◦ p3 = 1
Of course, immediately after each replacement, the aij is relabeled. This
means that we can arrange things so that :
p1 ◦ p2 = 0; p1 ◦ p3 = 1; p1 ◦ p4 = 1; p2 ◦ p3 = 1; p2 ◦ p4 = k = 0.
We also know that k = 1. For suppose that k = 1 and put q = p1+p2+p3+p4.
Then Then pi ◦ q = 0 for all i = 1, 2, 3, 4. If the four original vectors were
really linearly independent, then q = 0. But this would mean that the four
points would lie in a plane, an impossibility. Hence k = 0, 1.
The remaining points of M0 are of the form p3 + cp1, and the unique
point of M1 collinear in W (q) with p3 + cp1 is p4 + cp2. Choosing c = 0
and c = 1 we now have three ”horizontal” lines of R in W (q). These three
determine all the transversals M0 and 〈p2 +ap1, p4 +ap3〉. In general we have
〈p3 + cp1, p4 + cp2〉 meeting 〈p2 + ap1, p4 + ap3〉 at the point
adp1 + cp2 + ap3 + p4.
The question is now whether or not some transversal is in W (q). To see
this we compute: (p2 + ap1) ◦ (p4 + ap3) = k + a + a + a 2 = 0 if and only
k = a 2 . So for a = √ k we obtain the unique transversal of R that lies in
W (q).
p2 + ap1 is p4 + ap3.
9.5 Some GQ derived from Quadratic Forms
Let Q(¯x) = Q(x0, . . . , x4) = x0x1 + x2x3 + x 2 4 = ¯xA¯x T , where
⎛
⎜
A = ⎜
⎝
0 1 0 0 0
0 0 0 0 0
0 0 0 1 0
0 0 0 0 0
0 0 0 0 1
⎞
⎟
⎠ .

9.5. SOME GQ DERIVED FROM QUADRATIC FORMS 393
Then
⎛
B = A + A T ⎜
= ⎜
⎝
0 1 0 0 0
1 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 0 0 2
⎞
⎟
⎠ .
When q is odd, clearly B is nonsingular, so the quadratic form Q is
nondegenerate. When q is a power of 2, then it is easy to check that the
singular radical is still {0}, so Q is nondegenerate. This implies that there
is no totally singular plane, which means that if x is a singular point not on
a totally singular line L, then x cannot be orthogonal to each point of L, so
that the polar plane of x meets the line L in a unique point. If fQ(¯x, ¯y) = 0
for two singular points ¯x and ¯y, then the line through ¯x and ¯y is totally
singular.
Let Q(4, q) denote the geometry consisting of all singular points of Q
together with the totally singular lines. It is now a simple task to show that
Q(4, q) is a GQ(q,q). We want to investigate its collineation group. Recall
that
O(5, q) = {D ∈ GL(5, q) : DBD T = λB for some nonzero λ ∈ F }.
And P GO(5, q) is the group of linear projective collineations of Q(4, q).
We consider the action of P GO(5, q) on Q(4, q). Label three particular points
of Q(4, q) as follows.
Note that
Also,
p = (1, 0, 0, 0, 0), q = (0, 1, 0, 0, 0), e = (0, 0, 1, 0, 0). (9.4)
{y ∈ Q(4, q) : p ∼ y} = {(−(t 2 + dr), 1, d, t) : r, d, t ∈ F }. (9.5)
L∞ =< (0, 0, 1, 0, 0), (1, 0, 0, 0, 0) >; Lc =< (0, 0, −c 2 , 1, c), (1, 0, 0, 0, 0)},
(9.6)
for c ∈ F , are the lines of Q(4, q) through p.

394 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
For arbitrary r, d, t ∈ F put
⎛
⎜
D(r, d, t) = ⎜
⎝
1 0 0 0 0
−(t 2 + dr) 1 r d t
−d 0 1 0 0
−r 0 0 1 0
−2t 0 0 0 1
⎞
⎟
⎠
(9.7)
Lemma 9.5.1. (i) T = {D(r, d, t) : r, d, t ∈ F } is a group of translations
about the point p = (1, 0, 0, 0, 0), i.e., T acts regularly on the q 3 points not
collinear with p and fixes each line through p.
(ii) A(∞) = {D(r, 0, 0) : r ∈ F } is the full group of symmetries about
the line L∞, i.e., A(∞) fixes each line that meets L∞.
(iii) For each c ∈ F , A(c) = {D(−dc 2 , d, dc) : d ∈ F } is the full group
of symmetries about the line Lc.
(iv) A ∗ (∞) = {D(r, 0, t) : r, t ∈ F } is the stabilizer in T of the point e
on L∞.
(v) For c ∈ F , A ∗ (c) = {D(dc 2 − 2ct, d, t) : d, t ∈ F } is the stabilizer in
T of the point (0, 0, −c 2 , 1, c) on Lc.
We leave the proof of this lemma as a routine exercise.
Note: D(r, d, t) · A · DT (r, d, t) = A ′ ⎛
0 1 0 0
⎜ 0 0 0 r
= ⎜ 0 0 0 1
⎝ 0 −r 0 0
⎞
0
t ⎟
0 ⎟ = A, but A
0 ⎠
0 −t 0 0 1
and A ′ clearly determine the same quadratic form.
⎛
⎜
Lemma 9.5.2. D = ⎜
⎝
0 1
1 0
From this it is easy to show that
1
1
1
⎞
⎟ ∈ O(5, q), and D : p ↔ q.
⎠
Lemma 9.5.3. P GO(5, q) is transitive on the points of Q(4, q).

9.5. SOME GQ DERIVED FROM QUADRATIC FORMS 395
The next step is to determine the stabilizer of p and q in P GO(5, q). The
stabilizer of p and q must act on the set {p, q} ⊥ = {e} ∪ {(0, 0, −c 2 , 1, c) :
c ∈ F }. First note that
⎛
⎜
D = ⎜
⎝
1 0 0 0 0
0 1 0 0 0
0 0 −c 2 1 c
0 0 1 0 0
0 0 −2c 0 1
⎞
⎟ ∈ P GO(5, q), (9.8)
⎠
and x ↦→ xD fixes both p and q and maps e to (0, 0, −c 2 , 1, c), showing that
the stabilizer of p and q is transitive on {p, q} ⊥ . So we now ask what is the
stabilizer of p, q and e?
Matrices of the form
⎛
1 0 0 0 0
⎜
0
D = ⎜
⎝
d2 0
0
0
0
0
u
−u
0
1
0
c
h 2
d
d2 u h
0 0 −2u ⎞
h
d 0
⎟
⎠
d
(9.9)
belong to P GO(5, q), and x ↦→ xD fixes p, q and e and maps f := (0, 0, 0, 1, 0)
to (0, 0, −u
h 2 d , d
2
u , h) ≡ (0, 0, −u2h2 d4 , 1, uh
d2 ). This is for d, u, h ∈ F with
du = 0. This shows that the stabilizer of p and q is doubly transitive on the
points of {p, q} ⊥ . (This is most clearly seen by putting u = d = 1 in the
above matrix.)
Matrices of the form
⎛
⎜
D = ⎜
⎝
1 0 0 0 0
0 d2 0 0 0
0 0 u 0 0
0 0 0 d2
0 0 0
u
0
0
d
⎞
⎟
⎠
(9.10)
belong to P GO(5, q), and x ↦→ xD fixes p, q and e and f, and maps
(0, 0, −1, 1, 1) to (0, 0, −u, d2
−u2
, d) ≡ (0, 0, u d2 , 1, u).
It fixes (0, 0, −1, 1, 1) if
d
and only if u = d if and only if

396 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
⎛
⎜
D = ⎜
⎝
1 0 0 0 0
0 d 2 0 0 0
0 0 d 0 0
0 0 0 d 0
0 0 0 0 d
in which case x ↦→ xD fixes all lines through p.
We have now established the following result.
Theorem 9.5.4. The action of P GO(5, q) on Q(4, q):
(i) P GO(5, q) is transitive on the points of Q(4, q)
⎞
⎟ , (9.11)
⎠
(ii) The stabilizer P GO(5, q)p is transitive on the q 3 points of Q(4, q) not
collinear with p.
(iii) The stabilizer P GO(5, q)p,q of the noncollinear points p and q is triply
transitive on the points of {p, q} ⊥ .
(iv) Any collineation of P GO(5, q)p,q that fixes three lines through p fixes
all lines through p and has the form given in Eq. 9.11
(v) |P GO(5, q)| = [(1 + q)(1 + q 2 )]q 3 [(q + 1)q(q − 1)](q − 1)
= (q 4 − 1)(q 2 − 1)q 4 . (9.12)
Exercise 9.5.4.1. Show that each line of Q(4, q) is regular.
Exercise 9.5.4.2. Show that each point of Q(4, q) is regular if q is even,
and is antiregular if q is odd. (The definition of antiregular formally appears
later. Here it just means that each triad of points has either 0 or 2 centers.)
After we have introduced Plücker coordinates and the Klein correspondence
we will show that W (q) and Q(4, q) are point-line duals of each other.
9.6 Triads with Restricted Perp Size
We continue temporarily with the notation of Section 9.5 still with the assumption
that both s > 1 and t > 1.
Suppose there are three distinct integers α, β, γ, 0 ≤ α, β, γ ≤ t + 1,
for which i ∈ {α, β, γ} implies that ni = 0. Note that we allow ni = 0 for

9.6. TRIADS WITH RESTRICTED PERP SIZE 397
i ∈ {α, β, γ}. Then the equations in Lemmas 9.3.2, 9.3.3, and 9.3.4 may be
written in matrix form as
⎛
1 1
⎝ α β
1
γ
⎞⎛
⎞ ⎛
nα ts
⎠⎝
nβ ⎠ = ⎝
α(α − 1) β(β − 1) γ(γ − 1)
2 − st − s + t
st2 − s
t3 ⎞
⎠(9.13)
− t
The determinant of this linear system is ∆ = (α − β)(β − γ)(γ − α), and
we may use Cramer’s rule to solve for nα, nβ, nγ. As α, β, γ were given in
no particular order, it suffices to solve for just one:
nα = (ts2 − st − s + t)βγ − (t 2 − 1)s(β + γ) + (t 2 − 1)(s + t)
(α − β)(α − γ)
The equations 9.13 and 9.14 have the following easy corollaries.
nγ
(9.14)
Corollary 9.6.1. Suppose that nβ = nγ = 0, i.e., there is at most one index
α for which nα = 0. Then (t − 1)(s − 1)(s 2 − 1) = 0, and if s 2 = t = 1, then
α = s + 1 and
Of course, this was contained in Lemma 9.3.4
nα = s(s − 1)(s 2 + 1) (9.15)
Corollary 9.6.2. Suppose that nγ = 0. By the formula for nγ we have
(ts 2 − st − s + t)αβ − (t 2 − 1)s(α + β) + (t 2 − 1)(s + t) = 0.
If α = 0 and t > 1, then β = (s + t)/s. If α = 1, then β = (t 2 − 1)/(st −
s 2 + s + 1), which also forces s ≤ t if t = 1.
Corollary 9.6.3. If s > t > 1, then by Corollaries 9.3.7 and 9.3.8 it follows
that both n0 > 0 and n1 > 0.
Corollary 9.6.4. Suppose α = 0, β = 1, with s, t, γ > 1. Then
If γ = 2, then we have
nγ = t(t 2 − 1)/γ(γ − 1)
n0 = s 2 t − st 2 − st + (t + t 3 )/2
n1 = (t 2 − 1)(s − t)
n2 = (t 2 − 1)t/2.

398 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
If γ = t + 1, then we have
n0 = (s − t)t(s − 1)
n1 = (t 2 − 1)(s − 1)
nt+1 = t − 1.
If γ = (s + t)/s, then n1 = 0, and we are back in the situation considered in
Corollary 9.6.2
Corollary 9.6.5. Suppose 1 = α < β < γ = 1 + t. Then
n1 = (1 + t)(s − 1)(1 − t + β(s − 1))/(β − 1)
nβ = t(s − 1)(t + 1)(t − s)/(β − 1)(t + 1 − β)
nt+1 = (t 2 − 1 − β(st − s 2 + s − 1))/(t + 1 − β)
Now let x be a fixed point of S.
Theorem 9.6.6. If the triads contained in x ⊥ have a constant number α of
centers, then α = 1 + t/s, etc.
1. The triads {y1, y2, y3} contained in x ⊥ have a constant number of centers
iff the triads {x, u1, u2} containing x have exactly 0 or α (α a
constant) centers. If one of these equivalent situations occurs, then
(s + t) | s(t − 1) and the constants both equal 1 + t/s.
2. Let y ∈ P \ x ⊥ . Then no triad containing {x, y} has more than 1 + t/s
centers iff each such triad has exactly 0 or 1 + t/s centers iff no such
triad has α centers with 0 < α < 1 + t/s. In such a case there are
t(s − 1)(s 2 − t)/(s + t) acentric triads containing x and y, and (t 2 −
1)s 2 /(s + t) triads containing x and y with exactly 1 + t/s centers.
3. If s = q n and t = q m with q a prime power, and if each triad in x ⊥ has
1 + t/s centers, then there is an odd integer a for which n(a + 1) = ma.
Proof. For Part 1, suppose there is a constant α such that each triad {x, u1, u2}
containing x has 0 or α centers. By Corollary 9.3.13 we have α = 1 + t/s.
There are d = (t2−1)s3t/6 triads T1, . . . , Td contained in s⊥ . Let 1+ri be the
number of centers of Ti, so that ri = s2t(t+1)t(t−1)/6. Count the ordered
triples (Ti, u1, u2), where Ti is a triad in x⊥ and (x, u1, u2) is an ordered triad
in T ⊥
i , to obtain ri(ri−1) = s2tnα(1+t/s)(t/s)((t/s)−1)/6. Here nα is the

9.6. TRIADS WITH RESTRICTED PERP SIZE 399
number of triads containing (x, u1), x ∼ u1, and having exactly α = 1 + t/s
centers. From Lemma 9.3.3 it follows that nα = (t 2 − 1)s 2 /(s + t). Hence
d( r 2 i ) − ( ri) 2 = 0, implying that ri is the constant ( ri)/d = t/s.
Conversely, assume that the number ri + 1 of centers of Ti is a constant.
Then ri = s 2 t(t+1)t(t−1)/(t 2 −1)s 3 t = t/s. Fix y1 in x ⊥ \{x}. The number
of triads V1, . . . , Vd ′ containing x and having y1 as center is d ′ = t(t − 1)s 2 /2.
If 1 + ti denotes the number of centers of Vi, 1 ≤ i ≤ d ′ , it is easy to
check that ti = stt(t − 1)/2 and ti(ti − 1) = sts(t − 1)(t/s)(t/s − 1)/2.
Hence d ′ ( t 2 i ) = ( ti) 2 , and ti is the constant ( ti)/d ′ = t/s. It follows
immediately that each centric triad (x, u1, u2) has exactly 1 + t/s centers.
Suppose that these equivalent situations occur. Fix u1, u1 ∼ x, and let L
be a line which is incident with no point of {x, u1} ⊥ (since s = 1 such a line
exists). Then the number of points u2, u2 | L, for whch (x, u1, u2) is a centric
triad equals (t − 1)/(1 + t/s). Hence (s + t) | s(t − 1), and Part 1 is proved.
For Part 2, recall Corollary 9.3.8. The coefficient of nα in that equation is
nonnegative iff α ≥ 1+t/s, and equals 0 iff α = 1+t/s. Assume that nα = 0
for α > 1 + t/s. Since n1 ≥ 0, we must have nα = 0 for 1 ≤ α ≤ t/s.Hence
each triad containing (x, y) has exactly 0 or 1 + t/s centers. Conversely,
assume nα = 0 for 0 < α < 1 + t/s. Since n1 = 0, we necessarily have nα = 0
for α > 1 + t/s. Hence each triad containing (x, y) has exactly 0 or 1 + t/s
centers. Finally, if this last condition holds, it is easy to use Lemmas 9.3.2
and 9.3.3 to solve for n0 and n1+t/s, completing the proof of Part 2.
Given the hypotheses of Part 3, from Part 1 we have t ≥ s and (s + t) |
s(t − 1), from which it follows that (1 + q m−n ) | (q m − 1). Since q m − 1 =
(q m−n + 1)q n − q n − 1, there results (1 + q m−n ) | (1 + q n ). This forces n to be
an odd multiple of m − n, so n(a + 1) = ma with a odd.
For the remainder of this Section we suppose that each triad contained in
x ⊥ has exactly 1+t/s centers, so that each triad containing x has 0 or 1+t/s
centers. Let T = {x, u1, u2} be a fixed triad with T ⊥ = {y0, y1, . . . , yt/s}.
Each triad in T ⊥ also has 1 + t/s centers. For 0 ≤ i ≤ 1 + t/s let mi be the
number of points w such that w is collinear with exactly i points of T ⊥ but
not with any of x, u1, u2.
Theorem 9.6.7. Suppose that each triad contained in x ⊥ has exactly 1+t/s

400 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
centers.
mi = s 2 t − 2st − 2s + 3t − t/s (9.16)
imi = (s + t)(t − 2) (9.17)
i(i − 1)mi = (s + t)t(t − 2)/s 2
i(i − 1)(i − 2)mi = (s + t)t(t − s)(t − 2s)/s 4
(9.18)
(9.19)
Proof. The first sum just counts all points not collinear with any of x, u1, u2.
The second sum counts the number of ordered pairs (w, yi) with w ∼ yi but
w not collinear with any of x, u1, u2. The third sum counts the number of
ordered triples (w, yi, yj) with yi ∼ w ∼ yj, yi = yj, and s not collinear with
any of x, u1, u2. Finally, the fourth sum counts the number of ordered 4tuples
(w, yi, yj, yk) with w a center of the triad {yi, yj, yk} but not collinear
with any of x, u1, u2.
We end this section with a result on GQ of order (s, s 2 ), in which we have
seen that each triad of points must have exactly 1 + t/s = 1 + s centers. In
fact, when s = 3 this result is a corollary of the preceding theorem, but for
s > 3 it seems to require a special proof.
Theorem 9.6.8. In a GQ S = (P, B, I) of order (s, s 2 ), with s > 2, a triad
which nearly has a span of size 1 + s, in fact does have a span of size 1 + s.
Let S = (P, B, I) be a GQ of order (s, s 2 ), s > 2, with a triad T =
{x0, x1, x2} for which T ⊥ = {y0, . . . , ys}, and for which {x0, . . . , xs−1} ⊆
{x0, x1, x2} ⊥⊥ . Suppose that there is a point xs for which xi = xs ∼ yi for
0 ≤ i ≤ s − 1. Then xs ∼ ys.
Proof. The number of points collinear with ys and also with at least two
points of {y0, . . . , ys−1} is at most s(s − 1)/2 + s, and the number of points
collinear with ys and incident with some line xsyi, 0 ≤ i ≤ s − 1, is at most
s. Since s > 2, we have s(s − 1)/2 + 2s < s 2 + 1 = t + 1. Hence there
is a line L incident with ys, but not concurrent with xsyi, 0 ≤ i ≤ s − 1,
and not incident with an element of {yi, yj} ⊥ , i M = j, 0 ≤ i, j ≤ s − 1.The
point incident with L and collinear with yi is denoted by zi, 0 ≤ i ≤ s − 1.
Clearly all s points zi are distinct. Since S has no triangles, the point xs is
not collinear with any zi, forcing xs ∼ ys.
Exercise 9.6.8.1. Show that |T ⊥⊥ | = 4 for any triad in a GQ of order (3, 9).

9.7. THE INCIDENCE MATRIX 401
9.7 The Incidence Matrix
Let S = (P, B, I) be a GQ of order (s, t), and put | P |= v, | B |= b. So we
know
v = (1 + s)(1 + st); b = (1 + t)(1 + st) (9.20)
Suppose that the pointset and the lineset are ordered:
P = {x1, . . . , xv}; B = {L1, . . . , Lb}.
Let A be the line-point incidence matrix of S, i.e., A = (aij), where
aij = 1 if xjILi and aij = 0 otherwise. A is b × v and
(A T A)ij =
⎧
⎨
⎩
1 + t, if i = j;
1, if xi ∼ xj, i = j;
0, if xi ∼ xj.
The number of paths from xi to Lj with length 3 is
It follows that
(A T AA T )ij =
(Here J is the v × b matrix of 1’s.)
There are other easy results:
s + t + 1, xiILj;
1, otherwise.
(9.21)
(9.22)
A T AA T − (s + t)A T = Jv×b. (9.23)
tr(A T A) = (1 + s)(1 + t)(1 + st). (9.24)
Jb×vA T A = (1 + s)(1 + t)Jb×v. (9.25)
(A T A) 2 − (s + t)A T A = Jv×bA = (1 + t)Jv×v
(9.26)
Jv×v(A T A − (1 + s)(1 + t)I) = 0 (9.27)
A T A(A T A − (s + t)I)(A T A − (1 + s)(1 + t)I) = 0 (9.28)

402 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
Let A T A have eigenvalue θi with multiplicity di and eigenspace Wi, i =
0, 1, 2, with θ0 = (1+s)(1+t); θ1 = 0; θ2 = s+t. From Eq. 9.26, (θi) 2 −(s+t)θi
is an eigenvalue of (1 + t)Jv, i.e., it must be one of (1 + t)(1 + s)(1 + st), 0.
So θ0 = (1 + s)(1 + t) corresponds to (1 + t)(1 + s)(1 + st) and d0 = 1.
Let e be the column vector with as many entries as make sense, all equal
to 1.
W0 = 〈e〉 = {x : A T Ax = (1 + s)(1 + t)x} ⊆ col.sp.(A T ) (9.29)
In the next equation y is an arbitrary column vector with b rational
entries.
e T A T y = 0 iff (1 + s)e T y = 0 iff e T y = 0 (9.30)
Now define the following subspace U of column vectors with v rational
entries:
U = {A T y : y ∈ 〈e〉 ⊥ }
There is an immediate consequence:
U = e ⊥ ∩ col.sp.(A T ). (9.31)
Proof. As x ∈ U implies that x = A T y ∈ col.sp.(A T ) for some y with y T e = 0,
by equation 9.30 we have e T A T y = 0, so e T x = 0 and x ∈ e ⊥ ∩ col.sp.(A T ).
Conversely, let x ∈ col.sp.(A T ), say x = A T y. If e T x = 0, then by equation
9.30 e T y = 0, so x ∈ U by definition.
We have v = 1 + d1 + d2, and tr(A T A) = (1 + s)(1 + t)(1 + st) =
1 · (1 + s)(1 + t) + 0 · d1 + (s + t) · d2. Hence:
d1 = (1 + st)s 2 /(s + t) and d2 = (1 + s)(1 + t)st/(s + t) (9.32)
This gives an important restriction on the parameters s and t:
Theorem 9.7.1. (1 + s)(1 + t)st/(s + t) must be an integer.
Corollary 9.7.2. If s > 1, t > 1, s = t 2 and t = s 2 , then t ≤ s 2 − s and,
dually, s ≤ t 2 − t. In fact, t = s 2 − x with x = s, x = 1 + s or x ≥ 2s. Also,
t = 1 + s.

9.7. THE INCIDENCE MATRIX 403
Proof. Suppose s = 1 and t = s 2 . By Higman’s inequality we have t = s 2 − x
with x > 0. By Theorem 9.7.1 (s + s 2 − x)|s(s 2 − x)(s + 1)(s 2 − x + 1). Hence
modulo s+s 2 −x we have 0 ≡ x(−s)(−s+1) ≡ x(x−2s). If 0 < x < 2s, then
s+s 2 −x ≤ x(2s−x) forces x 2 −x(2s+1)+s(1+s) = (x−s)(x−(1+s)) ≤ 0,
from which we see x = s or x = 1 + s. Consequently, x = s, x = 1 + s, or
x ≥ 2s. Finally, if t = s + 1, then 2s + 1|(1 + s)(2 + s)s(s + 1), which can be
seen to force s = 1, a contradiction.
In the remainder of this section we use the incidence matrix and begin
to apply the theory of interlacing of eigenvalues (see the Appendix which is
Chapter 22) to generalized quadrangles.
Let S = (P, B, I) be a GQ with parameters (s, t). Let P = {x1, . . . , xv},
with v = (1 + s)(1 + st) be particular ordering of the points of S, and
B = {L1, . . . , Lb}, with b = (1 + t)(1 + st), a particular ordering of the lines
of S. Let A be the b × v (0,1)-matrix whose rows are indexed by the lines
of S and whose points are indexed by the points of S. Then A has a 1 in
the (i, j) position if and only if xj is incident with the line Li. We showed
that the real symmetric matrix AT A has eigenvalues θ0 = (1 + s)(1 + t),
θ1 = 0, θ2 = s + t, with multiplicities given by d0 = 1, d1 = s2 (1+st)
, and
s+t
d2 = (1+s)(1+t)st
, respectively.
s+t
Put B = tI + J − AT A, so B is a symmetric v × v (0,1)-matrix with a
1 in the (i,j) position if and only if i = j and xi and xj are NOT collinear.
The eigenvalues θ ′ i and their multiplicities d′ i = di are easily seen to be
θ ′ 0 = t + (1 + x)(1 + st) − (1 + s)(1 + t) = s 2 t; d ′ 0 = 1; (9.33)
θ ′ 1 = t + 0 − 0 = t; d′ 1 = s2 (1 + st)
;
s + t
θ ′ 2 = t + 0 − (s + t) = −s; d′ (1 + s)(1 + t)st
2 = .
s + t
We are going to apply Theorem 22.1.3 with B = BΓ , i.e., Γ is the identity
partion (i.e., each part has size 1). Let D = ∆1 + ∆2 be any partion of
{1, . . . , v} having two parts. So δ1 = |∆1|, δ2 = |∆2| = (1 + s)(1 + st) − δ1.
Here δ11 is the number of ordered pairs (i, j) for which i, j ∈ ∆1 and xi and
xj are not collinear in S. Also, δ12 is the number of ordered pairs (i, j) for
which i, j ∈ ∆1 × ∆2 and xi and xj are not collinear in S. Hence if e = δ11
δ1 ,
then δ12 = s2t − e and it follows readily that

404 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
B ∆ =
e s2 δ2
t − e
s2t − δ1(s2
t−e)
δ2
. (9.34)
δ1(s 2 t−e)
One eigenvalue of B∆ is clearly s2t, so the other is tr(B∆ ) − s2t = e −
δ1(s2t−e) . Since the eigenvalues of B δ2
∆ lie between the smallest and largest
roots of B, in particular
which simplifies to
−s ≤ e − δ1(s2t − e)
,
s
(δ1 − (1 + s)) ≤ e (9.35)
1 + s
Since e is the average number of points of ∆1 not collinear in S with a
fixed point of ∆1, it must be that δ1 − e is the average number of points
of ∆1 collinear in S with a fixed point of ∆1 (including that point itself).
Equation 9.35 may be rewritten to say that
δ1 − e ≤ s + δ1
1 + s
This proves the first part of the following theorem.
δ2
(9.36)
Theorem 9.7.3. Let ∆1 be any nonempty, proper subset of points of the GQ
S of order (s, t). Then the average number of points collinear in ∆1 with a
given point of ∆1 is less than or equal to s + δ1 . If equality holds, then each
1+x
point of ∆1 is collinear with exactly s + δ1
1+x points of ∆1. Moreover, this
holds if and only if each point of ∆2 is collinear with exactly δ1 points of
1+s
∆1.
Proof. The inequality of the theorem has been established. If equality holds,
i.e., −s = e − δ1(s2t−e) , then (δ1 − v, δ1) δ2
T is an eigenvector of B∆ associated
with the eigenvalue −s. Hence it must be that x = (δ1 − v, . . . , δ1 −
v, δ1, . . . , δ1) T is an eigenvector of B associated with eigenvalue −s, where
the first coordinate value is repeated δ1 times and the second δ2 = v − δ1
times.
Fix i ∈ ∆1 and let a be the number of points xj with j ∈ ∆1 for which xi
and xj are not collinear. Now consider the ith row of Bx = −sx. This says
δ1 j=1 Bij(δ1 − v) + δ2 j=1 Bi,δ1+jδ1 = a(δ1 − v) + (s2t − a)δ1 = −s(δ1 − v),

9.7. THE INCIDENCE MATRIX 405
which simplifies to a = δ1s
− s, which says that a = e is independent of
1+s
i ∈ ∆1. This implies that each point xi for i ∈ ∆1 is collinear with exactly
δ1 − e = s + δ1
1+s points xj for of j ∈ ∆1. Now let i ∈ ∆2 and let b be the
number of points xj for j ∈ ∆1 for which xi and xj are not collinear. Then
the ith row of Bx = −sx says that b(δ1 − v) + (s2t − b)δ1 = −sδ1, which
becomes b = δ1s
1+s for all i ∈ ∆2. This says that each point of ∆2 is collinear
with exactly δ1 − b = δ1 points of ∆1.
1+s
NOTE: There is a theory of “tight” sets for which the case of equality in
this theorem holds.
The preceding theorem deals with the simplest possible application of the
“interlacing of eigenvalues” techniques, but it has interesting applications.
The following result on GQ has the Higman inequality as a corollary.
Theorem 9.7.4. Let X = {x1, . . . , xm} and Y = {y1, . . . , yn} each be sets
of pairwise noncollinear points of S, m ≥ 2, n ≥ 2. Assume that X ∩ Y = ∅
and that yi is collinear with ki of the xj’s, 1 ≤ i ≤ n, 0 ≤ ki ≤ m. Then:
(1 + s)
n
ki ≤ mn + m2n2 + (s2 − 1)(m + n)mn + (s2 − 1) 2mn. (9.37)
i=1
Proof. As before, let Γ be the identity partition, so B = B Γ . Let Z =
P \ (X ∪ Y ) = {z1, . . . , zr}, r = v − (m + n). Let ∆ = ∆1 + ∆2 + ∆3 be the
partition of {1, . . . , v} determined by X, Y and Z, respectively. As δi = |∆i|,
we have δ1 = m, δ2 = n, δ3 = v − (m + n), δ11 = m(m − 1), δ22 = n(n − 1),
δ12 = δ21 = n i=1 (m − ki) = mn − Σ, where Σ = n i=1 ki. Since 3 δij
j=1 δi =
s2t, we also have δ13 = δ1s2t − δ12 − δ11 = s2tm − (mn − Σ) − m(m − 1).
Similarly, δ23 = s2tn − (mn − Σ) − n(n − 1). Using these results it is now
routine to complete the calculation of B∆ .
⎛
B ∆ = ⎝
m − 1 n − Σ s m
2t + 1 − m − n + Σ
m
m − σ n − 1 s n 2t + 1 − m − n + Σ
n
n[s2t+1−m−n]+Σ s v−m−n
2t − (m+n)[s2t+1−m−n]+2Σ v−m−n
m[s 2 t+1−m−n]+Σ
v−m−n
Let (x − s 2 t)(x − r1)(x − r2) be the characteristic polynomial of B ∆ with
the roots ordered so that r1 ≤ r2 ≤ s 2 t. The proof of the theorem amounts
⎞
⎠ .

406 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
to calculating r1 and using the inequality −s ≤ r1. Put (x − r1)(x − r2) =
x 2 − bx + c, so that 2r1 = b − √ b 2 − 4c. Hence −s ≤ r1 simplifies to
0 ≤ s 2 + bs + c; b = r1 + r2 = tr(B ∆ ) − s 2 t; c = det(B ∆ )/s 2 t. (9.38)
It is straightforward to calculate tr(B δ ) and then to write b as follows:
b =
(m + n)(s + st + 2) − 2v − 2Σ
. (9.39)
v − m − n
To calculate det(B ∆ ), add the first and second columns of B ∆ to the
third column and then subtract the first row from the second. At this point
det(B ∆ ) appears as follows.
det(B ∆ ) = s 2 t
m − 1 n − 0 Σ 1
m
1 − Σ
Σ
− 1 0
n
m . (9.40)
1
m[s 2 t+1−m−n]+Σ
v−m−n
n[s 2 t+1−m−n]+Σ
v−m−n
Expanding by the third column and simplifying, one may calculate c to
be as folows:
c = det(B ∆ )/s 2 t = (1 + s + st)(2Σ − m − n) + v − Σ2v/mn . (9.41)
v − m − n
Using the values for b and c obtained above, we can now write Eq. 9.38
as the following:
0 ≤ (s − 1)(m + n + s 2 − 1)mn + 2mnΣ − (1 + s)Σ 2 . (9.42)
Equality in Eq. 9.42 gives two roots Σ1 and Σ2 for which Eq. 9.42 says
Σ1 ≤ Σ ≤ Σ2, if Σ1 ≤ Σ2. But Σ2 is easiy evaluated.
Σ2 = mn + m2n2 + (s2 − 1)(m + n)mn + (s2 − 1) 2mn . (9.43)
1 + s
Clearly Σ ≤ Σ2 is just the inequality of the theorem.

9.7. THE INCIDENCE MATRIX 407
We now use Theorem 9.7.3 to investigate the case of equality in Theorem
9.7.4 in the special case that each ki = m, i.e., each xi is collinear with
each yj. Here Σ = mn, and Eq. 9.43 along with Σ ≤ Σ2 reduces to
(m − 1)(n − 1) ≤ s 2 . (9.44)
Suppose, moreover, that (m − 1)(n − 1) = s 2 , so −s is an eigenvalue of
A ∆ . Hence a nonzero eigenvector of A ∆ belonging to −s must span the null
space of A ∆ + sI.
⎛
A ∆ + sI = ⎝
m − 1 + s 0 s 2 t + 1 − m
0 n − 1 + s s 2 t + 1 − n
∗ ∗ ∗
where we need not bother to calculate the third row, since the rank must
equal 2. Clearly y = (y1, y2, 1) T spans the null space of A ∆ + sI, where
y1 =
⎞
⎠ ,
m − 1 − st
s + m − 1 ; y2 = n − 1 − s2t . (9.45)
s + n − 1
Let us assume that the pont of P are ordered (for the construction of A)
so that the first m points are those of X, the next n points are those of Y ,
and the last v − m − n ponts are those of Z. Then by Theorem 22.1.3 x must
be an eigenvector of A belonging to λ1 = −s, where x is as follows:
x = (y1, . . . , y1, y2, . . . , y2, 1, . . . , 1) T , (9.46)
where y1 appears m times, y2 appears n times, and 1 appears v − m − n
times.
For the first m + n rows of A this yields no new information. But let
z ∈ Z be the i th pont, i > m + n. Suppose z is not collinear with k1 points
of X, is not collinear with k2 points of Y , and hence is not collinear with
s 2 t − k1 − k2 points of Z. Then the product of the i th row of A with x, which
must equal −s, is actually
k1y1 + k2y2 + s 2 t − k1 − k2 = −s.
After a little simplification this becomes:
k1
s + m − 1 +
k2
= 1. (9.47)
s + n − 1

408 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
If z lies on a line joining a point of X with a point of Y , then k1 = m − 1
and k2 = n−1, i.e., since S has no triangles z is collinear with a unique point
of X and with a unique point of Y . On the other hand, if z is not on such
a line, either k1 = m or k2 = n. Suppose k1 = m, so z is collinear with no
point of X. Using Eq. 9.47 we find that the number of points of Y collinear
with z is
n − 1
n − k2 = 1 + . (9.48)
s
Similarly, any point of P collinear with no point of Y must be collinear
with 1 + m−1 points of X. If m = n = s + 1, this says each point not on a
s
line joining a point of X with a point of Y must be collinear with two points
of X and none of Y or must be collinear with two points of Y and none of
X. If 1 < m < s + 1, so 1 + m−1 is not an integer, then each point of P is
s
collinear with some point of Y . This implies that each point of P is either
on a line joining points of X and Y or is collinear with 1 + n−1
s
≥ 3 points of
Y . Clearly n ≤ 1 + t. Suppose n < 1 + t and let x1 ∈ X. Then there is some
line L through x1 not incident with any point of Y . But then any point z on
L, z = x1, cannot be collinear with any point of Y , a contradiction. Hence
it must be that n = 1 + t, from which it follows that m = 1 + s2
t .
This completes a proof of the following for a GQ S = (P, B, I) of order
(s, t).
Theorem 9.7.5. Let X = {x1, . . . , xm} and Y = {y1, . . . , yn} each be sets
of pairwise noncollinear points with each point of X collinear with each point
of Y , m ≥ 2, n ≥ 2. Then (m − 1)(n − 1) ≤ s2 . If equality holds, then one
of the following must occur:
(i) m = n = 1 + s, and each point of Z = P \ (X ∪ Y ) is collinear with
precisely two points of X ∪ Y .
(ii) m = n. If m < n, then s divides t, s < t, n = 1 + t, m = 1 + s2 , each t
point of S is collinear with either 1 or 1 + t
points of Y according as it is or
s
is not collinear with some point of X.
Note: (m − 1) divides s.
9.8 Regularity: Variations on a Theme
Let S = (P, B, I) be a GQ of order (s, t), with 1 ≤ s and 1 ≤ t. As before,
|P |= v, |B |= b. Let x, y be distinct points. There are several variations on

9.8. REGULARITY: VARIATIONS ON A THEME 409
the basic notion of regularity that was introduced earlier. The known GQ
have many points or point pairs that have some or all of the basic regularity
properties, so we want to consider the relationships between these various
notions, the restrictions forced on the parameters s, t by them, and the
associated structures that can often be constructed as a result of assuming
that one or more these conditions hold. For the convenience of the reader we
start by recalling the most basic notion, that of regularity.
regular The pair {x, y} of distinct points is said to be regular provided either
x ∼ y or x ∼ y and |{x, y} ⊥⊥ |= t + 1. The point x is regular provided
{x, z} is regular for all points z, z = x.The point x is coregular provided
each line incident with x is regular.
antiregular The pair {x, y} is antiregular provided x ⊥ y and |z ⊥ ∩{x, y} ⊥ |≤
2 for all z ∈ P \ {x, y}. The point x is antiregular provided (x, z) is
antiregular for all z ∈ P \ x ⊥ . The definition of “antiregular” was
designed to capture the property of failing to be regular as strongly as
possible. When s = t, this notion has turned out to be quite useful.
When s = t, however, it seems less useful.
net A (finite) net of order k (k ≥ 2) and degree r (r ≥ 2) is a point-line
incidence structure N = (P, B, I) satisfying the following properties:
1. Each point is incident with r lines, and two distinct points are
incident with at most one line.
2. Each line is incident with k points and two distinct lines are incident
with at most one point.
3. If p is a point and L is a line not incident with p, then there is a
unique line M incident with p and not concurrent with L.
A net of order k and degree r has k 2 points and kr lines.
Let x, y be a pair of noncollinear points, and recall the notation of Section
9.3. It is almost trivial to see that {x, y} is regular iff some pair {z, w}
of distinct points of {x, y} ⊥ is regular iff every pair {z, w} of distinct points
of {x, y} ⊥ is regular. It follows immediately that if x is regular then each
pair {u, v} of noncollinear points of x ⊥ is regular. (We shall see later that
there is an analogous result for antiregularity, but it requires a substantial
proof!)

410 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
Theorem 9.8.1. 1. The pair {x, y} is regular iff each triad {x, y, z} has
exactly 0, 1 or t + 1 centers. When s = t this is iff each triad {x, y, z}
is centric.
2. If the pair {x, y} is regular, then s = 1 or t ≤ s.
3. If s = t and each point in x ⊥ \ {x} is regular, then every point of S is
regular.
4. If s > 1, t > 1 and S has a regular point x and a regular pair {L0, L1}
of nonconcurrent lines for which x is incident with no line of {L0, L1} ⊥ ,
then s = t is even.
Proof. The first statement in Part 1 is just the definition of regular. The
second statemtnt of Part 1 and also all of Part 2 follow immediately from
Corollary 9.3.7. We remark that for s = 1 any pair of points is regular, and
that for t = 1 any pair of points is regular and any noncollinear pair of points
is antiregular.
Now let s = t and assume that each point in x ⊥ \ {x} is regular. Let
y ⊥ x and z1, z2 ∈ {x, y} ⊥ , z1 = z2. Since {z1, z2} is regular, clearly {x, y}is
regular. Hence x is regular. To complete the proof that each point is regular,
it suffices to show that if {x, u, u ′ } is a triad, then {u, u ′ } is regular. But
since x is regular, by Part 2 there is some point z ∈ {x, u, u ′ } ⊥ . By the
regularity of z, for any point z ′ ∈ {u, u ′ } ⊥ \ {z}, the pair (z, z ′ ) is regular,
forcing (u, u ′ ) to be regular. This completes the proof of Part 3.
Next suppose that x and {L0, L1} satisfy the hypotheses of Part 4, so
that by Part 2 of this Theorem we have s = t. If {L0, L1} ⊥⊥ = {L0, . . . , Ls},
then let yi be defined by x ⊥ yi | Li, i = 0, . . . , s. By Part 4 the elements
x, y0, y1, . . . , ys are s+2 points of the projective plane πx of order s defined by
x. It is easy to see that each line of πx through x contains exactly one point
of the set {y0, . . . , ys}. Suppose that the points yi, yj, yk, with i, j, k distinct,
are collinear in the plane πx. Then the triad {yi, yj, yk} has s+1 centers. Let
uj (resp., uk) be the point incident with Lj (resp., Lk) and collinear with yi.
Then uk ∈ {yi, yk} ⊥ , hence uk ⊥ yj, giving a triangle with vertices yj, uk, uj.
Consequently {y0, . . . , ys} is an oval of the plane πx. Since the s + 1 tangents
of that oval concur at x, the order s of πx is even.
Theorem 9.8.2. If the point x is coregular, then the number of centers of
any triad {x, y, z} has the same parity as does 1 + t.

9.8. REGULARITY: VARIATIONS ON A THEME 411
Proof. Assume that x is coregular. Let u1, . . . , um be all the centers of a
triad (x, y, z) with {x, y} ⊥ = {u1, . . . , um, um+1, . . . , ut+1}. We may suppose
m < t + 1. For i > m, let Li be the line through x and ui and Mi the
line through y and ui. Let K be the line through z meeting Li and N the
line through z meeting Mi. Let M be the line through y meeting K, and
L the line through x meeting N. Since the line Li through x is regular,
the pair (Li, N) must be regular, and it follows that M must meet L in
some point ui ′ ∈ {x, y}⊥ , m + 1 ≤ i ′ ≤ t + 1, i ′ = i. In this way with each
point ui ∈ {um+1, . . . , ut+1} there corresponds a point ui ′ ∈ {um+1, . . . , ut+1},
i = i ′ , and clearly this correspondence is involutory. Hence the number of
points of {x, y} ⊥ that are not centers of (x, y, z) is even.
Theorem 9.8.3. Coregularity plus regularity with s = t implies s is even.
1. If x is coregular and s = t, then x is regular iff s is even.
2. If each point of S is regular, then (t + 1) | (s 2 − 1)s 2 .
3. If S has a regular point x and a regular line L (with x |L), then s = t
is even.
4. If s = t is odd and if S contains two regular points, then S is not
self-dual.
Proof. Clearly Part 1 follows immediately from Theorem 9.8.2 and from
Corollary 9.3.9.
Now suppose that each point of S is regular. The number of hyperbolic
lines of S equals (1 + s)(1 + st)s 2 t/(t + 1)t. Hence (t + 1) | (1 + s)(1 + st)s 2 .
Since (1 + s)(1 + st)s 2 = (1 + s)(1 + s(t + 1) − s)s 2 , this divisibility condition
is equavilent to (t + 1) | (1 + s)(1 − s)s 2 , proving Part 2.
Suppose S has a regular point x and a regular line L not incident with
x. It is easy to construct a line Z, Z ∼ L, such that x is incident with no
line of {L, Z} ⊥ . Then from Part 4 of Theorem 9.8.1 it follows that s = t is
even, proving Part 3.
So suppose that s = t is odd and that S contains two regular points x
and y. If S admits an anti-automorphism θ, then x θ and y θ are regular lines.
Since at least one of these lines is not incident with at least one of x and y,
an application of Part 3 finishes the proof of Part 4.

412 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
Theorem 9.8.4. Antiregularity is equivalent to no unicentered triad, when
s = t.
1. If t ≤ s, then n1 = 0 iff either t = 1 or s = t and {x, y} is antiregular.
Hence for s = t the pair {x, y} is antiregular iff each triad {x, y, z} has
0 or 2 centers iff no triad {x, y, z} has a unique center iff (by part 1 of
Theorem 9.6.6) each triad contained in x ⊥ has exactly two centers.
2. If (x, y) is antiregular with s = t, then s is odd.
3. If x is coregular and s = t, then x is antiregular iff s is odd.
4. If x is coregular and t is odd, then | {x, y} ⊥⊥ |= 2 for all y ∈ x ⊥ .
Proof. Part 1 follows from Corollaries 9.3.8 and 9.3.12. Let (x, y) be antiregular
with s = t > 1 and {x, y} ⊥ = {u0, . . . , us}. For i = 0, 1, let xILiIuiIMiIy,
and let K ∈ {L0, M1} ⊥ , L1 = K = M0. The points of K not collinear
with x or y are denoted v2, . . . , vs. Let ui ∼ vj for some i, 2 ≤ i. Then
{x, y, vj} is a triad with center ui, and hence, by Part 1, with exactly one
other center ui ′. It follows that u2, . . . , us occur in pairs as centers of triads
of the form {x, y, vj}, each pair being uniquely determined by either of its
members. Hence s − 1 is even, and Part 2 is proved.
For Part 3 first note that if s = t and x is antiregular, then s is odd by
Part 2 of Theorem 9.8.4.
For Part 4, suppose that x is coregular, and y is an arbitrary point not
collinear with x. If z ∈ {x, y} ⊥⊥ \ {x, y}, and if z ′ Izu, z ′ ∈ {z, u}, for some
u ∈ {x, y} ⊥ , then u is the unique center of (x, y, z ′ ). Hence t is even by
Theorem 9.8.2.
Theorem 9.8.5. Nets (resp., projective planes) arise from regular (resp.,
antiregular) points.
1. Let x be a regular point of S. Then the incidence structure with pointset
x ⊥ \ {x}, with lineset the set of spans {y, z} ⊥⊥ , where y, z ∈ x ⊥ \ {x},
y not collinear with z, and with the natural incidence, is the dual of a
net of order s and degree t + 1. If in particular 1 < s = t, there arises
a dual affine plane of order s. Moreover, in this case the incidence
structure πx with pointset x ⊥ , with lineset the set of spans {y, z} ⊥⊥ ,
where y, z ∈ x ⊥ , y = z, and with the natural incidence, is a projective
plane of order s.

9.9. BAGCHI, BROUWER AND WILBRINK 413
2. Let x be an antiregular point of S with order s, s > 1.Let y ∈ x ⊥ \ {x}
with L being the line xy. A projective plane π(x, y) of order s may be
constructed as follows.
Points of π(x, y) are of three types:
(i) points of x ⊥ that are not on L;
(ii) lines through y different from L;
(iii) the point x.
Lines of π(x, y) are also of three types:
(a) points of y ⊥ that are not on L;
(b) lines through x different from L;
(c) the point y.
Incidence is defined as follows: A point x of type (i) is incident with a
line y of type (a) iff x ⊥ y in S, and is incident with a line L of type
(b) iff x|L in S. A point M of type (ii) is incident with the s lines of
type (a) with which it is incident in S and with the line y of type (c).
The point x of type (iii) is incident with all lines of type (b) and with
the line y of type (c). There are no other incidences.
Proof. We leave the proof of Part 1 as an easy exercise and give a few remarks
about the proof of Part 2.
So assume that x is antiregular and y ∈ x ⊥ \ {x}. If z and w are any two
noncollinear points of x ⊥ \y ⊥ , then {y, z, w} is a triad with x as center. Since
x is antiregular, each triad contained in x ⊥ has exactly two centers by Part
1 of Theorem 9.8.4. So there is a unique point u of y ⊥ \ x ⊥ which is collinear
in S with both z and w. Hence u is the unique line of π(x, y) through z and
w. The remaining details are easy.
9.9 Bagchi, Brouwer and Wilbrink
The main result of this section says that if a finite GQ S = (P, B, I) of order
s has an antiregular point (in which case s must be odd), then each point of
S must be antiregular. This is due to B. Bagchi, A. E. Brouwer and H. A.
Wilbrink [BBW91]

414 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
Let S = (P, B, I) be a GQ of order (s, t), s > 1, t > 1. Let V be the
vector space over F2 with basis P. Clearly the dimension of V over F2 is
dim(V ) = (1 + s)(1 + st).
Define π : V → V to be the linear map satisfying π : x ↦→ Γ(x) = x⊥ \{x}.
For a ∈ P, πa denotes π(a). For A ⊆ P, πA =
a∈A πa. The first lemma is
self-evident.
Lemma 9.9.1. For A ⊆ P, πA is the sum of all the points x ∈ P for which
|Γ(x) ∩ A| is odd.
For z ∈ P, put VZ = 〈x : x ∈ Z ⊥ 〉. So dim(VZ) = 1 + s + st. Define
PP = 〈πx : x ∈ P〉 and PZ = 〈πx : x ∈ Z ⊥ 〉, and KZ = 〈A ⊆ Z ⊥ : πA = 0〉.
Lemma 9.9.2. π : VZ ↦→ PZ is onto with kernel KZ, so dim(PZ) = 1 + s +
st − dim(KZ).
Proof. Immediate.
Lemma 9.9.3. Let Y, Z be distinct, noncollinear points. π Y ⊥ ∩Z ⊥ = πY +
πZ + {x : (x, y, z) is a triad with an odd number of centers.
Proof. Clearly if X ∈ Y ⊥ ∩ Z ⊥ , then X ∈ π Y ⊥ ∩Z ⊥, and X ∈ πy + πz if
s − 1 is even. And if X ∈ Γ(Y ) \ Γ(z), then X is collinear with a unique
point of y ⊥ ∩ z ⊥ . So (Γ(Y ) − Γ(Z)) ∪ (Γ(Z) − Γ(Y ) = πY + πZ ⊆ π Y ⊥. For
X ∈ Y ⊥ ∪ Z ⊥ , (X, Y, Z) is a triad and X is in π y ⊥ ∩z ⊥ if and only if (X, Y, Z)
has an odd number of centers. Finally, for X = Y or X = Z, X ∈ π y ⊥ ∩z ⊥
if and only if 1 + t is odd. But if X is not in the right hand side, then
(LHS)X = (RHS)X if and only if t is odd.
Lemma 9.9.4. Suppose that s is odd and fix Z ∈ P. Then dim(KZ) = t, so
dim(PZ = (s − 1)(t + 1) + 2.
Proof. We must first determine for which A ⊆ Z ⊥ it is true that πA = 0.
For each line ℓ through Z put ℓ ′ = ℓ ∗ \ {Z}. So suppose that ∅ = A ⊆ Z ⊥
and πA = 0. Suppose that some line ℓ through Z has points y, w with
y ∈ ℓ ′ ∩ A, w ∈ ℓ ′ \ A. Put e = 1 if Z ∈ A, e = 0 if Z ∈ A. Then
0 = (πA)Y ≡2 |(ℓ ′ ∩ A) \ {Y }| + e, and 0 = (πA)w ≡2 |ℓ ′ ∩ A| + e. This
impossibility shows that ℓ ′ ⊆ A or ℓ ′ ∩ A = ∅.
Since s is odd, 0 = (πA)Z ≡2 |{ℓ ′ : ℓ ′ ⊆ A}|, implying that A contains
i = j, ℓ ′ i ∪ ℓ′ j = D satisfies πD = 0. One way to see this is: s odd implies

9.9. BAGCHI, BROUWER AND WILBRINK 415
πℓi = P. So πℓi+ℓj = π ℓ ′ i +ℓ′ j
= πℓi + πℓj = P + P = 0. Starting with any
A ⊆ Z ⊥ for which πA = 0, we may add a pair of lines (if necessary) so as to
suppose that some particular point W ∈ Z ⊥ \ {Z} satisfies W ∈ A. Then
the only possible point of A collinear with W would be Z. So if Z ∈ A,
(πA)W = 1, an impossibility. Hence Z ∈ A. Since Z is not in the sum of two
lines through Z, Znot ∈ A for every A ⊆ Z ⊥ with πA = 0. It is now easy to
see that
{ℓ0 + ℓ1, ℓ0 + ℓ2, . . . , ℓ0 + ℓt}
is a basis for KZ.
Lemma 9.9.5. Assume s is odd and let Z be any point of P. Then dim(P) ≥
(s − 1)(t + 1) + 2, with equality if and only if all triads (X, Y, Z) have an
even number of centers if and only if each triad of points of P has an even
number of centers.
Proof. Clearly P ⊇ PZ, so dim(P) ≥ dim(PZ) = (s − 1)(t + 1) + 2, with
equality if and only if for each y ∈ P \ Z ⊥ there is some Y ⊆ Z ⊥ with
πy = πY . First, if all triads (z, x, y) have an even number of centers, then
for any y ∈ P \ z ⊥ by Lemma 9.9.3 we have πy = π y ⊥ ∩z ⊥ + πz = πY , with
Y = {z} ∪ (y ⊥ ∩ z ⊥ ). So P = Pz and dim(P) = (s − 1)(t + 1) + 2.
Conversely, suppose dim(P) = (s − 1)(t + 1) + 2. Let y ∈ P \ z ⊥ . There
must be some Y ⊆ z ⊥ with πy = πY . Suppose zIℓIIm ∼ y. We may add
the sum of ℓ and one other line through z to Y to obtain Y0 ⊆ z ⊥ with
πY0 = πY = πy. So we may assume m ∈ Y . Then let n be any point of ℓ ′
with n = m. If n ∈ Y , we would have
1 = (πy)m = (πY )n = (πy)n = 0,
where (πY )m = (πY )n because m and n are collinear with the same number
of points of Y , all on ℓ. This impossibility shows that if m ∈ Y , then n ∈ Y .
It follows readily that for each line ℓ through z, either ℓ ′ ∩ Y = {m} or
ℓ ′ ∩ Y = ℓ ′ \ {m}.
So we may assume that Y ∩ ℓ ′ i = {mi}, 0 ≤ i ≤ t − 1, and Y ∩ ℓ ′ t = {mt}
or Y ∩ ℓ ′ t = ℓ ′ t \ {mt}. So Y ⊆ 〈y⊥ ∩ z⊥ , ℓt, z〉. But since πℓt = 1, we may
write πy = πY = 〈πy⊥∩z ⊥, 1, πz〉.
Suppose πy = aπy⊥∩z ⊥ + bπz + c · 1, and consider the coefficients on points
p, m, n where zIℓIm, n ∈ ℓ ′ \ {m}, mIKIy, KIp, m = p = y.

416 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
(πy)p = 1 (π y ⊥ ∩z ⊥)p = 1 (πz)p = 0 1p = 1
(πy)m = 1 (π y ⊥ ∩z ⊥)m = 0 (πz)m = 1 1m = 1
(πy)m = 0 (π y ⊥ ∩z ⊥)n = 1 (πz)n = 1 1n = 1.
The top row of equalities sayw 1 = a+c; the second row implies 1 = b+c;
and the third row says 0 = a + b + c. The first row minus the second says
a = b, so 0 = a + b implies c = 0, and then a = b = 1. So πy = π y ⊥ ∩z ⊥ + πz.
By Lemma 9.9.3, each triad (z, x, y) has an even number of centers.
Theorem 9.9.6. Let s be odd and suppose each triad of ponts has an even
number of centers. Then t is odd and t ≥ s. If s = t, then each triad of
points has 0 or 2 centers.
Proof. Fix y, z ∈ P, y ∼ z, y ⊥ ∩ z ⊥ = {m0, m1, . . . , mt}. Let L be a line
through m0 not through y or z, and suppose that x − 1, x2, ldots, xs are the
points of L ∗ \ {m0}. So (xi, y, z) is a triad with m0 as one center. Moreover,
△i = x ⊥ i ∩y ⊥ ∩z ⊥ \{m0}, 1 ≤ i ≤ s, partitions {m1, m2, . . . , mt} into an odd
number of sets, each having an odd number of elements. So t must be odd.
And t ≥ s, with equality if and only if each centric triad (x, y, z) has exactly
2 centers. So if s = t, then each triad of points has 0 or 2 centers.
Corollary 9.9.7. If s = t and some point is antiregular, then all points are
antiregular (and s is odd by Theorem 9.8.4, part 2.).
Corollary 9.9.8. If s = t is odd and some point x is coregular (i.e., each
line through x is regular), then all points are antiregular (by Theorem 9.8.4,
part 3.).
9.10 Ovoids, Spreads and Polarities
Recall the notion of k-arc, and note that the number of lines through the
points of a k-arc is (1+t)k, which must be less than or equal to (1+t)(1+st).
It follows that k ≤ 1 + st, with equality if and only if the k-arc effects a
partition of the lines. We give these absolutely maximal k-arcs a name.
ovoid An ovoid of S is a set O of points of S such that each line of S is
incident with a unique point of O. Clearly an ovoid is just a k-arc with
|k| = 1 + st.

9.10. OVOIDS, SPREADS AND POLARITIES 417
spread, larc Dually, a spread of S is a set M of lines of S for each each
point of S is incident with a unique line of S. The point-line dual
notion to that of arc apparently does not have a standard name, but
we shall use the term line-arc or larc. So a k-larc is a set of k lines, no
two concurrent. Clearly a spread is just a (1 + st)-larc.
polarity Suppose s = t and let ρP : P → B and ρB : B → P be bijections
with the following properties: (i) x and y are collinear points if and
only if ρp(x) and ρp(y) are concurrent lines.
(ii) L and M are concurrent lines if and only if ρB(L) and ρB(M) are
collinear points.
(iii) The maps ρP and ρB are inverses of each other.
Then the map ρ : P ∪ B → B ∪ P defined so that the restriction of ρ
to P (respectively, B) is just ρP (respectively, ρB), is called a polarity
of S. We often denote the image of a point x under a polarity ρ by
x ρ . Then x is called an absolute point with respect to ρ provided xIx ρ .
Absolute lines are defined dually.
duality If s = t and ρP : P → B satisfies (i), then ρP induces a unique map
ρB : B → P that satisfies (ii), and ρ defined as above is an isomorphism
between S and its point-line dual S × . The map ρ is called a duality.
A great deal is known about which GQ have ovoids and/or spreads, however,
there still are some open problems. When it is known that a GQ does
not have an ovoid, for example, it usually is not known just how large k
can be for which a k-arc exists. The following argument, which is the one
originally used by E. Shult [SS79], shows that no GQ with t = s 2 can have
an ovoid. This latter result was also proved by J. A. Thas [Th81].
Theorem 9.10.1. If s > 1 and t > s 2 − s, then S has no ovoid.
Proof. Let O be an ovoid of S and suppose that s > 1. Fix a point x ∈ O
and let V = {y ∈ O : y ∼ x}. Let Z = {z1, . . . , zd}, d = t(s2 − s + 1), be the
set of points not in O and not collinear with x. For each i, 1 ≤ i ≤ d, let ti
be the number of points of V collinear with zi. If we count the ordered pairs
(y, z) for which y ∈ V , z ∈ Z and y ∼ z, we find that d i=1 ti = (1 + t)ts. If
we cound the ordered triples (y1, y2, z) ∈ V ×V ×Z for which y1 ∼ z ∼ y2 and
y1 ∼ y2, we find that
i ti(ti − 1) = (1 + t)t 2 . Since d
i t2 i
− (
i ti) 2 ≥ 0,

418 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
it follows that t(s 2 − s + 1)(1 + t)t(t + s) − (1 + t) 2 t 2 s 2 ≥ 0. After a little
simplification we find (s − 1)(s 2 − s − t) ≥ 0, from which t ≤ s 2 − s.
Theorem 9.10.2. Let s = t and suppose that S has a regular pair (x, y) of
noncollinear points. If O is an ovoid of S, then |O∩{x, y} ⊥⊥ |, |O∩{x, y} ⊥ | ∈
{0, 2}, and O ∩ ({x, y} ⊥ ∪ {x, y} ⊥⊥ )| = 2. If s = 1 and S has a regular point
not on O, then s is even.
Proof. Let O ∩ ({x, y} ⊥ ∪ {x, y} ⊥⊥ ) = {y1, . . . , yr}. If u ∈ P \ ({x, y} ⊥ ∪
{x, y} ⊥⊥ ), then u is on just one line joining a point of {x, y} ⊥ to a point of
{x, y} ⊥⊥ . We count the number of pairs (L, u), with u a point of O which
is incident with L. We obtain (s + 1) 2 = s 2 + 1 − r + r(1 + s). Hence
r = 2. Since no two points of O are collinear, it must be that |O ∩ {x, y} ⊥ |,
|O ∩ {x, y} ⊥⊥ | ∈ {0, 2}.
Let z be a regular point of S not on the ovoid O, and keep in mind that
we are assuming s = t. Let y ∈ O, z ∼ y = z. The points of O collinear with
y are denoted by z0, . . . , zs, with z0Izy. By the first part of the theorem, for
each i = 1, . . . , s, {z, zi} ⊥⊥ ∩O = {zi, zj} for some j = i. Hence |{z1, . . . , zs}|
is even, proving the second part of the theorem.
There is an immediate corollary.
Corollary 9.10.3. If a GQ of order s, s even, has a regular pair of noncollinear
points, then its pointset cannot be partitioned into ovoids.
Proof. Let {x, y} be a regular pair of noncollinear points of the GQ S of
order s. By Theorem 9.9.2 |{x, y} ⊥⊥ | is even, forcing s to be odd.
Theorem 9.10.4. If s = t and S admits a polarity ρ, then 2s is a square.
Moreover, the set of all absolute points (resp., lines) is an ovoid (resp.,
spread) of S.
Proof. Let ρ be a polarity of S. We first prove that each line of S is incident
with at most one absolute point of ρ. Let x and y be distinct absolute points
incident with the line L. Then xIx ρ , yIy ρ , and x ρ ∼ y ρ since x ∼ y. Hence
L ∈ {x ρ , y ρ }, since otherwise there arises a triangle with sides L, x ρ , y ρ . So
suppose L = x ρ . As yIx ρ , we have xIy ρ . Since yIy ρ , we have y ρ = xy =
L = x ρ , implying x = y, a contradiction. Hence each line of S is incident
with at most one absolute point. Dually, each point is incident with at most
one absolute line. Also note that L is absolute if and only if LIL ρ if and

9.11. SUBQUADRANGLES 419
only if L ρ is absolute. This shows that any absolute line is incident with
a (necessarily unique) absolute point. Now let L be a non-absolute line, so
L IL ρ . If L ρ IMIuIL, then L ρ Iu ρ IM ρ IL. Hence u ρ is the line through L ρ
concurrent with L, i.e., u ρ = M, and M ρ = u. Consequently u and M are
absolute, proving that each line (resp., point) is incident with a (necessarily
unique) absolute point (resp., line). It follows that the set of absolute points
is an ovoid of S, and that dually the set of absolute lines is a spread.
Denote the absolute points of ρ by x1, . . . , xs2 +1. It is clear that the
absolute lines of ρ are the images x ρ
i = Li, 1 ≤ i ≤ s2 + 1. Suppose that the
points and lines of S are ordered so that P = {x1, . . . , xs2 +1, . . . , xv},
B = {L1, . . . , Ls2 +1, . . . , Lb}, with Li = x ρ
i , for 1 ≤ i ≤ v = b. It follows
that for all i, j, 1 ≤ i, j ≤ v, xiILj if and only if LiIxj. Hence the linepoint
incidence matrix A defined in Section 9.7 is symmetric and by the
proof of Theorem 9.7.1 A2 has eigenvalues θi with multiplicity di, as follows:
θ0 = (1+s) 2 , d0 = 1; θ1 = 0, d1 = (1+s2 )s
; θ2 = 2s, d2 = 2
(1+s)2s . Since A has
2
constant row sum equal to s + 1, it clearly has 1 + s as an eigenvalue. Hence
A has eigenvalues s + 1, 0, √ 2s, and − √ 2s, with respective multiplicities
1, s(s 2 +1)/2, a1 and a2, where a1 +a2 = s(s+1) 2 /2. Consequently tr(D) =
s + 1 + (a1 − a2) √ 2s. But tr(D) is also the number of absolute points of ρ,
i.e., tr(D) = 1 + s 2 . So s 2 + 1 = s + 1 + (a1 − a2) √ 2s, implying that 2s is a
square.
There is a related result by W. Haemers [Ha03] which says that if there
is a self-dual GQ with order s ≡ 2 (mod 4), then 2s is a square.
9.11 Subquadrangles
The GQ = (P ′ , B ′ , I ′ ) of order (s ′ , t ′ ) is called a subquadrangle of S provided
P ′ ⊆ P , B ′ ⊆ B and I ′ is the restriction of I to P ′ × B ′ ∪ B ′ × P ′ . If S ′ = S,
we say S ′ is a proper subquadrangle of S. From |P | = |P ′ | it follows readily
that s ′ = s and t ′ = t, and hence S ′ = S. Similarly, if |B ′ | = |B|, then
S ′ = S.

420 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
Let L ∈ B. Then precisely one of the following occurs:
(i) L ∈ B ′ , i.e., L belongs to S ′ ;
(ii) L ∈ B ′ and L is incident with a unique point x of S ′ ;
i.e., L is tangent to S ′ at x;
(iii) L ∈ B ′ and L is incident with no point of S ′ ;
i.e., L is external to S ′ .
Dually, one may define tangent point and external point of S. From the
definition of a GQ it follows easily that no tangent point can be incident with
a tangent line.
Theorem 9.7.3 deals with the simplest possible application of the “interlacing
of eigenvalues” techniques, but it has an interesting application as
follows.
Theorem 9.11.1. Let S ′ be a subquadrangle of S having parameters (s ′ , t ′ ).
Clearly 1 ≤ s ′ ≤ s and 1 ≤ t ′ ≤ t. Then either s = s ′ or s ≥ s ′ t ′ . Dually,
t = t ′ or t ≥ s ′ t ′ . Moreover, if s = s ′ , then each point of S \ S ′ is collinear
with exactly the 1 + st ′ points of an ovoid of S ′ . Dually, if t = t ′ , then each
line of S \ S ′ is concurrent with the 1 + s ′ t lines of a spread of S ′ . If s = s ′ t ′ ,
then each line of S \ S ′ is concurrent with 1 + s ′ lines of S ′ .
Proof. Let ∆1 be the set of indices of the points of S ′ , and let ∆2 be the set
of the remaining indices of points. So δ1 = 1 + s ′ + s ′ t ′ ≤ s + (1+s′ )(1+s ′ t ′ )
. 1+s
This is equivalent to 0 ≤ (s − s ′ t ′ )(s − s ′ ), which proves the main inequality.
If s = s ′ , then any line of S with at least two points of S ′ must have all of its
points in S ′ . By Theorem 9.7.3, a point x outside S ′ must be collinear with
1 + s ′ t ′ points inside S ′ , with at most one on each line through x. The ovoid
of S ′ obtained this way is said to be subtended by x. The remainder of the
theorem is obtained by point-line duality.
The previous proof was given by S. E. Payne [Pa73], but see FGQ for
references to works of J. A. Thas with simple combinatorial proofs of this
result as well as of those that follow.
Theorem 9.11.2. Let S ′ = (P ′ , B ′ , I ′ ) be a proper subquadrangle of S =
(P, B, I), with S having order (s, t) and S ′ having order (s, t ′ ), i.e., s = s ′
and t > t ′ . Then we have:

9.11. SUBQUADRANGLES 421
(i) t ≥ s; if t = s, then t ′ = 1.
(ii) If s > 1, then t ′ ≤ s; if t ′ = s ≥ 2, then t = s 2 .
(iii) If s = 1, then 1 ≤ t ′ < t is the only restriction on t ′ .
(iv) If s > 1 and t ′ > 1, then √ s ≤ t ′ ≤ s, and s 3/2 ≤ t ≤ s 2 .
(v) If t = s 3/2 > 1 and t ′ > 1, then t ′ = √ s.
(vi) Let S ′ have a proper subquadrangle S ′′ of order (s, t ′′ ), s > 1. Then
t ′′ = 1, t ′ = s, and t = s 2 .
Proof. All these are easy consequences of Theorem 9.11.1 and Higman’s inequality.
We give two examples. Start with part (ii). Suppose that s > 1.
By Higman’s inequality we have t ≤ s 2 . Using the point-line dual of Theorem
9.11.1 st ′ = s ′ t ′ ≤ t ≤ s 2 , so t ′ ≤ s. If t ′ = s, then clearly t = s 2 .
Now consider part (vi). Let S ′ have a proper subquadrangle S ′′ of order
(s, t ′′ ), s > 1. Then by part (ii) t ′ ≤ s and by the previous theorem st ′′ ≤ t ′ .
Hence t ′′ ≤ t ′ /s ≤ s/s, implying t ′′ = 1 and t ′ = s. Again using (ii) we have
t = s 2 .
The following theorem can be helpful in recognizing subquadrangles.
Theorem 9.11.3. Let S ′ = (P ′ , B ′ , I ′ ) be a substructure of the GQ S =
(P, B, I) of order (s, t) for which the following conditions are satisfied:
(i) If x, y ∈ P ′ (x = y) and xILIy, then L ∈ B ′ .
(ii) Each element of B ′ is incident with 1 + s elements of P ′ .
Then there are four possibilities:
(a) S ′ is a dual grid (and then s ′ = 1).
(b) The elements of B ′ are lines which are incident with a distinguished
point of P , and P ′ consists of those points of P which are incident with those
lines.
(c) B ′ = ∅ and P ′ is a set of pairwise noncollinear points of P .
(d) S ′ is a subquadrangle of order (s, t ′ ).
Proof. Suppose that S ′ = (P ′ , B ′ , I ′ ) satisfies (i) and (ii) and is not of type
(a), (b) or (c). Then P ′ and B ′ are not empty and s > 1. If L ′ ∈ B ′ , then
there exists a point x ′ ∈ P ′ such that x ′ is not I-incident with L ′ . Let x
and L be defined by x ′ ILIxIL ′ . By (i) and (ii) we have x ∈ P ′ and L ∈ B ′ .
Hence S ′ satisfies Axiom 2 in the definition of GQ. We now show that each

422 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
point is incident with the same number of points. So suppose that x ′ ∈ P ′
and suppose that x ′ is incident with t ′ + 1 lines of B ′ . Since B ′ = ∅, t ′ ≥ 0.
Let y ′ ∈ P ′ be a point which is not collinear with x ′ and suppose it is incident
with t ′′ + 1 lines of B ′ . By Axiom 2 t ′ = t ′′ . Hence t ′ + 1 is the number of
lines of B ′ which are incident with any point not collinear with at least one
of the points x ′ , y ′ . So we consider a point z ′ ∈ P ′ which is in {x ′ , y ′ } ⊥ .
First suppose that t ′ = 0. Let L ′ = x ′ z ′ , L ′′ = y ′ z ′ , and let L ∈ B ′ \
{L ′ , L ′′ }. Then neither x ′ nor y ′ is incident with L. Since there exists a line
of B ′ which is incident with x ′ (resp., y ′ ) and concurrent with L, it follows
that L and L ′ (resp., L and L ′′ ) are concurrent. Hence z ′ IL and S ′ is of type
(b), a contradiction.
Now suppose that t ′ > 0. Consider a line L ∈ B ′ for which x ′ IL and z ′
is not incident with L. On L there is a point u ′ with u ′ ∼ y ′ and u ′ ∼ z ′ .
Then the number of lines of B ′ which are incident with z ′ equal the number
of lines of B ′ which are incident with u ′ , which equals t ′′ + 1 since y ′ ∼ u ′ ,
and hence equals t ′ + 1. We conclude that each point of B ′ is incident with
t ′ + 1 ≥ 2 lines of B ′ . This completes a proof of the theorem.
9.12 Collineations
A collineation θ of a GQ(s, t) = S is formally defined as a pair of permutations
θ = (α, β), where α is a permutation of the points and β is a
permutation of the lines, such that a point x is incident with a line ℓ if and
only if x α is incident with the line ℓ β . However, since there are no triangles
in S, a collineation is completely determined by a permutation of points that
preserves collinearity or a permutation of the lines that preserves concurrency.
So we usually describe a collineation as a permutation of the points
with the induced permutation of lines taken for granted.
Theorem 9.12.1. The substructure Sθ = (Pθ, Bθ, Iθ) of the elements fixed
by θ must be given by at least one of the following:
(i) Bθ = ∅ and Pθ is a set of pairwise noncollinear points.
(i’) Bθ = ∅ and Bθ is a set of pairwise nonconcurrent lines.
(ii) Pθ contains a point x such that x ∼ y for every point y ∈ Pθ, and
each line of Bθ is incident with x.
(ii’) Bθ contains a line L such that L ∼ M for each line M ∈ Bθ, and
each point of Pθ is incident with L.

9.12. COLLINEATIONS 423
(iii) Sθ is a grid.
(iii’) Sθ is a dual grid.
(iv) Sθ is a subquadrangle of order (s ′ , t ′ ), s ′ ≥ 2, t ′ ≥ 2.
Proof. Suppose Sθ is not of any of the first six types. Then Pθ = ∅ = Bθ.
Bθ. Dually, if L, B ∈ Bθ, L = M, L ∼ M, then the point common to
L and M belongs to Pθ. Next, let L ∈ Bθ and consider a point x ∈ Pθ
with x not incident with L. further, let y and M be defined by xIMIyIL.
Then x θ IM θ Iy θ IL θ , i.e., xIM θ Iy θ IL. Hence M = M θ and y = y θ . It
follows that Sθ satisfies Axiom 2 in the definition of GQ. At this point the
proof of Theorem 9.11.3 can be adapted to complete a proof of the present
theorem.
Let S = (P, B, I) be a GQ of order (s, t) with P = {x1, . . . , xv} and
B = {L1, . . . , Lb}. Let A be the b × v incidence matrix given in Section 9.7.
Then A T A = B + (t + 1)I, where B is an adjacency matrix of the point
graph of S. If M = A T A, then (see Eq. 9.32) M has eigenvalues θ0 =
(1 + s)(1 + t); θ1 = 0; θ2 = s + t, with respective multiplicities d0 = 1; d1 =
s 2 (1 + st)/(s + t); θ2 = st(1 + s)(1 + t)/(s + t).
Let θ be an automorphism (i.e., collineation) of S and let Q = (qij)
(resp., R = (rij)) be the v × v matrix (resp., b × b matrix) with qij = 1 if
x θ i = xj (resp., L θ i = Lj) and qij = 0 (resp., rij = 0) otherwise. Then Q
and R are permutation matrices for which AQ = RA (which is equivalent to
A T R = QA T since Q T = Q −1 and R T = R −1 ). It follows that
QM = QA T A = A T RA = A T AQ = MQ.
Theorem 9.12.2. (C.T. Benson [Be70]) If f is the number of points fixed by
the automorphism θ and if g is the number of points x for which x ∼ x θ = x,
then
tr(QM) = (1 + t)f + g and (1 + t)f + g ≡ 1 + st (mod s + t).
Proof. Suppose that θ has order n, so that (QM) n = Q n M n = M n . It
follows that the eigenvalues of QM are the eigenvalues of M multiplied by the
appropriate roots of unity. Since MJ = (1+s)(1+t)J (where in this case J is
the v×v matrix with all entries equal to 1), we have (QM)J = (1+s)(1+t)J,
so (1 + s)(1 + t) is an eigenvalue of QM. Since d0 = 1 it follows that this
eigenvalue of QM has multiplicity 1. Further, it is clear that 0 is an eigenvalue

424 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
of QM with multiplicity d1 = s 2 (1+st)/(s+t). For each divisor d of n, let Ud
denote the sum of all primitive dth (complex) roots of unity. Then Ud is an
integer. For each divisor d of n, the primitive dth roots of unity all contribute
the same number of times to eigenvalues of QM having absolute value equal
to s + t. Let ad denote the multiplicity of ζd(s + t) as an eigenvalue of QM,
with d|n and ζd a primitive dth root of unity. Then we have
tr(QM) =
ad(s + t)Ud + (1 + s)(1 + t).
d|n
Hence tr(QM) ≡ 1 + st (mod s + t). Let f and g be as given in the theorem.
Since the entry on the ith row and ith column opf QM is the number of
lines incident with xi and x θ i , we have tr(QM) = (1 + t)f + g, completing
the proof.
Corollary 9.12.3. If fP (resp., fB) is the number of points (resp., lines)
fixed by the automorphism θ, and if gP (resp., gB) is the number of points x
(resp., lines L) for which x ∼ x θ = x (resp., L ∼ L θ = L), then
tr(QA T A) = (1 + t)fP + gP = (1 + s)fB + gB = tr(RAA T ).
Proof. The last equality of just the dual of the first, which was established
in the preceding theorem. To obtain the middle equality we introduce some
additional notation.
FP = {x ∈ P : x θ = x}; fP = |FP |.
GP = {x ∈ P : x ∼ x θ = x}; gP = |GP |.
For each x ∈ GP there is a unique line Lx = 〈x, x θ 〉. We split GP into two
subsets.
G 1 P = {x ∈ GP : L θ x = Lx}; g 1 P = |G1 P |.
G 2 P = {x ∈ GP : L θ x = Lx}; g 2 P = |G2 P |.
M = {(x, L) ∈ P × B : xIL, x ∼ x θ , L ∼ L θ }.
N = {(x, L) ∈ P × B : xIL, x ∼ x θ = x; L ∼ L θ = L}.
Clearly N ⊆ M.
Suppose that x is a point such that there is a line L with (x, L) ∈ M.
Then x ∈ FP ∪ GP . If x ∈ FP , then (x, L) ∈ M if and only if xIL. So s

9.12. COLLINEATIONS 425
belongs to 1 + t pairs (x, L) ∈ M, none of which are in N . If x ∈ G1 P , then
the unique pair (x, L) ∈ M is (x, Lx), and this pair is not in N . If x ∈ G 2 P ,
then the only pairs (x, L) ∈ M are (x, Lx) and (x, L θ−1
x ), and they are both
in N . Hence we have the following:
|mN| = 2 · g 2 P ;
|M| = (1 + t)fP + g 1 P + 2 · g 2 P
= (1 + t)fP + gP + g 2 P
= (1 + t)fP + gP + 1
|N |.
2
Clearly the dual of this result is: |M| = (1 + s)fB + gB + 1
|N |. So the
2
desired middle equality of the corollary holds.
whorl A whorl about a point p is a collineation of S that fixes each line
incident with p. A whorl about a line is defined dually.
elation Traditionally an elation θ about a point p was defined to be a whorl
about p such that θ = id or such that θ fixes no point of P \ p ⊥ . It
gradually has become clear that a better definition would require that
an elation act semiregularly on the points of P \ p ⊥ , i.e., if some power
of θ fixes a point of P \ p ⊥ , then θ is the identity. (For a thorough
discussion of these ideas see [PT03] and [TP06].) An elation about a
line is defined dually.
(p, l)-elation Let (p, L) ∈ P × B be a flag, i.e., the point p is incident
with the line L. A (p, L)-elation is an automorphism θ of S that is
simultaneously a whorl about p and a whorl about L. The name is
justified, since using Theorem 9.12.1 it is easy to check that a (p, L)elation
is indeed an elation about p and also an elation about L.
Theorem 9.12.4. Let (p, L) be a flag, and let G(p, L) be the group of all
(p, L)-elations. Then G(p, L) has order dividing st.
Proof. For, if N is a line through p different from L, then G(p, L) acts
semiregularly on the st lines meeting N at points other than p. Similarly, if
u is a point of L different from p, then G(p, L) acts semiregularly on the st
points of u ⊥ not on L. Hence, G(p, L) has order st if and only if G(p, L) acts
regularly on the set of line of N ⊥ not through p if and only if G(p, L) acts
regularly on the set of points of u ⊥ not on L.

426 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
(p, L)-transitive If G(p, L) has order st we say S is (p, L)-transitive.
panel A panel of S is an ordered triple (x, L, y) where x, y are distinct points
on L, or an ordered triple (L, p, M) where L and M are distinct lines
incident with p.
(x, L, y)-elation For a given panel (x, L, y), let H(x, L, y) = G(x, L)∩G(y, L),
i.e., θ ∈ J(x, L, y) if and only if x fixes x and y linewise and L pointwise.
An element θ of H(x, L, y) is called an (x, L, y)-elation. Note
that θ is an elation about each of x, L, y.
Moufang panel Clearly the set H(x, L, y) of (x, L, y)-elations is a group.
If M is a line through x different from L, then H(x, L, y) acts semiregularly
on the points of M different from x. Hence |H(x, L, y)| divides
s. If |H(x, L, y)| = s, we say that the panel (x, L, y) is Moufang.
(x, L, y)-symmetry By definition each θ ∈ H(x, L, y) fixes both x and y
linewise. Let M be a line fixed by the nonidentity automorphism θ ∈
H(x, L, y) and suppose that M is not incident with either x or y. An
easy exercise shows that M must meet L at some point z different from
x and y. However, in general we do not know whether or not θ is a whorl
about z. If it is true for each point z incident with L that θ is a whorl
about z whenever θ fixes some line through z different from L, then we
call θ an (x, L, y)-symmetry. It is not clear from the definition that
the set of all (x, L, y)-symmetries is a group or that the group H(x, L, y)
of (x, L, y)-elations is in fact a group of (x, L, y)-symmetries. We shall
study this situation in some detail.
symmetry A symmetry about the line L is an automorphism of S that
fixes each line concurrent with L. Clearly, if θ is a symmetry about L
then it is an (x, L, y)-symmetry for each pair (x, y) of distinct points
incident with L, and hence θ is also an elation about L. The symmetries
about L form a group that acts semiregularly on the set of s points
incident with any line M meeting L but not on L. Hence the group of
symmetries about L has order dividing s. It is also clear that if L is
regular, then any (x, L, y)-elation is an (x, L, y)-symmetry.

9.13. A COSET GEOMETRY 427
9.13 A Coset Geometry
Let G be a group with |G| = s 2 t, s > 1, t > 1. Let F = {A0, . . . , Ar} be a
collection of r + 1 subgroups, each of order s. Let F ∗ = {A ∗ 0, . . . , A ∗ r} be a
family of r + 1 subgroups, each of order st, with the property that Ai ≤ A ∗ i
for 0 ≤ i ≤ r. Construct a coset geometry Q(G, F, F ∗ ) as follows:
Points are of three types:
(1) Elements g of G.
(2) Cosets A∗ i g, 0 ≤ i ≤ r, g ∈ G.
(2) A symbol (∞).
Lines are two types:
(a) Cosets Aig, 0 ≤ i ≤ r, g ∈ G.
(b) Symbols [Ai], 0 ≤ i ≤ r.
Incidence is defined as follows:
• (∞) is incident with each [Ai], 0 ≤ i ≤ r.
• [Ai] is incident with (∞) and with each of the s cosets A ∗ i g of A∗ i
0 ≤ i ≤ r.
in G,
• A ∗ i g is incident with [Ai] and with each of the t cosets of Ai contained
in A ∗ i g.
• Aig is incident with A ∗ i and with each of the s elements of G in Aig.
• g is incident with each coset Aig, 0 ≤ i ≤ r.
We now consider what properties of (G, F, F ∗ ) correspond to properties
of Q(G, F, F ∗ ) that are required for Q(G, F, F ∗ ) to be a GQ (necessarily of
order (s, t)).
We first consider whether or not there are two points on two lines. It will
follow that this cannot happen if and only if the following property holds:
P1. Ai ∩ Aj = {e} for 0 ≤ i < j ≤ r.
(i) It is clear that [Ai] and [Aj] have only the point (∞) in common.

428 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
(ii) [Ai] and Ajg have only A∗ jg in common, and then only when i = j.
(iii) Suppose some two lines Aig and Ajh have two points in common.
Without loss of generality we may suppose that these are the points g and h
(since otherwise i = j and A∗ i g = A∗i h and |Aig∩Aih| ≥ 1 implies Aig = Aih).
This implies gh−1 ∈ Ai ∩ Aj. Hence we have the following: Two points are
on at most one line if and only if Ai ∩ Aj = {e} for 0 ≤ i < j ≤ r. From now
on we suppose that this condition P1 is satisfied.
The next property to consider is that there should be no triangles. It
is easy to check that there is no triangle with (∞) as one of its vertices.
Similarly, there is no triangle with two vertices of the type A ∗ i g, A ∗ i h.
P2. A ∗ i ∩ Aj = {e} if i = j.
Suppose there is a triangle with distinct vertices g, h ∈ G and whose
third vertex is A ∗ i g = A ∗ i h. To have a triangle, the line through g and h must
be of the form Ajg = Ajh, for some i = j. Then gh−1 ∈ Aj ∩ A∗ i . Conversely,
if |Aj ∩ A∗ i | ≥ 2, such a triangle arises. Hence we have the following: There is
no triangle having a vertex on a line of the type [Ai] if and only if property
P2 holds. From now on we suppose that this condition P2 is satisfied.
P3. AiAj ∩ Ak = {e} if i, j, k are distinct.
Suppose that g, h and x are three distinct points of type (1) that are
the vertices of a triangle. Necessarily there are three distinct indices i, j
and k for which Aih = Aix, Ajx = Ajg and Akg = Akh. This forces
hx −1 ∈ Ai, xg −1 ∈ Aj and gh −1 ∈ Aj and gh −1 ∈ Ak, from which it follows
that hg −1 ∈ AiAj ∩ Ak. Conversely, suppose e and x are distinct elements of
AiAj ∩ Ak, with x = ab, a ∈ Ai, b ∈ Aj. Then e, a and b −1 are the vertices
of a triangle with sides Aia = Aie, Aje = Ajb −1 and Aka = Akb −1 . It follows
that no triangle exists with vertices of type (1) if and only if property P3
holds. From now on we suppose that this condition P3 is satisfied.
Let p be a point not incident with a line L. We now consider whether or
not there really is a point x incident with L and collinear with p.
P4. G = A∗ i Aj for i = j.
The point (∞) is always collinear with the point A∗ i g on an arbitrary line
Aig not incident with (∞). Consider the case where p = A∗ i g and L = Ajh. If
i = j, then A∗ i h is collinear with A∗ i g on [Ai. So suppose i = j. A point x on
L collinear with p would have to be an element x ∈ G for which Ajh = Ajx
and A∗ i g = A∗i x, i.e., x ∈ A∗i g ∩ Ajh. This says that for any g, h ∈ G, there

9.13. A COSET GEOMETRY 429
would have to be elements a∗ i ∈ A∗i and aj ∈ Aj for which a∗ i g = ajh, which
is equivalent to gh−1 = (a∗ i ) −1aj ∈ A∗ i Aj. So our GQ property holds with p
of the form A∗ i g if and only if property P4 holds.
To consider the final case with p = g ∈ G, first define
Ω = ∪{Ai : 0 ≤ i ≤ r}.
Let i = j, 0 ≤ i, j ≤ r. Since P1 and P2 are assumed to hold, each coset
of Ai different from Ai but contained in A ∗ i is disjoint from Aj, and we have
the following
A ∗ i ⊆ Ai ∪ {Aig : Ai ∩ Ω = ∅}.
Clearly g ⊥ A∗ i g for each i, so we need to consider only those cases where
p = g and L = Ajh with g ∈ Ajh. Multiplying on the right by g−1 we may
without loss of generality assume that p = g = e and L = Ajg with g ∈ Aj.
First consider the case where g ∈ A∗ j \ Aj. Here A∗ jg = A∗ j is a point on Ajg
collinear with e on the line Aj. So we may assume that g ∈ A∗ j . So we need
a point x on Aj different from A∗ j collinear with e on some line Ai = Aix,
i = j. Hence there must be an i, i = j, with Ai ∩ Ajg = ∅. Since there are
to be no triangles, in fact |Ai ∩ Ajg| must equal 1. In other words, if Ajg is
a coset of Aj disjoint from Ω it must be the case that Ajg ⊂ A∗ j . Hence, in
the presence of the properties P1 through P4, if p is a point not on a line L,
there will be a (necessarily unique) point x on L collinear with p if and only
if the following property holds:
P5. A ∗ i = Ai ∪ {Aig : Ai ∩ Ω = ∅}.
Clearly P2 implies P1, and since G is finite, a combinatorial argument
shows that P2 implies P4. Hence Q(G, F, F ∗ ) is a GQ if and only if P2, P3
and P5 hold.
For the following three lemmas suppose that (G, F, F ∗ ) satisfies P2 and
P3.
Lemma 9.13.1.
Proof. If g ∈ Ai, the result is true by P2. So suppose g ∈ Ai and let
x, y ∈ Aig ∩ Aj. Say x = aj = aig and y = a ′ j = a′ ig with aj, a ′ j ∈ Aj and
ai, a ′ i ∈ Ai. Then g = a −1
i aj = (a ′ i )−1a ′ j , which implies that a′ ia−1 i = a′ ja−1 j ∈
Ai ∩ Aj = {e}. Hence x = y, proving the lemma.

430 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
Lemma 9.13.2. For g ∈ Ω \ Aj, Ajg ∩ Ω = {g}.
Proof. Clearly g ∈ Ajg ∩ Ω. And g ∈ Ai for a unique i, i = j. By the
previous lemma, |Ajg ∩ Ai| = 1. If x ∈ Ajg ∩ Ak for some k distinct from i
and j, then x ∈ AjAi ∩ Ak, contradicting P3.
Lemma 9.13.3. Under the existing hypotheses, r ≤ t, with equality holding
if and only if P5 holds.
Proof. Clearly |Ω \ Aj| = r(s − 1). So the 1 + r(s − 1) distinct cosets of Aj
that meet Ω leave st − (1 + r(s − 1)) = st − 1 − r(s − 1) cosets of Aj disjoint
from Ω. P5 says that |A ∗ j | = st = s + s(st − 1 − r(s − 1)) = s2 t − rs(s − 1),
i.e., rs(s − 1) = s 2 t − st = ts(s − 1), so that P5 holds if and only if r = t.
Theorem 9.13.4. (W. M. Kantor) If (G, F, F ∗ ) is given as as originally
defined but with r = t, then Q(G, F, F ∗ ) is a GQ of order (s, t) provided P2
and P3 hold.
For the remainder of this book we revert to the usual notation
for P2 and P3. Relabel P3 as K1 and P2 as K2.
Lemma 9.13.5. Let (G, F, F ∗ ) be given as as originally defined but satisfying
K1 and K2, so that r ≤ t. If i, j, k are distinct indices between 0 and r,
then A ∗ i , AiAj \ Ai, AiAk \ Ai are disjoint.
Proof. It is easy to see that K2 forces A∗ i ∩AiAj = Ai, so that A∗ i ∩AiAj = ∅.
Suppose x ∈ AiAj \ Ai ∩ AiAk \ Ai. Then x = aiaj = a ′ iak, aj = e = ak, with
the obvious notation. Then a −1
i a′ i = aja −1
k ∈ Ai ∩ AjAk = {e}. This forces
ai = a ′ i and aj = ak ∈ Aj ∩ Ak = {e}, and hence x ∈ Ai, a contradiction.
From the above Lemma we see that since A ∗ 0 , A0A1 \ A0, . . . , A0Ar \ A0
are pairwise disjoint, st+r(s 2 −s) ≤ s 2 t, with equality if and only if these sets
partition G if and only if r = t. We record this information in the following
theorem.
Theorem 9.13.6. If Q(G, F, F ∗ ) is a GQ (of order (s, t)), then we have the
following two partitions:

9.14. WEAK 4-GONAL FAMILIES 431
(1) A ∗ i ∪j=i(AiAj \ Ai) is a partition of G.
(2) Ai ∪ (Aig : Aig ∩ Ω = ∅) is a partition of A ∗ .
When (G, F, F ∗ ) satisfies r = t and satisfies properties K1 and K2, we
refer to (G, F, F ∗ ) (or sometimes just F) as a Kantor family or a 4-gonal
family. It is clear that F ∗ is completely determined as a collection of sets by
F, but it is not clear if only F is given that the members of F ∗ are actually
groups. We will have more to say on this matter later.
Let (G, F, F ∗ ) be a 4-gonal family, and let S = Q(G, F, F ∗ ) = (P, B, I)
be the associated generalized quadrangle. For each h ∈ G define θh by
g θh = gh; (Aig) θh = Aigh; (A ∗ i g) θh = A ∗ i gh; [Ai] θh = [Ai]; (∞) θh = (∞); for
g ∈ G, Ai ∈ F, A ∗ i ∈ F ∗ . Then θh is an automorphism of S which fixes the
point (∞) and fixes each line [Ai], 0 ≤ i ≤ t. Moreover, G ′ = {θh : h ∈ G} is
a group (under composition of maps) isomorphic to G, and G ′ acts regularly
on the points g ∈ G of type (1). Hence each θh is an elation about (∞).
We say that G (or G ′ ) is an elation group and that (S (∞) , G) is an elation
generalized quadrangle (EGQ).
9.14 Weak 4-Gonal Families
Let G be a group of order s 2 t, s > 1, t > 1. Let F = {Ai : 0 ≤ i ≤ t} be a
family of 1 + t subgroups of G, each of order s, for which K1 holds. Then we
say F is a weak 4-gonal family for G. Put
A ∗ i = Ai ∪ (Aig : g ∈ G and Aig ∩ Aj = ∅∀j = i). (9.49)
Then for each i, 0 ≤ i ≤ t, A ∗ i
F is a 4-gonal family for G if and only if A ∗ i
is a set of st elements of G. It follows that
is a subgroup of G for all i.
Problem: When is a weak 4-gonal family necessarily a 4-gonal family?
Later, after we have studied TGQ and STGQ a bit more we will give
two theorems that provide partial answers to this problem. For the moment
we show that some additional hypothesis is required by giving some examples
due to X. Chen (the two smallest caes, 1987) and D. Frohardt (private
communication, 1988).
Let D2n be the dihedral group of order 4n. Say

432 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
D2n =< a, b : a 2 = b 2n = 1, ab = b −1 a > .
So D2n = {1, b, b 2 , . . . , b 2n−1 , ab, ab 2 , . . . , ab 2n−1 }, and b i a = ab −i for all
i ∈ Z.
Case 1 n = 2. Put A0 =< a >, A1 =< ab >, A2 =< b 2 >. It is an
easy exercise to show that F = {A0, A1, A2} is a weak 4-gonal family for
D2n = D4 that is not a 4-gonal family.
Case 2 n ≥ 3. Put A0 =< a >, and for 1 ≤ i ≤ n, put Ai =< ab 2i−1 >.
It is again a straightforward exercise to show that F = {A0, A1, . . . , An} is a
weak 4-gonal family for D2n. Since there is no GQ(2, n) for n = 3 or n ≥ 5,
clearly F cannot be a 4-gonal family in these cases. We give a computational
argument that shows this for all n ≥ 3. So let 1 ≤ i, t, k ≤ n with t, k, −i
distinct modulo n. Put j ≡ i + t − k (mod n), with 1 ≤ j ≤ n. Then
Aib 2k+1 ∪Aib 2t+1 ⊆ A ∗ i , but ab2j−1 = b 2k+1 ·ab 2i−1 ·b 2t+1 = ab −2k−1+2i−1+2t+1 ∈
Aib 2k+1 · Aib 2t+1 ∩ Aj ⊆ A ∗ i ∩ Aj, so F is not a 4-gonal family.
9.15 Elation Generalized Quadrangles
Let (S p , G) = (P, B, I) be an elation generalized quadrangle (EGQ) of order
(s, t), s > 1, t > 1, i.e., G is a group of elations about the point p acting
regularly on the s 2 t points not collinear with p. Let y be a fixed point of
P \ p ⊥ . Let L0, . . . , Lt be the lines incident with the point p, and define
zi and Mi by LiIziIMiIy, 0 ≤ i ≤ t. Put Si = {θ ∈ G : M θ i = Mi},
S ∗ i = {θ ∈ G : zθ i = zi}, and F = {Si : 0 ≤ i ≤ t}. Then |G| = s 2 t; F is a
collection of 1 + t subgroups of G, each of order s; for each i, 0 ≤ i ≤ t, S ∗ i is
a subgroup of order st containing Si as a subgroup. Moreover, the following
two conditions are satisfied:
K1. SiSj ∩ Sk = {e}, for i, j, k distinct.
K2. S ∗ i ∩ Sj = {e}, for distinct i, j.
Hence if we also put F ∗ = {S ∗ i : 0 ≤ i ≤ t}, (G, F, F ∗ ) is a Kantor family,
and Q(G, F, F ∗ ) is a GQ(s, t).
Exercise 9.15.0.1. Show that if we start with an EGQ (S p , G) and obtain
the Kantor family (G, F, F ∗ ) as above, the associated generalized quadrangle
Q(G, F, F ∗ ) is naturally isomorphic to the original EGQ (S p , G).

9.15. ELATION GENERALIZED QUADRANGLES 433
Most known examples of GQ are EGQ, the notable exceptions being
those of order (s − 1, s + 1) and their duals. Moreover, the known examples
of EGQ fall into one of two categories, in the first of whcih G contains
a full group of symmetries about each line through the base point p. In
this case we shall soon see that G is elementary abelian and S is called
a translation generalized quadrangle (TGQ) with base point p and
translation group G. Briefly, we say S (p) , G) (or S (p) ) is a TGQ.
A TGQ of order (s, t) must have s ≤ t since it has some regular line. The
second category of EGQ is at the opposite end of the hypothetical spectrum.
Here G contains a full group C of symmetries about the point p. Here we say
S (p) , G) (or S (p) ) is a skew-translation generalized quadrangle (STGQ)
with base point p and skew-translation group G. If S (p) is an STGQ, then
t ≤ s, since S has a regular point.
The following simple result is occasionally helpful.
Lemma 9.15.1. Let σ, θ be nonidentity symmetries about distinct lines L,
M, respectively, in an arbitrary finite generalized quadrangle. Then
(i) σθ = θσ if and only if L ⊥ M.
(ii) σθ is not a symmetry about any line (or point).
(iii) The existence of even one nonidentity symmetry σ about some line L
means that σ has f = s + 1 fixed points and g = (1 + s)st points moved
by σ to collinear points. So by Theorem 9.12.2 we have (1 + t)f +
g = (1 + t)(1 + s) + (1 + s)st ≡ 1 + st (mod s + t), which implies
(1 + s)st ≡ 0 (mod s + t.
Proof. First suppose that L and M meet at a point x, and let y ∈ P \ x ⊥ .
Let L ′ be the line through y meeting L and M ′ the line through y meeting
M. It follows readily that both y σθ and y θσ must be the point at which (M ′ ) σ
meets (L ′ ) θ . But if σθ and θσ have the same effect on points of P \x ⊥ , clearly
σθ = θσ. Now suppose that L ⊥ M. Clearly L θ ⊥ L, so that L θσ = L θ , but
L σθ = L θ . This proves (i).
For the proof of (ii), note that if LIxIM, then x σθ = x, y σθ = y = y σθ if
and only if y ∈ x ⊥ \ {x}, and y σθ ⊥ y if and only if y ∈ x ⊥ . And if l ⊥ M,
then y σθ ⊥ y for all y not incident with any line of {L, M} ⊥ . It follows
readily that ξθ is not a symmetry about any line or point.

434 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
For the remainder of this section let (S p , G) be an EGQ with Kantor
family (G, F, F ∗ ), where F = {A0, . . . , At}, etc., and identify (S p , G) with
the coset geometry version Q(G, F, F ∗ ). We start by collecting a variety of
elementary facts most of whose proofs are either already clear or are left to
the reader as routine exercises.
Theorem 9.15.2. Let (S p , G) = Q(G, F, F ∗ ) be an EGQ with Kantor family
F = {A0, . . . , At}, etc.
(i) G acts by right multiplication as a (maximal) group of elations about
p = (∞).
(ii) Ai is the subgroup of G fixing the line Ai of Q(G, F, F ∗ ). More generally,
A h is the stabilizer in G of the line Ah.
(iii) A ∗ i h is the stabilizer in G of the point A ∗ i h.
(iv) Each automorphism of G leaving F invariant induces a collineation of
Q(G, F, F ∗ ) fixing (∞).
(v) G =< Ai : 0 ≤ i ≤ t >.
(vi) Ai is a group of symmetries about [Ai] if and only if Ai ✁ G only if [Ai]
is a regular line if and only if AB = BA for all B ∈ F, in which case
s ≤ t.
(vii) C = ∩(A ∗ : A ∈ F) is a group of symmetries about (∞) if and only if
C ✁ G. Moreover, in this case G contains a full group of symmetries
about (∞) if and only if |C| = t, in which case (∞) is a center of
symmetry and is regular, implying that s ≥ t. Recall that in this case
S (∞) is called a skew-translation generalized quadrangle (STGQ).
(viii) Ai is a group of whorls about [Ai] if and only if Ai G ⊆ A ∗ i .
(ix) Ai is a group of whorls about the point A∗ i if and only if Ai ✁ A∗ i if and
only if Ai h is a group of whorls about A∗ i h for each h ∈ G.
(x) A ∗ i is a group of whorls about [Ai] if and only if A ∗ ✁ G.
(xi) If G is abelian, then Q(G, F, F ∗ ) is a TGQ.

9.15. ELATION GENERALIZED QUADRANGLES 435
Proof. The details are fairly straightforward, so we leave them as exercises,
except that we give the proof of part (vi) assuming that the previous parts
have been proved. Then h ∈ G determines a symmetry about [Ai] if and only
if the collineation it determines by right multiplication fixes each line of the
form Aig if and only if Aigh = Aig for all g ∈ G if and only if ghg −1 ∈ Ai
for all g ∈ G. Hence h is a symmetry about [Si] if and only if all conjugates
of h lie in Ai. It follows that Ai is a group of symmetries about [Ai] if and
only if Ai ✁ G, in which case [Ai] is a regular line. Now let g be an arbitrary
point not collinear with (∞). The set AiAjg consists of those points not
collinear with (∞) which lie on lines of {[Ai], Ajg} ⊥ , i = j. Similarly, the
set AjAig consists of those points not collinear with (∞) which lie on lines
of {[Aj], Aig} ⊥ . Hence ([Ai], Ajg) is regular if and only if AiAjg = AjAig if
and only if AiAj = AjAi. So [Ai] is regular if and only if AiAj = AjAi for
all j = 0, . . . , t.
Exercise 9.15.2.1. Let Q(G, F, F ∗ ) be an EGQ with s = t > 1. Let A and
B be distinct members of F. Show that the pair ([A], B) of nonconcurrent
lines is antiregular if and only if both the following hold:
(i) AB ∩ BA = A ∪ B, and
(ii) For each D ∈ F \ {A, B}, |AB ∩ Da| = 1 for each a ∈ A.
Note: In the above context let A be a fixed member of F. Suppose that
whenever D and B are (not necessarily distinct) members of F \ {A} it is
true that |AB ∩ DA| = 2s − 1. It follows from the above exercise that the
line [A] must be antiregular.
PROBLEM 1. Let Q(G, F, F ∗ ) be an EGQ with s = t > 1. Suppose that
whenever A and B are distinct members of F it is true that AB∩BA = A∪B.
Does it follow that each line through the point (∞) must be antiregular?
Exercise 9.15.2.2. An EGQ(p, p) (with p a prime) must be classical.
The sequence of steps in this exercise provide a fairly elementary proof
that an EGQ with order (p, p) with p a prime must be classical. Let G be
a group with order p 3 , and let F = {A0, . . . , Ap} be a 4-gonal family with
F ∗ the appropriate set of 1 + p subgroups of order p 2 , etc. Since we already
know that there is only one GQ(2, 2), we assume throughout that p is an odd
prime.

436 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
(i) Use K2 to show that no A ∗ ∈ F ∗ may be cyclic and hence each A ∗
must be elementary abelian.
(ii) Show that for distinct A ∗ , B ∗ ∈ F ∗ it must be true that |A ∗ ∩ B ∗ | = p.
(iii) Let G be abelian. For distinct A, B ∈ F, AB = BA is a subgroup
of order p 2 . Use Theorem 9.13.6 to show that each element of G has
order dividing p. Hence G is elementary abelian.
(iv) Let G be abelian. Show that G may be viewed as a vector space over
Zp and that F may be viewed as a p+1-arc in the associated projective
plane P G(2, p). By a famous theorem of B. Segre F is a conic and hence
is unique up to isomorphism. The associated GQ, which we denote
at this time by Q(G, F, F ∗ ), is isomorphic to the orthogonal geometry
Q(4, p) and also to the point-line dual of the symplectic geometry W (p).
(v) Let G be non-abelian. Recall (cf. [Ha59]) that there are only two nonabelian
groups (up to isomorphism) of order p 3 and that in one of them
the subgroups of order p 2 are cyclic. Hence this group cannot admit a
4-gonal family. We have the following construction of the other nonabelian
group of order p 3 . G = {(a, c, b) : a, c, b ∈ Zp. The binary
operation of G is given by
(a, c, b) · (a ′ , c ′ , b ′ ) = (a + a ′ , c + c ′ + ba ′ , b + b ′ ),
where addition and multiplication here are those in Zp.
(vi) Put C = {(0, c, 0) : c ∈ Zp}. Then C is the center of G and A ∗ ∩B ∗ = C
for distinct A ∗ , B ∗ ∈ F ∗ . C ∪ (A ∗ \ C) : A ∗ ∈ F ∗ ) is a partition of
G. Two elements of G belong to subgroup of order p 2 if and only if
they commute. If g ∈ (G \ C), then the unique subgroup of order
p 2 containing g is < g > ·C. Conclude that there are exactly p + 1
subgroups of G having order p 2 . Hence F ∗ is uniquely determined.
(vii) Let A ∗ , B ∗ be distinct subgroups of order p 2 , i.e., elements of F ∗ .
Then A ∗ is normal in G, A ∗ has p subgroups of order p other than C.
Conjugation by elements of B fixes C and permutes the p other groups
of order p in A ∗ regularly in a cycle of order p. Let A ′ be any subgroup
of order p in A ∗ different from C. Similarly for B ′ in B ∗ . By first
letting B ′ act on G by conjugation we may choose any subgroup of A ∗

9.15. ELATION GENERALIZED QUADRANGLES 437
(other than C) we want for the group A. Then conjugating by A we
may choose any subgroup of B ∗ (other than C) to be the group B.
(viii) Put A∞ = {(0, 0, b) : b ∈ Zp} and then necessarily A∗ ∞ = {(0, c, b) :
c, b ∈ Zp}. Similarly, put A0 = {(a, 0, 0) : a ∈ Zp} and A∗ 0 = {(a, c, 0) :
a, c ∈ Zp}. For 0 = t ∈ Zp, put A∗ t = {(a, c, ta) : a, c ∈ Zp}. At
this point we do not know what At is, but we will show that it is now
uniquely determined. By K1, we know that A0A∞ ∩ At = {(0, 0, 0)}. If
i, j are distinct, nonzero elements of Zp, then (i, 0, ti) and (j, 0, tj) are
p − 1 distinct elements of A0A∞ ∩ At and do not belong to a subgroup
of order p. These p − 1 elements generate p − 1 distinct subroups
of A∗ t
of A∗ t different from C. Hence there is only one remaining subgroup of
A∗ t of order p that can be At. Hence F is uniquely determined as soon
as we have chosen A0 and A∞.
(ix) Show that for 0 = t ∈ Zp, it must be that
At = {(a, t
2 a2 , ta) : a ∈ Zp}.
Theorem 9.15.3. Let S = (P, B, I) be a GQ of order (s, t) with 1 ≤ s ≤ t,
and let p be a point for which {p, x} ⊥⊥ = {p, x} for all x ∈ P \ p ⊥ . Let G be
a group of whorls about p.
(i) If y ⊥ p, y = p, and θ is a nonidentity whorl about both p and y, then
all points fixed by θ lie on the line py and all lines fixed by θ meet py.
(ii) If θ is a nonidentity whorl about p, then θ fixes at most one point of
P \ p ⊥ .
(iii) If G is generated by elations about p, then G is a group of elations, i.e.,
the set of elations about p is a group.
(iv) If G is transitive on P \ p ⊥ and |G| > s 2 t, then G is a Frobenius group
on P \ p ⊥ , so that the set of all elations about p is a normal subgroup
of G of order s 2 t acting regularly on P \ p ⊥ , i.e., S (p) is an EGQ with
some normal subgroup of G as elation group.
(v) If G is transitive on P \p ⊥ and G is generated by elations about p, then
(S (p) , G) is an EGQ.

438 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
Proof. Both (i) and (ii) are easy consequences of Theorem 9.12.1 Suppose
there is some point x ∈ P \ p ⊥ for which |G| = |Gx| = 1. Then by (ii) G
is a Frobenius group on x G (cf. [Ja80]). So the Frobenius kernel of G acts
regularly on x G . If G is generated by elations about p (so trivially |G| = |Gx|
if |G| > 1), then G itself must act regularly on x G . Since this holds for each
x ∈ P \ p ⊥ , each element of G is an elation about p. Parts (iii), (iv) and (v)
of the theorem are now easy consequences.
9.16 Elation Groups as p-Groups
Let G be a group with |G| = s 2 t, s > 1, t > 1. Let F = {A0, A1, . . . , Ar}
be a set of 1 + r subgroups of G of order s. Suppose that Ai ≤ A ∗ i with
|A ∗ i | = st. Assume that F satisfies K1 and K2. We have seen that r ≤ t and
if A, B and C are distinct members of F, then A ∗ , AB \ A, and AC \ A are
disjoint. In this general a setting we need one more preliminary lemma.
Lemma 9.16.1. For 0 ≤ i, j ≤ r, i = j, and g ∈ G, we have A ∗ i
∩ Ag
j
= {e}.
Proof. |A∗ i Aj| = s2t = |G|, so A∗ i Aj = G. (Notation: ag = g−1ag.) Suppose
a ∈ A∗ i ∩ Agj
, so ag−1 ∈ Aj. Write g−1 = hb, h ∈ A∗ i , b ∈ Aj. Then
ag−1 = gag−1 = b−1h−1ahb ∈ Aj, so ah ∈ A∗ i ∩ Aj = {e}, forcing a = e.
For the remainder of this section we assume that r = t, so Q(G, F, F ∗ ) is
a GQ(s, t). If p is a prime and s any positive integer, we write p e ||s to mean
that p e divides s but p e+1 does not divide s. And we write sp = p e to mean
that pe s
||s. So sp = 1 means that p does not divide s. Also define sp ′ := sp .
Lemma 9.16.2. If sp > 1, then tp ′ < sp.
Proof. Choose A ∈ F. Let P ∈ Sylp(G) contain A∗ p ∈ Sylp(A∗ ). For each
B ∈ F \ A choose Bp ∈ Sylp(B). Then Bp is G-conjugate to a subgroup QB
of P . From Lemma 9.16.1, if B, C ∈ F \ {A}, B = C, then A∗ p ∩ QB = {e},
and if QB = Bg1 p , QC = Cg2 p , then |QB
−1
g1g ∩ QC| = |B 2 ∩ C∗ | = 1. So A∗ p ,
QB : B ∈ F \ {A} pairwise intersect in {e}. So |A∗ p| +
B∈F\{A} (|Qp| − 1) =
sptp + t(sp − 1) ≤ |P | = s2 ptp. Hence tp ′tp(sp − 1) ≤ sptp(sp − 1), forcing
tp ′ ≤ sp.
Notation: Let π(s) denote the set of distinct primes that divide s.
The next two theorems are due to D. Frohardt (cf., [Fr88]).

9.16. ELATION GROUPS AS P -GROUPS 439
Theorem 9.16.3. Let k = |π(s)|. Then t k−1 < s. If s ≤ t, then s is a prime
power.
Proof. If k = 1, clearly s is a prime power. So suppose k ≥ 2. By
Lemma 9.16.2, tk−1 ≤
p∈π(s)
tp ′ < p∈π(s) sp = s, implying t < s.
Theorem 9.16.4. If s is a prime power, say s = p e , then G is a p-group.
Proof. For each A ∈ F, choose SA ∈ Sylp(G) such that A ≤ SA and SA
contains a Sylow p-subgroup of A∗ . Since G has at most |G|p ′ = tp ′ distinct
Sylow p-subgroups and |F| = ⊔√⊔√ ′ + ∞, by the pigeonhole principle there
is an S ∈ Sylp(G) with S = SA for at least tp + 1 members A of F. Fix
such an S, and put FS = {A ∈ F : SA = S}. Put A + := A∗ ∩ S for
all A ∈ FS and F ∗ S = {A+ : A ∈ FS}. It follows that (S, FS, F ∗ S ; s, tp)
satisfies K1 and K2. Here |A| = s, |A∗ | = |A∗ ∩ S} = (st)p = stp. By
the observation recalled at the beginning of this section, |FS| = tp + 1 and
S = (A∗ ∩ S) ∪ {AB \ A : B ∈ FS \ {A}} for each A ∈ FS. (Keep in mind
that A is a p-group, A ≤ S, but A∗ meets S in a Sylow p-subgroup of A∗ .)
Fix a group A ∈ FS. Let C ∈ F. Then |CS| ≤ |G| = s2t = tp ′ · tps2 <
stps2 = |C|·|S|, since tp < s by Lemma 9.16.2. Then |CX| = |C|·|S|
< |C|·|S|,
|C∩S
forcing C ∩ S = {e}. So either C ∩ A∗ = {e} (implying C = A), or there
is some B ∈ FS \ {A} with C ∩ (AB \ A) = {e} (implying that C = B).
In any case C must be one of the A ′ is contained in FS, so C ≤ S. Hence
F = FS. Then t = |F| − 1 = |FS| − 1 = tp, implying that G is a pp-group
(and G = S).
Corollary 9.16.5. Let S = (S (p) , G) be an EGQ of order (s, t) with s > 1,
t > 1. If s ≤ t, then s and t are powers of the same prime. If s is a prime
power, then s and t are powers of the same prime. And in any case, since
s ≤ t 2 , by Theorem 9.16.3s is divisible by at most two distinct primes.
The next theorem is due to D. Hachenberger(cf. [Ha96]). Note: This
theorem was also proved by X. Chen, around 1990 ([Ch90]), but I cannot
find a specific published reference.
Let F be a 4-gonal family for G with the same notation as above. Suppose
that G contains a subgroup C of order t of symmetries about (∞), so (∞)
is a regular point. So far we have just said that if S = S(G, F) is the

440 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
associated EGQ, then (S (p) , G) is an STGQ. This implies that s ≥ t. It is
clear that C must be normal in G and that C ⊂ A ∗ for each A ∈ F, so that
C ⊆ ∩(A ∗ i : Ai ∈ F). Since |A ∗ ∩ B ∗ | = t whenever A and B are distinct
members of F, it follows that C = ∩{A ∗ i : Ai ∈ F} = A ∗ ∩ B ∗ for disinct
A, B ∈ F. It now follows easily that A ∗ = AC for each A ∈ F.
Hachenberger’s theorem says that in the situation just described, s and t
are powers of the same prime p, i.e., G is a p-group.
Theorem 9.16.6. If S is an STGQ with parameters (s, t), then s and t are
powers of the same prime.
Proof. Assume the notation developed previously. Clearly G = A∗ i · Aj, so
for distinct
|A ∗ i · A∗ j | = s2 t = (st)2
|A ∗ i ∩A∗ j |, so |A∗ i ∩ A∗ j | = t, forcing C = A∗ i ∩ A∗ j
i, j. It is clear that the group of symmetries about (∞), i.e., C, must be
normal in G, so that A ∗ i = AiC for all i.
For distinct A, B ∈ F and any g1, g2 ∈ G, suppose that gC ∈ (AC) g1 /C ∩
(BC) g2 g1 g2 g1 g2 /C = A C/C ∩ B C/C. So gC = a C = b C for some a ∈ A and
−1
−1
−1
g1g b ∈ B. Then a 2 g1g C = bC, implying a 2 g1g ∈ A 2 ∩ B∗ = {C}. Hence
a = e, implying gC = C. This says that (AC) g1 g2 /C ∩ (BC) /C = {id}.
Suppose p is a prime with sp = pe > 1. Fix A ∈ F. Let Ap ∈
Sylp(AC/C), so |Ap| = pe . Then there exists P ∈ Sylp(G/C) such that
Ap ≤ P , and |P | = p2e . For each B ∈ F \ {A}, choose ¯ Bp ∈ Sylp(BC/C).
There exists gB ∈ G such that Bp = ¯ BgB p ≤ P . Then AC/C ∩ Bp = {id},
and for distinct B, D ∈ F \ {A}, it must be the case that Bp ∩ Dp = {id}.
Hence SC = {Bp : B ∈ F} is a partial congruence partition of type (pe , t+1)
in P . (This just means that SC is a set of t + 1 subgroups of P , each of order
pe , and such that each two of them intersect in the identity element.) This
forces (t + 1)(pe − 1) + 1 ≤ p2e , which forces t ≤ pe .
Suppose there is a second prime q with sq = qf > 1. (Frohardt’s results
say that q is unique and s > t.) We may suppose that pe < qf , so pe < √ s.
But now t ≤ pe < √ s. This forces t2 < s, which contradicts the inequality
of D. G. Higman (cf. [PT84]). This proves that s is a prime power. Then by
Theorem 9.16.4 G must be a p-group.
9.17 A Model for STGQ
There is always the hope that examples of GQ will be found that are truly different
from the known ones. It is possible that such GQ may be constructed

9.17. A MODEL FOR STGQ 441
using models that are suggested by but are more general than the known
ones. With this thought as inspiration, we develop the following abstract
model for skew translation Gq (STGQ).
Let F = Fq and ˜ F = F ∪ {∞}. For m and n positive integers, let
f : F m × F m → F n be a fixed biadditive map. Put G = {(α, β, c) : α, β ∈
F m ; c ∈ F n }. Define a binary operation on G by
(α, β, c) · (α ′ , β ′ , c ′ ) = (α + α ′ , β + β ′ , c + c ′ + f(β, α ′ )).
This makes G into a group that is abelian if f is trivial and whose center
is C = {(0, 0, c) : c ∈ F n } if f is nonsingular. Suppose that for each t ∈ F n
there is an additive map δt : F m → F m and a map gt : F m → F n . Put
For each t ∈ F n put
A(∞) = {(0, β, 0) : β ∈ F m }.
A(t) = {(α, α δt , gt(α)) : α ∈ F m }.
We want A(t) to be closed under the product in G, so that it will be a
subgroup of order q m . Writing out the product of two elements of A(t) we
find that A(t) is a subgroup if and only if
gt(α + β) − gt(α) − gt(β) = f(α δt , β) for all α, β ∈ F m , t ∈ F n . (9.50)
Put β = 0 in Eq. 9.50 to obtain
gt(0) = 0 for all t ∈ F n . (9.51)
From now on we suppose that the condition Eq. 9.50 holds, and put
J = {A(t) : t ∈ ˜ F }.
With A ∗ = AC for A ∈ J , we seek conditions on J and J ∗ = {A ∗ : A ∈
J } that will force K1 and K2 to hold, i.e., that will force J to be a 4-gonal
family. Clearly A ∗ is a group of order q m+n containing A as a subgroup, for
each A ∈ J . We note that
A ∗ (∞) = {(0, β, c) ∈ G : β ∈ F m , c ∈ F n },
and (9.52)
A ∗ (t) = {(α, α δt , c) ∈ G : α ∈ F m , c ∈ F n }, t ∈ F n .

442 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
It is easy to check that A ∗ (∞) ∩ A(t) = 1 = A ∗ (t) ∩ A(∞) for all t ∈ F n .
An element of A ∗ (t) ∩ A(u) has the form (α, α δt , c) = (α, α δu , gu(α)). For
t = u this must force α = 0, so
A ∗ (t) ∩ A(u) = 1
iff (9.53)
δ(t, u) : α ↦→ α δt − α δu is nonsingular whenever t = u.
From now on we assume that Eq. 9.53 holds. Then J will be a 4-gonal
family for G iff AB ∩ D = 1 for distinct A, B, D ∈ J . Before investigating
this condition we need a little more information about gt. Put β = −α in
Eq. 9.50 to obtain
implying
−gt(α) − gt(−α) = f(α δt , −α) = −f(α δt , α) = −gt(2α) − 2gt(α),
Using Eqs. 9.50 and 9.54 we obtain
gt(2α) = 3gt(α) + gt(−α). (9.54)
gt((n + 1)α) = (n + 1)gt(α) + gt(nα) + ngt(−α),
from which an induction argument may be used to show that
n + 1 n
gt(nα) = gt(α) + gt(−α). (9.55)
2
2
Note: If gt(−α) = −gt(α), then gt(nα) = ngt(α).
If gt(−α) = gt(α), then gt(nα) = n 2 gt(α).
Let g ∈ A(∞) · A(t) ∩ A(u), t = u, so g has the form
g = (0, β, 0) · (α, α δt , gt(α)) = (α, β + α δt , gt(α) + f(β, α))
= (α, α δu , gu(α)).
So gt(α) − gu(α) = −f(β, α), with β = α δ(u,t) , should imply α = 0. That is:
gu(α) − gt(α) = f(α δ(u,t) , α)
= f(α δu , α) − f(α δt , α)
= (gu(2α) − 2gu(α)) − (gt(2α) − 2gt(α))
= (gu(α) + gu(−α)) − (gt(α) + gt(−α))

9.17. A MODEL FOR STGQ 443
should imply that α = 0. This last condition says that gt(−α) = gu(−α)
should imply that α = 0, and this holds for any α ∈ F 2 . Hence A(∞)·A(t)∩
A(u) = 1 (for t = u) if and only if
gt(α) = gu(α), t = u, implies α = 0. (9.56)
It is also routine to show that A(∞) · A(t) ∩ A(u) = 1 iff A(t) · A(∞) ∩
A(u) = 1 iff A(t) · A(u) ∩ A(∞) = 1. Hence we assume that Eq. 9.56 holds
and proceed to the hard case: What does A(t) · A(u) ∩ A(v) = 1 mean when
t, u, v are distinct elements of F ? An element of this intersection would be
of the form
(α + β, α δt + β δu , gt(α) + gu(β) + f(α δt , β)) = (α + β, (α + β) δv , gv(α + β)).
Hence the intersection is trivial provided
gt(α) + gu(β) + f(α δt , β) = gv(α + β)
α δt + β δu = (α + β) δv
t, u, v distinct
⎫
⎬
⇒ α = β = 0. (9.57)
⎭
Use the second line in Eq. 9.57 to solve for β = α δ(t,v)δ−1 (v,u) and put
γ = α δ(t,v) = β δ(v,u) . Then the first line of Eq. 9.57 becomes
0 = gt(α) + gu(β) + f(α δt , β) − gv(α) − gv(β) − f(α δv , β)
= gt(α) − gv(α) + gu(β) − gv(β)f(α δ(t,v) , β)
= gt(α) − gv(α)gu(β) − gv(β) + f(β δ(v,u) , β)
= gt(α) − gv(α) + gu(β) − gv(β) + (gv(β) + gv(−β)) − (gu(β) + gu(−β))
= gt(α) − gv(α) + gv(−β) − gu(−β).
Hence Eq. 9.57 is equivalent to
gt(γ δ−1 (t,v) ) − gv(γ δ−1 (t,v) ) + gv(−γ δ−1 (v,u) ) − gu(−γ δ−1 (v,u) ) = 0
implies γ = 0 if t, u, v are distinct elements of F.
(9.58)
We summarize these results (which first appeared in [Pa80]) as follows.
Theorem 9.17.1. J is a 4-gonal family for G provided the following hold:

444 CHAPTER 9. FINITE GENERALIZED QUADRANGLES
(i) gt(α + β) − gt(α) − gt(β) = f(α δt , β) = f(β δt , α) for all α, β ∈ F m , t ∈
F n .
(ii) δ(t, u) : α ↦→ α δt − α δu is nonsingular for t = u.
(iii) gt(α) = gu(α), t = u, implies that α = 0.
(iv) Eq. 9.58 holds.
If J is a 4-gonal family, the resulting GQ S = S(G, J ) has order (s, t) =
(q m , q n ). As C is a group of t symmetries about (∞), it follows that (S (∞) , G)
is an STGQ and m ≥ n. By (the point-line dual of ) part (iii) of Lemma 9.15.1
we have (1+t)st ≡ 0 (mod s+t). In this case this means that q m+n (1+q n ) ≡
0 (mod q m + q n ). Divide by q n to get 1 + q n ≡ 0 (mod 1 + q m−n ) (if m > n).
From Lemma 20.18.7 we must have that m − n divides n and the quotient
a is odd. This means that there is some positive integer v (and some odd
integer a) for which m = v(1 + a) and n = va, so s = (q v ) (1+a) and t = (q v ) a .
The known examples of STGQ all have either s = t or s = t 2 .

Chapter 10
GQ Associated with an Oval or
Ovoid
Starting with an oval O in a plane π embedded in P G(3, q) J. Tits constructed
a GQ T2(O) with parameters s = t = q that is isomorphic to Q(4, q) when
O is a conic. Similarly, starting with an ovoid Ω in P G(3, q) embedded
as a hyperplane in P G(4, q), he constructed a GQ T3(Ω) with parameters
(s, t) = (q, q 2 ) that is isomorphic to Q(5, q) when Ω is an elliptic quadric.
These constructions first appeared in Dembowski [De68]. In this chapter
we start with T2(O) and along with it we study closely related GQ with
paramteters (q − 1, q + 1) and (q + 1, q − 1). Our primary interest will be in
the case that q = 2 e , but to start with we let q be an arbitrary prime power.
Before considering the actual constructions we show how to use a regular
point in a GQ with parameters (q, q) to produce a GQ with parameters
(q − 1, q + 1).
10.1 Expansion about a Regular Point
Let S = (P, B, I) be a GQ with order s (i.e., we want s = t ≥ 2). The
following construction first appeared in [Pa71b].
Theorem 10.1.1. To each regular point x of S there corresponds a GQ
S ′ = (P ′ , B ′ , I ′ ) (often denoted by P(S, x)) with parameters (s − 1, s + 1).
Proof. Let x be regular and put P ′ = P \ x ⊥ . The elements of B ′ are of two
types: The lines of B ′ of type (a) are the lines of B which are not incident
445

446 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
with x; the lines of B ′ of type (b) are the hyperbolic lines {x, y} ⊥⊥ , y ∈ P ′ .
The incidence I ′ is defined as follows: for y ∈ P ′ and L ∈ B ′ of type (a), put
yI ′ L iff yIL; if y ∈ P ′ and L ∈ B ′ is of type (b), then yI ′ L iff yIL. We now
show that S ′ is a GQ with parameters (s − 1, s + 1).
An easy check shows that any two points of S ′ are incident (with respect
to I ′ ) with at most one line of S ′ . Also, each point of P ′ is incident with
exactly s + 2 lines of B ′ , and each line of B ′ is incident with exactly s − 1
points of P ′ . Consider a point y ∈ P ′ and a line L of type (a) with y I ′ L.
Let z ∈ P ′ be defined by x ∼ z and zIL. If y ∼ z, then no line of type (a) is
incident with y and concurrent with L. But then, by the regularity of x, there
is a point of P ′ which is incident with the line {x, y} ⊥⊥ of type (b). If y ∼ z,
then there is just line of type (a) with is incident with y and concurrent with
L. By the regularity of x, the line of type (b) containing y is not concurrent
with L. Finally, consider a point y ∈ P ′ and a line L = {x, u} ⊥⊥ , x ∼ u, of
type (b) with y ∈ L. It is clear that no line of type (b) is incident with y and
concurrent with L. If y is collinear with at least two points of L, then by
the regularity of x we have that y ∼ x, i.e., y ∈ P ′ , a contradiction. Hence
y is collinear with at most one point of L. Suppose u ∼ y ∈ P ′ . Since x
is regular, the triad (x, u, y) must have a center v, and each line through v
must be incident with a point of L. Hence the line M of type (a) defined
by vIMIy is incident with y and concurrent with L. We say that the GQ
P(S, x) of order (s − 1, s + 1) is obtained by expanding S about the point
x.
If L is a regular line of the GQ S of order s, we can also apply the pointline
dual construction and expand S about the regular line L to obtain a
GQ P(S, L) of order (s + 1, s − 1).
10.2 The Construction of T2(O)
Let Σ = P G(3, q) and let π be a hyperplane of Σ containing an oval O.
Construct the following incidence geometry T2(O) = (P, B, I), with point
set P and line set B as follows.
The points are of three types:
(i) points of Σ \ π;
(ii) planes of Σ different from π and meeting π in a line tangent to O;
(iii) (∞).

10.2. THE CONSTRUCTION OF T2(O) 447
The lines are of two types:
(a) lines of Σ not in π and meeting π in a point of O;
(b) lines of π tangent to O.
Incidence is as follows. The point (∞) is incident with all lines of type
(b). A point of type (ii) is incident with the line of type (b) contained in it
and all lines of type (a) contained in it. A point of type (i) is incident with
the lines of type (a) containing it. There are no other incidences.
It is a good exercise to show that T2(O) really is a GQ with parameters
(q, q).
Lemma 10.2.1. Each line of T2(O) of type (b) is regular in T2(O).
Proof. Let LX be the line of π tangent to O at the point X, and let M be
a line of T2(O) not concurrent with LX. This means that M is a line of Σ
meeting π at some point Y of O different from X. Let π ′ = 〈X, M〉. Then
and
{LX, M} ⊥ = {LY } ∪ {L : L is a line through X in π ′ },
{LX, M} ⊥⊥ = {LX} ∪ {L : L is a line through Y in π ′ }.
This means that π is a coregular point of T2(O). Hence by Lemma 9.8.3
we know that π is antiregular when q is odd and regular when q is even.
When q is odd, up to duality the only known GQ of order q is this T2(O)
where O must be a conic. Since our main interest is in the case where q
is even, until further notice we suppose that q = 2 e . In this case a
slightly modified description of T2(O) is sometimes more useful. It should be
clear to the reader that this construction is equivalent to the original in this
case.
Let Σ = P G(3, q) with q = 2 e and let π be a hyperplane of Σ containing an
oval O. Construct the point-line incidence geometry T2(O) = S = (P, B, I)
with point set P and line set B as follows.
The points of S are of two types:
(i) points of Σ \ π;
(ii) planes of Σ containing the nucleus N of O.

448 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
The lines are just the lines of Σ that meet O in a unique point. Incidence
is that induced by incidence in Σ. Note that the plane π considered as a point
of S is incident in Σ with the lines of π incident with N.
The geometry T2(O) is a generalized quadrangle (GQ) of order q with a
regular point π and regular lines. So we have two contexts in which to apply
the construction of Theorem 10.1.1. But since we want to become quite
familiar with these GQ we first give a direct proof that π is a regular point.
The process of expanding about a regular point and also about a regular line
will be discussed in Section 10.6.
Lemma 10.2.2. The point π of T2(O) is regular.
Proof. First note that the lines of T2(O) through the point π are the lines
through N in the plane π. Hence π ⊥ is just the set of planes of Σ through
the nucleus N, including π itself. Let π1 and π2 be distinct planes through N
viewed as points of T2(O). Then π1 ∩ π2 = L is a line of Σ through N. If L is
in π, then L is a line of T2(O) through both π1 and π2. Suppose that L does
not lie in π. Let Qi be the unique point of O on πi, i = 1, 2. If X is a point
of L not in π, then X is a point of 〈Qi, X〉, which is a line of T2(O) incident
with πi. It follows that {π1, π2} ⊥ is the set of points of L \ π together with
the point π. It then follows that {π1, π2} ⊥⊥ is the set of planes through L.
This is enough to show that π is regular.
If we view π as a regular point of T2(O), we can describe the projective
plane Π(π) derived from the regularity of π (see Theorem 9.8.5) as follows.
The points of Π(π) are just the planes of Σ through the point N. The
lines of Π(π) are the lines of Σ through N. Incidence is just that induced by
incidence in Σ. It is clear that Π(π) is isomorphic to P G(2, q).
By Lemma 10.2.1 we know that each line L through π in T2(O) is a
regular line. So we now describe the projective plane Π(L). However, it is
more natural to describe the point-line dual plane Π(L) ˆ .
The “points” of Π(L) ˆ are the planes of Σ through Q. The “lines” of Π(L) ˆ
are the lines of Σ through Q not in π plus the lines through N in π. Incidence
is just that inherited from Σ, except for the lines of π through N different
from L. Suppose that L ′ is a line of π through N and the point Q ′ of O,
Q ′ = Q. Then L ′ is incident with each of the q + 1 planes through the line
〈Q, Q ′ 〉. If we replace the line L ′ = 〈N, Q ′ 〉 with the line 〈Q, Q ′ 〉, then we may
describe Π(L) ˆ as follows: the points of Π(L) ˆ are just the planes through Q,

10.2. THE CONSTRUCTION OF T2(O) 449
and the lines of Π(L) ˆ are just the lines of Σ through Q. Incidence is exactly
that inherited from Σ. Hence Π(L) ˆ is clearly a Desarguesian plane.
Now we suppose that O is a translation oval in π with translation axis
L∞. (If O is not a conic, then L∞ is the unique translation axis of O.) In
this case it follows that each point of L∞ is regular in T2(O). We want to
interpret this geometrically.
Theorem 10.2.3. If O is a translation oval in π with translation axis L∞,
then each point of L∞ is regular in T2(O).
Proof. Suppose that O = {Q1, Q2, . . . , Qq, Q∞} with nucleus N and axis
L∞ = 〈Q∞, N〉. Let π∞ be a point of T2(O) different from π on the line L∞.
So in Σ, π∞ is a plane different from π passing through the line L∞. The
points of π∞ ⊥ are of two types:
(i) the points of π∞ \ π, collinear with π∞ on the lines of π∞ through the
point Q∞;
(ii) the planes of Σ containing L∞ (each incident with L∞ in T2(O)).
To show that π∞ is regular in T2(Oα) we need to look at point-pairs
(π∞, Z) where Z is a point of T2(Oα) not collinear with π∞. If Z is a point of
the type that it is a plane through N meeting π∞ in a line L different from
L∞, then {π∞, Z} ⊥⊥ is the set of planes through the line L, implying that
the pair is regular. So we may concentrate on the case where Z is a point
X ∈ P G(3, q) \ π.
Fix a point X ∈ Σ \ (π ∪ π∞). Put Ri = 〈X, Qi〉 ∩ π∞, 1 ≤ i ≤ q. Then
in T2(O)
{X, π∞} ⊥ = {〈X, Q∞, N〉} ∪ {Ri : 1 ≤ i ≤ q}.
We have the ordinary quadrangle:
X I 〈X, Q∞〉 I 〈X, Q∞, N〉 I L∞ I π∞ I 〈Ri, Q∞〉 I Ri I 〈X, Qi〉 I X.
Clearly {X, π∞} ⊥ \ {〈X, Q∞, N〉} is the projection from X onto π∞ of the
q-arc O \ {Q∞}. Suppose P = 〈Ri, Rj〉 ∩ L∞ for distinct i, j. If P were
either Q∞ or N, then the plane 〈X, Ri, Rj〉 would contain Qi, Qj and either
Q∞ or N, a contradiction. Let πX denote the plane πX = 〈X, Q∞, N〉, and
let M1, M2, . . . , Mq−1 be the lines of πX through X meeting L∞ in the q − 1
points different from Q∞ and N.

450 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
For 1 ≤ i ≤ q − 1 and 1 ≤ j ≤ q, put πi,j = 〈Mi, Rj〉. Then πi,j ∩ O =
{Qj, Qi ′} for some point Qi ′ of O different from Qj and Q∞ and apparently
depending on both i and j. Put yi,j = 〈Qi ′, Rj〉 ∩ πX.
Then {〈X, Q∞, N〉, 〈X, Qj〉 ∩ π∞ = Rj} ⊥ = {y1,j, . . . , yq−1,j} ∪ {π∞, X}.
This must be the span of {π∞, X}, so as a set it must be independent of Rj.
We are about to introduce coordinates to show that this is indeed the case.
Let α be a generator of the Galois group of Fq. We may choose coordinates
of Σ, π, etc., so that π = [0, 0, 0, 1] and Oα = {(x0, x1, x2, x3) : x3 = xα 1 +
x α−1
0 x2 = 0. So
Oα = {(1, t, t α , 0) : t ∈ Fq} ∪ {(0, 0, 1, 0)} with nucleus N = (0, 1, 0, 0).
Keep in mind that Q∞ = (0, 0, 1, 0) and N = (0, 1, 0, 0).
Let L∞ be the line 〈(0, 0, 1, 0), (0, 1, 0, 0)〉, so that L∞ is a translation axis
of Oα, and the unique one if α = 2.
The planes that pass through L∞ other than π are those of the form
πa = [1, 0, 0, a], a ∈ Fq. Fix πa. We want to see algebraically that πa is a
regular point of T2(Oα).
Let a = a ′ ∈ Fq, b ′ , c ′ ∈ Fq, so X = (a ′ , b ′ , c ′ , 1) is a typical point of
Σ \ (π ∪ πa). For each t ∈ Fq, the line 〈(1, t, t α , 0), X〉 meets the plane πa in
the point Rt = (a, b ′ + (a + a ′ )t, c ′ + (a + a ′ )t α , 1). For each b ∈ Fq the point
Yb = (a ′ , b ′ + b, c ′ + (a + a ′ ) 1−α b α , 1) lies in the plane 〈X, L∞〉, and the line
〈Rt, Yb〉 meets the oval Oα in the point (1, b
a+a ′ + t, b
a+a ′ + t α , 0). Note that
Y0 = X. Then in T2(Oα) we have {X, πa} ⊥⊥ = {πa} ∪ {Yb : b ∈ Fq} (which
is independent of t).
Perhaps a simpler way to show that πa is a regular point of T2(Oα) is the
following.
For any b ∈ Fq let θb be the homography of Σ with matrix
⎛
⎞
[θb] =
⎜
⎝
1 b b α 0
0 1 0 0
0 0 1 0
0 ab ab α 1
Then θb is an elation with axis πa and center (0, 1, b α−1 , 0) and leaving the oval
Oα invariant. Hence θb induces a collineation of T2(Oα) that is a symmetry
about πa, showing that πa is a center of symmetry and hence regular as a
point of T2(Oα). One can check that θd : Yb ↦→ Yb+(a+a ′ )d.
⎟
⎠ .

10.2. THE CONSTRUCTION OF T2(O) 451
At this point it is easy and worthwhile to describe the affine plane A(πa, L∞)
obtained from the regular point πa of T2(Oα) by considering the line L∞ as
the line at infinity.
The points of A(πa, L∞) are just the points of πa \ L∞. The lines of
A(πa, L∞) are of three types:
(a) The perps {πa, X} ⊥ , X a point of Σ\(πa∪π). These are the projections
of the affine part of Oa from the point X to the plane πa. Using coordinates
as above we see that they are the q-arcs of the form
Ob,c,m = {(a, b + mt, c + mt α , 1) : t ∈ Fq} for b, c, m ∈ Fq, m = 0.
Note: These are all the translation ovals isomorphic to Oα containing
Q∞, having L∞ as translation axis and N as nucleus. The q-arc Ob,c,m is the
same as the q-arc O0,c+m 1−α b α ,m, so there really are only q(q − 1) such arcs.
Warning: In the plane πa all points look like (ax, y, z, x), so we may
omit the first coordinate. But then the ovals being discussed all appear in
slightly modified form. The usual homogeneous coordinates (x, y, z) have
been rotated to (y, z, x). So the standard form for an oval in this context is
to have points (t, f(t), 1) for some o-permutation f along with point Q∞ =
(0, 1, 0) and nucleus (1, 0, 0).
(b) The lines of Σ lying in πa through the point Q∞.
(c) The lines of πa \ L∞ with point at infinity N.
Incidence is determined as follows. Consider two points Y and Z of
πa \ L∞. If they are collinear in T2(Oα), then they lie on a line meeting L∞
in the point Q∞, and are incident with this line in A(πa, L∞). If Y and Z
are not collinear in T2(Oα) but the line of Σ through them meets L∞ in the
point N, then the span {Y, Z} ⊥⊥ contains the affine points of the line Y Z of
Σ plus the point π of T2(Oα). Hence the line through them in A(πa, L∞) is
the line Y Z, which is then incident with just its affine points. If the line Y Z
of Σ meets L∞ in a point different from Q∞ and N, then {Y, Z} ⊥⊥ is equal
to {X, πa} ⊥ for any point X of {Y, Z} ⊥ \ {πa}. Here X ∈ Σ \ (πa ∪ π), and
{X, πa} ⊥ in T2(Oα) must be the projection of the q-arc Oα \ {Q∞} from X
onto πa plus the plane 〈X, L∞〉. So the line through Y and Z in A(πa, L∞)
is the q-arc just described. This proves the following:

452 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
Lemma 10.2.4. The affine plane A(πa, L∞) is obtained by α-derivation of
πa \L∞ with respect to (Q∞, N). (See the chapter on ovals for an explanation
of this notation.)
10.3 T2(Oα) is Self-dual when α is Additive
J. Tits [Ti62] first showed that W (q) (which in the next section we show is
isomorphic to T2(C) if C is a conic) is self-dual when q = 2 e and even selfpolar
when e is odd. This was generalized in S. E. Payne [Pa70] to show that
when O is a translation oval T2(O) is self-dual and is self-polar if and only if
e is odd. It was implicit in [Ti62] that when q = 2 e W (q) is isomorphic to the
point-line dual of Q(4, q). Using the Klein correspondence (see Chapter 12)
C. T. Benson [Be65] also showed explicitly that for all q W (q) is the point-line
dual of Q(4, q) and is self-dual if and only if q is even. Some of his results did
not appear in his published paper [Be70], and in 1972 J. A. Thas [Th72] gave
an explicit treatment of this material (and more) with basically the same
maps that appeared in [Ti62]. Some of this material again appeared in the
survey paper [TP76] along with a good bit more.
While we have a description of T2(Oα) handy we give an explicit duality
δ of T2(Oα).
Lemma 10.3.1. The map δ defined below is a duality of T2(Oα).
δ :
(a, b, c, 1) ↦→ 〈(1, a, a α , 0), (0, c, b α , 1)〉;
[t, 0, 1, w] ↦→ 〈(0, 0, 1, 0), (t, w, 0, 1)〉;
[1, 0, 0, a] ↦→ 〈(1, a, a α , 0), (0, 1, 0, 0)〉;
[0, 0, 0, 1] ↦→ 〈(0, 0, 1, 0), (0, 1, 0, 0)〉;
〈(1, t, t α , 0), (0, u, w, 1)〉 ↦→ (t α , w, u α , 1);
〈(a, b, 0, 1), (0, 0, 1, 0)〉 ↦→ [a α , 0, 1, b α ];
〈(1, t, t α , 0), (0, 1, 0, 0)〉 ↦→ [1, 0, 0, t α ];
〈(0, 0, 1, 0), (0, 1, 0, 0)〉 ↦→ [0, 0, 0, 1].
Proof. It is a routine but somewhat tedious task (that we leave to the reader)
to verify all the details. However, it is certainly easier to verify that δ really
is a duality as claimed than it is to discover δ in the first place.
When q is not a square we can say more. (See Theorem 9.10.4.)

10.4. THE ISOMORPHISM FROM T2(C) TO W (Q) 453
Lemma 10.3.2. If α is an automorphism of maximal order and q is not a
square, there is a field automorphism σ for which α · σ 2 = 1. In this case
T2(Oα) is actually self-polar. In fact, the map δ ◦ σ 2 = σ 2 ◦ δ is a polarity.
Proof. The routine argument is left as an exercise.
10.4 The isomorphism from T2(C) to W (q)
The classical geometry W (q) is defined as the singular points and lines of
a symplectic polarity of P G(3, q). When q = 2 e , W (q) is a self-dual GQ of
order q with all points and lines regular. When Q is a non-singular (parabolic)
quadric in P G(4, q), its points and lines form a GQ of order q which is denoted
by Q(4, q). In general it is isomorphic to the point-line dual of W (q), and
when q = 2 e it is isomorphic to W (q). For q even and C an irreducible conic
in the plane π, we now compose an isomorphism from T2(C) to Q(4, q) with
one from Q(4, q) to W (q) to get one from T2(C) to W (q), which we then
slightly adjust to be in a more useful form.
Always q = 2 e . Let Σ = P G(3, q) be embedded as the hyperplane
[0, 0, 0, 0, 1] of P G(4, q). So Σ = {x0, x1, x2, x3, 0) ∈ P G(4, q) : xi ∈ Fq}.
Let Q be the non-singular (parabolic) quadric in P G(4, q) with equation
x0x1 + x 2 2 + x3x4 = 0, so
with matrix
Q = {(x0, x1, √ x0x1 + x3x4, x3, x4) ∈ P G(4, q) : xi ∈ Fq}
⎛
⎜
AQ = ⎜
⎝
0 1 0 0 0
0 0 0 0 0
0 0 1 0 0
0 0 0 0 1
0 0 0 0 0
having polar form matrix BQ = AQ + A T Q
⎛
⎜
BQ = ⎜
⎝
⎞
⎟
⎠
given by
0 1 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 1
0 0 0 1 0
⎞
⎟
⎠

454 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
Clearly (0, 0, 1, 0, 0) is the nucleus of Q, and if ⊥ denotes the polarity
determined by Q,
(x0, x1, x2, x3, x4) ⊥ = BQ(x0, x1, x2, x3, x4) T = [x1, x0, 0, x4, x3]
with polar form
B((xi), (yi)) = x0y1 + x1y0 + x3y4 + x4y3 = 0.
Let π be the plane of Σ defined by x3 = 0 and C = π ∩ Q, so
C = {(1, t, t 1/2 , 0, 0) : t ∈ Fq} ∪ {(0, 1, 0, 0, 0)} with nucleus N = (0, 0, 1, 0, 0).
Construct T2(C) from Σ, π and C as usual. Note that C ⊂ Q. Moreover,
C ⊂ (0, 0, 0, 0, 1) ⊥ , that is, (0, 0, 0, 0, 1) is collinear in Q(4, q) with every point
of C.
We may now define an isomorphism φ1 : T2(C) → Q(4, q) (the GQ obtained
from Q) whose inverse map essentially acts by projecting Q from
(0, 0, 0, 0, 1) onto Σ. It suffices to define φ1 on points. (x0, x1, x2, 1, 0) is a
typical point of T2(C) in Σ \ π.
Here
φ1 : (x0, x1, x2, 1, 0) ↦→ (x0, x1, x2, 1, x0x1 + x 2 2).
(x0, x1, x2, 1, 0) ∼ (y0, y1, y2, 1, 0) in T2(C)
iff (x0 + y0, x1 + y1, x2 + y2, 0, 0) ∈ C
iff (x0 + y0)(x1 + y1) + (x2 + y2) 2 = 0
iff B((x0, x1, x2, 1, x0x1 + x 2 2), (y0, y1, y2, 1, y0y1 + y 2 2)) = 0
iff (x0, x1, x2, 1, x0x1 + x 2 2 ) ∼ (y0, y1, y2, 1, y0y1 + y 2 2 ) in Q.
The other type of point of T2(C) is a plane of Σ meeting π in a tangent to C.
Fix t ∈ Fq. Then for each s ∈ Fq, [t, 1, 0, s, 0]∩[0, 0, 0, 0, 1] = (1, t, t 1/2 , 0, s) ⊥ ∩
Σ is a plane of Σ meeting π in a tangent to C at (1, t, t 1/2 , 0, 0). So we put
φ1 : [t, 1, 0, s, 0] ∩ [0, 0, 0, 0, 1] ↦→ (1, t, t 1/2 , 0, s).
Now (0, 1, 0, 0, s) ⊥ = [1, 0, 0, s, 0] and [1, 0, 0, s, 0]∩π ∩Q = (0, 1, 0, 0, 0). So a
typical plane of Σ meeting π in a tangent to C at (0, 1, 0, 0, 0) is [1, 0, 0, s, 0]∩
[0, 0, 0, 0, 1]. So we put
φ1 : [1, 0, 0, s, 0] ∩ [0, 0, 0, 0, 1] ↦→ (0, 1, 0, 0, s).

10.4. THE ISOMORPHISM FROM T2(C) TO W (Q) 455
Finally put
φ1 : π = (∞) :↦→ (0, 0, 0, 0, 1) ∈ Q.
Since a map from one GQ of order q to a second one of order q is completely
determined by a map just on points preserving collinearity, φ1 is an
isomorphism from T2(C) to Q(4, q).
Let W (q) be the GQ defined as the singular points and lines of the symplectic
polarity of P G(3, q) = [0, 0, 1, 0, 0] with form x0y1+x1y0+x3y4+x4y3 =
0. We consider an isomorphism φ2 from Q(4, q) to W (q). Recall that the
point (0, 0, 1, 0, 0) is the nucleus of Q. Let H be the hyperplane of P G(4, q)
with coordinates [0, 0, 1, 0, 0] and note that (0, 0, 1, 0, 0) ∈ H. Projecting from
(0, 0, 1, 0, 0) onto H maps the points (respectively, lines) of Q(4, q) onto the
points (respectively, lines) of the GQ W (q) just defined. This map defined
on the points of Q(4, q) is given by
φ2 :
φ2 : (x0, x1, x2, x3, x4) ↦→ 〈(x0, x1, x2, x3, x4), (0, 0, 1, 0, 0)〉 ∩ H
= (x0, x1, 0, x3, x4).
So φ2 maps the points of Q(4, q) specifically as follows:
(x0, x1, x2, 1, x0x1 + x 2 2 ) ↦→ (x0, x1, 0, 1, x0x1 + x 2 2 ), for x0, x1, x2 ∈ Fq,
(1, t, t 1/2 , 0, s) ↦→ (1, t, 0, 0, s), for s, t ∈ Fq,
(0, 1, 0, 0, s) ↦→ (0, 1, 0, 0, s), for s ∈ Fq,
(0, 0, 0, 0, 1) ↦→ (0, 0, 0, 0, 1).
Note that for points (xi), (yi) of Q, (xi) ∼ (yi) in Q if and only if
0 = B((xi), (yi)) = x0y1 + x1y0 + x3y4 + x4y3.
So φ2 : Q(4, q) → W (q) is an isomorphism, where W (q) is the symplectic
geometry in H with the given form.
Now define the map φ3 : W (q) → W (q) : (x0, x1, 0, x3, x4) ↦→ (x0, x1, x4, x3).
Finally, put φ = φ3◦φ2◦φ1 : T2(C) → W (q). Before expressing this succinctly
in the following theorem we give a little morenotation. 0 1
P 0
Let P = and put C = . Then
1 0
0 P
⎡ ⎤ ⎡ ⎤
C
⎢
⎣
y0
y1
y2
y3
⎥
⎦ =
⎢
⎣
y1
y0
y3
y2
⎥
⎦ .

456 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
Then ν : points of P G(3, q) ↔ planes of P G(3, q) : (yi) ↦→ C(yi) T is a
symplectic polarity of P G(3, q). So we may now view W (q) as the geometry
of singular points and lines of ν.
Theorem 10.4.1. Let π be the plane [0, 0, 0, 1] of Σ = P G(3, q), and let C
be the conic in π defined by the equations x3 = x0x1 + x 2 2 = 0, that is,
C = {(1, t, t 1/2 , 0) : t ∈ Fq} ∪ {(0, 1, 0, 0)}
with nucleus N = (0, 0, 1, 0). Construct T2(C) from Σ, π and C. Recall
that the points of T2(C) are the points (x0, x1, x2, 1) ∈ Σ \ π and the planes
[x0, x1, 0, x2] containing the point N = (0, 0, 1, 0). Let W (q) be the GQ defined
as the singular points and lines of the symplectic polarity ν. Then there is
an isomorphism φ : T2(C) → W (q) defined on the points of T2(C) by:
φ : (x0, x1, x2, 1) ↦→ (x0, x1, x0x1 + x 2 2 , 1), for x0, x1, x2 ∈ Fq,
[x0, x1, 0, x2] ↦→ (x1, x0, x2, 0).
10.5 Coordinates for the general T2(O)
Review the coordinatization of ovals in Chapter 4. If O is any oval of π∞ =
P G(2, q) we may rearrange coordinates so that the nucleus is N = (0, 1, 0)
and any three particular particular points of O have coordinates Y = (0, 0, 1),
X = (1, 0, 0), and Z = (1, 1, 1). Then there is an o-polynomial f(x) for which
O = {(1, t, f(t)) : t ∈ Fq} ∪ {(0, 0, 1)}.
For notational convenience we write f(t) = t α . Then we embed π∞ into
Σ = P G(3, q) as the plane [1, 0, 0, 0]. This means that any particular line of
T2(O) through the point π∞ may be chosen to be the line ℓ∞ = 〈N, Y 〉. From
the previous sections of this chapter we know that π∞ is a regular point and
each line through it is regular. Moreover, if α is additive, we know that each
point of ℓ∞ is also regular. We want to consider the converse situation: What
if some additional point of T2(O) is regular? We approach this problem by
first assigning coordinates to the points and lines of T2(O) as follows.

10.5. COORDINATES FOR THE GENERAL T2(O) 457
Coordinates first for points and then for lines:
π∞ ↔ (∞)
[a, 1, 0, 0] ↔ (a)
[v, u α , 0, 1] ↔ (u, v)
(1, a, b, c) ↔ (a, b, c)
〈Y, N〉 ↔ [∞]
〈(0, 1, u, u α ), N〉 ↔ [u]
〈(1, a, b, 0), Y 〉 ↔ [a, b]
〈(1, a, b, c), (0, 1, u, u α )〉 ↔ [u, au α + c, au + b]
(10.1)
In terms of the coordinates we now have the following description of
T2(O). x ↦→ x α is an o-polynomial with 0 α = 0; 1 α = 1.
The points of T2(O) are:
The lines of T2(O) are:
(i) (∞)
(ii) (a), a ∈ Fq
(iii) (u, v), u, v ∈ Fq
(iv) (a, b, c), a, b, c ∈ Fq (10.2)
(a) [∞],
(b) [u], u ∈ Fq
(c) [a, b], a, b ∈ Fq
(d) [u, v, w], u, v, w ∈ Fq (10.3)
Incidence in T2(O) with these coordinates is as follows:
The point (∞) is incident with [∞] and with each [u] for u ∈ Fq. The
point (a) is incident with [∞] and with [a, b] for all a, b ∈ Fq. The point
(u, v) is incident with [u] and with [u, v, w] for all u, v, w ∈ Fq. The point
(a, b, c) is incident with [a, b] and with [u, v, w] if and only if b + w = au and
c + v = au α , for a, b, c, u, v, w ∈ Fq.
For convenience we note the following:

458 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
(a, au + w, au α ) is on [u, v, w] for all a ∈ Fq;
(a, b, c) is on [u, au α + c, au + b] for all u ∈ Fq;
(a1, b1c1) ∼ (a2, b2, c2) iff (i) a1 = a2 and b1 = b2 or
(ii) a1 = a2 and ((b1 + b2)/(a1 + a2)) α = (c1 + c2)/(a1 + a2);
(a1, b1, c1) ∼ (a2, b2) iff c1 = a1a α 2 + b2. (10.4)
It is easy to see that each of the q 3 elations of Σ with axis π∞ induces
a collineation (called a translation) of T2(O) which is given in terms of the
coordinates of T2(O) by the following. For (σ1, σ2, σ3) ∈ F 3 q , define θ =
θ(σ1, σ2, σ3) by:
(x, y, z)
(x, y)
(x)
(∞)
[u, v, w]
[u, v]
[u]
[∞]
θ
↦→ (x + σ1, y + σ2, z + σ3)
θ
↦→ (x, y + σ1x α + σ3)
θ
↦→ (x + σ1)
θ
↦→ (∞)
θ
↦→ [u, v + σ1u α + σ3, w + σ1u + σ2]
θ
↦→ [u + σ1, v + σ2]
θ
↦→ [u]
θ
↦→ [∞]
(10.5)
Note that the lines meeting [∞] are precisely the lines of the form [a]
and [a, b] for all a, b ∈ Fq, and the translations of the form θ(0, 0, σ3) for
σ3 ∈ Fq form a group fixing each of these lines. These are symmetries about
the line [∞], so that [∞] is an axis of symmetry. Similarly, for fixed u ∈ Fq
the lines meeting [u] are the lines of the form [∞], [a], and [u, v, w] for all
a, v, w ∈ Fq. The q translations of the form θ(σ1, σ1u, σ1u α ), σ1 ∈ Fq are
then symmetries about [u] (i.e., fixing each line meeting [u]), so that [u] is
also an axis of symmetry. It follows that each line incident with (∞) is an
axis of symmetry.
It also follows that (∞) is a center of symmetry, since the translations of
the form θ(0, σ2, 0), σ2 ∈ Fq are symmetries about the point (∞).

10.5. COORDINATES FOR THE GENERAL T2(O) 459
For each nonzero t ∈ Fq there is a collineation φ = φt defined as follows
(x, y, z)
(x, y)
(x)
(∞)
[u, v, w]
[u, v]
[u]
[∞]
φ
↦→ (tx, ty, tz)
φ
↦→ (x, ty)
φ
↦→ (tx)
φ
↦→ (∞)
φ
↦→ [u, tv, tw]
φ
↦→ [tu, tv]
φ
↦→ [u]
φ
↦→ [∞]
(10.6)
If some point (1, a, b, c) is regular, then we see that all points (1, a ′ , b ′ , c ′ )
are regular. From this it follows easily that all points of T2(O) are regular
and T2(O) is essentially just T2(C) for a conic C. So suppose that O is not
a conic, so that no point of the form (1, a, b, c) is regular.
Now suppose that some point collinear with but distinct from (∞) is
regular. We have seen that without loss of generality we may assume that
such a point is on [∞]. And the translations are transitive on the points
of [∞] \ {(∞)}, so we may suppose that the point is (0). Because (∞) is
regular, each pair of the form {(0), (u, v)} is regular. Hence we need only
consider pairs of the form {(0), (a, b, c)} with a = 0, so (0) and (a, b, c) are
not collinear. Then by applying a translation of the form θ(0, σ2, σ3) we may
assume that (a, b, c) = (a, 0, 0). But then by applying the collineation φ a −1
we may assume that (a, 0, 0) = (1, 0, 0). So we want to examine the case that
{(0), (1, 0, 0) is a regular pair of points. First note that
and
{(0), (1, 0, 0)} ⊥ = {(1)} ∪ {(0, x, x α ) : x ∈ Fq},
{(1), (0, 0, 0)} ⊥ = {(0)} ∪ {(1, y, y α ) : y ∈ Fq}.
It follows that the pair {(0), (1, 0, 0)} is regular if and only if (0, x, x α ) ∼
(1, y, y α ) for all x, y ∈ Fq. By Eq. 10.4 this is if and only if (x+y) α = x α +y α
for all s, y ∈ Fq. In this case α is additive, so by Cor. 4.9.4 α is a generator

460 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
of the automorphism group of Fq, and we have seen that all points of [∞]
are regular and T2(O) is even self-dual. We record this as follows.
Theorem 10.5.1. Some point W of (∞) ⊥ \ (∞) is regular if and only if O
is a translation oval if and only if each point of 〈W, (∞)〉 is regular if and
only if T2(O) is self-dual.
Now suppose that some line L not incident with (∞) is regular. L meets
some line M incident with (∞), say at the point X. Since the translations are
(sharply!) transitive on the points not collinear with (∞) it follows readily
that each line meeting M is also regular. Suppose this is the case. Now
suppose that {L1, L2} is any pair of skew lines not incident with (∞). If either
of them meets M it is regular, so the pair {L1, L2} is regular. So suppose
that neither of them meets M. Since M is regular, any triad containing M
has a transversal, so there is a line L that meets each of M, L1, L2. Since
L is regular, the pair {L1, L2} must also be regular. Hence every line is
regular. Since q = 2 e , this means that every point is regular and T2(O) is
again isomorphic to T2(C) for a conic C. (Actually, it forces O to be a conic.)
Now suppose that each of two lines M1, M2 through (∞) has a regular
point in addition to (∞), so each point of Mi is regular, i = 1, 2. Without
loss of generality we may assume that M1 = [∞] and M2 = [0]. Hence α is
additive and T2(O) is self-dual by a duality interchanging (∞) and [∞], and
mapping [0] to a point of [∞] different from (∞) and the regular point of [0]
to a regular line (not [∞]) through the image of [0]. Hence T2(O) is classical.
This means that we have the following possibilities for regular points or lines.
Theorem 10.5.2. In any T2(O) the point (∞) and the lines through it are
regular. If T2(O) is not classical, then no other line can be regular and no
point not collinear with (∞) can be regular. In this case it follows that every
collineation of T2(O) must fix the point (∞). If some point X (not (∞)) on
a line M through (∞) forms a regular pair with even one point not in (∞) ⊥ ,
then we may assume that O is a translation oval Oα with α a generator of
the automorphism group of Fq. If T2(Oα) is not classical, the only regular
points are those on [∞] and the only regular lines are those incident with
(∞). In this case it follows that every collineation of T2(O) must fix the line
[∞].

10.6. EXPANDING T2(O) ABOUT A REGULAR ELEMENT 461
10.6 Expanding T2(O) about a Regular Element
Let O be an oval in π = P G(2, q) which is embedded as a hyperplane in
Σ = P G(3, q). We expand about the regular point π. From the proof
of Theorem 10.1.1 we see that we get the following construction of a GQ
T ∗ 2 (O+ ) of order (q − 1, q + 1).
10.6.1 P(T2(O), π)
Let T ∗ 2 (O+ ) = (P ∗ , B ∗ , I ∗ ) be the point-line geometry with pointset P ∗ equal
to the set of points of Σ \ π, and with lineset B ∗ equal to the set of lines of
Σ not in π that meet π in a point of the unique hyperoval O + containing O.
Then T ∗ 2 (O+ ) is a GQ of order (q − 1, q + 1).
This example was first given independently by R. W. Ahrens and G.
Szekeres [AS69] and by M. Hall, Jr. [Ha71]. In the former paper an
analogous construction was given for q odd by means of coordinates.
Let x be a fixed point of O + . The q 2 lines of Σ incident with x but not in
π form a spread of T ∗ 2 (O+ ). The set of all these spreads is a packing . Later
we will consider the process of starting with such a packing and recapturing
T2(O). For the present we merely realize that we could have started with
any point x of O + as the nucleus of an oval Ox in π. Then expanding T2(Ox)
about the regular point π will give the GQ T ∗ 2 (O+ ).
10.6.2 P(T2(O), L)
Now we try expanding in the “other” direction. For this construction we
do not need to restrict q to be even. Let L be the line of π tangent to
O at the point Q. For all q, L is a regular line of T2(O). If we expand T2(O)
about the regular line L we obtain the following GQ S ′′ = (P ′′ , B ′′ , I ′′ ) of
order (q + 1, q − 1).
The pointset P ′′ has three types of points:
(i) The points of Σ \ π;
(ii) The planes that meet π in a line tangent to O at a point different
from Q;
(iii) The planes through Q not containing L.

462 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
The lineset B ′′ consists of all lines of Σ\π meeting π at a point of O\{Q}.
When q = 2 e and N is the nucleus of O, we can describe the points of type
(ii) as the planes of Σ that contain N but not Q, and the points of type (iii)
are the planes of Σ that contain Q but not N.
One could also start with an arbitrary point P of W (q) and expand about
P to get a GQ with parameters (q − 1, q + 1), to get the point-line dual of
the example just given. In this form the GQ is isomorphic to the examples
first found by Ahrens and Szekeres [AS69].
At this point we want to establish a more uniform notation for the three
types of GQ that we have associated with an oval in case q = 2 e . Since
each q-arc of P G(2, q) extends uniquely to a hyperoval, we suppose that O +
is a hyperoval of π∞ embedded as a hyperplane in Σ = P G(3, q). Let P +
denote the set of points of Σ \ π∞ and let B + denote the set of lines of Σ not
contained in π∞ and meeting π∞ in a point of O + .
Construction 10.6.3. S + = S(O + ) = (P + , B + , I + ) is the GQ with parameters
(q − 1, q + 1) whose points are the points of P + and whose lines are the
lines of B + . Incidence I + is just that inherited from incidence in Σ.
Let x be any point of O + and put Ox = O + \ {x}. Then Ox is an oval of
π∞ with nucleus x. Let Ox denote the set of planes of Σ containing the point
x, and let Bx denote the set of lines of Σ not contained in π∞ and meeting
π∞ in a point of Ox. Then we have the following construction.
Construction 10.6.4. Sx = S(Ox) = (Px, Bx, Ix) is the GQ with parameters
(q, q) whose points are the points of Px = P + ∪ Ox and whose lines are the
lines in Bx. Incidence Ix is the one naturally induced by incidence in Σ.
(Usually the nucleus x of Ox is understood and we write S = S(O) in place
of Sx = S(Ox).)
Now let x and y be any two points of O + and put O − xy = O+ \ {x, y}.
Then O − xy is a q-arc contained in a unique hyperoval O + . Let Oxy denote the
set of planes of Σ containing x but not y, so naturally Oyx denotes the set of
is the set of lines of Σ not contained
planes of Σ containing y but not x. B− xy
in π∞ and meeting π∞ in a point of the q-arc O− xy.
Construction 10.6.5. S− xy = S(Oxy) = (P − xy , B− xy , I− xy ) is the GQ with parameters
(q + 1, q − 1) whose points are the points of P − x,y = P + ∪ Oxy ∪ Oyx
and whose lines are the lines of B− xy . Incidence I− xy is just that inherited from
incidence in Σ.

10.7. CONSTRICTING ABOUT A REGULAR SPREAD 463
It should be clear that if we start with Sx and expand about the regular
point π∞ we obtain the Gq S + . Also, if we start with Sx and expand about
the line L where L is the line of π∞ through x meeting Ox in the point y,
. We now want to consider the reverse problem:
then we obtain the GQ S− xy
How to recover Sx from either S + or S− xy.
10.7 Constricting about a Regular Spread
Let S = (P, B, I) be a GQ of order s with a regular point x. Put P + x = P \x⊥ ,
B1 x = {L ∈ B : x is not incident with L}, B2 x = {{x, y} ⊥⊥ : y ∈ P \ x⊥ },
B + x = B1 x ∪ B2 x . If I+ x is the natural incidence on P + x × B+ x induced by I and
containment, then S + x = (P + x , B+ x , I+ x ) is the GQ P(S, x) of order (s−1, s+1)
obtained by expanding S about the point x.
Let L0, L1, . . . , Ls be the lines of S incident with x. Put
Mi = {L ∈ B 1 x : L ∼ Li}, 0 ≤ i ≤ s.
Then each Mi is a spread of S + x , i.e., a set of 1 + (s − 1)(s + 1) = s 2 lines
that partitions the points. Similarly, write M∞ for B 2 x . Then M∞ is also a
spread, so that
M = {M∞, M0, M1, . . . , Ms}
is a packing of S + x , that is a partition of the lines of S+ x into spreads. Moreover,
M∞ is a regular spread. This means that each pair {L, M} of distinct lines
of M∞ is a regular pair, so |{L, M} ⊥⊥ | = s, the maximum possible size,
and {L, M} ⊥⊥ ⊂ M∞. But M∞ is even pivotal for the packing M: For
L, M ∈ M∞, L = M, {L, M} ⊥ ⊂ Mk for some k ∈ {0, 1, . . . , s}.
The fact that M∞ is a regular spread which is pivotal for the packing M
can be viewed in another way.
Consider the elements of M∞ as “points” and the spans {L, M} ⊥⊥ of distinct
elements of M∞ as “lines.” Then M∞ with these “points” and “lines”
is an affine plane π(M∞) of order s. The spreads M0, . . . , Ms determine the
parallel classes of lines of π(|mM∞) in the following way. For distinct M1,
N1 ∈ M∞, let T1 = {M1, N1} ⊥⊥ be the “line” of π(M∞) through M1 and
N1. Then T ⊥ 1 = {M1, N1} ⊥ is a subset of some Mk, 0 ≤ k ≤ s. Moreover,
if Ti = {Mi, Ni} ⊥⊥ , 1 ≤ i ≤ s, are the distinct “lines” of the parallel class of
π(M∞) containing T1, then each T ⊥
i is contained in Mk, and {T ⊥ 1 , . . . , T ⊥ s }

464 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
partitions Mk. Two spans (i.e., “lines” of π(M∞)) of M∞ have an element
in common if and only if their perps belong to different members of
{M0, . . . , Ms}. In this way each spread Mk, 0 ≤ k ≤ s, determines one
parallel class of π(M∞).
We are now ready to consider the reverse process of recovering S from
S + x by constricting S+ x
about a regular spread M∞.
Let S + = (P + , B + , I + ) be a GQ with parameters (s − 1, s + 1). Let
M∞ be a regular spread of S + . So M∞ is a set of s 2 lines, no two concurrent,
such that each pair {L, M} of distinct lines of M∞ is regular and
{L, M} ⊥⊥ ⊂ M∞. It follows immediately that the lines and spans of M∞
form the “points” and “lines,” respectively, of an affine plane π(M∞) of
order s. If Ti = {Mi, Ni} ⊥⊥ , 1 ≤ i ≤ s, are the distinct “lines” in one
parallel class Ek of π(M∞), then Mk = ∪s ⊥
i=1Ti = ∪si=1 {Mi, Ni} ⊥ = Ek is
a spread of S + . If M0, . . . , Ms are the spreads obtained in this way, then
M = {M∞, M0, . . . , Ms} is a packing of S + for which M∞ is pivotal.
Construction 10.7.1. Constriction about a regular spread:
Let S + = (P + , B + , I + ) be a GQ with parameters (s − 1, s + 1). Let M∞ be
a regular spread of S + for which M = {M∞, M0, . . . , Ms} is the associated
packing for which M∞ is pivotal.
The lines of B are to be given by
B = (B + \ M∞) ∪ {[M0], . . . , [Ms]}.
The elements of P are to be of three types:
(i) the elements of P + ;
(ii) the perps {L, M} ⊥ with L, M ∈ M∞, L = M;
(iii) the symbol ∞.
Incidence is defined as follows: If x ∈ P + , L ∈ B + \ M∞, then xIL if and
only if xI + L. If x ∈ P + and L = [Mi], then it is NOT true that xIL. If
x = {L, M} ⊥ , L, M ∈ M∞, L = M, and N = [Mi], then xIN if and only
if {L, M} ⊥ ⊂ Mi. Finally, (∞)I[Mi] for each i = 0, 1, . . . , s. Then S is a
GQ with order s for which (∞) is a regular point. In addition, Mj is pivotal
for M if and only if [Mj] is regular as a line of S.
Proof. It should be fairly clear that S is a GQ with order s for which (∞) is
a regular point, and that if S is expanded about the point (∞), the GQ S +

10.8. CHARACTERIZING T ∗ 2 (O + ) WITH (0, 2)-SETS 465
is obtained. So we want to consider what it means for [Mj] to be regular as
a line of S. Without loss of generality we may consider what it means for
[M0] to be regular. Let L be a line of S not concurrent with [M0], i.e., L
belongs to Mk for some k with 1 ≤ k ≤ s. Let x and y be distinct points of L
in S + . Let Mx and My, respectively, be the lines through x and y belonging
to M0. Let z be a point of My not on L or [M0], and let Lz be the line
through z meeting Mx in a point w. To be able to guarantee that Lz meets
[Mk] is equivalent to saying that {Mx, My} is regular and that {Mx, My} ⊥⊥
is contained in a single member of M, which would have to be Mk since L
is in Mk. It follows that [M0] is regular in S if and only if M0 is regular
and pivotal in S + .
10.8 Characterizing T ∗ 2 (O+ ) with (0, 2)-Sets
We are going to identify some properties of T ∗ 2 (O + ) and show that these
properties characterize such a GQ. This work derives primarily from [DST84]
and [DST86].
Let O + be a hyperoval in π∞ embedded as a hyperplane in Σ = P G(3, q).
Recall that T ∗ 2 (O+ ) was obtained from T2(Ox) (for any point x of O + ) by
expanding about the regular (and coregular) point π∞. Hence each of the
spreads of M = {M∞, M0, . . . , Mq} is (regular and) pivotal for M.
Let x and y be any two points of Σ \ π∞ and put Lxy = 〈x, y〉. If Lxy
meets π∞ in a point of O + , then Lxy is a line of T ∗ 2 (O). But what if Lxy
meets π∞ at a point not on O + ? Let z be any point of Σ \ (Lxy ∪ π∞). The
plane π = 〈z, Lxy〉 meets π∞ in a line that meets O + in 0 or 2 points, which
means that in T ∗ 2 (O+ ) z is collinear with 0 or 2 points of Lxy. For this reason
we call the set of points on Lxy a (0, 2)-set. The basic observation here is that
there is a basic collection B1 of (0, 2)-sets such that each pair of noncollinear
points of T ∗ 2 (O+ ) belongs to a unique member of B1.
Until further notice we let S = (P, B, I) be a GQ with parameters (s, t),
s > 1, t > 1.
Definition 10.8.1. A nonempty subset K of pairwise noncollinear points
of P is said to be a (0, 2)-set provided |x ⊥ ∩ K| ∈ {0, 2} for each point
x ∈ (P \ K).
Lemma 10.8.2. If K is a (0, 2)-set in S, then |K| = s + 1.

466 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
Proof. Fix a line L ∈ B on a point x of K. By the definition of a (0, 2)-set
and the fact that there are no triangles in S, different points of K \ {x} are
collinear with different points of L, and each point of L\{x} is collinear with
a unique point of K different from x. Hence we have a bijection from the
points of K to the points of L and |K| = s + 1.
Lemma 10.8.3. If S has a (0, 2)-set, then s is odd.
Proof. The number of lines having no point in common with K is (1 + t)(1 +
st) − (1 + t)(1 + s) = (1 + t)s(1 − t) > 0, so there is a line L disjoint from
K. Each point of K is collinear with a unique point of L, which in turn is
collinear with a second point of K. Hence the points of K are split into pairs
with the two points in a pair being collinear with the same point of L. Hence
s + 1 is even. We also see that half the points of L are collinear with two
points of K and half are collinear with no point of K.
Example 10.8.3.1. The following examples show that the definition is not
vacuous.
(a) In the GQ W (q) there can be no (0, 2)-set. First, by Lemma 10.8.3 if
there were a (0, 2)-set K, q would have to be odd. But then any three
points of K would be a triad. Since all points of W (q) are regular, each
triad must be centric, contradicting the definition of (0, 2)-set.
(b) Consider the GQ Q = Q(4, q) with q odd. Let x, y, z be pairwise
noncollinear points of Q for which x ⊥ ∩ y ⊥ ∩ z ⊥ = ∅. (Reminder: all
lines of Q are regular, so all points are antiregular, so that each triad of
points has 0 or 2 centers.) The plane π = 〈x, y, z〉 meets Q in a conic
C. Note that |C| = q + 1. We claim that C is a (0, 2)-set in Q. Since
x ⊥ ∩y ⊥ ∩z ⊥ = ∅, the polar line L of π has no point in common with Q.
Let u be an arbitrary point of Q \ C. Suppose first that |u ⊥ ∩ C| ≥ 3.
Then u ⊥ contains π and hence contains C. This would contradict the
choice of the points x, y, z so that x ⊥ ∩y ⊥ ∩z ⊥ = ∅. Hence we see that
|u ⊥ ∩C| ≤ 2. Suppose that |u ⊥ ∩C| = 1, say u ⊥ ∩C = {w}. Then each
point of C \ {w} must be collinear with a unique point of uw \ {u, w}.
Hence q = |C \ {w}| ≤ |uw \ {u, w}| = q − 1, an impossibility. Hence
|u ⊥ ∩ C| = 0 or 2, showing that C is a (0, 2)-set.
(c) Let O be a hyperoval in the plane π embedded as a hyperplane of Σ =
P G(3, q). Consider the GQ S = T ∗ 2 (O) of order (q − 1, q + 1), where

10.8. CHARACTERIZING T ∗ 2 (O + ) WITH (0, 2)-SETS 467
necessarily q is even. Let L be a line of Σ not in π and meeting π in a
point that is not in O. We claim that the q points of K = L \ π form
a (0, 2)-set in S. Let x ∈ P = Σ \ π be a point of S not on L. Then in
Σ, the plane 〈x, L〉 meets π in a line M. If M is a secant of O, then x
is collinear in S with 2 points of K. If M is exterior to O, then x is
collinear in S with no point of K.
Let B1 be the set of all (0, 2)-sets of S of the type just constructed. So
any two noncollinear points of S lie in a unique member of B1.
(d) Consider the special case of the previous example where O is a hyperconic
C (a conic C ′ plus its nucleus N). We determine all (0, 2)-sets
of T ∗ 2 (C). Since q must be even, we know that T2(C ′ ) ∼ = W (q), so
that T ∗ 2 (C) ∼ = P(W (q), x), where x is any point of W (q). Let K be a
(0, 2)-set in the GQ S = P(W (q), x), and let z1, z2, z3 be three distinct
points of K. Then z1, z2, z3 are pairwise noncollinear in S, but they
could lie on a line of Σ. Suppose they lie on a line L of Σ. Let ν be the
symplectic polarity of Σ defining W (q). Since L does not contain x,
the polar line L ′ (with respect to ν) of L is not contained in the polar
plane x ν . If y ∈ L ′ \ x ν , then in P(W (q), x), y is collinear with the
three points z1, z2, z3 of K, a contradiction. So z1, z2, z3 define a
unique plane H of Σ. The point H ν is conjugate (with respect to ν) to
all points of H, especially to z1, z2, z3. So if H ν were in P , it would
be a point of S collinear with three points of K, an impossibility. It
follows that H ν must be collinear in W (q) to x, i.e., H ν ∈ x ν , implying
x ∈ H. It follows that H is already determined just by z1 and z2 (i.e.,
H = 〈x, z1, z2〉). As z3 varies over the other points of K we see that
K ⊂ H. We know already that no three points of K are collinear in Σ,
and since no two points of K are collinear in P(W (q), x), the line of Σ
defined by any two points of K does not contain x or H ν . We conclude
that K ∪ {x, H ν } is a hyperoval of the plane H.
Conversely, let O∗ be a hyperoval of a plane H = xν of Σ, with x, Hν ∈
O∗ . We show that K = O∗ \ {x, Hν } is a (0, 2)-set of P(W (q), x).
Clearly no two ponts of K are collinear in P(W (q), x). We now claim
that z ⊥ 1 ∩ z⊥ 2 ∩ z⊥ 2
= ∅ for any three points of K. So suppose that
u ∈ (z ⊥ 1 ∩ z ⊥ 2 ∩ z ⊥ 2 ). If uz1, uz2, uz3 are lines of P(W (q), x), then they
are contained in the plane u ν containing u. But this means the plane u ν
is H and u = H ν . This is impossible since H ν is not in P(W (q), x).

468 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
So suppose, for example, that uz1 is a line of W (q) not in P(W (q), x),
so it must go through the point x. In this case uz2 and uz3 cannot pass
through x and must be lines in P(W (q), x). Since u is collinear with
all three of z1, z2, z3 in W (q), it must be that u ∈ 〈z1, z2, z3〉 = H, so
uν = H and u = Hν , which is still impossible since u ∈ P(W (q), x) but
H ν ∈ P(W (q), x). Hence z ⊥ 1 ∩z⊥ 2 ∩z⊥ 2
= ∅. This means that each point
of P \ K is collinear in P(W (q), x) with at most two elements of K.
So suppose that u ⊥ ∩K = {w}. Each of the q −1 points z of K \{w} is
collinear in P(W (q), x) with a unique one of the q − 2 points (distinct
from u and w) of the line uw of P(W (q), x). And each of the points
of uw different from u and w is collinear with just one of the points of
K \ {w}, so we have an impossibility. This means that K is a (0, 2)-set
of P(W (q), x).
Let H be any plane containing x and for which H ν = x. For each pair
(z1, z2) of points of H such that {x, H ν , z1, z2} is a 4-arc, the number
of (0, 2)-sets of P(W (q), x) containing z1 and z2 is the number of
hyperovals in P G(2, q) containing a given 4-arc.
(e) Consider again the GQ T ∗ 2 (O) where O is a hyperoval in the plane H
of Σ. Let H ′ be a plane different from H and meeting H in a line
disjoint from O. Let O ′ be a hyperoval of H ′ disjoint from H. For a
fixed point x ∈ O, denote by Li, 1 ≤ i ≤ q + 2, the lines of Σ joining
x to the points of O ′ . We claim that the set K consisting of the lines
L∗ i = Li \H of T ∗ 2 (O), 1 ≤ i ≤ q+2, is a (0, 2)-set of the point-line dual
of T ∗ 2 (O). This means that we need to show that a line of T ∗ 2 (O) \ K
meets 0 or 2 lines of K. Such a line through x clearly meets 0 lines
of K. Let L be a line of Σ meeting H in the point y, y = x. Suppose
L meets the line M ∈ K and that M meets O ′ in the point z. So the
plane 〈L, M〉 meets H ′ in a secant line to O ′ meeting O ′ at z and at
some other point z ′ . So L meets the lines M and M ′ = 〈x, z ′ 〉 of K.
Since no three of the lines of K can be coplanar, the claim is proved.
10.9 The Basic Assumptions
Until notified otherwise we assume that S = (P, B, I) is a GQ with order
(s, t), s > 1, t > 1. If either s ≤ 3 or t ≤ 3, all possibilities for S are
known. Hence we also assume that s ≥ 4 and t ≥ 4. We are going to assume

10.9. THE BASIC ASSUMPTIONS 469
two basic axioms that will allow us eventually to characterize T ∗ 2 (O) where
O is a hyperoval. To begin, we assume that there are appropriately many
(0, 2)-sets.
Basic Assumption 1. (BA1) S contains a collection B1 of (0, 2)-sets such
that each pair of non-collinear points is contained in a unique element of B1.
Lemma 10.9.1. Let x ∈ P , L ∈ B, with x ∈ L. There is one point y0 of L
for which xy0 ∈ B. Let y1, . . . , ys be the other points of L and let Li be the
(0, 2)-set in B1 guaranteed by Basic Assumption 1 to contain x and yi. Then
for each choice i, j of two different indices i, j ∈ {1, 2, . . . , s} there exists a
line M ∈ B such that M ∩ L = ∅ and M ∩ Li = ∅ = M ∩ Lj.
Proof. Fix i, j ∈ {1, . . . , s}, i = j. The set y⊥ i ∩ Lj contains the point yj.
Hence by the definition of a (0, 2)-set, |y⊥ i ∩ Lj| = 2. So there exists a second
point y ′ j ∈ Lj, y ′ j = yj, such that y ′ j ∈ y⊥ i . Again from the definition of
(0, 2)-set it follows that there is a second point y ′ i ∈ y ′ j ∩ Li, y ′ i = yi. Then
y ′ iy′ j ∈ B and y′ iy′ j ∩ L = ∅.
Basic Assumption 2. (BA 2) (Usually known as Axiom (T)): Let x ∈ P ,
L ∈ B with x ∈ L. Then there exists s (0, 2)-sets in B1 containing x and a
point of L. We denote these sets by L1, . . . , Ls. If M ∈ B, M ∩ L = ∅, and
if for at least two different indices i, j ∈ {1, . . . , s} we have both M ∩ Li = ∅
and M ∩ Lj = ∅, then M ∩ Lk = ∅ for all k ∈ {1, . . . , s}.
From now on we assume that both BA 1 and BA 2 are satisfied. Then we
define an incidence structure S ′ = (P ′ , B ′ , ∈) where P ′ = P , B ′ = B ∪ B1.
By Lemmas 10.8.2 and 10.8.3, S ′ is a 2−((s+1)(st+1), s+1, 1)-design with
s odd. The elements of B ′ will be called blocks. In the case where s = q − 1
and t = q + 1, such a design has the parameters of affine 3-space with q even.
A major goal is to prove that this design is the affine space AG(3, s + 1).
Lemma 10.9.2. Axiom (T) is equivalent to the axiom arising from Axiom
(T) when the condition M ∩ L = ∅ is omitted.
Proof. Let x, L, L1, . . . , Ls, y1, . . . , ys be given as in the hypothesis of
Lemma 10.9.1. Let M ∈ B be such that for certain indices i and j, M ∩ Li =
{yi}, M ∩ Lj = {y ′ j } and y′ j = yj. So in this case M ∩ L = ∅. Suppose that
for an index r with i = r = j we have M ∩ Lr = ∅. Since yr ∈ y⊥ j ∩ Lr, it
must be true that |y⊥ j ∩Lr| = 2. So there exists a point y ′ r ∈ Lr, y ′ r = yr, such
that y ′ r ∈ y⊥ j and yjy ′ r = M. Since y′ ⊥
r ∩ Lj contains the point yj ∈ Lj, it

470 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
follows that y ′ ⊥ ′
r contains a second point y j ∈ Lj, y ′ j = yj. We have y ′ jy′ r ∈ B,
y ′ jy ′ r ∩ L = ∅, and y ′ jy ′ r ∩ M = ∅. Using Axiom (T) we have y ′ jy ′ r ∩ Lk = ∅, for
all k ∈ {1, . . . , s}. If we consider the point x and the line y ′ jy′ r , we have three
(0, 2)-sets Lj, Lk, Lr in B1 joining x to a point of y ′ jy′ r . Since M ∩ y′ jy′ r = ∅
and M ∩ Lj = ∅, M ∩ Lk = ∅, it follows from Axiom (T) that M ∩ Lr = ∅, a
contradiction. Hence we may conclude that M ∩Lk = ∅ for all k ∈ {1, . . . , s},
completing the proof.
From now on S satisfies BA 1 and BA 2, and we let x, L, L1, . . . , Ls be as
in the statement of Axiom (T), and y1, . . . , ys as in the proof of Lemma 10.9.1.
Lemma 10.9.3. Using the notation of Axiom (T) suppose that L ∩ M = ∅.
Then {L, M} is a regular pair.
Proof. Then M ∩ Lk = ∅ for all k ∈ {1, . . . , s}. Since y⊥ k ∩ Lr = ∅, for
all r = k, by the definition of (0, 2)-set Lr contains a point ykr such that
yk ∼ ykr. As we saw in the proof of Lemma 10.9.2, the s − 1 points ykr,
1 ≤ r ≤ s, r = k, are collinear. The only way this can happen is for there
to be a line Nk that contains the points yk and ykr, r = k. So there arise
s lines, each of them having a point in common with each of L1, . . . , Ls. In
this way we get s2 points that are exactly the points of Li \ {x}, 1 ≤ i ≤ s.
This means that we have such a line Nk through each common point of M
and Lk, 1 ≤ k ≤ s. Hence N1, . . . , Ns ∈ {L, M} ⊥ .
Let Ns+1 be the unique line of B through x meeting L, say at the point z,
and let zk = M ∩ Nk, 1 ≤ k ≤ s. We show that Ns+1 ∩ M = ∅. Since for each
k we have zk ∈ L1 ∪ L2 ∪ · · ·∪Ls, it must be that zk ∼ x. On the other hand,
a point zk cannot be collinear with z ∈ Ns+1 ∩ L without forcing a triangle
in S, so we only have s − 1 points on Ns+1 which might be collinear with a
point zk ∈ M. If Ns+1 ∼ M, then any of these s − 1 points is collinear with
at most one of the points z1, . . . , zs, a contradiction. Hence M ∩ Ns+1 = ∅.
Hence Ns+1 ∈ {L, M} ⊥ , forcing {L, M} ⊥ = {N1, . . . , Ns, Ns+1}.
Now consider an arbitrary line Nk ∈ {L, M} ⊥ , k = s + 1. With the
previous notation we have {ykr} = Nk ∩Lr, r ∈ {1, . . . , s}, r = k, {ykr, yk} ⊂
Nk, {yk} = Nk ∩ Lk ∩ L. From the definition of a (0, 2)-set it follows that for
each r = k there exists another point y ′ kr ∈ Lk such that ykr ∼ y ′ kr . By Axiom
(T) the line ykry ′ kr has one point in common with each of the sets Li \ {x},
It follows that there are s − 1 lines ykry ′ kr . Label these lines as K1, . . . , Ks−1,
so that |Ki ∩ Nj| = 1, 1 ≤ i ≤ s − 1, 1 ≤ j ≤ s. Hence applying Lemma 9.3.5
of TFG we see that |Ki ∩ Ns+1| = 1. Consequently L and the s − 1 lines Ki

10.9. THE BASIC ASSUMPTIONS 471
belong to {L, M} ⊥⊥ (since M must be one of the Ki). Through the point x
on Ns+1 there is a line of B meeting Nk, for 1 ≤ k ≤ s. But there is only
one point left on each of the Nk available to be collinear with x and these
points must be collinear with each other, so there is a line K meeting Ns+1
at x and meeting each of the lines Nk, 1 ≤ k ≤ s. It follows that
and {L, M} is a regular pair.
{L, M} ⊥⊥ = {L, K, K1, . . . , Ks−1},
Note: The grid defined by x and L will be denoted by G(x, L).
We continue with the same notation as above: x ∈ P ; L ∈ B; x is not
incident with L in S. Using BA 1 and BA 2 we construct a unique grid
G(x, L) with lines N1, . . . , Ns+1 in one ruling of G(x, L) with xINs+1 and
lines {L, K, K1, . . . , Ks−1} in the other ruling with xIK. Define an incidence
structure S ∗ = (P ∗ , B ∗ , ∈) as follows. Put P ∗ = {y ∈ P : y ∈ Ni, Ni ∈
{L, M} ⊥ , 1 ≤ i ≤ s + 1}; B ∗ = {L, M} ⊥ ∪ {L, M} ⊥⊥ ∪ B ∗ 1, where B ∗ 1 is the
set of all (0, 2)-sets in B1 containing at least two points of P ∗ .
Lemma 10.9.4. The structure S ∗ = (P ∗ , B ∗ , ∈) is an affine plane of order
s + 1, s odd.
Proof. Clearly |P ∗ | = (s+1) 2 , and by Lemma 10.8.2 each block (i.e., element
of B ∗ ) contains contains s + 1 points of P . We claim that for any two points
u, v ∈ P ∗ , the block uv of S ∗ contains s + 1 points of P ∗ . The case u ∼ v
is trivial, so we may assume that at least one of u, v is not on L, say u ∈ L.
If we can show that G(x, L) = G(u, L), it would follow that either v ∈ L,
in which case uv is one of the (0, 2)-sets used in defining P ∗ , or u ∈ L. In
the latter case we may assume that u ∼ v. Since the blocks uw, w ∈ L
cover all the points of P ∗ , including v, we have that uv is one of the blocks
defining G(u, L) = G(v, L) and contained in P ∗ . Clearly we may assume
that u = x = v. There are three cases.
Case 1 u ∼ x. In this case the (0, 2)-set Lk ∈ B1 defined by u and x
belongs to the point sets of the grids G(x, L) and G(u, L). If u ∈ Nr, then
clearly Nr belongs to both grids. If y ′ r ∈ Nr, y ′ r = u, y′ r = yr, then y ′ ⊥
r
⊥
contains a
contains the point u of Lk. By the definition of a (0, 2)-set, y ′ r
second point of Lk, for example u ′ . The lines L, Nr, y ′ u ′ belong to the grids
G(x, L) and G(u, L). Consequently we may conclude that the two grids
coincide.

472 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
Case 2 u ∼ x with ux ∩ L = ∅. In this case choose a point w ∈ P ∗
such that w ∼ x, w ∼ u, w ∈ L. (This is always possible since since are are
assuming that s ≥ 4.) By Case (1) we have G(x, L) = G(w, L). Using the
same reasoning as in Case (1) it follows that G(u, L) = G(w, L). Hence we
have G(x, L) = G(u, L).
Case 3 u ∼ x and |ux ∩ L| = 1. In this case choose a point w ∈ P ∗
such that w ∼ x, w = x and wx ∩ L = ∞. From the previous cases it
follows that G(x, L) = G(w, L). On the other hand, since u is a point of
G(x, L) = G(w, L) and w ∼ u, we have G(u, L) = G(w, L). Again we may
conclude that G(x, L) = G(w, L).
It is now clear that S ∗ is an affine plane of order s + 1 (with s odd).
Corollary 10.9.5. If x ∈ P , L ∈ B with x ∈ L, then the substructure of S ′
generated by x and L is an affine plane of order s + 1.
Proof. Let N be the unique line of B incident with x and meetiing L, say
in the point z. With the s remaining points of L we define s (0, 2)-sets
L1, . . . , Ls of B1, each of them containing the point x. Lemmas 10.9.1, 10.9.3
and 10.9.4 finish the proof.
Corollary 10.9.6. If x ∈ P , L ∈ B1, with x ∈ L and |x ⊥ ∩ L| = 2, then the
substructure of S ′ generated by x and the line L, is an affine plane of order
s + 1.
Proof. If x ⊥ ∩ L = {y1, y2}, then xyi ∈ B. By Cor. 10.9.5 the substructure of
S ′ generated by y1 and the line xy2 is an affine plane of order s+1. This affine
plane, , containing x and L = y1y2, is also the substructure of S ′ generated
by x and L.
Definition 10.9.7. The affine plane of Lemma 10.9.4 will be called an affine
plane of type 1. Each point in such a plane is incident with two lines of B
and s (0, 2)-sets of B1. The lines of B in such a plane form a (s+1)×(s+1)
grid. Moreover, for each point x and each (0, 2)-set L ∈ B1, x ∈ L, in the
plane, we have |x ⊥ ∩ L| = 2.
Lemma 10.9.8. Let α be an affine plane of type 1 in S ′ with pointset Pα.
Let u ∈ P ′ \ Pα and let u ⊥ ∩ Pα = {x1, . . . , xs+1}. If x ′ and x ′′ are the points
of α (the projective closure of α) defined by the two parallel classes of lines
of B in α. Then the set O = {x1, . . . , xs+1, x ′ , x ′′ } is a hyperoval in α.

10.9. THE BASIC ASSUMPTIONS 473
Proof. Let u ∈ P ′ \ Pα. The main axiom for GQ applies to u and the lines of
the grid in α. So we can put u ⊥ ∩ Pα = {x1, . . . , xs+1}. The lines of the grid
give rise to two parallel classes X ′ and X ′′ in α, which determine two points
x ′ and x ′′ in α. We claim that O = {x1, . . . , xs+1, x ′ , x ′′ } is a hyperoval in α.
As the set contains s + 3 points, we have only to prove that each line of α
meets O in 0 or 2 points.
Clearly the line x ′ x ′′ of α has only the points x ′ and x ′′ in common with
O. Next consider a line R ∈ B in α. By the main axiom for GQ (applied to
S) we have |u ⊥ ∩ R| = 1, i.e., u ⊥ ∩ R = {xi}, for a certain i ∈ {1, . . . , s + 1}.
Moreover, R belongs to exactly one of the two parallel classes, say X ′ . Hence
x ′ belongs to R, with R the projective closure of R in α. Consequently
|R ∩ O| = |{xi, x ′ }| = 2.
Finally we suppose that R is a line of α belonging to B1. Then |u ⊥ ∩R| ∈
{), 2} and hence |R∩(O \{x ′ , x ′′ })| ∈ {0, 2}. Since R is an element of neither
of the parallel classes X ′ , X ′′ , there holds for R, the projective closure of R
in α, |R ∩ O| ∈ {0, 2}.
Hence we conclude that O is a hyperoval in α.
Theorem 10.9.9. Let S be a GQ with parameters (s, s + 2) satisfying BA
1 and BA 2. Then S has a packing of spreads for which each member is
pivotal. Hence a GQ S1 may be constructed with a biregular point (∞) for
which S ∼ = P(S1, (∞)). (This just means that (∞) is regular and each line
through (∞) in S1 is regular.)
Proof. We define a relation “” (“is parallel to”) on the set B of lines of S.
For L, M ∈ B, put LM iff there exists an affine plane α of type 1 such that
L and M are parallel lines in α, in which case we may also write LαM.
Claim 1: Given any point x and any line L ∈ B, there is exactly one line
through x which is parallel to L. To see this, count the number of lines of B
which are parallel to L. First note that |P | = (1 + s)(1 + s(s + 2)) = (1 + s) 3 .
If L is a fixed line, there are (s+1)(s+2)s points x not on L in P , each giving
a plane whose points are those of the grid G(x, L) with (s+1)s points not on
L and giving the same plane. Hence we obtain (s+1)(s+2)s/(s+1)s = s+2
planes on L, each two intersection precisely in L. Each of these planes has
s lines parallel to L and distinct from L, so together we have s(s + 2) + 1 =
(s + 1) 2 lines parallel to L (including L). If two lines M and N are both
parallel to L, M = N, then M and N are disjoint. Hence the lines parallel
to L cover exactly (s + 1) · (s + 1) 2 points, i.e., the set of lines parallel to L
is a spread of S, proving the Claim 1.

474 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
Claim 2. is an equivalence relation in B. Since reflexivity and symmetry
are immediate, we have only to verify transitivity. So suppose L, M, N ∈ B
with LαM and MβN. If we exclude the trivial cases, we may suppose that
L, M and N are distinct and that α = β. We must show that there is an
affine plane γ of type 1 such that LγN.
Let x ∈ N. Applying Cor. 10.9.5 we see that the substructure of S ′
generated by x and L is an affine plane γ of type 1. In γ there is just one
parallel N ′ to L through x. We prove that N = N ′ . Suppose the contrary
and let K ∈ {L, M} ⊥ , so K is a line of α. By the preceding, x is incident
with just one parallel to K. Since x is not a point of α (for otherwise α would
coincide with β), x ⊥ contains s + 1 points of α. We show that no one of the
s + 1 lines of B joining x to a point of α can be parallel to K.
So let R be a line of B through x meeting the plane α at the point y. In
α, y is incident with just one parallel K ′ to K. If RK, y would be incident
with two parallels K ′ and R to K, a contradiction. Also, the line N (resp.,
N ′ ) cannot be parallel to K, otherwise we would have two parallels, namely
M and K (resp., L and K) through M ∩ K (resp., L ∩ K) to N (resp.,
N ′ ). Besides, we notice that the line N (resp., N ′ ) has no point in common
with α, since otherwise α and β would coincide. So none of the s + 3 lines
through x, described above, is parallel to K. Since x is incident with exactly
s + 3 lines and since one of these lines has to be a parallel to K, we have
a contradiction. Hence N = N ′ , so is an equivalence relation with s + 3
equivalence classes in B, each consisting of (s+1) 2 elements. In other words,
is a packing of spreads.
Claim 3: Each equivalence class of is a regular spread, which is in fact
pivotal for the packing . Let R be an equivalence class of lines with respect
to . We must prove for L, M ∈ R that {L, M} ⊥⊥ ⊂ R and |{L, M} ⊥⊥ | =
s + 1. Since L, M ∈ R, there exists an affine plane α of type 1 such that
LαM. The plane α contains s + 1 parallels to L, all of them belonging
to {L, M} ⊥⊥ . In other words, |{L, M} ⊥⊥ | = s + 1, and from the definition
of parallelism in B it follows that {L, M} ⊥⊥ ⊂ R. This shows that R is a
regular spread.
If L1 and L2 are distinct lines of R, say contained in the plane α, the
s + 1 lines of {L1, L2} ⊥ are also contained in α and hence belong to one
equivalence class of . It follows that is a packing of spreads of S for which
each member of the packing is pivotal. Hence we can constrict about any
one of the equivalence classes and get a GQ S1 of order s + 1 with a regular
point (∞) for which S ∼ = P(S1, (∞)). Moreover, each line through (∞) in

10.10. PARALLELISM 475
S1 is regular.
10.10 Parallelism
We continue to assume that S is a GQ with parameters (s, s+2) satisfying the
basic BA 1 and Axiom (T). In the previous section we defined an equivalence
relation on the set B of lines of S so that each equivalence class is a regular
spread, the collection of equivalence classes was a packing of spreads, and
each member of the packing is pivotal for the packing. We now extend the
notion of parallelism.
10.10.1 Parallelism in the set of affine planes of type 1
Consider an affine plane α of type 1 in S ′ and let x ∈ P be such that x
does not belong to α. The set x ⊥ contains s + 1 points of α, so there exist
two lines L, M ∈ B incident with x but having no point in common with α.
If follows from Cor. 10.9.5 that the substructure of S ′ generated by L and
M is an affine plane β of type 1. We claim that α and β have no point in
common. To prove this, suppose that y ∈ (α ∩ β). The elements of B in β
belong to two different parallel classes. These parallel classes coincide with
the parallel classes containing the elements of B in α (the s + 1 elements
of B which are incident with x and meet α, define the other s + 1 parallel
classes). Constructing the two lines N1, N2 through y which belong to these
classes, we note that N1, N2 are lines of both α and β. Consequently α and
β coincide and x is a point of α, a contradiction. So we conclude that α and
β have no points in common.
Now we are able to define a parallel relation for affine planes of type 1.
Definition 10.10.2. Parallelism in the set of affine planes of type 1.
Two affine planes α and β of type 1 are said to be parallel (write αβ) if and
only if there exist lines L, M ∈ B in α, L ′ , M ′ ∈ B in β, with |L ∩ M| = 1,
|L ′ ∩ M ′ | = 1, LL ′ and MM ′ .
Since “” is an equivalence relation in B, it follows that the new parallelism
on planes of type 1 is also an equivalence relation. This relation defines
a partition in the set of all affine planes of type 1 in S ′ . It follows from the
preceding paragraphs that each equivalence class determines a partition of
the point set of S ′ .

476 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
Suppose that αβ, α = β, and that x is a point of β. There exist s + 1
lines of B containing x and a point of α. The two remaining lines of B
through x lie in β. The point x is joined to the points of α which are not in
x ⊥ by s 2 + s (0, 2)-sets of B1, while the s remaining (0, 2)-sets through x lie
in β.
Lemma 10.10.3. If two type 1 planes in S do not have a common block,
then they are parallel.
Proof. Let α, β be non parallel planes of type 1. The planes parallel to α
are denoted by α = α1, α2, . . . , αs+1. Clearly β = αi for all i so the points
of αi ∩ β are collinear in S ′ . Hence αi and β have at most s + 1 points in
common. Since β has (s + 1) 2 points and since α1, . . . , αs+1 partition the
points of P ′ , it follows that αi and β have exactly s + 1 points in common
for 1 ≤ i ≤ s + 1. Hence αi and β have a block in common. In particular,
|α ∩ β| = s + 1.
Let αα ′ , α = α ′ , and let α β. Further, let L, L ′ be the common blocks
of α, α ′ , respectively. Since L ∩ L ′ = ∅, the blocks L, L ′ are parallel in β.
Hence they both belong to B or to B1.
Let α β, αα ′ , ββ ′ with α, α ′ , β, β ′ planes of type 1. The common
block of α and β (resp., α ′ and β ′ ) is denoted by L (resp., L ′ ). Let γ,
α = γ = β, be a plane of type 1 containing L. We claim that L ′ is contained
in a plane γ ′ parallel to γ. Let x ∈ L ′ and let γ ′ be the plane containing x
and parallel to γ. The block common to γ ′ and α (resp., γ ′ and β) is denoted
by M (resp., N). We have LαM and LβN. If γ = γ ′ then M = N; if
γ = γ ′ , then M ∩ N = ∅. Hence M and N are parallel in γ ′ . Let M ′ (resp.,
N ′ ) be the common block of γ ′ and α ′ (resp., γ ′ and β ′ ). Then Nγ ′N ′ and
Mγ ′M ′ . Hence M ′ γ ′N ′ . Since x ∈ M ′ ∩ N ′ we have M ′ = N ′ . Now it is
clear that M ′ = N ′ = L ′ , so γ ′ contains L ′ .
We have proved the following lemma (where all planes are of type 1).
Lemma 10.10.4. Let α and α ′ be distinct parallel planes, β and β ′ be distinct
parallel planes, and α β. Put α ∩ β = L and α ′ ∩ β ′ = L ′ . If γ is a plane
through L distinct from both α and β, then there is a plane γ ′ through L ′ that
is parallel to γ.
This permits us to define a parallelism in B1. For L, M ∈ B1, define
LM provided there exist affine planes α, α ′ , β, β ′ of type 1 such that
α ∩ β = L, α ′ ∩ β ′ = M, αα ′ and ββ ′ . Clearly the relation “” is reflexive

10.10. PARALLELISM 477
and symmetric. Transitivity can be seen as follows. Let L, M, N ∈ B1 with
LM, MN, in which case there are type 1 affine planes α, α ′ , β, β ′ , π, pi ′ ,
ν, ν ′ with αα ′ , ββ ′ , ππ ′ , νν ′ , such that α ∩ β = L, α ′ ∩ β ′ = M = π ∩ ν,
N = π ′ ∩ ν ′ . Letting π, ν, α ′ play the respective roles of α, β, γ in the
previous lemma forces the existence of some type 1 plane α ′′ which is parallel
to α ′ and which contains N. Likewise, letting β ′ play the role of γ forces the
existence of some type 1 affine plane β ′′ which is parallel to β ′ and which
contains N. In fact N = α ′′ ∩ β ′′ . By the transitivity of “” on type 1 affine
planes, αα ′′ and ββ ′′ . Hence LN and we have the following:
Lemma 10.10.5. “” is an equivalence relation on B1.
Corollary 10.10.6. Each parallel class of B1 partitions the points ofP ′ .
Proof. Let L ∈ B1, x ∈ P ′ , and let let α, β be distinct type 1 affine planes
which meet at L. Furthermore, let α ′ , β ′ be distinct type 1 affine planes
containing x such that such that αα ′ , ββ ′ . Put L ′ = α ′ ∩ β ′ . So x ∈ L ′
and LL ′ . Suppose there is some L ′′ ∈ B1 which contains s and which is
parallel to L. this forces the existence of planes γ and δ through LO ′′ with
γα, δβ. but as classes of parallel planes partition P ′ , γ = α ′ , δ = β ′ .
Hence L ′′ = L ′ . Hence there is exactly one member of B1 parallel to L which
contains x (namely L ′ ). Observe that if x ∈ L, then L = L ′ .
Corollary 10.10.7. If two members of B1 are parallel as lines in a type 1
affine plane, then they are parallel under as defined above.
Proof. Let L, M ∈ B1 with LαM for some type 1 affine plane α. Let β
be some other type 1 plane through L; let y be a point of M, let β ′ be the
type 1 affine plane through y and parallel to β, and let M ′ be the common
block on α and β ′ . Ovserve then that M ′ L, and hence M ′ αL. This forces
M ′ = M, and hence LM.
Lemma 10.10.8. Let α be a type 1 affine plane which does not contain the
block L. Then α and L have no common point if and only if L belongs to
one of the parallel classes defined by the blocks of α.
Proof. First assume that L is parallel to some block M ⊂ α. If L, M ∈ B,
then they are parallel in an affine plane which meets α exactly at M and
hence L has no point of α. If L, M ∈ B1 then there are type 1 affine planes
γ, γ ′ , δ, δ ′ , with γγ ′ , δδ ′ , L = γ ∩ δ, M = γ ′ ∩ δ ′ . Suppose L met α in a

478 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
point x. This would force γ to meet α in a block N through x. If N met M
then γ could not be parallel to γ ′ , but NαM implies NM, which implies
NL, a contradiction. Hence L must miss α.
So now suppose that α and L have no common points. Then L is in a
parallel class containing a block of α as follows. If L ∈ B, choose x to be any
point of α. If L ∈ B1, choose x to be a point of α collinear in S with two
points of L. Let β be the type 1 affine plane generated by L and x, and let
M = α ∩ β. Because L misses σ, LβM, and hence LM.
This lemma has the following immediate corollary.
Corollary 10.10.9. Parallel affine planes of type 1 define the same s + 2
parallel classes of blocks.
Corollary 10.10.10. If α and β are type 1 affine planes such that two
distinct intersecting blocks L and M of α are parallel, respectively, to two
distinct intersecting blocks L ′ , M ′ of β, then αβ.
Recall that since all GQ with parameters (3, t) are known, we are assuming
that s > 3.
Lemma 10.10.11. Let y1z1u1 and y2z2u2 be two triangles of a type 1 affine
plane α which are perspective from the point w, with w, y1, z1, u1, y2, z2, u2
all distinct points. Also assume that y1z1, y1u1, z1u1, y2z2, y2u2, z2u2, wy1,
wz1, wu1 are all distinct lines. If y1z1αy2z2 in B and z1u1αz2u2 in B, then
u1y1αu2y2.
Proof. Since members of B are lines of S, no three of them can form a
triangle. This forces u1y1, u2y2 ∈ B1. Let L ∈ B be a line meeting α at
the point w and construct a type 1 affine plane β1 through u1y1 that meets
L at some point w ′ different from w. Count the different planes that could
be constructed in this manner. For p ∈ u1y1 there is a unique point w1 of L
which is collinear in S with p1. As u1y1 is a (0, 2)-set, there is a unique other
point p2 ∈ u1y1 which is also collinear with w1. If w is collinear in S with
both u1 and y1, this leaves s−1
2 pairs of points in u1y1 which are collinear
with a point of L \ {w}. If w is not collinear in S with both u1 and y1, this
leaves w+1
2 pairs of points in u1y1 which are collinear with a point of L \ {w}.
This means that there are at least s−1
2 different β1 which can be constructed
in this manner. Similarly there are at least s−3 ways to construct a type 1
2
affine plane β2 through u2y2 which meets L \ {w, w1} in a point w2.

10.10. PARALLELISM 479
If β1β2, then u1y1u2y2. Assume that β1 and b2 are not parallel. Suppose
(hoping for a contradiction) that w2z2w1z1. This would force w2z2y2w1z1y1
and w2z2u2w1z1u1. It has been shown that when two parallel planes are
intersected with a third plane, the resulting lines of intersection are parallel.
Specifically, (Ly1 ∩ w1z1y1)(Ly1 ∩ w2z2y2), i.e., y1w1y2w2. Similarly,
u1w1u2w2. By the previous corollary, β1β2, a contradiction. Hence
w2z2 w1z1. As w, z1, z2 are all collinear in the plane α, this argument shows
also that w1y1 w2y2 and w1u1 w2u2, because either of these parallelisms
would imply w1z1w2z2.
Label these points of intersection: z3 = w1z1 ∩ w2z2; u3 = w1u1 ∩ w2u2,
y3 = w1y1 ∩ w2y2. Recall that by hypothesis u1z1u2z2, and observe that
u3z3 is the line of intersection of the planes u1z1w1 and u2z2w2. From this
u3z3 ∩ u1z1 = ∅ = u3z3 ∩ u2z2; i.e., u3z3u1z1w1u1z1 and u3z3u2z2w2u2z2. Thus
u1z1u3z3, and similarly y1z1y3z3.
This forces u3z3 and y3z3 to be members of B, and hence by the previous
corollary, αu3z3y3. Finally, observe that the plane u1y1w1 hits the parallel
planes α and u3z3y3 in the respective parallel lines y2u2 and y3u3. By
transitivity, y1u1y2u2.
Lemma 10.10.12. Let α and β be distinct parallel type 1 affine planes. Let
u be a point not in α or β, and let L1, L2 be distinct blocks through u which
do not belong to the parallel classes of blocks defined by α or β. Let yi be
the common point of α and Li, and let zi be the common point of Li and β.
Then y1y2z1z2.
Proof. Since αβ, y1y2∩z1z2 = ∅. If y1y2 ∈ B, then the points u, y1, y2, z1, ζ2
all belong to a type 1 affine plane, in which case y1y2z1z2. Similarly, z1z2 ∈ B
implies y1y2z1z2. Assume that neither y1y2 nor z1z2 is in B. For i = 1, 2 let
Mi ∈ B be a line through y1 such that M1 meets M2 at some point w. Let w ′
be the point of β on uw, and observe that M1u ∩ β = z1w ′ , M2u ∩ β = z2w ′ .
As αβ, it must be that M1M1uz1w ′ and M2M2uz2w ′ . Since M1, M2 ∈ B,
it follows that z1w ′ , z2w ′ ∈ B.
Now choose a line L ∈ B through u which meets both α and β but misses
both z1w ′ and z2w ′ . If L meets z1z2, then the type 1 affine plane containing
L and z1z2 must meet α at exactly the line y1y2 (as αβ). In this case
y1y2z1z2, and the proof is completed. So now assume that L misses z1z2
and let u ′ = L ∩ β. As shown with β2 in the previous lemma, there are s−3
2
planes different from α which contain y1y2 and a pont of L \ {u ′ }. Let γ be
such a plane.

480 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
If u ∈ γ, then as above u, y1, y2, z1, z2 are all coplanar (here in γ) and
hence y1y2z1z2. So further assume that u is not in γ, and let u ′′ = L ∩ γ.
Let v1 = y1u ′′ ∩ β, v2 = y2u ′′ ∩ β, w ′′ = wu ′′ ∩ β. To see that such points
actually exist, use Lemma 10.10.8 applied to β and observe that the parallel
classes of β are the same as those of α. The plane u ′′ wy1 contains the lines
M1 of α and v1w ′′ of β. Hence M1v1w ′′ and likewise M2v2w ′′ , forcing v1w ′′ ,
v2w ′′ ∈ B. By transitivity, z1w ′ v1w ′′ and z2w ′ v2w ′′ .
Now y1 ∈ z1u implies y1 ∈ Lz1 which implies y1u ′′ ⊂ Lz1, and hence
v1 ∈ Lz1. Clearly u ′ and z1 are in Lz1. but as these three points (v1, z1 and
u ′ ) are also in β., they must lie in a common block. Similarly, v2, z2, u ′ are
on a common block, as are w ′′ , w ′ , u ′ .
The two triangles frome by v1, v2, w ′′ and z1, z2, w ′ are perspective from
the point u ′ . Since z1w ′ v1w ′′ and z2w ′ v2w ′′ , Lemma 10.10.11 indicates that
z1z2v1v2. As γ meets α and β in parallel lines, u1yu2v1v2. Therefore
y1y2z1z2.
We are finally ready for the main theorem giving a characterization of
T ∗ 2 (O), s + 1 even, which we give for both s and t greater than 3, since
otherwise the GQ is known to be unique (up to duality).
10.11 A Characterization of T ∗ 2 (O), s + 1 even
Theorem 10.11.1. Let S = (P, B, I) be a GQ with parameters (s, s + 2),
s > 3, t > 3. Suppose that S has a set B1 of (0, 2)-sets such that every
pair of non-collinear points is contained in a unique member of B1, and such
that Axiom (T) holds. It follows that S is isomorphic to T ∗ 2 (O+ ) for some
hyperoval O + .
Proof. The hypotheses of the theorem have already been used to show that
S satisfies a long list of properties, and it particular s is odd. Let S ′ =
(P ′ , B ′ , I ′ ) be defined as above. The main goal is to show that S ′ is affine
three-space over Fq with q = 2 e .
From the construction, we know that for any two points x, y of P , either
x and y lie in a unique line of B or a unique (0, 2)-set of B1. The two
block parallelisms have been shown to be equivalence relations on B and B1,
respectively. Jointly they form an equivalence relation on B ′ . Let x ∈ P ′ ,
L ∈ B ′ . We have shown that there exists a unique M ∈ B ′ through x which
is parallel to L.

10.12. REGULAR OVOIDS IN P(T2(O), L) 481
Let L and M be parallel blocks with x1 ∈ L, x2 ∈ M. Let w ∈ x1x2 \
{x1, x2}, x ′ 1 ∈ L \ {x1}. Let α be an affine plane of type 1 through x1x ′ 1 but
not through w. Let β be the type 1 affine plane through x2 parallel to α.
Using Lemma 10.10.8 we see that M ⊂ β and there is a point x ′ 2 = wx′ ∩ β.
Since M is the unique line of β which is parallel to L and which contains
x2, it follows that M = x2x ′ 2 , i.e., it is shown that for any x′ ∈ L \ {x1} and
w ∈ x1x2 \ {x1, x2}, the block wx ′ meets M. This shows that S ′ satisfies (an
affine version of) the Axiom of Veblen.
Clearly each block contains more than 2 points.
By Section 1.1.3 we see that S ′ is the design of points and lines of an
affine space. Since |P ′ | = (s + 1) 3 , the design S ′ is the design of points and
lines of AG(3, s + 1). Hence S is embedded in the affine space AG(3, s + 1),
s + 1 odd. By a result of J. A. Thas (Section 7.3 in [PT84]) we know that
either S ∼ = P(W (s + 1), x) or S ∼ = T ∗ 2 (O) for some hyperoval O. Since
s + 1 is even, we have W (s + 1) ∼ = T2(O ′ ) with O ′ a conic. It follows that
P(W (s + 1), x) ∼ = P(T2(O ′ ), (∞)) ∼ = T ∗ 2 (O′ ∪ {N}) with N the nucleus of
the conic O ′ . We conclude that S is always isomorphic to a T ∗ 2 (O) with O a
hyperoval.
10.12 Regular Ovoids in P(T2(O), L)
Let O + be a hyperoval (so q = 2 e ) in the plane π∞ embedded as a hyperplane
in Σ = P G(3, q), and let P ′ be the set of points Σ \ π∞. Let x, y be distinct
points of O + , and put Ox = O + \ {x}, O − xy = O + \ {x, y}. Then O − xy is a
q-arc contained in a unique hyperoval O + . Also put L = xy. Then π∞ is a
biregular point of T2(Ox). Let Oxy denote the set of planes of Σ containing
x but not y, so naturally Oyx denotes the set of planes of Σ containing y but
not x. B − xy is the set of lines of Σ not contained in π∞ and meeting π∞ in a
point of the q-arc O − xy. If we expand Sx = T2(Ox) about the regular line L,
we get a GQ(q + 1, q − 1) with the following description.
Construction 10.12.1. S− xy = S(Oxy) = (P − xy , B− xy , I− xy ) is the GQ with
parameters (q + 1, q − 1) whose points are the points of P − x,y = P + ∪ Oxy ∪ Oyx
and whose lines are the lines of B− xy . Incidence I− xy is just that inherited from
incidence in Σ.
We now want to consider the reverse problem: How to recover Sx from
S − xy by constricting about a certain ovoid. It is easy to see that Oxy and Oyx

482 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
are both ovoids of S − xy . Let π1, π2 ∈ Oxy. So π1 ∩ π2 = M is a line through
x but not through y. First suppose that M is not contained in π∞. Then
πi ∩ O − xy = zi, i = 1, 2, with z1 = z2. Each of the q points of M ∗ = M \ π∞ is
collinear with πi on the line containing it and zi. It is easy to see that if M is
any line through x but not contained in π∞, then the q planes through M not
containing y are all collinear with the q points of M ∗ = M \ π∞. Analogous
remarks could be made about lines M ′ through y but not contained in the
plane π∞. Now suppose that M = π1 ∩ π2 lies in π∞ and contains the point
z ∈ O − xy , i.e., M = zx. And let M ′ be the line M ′ = zy. Then the planes
different from π∞ through M form a span whose trace is the set of planes
different from π∞ through M ′ .
Let π1, π2, . . . , πq be the planes of Σ meeting π∞ at the line L = xy,
and let Oi be the set of points of Σ \ π∞ contained in πi, 1 ≤ i ≤ q. It is
easy to see that each Oi is an ovoid and Ω = {Oxy, Oyx, O1, . . . , Oq} is a fan
of ovoids, i.e., a partition of the points of Sxy into disjoint ovoids. It is also
the case that both Oxy and Oyx are pivotal for Σ. This means that each pair
of points of Oxy (resp., Oyx) is a regular pair whose span is contained in Oxy
(resp., Oyx) and whose perp is contained in some member of Ω. In fact, by
construction, Oyx is pivotal for the fan Ω. And Oxy is regular since π∞ is
a regular point of Sx (Oxy is the set of points of S− xy that were points of Sx
collinear with π∞.
If even one of the ovoids Oi is pivotal for the fan Ω, then the corresponding
point of the line L∞ = L in Sx is regular in Sx. We saw earlier that this
means that each point of L is regular in Sx and Ox is a translation oval.
10.13 A Regular Ovoid in GQ(q + 1, q − 1)
At this stage we let q be any positive integer, not necessarily even a prime
power. We want to find a characterization of the GQ Sx ∼ = P(T2(Ox), L).
So first merely assume that S is a GQ with parameters (q + 1, q − 1) and
that S has a regular ovoid O∞. This means that O∞ is a set of q 2 pairwise
noncollinear points of S such that for distinct points x, y ∈ O∞ the pair
{x, y} is regular and {x, y} ⊥⊥ ⊂ O∞.
Lemma 10.13.1. Let π(O∞) be the point-line geometry whose points are
the points of O∞, and whose lines are the hyperbolic lines {x, y} ⊥⊥ , x = y,
x, y ∈ O∞, and whose incidence is containment. Then π(O∞) is an affine

10.13. A REGULAR OVOID IN GQ(Q + 1, Q − 1) 483
plane of order q. It follows that each hyperbolic line is in a “parallel class”
of pairwise disjoint hyperbolic lines that partition O∞.
Proof. Easy exercise.
Let T1, T2, . . . , Tq be the pairwise disjoint lines of one parallel class of
π(O∞. The points of these lines account for all the points of the ovoid O∞.
Hence T1 ∪ · · · ∪ Tq = O∞. The next result shows how each such parallel
class of π(O∞) determines an ovoid.
Lemma 10.13.2. T ⊥ 1 ∪ T ⊥ 2 ∪ · · · ∪ T ⊥ q is an ovoid of S.
Proof. Let Ti = {xi, yi} ⊥⊥ . Recall that T ⊥
i is the set of points collinear in S
with every point of Ti. If z ∈ T ⊥ 1 ∩ T ⊥ 2
, then every line of S through z must
contain a point of T1 and a point of T2. This is clearly impossible since O∞
is an ovoid and T1 ∩ T2 = ∅, Hence T ⊥ 1 ∪ T ⊥ 2 ∪ · · · ∪ T ⊥ q is certainly a set of
q 2 distinct points. If z1 and z2 are distinct points of T ⊥ 1 , then both z1 and
z2 are collinear with all q points of T1. Clearly z1 ∼ z2. Now suppose that
zi ∈ T ⊥
i , i = 1, 2. So each line through zi contains a point of Ti. This leaves
no line through z1 available to pass through z2. It follows that no two points
of T ⊥ 1 ∪ T ⊥ 2 ∪ · · · ∪ T ⊥ q are collinear. Since this set has q 2 points, it must be
an ovoid.
The previous lemma shows that for each parallel class of π(O∞) there is a
corresponding ovoid which is the union of the perps of the hyperbolic lines in
that parallel class. Let E0, 1, . . . , Eq be the distinct parallel classes of lines
in π(O∞), and let Oi be the ovoid consisting of the perps of the hyperbolic
lines in Ei, 0 ≤ i ≤ q. Then the following result is clearly already proved.
Corollary 10.13.3. M = {O∞, O0, . . . , Oq} is a fan of ovoids for which
O∞ is pivotal.
As an aside we show that in fact many more ovoids can be obtained from
the above construction.
Lemma 10.13.4. If T1, . . . , Tq are the hyperbolic lines in one parallel class
of π(O∞, let T ′
i ∈ {Ti, T ⊥
′
i }, 1 ≤ i ≤ q. Then ∪q T i is an ovoid.
Proof. For distinct i and j we have already noted that no point of Ti (resp.,
T ⊥
i ) is collinear with any point of Tj (resp., T ⊥ j ). Let y1 ∈ T1, z1 ∈ T ⊥ 1 ,
z2 ∈ T ⊥ 2 . If z2 ∈ T1, then z2 ∼ z1, an impossibility. So suppose that y1 ∼ z2.
i=1

484 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
Since every line through z2 contains a point of T2, there is some z ′ ∈ y1z2
with z ′ ∈ T2. But then z ′ and y1 are two points of O∞ that are collinear in S,
again an impossibility. Therefore no two points of ∪ q ′
i=1T i are collinear.
Later we will consider the possibility that ovoids in M \ {O∞} might be
regular or even pivotal.
Put I = {0, 1, . . . , q} and Ĩ = I ∪ {∞}.
Lemma 10.13.5. Let b, d ∈ Oj, with j ∈ I, b = d. If {b, d} ⊥ ∩ O∞ = ∅,
then {b, d} ⊥ ⊂ O∞. And the perps of the hyperbolic lines in the parallel class
of {b, d} ⊥ partition the ovoid Oj.
Proof. Let b, d ∈ Oj with j ∈ I, b = d. Suppose a ∈ {b, d} ⊥ ∩ O∞ and
c ∈ {b, d} ⊥ with c ∈ O∞. As each line meets each ovoid of M exactly once,
let a ′ = bc ∩ O∞ and let f be the unique point of ad collinear with a ′ . As
a, a ′ ∈ O∞ (which is pivotal for M) and b ∈ {a, a ′ } ⊥ , then all of {a, a ′ } ⊥
is contained in Oj. Specifically, f ∈ Oj. But d is assumed to be in Oj, a
contradiction as d ∼ f.
Lemma 10.13.6. Let O∞, Oi, Oj be distinct ovoids of M.
(i) x ∈ Oi, y ∈ Oj, and x ∼ y implies that |{x, y} ⊥ ∩ O∞| = 1.
(ii) a ∈ O∞, b ∈ Oi, and a ∼ b implies that |{a, b} ⊥ ∩ Oj| = 1.
Proof. For each a ∈ O∞, a ⊥ contains q(q − 1) noncollinear pairs (z, w) ∈
Oi × Oj. As O∞ is pivotal, if a and a ′ are distinct points in O∞, then a ⊥
and a ′⊥ can never contain the same noncollinear pair (z, w) ∈ Oi × Oj. Thus
there are (|O∞|)q(q − 1) = q 3 (q − 1) such pairs as a runs over the elements
of O∞.
On the other hand, counting such pairs shows that there are q 2 choices
for z ∈ Oi and q 2 − q choices for w ∈ Oj with z ∼ w. This gives q 3 (q − 1)
noncollinear pairs (z, w)Oi × Oj, which proves part (i).
For a fixed a ∈ O∞, x ∈ Oi, {a, x} ⊥ has exactly q points. If any two
of these were in the same ovoid of M, then the previous lemma would be
contradicted. This proves (ii).

10.14. CONSTRICTING ABOUT A REGULAR OVOID 485
10.14 Constricting about a Regular Ovoid
As in the previous section we assume that S is a GQ with parameters (q +
1, q − 1) and that S has a regular ovoid O∞ with associated fan M for which
O∞ is pivotal. At this point we construct a GQ S∞ = (P∞, B∞, I∞) of order
q via a method called constriction about O∞.
• Points of S∞ : P∞ = P \O∞ together with the symbols (O0), . . . , (Oq).
• Lines of S∞ : B∞ = B ∪ {T ⊥ : T is a line of π(O∞)} ∪ {L∞}.
• Incidence I∞ of S∞: A line of B is incident with a point of P∞
provided the two were incident in S. If T is a hyperbolic line of O∞,
then T ⊥ is incident with the q points it contains, and T ⊥ I∞ (Oi)
provided Oi is the (unique) ovoid in M containing T ⊥ . Finally, L∞ is
incident with each of the q + 1 points (Oi), 0 ≤ i ≤ q.
Of course this is just the point-line dual version of constricting about a
regular spread. Hence it follows that L∞ is a regular line of S∞, and (Oj),
j ∈ I, is a regular point of S∞ if and only if Oj is pivotal for M. (See
Construction 10.7.1.)
For the moment, consider the plane π(O∞). Let E0 = {T1, T2, . . . , Tq}
and E1 = {R1, R2, . . . , Rq} be distinct parallel classes of lines in π(O∞).
Obs. 10.14.1. Each point of π(O∞) is on a unique hyperbolic line Ti and a
unique hyperbolic line Rj, so we may label these points as xi,j = Ti ∩ Rj.
Hence T ⊥
i = {xi,j, xi,h} ⊥ for any h = j, and R ⊥ j = {xi,j, xk,j} ⊥ for any
k = i. We now wish to abuse the notation bit, using the “perp” notation for
whichever GQ we find convenient. The context should make it clear what
the notation means. The next observation follows with a little thought.
Obs. 10.14.2. In the GQ S∞, {T ⊥
i , R⊥ j }⊥ is the set of lines of S which
contain xi,j, together with the line L∞.
Moreover,
Obs. 10.14.3. {T ⊥
i , R ⊥ j } ⊥⊥ = {{xi,j, xi,h} ⊥ , {xi,j, xk,j} ⊥ } ⊥⊥ for any h = j
and k = i. In other words,
{T ⊥
i , R⊥ j }⊥⊥ = {ℓ ⊥ : ℓ is incident with xi,j in π(O∞)}.

486 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
From the regularity of L∞ in S∞ there is an affine plane π(L∞) =
(P(L∞, L(L∞)) whose points are the spans (in S∞) of lines meeting L∞
at distinct poionts, and whose lines are the lines of S∞ different from L∞ but
meeting it at distinct points. Incidence is containment. Hence
P(L∞) = {{ℓ ⊥ , m ⊥ } ⊥⊥ : ℓ, m are lines in distinct parallel classes of π(O∞)}
Also,
= {{T ⊥
i , R⊥ j }⊥⊥ : 1 ≤ i, j ≤ q}.
L(L∞) = {ℓ ⊥ : ℓ is a line of π(O∞)}.
Probably the next result is intuitively clear, but we give a detailed proof.
Theorem 10.14.4. The affine planes π(O∞) and π(L∞) are isomorphic.
Proof. Define ψ from P(O∞) to P(L∞) by ψ(xi,j) = {T ⊥
i , R ⊥ j } ⊥⊥ . To see
that ψ preserves collinearity, first suppose that ℓ = Ti = {xi,1, . . . , xi,q}.
So, ψ(ℓ) = {{T ⊥
i , R⊥ 1 }⊥⊥ , {T ⊥
i , R⊥ 2 }⊥⊥ , . . . , {T ⊥
i , R⊥ q }⊥⊥ }, i.e., exactly the
set of points in P(L∞) on the line T ⊥
i . Now suppose that ℓ ∈ E0, say
ℓ = {x1,r1, x2,r2, . . . , xq,rq}. So
ψ(ℓ) = {ψ(x1,r1), ψ(x2,r2), . . . , ψ(xq,rq)}
= {{T ⊥ 1 , R⊥ r1 }⊥⊥ , {T ⊥ 2 , R⊥ r2 }⊥⊥ , . . . , {T ⊥ q , R⊥ rq }⊥⊥ }
= {{T ⊥ 1 , ℓ⊥ } ⊥⊥ , {T ⊥ 2 , ℓ⊥ } ⊥⊥ , . . . , {T ⊥ q , ℓ⊥ } ⊥⊥ }.
But this is exactly the set of points in P(L∞) on the line ℓ ⊥ ∈ L(L⊥).
So ψ maps lines to lines. Thus ψ is an isomorphism between π(O∞) and
π(L∞).
Now suppose that the ovoid O0 is also pivotal for the fan M, so that
the point (O0) of L∞ is regular in S∞. From the regularity of (O0) there
is a standard construction of a projective plane π ∗ ((O0)) whose points are
the points of S∞ collinear with (O0), and whose lines are the spans {x, y} ⊥⊥
for which x and y are distinct points of S∞ collinear with (O0). Now form
an affine plane π((O0)) = (P((O0)), L((L0)) by removing the line L∞ and
all of its points from the plane π ∗ ((O0)). This new affine plane has pointset
P((O0)) = {x : x ∈ O0} and lineset L((O0)) = {{x, y} ⊥⊥ : x, y ∈ O0}. But
these are exactly the point- and linesets of π(O0), and the incidences are the
same. So here we see that the following theorem is proved.

10.14. CONSTRICTING ABOUT A REGULAR OVOID 487
Theorem 10.14.5. Suppose that both O∞ and O0 are pivotal for the fan
M. The two affine planes π((O0)) and π(O0) are identical (not merely isomorphic).
Continue with the basic hypothesis being that S is a GQ of order (q +
1, q − 1) with the regular ovoid O∞, which means it is pivotal for a unique
fan M.
Theorem 10.14.6. Suppose that {x1, x2} is a regular pair of points in S
with x1 ∼ x2. Put T = {x1, x2} ⊥⊥ = {x1, . . . , xq} and T ⊥ = {y1, . . . , yq}.
Then exactly one of the following must occur:
(i) There are distinct i, j ∈ Ĩ such that T ⊆ Oi and T ⊥ ⊆ Oj.
(ii) Exactly one of T , T ′ has a unique point in common in O∞. Without loss
of generality we may assume that T ⊥ ∩ O∞ = {y1} and T ∩ O∞ = ∅.
In this case there is a unique Oi for which T ⊥ \ {y1} ⊆ Oi. Also,
|T ∩ Oj| = 1 for each j = ∞, i.
Proof. Since O∞ is pivotal for M, if either T or T ⊥ has two points in common
with O∞, then it is contained in O∞ and its perp is contained in some Oi.
If T ∩ O∞ = ∅ = T ⊥ ∩ O∞, then by part (i) of Lemma 10.13.6 there must
be distinct i, j ∈ I for which T ⊂ Oi and T ⊥ ⊂ Oj. In the remaining case,
exactly one of T , T ⊥ has a unique point in common with O∞. Without loss
of generality we assume that T ⊥ ∩ O∞ = {y1}. If T had two points in the
same Oi, 0 ≤ i ≤ q, then by Lemma 10.13.5, T ⊥ would have to be contained
in O∞, a contradiction. So T meets some q of the ovoids each in one point
and there is a unique i left for which T ⊥ \{y1} ⊂ Oi. Also, y2, . . . , yq must be
in perps of distinct hyperbolic lines of O∞, which means they are on distinct
lines (= L∞) through (Oi) in S∞. By part (ii) of Lemma 10.13.6, |T ∩Oj| = 1
for each j = ∞, i.
Hence in S∞, {y2, . . . , yq} ⊆ {(Oi), x1, . . . , xq} ⊥ and {y2, . . . , yq} ⊥ =
{(Oi), x1, . . . , xq}. Let T0 be the line of S∞ through (Oi), T0 = L∞, and
T0 ∩ {y2, . . . , yq} = ∅. The lines of S through y1 together with L∞ form one
ruling of a grid Γ. For ease of notation, suppose that Oi = O0, and xj ∈ Oj,
1 ≤ j ≤ q. Then the other ruling of Γ consists of y ⊥ 1 ∩ Ok = Tk, 0 ≤ k ≤ q,
and xj ∈ Oj ∩ y ⊥ 1 .
Note 10.14.7. A consequence of the preceding proof is the following: If y1 ∈
O∞ and y2 ∈ Oi, and if {y1, y2} is regular in S, then {y1, y2} ⊥⊥ \ {y1} ⊂ Oi.

488 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
Corollary 10.14.8. With the notation adopted above, if for some i ∈ I, Oi
is regular, then it is pivotal for M.
Proof. If Oi is regular for some i ∈ I, and x, y are distinct points of Oi, then
{x, y} is regular and {x, y} ⊥⊥ ⊂ Oi. This does not permit the second case
of Theorem 10.13.6 to hold, hence the first case must hold.
Now suppose that O is a regular ovoid of S different from O∞. Recall
that π(O∞) and π(O) are affine planes whose “lines” are the hyperbolic lines
contained in them. Clearly π(O∞) ∩ π(O) is a subspace of π(O∞) and of
π(O). Hence we have proved the following.
Theorem 10.14.9. Any regular ovoid of S different from O∞ meets O∞ in
exactly 0, 1, or q points.
Theorem 10.14.10. If O is a regular ovoid with |O ∩ O∞| = q, then O =
T1 ∪ T ⊥ 2 ∪ · · · ∪ T ⊥ q , where O ∩ O∞ = T1, and O∞ = T1 ∪ T2 ∪ · · · ∪ Tq is a
parallel class partition of the points of O∞, so Ok = T ⊥ 1 ∪ T ⊥ 2 ∪ · · · ∪ T ⊥ q for
some k ∈ I.
Proof. Suppose that O∞ = {z1, . . . , zq, xa+1, . . . , x2q, . . . , x q 2}, where T1 =
{z1, . . . , zq}, T2 = {xq+1, . . . , xq}, . . . , Tq = {x(q−1)q+1, . . . , x q 2} are the disjoint
hyperbolic lines of one parallel class of O∞, and
O = {z1, . . . , zq, yq+1, . . . , y q 2},
with O ∩ O∞ = T1. Let y ∈ O ∩ Ok for some k = ∞. Then {zi, y} ⊥⊥ \
{zi} ⊂ Ok by Theorem 10.13.6, so ∪ q
i=1 {zi, y} ⊥⊥ \ {zi} gives 1+q(q −2) =
q2 − 21 + 1 points of O ∩ Ok. Since 2(q2 − 2q + 1) > q2 − q for q ≥ 3,
O \ O∞ must be contained in just one Ok, i.e., |O ∩ Ok| = q2 − q, and
hence O ∩ Ok = {yq+1, . . . , yq2}. Clearly no yj is collinear with any zi. So
y⊥ j ∩ O∞ = {xi1, . . . , xiq} is disjoint from T1. So the parallel class of O∞
containing T1 must contain q − 1 hyperbolic lines T2, . . . , Tq whose perps
cover O ∩ Ok. Let T ′ 1 be the points of Ok \ (O ∩ Ok). Then T ′ 1 must meet
the same lines that T1 does, forcing T ′ 1 = T ⊥ 1 .
So we have the following:
O∞ = T1 ∪ · · · ∪ Tq,
Ok = T ⊥ 1 ∪ · · · ∪ T ⊥ q ,
O = T1 ∪ T ⊥ 2 ∪ · · · ∪ T ⊥ q .

10.14. CONSTRICTING ABOUT A REGULAR OVOID 489
This completes a proof of the theorem.
Note 10.14.11. Continuing with the notation of the above proof, let x1 ∈ T1,
x2 ∈ T ⊥ 2 . By hypothesis {x1, x2} ⊥⊥ ⊂ O. If two points of {x1, x2} ⊥⊥ belong
to the same T ⊥ j , then of course {x1, x2} ⊥⊥ would have to equal T ⊥ j . So
T = {x1, x2} ⊥⊥ has one point in each of T1, T ⊥ 2 , . . . , T ⊥ q . Now let {x1, x2} ⊥ =
{w1, . . . , wq} = T ⊥ . Clearly T ⊥ ∩ O∞ = T ⊥ ∩ Ok = ∅, since both O∞ and Ok
are ovoids. If two points of T ⊥ belong to the same Oj, then since T ∩O∞ = ∅,
it would have to be that T ⊂ O∞. Hence |T ⊥ ∩ Oj| = 1 for each j = ∞, k.
Theorem 10.14.12. If O is a regular ovoid with |O ∩ O∞| = 1, then |O ∩
Oi| = q − 1 for 0 ≤ i ≤ q. If O ∩ O∞ = {z} and x ∈ O ∩ Oi, then
{z, x} ⊥⊥ \ {z} = O ∩ Oi.
Proof. Suppose O ∩O∞ = {z}. From Note ?? we see that if x ∈ O \O∞, say
x ∈ Oi, then {z, x} ⊥⊥ \{z} ⊂ Oi. First suppose that {z, x} ⊥⊥ \{z} ⊂ O∩Oi,
{z, y} ⊥⊥ \ {z} ⊂ O ∩ Oi, and z ∈ {x, y} ⊥⊥ , so {x, y} ⊥⊥ ∩ O∞ = ∅. Say:
{z, x} ⊥⊥ = {z, x = x2, . . . , xq}; {z, y} ⊥⊥ = {z, y = y2, . . . , yq}.
Since {x, y} ⊥⊥ ∩ O∞ = ∅, if {x, y} ⊥ ∩ O∞ = ∅ also, then {x, y} ⊥ ⊂ Ok for
some k = i, ∞, and {x, y} ⊥⊥ ⊂ Oi. On the other hand, if {x, y} ⊥ ∩ O∞ = ∅,
then {x, y} ⊂ Oi implies {x, y} ⊥ ⊂ O∞ and {x, y} ⊥⊥ ⊂ Oi. Either way we
have the following: For x, y ∈ O ∩ Oi, {x, y} ⊥⊥ ⊂ (O ∩ Oi) ∪ {z} = O. Then
the lines covered by z1 are the same lines as those covered by z, an obvious
impossibility. Hence it must be that |(O∩Oi)| = q−1 for each i = 0, 1, . . . , q.
And if x ∈ O ∩ Oi, then {z, x} ⊥⊥ \ {z} = O ∩ Oi for each i = 0, . . . , q.
Theorem 10.14.13. Let O be a regular ovoid with |O ∩ O∞| = 0. Then
exactly one of the following two possibilities must hold:
(i) O is some Oi ∈ M, in which case Oi is pivotal for the fan M;
(ii) There is a unique Oi, 0 ≤ i ≤ q, for which |O ∩ O∞| = |O ∩ Oi| = 0,
and |O ∩ Oj| = q for j = i, ∞. In this case O ∪ {(Oi)} is an ovoid of
S∞.
Proof. Suppose O ∩ O∞ = ∅. Let x, y be distinct points of O, so T =
{x, y} ⊥⊥ ⊂ O, and T ∩ O∞ = ∅. If T ⊥ ∩ O∞ = ∅, then T ⊂ Oi for some

490 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
i = ∞, and T ⊥ ⊂ Ok for some k = i, ∞. If T ⊥ ∩ O∞ = ∅, then either
T ⊥ ⊂ Oinfty and T ⊂ Oi for some i, or |T ⊥ ∩ O∞| = 1 and T ⊥ \ O∞ ⊂ Oi
for some i ∈ I while |T ∩ Oj| = 1 for each j = i, ∞. In all cases, if x, y are
distinct points of O ∩ Oi for some i, then {x, y} ⊥⊥ ⊂ O ∩ Oi. Hence O ∩ Oi
is a subspace of O, and must have size 0, 1, q or q 2 . One possibility is that
O = Oi ∈ M, for some i with 0 ≤ i ≤ q.
Suppose that O ∈ M, so |O ∩ Oi| = 0, 1 or q for each i = ∞. Let ai be
the number of Oj with |O ∩ Oj| = i, i = −.1.q, for 0 ≤ j ≤ q. Clearly each
ai ≥ 0, and we have
(i) a0 + a1 + aq = q + 1,
(ii) 0 · a0 + 1 · a1 + q · aq = q 2 ,
from which a1 = q(q − aq). Put k = q − aq, so a1 = kq. It follows that
q − k = aq = q + 1 − a0 − a1, implying a1 + a0 = k + 1 = a0 + kq, or k(q − 1) =
1−a0, which must be nonnegative. This allows only two possibilities: a0 = 1
or 0. If a0 = 1, then ik = 0 = a1 and aq = q. In this case O is disjoint
from O∞ and from one other member of M, say O0, and meets each of the
others O1, . . . , Oq in q points. For the other case, suppose a0 = 0. Here
k = 1 and q = 2. So for q ≥ 4 we have only the first case, i.e., O meets
each of O1, . . . , Oq in a “line” of O. It is now straightforward to check that
O ∪ {(Oi)} is an ovoid of S∞.
At this point we suppose that in addition to having O∞ pivotal
for the fan M, the ovoid O0 is regular, so that in fact it is also
pivotal for M.
Theorem 10.14.14. Suppose {x, y} is a regular pair of points with x ∼ y.
Put T = {x, y} ⊥⊥ , T ⊥ = {x, y} ⊥ . Then there must exist distinct i, j ∈ Ĩ for
which T ⊂ Oi, T ⊥ ⊂ Oj.
Proof. Suppose not. Then one of T ⊥ , T has a unique point in common with
O∞, say |T ⊥ ∩O∞| = 1. Then there is some i = ∞ for which |T ⊥ ∩Oi| = q−1.
Clearly Oi is not regular, so i = 0. Then |T ∩ Oj| = 1 for each j = i, ∞,
including j = 0. But |T ∩ O0| = 1 says there must be some k for which
|T ∩ Ok| = q − 1, an impossibility for q ≥ 4.
Theorem 10.14.15. Suppose O is a regular ovoid of S. Then O ∈ M.

10.15. QUIVERS - AND A3 491
Proof. Suppose that O is a regular ovoid of S with O ∈ M. Then |O∩O∞| =
0, 1 or q, and |O ∩ O0| = 0, 1 or q.
Case (i) |O ∩ O0| = q. In this case there is a k with O ∩ Ok| = q 2 − q,
so k = 0 and |O ∩ O0| = 0. But Theorem 10.14.13 applied to O0 in place of
O∞ says there is some j with |O ∩ Oj| = 0 and O ∩ Om| = q for m = 0, j.
Hence this case cannot arise.
Case (ii) |O ∩ O∞| = 1. Here |O ∩ Oi| = q − 1 for 0 ≤ i ≤ q. But
|O ∩ O0| = q − 1 if q > 2, so this case cannot arise.
Case (iii) |O ∩ O∞| = 0. By symmetry we may suppose that we also have
|O ∩ O0| = 0. But if O ∈ M then also O ∩ Oj| = q for j = 0, ∞. But
suppose x ∈ O ∩ Oj, y ∈ O ∩ Ok, j = k, and j, k = i, ∞. Then {x, y} ⊥⊥
does not belong to a single member of M, contradicting Theorem 10.14.14.
Hence this case does not arise, completing the proof.
10.15 Quivers - and A3
Continue to assume that O∞ is pivotal for the fan M as above. Then for
a1, a2 ∈ O∞ put a1 ≡ a2 provided a⊥ 1 ∩ O0 = a − 2⊥ ∩ O0. Clearly “ ≡ ” is
an equivalence relation defined on O∞ by O0. Let [a] denote the equivalence
class containing the element a, so that [a1] = [a2] if and only if a⊥ 1 ∩Oo ∩a⊥ 2 =
∅. There are q equivalence classes in O∞ each with q elements. Similarly, for
b ∈ O0, put [b] = {b1 ∈ O0 : b ⊥ ∩ O∞ ∩ b ⊥ 1
= ∅}.
For a ∈ O∞ let [a] ⊥ = [b]. Then the q 2 lines joining points of [a] with
points of [b] form a set of lines called a quiver. Two lines of a quiver are
concurrent if and only if they meet at a point of O∞ ∪O0. Two lines meeting
at a point of O∞ ∪ O0 are in a unique quiver. Two noncurrent lines L and M
are in the same quiver if and only if the point of O∞ on L is collinear with
the point of O0 on M, and the point of O0 on L is collinear with the point
of O∞ on M. Each point of O∞ (resp., O0) determines a unique quiver. It
follows that there are q quivers, each containing q 2 lines, and each line is in
a unique quiver. Each line not in a given quiver is concurrent with exactly q
pairwise noncurrent lines of that quiver, and each point of P \ (O∞ ∪ O0 is
on a unique line in each quiver.
In the paper [Pa85b] where quivers were first defined, there were properties
A1, . . . , A7 that were shown to characterize the GQ P(T2(O), L) where
O is an oval in P G(2, 2 e ) and L is one of the lines (through π∞) known to

492 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
be regular. In the present situation S is already assumed to satisfy A1 and
A2 merely because S has a regular ovoid O∞ that is pivotal for the fan M.
Since this notation has been used in a variety of published accounts of this
material, we continue to use it. It is time to discuss property A3.
In fact, for each i, 0 ≤ i ≤ q, there is a property A3(i). To simplify
notation, we write A3 in place of A3(0), and state this property for i = 0.
A3 Let L1, M1 be nonconcurrent lines of S meeting lines L2, M2 at four
distinct points belonging to members of M ′ = M \ {O∞, O0}. Let aj
be the point of O∞ incident with Lj, j = 1, 2, and let bj be the point
on Mj collinear with aj, j = 1, 2. Then b1 ∈ O0 iff b2 ∈ O0.
We now suppose that S also satisfies A3 (so the ovoid O0 is playing a
special role), and thus it satisfies A1, A2 and A3.
Lemma 10.15.1. Let L and M be nonconcurrent lines belonging to some
quiver Q. Then the following hold:
(i) The lines of {L, M} ⊥ not in Q (i.e., not incident with the points of
O∞ ∪ O0 on L and M) are all in a common quiver Q ′ .
(ii) L and M belong to a q × q grid G having q lines (of {L, M} ⊥ ) in Q ′
and q lines in Q. The points of G are precisely those points on lines
{L, M} ⊥ not in O∞ ∪ O0.
(iii) If (L, M, K) is a triad of lines in Q, then K belongs to the q × q grid
containing L and M iff (L, M, K) is centric.
Proof. Let L and M be nonconcurrent lines belonging to a same quiver Q.
This means that the point of O∞ on L (resp., M) is collinear with the point
of O0 on M (resp., L). Let K1, . . . , Kq be the lines of {L, M} ⊥ meeting L and
M at points of P \ (O∞ ∪ O0). For 1 ≤ i, j ≤ q, i = j, by A3 the point of O∞
on Ki must be collinear with the point of O0 on Kj. It follows that K1, . . . ,
Kq must all belong to a quiver Q ′ , proving (i). Let Ki, Kj, Kk be any three
distinct lines of {L, M} ⊥ meeting L and M at points of P \(O∞ ∪O0). Let x
be any point of Ki not on L or M and not in O∞ ∪ O0. To complete the grid
it is suffices to know that the line Nj through x and meeting Kj is the same
as the line Nk through x meeting Kk. But there is only one line through x in
the quiver Q, and applying part (i) to Ki and Kj (and independently to Ki
and Kk) we see that all the lines of {Ki, Kj} ⊥ (including L and M and Nj)

10.15. QUIVERS - AND A3 493
belong to one quiver, i.e., Q. Similarly all the lines of {Ki, Kk} ⊥ (including
L and M and Nk) belong to the same quiver. Hence Nj = Nk. It is now
clear how to complete the grid.
By the point-line dual of Theorem 9.7.5 (with m = n = q in this case)
each line not in the grid is concurrent withy exactly two lines of the grid.
This means that each line of S not in Q or Q ′ meets the grid in a unique
point.
For the proof of part (iii), let L = L1, M = L2, L3, . . . , Lq be the lines
of Q belonging to the grid containing L and M. If K is one of L3, . . . , Lq,
then (L, M, K) has q centers (i.e., transversals). Otherwise, K must be the
line through the point of O∞ on one of L1, . . . , Lq and through the point of
O0 on one of the other lines of L1, . . . , Lq. In this case it is easy to check
that (L, M, K) has no center.
Lemma 10.15.2. In the GQ S∞ the point (O0) is coregular.
Proof. L∞ is already known to be regular in S∞. What remains to be seen is
that each line K of the form K = {a1, a2} ⊥ such that K ⊂ O0 and K ⊥ ⊂ O∞
is regular as a line of S∞. As a set of points of S, K determines a quiver
Q for which a line L of S is in Q iff it is concurrent with K in S∞. Let M
be a line of S∞ not concurrent with K. If M is concurrent with L∞ in S∞,
then (I, M) is regular in S∞ because L∞ is regular. If M is not concurrent
with L∞ in S∞, then M ∈ B, i.e., M is also a line of S and belongs to some
quiver Q ′ , Q ′ = Q. Through each point of O1 ∪ · · · ∪ Oq on M (in S) there
is a unique line of Q. These lines form part of a q × q grid G of lines and
points in S (by Lemma ??). In S∞, G is completed to a (q + 1) × (q + 1) grid
G∞ by including the point (O0), the line K = {a1, α2} ⊥ determining Q, the
line {a ′ 1 , a′ 2 }⊥ determining Q ′ , and the points on these latter two lines.
It is easy to see in view of Lemmas 10.15.1 and 10.15.2 that A3(i) is
equivalent to having (Oi) be a coregular point of S∞. If each line meeting
L∞ is regular, then each line of S∞ is regular. (Proof: If {L1, L2} is any
pair of noncurrent lines, they form a regular pair if either of them meets
L∞. If neither meets L∞, since L∞ is regular, the triad {L∞, L1, L2} must
be centric, say with transversal M. But then since M ∈ {L1, L2} ⊥ , the pair
{L1, L2} must be regular. Hence all lines are regular.) By the theorem of
Benson, S∞ is then isomorphic to the dual of W (q), i.e., to the GQ Q(4, q)
arising from a nonsingular (parabolic) quadric in P G(4, q). Here q is any
prime power. This essentially completes a proof of the following theorem.

494 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
Theorem 10.15.3. Let S = (P, B, I) be a GQ of order (q + 1, q − 1), q ≥ 4.
Then S is isomorphic to the GQ obtained by expanding Q(4, q) about any
(necessarily regular) line if and only if there is a partition of the pointset
P = O∞ ∪ O0 ∪ · · · ∪ Oq into ovoids in such a way that the following hold:
(i) O∞ is pivotal for the fan M = {O∞, O0, . . . , Oq}, i.e., x, y ∈ O∞,
x = y, implies that {x, y} ⊥ is contained in a single member of M.
(ii) If L1 and M1 are nonconcurrent lines meeting nonconcurrent lines
L2 and M2 in four points of O0 ∪ · · · ∪ Oq, let ai be the point of O∞ on Li,
i = 1, 2. If ai ∼ bi with biIMi, i = 1, 2, then b1 and b2 belong to the same
member of M \ {O∞}.
Here condition (i) says that we may constrict S about O∞ to obtain a
GQ S∞ of order s with a regular line L∞. Condition (ii) says that each point
(Oi), 0 ≤ i ≤ q, is coregular in S∞. We know that this means that the point
(Oi) itself is regular if q is even and is antiregular if q is odd. So if even one
of the coregular points (Oi), 0 ≤ i ≤ q, is pivotal for Ω, forcing (Oi) to be
regular as a point of S∞, then s is even and all of the points as well as all of
the lines of S∞ are regular.
10.16 A1, A2, A3 and A4
Our next goal is to characterize the GQ of order (q + 1, q − 1) arising from
a q-arc in P G(2, q), q = 2e . So in addition to the basic properties A1, A2
and A3 used in the preceding results, we now assume that q is even, i.e.,
that A4 holds.
Continuing with the notation from above and assuming that Ai holds for
1 ≤ i ≤ 4, we are now essentially assuming that the coregular point (O0)
of S∞ is also regular, so that the ovoid O0 is also pivotal for M. But this
means that in all the results obtained so far (excluding Theorem 10.15.3) the
roles of O0 and O∞ are interchangeable.
Since (O0) is a regular point of S∞, we may expand S∞ about the point
(O0) to obtain a GQ S0 ∞ of order (q − 1, q + 1) as follows:
S0 ∞ = (P 0 ∞ , B0 ∞ , I0 ∞ ), where P 0 ∞ = O1 ∪ · · · ∪ Oq, and B0 ∞ = B ∪ {x, y}⊥ :
x = y; x, y ∈ O∞ or x, y ∈ O0 and {x, y} ⊥ ∩ (O∞ ∪ O0) = ∅}. Here and
(unless noted otherwise) in all that follows “⊥” denotes the perp in S. For
x ∈ P 0 ∞ , L ∈ B, we have xI0 ∞L iff xIL. For z ∈ P 0 ∞ , L = {x, y}⊥ ∈ B0 ∞ \ B,
L if and only if z ∈ L.
we have zI 0 ∞

10.17. PSEUDOLINES: A1, · · · , A6 495
Let x, y ∈ P 0 ∞ . To denote that x and y are collinear in S0 ∞ , we write
x ′′
∼ y. Note that x ′′
∼ y iff x ∼ y or {x, y} ⊥ ⊂ O∞ or {x, y} ⊥ ⊂ O0.
Index the quivers of lines of S as Q1, . . . , Qq, and put
and
Q∞ = {{a1, a2} ⊥ : a1, a2 ∈ O∞ and [a1] = [a2]},
Q0 = {{b1, b2} ⊥ : b1, b2 ∈ O0 and [b1] = [b2]}.
Then Q = {Q∞, Q0, . . . , Qq} is the partition of the lines of S 0 ∞ into
spreads arising from expanding S∞ about the point (O0). Because (O0)
is coregular in S∞, each spread in Q is a pivotal member of Q. Note also
that if the roles of O∞ and O0 were interchanged so as to first construct a
GQ S0 of order q and then a GQ S ∞ 0 of order (q − 1, q + 1), S∞ 0 and S0 ∞
would be exactly the same GQ.
10.17 Pseudolines: A1, · · · , A6
Let x ∈ Oi, y ∈ Oj, 1 ≤ i, j ≤ q, i = j, x ∼ y. By Lemma ?? there is
a unique a ∈ O∞ ∩ {x, y} ⊥ and there is a unique b ∈ O0 ∩ {x, y} ⊥ . Put
xy = {a, b} ⊥ . Then xy is called the pseudoline through x and y. Clearly
xy is a set of q points containing x and y, uniquely determined by any two
points of xy, and with |xy ∩ Oi| = 1 for each i = 1, 2, . . . , q. Each point x of
P 0 ∞ is on q2 pseudolines of this type determined by the q 2 pairs in
(x ⊥ ∩ O∞) × (x ⊥ ∩ O0).
The next property we introduce for S is A5.
A5 : Let (L1, L2, L3) be a centric triad of lines for which the points of O∞
on L1, L2, L3 are each collinear with the points of O0 on L1, L2, L3. If for
some (a, b) ∈ O∞ × O0, a ∼ b, both L1 and L2 are incident with points of
{a, b} ⊥ , then L3 is also incident with a point of {a, b} ⊥ .
Property A5 is easily seen to be equivalent to the following.
Lemma 10.17.1. Let x and y be noncollinear points in distinct members of
Ω \ {O∞, O0}. If G is any grid containing x and y, then G contains all q
points of the pseudoline xy.

496 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
Finally, we state A6.
A6: Let x1, x2, x3 be distinct points of some Oj, 1 ≤ j ≤ q. Let (L1, L2, L3)
and (M1, M2, M3) be two triads of lines such that Li meets Mi at xi, i =
1, 2, 3. Suppose that the three points of O∞ on L1, L2, L3 (respectively, M1, M2, M3)
are each collinear with the three points of O0 on L1, L2, L3 (resp., M1, M2, M3).
Then (L1, L2, L3) is centric if and only if (M1, M2, M3) is centric.
Property A6 also has a restatement in terms of grids. (Throughout
this section the word “grid” will refer to one of the grids constructed in
Lemma 10.15.1.)
Theorem 10.17.2. Let Gi be a q × q grid with lines from quivers Qi and
Q ′ i , i = 1, 2. Suppose that G1 and G2 both have lines incident with x and y,
with x, y ∈ Oj, 1 ≤ j ≤ q, x = y. Then the set of q point of Oj incident with
the lines of G1 is the same set of q points of Oj incident with G2.
Proof. This is a rather immediate restatement of A6, since in the statement
of A6 the triad (L1, L2, L3) is centric if and only if it is part of a grid G1 if
and only if (M1, M2, M3) is centric if and only if it is part of a grid G2.
The following corollary just collects some of the information already obtained.
Corollary 10.17.3. Let x, y ∈ Oj, x = y, 1 ≤ j ≤ q. There are three
possibilities:
(i) {x, y} ⊥ ∩ O∞ = ∅, iin which case {x, y} ⊥ ⊂ O∞ and {x, y} ⊥⊥ is a set
of q points of Oj that form a line through the point (O∞) of S 0 ∞. No
grid contains two points of {x, y} ⊥⊥ , but for each a ∈ {x, y} ⊥ , the lines
through a are the lines of one quiver through the points of {x, y} ⊥⊥ .
(ii) {x, y} ⊥ ∩ O0 = ∅, in which case {x, y} ⊥ ⊂ O0 and {xy} ⊥⊥ is a set
of q points of Oj that form a line through the point (O0) of S0 ∞ . No
grid contains two points of {x, y} ⊥⊥ , but for each b ∈ {x, y} ⊥ , the lines
through b are the lines of one quiver through the points of {x, y} ⊥⊥ .
(iii) {x, y} ⊥ ∩(O∞ ∪O0) = ∅. In this case let Lx and Ly be the lines through
x and y, respectively, in some quiver Q. Let G be the grid containing
Lx and Ly. The lines of G lie in Q and in a second quiver Q ′ and
meet Oj in a set xy of q points. By Theorem 10.17.2 there are q/2
grids containing x and y and all meeting Oj in the same set xy of q

10.17. PSEUDOLINES: A1, · · · , A6 497
points. Here xy is called the pseudoline through x and y and is clearly
determined uniquely by any two points of xy.
At this point a pseudoline xy is defined for all pairs of noncolinear points
in different members of M \ {O∞, O0} and for distinct points x, y in Oj,
1 ≤ j ≤ q when {x, y} ⊥ ∩ (O∞ ∪ O0) = ∅. Fix x ∈ P 0 ∞ . Then (x⊥ ∩ (O∞)) ⊥
and (x ⊥ ∩ (O0)) ⊥ are the unique lines through (O∞) and (O0), respectively,
in S 0 ∞ that are incident with x. If x ∈ Oj, 1 ≤ j ≤ q, then there are q − 1
pseudolines through x and lying in Oj.
Lemma 10.17.4. Let x and y be distinct points of P 0 ∞ = O1 ∪· · ·∪Oq. Then
there are four possibilities for the pair (x, y).
(i) x ∼ y in S, in which case x ′′
∼ y.
(ii) If x ∼ y where x and y are in distinct members of M \ {O∞, O0},
then xy is a pseudoline with xy = {a, b} ⊥ = {x1, x2, . . . , xq}, where
a ∈ O∞, b ∈ O0, xi ∈ Oi, 1 ≤ i ≤ q are uniquely determined. For
each a ′ ∈ O∞ \ [a] (resp., b ′ ∈ O0 \ [b]) there is a unique i ∈ {1, . . . , q}
(resp., j ∈ {1, . . . , q} for which a ′ ∈ x ⊥ i (resp., b′ ∈ x ⊥ j ).
(iii) If x and y are both in Oj, 1 ≤ j ≤ q, with {x, y} ⊥ ∩ (O∞ ∪ O0) = ∅,
then xy is a pseudoline with xy ⊂ Oj.
(iv) Suppose x and y are in the same ovoid Oj with {x, y} ⊥ ⊂ O∞ (resp.,
{x, y} ⊥ ⊂ O0). In this case {x, y} ⊥⊥ is a line of S 0 ∞ .
Proof. The only part that might still need verification is part (ii). For case
(ii), let xy = {a, b} ⊥ = {x1, . . . , xq} as in the statement of the lemma. For
1 ≤ i < j ≤ q, (x⊥ i ∩ O∞) ∩ (x⊥ j ∩ O∞) = {a}. So ∪ q
⊥
i=1 (xi ∩ O∞) \ {a}
consists of q(q−1) points of L∞\{a}. Moreover, if x⊥ i ∩O∞ contains a point a ′
of [a], a ′ = a, then xi ∈ {a, a ′ } ⊥ ⊂ Oi, an impossibility. As |O∞\[a]| = q2 −q,
for each a ′ ∈ O∞ \ [a], there is a unique i for which a ′ ∈ x⊥ i . As the roles of
O∞ and O0 may be interchanged, the proof of (ii) is complete.
Now consider the two cases where the pseudoline xy is defined. Define
the slope of the pseudoline xy as follows. If xy = {a, b} ⊥ where a ∈ O∞ and
b ∈ O0, then xy is said to have slope s(xy) = [a]. If xy ⊂ Oj, then xy is said
to have slope s(xy) = ∞. This gives q + 1 different slopes.

498 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
Lemma 10.17.5. Let x, y ∈ P 0 ∞ be in the same ovoid Oj of M, so that
xy = {x1, . . . , xq} ⊂ Oj. For each a ′ ∈ O∞ (resp., b ′ ∈ O0) there is a unique
xi (resp., xj) such that a ′ ∈ x ⊥ i (resp., b′ ∈ x ⊥ j ).
Proof. For x1, x2, distinct members of xy, observe that (x⊥ 1 ∩ O∞) ∩ (x2 ∩
O∞) = ∅, since otherwise x1x2 would not be defined (but x1x2 = xy). This
forces | ∪ q
i=1 (x⊥i ∩ O∞)| = q2 , in which case ∪ q
i=1 (x⊥i ∩ O∞) = O∞. A similar
argument holds with O0 in place of O∞.
Corollary 10.17.6. Let z ∈ P 0
infty \ xy, where xy is a pseudoline with slope
m, and suppose that z is not collinear in S0 ∞ with w for each w ∈ xy. For
each slope m ′ = m there is a unique w ∈ xy for which s(zw) = m ′ .
The following lemma demonstrates the connection between pseudolines
of S and (0, 2)-sets of S 0 ∞ .
Lemma 10.17.7. Let x, y ∈ P 0 ∞ be such that x and y are not collinear in
S0 ∞. Then xy is a (0, 2)-set in S0 ∞.
Proof. First consider the case where xy ⊂ Oi ∈ M, 1 ≤ i ≤ q. Let z be a
point of P 0 ∞ \ xy which is collinear in S0 ∞ with some point of xy. As xy is
determined by any two of its points, assume this point to be x. Here there
are two subcases.
Begin by assuming z ∈ Oi and hence {x, z} ⊥ ⊂ O∞ or {x, z} ⊥ ⊂ O0.
Without loss of generality suppose that {x, z} ⊥ ⊂ O∞. There is a particular
w which is collinear with at least one point of z⊥ ∩ O0. But as z and w are
both in Oi it follows that {z, w} ⊥ ⊂ O0 and hence z is collinear with w in
S0 ∞ ( and also w = x). Furthermore w is uniquely determined this way. If
there were some u ∈ xy with z ′′
∼ u this would force z ∼ y, a contradiction
as z, u ∈ Oi.
Next assume z ∈ Oj = Oi. In this case z ′′
∼ x implies z ∼ x, in which
case there is a line L ∈ B through x and z. Let Q be the quiver containing
L, and let M be the line of Q through y. If G is the grid containing Q,
there is another quiver Q ′ in G. Let K be the unique line of Q ′ through z
in G. Observe that K contains a unique point w of xy. On the other hand
if xILIzIKIw with w ∈ xy, L = K, then L and K are in the unique grid
determined by L and y. Hence xy is a (0, w)-set of S0 ∞ .
Now consider the case where x ∈ Oj, y ∈ Ok, j = k with x ∼ y. Write
xy = {a, b} ⊥ with a ∈ O∞, b ∈ O0. Choose z ∈ P 0 ′′
∞ − xy with z ∼ x. Begin

10.17. PSEUDOLINES: A1, · · · , A6 499
by assuming z ∈ Oj; which would mean that x ∼ z on some line L ∈ B with
L in the quiver Q. Let M be the line through y in Q. Let Q ′ be the other
quiver containing lines of the grid determined by L and M. Let K be the
unique line of Q ′ through z. If G is any grid containing x and y, then G
contains all q points of xy. So K has a unique point w of xy. Observe that
w is the unique other point of xy collinear in S0 ∞ with z.
Finally assume z ′′
∼ x with z ∈ Oj. Once again assume that {x, z} ⊥ ⊂
O∞. Thus a ∈ {x, z} ⊥ . Since x is the unique point of Oj on xy, if w is a
point different from x with z ′′
∼ w, it follows that z ∼ w. Let K be the line
of B through a and z. The point b is collinear with a unique point w on
K. This forces w ∈ {a, b} ⊥ , i.e., w ∈ xy. Observe that this makes w the
unique point of xy other than x with which z is collinear in S0 ∞ . Hence xy is
a (0, 2)-set.
At this point we see that if x and y are points of P 0 ∞ that are not collinear
in S0 ∞, then the pseudoline xy provides a uniquely defined (0, 2)-set through
x and y. This means that Axiom 1 of Section 2 is satisfied.
If β is a set of q2 points of P 0 ∞ , we call β a special plane provided that
for x, y ∈ β, x ′′
∼ implies that {x, y} ⊥⊥ ∩ P 0 ∞ ⊂ β, and if it is not true that
x ′′
∼ y, then xy ⊂ β. Note that a special plane β together with the lines and
pseudolines it contains is actually an affine plane of order q. Moreover, two
distinct special planes intersecting in at least three distinct points must have
a line of S0 ∞ or a pseudoline of S as their intersection.
Lemma 10.17.8. Three types of special planes immediately arise.
(i) Each ovoid of M \ {O∞, O0} is a special plane.
(ii) The q 2 points on a q × q grid G of S (restricted to points of P 0 ∞) is
a special plane denoted βG. This follows from A5 and A6 (see their
restatements above).
(iii) For a ∈ O∞ ∪ O0, β = a ⊥ ∩ P 0 ∞ is a special plane.
Proof. Parts (i) and (ii) are more or less obvious. For part (iii), let a ∈ O∞
with x, y distinct points in β. Consider first the case x ′′
∼. If x ∼ y by some
line L ∈ B, then {x, y} ⊥⊥ = L. So {x, y} ⊥⊥ ∩ P 0 ∞ = L \ {a} ⊂ a⊥ ∩ P 0 ∞.
If x ′′
∼ but x ∼ y, then x and y belong to the same ovoid Oi of M. As

500 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
a ∈ {x, y} ⊥ , it follows that {x, y} ⊥ ⊂ O∞ and hence {x, y} ⊥⊥ ⊂ Oi. In fact,
{x, y} ⊥⊥ ⊂ Oi ∩ a⊥ ⊂ β.
Finally, consider the case that x is not collinear with y in S0 ∞ . Then
x and y are in distinct ovoids of M, and there is a unique b ∈ O0 with
{x, y} ⊥ ∩ O0 = {b}. In this case xy = {a, b} ⊥ , which is clearly contained in
β.
Let Π∗ be the set of special planes formed in one of these three ways.
Let (x, L) be a nonincident point-line pair from S0 ∞. First consider the case
where L ∈ B. There is a unique point on L which is collinear with x in S.
If this point is in O∞ ∪ O0, call it a and observe that a⊥ ∩ P 0 ∞ is the unique
plane of Π∗ containing x and L. If this poiont is in P 0 ∞, call it y and let
K ∈ B be the line through x and y. Since L and K determine a unique grid
G, βG is the unique plane of Π ∗ through x and L.
Now consider L ∈ B0 ∞ \ B, in which case there is some i ∈ {1, . . . , q} with
L ⊂ Oi. Suppose that L⊥ ⊂ O∞. If x ∈ Oi, then Oi is the unique plane of
Π∗ containing x and L. If x ∈ Oj = Oi, then there is a unique y incident in
S0 ∞ with L such that x ′′
∼ y on some line M ∈ B0 ∞. As x and y are in different
ovoids of M, M is actually in B. Let M ∩ O∞ = {a}. Then a⊥ ∩ P 0 ∞ is the
unique plane of Π∗ containing x and L.
The following important lemma has now been proved.
Lemma 10.17.9. If (x, L) is a nonincident point-line pair from S0 ∞ , then x
and L lie in a unique plane of Π∗ .
If β is the plane of Π ∗ determined by the nonincident point-line pair (x, L)
from S 0 ∞, then by this lemma there exist q −1 pseudolines in β which contain
x and a point of L. Let L = {L1, . . . , Lq−1} be the set of these pseudolines.
This leads to the following lemma which is the penultimate (and perhaps
most significant) claim of this section.
Lemma 10.17.10. Let x, L, β and L be as described above with x ∈ M ∈ B 0 ∞
and L ∩ M = ∅. If at least two distinct members, say Li and Lj of L meet
M, then every member of L meets M.
Proof. Let yi = Li ∩M, yj = Lj ∩M. Obviously x, yi and yj are in β. By the
previous lemma (x, M) determine a unique plane β ′ ∈ Π ∗ which also contains
x, yi, yj, i.e., {x, yi, yj} ⊂ β ∩ β ′ . But the intersection of two distinct planes
of Π ∗ must be a line. Since x, yi, and yj are not mutually collinear (by an

10.17. PSEUDOLINES: A1, · · · , A6 501
affine line), it follows that β ′ = β. Hence M is parallel to L in β. Therefore
each affine line which meets L in β must also meet M in β. Namely, each of
the pseudolines of L meets M.
The importance of the preceding lemma lies in the fact that it shows that
Axiom (T) (also called Axiom 2) of Section 10.9 is satisfied in S0 ∞ . Hence it
follows fromTheorem 10.11.1 that
Corollary 10.17.11. S 0 ∞ is isomorphic to T ∗ 2 (O+ ) for some hyperoval O + .
But then it is clear that S is obtained by first constricting about a pivotal
spread to obtain a GQ T2(O) for some oval O contained in O + , and then
expanding about a regular line, i.e., S is the GQ of order (q + 1, q − 1)
derived from a q-arc in P G(2, q).
Recapitulatiion: Suppose that S is a GQ with parameters (q +1, q −1),
q = 2 e . Suppose that S has an ovoid O∞ that is regular, i.e., each pair {x, y}
of distinct points of O∞ is a regular pair with {x, y} ⊥⊥ ⊂ O∞. This gives an
affine plane π(O∞) whose points are the points of O∞ and whose lines are the
spans of pairs of points of O∞. This is enough to complete the construction
of a fan M = {O∞, O0, O1, . . . , Oq} of ovoids for which O∞ is pivotal. This
means that the perp of each pair of points of O∞ is contained in some Oi,
and each Oi is the disjoint union of the perps of the affine lines in one parallel
class of π(O∞), for 0 ≤ i ≤ q. At this point we can say that S is isomorphic
to a GQ obtained from a q-arc in P G(2, q) provided that properties A3, A5
and A6 hold.

502 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
10.18 The Construction of T3(Ω)
Let Ω be an ovoid of Σ = P G(3, q) which is embedded as a hyperplane in
P G(4, q). Construct the point-line incidence geometry S = (P, B, I) with
pointset P, lineset B, and incidence I, as follows:
The points of S, i.e., elements of P are of three types:
(i) The points of P G(4, q) \ Σ;
(ii) The hyperplanes of P G(4, q) meeting Σ in a plane tangent to O.
(iii) The symbol (∞).
The lines of S are of two types:
(a) The lines of P G(4, q) that are not contained in Σ and meet Σ in a
(necessarily unique) point of Ω.
(b) The points of Ω.
It is routine to verify that S is a GQ with parameters (q, q 2 ). When Ω
is an elliptic quadric, it is possible to show that T3(Ω) is isomorphic to the
point-line dual of Q(5, q). When q is odd, there is, of course, only the classical
elliptic quadric ovoid. In all cases there is an elementary abelian group of
collineations of S acting sharply transitively on the points of type (i) (i.e.,
points not collinear with (∞)), and fixing each line through (∞). A GQ
admitting such a group of collineations is called a translation GQ with base
point (∞). If S is a GQ with a fixed point P for which there is a group G
of collineations fixing each line through P and acting sharply transitively on
the set of points not collinear with P , then S is said to be an elation GQ with
elation group G at the base point P . We then write that S = (S (P ) , G) is an
elation GQ, or (S (P ) , G) is an EGQ. See [PT84] and [KT04] for a much more
thorough introduction to finite EGQ. For our present purposes it suffices to
be able to view some of the known examples as EGQ. In the next section we
consider the point-line dual of T3(Ω) in the special case where q = 2 e , e odd,
and Ω is the Tits ovoid.
10.19 The Point-Line Dual of T3(Ω)
The GQ T3(Ω) where Ω is the Tits-ovoid is naturally viewed as a TGQ of
order (q, q 2 ), where q is an odd power of 2. In this section we view it as
an EGQ with parameters (q 2 , q). The fact that the elation group G admits
this one four-gonal family at least suggests the possibility that it may admit

10.19. THE POINT-LINE DUAL OF T3(Ω) 503
another one. We give an explicit description of this group G with the hope
that this view of it may suggest some other way to construct a four-gonal
family. However, this approach has not led to anything new so far as we
know.
Let Fq = GF (q), q = 2 e , e odd. Let σ ∈ Aut(Fq) be chosen so that
σ 2 = 2. Define
f : F 2 q → Fq : (a, b) ↦→ a σ+2 + ab + b σ .
The Tits-ovoid Ω of Σ = P G(3, q) is given by
Ω = {(0, 1, 0, 0)} ∪ {(1, f(a, b), a, b) : a, b ∈ Fq}.
The GQ T3(Ω) with parameters (q, q 2 ) is constructed as follows. First embed
Σ into P G(4, q) by (x, y, z, w) ↦→ (0, x, y, z, w).
Points of T3(Ω) are of three types:
(i) points of P G(4, q) \ Σ;
(ii) solids of P G(4, q) \ Σ meeting Σ in a plane tangent to Ω;
(iii) a symbol (∞).
Lines of T3(Ω) are of two types:
(a) lines of P G(4, q) meeting Σ in a point of Ω.
(b) points of Ω.
Incidence in T3(Ω) is defined by the following:
The point (∞) is incident with the 1 + q 2 lines of type (b). Suppose △
is a solid of P G(4, q) meeting Σ in the plane Tp tangent to Ω at the point p.
Then △ is incident with p (as a line of type (b)) and with the q 2 lines of △
not in Σ. The point x of P G(4, q) \ Σ is incident with the 1 + q 2 lines px,
p ∈ Ω.
This construction gives T3(Ω) as a translation GQ (TGQ) of order (q, q 2 )
whose point-line dual T3(Ω) ˆ is an elation GQ (EGQ) of order (q 2 , q).
Define θ(a, b, c, d, e) : (u, x, y, z, w) ↦→ (u, x, y, z, w)[a, b, c, d, e], where
⎛
⎜
[a, b, c, d, e] = ⎜
⎝
1 0 c d e
0 1 f(a, b) a b
0 0 1 0 0
0 0 a σ+1 + b 1 a σ
0 0 a 0 1
⎞
⎟ ; for a, b, c, d, e ∈ Fq.
⎠

504 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
A routine check shows that G = {θ(a, b, c, d, e) : a, b, c, d, e ∈ Fq} is a
group of order q 5 with binary operation
[a, b, c, d, e] · [a ′ , b ′ , c ′ , d ′ , e ′ ] =
[a + a ′ , b + b ′ + a · a ′σ , c + c ′ + d(a ′σ+1 + b ′ ) + ea ′ , d + d ′ , e + e ′ + da ′σ ].
Moreover, G leaves Ω invariant. In fact, G fixes the “line” P = (0, 0, 1, 0, 0) ∈
Ω and is sharply transitive on the q 5 lines of T3(Ω) not concurrent with P
(i.e., the q 5 lines of P G(4, q) meting Σ at a point of Ω different from P ).
Let Q be the “line” Q = (0, 1, 0, 0, 0) ∈ Ω. The plane TQ = {(0, z1, 0, z3, z4) :
z1, z3, z4 ∈ Fq} is the plane of Σ tangent to Ω at Q. Let L∞ be the
line joining the point (1, 0, 0, 0, 0) with the “point” TQ, literally the line
L∞ = 〈(1, 0, 0, 0, 0), (0, 1, 0, 0, 0)〉. So L∞ is an “arbitrary” line of T3(Ω) not
meeting P . The points of T3(Ω) with which it is incident are the points of
P G(4, q) \ Σ incident in Σ with L∞ and the solid 〈Q, TQ〉. The subgroups of
G fixing the various points of L form a 4-gonal family for G. We are about
to calculate these groups.
First note that the center Z(G) = {[0, 0, c, 0, 0] : c ∈ Fq} of G acts as a
group of symmetries about the “line” P (i.e., fixing all lines meeting P ).
For t ∈ Fq, put
A(t) = the stabilizer in G of (1, t, 0, 0, 0)
= {[a, b, tf(a, b), ta, tb] : a, b ∈ Fq};
A ∗ (t) = {[a, b, c, ta, tb] : a, b, c ∈ Fq};
A(∞) = the stabilizer of the remaining point of L∞
= {0, 0, 0, d, e] : d, e ∈ Fq};
A ∗ (∞) = {[0, 0, c, d, e] : c, d, e ∈ F + q}.
Now recall the following well-known facts about the ovoidal function f:
f(a, b) + f(a ′ , b ′ ) + ab ′ + a ′ b = 0 iff (a, b) = (a ′ , b ′ ). (10.7)
In particular, f(a, b) = 0 if and only if a = b = 0.
f(a, b) σ+1 = b σ+2 + abf(a, b) σ + a σ f(a, b) 2 . (10.8)

10.19. THE POINT-LINE DUAL OF T3(Ω) 505
1
b
+ f
f(a, b) f(a, b) ,
a
, for (a, b) = (0, 0). (10.9)
f(a, b)
f(a, b) + f(a ′ , b ′ ) + aa ′σ+1 + ab ′ + a ′ b = f(a + a ′ , b + b ′ + aa ′σ ) (10.10)
f(a, b) = f(a, y) iff y = b or y = b + a σ+1 . (10.11)
We want to consider the problem of showing directly that the groups
A(t), t ∈ Fq ∪ {∞} actually give a 4-gonal family for G.
It is routine to check that A ∗ (s) ∩ A(t) = {id} for distinct s, t ∈ Fq ∪
{∞}. It is also easy to check that A(s)A(t) ∩ A(u) = {id} for distinct
s, t, u ∈ Fq ∪ {∞}, as long as ∞ ∈ {s, t, u}. But suppose that s, t, u are
distinct elements of Fq. We sketch the steps involved in showing directly
that A(s)A(t) ∩ A(u) = {id}. So suppose that
[a1 + a2, b1 + b2 + a1a σ 2, s(a σ+2
1
+ a1b1 + b σ 1) + t(a σ+2
2
a1s + a2t, b1s + b2t + sa1a σ 2 ]
?
= [a3, b3, u(a σ+2
3 + a3b3 + b σ 3 ), a3u, b3u].
+ a2b2 + b σ 2),
Equality holds if and only if the following five conditions are satisfied:
(i) a1 + a2 = a3;
(ii) b1 + b2 + a1aσ 2 = b3;
(iii) a1s + a2t = a3u;
= b3u;
(iv) b1s + b2t + sa1a σ 2
(v) s(a σ+2
1
+ a1b1 + b σ 1 ) + t(a σ+2
2
Using (i) and (iii) we get:
Using (ii) and (iv) we get:
Then (i), (ii) and (v) imply that
+ a2b2 + b σ 2) = u(a σ+2
3
+ a3b3 + b σ 3).
a1s + a2t = (a1 + a2)u. (10.12)
b1s + b2t + sa1a σ 2 = (b1 + b2 + a1a σ 2 )u. (10.13)

506 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
3. s(a σ+2
1
+ a1b1 + b σ 1 ) + t(aσ+2 2 + a2b2 + b σ 2
= u (a1a + 2) σ+2 + (a1a2) σ+2 + (a1 + a2)(b1 + b2 + a1a σ 2 ) + (b1 + b2 + a1a σ 2 )σ .
At this point it is quite straight-forward to reach the desired conclusion
if a1a2(a1 + a2) = 0. So suppose that a1a2(a1 + a2) = 0. Solve for u:
u = a1s + a2t
a1 + a2
=
= b1s + b2t + sa1a σ 2
b1 + b2 + a1a σ 2
sf(a1, b1) + tf(a2, b2)
f(a1, b1) + f(a2, b2) + a1a σ+1
2
+ a1b2 + a2b1
Use the first line, cross multiply, cancel repeated terms, then organize so the
factor s + t can be cancelled, to obtain
a1a σ+1
2 = a1b2 + a2b1. (10.14)
This equation says that the last denominator above is equal to f(a1, b1)+
f(a2, b2). So now use
a1s + a2t
a1 + a2
Simplify and cancel s + t to obtain
= sf(a1, b1) + tf(a2, b2)
f(a1, b1) + f(a2, b2) .
f(a, b)
a = f(a2, b2)
. (10.15)
a2
Recall: If α is an automorphism of Fq of maximal order, and if a·b·c = 0,
then axα + bx + c = 0 has two solutions if
tr
a 1
α−1 c
b α
α−2
= 0,
and no solution otherwise. So for α = σ with σ 2 = 2, we have
ax σ σ+1 a c
+ bx + c = 0 has no solution if and only if tr
bσ+2
= 1. (10.16)
.

10.19. THE POINT-LINE DUAL OF T3(Ω) 507
Solve for b1 in Eq. 10.14 and put this value of b1 in Eq. 10.15. After
clearing terms, we find
(a σ 1 a1−σ
2
+ a1)b σ 2 + a1(a1 + a2)b2 + a1a σ+2
2
σ+2
+ a2a1 + aσ1 a32 + a21 aσ+1 2 = 0,
which we view as saying that b2 is a solution of an equation of the type in
Eq. 10.16, so it is exactly what we want not to happen.
+ a1 = 0. Also, f(a1, a2) = 0 can be
+ a2a σ+2
1 + aσ 1a32 + a21 aσ+1 2 = 0.
So A(s)A(t) ∩ A(u) = {id} if and only if
Now a1 = a2 implies that a σ 1a 1−σ
2
used to show that the “constant” term a1a σ+2
2
= tr
σ (a1a 1 = tr
1−σ
2
a σ+1
1
⎜
= tr ⎝
+ a1) σ+1 (a1a σ+2
2
σ+2
+ a2a1 (a1a2 + a 2 1) σ+2
+ a2 1 aσ+1
2
(a σ−1
1 a 1−σ
2 + 1) σ+1 · a1 · (a σ+2
2 + a2a σ+1
1 + a1a σ+1
2
⎛
a σ+2
1 (a1 + a2) σ+2
σ−1
a1 +aσ−1
2
a σ−1
σ+1 · a2 · (a1 + a2)(a
2
σ 2
σ−1
(a1 = tr
σ+1
a1 + a
= tr
σ+1
2
= 1 + tr
= 1 + tr
+ a σ−1
2
+ a 2−σ
1 a 2σ−1
2
2−σ
a1 a2σ−1 2
2−σ
a1 a 2σ−1
2
+ a 2σ−1
1
+ a 2σ−1
1
(a1 + a2) σ+2
) σ+1 (a σ 2 + a2a σ−1
1
(a1 + a2) σ+1
+ a 2σ−1
1
a 2−σ
2
(a1 + a2) σ+1
a 2−σ
2
+ a2σ−2
1
(a1 + a2) σ+1
a 2−σ
2
+ a2σ−2
1
(a1 + a2) σ+1
We used the fact that (see Lemma 4.13.4(b))
σ+1
a1 + a
tr
σ+1
2
(a1 + a2) σ+1
= 1.
This means that we need to show that
+ a2a σ−1
1
+ a σ 1 )
+ aσ 1 a3 2 )
+ a σ−1
1
+ a σ 1 )
+ a 2σ−2
1 a 3−σ
2
a 3−σ
2
a 3−σ
2
+ a2 1 aσ−1
2
+ a2 1 aσ−1
2
⎞
⎟
⎠
+ a2a σ−1
2
.
a3
2)

508 CHAPTER 10. GQ ASSOCIATED WITH AN OVAL OR OVOID
tr
2−σ
a1 a 2σ−1
2
Write this last trace as
tr
a 2−σ
1
+ a 2σ−1
1
aσ−1
2
a 2−σ
2
+ a 2σ−2
1
(a1 + a2) σ+1
(a1 + a2) σ + a 2σ−2
1
a 2−σ
1
(a1 + a2) σ+1
a σ−1
2
= tr
(a1 + a2)
2−σ
a1 a
= tr
σ−1
2
+
(a1 + a2)
a2σ−2 1
+
a 3−σ
2
a 2−σ
2
a 2−σ
2
+ a 2 1a σ−1
2
(a1 + a2)
(a1 + a2) σ
2−σ
a1 a σ−1 σ
2
= 0.
(a1 + a2)
?= 0.
Hence we have independent confirmation that we have a 4-gonal family.

Chapter 11
Classifiying Ovoids - Part 2
The bulk of this chapter is adapted more or less directly from the two papers
by M. Brown [Br00a] and [Br00b].
11.1 Ovoids
If Ω is an ovoid of P G(3, q), q = 2 e , then Ω defines a polarity ν of P G(3, q)
as follows. If π is the tangent plane to Ω at the point P , then ν interchanges
π and P . If π meets Ω in the oval O with nucleus N, then the polarity ν
interchanges π and N. This polarity is necessarily a symplectic polarity of
P G(3, q). The points of P G(3, q) (all of which are singular points with respect
to ν) and the lines of P G(3, q) that are tangent to Ω (i.e., the singular lines
with respect to ν – singular means incident with its image under ν) form a
generalized quadrangle denoted W (q). It then follows that for any ovoid Ω of
P G(3, q), Ω is also an ovoid of the generalized quadrangle W (q) constructed
from the symplectic polarity that it defines. Conversely, if Ω is an ovoid of
the GQ W (q), then by Thas, Ω is also an ovoid of P G(3, q). (This is our
Theorem 7.6.2.)
Let π be a plane of Σ = P G(3, q) and let O be an oval in π. Let T2(O)
be the GQ constructed from Σ, π and O, and let O be an ovoid of T2(O). If
π = (∞) is not a point of Ω then of the q 2 + 1 points of Ω, q 2 − q are affine
points (that is, points of Σ \ π) and q + 1 points are planes of Σ meeting π in
distinct lines tangent to O. If the q 2 − q affine points of Ω have the property
that no three are collinear in Σ, then we say that Ω is a projective ovoid of
T2(O).
509

510 CHAPTER 11. CLASSIFIYING OVOIDS - PART 2
Lemma 11.1.1. Let Ω be a projective ovoid of the GQ T2(O) and let A be
the set of q 2 − q affine points of Ω. Then A ∪ O is an ovoid of Σ = P G(3, q).
Conversely, let Ω be an ovoid of Σ and let the oval O be the intersection
of the plane π with Ω. Then
Ω = (Ω \ O) ∪ {πP ; πP is the tangent plane to Ω at a point P ∈ O}
is a projective ovoid of the GQ T2(O) constructed from O, π and Σ.
Proof. Start with the projective ovoid Ω of T2(O). We check that A ∪ O has
the property that no three of its points are collinear in Σ. Let P , Q, R be
three distinct points of A ∪ O. If {P, Q, R} ⊂ A or {P, Q, R} ⊂ O, then
clearly they are not collinear in Σ. First suppose P, Q ∈ O and R ∈ A. Since
〈P, Q〉 is a line of π and R ∈ π, it follows that P, Q, R are not collinear in Σ.
So suppose P ∈ O and Q, R ∈ A. Since any line of S incident with P and
not contained in π is a line of T2(O), and hence cannot contain two points
of the ovoid Ω, clearly P, Q, R cannot be collinear in Σ. Thus A ∪ O is an
ovoid of Σ.
For the converse, suppose that Ω is an ovoid of Σ with O = π ∩ Ω and
let P, Q, be two points of Ω. If P, Q ∈ Ω \ O, they are not collinear in T2(O)
because otherwise they would lie on a line of Σ through a point R of O giving
three points of Ω collinear in Σ. If πP and πQ are distinct points of T2(O),
the only lines of T2(O) incident with them that have a point in common in S
are their lines in π which meet at the nucleus N of O, and N is not a point of
T2(O). So no two points of Ω of this type are collinear in T2(O). So suppose
P ∈ Ω \ O and πQ are collinear points of T2(O). But πQ is the tangent plane
to Ω at Q and cannot contain the point P , so no line through P could lie in
πQ and be incident with both points. So no two points of Ω are collinear in
T2(O).
Let π be a plane of Σ = P G(3, q) and Oα a translation oval of π with
nucleus N, axis L and Q the point of Oα on L. Let Ω be an ovoid of Σ
containing Oα. Let πa be a plane of Σ meeting π in the line L and meeting Ω
in the oval Oa. The line L is tangent to Oa and the nucleus of Oa is not the
point N. This follows from the facts that N is the nucleus of Oα; both Oα
and Oa are secant plane sections of the ovoid Ω of Σ; and the polarity defined
by Ω ensures that any point of Σ\Ω is the nucleus of exactly one secant plane
section of Ω. Construct T2(Oα) from Σ, π, Oα as usual. Let Ω be the ovoid
of T2(Oα) consisting of the set of tangent planes to Ω meeting Ω at a point of

11.1. OVOIDS 511
Oα and the affine points of Ω\Oα. Recall that the plane πa is a regular point
of T2(Oα). It follows that any pair (Y, Z) of distinct points with Y, Z ∈ π ⊥ a
in T2(Oα) is also a regular pair in T2(Oα). Thus Theorem 9.10.2 may be
applied to such a regular pair and the ovoid Ω of T2(Oα). It follows that the
intersection of the span {Y, Z} ⊥⊥ and Ω has size 0 or 2. Now interpret this
fact in the construction of the affine plane A(πa, L) constructed in Lemma ??
(and the paragraph preceding the lemma). It says that any line of A(πa, L)
meets the set Oa \ L in at most 2 points. Hence Oa \ L is a q-arc of A(πa, L).
We summarize this discussion in the following lemma.
Lemma 11.1.2. Let Ω be an ovoid of Σ = P G(3, q), with π a plane of Σ
such that π ∩ Ω is a translation oval Oα. Let L be an axis of Oα and let Q
be the point of Oα on L. Let T2(Oα) be the GQ constructed from Σ, π, Oα,
and let πa be a plane of Σ intersecting π in L, and so also a regular point of
T2(Oα). Let A(πa, Q) be the affine plane constructed from Π(πa) by taking L
to be the line at infinity. If πa, as a plane of Σ, meets Ω in an oval Oa, then
Oa \ L is a q-arc in both πa and in A(πa, Q).
Recall that the affine plane πa \ L can be recovered from A(πa, Q) by
α −1 -derivation.
At this point we want to explain
how to normalize the coordinates for
0 1
P 0
the ovoid Ω. Let P = and let C be the matrix C = , so
1 0
0 P
C is skew-symmetric (with 0’s on the diagonal). For ¯x = (x0, x1, x2, x3) and
¯y = (y0, y1, y2, y3) in F 4 q , put
¯x ∗ ¯y = ¯xC ¯y T .
Then (¯x, ¯y) ↦→ ¯x ∗ ¯y is a nonsingular, bilinear alternating form. If we put
¯x ⊥ = {¯y : ¯x ∗ ¯y = 0}, then for nonzero ¯x, ¯x ⊥ is a plane containing ¯x, and
the map ν : ¯x ↦→ ¯x ⊥ is a symplectic polarity. Recall that any two symplectic
polarities are equivalent. So if Ω ′ is an ovoid with a plane section being
an oval projectively equivalent to some Of, then there is an homography of
P G(3, q) mapping Ω ′ to an ovoid Ω whose symplectic polarity is exactly ν,
and Ω will also contain a plane section that is an oval O equivalent to Of.
If W (q) is the symplectic geometry (i.e., generalized quadrangle) defined by
Ω, then the group of homographies of P G(3, q) leaving W (q) invariant is
known as the symplectic group Sp(4, q) and is transitive on the points of
W (q). Hence there is an homography preserving Ω and mapping the nucleus

512 CHAPTER 11. CLASSIFIYING OVOIDS - PART 2
of O to the point N = (0, 0, 1, 0). Then by our assumption that ν is the
polarity defined by Ω, we know that the plane containing O is [0, 0, 0, 1],
i.e., O = Ω ∩ [0, 0, 0, 1]. Furthermore, the subgroup of Sp(3, q) fixing N is
transitive on the points collinear with N in W (q), and hence also transitive
on the q + 1 lines of W (q) through N. So if O were a translation oval, for
example, we could without loss of generality assume that the line x0 = x3 = 0
is a translation axis of O.
Let N be the point of P G(3, q) given above, with ν : N ↔ N ⊥ . The
subgroup of Sp(4, q) leaving N ⊥ invariant induces a group of collineations
of N ⊥ containing the group of elations of N ⊥ with center N. (To see this,
consider a line ℓ of N ⊥ and a point Q different from N on ℓ. The group
of elations with center Q and axis Q⊥ induces in N ⊥ a group containing all
elations of N ⊥ with center Q and axis ℓ. Letting Q range over all points
collinear in W (q) with N, the group generated contains all elations of N ⊥
with center N.) For points in N ⊥ = [0, 0, 0, 1] we may temporarily drop the
final coordinate, which is always 0. Then for the record, the homography
with matrix
⎛ ⎞
1 0 c
⎝ 0 1 bc ⎠ ,
0 0 1
for b, c ∈ Fq with c = 0, is an elation of [0, 0, 0, 1] with center N = (0, 0, 1)
and axis [1, b, 0]. The homography with matrix
⎛
⎝
1 0 0
0 1 b
0 0 1
for b = 0, is an elation of [0, 0, 0, 1] with center N = (0, 0, 1) and axis [0, 1, 0].)
Now the group of elations of N ⊥ with center N fixes each of the q + 1 lines
of W (q) through N and is transitive on the q points different from N on
each of the lines. Hence we may assume that O contains the point Q∞ =
(0, 1, 0, 0), that the point N is the nucleus, and that the line L∞ = 〈Q∞, N〉
is a translation axis of O. This means that there is an o-permutation f such
that
O = {(1, f(t), t, 0) : t ∈ Fq} ∪ {(0, 1, 0, 0)} and has nucleus N = (0, 0, 1, 0).
⎞
⎠

11.2. A RESTRICTION ON PLANE SECTIONS OF AN OVOID 513
Now consider the homography of P G(3, q) with matrix
⎛
⎞
1 f(0)/A 0 0
⎜
H = ⎜ 0 1/A 0 0 ⎟
⎝ 0 0 1 0 ⎠
0 0 0 1/A
where A = f(0) + f(1). It may be checked that HCH T =
1 C, so that
A
the homography commutes with the symplectic polarity. The nucleus N is
left invariant, so of course the plane N ⊥ = [0, 0, 0, 1] is left invariant. The
oval point Q∞ is fixed and the general oval point (1, f(t)t, t, 0) is mapped to
(1, f(0)+f(t)
, t, 0). But if the oval is really a translation oval with translation
A
axis L∞, then since the function t ↦→ f(0)+f(t)
maps 0 to 0 and 1 to 1, this
A
must be the oval Oα for α a generator of the automorphism group of Fq,
where Oα = {(1, tα , t, 0) : t ∈ Fq} ∪ {(0, 1, 0, 0)} with nucleus N = (0, 0, 1, 0).
We state this as a theorem.
Theorem 11.1.3. Let Ω be an ovoid of P G(3, q) containing an oval section
O equivalent to the oval O(α) = {(1, t, t α ) : t ∈ Fq} ∪ {(0, 0, 1)} with nucleus
(0, 1, 0), where α is a generator of Aut(Fq). Then coordinates of P G(3, q)
may be chosen so that the symplectic polarity of P G(3, q) defined by Ω has the
alternating form x0y1 + x1y0 + x2y3 + x3y2 and contains the oval {(1, t α , t, 0) :
t ∈ Fq} ∪ {(0, 1, 0, 0)} with nucleus (0, 0, 1, 0).
At this point the reader should review the proof of the “plane representation
theorem” of Glynn and Penttila, Theorem 7.6.1
11.2 A restriction on plane sections of an ovoid
Fix the notation essentially as in the preceding section. For any a ∈ Fq,
a = 0, πa = [1, 0, 0, a] is a secant plane to Ω. Recall π0 = [1, 0, 0, 0] and
suppose that π0 ∩ Ω = Oα = {(0, t, t α , 1) : t ∈ Fq} ∪ {(0, 0, 1, 0)} with
nucleus N = (0, 1, 0, 0), where α is a generator of the Galois group of Fq.
So Oα contains the point Q∞ = (0, 0, 1, 0) and has nucleus N = (0, 1, 0, 0)
. Let ν be the symplectic polarity defined by Ω. We may assume that ν
has the form x0y1 + x1y0 + x2y3 + x3y2 = 0 (by applying the homography
(x0, x1, x2, x3) ↦→ (x3, x2, x1, x0), which commutes with the polarity ν, to the
coordinatization given in Theorem 11.1.3. And we know that
x ∈ Ω ⇐⇒ x ν is the tangent plane to Ω at x,

514 CHAPTER 11. CLASSIFIYING OVOIDS - PART 2
and
x ∈ Ω ⇐⇒ x is the nucleus of the oval x ν ∩ Ω.
Then ν : πa ↔ (0, 1, a, 0), which must be the nucleus Na of Oa = Ω ∩ πa.
Recall the affine plane A(πa, L∞) obtained from the regularity of the point πa.
The map θα : (ax, y, z, x) ↦→ (ax, y α , z, x) maps the affine plane A(πa, L∞) to
the plane πa in standard form.
Since Oa is an oval in πa containing Q∞ and having nucleus Na =
(0, 1, a, 0), there must be some o-permutation fa such that :
Oa = {(a, y, ay + fa(y), 1) : y ∈ F } ∪ {(0, 0, 1, 0)}. (11.1)
The map θα maps the affine part of Oa to {(a, y α , ay + fa(y), 1) : y ∈ Fq}.
But we know that the q-arc Ω ∩ πa \ {(0, 0, 1, 0)} is a q-arc in both the
affine plane πa minus the line L∞ = x0 = x3 = 0, and in the affine plane
A(πa, L∞) with the point Q∞ extending the q-arc to an oval in both cases.
In the projective completion Āa of A(πa, L∞) we also have that Oa is
an oval with some nucleus not the parallel class corresponding to (0, 1, 0, 0).
Hence under the isommorphism from Āα to πa we have that
O ′ a = O θα
a = {(a, yα , ay + fa(y), 1) : y ∈ Fq} ∪ {(0, 0, 1, 0)}
is an oval with nucleus (0, 1, ā, 0) for some ā = 0. This means that there is
an o-permutation ga such that
The affine part of O ′ a
which is also
= Oθα
a is {(a, y α , āy α + ga(y α ), 1) : y ∈ Fq} (11.2)
= {(a, y α , ay + fa(y), 1) : y ∈ Fq}.
This completes a proof of the following:
Theorem 11.2.1. If Ω is an ovoid of P G(3, q) with a plane section π0 ∩ Ω
that is a translation oval with axis a tangent to Ω, then there are two opermutations
f and g and nonzero constants a, ā ∈ Fq such that
ay + f(y) = āy α + g(y α ) ∀y ∈ F. (11.3)
Note that if we replace y with yα−1, this equation looks like

11.2. A RESTRICTION ON PLANE SECTIONS OF AN OVOID 515
āy + g(y) = ay α−1
+ f(y α−1
), (11.4)
i.e., the roles of f and g are interchanged and α is interchanged with α −1 .
For a given nonzero a the o-permutations f and g were denoted above by
fa and ga, respectively. When a = 0, we have f0 = α, i.e., f0(x) = x α .
Recall the homography µs, for any s ∈ Fq. For each s,
µs : πs ∩ Ω ↦→ Ōs = {(0, y, fs(y), 1) : y ∈ Fq} ∪ {(0, 0, 1, 0)}
with nucleus N0 = (0, 1, 0, 0).
We know that for distinct s, t ∈ Fq, Ōs and Ōt are compatible at Ns+t =
(0, 1, s+t, 0). This means that a line of the form 〈(0, x, fs(x), 1), (0, y, ft(y), 1)〉
cannot contain the point (0, 1, s + t, 0). This is equivalent to the following:
Result 11.2.2. fs(x) + ft(y) = (s + t)(x + y) whenever s = t.
In the special case where t = 0 = s, we have fs(x) + y α = s(x + y). This
says that for nonzero s and each x ∈ Fq,
This proves the following:
Corollary 11.2.3. tr
y α + sy + fs(x) + sx = 0 has no solution y.
fs(x)+sx
s α
= 1 ∀x ∈ Fq.
α−1
Note that if fs(x) + sx = fs(y) + sy, then fs(x)+fs(y)
x+y = s = 0. Since
fs is an o-permutation, there cannot be a third element z ∈ Fq for which
fs(z) + sz = fs(x) + sx. Cor. 11.2.3 says that the function fs(x) + sx must
take on exactly half the elements of Fq.
Let K be the set of elements of Fq with trace 0 and δ a fixed element
with trace 1.
If we put R = {fs(x) + sx : x ∈ Fq}, then Cor. 11.2.3 says that
s −α
α−1 · R = K + δ, i.e., R = s α
α−1 K + s α
α−1 δ.

516 CHAPTER 11. CLASSIFIYING OVOIDS - PART 2
11.3 Ovoids of P G(3, q) containing a conic
We now restrict our attention to the case of an ovoid Ω of P G(3, q) that
contains a conic section C. In Section 11.1 we saw how to obtain an associated
ovoid Ω of the GQ T2(C). By using the isomorphism φ from T2(C) to W (q) we
may map Ω onto an ovoid Ω (1) of W (q). It follows from Theorem 7.6.2 that
Ω (1) is also an ovoid of P G(3, q). We now start with Ω in a canonical form
and apply the explicit form of φ to construct the ovoid Ω (1) of P G(3, q). Note
that while we shall eventually prove that Ω (1) and Ω are elliptic quadrics it
is not immediately clear at this point that Ω (1) and Ω are even projectively
equivalent in P G(3, q).
Let Ω be an ovoid of P G(3, q) meeting the plane π in a conic C. We
may assume that the symplectic polarity of P G(3, q) defined by Ω has the
form x0y1 + x1y0 + x2y3 + x3y2 = 0 (by applying the Klein correspondence
to theorem 22.6.6 of GGG). This means that the symplectic polarity is the
same ν : (x0, x1, x2, x3) ↔ [x1, x0, x3, x2] considered above. Moreover, we
know that x ∈ Ω ⇐⇒ x ν is the tangent plane to Ω at x and x ∈ Ω ⇐⇒ x
is the nucleus of the oval x ν ∩ Ω.
We may also assume that the plane π has equation x3 = 0 and that
the conic C has equation x3 = x0x1 + x 2 2 = 0, that is, C = {1, t, t 1/2 , 0) :
t ∈ Fq} ∪ {(0, 1, 0, 0)} and has nucleus N = (0, 0, 1, 0) (see Brown, O’Keefe,
Penttila, Triads ... etc.). Then Ω gives rise to an ovoid Ω of T2(C) as follows:
Ω = (Ω \ C) ∪ {[t, 1, 0, t 1/2 ] : t ∈ Fq} ∪ {[1, 0, 0, 0]}.
Let φ : T2(C) → W (q) be the isomorphism given in Theorem 10.4.1 and
let φ(Ω) = Ω (1) . We don’t know just which points are in Ω \ C = Ω \ C,
but they all have the form (x0, x1, x2, 1) for some choices of xi ∈ Fq, and
φ : (x0, x1, x2, 1) ↦→ (x0, x1, x0x1 + x 2 2 , 1). Also, φ : [t, 1, 0, t1/2 ] ↦→ (1, t, t 1/2 , 0)
and φ : [1, 0, 0, 0] ↦→ (0, 1, 0, 0). It follows that
Ω (1) = {(x0, x1, x0x1+x 2 2, 1) : certain xi ∈ Fq}∪{(x0, x1, √ x0x1, 0) : x0, x1 ∈ Fq}.
We now determine the symplectic polarity given by Ω (1) .
Lemma 11.3.1. The tangent planes to Ω (1) are given as follows:
(a) If (x0, x1, x0x1 + x 2 2 , 1) ∈ Ω(1) , the tangent plane to Ω (1) at X =
(x0, x1, x0x1 + x 2 2, 1) is πX = [x1, x0, 1, x0x1 + x 2 2].
(b) The tangent plane to Ω (1) at X = (x0, x1, √ x0x1, 0) is
πX = [x1, x0, 0, √ x0x1].

11.4. RESTRICTIONS ON A SECANT PLANE SECTION OF Ω 517
Proof. Suppose (x0, x1, x0x1+x2 2, 1) and (y0, y1, y0y1+y 2 2, 1) are distinct points
of Ω\C. The plane [x1, x0, 1, x0x1 +x2 2 ] contains the point (y0, y1, y0y1 +y 2 2 , 1)
iff 0 = x1y0 + x0y1 + y0y1 + y2 2 + x0x1 + x2 2 = (x0 + y0)(x1 + y1) + (x2 + y2) 2
iff (x0 + y0, x1 + y1, x2 + y2, 0) ∈ C iff the line 〈(x0, x1, x2, 1), (y0, y1, y2, 1)〉
through two points of Ω \ C contains (x0 + y0, x1 + y1, x2 + y2, 0) ∈ C. This is
not possible since Ω is an ovoid of Σ containing C. Clearly πX does contain
the point X and does not contain N = (0, 0, 1, 0), so X is the unique point
of Ω contained in πX.
Similarly, if X = (x0, x1, √ x0x1, 0), then X ∈ C and it is on the plane
πX = [x1, x0, 0, √ x0x1]. If πX is secant to Ω (1) , then each line in πX through
X must contain a point of Ω different from X. In particular, πX must contain
a point of πX ∩ π on C. Suppose πX contains the point (y0, y1, √ y0y1, 0).
Then 0 = x1y0 + x0y1. This implies that there is a nonzero λ for which
(x0, x1) = λ(y0, y1) and (x0, x1, √ x0x1, 0) = λ(y0, y1, √ y0y1, 0). This completes
the proof.
We now have the polarity associated with Ω (1) determined on all the
points of Ω (1) . This is enough to establish that it must be the same polarity
ν given by Ω.
Since C ⊂ Ω (1) and Ω (1) defines the same symplectic polarity as Ω, it
follows that we may repeat the above process to generate another ovoid of
P G(3, q) from Ω (1) . In fact we may repeat this process ad infinitum. However,
for our purposes here one iteration will be sufficient.
We will now use φ and the fact that φ(Ω) = Ω (1) is an ovoid of P G(3, q)
to generate conditions on some of the secant plane sections of Ω.
11.4 Restrictions on a secant plane section of
Ω
In order to derive algebraic conditions on some secant plane sections of an
ovoid Ω of P G(3, q) containing a conic we use the o-polynomial representation
of hyperovals of P G(2, q). We have seen that any hyperoval of P G(2, q) is
isomorphic to a hyperoval H of the form
H = {(f(t), t, 1) : t ∈ Fq} ∪ {(1, 0, 0), (0, 1, 0)},

518 CHAPTER 11. CLASSIFIYING OVOIDS - PART 2
where f is a permutation polynomial over Fq with f(0) = 0, f(1) = 1 and
the degree of f(t) is at most q − 2. Recall the following theorem.
Theorem 11.4.1. If q > 2, a permutation polynomial f with f(0) = 0 and
f(1) = 1 is an o-polynomial if and only if for each element s ∈ Fq, the
function
fs(x) = [f(x+s)+f(s)]/x, with fs(0) = 0, is a permutation polynomial over Fq.
Let Ω be the ovoid of P G(3, q) as given early in Section 11.3, and let πa
be the plane defined by the equation x0 = ax3, for a = 0. (Note that in
the case where a = 0, the plane π0 is the tangent plane to Ω at the point
(0, 1, 0, 0) and hence is a point of the ovoid Ω of T2(C).)
Let φ, which is an isomorphism from T2(C) to W (q), also represent the
map on AG(3, q) = P G(3, q) \ π that it induces. Thus
φ : (x0, x1, x2, 1) ↦→ (x0, x1, x0x1 + x 2 2, 1).
We claim that φ leaves invariant the set of points of any plane of AG(3, q)
whose projective completion contains N. For let π = [y1, y0, 0, y1x0 + y0x1]
be any such plane. It contains the point (y0, y1, √ y0y1, 0) of C and the point
(x0, x1, x2, 1). Then φ : (x0, x1, x2, 1) ↦→ (x0, x1, x0x1 + x2 2 , 1) ∈ π. Hence
the affine part of π (i.e., π \ π) is mapped to itself by φ. (Note also that
φ : [y1, y0, 0, x0y1 + x1y0] ↦→ (y0, y1, x0y1 + x1y0, 0) ∈ π.)
We observe that πa = [1, 0, 0, a] contains the nucleus N = (0, 0, 1, 0) of C.
The planes πa for 0 = a ∈ Fq are the secant planes to Ω meeting π in the
line 〈(0, 0, 1, 0), (0, 1, 0, 0)〉. Also the restriction of φ to the affine part of πa
is given by
φ|πa : (a, x1, x2, 1) ∈ πa ↦→ (a, x1, ax1 + x 2 2, 1).
Now let Oa = Ω∩πa, so Oa is an oval of πa containing (0, 1, 0, 0) and with
nucleus (0, 1, a, 0) (since (0, 1, a, 0) is the image of πa under the polarity ν
defined by Ω). For convenience we represent the points of the plane πa using
the coordinates x1, x2, x3 (i.e., omitting the x0 coordinate). Thus we write
Oa = {(F (t), t, 1) : t ∈ Fq} ∪ {(1, 0, 0)} with nucleus Na = (1, a, 0),
for some o-polynomial F . Since the line 〈(F (1), 1, 1), (F (0), 0, 1) does not
contain Na,
⎛
F (1) 1
⎞
1
det ⎝ F (0) 0 1 ⎠ = A
1 a 0
′ = 1 + aF (1) + aF (0) = 0.

11.4. RESTRICTIONS ON A SECANT PLANE SECTION OF Ω 519
Also put B ′ = aF (0) and consider the homography of P G(2, q) given by
⎛
⎞
(x1, x2, x3) ↦→ (x1, x2, x3) ⎝
a/A ′ 0 0
1/A ′ 1 0
B ′ /A ′ 0 1
It follows that {(f(t), t, 1) : t ∈ Fq} ∪ {(1, 0, 0)}, where
f(t) =
⎠ .
af(t) + t + B′
A ′ , f(1) = 1 and f(0) = 0, (11.5)
is an oval projectively equivalent to Oa and on the fundamental quadrangle.
So f is an o-polynomial.
Now φ|πa : (F (t), t, 1) ↦→ (F (t), t 2 + aF (t), 1), implying
O (1)
a = {(F (t), t2 + aF (t), 1) : t ∈ Fq} ∪ {(1, 0, 0)}
= {(F ( √ t, t + aF ( √ t), 1) : t ∈ Fq} ∪ {(1, 0, 0)}
is the oval that is the intersection of πa with the ovoid Ω (1) , which also
defines the symplectic polarity ν. As ν : πa ↦→ (1, a, 0) (for either Ω or Ω (1) ),
it follows that O (1)
a also has nucleus (1, a, 0).
Consider the map
⎛
(x1, x2, x3) ↦→ (x1, x2, x3) ⎝
a 0 0
1 1/A ′ 0
0 B ′ /A ′ 1
This homography maps (1, 0, 0) to itself and (F ( √ t), t+aF ( √ t), 1) ↦→ (t, g(t), 1)
where
g(t) = t + aF (√t) + B ′
A ′
,
The important thing here is that
g(1) = 1 and g(0) = 0. (11.6)
g(t) is an o-polynomial!
Putting together Eqs. 11.5 and 11.6 and switching to the indeterminate
x we obtain
⎞
⎠ .
g(x) = A(x + √ x) + f( √ x), A = 1/A ′ = 0. (11.7)

520 CHAPTER 11. CLASSIFIYING OVOIDS - PART 2
So far we have only considered planes of P G(3, q) that meet π in a line
tangent to C at the point (0, 1, 0, 0). We now extend our considerations to any
plane meeting π in a line tangent to C. Let ψt be the homography of P G(3, q)
that acts on points of P G(3, q) by (x0, x1, x2, x3) ↦→ (x0, x1, x2, x3)[ψt], where
[ψt] =
Hence det[ψt] = 1 and
⎛
⎜
⎝
t t 2 + 1 t 1
2 (t + 1) 0
1 t t 1
2 0
0 0 1 0
t 1
2 t 1
2 (t + 1) t 1
ψt : (x0, x1, x2, x3) ↦→
⎞
⎟
⎠ .
(tx0 + x1 + t 1
2 x3, (t 2 + 1)x0 + tx1 + t 1
2 (t + 1)x3,
t 1
2 (t + 1)x0 + t 1
2 x1 + x2 + tx3, x3).
It is a routine exercise to show that ψt interchanges the point (0, 1, 0, 0)
with the point (1, t, t 1
2 , 0). If s = t, it maps (1, s, s 1
2 , 0) to the point (1, b, b 1
2 , 0)
where b = t2 +ts+1.
Hence ψt leaves invariant the conic C.
t+s
More remarkably, ψt commutes with the symplectic polarity ν. To see
0 P
this, recall the matrix C =
that effects the polarity ν. A straight
P 0
forward calculation shows that [ψt]C[ψt] T = C, showing that ψt really does
commute with ν. Hence if we let Ωt = ψt(Ω), then we may apply the discussion
of this section to the ovoid Ωt. Consequently we have the following
lemma.
Lemma 11.4.2. Let Ω be an ovoid of P G(3, q) and let π be a plane of
P G(3, q) such that π ∩ Ω is a conic C. If π ′ is any plane of P G(3, q) such
that π ∩ π ′ is a line tangent to C but π ′ is not tangent to Ω, then the oval
π ′ ∩ Ω is projectively equivalent to an oval
{(f(t), t, 1) : t ∈ Fq} ∪ {(1, 0, 0)} with nucleus {(0, 1, 0)},
where f is an o-polynomial satisfying the equation
for some o-polynomial g and nonzero A ∈ Fq.
g(x) = A(x + √ x) + f( √ x), (11.8)

11.4. RESTRICTIONS ON A SECANT PLANE SECTION OF Ω 521
Our long term goal is to show that if Eq. 11.8 is satisfied then g(x) = √ x
and f(x) = x 2 .
Now consider o-polynomials f and g satisfying Eq. 11.8. Since f and g
have zero constant term and no odd degree terms we may write
From
we see that from Eq. 11.8
f(x) = f2x 2 + f4x 4 + · · · + fq−2x q−2 ,
g(x) = g2x 2 + g4x 4 + · · · + gq−2x q−2 .
f( √ x) = f2x + f4x 2 + f6x 3 + · · · + fq−2x q−2
2
g(x) = (A + f2)x + f4x 2 + f6x 3 + · · · + fq−2x q−2
2 + Ax q
2 .
From this we see that A = f2 = 0 and similarly that gq/2 = A = 0, that
is, the coefficient of √ x = x q
2 in g(x) is not zero. Further, since g has no odd
degree terms, it follows that
f6 = f10 = · · · = fq−2 = 0,
that is, f has no terms of degree 2n where n is odd. And since f has no term
with degree greater than q − 2, it follows that
gq/2+2 = gq/2+4 = · · · = gq−2 = 0,
that is, g has no term of degree greater than q/2. Thus
f(x) = f2x 2 +f2·2x 2·2 +f2·4x 2·4 +· · ·+fq−4x q−4 = Ax 2 (q−4)/4
+
where A = 0 and
g(x) = g2x 2 + g4x 4 + g6x 6 + · · · + gq/2x q/2 q/4
=
where gq/2 = A = 0.
Putting these equations together we find the following:
(i) f2 = A = gq/2 = 0;
(ii) fi·4 = gi·2; 1 ≤ i ≤
i=1
i=1
f4ix 4i , (11.9)
g2ix 2i , (11.10)
q − 4
. (11.11)
4

522 CHAPTER 11. CLASSIFIYING OVOIDS - PART 2
11.5 Sparse and spare o-polynomials
Define a partial ordering ≪ on the set of integers n where 0 ≤ n ≤ q − 1 and
q = 2h . If
h−1
b = bi2 i h−1
and c = ci2 i
i=0
(where each bi and each ci is either 0 or 1), then b ≪ c if and only if bi ≤ ci
for all i. Glynn defined this partial order to express the following theorem:
Theorem 11.5.1. Let f be a polynomial over Fq of degree at most q − 2
satisfying f(a) = 0 if and only if a = 0 and f(1) = 1. Then f is an opolynomial
if and only if the coefficient of x c in f(x) b (mod x q − x) is zero
for all pairs of integers (b, c) satisfying 1 ≤ b ≪ c ≤ q − 1, b = q − 1.
(Note that b = 1 in the theorem shows that an o-polynomial has no odd
degree terms.)
Let q = 2 h , h ≥ 3, and let f be an o-polynomial over Fq. We say that f
is 2 k -sparse provided 1 ≤ k ≤ ⌊(h − 1)/2⌋ + 1 and
f(x) = f2x 2 +
2h−2k −1
i=1
i=0
fi·2k+1x i·2k+1
.
This says f(x) = f2x2 + a polynomial in x2k+1 with degree at most
2h−k+1 − 2k+1 in x. We note that the f of Eq. 11.9 is 21-sparse. Also,
since x2k+2 = (x2k+1) 2 , if f is 2k+1-sparse then it is 2k-sparse (for 1 ≤ k ≤
⌊(h − 1)/2⌋).
Let g be another o-polynomial. We say that g is 2 k -spare provided 1 ≤
k ≤ ⌊(h − 1)/2⌋ + 1 and
g(x) =
2h−2k −1
i=1
gi·2kx i·2k
+ gq/2x q/2 .
This says that g(x) is a polynomial in x2k has degree at most q/2. We note that the g(x) of Eq. 11.10 is 21-spare. + a single term gq/2x q/2 and
Lemma 11.5.2. Let f be an o-polynomial over Fq, q ≥ 8, and suppose that
f is 2 k -sparse with k = ⌊(h − 1)/2⌋ + 1. Then f(x) = x 2 . (So if f and g
satisfy Eq. 11.8, then g(x) = x 1/2 .)

11.5. SPARSE AND SPARE O-POLYNOMIALS 523
Proof. Start with
f(x) = f2x 2 +
2h−2k −1
i=1
fi·2k+1x i·2k+1
. (11.12)
If h = 2j +1, then k = ⌊ h−1
2 ⌋+1 = j +1 =⇒ 2k = h+1 =⇒ 2h−2k −1 =
− 1
2 . Hence the sum in Eq. 11.12 is empty and f(x) = f2x 2 .
If h = 2j, then k = ⌊ h−1
2 ⌋ + 1 = j =⇒ 2k = h =⇒ 2h−2k − 1 = 0.
Hence the sum in Eq. 11.12 is empty and f(x) = f2x 2 .
Since f(1) = 1, A = 1, and by Eq. 11.8 f(x) = x 2 and g(x) = √ x.
Note: In the proof we showed that
if k < ⌊
h − 1
⌋ + 1, then in all cases 2k + 1 ≤ h. (11.13)
2
We now know that f is 2 1 -sparse and g is 2 1 -spare. We proceed inductively
to the case where f is 2k-sparse and g is 2k-spare with k = ⌊ h−1⌋
+ 1. Hence
2
by Lemma 11.5.2 it will follow that f(x) = x2 and g(x) = √ x.
Lemma 11.5.3. Let f and g be o-polynomials over Fq with q = 2h ≥ 8 and
satisfying Eq. 11.8. If 1 ≤ k ≤ ⌊ h−1
2 ⌋, and if f is 2k-sparse and g is 2k-spare, then f is 2k+1-sparse and g is 2k+1-spare. Proof. Assume that f is 2 k -sparse and g is 2 k -spare. First we shall apply
Glynn’s criterion to f for particular values of b and c. Let
b = 2 k+1 − 1 and c = 2 k+1 − 1 + α2 2k+1 ; 0 ≤ α ≤ 2 h−2k−1 − 1. (11.14)
(These values of α ensure that c ≤ q − 1.)
We consider the expansion of f(x) b before reduction modulo x q −x. Note
that
deg(f(x) b ) ≤ (2 k+1 − 1)(2 h−k+1 − 2 k+1 )
= 2 h+2 + 2 k+1 − 2 2k+2 − 2 h−k+1 . (11.15)
Now we consider which x d ≡ x c (mod x q − x) have degree small enough
to appear in f(x) b . By Eq. 11.15 we have
d = c + γ(2 h − 1) ≤ 2 h+2 + 2 k+1 − 2 2k+2 − 2 h−k+1 (for γ ≥ 0). (11.16)

524 CHAPTER 11. CLASSIFIYING OVOIDS - PART 2
Now
d = c + γ(2 h − 1) = 2 k+1 − 1 + γ(2 h − 1) + α2 2k+1
≤ 2 h+2 + 2 k+1 − 2 2k+2 − 2 h−k+1 (from Eq. 11.16)
=⇒ γ(2 h − 1) ≤ 2 h+2 + 1 − α2 2k+1 − 2 2k+2 − 2 h−k+1
≤ 1 + 2 h+2 − 2 2k+2 − 2 h−k+1 (α ≥ 0)
=⇒ γ ≤ 4(2h − 1) + 5 − 22k+2 − 2h−k+1 2h − 1
≤
4(2h − 1) + 4 − 24 − (2h − 1)
2h = 3 −
− 1
12
2h < 3
− 1
=⇒ γ = 0, 1, or 2.
If γ is even, then d = 2 k+1 − 1 + γ(2 h − 1) + α2 2k+1 is odd. But since f(x)
has no odd degree term, neither does f(x) b . Hence if γ is even, then x d does
not appear in f(x) b .
This forces γ = 1, so
d = c + 2 h − 1 = 2 k+1 − 1 + α2 2k+1 + 2 h − 1 ≡ 2 k+1 − 2 (mod 2 k+1 ). (11.17)
Since f is 2 k -sparse,
f(x) = f2x 2 +
so each term of f(x) b looks like
where
f2 · x 2 t ·
2h−2k −1
i=1
2h−2k −1
i=1
fi·2k+1x i·2k+1
, (11.18)
f i·2 k+1x i·2k+1 ti
, (11.19)
b = 2 k+1 2
− 1 = t +
h−2k −1
2
ti and d = 2t +
h−2k −1
i · 2 k+1 · ti ≡ 2t (mod 2 k+1 ).
i=1
(11.20)
(Actually some of the terms in the sum would be missing if the corresponding
f i·2 k+1 were zero, but this does not affect the congruence modulo 2 k+1 .)
Eqs. 11.17 and 11.20 imply that 2 k+1 − 2 ≡ 2t (mod 2 k+1 ), which implies
i=1
t ≡ 2 k − 1 (mod2 k ) =⇒ t ≥ 2 k − 1. (11.21)

11.5. SPARSE AND SPARE O-POLYNOMIALS 525
This implies that
f 2k −1
2
· (x 2 ) 2k −1 divides any term of f(x) b .
Suppose that T (x) is such a term of degree d = c + 2 h − 1. So
f 2k −1
2
· (x 2 ) 2k −1 |T (x).
Note that b = 2 k+1 − 1 = 1 + 2 + 2 2 + · · · 2 k−1 + 2 k . Since t ≤ b = 2 k+1 − 1,
by Eq. 11.21 t = 2 k − 1 or t = 2 k − 1 + 2 k = 2 k+1 − 1. But if t = b, then
x d = f2 · x 2 b = f b 2 · (x 2(2k+1 −1) ).
This says d = 2 k+2 −2 ≥ 2 k+1 +2 h −2 by Eq. 11.17, since α ≥ 0. And this
inequality implies 2 k+1 ≥ 2 h , which is clearly impossible. Hence t = 2 k − 1
and the unique 2-adic expansion of t is
Note that
t = 2 k − 1 = 1 + 2 + 2 2 + · · · + 2 k−1 . (11.22)
f(x) b = f(x) 20
· f(x) 21
· f(x) 22
· f(x) 2k
. (11.23)
Since the only way to write t = 2 k − 1 in base 2 is given in Eq. 11.22, we
must have only one ti = 0 (see Eq. 11.20) and
Then (see Eq. 11.17)
T (x) = f 2k −1
2
· (x 2 ) 2k −1 ·
f u·2 k+1 · x u·2k+1 2 k
. (11.24)
d − 2(2 k − 1) = u · 2 2k+1 = 2 k+1 − 1 + 2 h − 1 + α · 2 2k+1 − 2 k+1 + 2
= 2 h + α · 2 2k+1 .
Hence (u − α)2 2k+1 = 2 h , so
u = α2 h−(2k+1) is an integer, as it should be. (11.25)
At this point we have shown that
T (x) = (f2 · x 2 ) 2k −1 ·
f α2 k+1 +2 h−k · x α·2k+1 +2 h−k 2 k
. (11.26)

526 CHAPTER 11. CLASSIFIYING OVOIDS - PART 2
is the unique term of degree d = c + 2 h − 1 in the expansion of f(x) b before
reduction modulo x q − x. By Glynn’s theorem
Since f2 = 0, we must have
This says
f 2k −1
2
· f 2k
α·2 k+1 +2 h−k = 0.
f α·2 k+1 +2 h−k = 0 for 0 ≤ α ≤ 2 h−2k−1 − 1. (11.27)
f 2 h−k = f 2 h−k +1·2 k+1 = f 2 h−k +2·2 k+1 = · · · = f 2 h−k+1 −2 k+1 = 0. (11.28)
By Eq. 11.11,
g 2 h−k−1 = g 2 h−k−1 +1·2 k = g 2 h−k−1 +2·2 k = · · · = g 2 h−k −2 k = 0. (11.29)
Note: This says that g i·2 k = 0 for all i ≥ 2 h−2k−1 (except for gq/2). Since
g is 2 k -spare, at this point we have
g(x) =
2h−2k−1 −1
i=1
g i·2 kx i·2k
+ gq/2x q/2 . (11.30)
At this point we want to apply Glynn’s theorem to g with
b = 2 k+1 − 1 and c = 2 k+1 − 1 + α · 2 k+1 , 0 ≤ α ≤ 2 h−2k−1 − 1.
deg(g(x) b ) = q
2 b = 2h+k − 2 k−1 . (11.31)
We consider which x d ≡ x c (mod x q − x) have d ≤ deg(g(x) b ). By Eq. 11.31
we have
d ≤ 2 h+k − 2 k−1 .
Here d = c + γ(2 h − 1) for some γ ≥ 0, so
d = 2 k+1 − 1 + α · 2 k+1 + γ(2 h − 1) ≤ 2 h+k − 2 h−1
=⇒ γ(2h − 1) ≤ 2 h+k − 2 h−1 − 2 k+1 + 1 = 2 k (2 h − 1) + 2 k − 2 h−1 − 2 k+1 + 1
=⇒ γ ≤ 2 k + 2k − 2h−1 − 2k+1 + 1
2h < 2
− 1
k .

11.5. SPARSE AND SPARE O-POLYNOMIALS 527
Hence
Now
γ ≤ 2 k − 1. (11.32)
d = c + γ(2 h − 1) = 2 k+1 − 1 + α · 2 k+1 + γ(2 h − 1) ≡ −(γ + 1) (mod 2 k ).
Since every term of g(x) has degree congruent to 0 modulo 2 k , so does every
term of g(x) b (before reduction modulo x q − x). So for x d to appear in g(x) b
we must have d ≡ −(γ +1) ≡ 0 (mod 2 k ), i.e., γ = u·2 k −1. Since γ ≤ 2 k −1,
clearly u = 1, γ = 2 k − 1 and
d = 2 k+1 −1+α·2 k+1 +(2 k −1)(2 h −1) = (α+1)2 k+1 +2 h+k −2 h −2 k . (11.33)
We now show that most of the terms of g(x) b have degree less than d.
First consider a term of the form
(gq/2x q/2 ) 2k+1−3 (gi·2kx i·2k
) 2 , 1 ≤ i ≤ 2 h−2k − 1. (11.34)
We know from Eq. 11.29 that g i·2 k = 0 for i ≥ 2 h−2k−1 . So the largest possible
degree of the form in Eq. 11.34 is
2 h−1 (2 k+1 − 3) + (2 h−2k−1 − 1)2 k+1
= 2 h+k − 2 h + 2 k (2 h−2k − 2 h−k−1 − 2)
< 2 h+k − 2 h (since k ≥ 1 =⇒ 2 h−2k ≤ 2 h−k−1 )
< 2 h+k − 2 h + 2 k+1 − 2 k
≤ (α + 1)2 k+1 + 2 h+k − 2 h − 2 k
= d (by Eq. 11.33).
Thus any term of g(x) b of the form in Eq. 11.34 has degree less than d. So we
seek terms in g(x) b that have degree greater than all the terms of the form
in Eq. 11.34.
Consider the term
(gq/2x q/2 ) 2k+1 −1 = g 2 k+1 −1
q/2
· (x 2h−1
) 2k+1−1 . (11.35)
Since the degree of this term is congruent to 0 modulo 2 h−1 , we also need
d = (α + 1)2 k+1 + 2 h+k − 2 h − 2 k ≡ (α + 1)2 k+1 − 2 k ≡ 0 (mod 2 h−1 ).

528 CHAPTER 11. CLASSIFIYING OVOIDS - PART 2
Since α ≤ 2 h−2k−1 − 1, also
(α + 1)2 k+1 − 2 k ≤ 2 h−k − 2 k < 2 h−1 (for k ≥ 1).
This means that d cannot be congruent to 0 modulo 2 h−1 . Hence the term
in Eq. 11.35 cannot have degree d = (α + 1)2 k+1 + 2 h+k − 2 h − 2 k for any α.
By hypothesis g(x) is 2 k -spare, so
g(x) =
2h−2k −1
i=1
gi·2kx i·2k
+ gq/2x q/2 , with gq/2 = 0. (11.36)
However, i = 2 h−2k − 1 gives i · 2 k = 2 h−k − 2 k , and g 2 h−k −2 k = 0 (see
Eq. 11.29), and by Eq. 11.30 the upper limit on the sum in Eq. 11.36 may
be taken to be 2 h−2k−1 − 1. This just says that for g i·2 k = gq/2, g i·2 k = 0 for
i ≥ 2 h−2k−1 .
Note that
g(x) b = (g(x)) 20
· (g(x)) 21
· (g(x)) 22
· · · (g(x)) 2k
A typical term in this expansion is
(gq/2x q/2 ) t 2
·
h−2k−1−1 i=1
(gi·2kx i·2k
) ti k+1
, b = 2 − 1 = t +
2h−2k−1 −1
i=1
(11.37)
ti. (11.38)
The degree of such a term is at most 2h−1t + 2h−2k−1−1 i=1 i · 2kti, and this is
maximized by putting t1 = · · · t2h−2k−1−2 = 0 and t2h−2k−1−1 = 2k+1 − 1 − t. It
is also clear that this is increased by increasing the value of t. We want this
to be strictly greater than the largest degree of any term in Eq. 11.34 which
is 2h+k − 2h − 2h−1 + 2h−k − 2k+1 . Hence we need
2 h−1 t+(2 h−2k−1 −1)2 k (2 k+1 −1−t) > 2 h+k −2 h −2 h−1 +2 h−k −2 k+1 . (11.39)
If t = 2 k+1 − 3, then we get equality in Eq. 11.39. So 2 k+1 − 2 ≤ t ≤
2 k+1 − 1, and we have already ruled out 2 k+1 − 1. Hence the only possibility
is t = 2 k+1 − 2 and we get a term of the form
(gq/2x q/2 ) 2k+1−2 · (ti·2kx i·2k
) 1 = g 2k+1−2 q/2 · gi·2k · x 2h+k−2h +i·2k , (11.40)

11.5. SPARSE AND SPARE O-POLYNOMIALS 529
for some choice of i. This term has degree d if and only if
if and only if
d = (α + 1)2 k+1 + 2 h+k − 2 h − 2 k = 2 h+k − 2 h + i · 2 k
i = 2α + 1.
Thus the coefficient of x d in g(x) b (and of x c when g(x) b is reduced mod
x q − x) is
g 2k+1 −2
q/2
· g α2 k+1 +2 k.
Since gq/2 = 0, by Glynn’s criterion we have
g α·2 k+1 +2 k = g (2α+1)·2 k = 0 for 0 ≤ α ≤ 2 h−2k−1 − 1. (11.41)
This says g i·2 k = 0 if i is odd. Hence (compare Eq. 11.30) we have
=
2h−2k−2 −1
i=1
g(x) =
2h−2k−1 −1
i even
g i·2 k+1 · x i·2k+1
This says g(x) is 2 k+1 -spare!
By Eq. 11.11 we then also have
This says that
g i·2 k · x i·2k
+ gq/2x q/2
+ gq/2 · x q/2 , and gq/2 = 0.
(11.42)
f α·2 k+2 +2 k+1 = 0 for 0 ≤ α ≤ 2 h−2k−1 − 1. (11.43)
f i·2 k = 0 for all odd i with 1 ≤ i ≤ 2 h−2k − 1.
(Compare Eq. 11.12.) This shows that
f(x) = f2x 2 +
2h−2k−1 −1
i=1
fi·2k+2 · x i·2k+2
, f2 = 0. (11.44)
And, finally, this is just the statement that f is 2 k+1 -sparse!
Lemmas 11.5.2 and 11.5.3 have the following theorem as an immediate
corollary.

530 CHAPTER 11. CLASSIFIYING OVOIDS - PART 2
Theorem 11.5.4. Let f(x) = q/2−1
i=1
o-polynomials over Fq with q = 2 h ≥ 8 and satisfying
Then f(x) = x 2 and g(x) = √ x.
g(x) = A(x + √ x) + f( √ x), A = 0.
f2ix2i and g(x) = q/2−1 i=1
g2ix2i be two
Proof. Since f and g satisfy A(x + √ x) + f( √ x) = g(x), it follows that
f2 = 0 = gq/2, f is 2 1 -sparse and g is 2 1 -spare. From Lemmas 11.5.2 and
11.5.3 we see that f(x) = x 2 and g(x) = √ x.
11.6 The classification of ovoids containing a
conic
Lemma 11.4.2 says the following. Let Ω be an ovoid of P G(3, q) and let π
be a plane of P G(3, q) such that π ∩ Ω is a conic C. Let π ′ be any plane of
P G(3, q) such that π ∩ π ′ is a line tangent to C but not tangent to Ω (i.e.,
π ′ is any plane secant to Ω but meeting C in a unique point). Then the oval
π ′ ∩ Ω is projectively equivalent to an oval
{(f(t), t, 1) : t ∈ Fq} ∪ {(1, 0, 0)} with nucleus (0, 1, 0),
where f is an o-polynomial satisfying the equation
g(x) = A(x + √ x) + f( √ x),
for some o-polynomial g and nonzero A ∈ Fq. Hence by Theorem 11.5.4 we
see that π ′ ∩ Ω is a conic. This proves the following theorem.
Theorem 11.6.1. Let Ω be an ovoid of P G(3, q), q even, and let π be a
plane of P G(3, q) such that π ∩ Ω is a conic C. Then every secant plane
section of Ω that intersects C in exactly one point is a conic.
This theorem allows us to prove the following major classification theorem
of M. Brown [Br00a].
Theorem 11.6.2. Let Ω be an ovoid of P G(3, q), q even, and suppose there
is at least one plane π of P G(3, q) such that π ∩ Ω is a conic. Then Ω is an
elliptic quadric.

11.7. TRANSLATION OVALS AGAIN 531
Proof. Let π ∩ Ω be the conic C, and let O be a secant plane section of Ω not
meeting C in exactly one point (so O meets C in exactly zero or two points).
Let the plane containing O be πO and let the nucleus of O be NO. Let P
be a point of π ∩ πO not contained in Ω. The plane 〈P, NO, N〉 (where N is
the nucleus of C) meets Ω in an oval O ′ such that |O ′ ∩ C| = |O ′ ∩ O| = 1.
Theorem 11.6.1 implies that O ′ is a conic, which then by the same theorem
implies that O is a conic. This shows that all secant plane sections of Ω must
be conics. Hence by Theorem 7.2.7 of Barlotti and Panella Ω must be an
elliptic quadric.
11.7 Translation ovals again
Recall the situation in Section 11.6, but let the polynomials fa(x) and ga(x)
be denoted by f(x) and g(x), respectively. Then we have two o-permutations
f and g that satisfy the equation
ay + f(y) = āy α + g(y α ) ∀y ∈ F. (11.45)
Putting y = 0 in this equation shows that f(0) = g(0) = A.
Replace f and g with the functions
¯f(y) = f(y) + A and ¯g(y) = g(y) + A,
respectively. It follows that ¯ f and ¯g are o-permutations satisfying the equation
Eq. 11.45 with ¯ f(0) = ¯g(0) = 0. For convenience from now on (until
notice otherwise) we use f and g to represent the o-permutations satisfying
Eq. 11.45.
Since f and g are o-permutations with f(0) = g(0) = 0, we know they
can be written as follows:
f(t) =
2
f2i · t 2i ; g(t) =
q−2
i=1
2
g2i · t 2i . (11.46)
Let α : t ↦→ t2k, where 1 ≤ k ≤ e − 1 and gcd(k, e) = 1. Then Eq. 11.45
becomes
at + āt 2k
q−2
2
+ f2i · t 2i q−2
2
+ g2i · t 2i·2k
= 0. (11.47)
i=1
i=1
q−2
i=1

532 CHAPTER 11. CLASSIFIYING OVOIDS - PART 2
Since this equation must hold for all t ∈ Fq, when exponents are reduced
modulo q − 1 it must be that all coefficients are 0.
Suppose that for some i with 1 ≤ i ≤ q−2
2 = 2e−1 −1 the binary expansion
of i is given by
i = b0 + b12 1 + b22 2 + · · · + be−22 e−2 .
It then follows that
i · 2 k+1 = b0 · 2 k+1 + b1 · 2 k+2 + · · · + be−k−2 · 2 e−1
+ be−k−1 · 2 e + be−k · 2 e+1 + · · · + be−2 · 2 e+k−1
≡ be−k−1 · 2 0 + · · · + be−2 · 2 k−1
+ b0 · 2 k+1 + · · · + be−k−2 · 2 e−1 (mod 2 e − 1).
(11.48)
Consider the coefficient on t in Eq. 11.47 From Eq. 11.48 it follows that
i · 2 k+1 ≡ 1 (mod q − 1) if and only if be−k−1 = 1 and all other bj equal zero.
Hence i = 2 e−k−1 , and 2i = 2 e−k . Also, clearly 2i can never be congruent to
1 mod q − 1 for i in the restricted range. Hence we have
g 2 e−k = a. (11.49)
Similarly, if we consider the coefficient on 2 k we see that
f 2·2 k−1 = f 2 k = ā. (11.50)
It is clear that 2i will never be odd modulo q−1, and from Eq. 11.48 i·2 k+1
will be odd and greater than 1 (after reduction) if and only if be−k−1 = 1 and
i = 2 e−k−1 . Hence:
If 2 e−k−1 ≪ i and 2 e−k−1 < i, then g2i = 0. (11.51)
Interchanging the roles of f and g and k with e − k we have also that
If 2 k−1 ≪ i and 2 k−1 < i, then f2i = 0. (11.52)
If we consider the coefficients on powers of t other than t and t2k, we see
that 2i ≡ 2j · 2k (mod q − 1) if and only if i ≡ j · 2k (mod q − 1). Therefore
f 2j·2 k = g2j, or, equivalently, g 2j·2 e−k = f2j, 1 ≤ j ≤ (q − 2)/2. (11.53)
By studying Eq. 11.48 we see that all cases are covered by the following:

11.7. TRANSLATION OVALS AGAIN 533
and
Dually,
f2j =
g2i =
Keep in mind:
⎧
⎪⎨
⎪⎩
⎧
⎪⎨
⎪⎩
At this stage we have that
ā, if j = 2 k−1 ;
0, if 2 k−1 ≪ j and 2 k−1 < j;
g2i, if 2j = 2i · 2 k , which holds for i ≡ j · 2 e−k
iff it is not true that 2 k−1 ≪ j.
a, if i = 2 e−k−1 ;
0, if 2 e−k−1 ≪ i and 2 e−k−1 < i;
f2j, if 2i = 2j · 2 e−k , which holds for j ≡ i · 2 k
iff it is not true that 2 e−k−1 ≪ i.
f 2i·2 k = g2i, and g 2j·2 e−k = f2j.
f(t) = ā · t 2k
+
= ā · t 2k
+
= ā · t 2k
+
g(t) = a · t 2e−k
+
= a · t 2e−k
+
= a · t 2e−k
+
2e−1 −1
i=1
2e−1 −1
j=1
2 k−1 ≪j
2e−1 −1
j=1
2 k ≪2j
2e−1 −1
j=1
2e−1 −1
i=1
2 e−k−1 ≪i
2e−1 −1
i=1
2 e−k ≪2i
f 2i·2 kt 2i·2k
f2j · t 2j .
f2j · t 2j .
g 2j·2 e−kt 2j·2e−k
g2i · t 2i .
g2i · t 2i .
(11.54)
(11.55)
(11.56)
(11.57)

534 CHAPTER 11. CLASSIFIYING OVOIDS - PART 2
2 e−k ≪ 2i ⇐⇒ 2i ≡ 2j · 2 e−k has a solution j;
2 k ≪ 2j. ⇐⇒ 2j ≡ 2i · 2 k has a solution i.
Note that 2 e − 2 k = 2 k + 2 k+1 + · · · + 2 e−1 ≤ 2j ⇐⇒ 2 r ≪ 2j for all
r = k, k + 1, . . . , , e − 1. Hence in particular, f2j = 0 for 2j ≥ 2 e − 2 k . Thus
f(t) = ā · t 2k
+
2 e −2 k −2
2
j=1
2 k ≪2j
f2jt 2j . (11.58)
Dually, 2e − 2e−k = 2e−k + 2e−k+1 + · · · + 2e−1 ≤ 2i ⇐⇒ 2r ≪ 2i for all
r = e − k, e − k + 1, . . . , e − 1. Hence in particular, g2i = 0 for 2i ≥ 2e − 2e−k .
Thus
g(t) = a · t 2e−k
+
2 e −2 e−k −2
2
i=1
2 e−k ≪2i
g2it 2i . (11.59)
Without loss of generality we may assume that e ≥ 4 and 2 ≤ k ≤ e/2,
since the cases with 1 ≤ e ≤ 3 or with k = 1 were already taken care of in
the earlier sections. It is then easy to check that
deg(f(t)) ≤ 2 e − 2 k − 2, and deg(g(t)) ≤ 2 e − 2 e−k − 2. (11.60)
Keep in mind:
Dually,
2 ≤ 2j ≤ 2 e − 2 k − 2 and 2 k ≪ 2j ⇐⇒
2j = 2j0 + 2 k+1 j1, where 0 ≤ j0 ≤ 2 k−1 − 1, (11.61)
0 ≤ j1 ≤ 2 e−k−1 − 1, not both j0 = 0 and j1 = 0.
2 ≤ 2i ≤ 2 e − 2 e−k − 2 and 2 e−k ≪ 2i ⇐⇒
2i = 2i0 + 2 e−k+1 i1, where 0 ≤ i0 ≤ 2 e−k−1 − 1, (11.62)
0 ≤ i1 ≤ 2 k−1 − 1, not both i0 = 0 and i1 = 0.

11.7. TRANSLATION OVALS AGAIN 535
And with this notation,
2i · 2 k = (2i0 + 2 e−k+1 i1)2 k = 2 e+1 i1 + 2 k+1 i0 ≡ 2i1 + 2 k+1 i0,
with the roles of i0 and i1 interchanged exactly right and the congruence is
modulo 2 e − 1.
Dually,
2j · 2 e−k = (2j0 + 2 k+1 j1)2 e−k = 2 e+1 j1 + 2 e−k+1 j0 ≡ 2j1 + 2 e−k+1 j0.
f(x) − ā · x 2k
g(x) − a · x 2e−k
= f2jx 2j = f 2i·2 kx 2i·2k
2j = 2j0 + 2 k+1 j1
2i = 2i0 + 2 e−k+1 i1
0 ≤ j0 ≤ 2 k−1 − 1 0 ≤ i0 ≤ 2 e−k−1 − 1
0 ≤ j1 ≤ 2 e−k−1 − 1 0 ≤ i1 ≤ 2 k − 1
(0 ≤ 2j ≤ 2 e − 2 k − 2) (0 ≤ 2i ≤ 2 e − 2 e−k − 2)
(11.63)
= g2ix 2i = g 2j·2 e−kx 2j·2e−k
2i = 2i0 + 2 e−k+1 i1
2j = 2j0 + 2 k+1 j1
0 ≤ i0 ≤ 2 e−k−1 − 1 0 ≤ j0 ≤ 2 k−1 − 1
0 ≤ i1 ≤ 2 k−1 − 1 0 ≤ j1 ≤ 2 e−k − 1
(0 ≤ 2i ≤ 2 e − 2 e−k − 2) (0 ≤ 2j ≤ 2 e − 2 k − 2)
(11.64)
******************************************************************************************
Glynn’s criterion for o-polynomials says that if (b, c) is a pair of integers
satisfying 1 ≤ b ≪ c ≤ q − 1 with b = q − 1, then the coefficient of x c in
f(x) b (mod x q − x) is zero. We want to apply this with b = 2 k+1 − 1 and
c = b + 2 k+1 α, 0 ≤ α ≤ 2 e−k−1 − 1. This form of c guarantees that b ≪ c
and keeps c in the range c ≤ 2 e − 1.
From Eq. 11.60 we see that deg(f(x)) ≤ 2 e − 2 k − 2. Hence
deg(f(x) b ) ≤ (2 k+1 − 1)(2 e − 2 k − 2) = 2 e+k+1 − 2 e − 2 2k+1 − 2 k+2 + 2 k + 2.
We now determine which x d ≡ x c (mod x q −x) have degrees small enough
to appear in f(x) b , where we consider f(x) b before reduction modulo x q − x.
d = c + γ(2 e − 1) ≤ 2 e+k+1 − 2 e − 2 2k+1 − 2 k+2 + 2 k + 2
=⇒ γ(2 e − 1) ≤ 2 e+k+1 − 2 e − 2 2k+1 − 2 k+2 − 2 k+1 + 2 k + 3 − 2 k+1 α
=⇒ γ ≤ (2 k+1 − 1) − (22k+1 − 1) + (2 k+2 − 1) − 2 k
2 e − 1
< 2 k+1 − 1.

536 CHAPTER 11. CLASSIFIYING OVOIDS - PART 2
If γ is even, then d = 2 k+1 − 1 + 2 k+1 α + γ(2 e − 1) is odd. But since
f has no odd degree term, neither does f(x) b . Hence γ must be odd. Say
γ = 2m + 1 ≤ 2 k+1 − 3, so 0 ≤ m ≤ 2 k − 2.
d = 2 k+1 − 1 + 2 k+1 α + (2m + 1)(2 e − 1)
≡ 2 k+1 − 1 − (2m + 1) (11.65)
≡ 2 k+1 − 2 − 2m (mod 2 k+1 ).
Consider a term of f(x)−a·x 2k as in the far right hand side of Eq. 11.63.
It looks like f2i·2kx2i·2k , where 2i · 2k = 2i0 · 2k + 2e+1i1 has been reduced
modulo 2e − 1 to 2i · 2k ≡ 2i0 · 2k + 2i1. But we are first considering f(x) b
before reduction modulo xq − x, so we may consider the polynomial ¯ f(x) b ,
where ¯ f ≡ f (mod xq − x) is obtained by replacing 2i · 2k ≡ 2i0 · 2k + 2i1
with 2i · 2k = 2i0 · 2k + 2e+1i1.

Chapter 12
The Klein Correspondence
12.1 Plücker Coordinates
The so-called Klein Correspondence is a bijection between the set of lines
of P G(3, q) and the points of a nonsingular hyperbolic quadric in P G(5, q).
This is an extremely useful correspondence that will be applied in many of
the results to come later. To establish the correspondence we need first to
discuss “Plücker coordinates” of lines in P G(3, q).
Let x = (x0, x1, x2, x3) and y = (y0, y1, y2, y3) be distinct points determining
a line ℓ of P G(3, q).
The coordinates of these two points determine
six field elements pij =
xi
xj
yi yj , 0 ≤ i < j ≤ 3, which are the Plücker
coordinates of the line ℓ. Only the ratios of the pij are uniquely determined,
and any other two points of ℓ determine the same set of line coordinates,
since the ratios of the pij’s are unchanged if (xi) is replaced with (xi + λyi).
Expand the following identity in terms of two-rowed minors:
x0 x1 x2 x3
y0 y1 y2 y3
x0 x1 x2 x3
y0 y1 y2 y3
≡ 0.
When the characteristic of the field is not 2, we see that the Plücker
coordinates satisfy
p01p23 + p02p31 + p03p12 = 0. (12.1)
537

538 CHAPTER 12. THE KLEIN CORRESPONDENCE
When the characteristic is 2, a patient expanding of the expression in
Eq. 12.1 also easily yields the same conclusion. (Of course this argument
works for any characteristic.) Hence the point
G(ℓ) = L = (p01, p23, p02, p31, p03, p12) ∈ P G(5, q), (12.2)
which is uniquely determined by ℓ, lies on the Klein quadric
H5 = {¯x = (x0, . . . , x5) : x0x1 + x2x3 + x4x5 = 0}. (12.3)
It is clear that H5 is nonsingular and contains the plane x0 = x2 = x4 = 0,
so it must be hyperbolic.
The point z = (zi) is on ℓ iff the matrix ⎝
⎛
z0 z1 z2 z3
x0 x1 x2 x3
y0 y1 y2 y3
⎞
⎠ has rank 2,
which holds iff each 3 × 3 subdeterminant is zero. Expanding by minors of
the first row yields
z is on ℓ iff zipjk − zjpik + zkpij = 0 whenever 0 ≤ i < j < k ≤ 3. (12.4)
This is equivalent to:
⎛
⎜
⎝
0 p23 p31 p12
−p23 0 p03 −p02
−p31 −p03 0 p01
−p12 p02 −p01 0
which is equivalent to having
(zi) in the row space of
⎛
⎜
⎝
⎞ ⎛
⎟ ⎜
⎟ ⎜
⎠ ⎝
z0
z1
z2
z3
⎞
⎟
⎠ =
⎛
⎜
⎝
0 p01 p02 p03
0
0
0
0
−p01 0 p12 −p31
−p02 −p12 0 p23
−p03 p31 −p23 0
⎞
⎟
⎠ , (12.5)
⎞
⎟
⎠ . (12.6)
The preceding argument makes it clear that the matrix in Eq. 12.5 has
rank 2. It is easy to check that the rows of the matrix in Eq. 12.6 (when
transposed) belong to the null space of the matrix in Eq. 12.5. But if any
one of the pij’s is nonzero, the matrix of Eq. 12.6 has rank at least 2. And
since ¯x and ¯y are distinct points, clearly some pij is not zero. So the matrix
in Eq. 12.6 must have rank 2.

12.1. PLÜCKER COORDINATES 539
Conversely, let (p01, p23, p02, p31, p03, p12) be a point of H5, i.e., some pij is
nonzero and Eq. 12.1 is satisfied. It is now possible to show that the matrices
in Eqs. 12.5 and 12.6 each have rank 2, and rows of one are in the (right)
null space of the other. Each two lines in either matrix have a unique pij
(plus or minus) in common, and they are independent iff that pij is nonzero.
Morever, the line determined by two distinct points of P G(3, q) given by
independent vectors in the row space of Eq. 12.6 has exactly the Plücker
coordinates (p01, p23, p02, p31, p03, p12). This proves that each point of H5 is
the image of some line in P G(3, q) and provides us with the inverse of the
map G. Specifically, let ā = (a0, a1, a2, a3, a4, a5) be an arbitrary point of H5.
Define H(ā) to be the line of P G(3, q) given by
H(ā) = the row space of
⎛
⎜
⎝
0 a0 a2 a4
−a0 0 a5 −a3
−a2 −a5 0 a1
−a4 a3 −a1 0
⎞
⎟
⎠ .
Then H is the inverse of G. This shows that the correspondence between
lines of P G(3, q) and the points of H5 is a bijection. We state this as a
theorem.
Theorem 12.1.1. The correspondence
ℓ = 〈(xi), (yi)〉 ↔ L = (p01, p23, p02, p31, p03, p12)
is a bijection from the set of lines of P G(3, q) to the set of points of the Klein
quadric H5.
We shall refer to the bijection established in Theorem 12.1.1 as the Klein
correspondence.
The Klein correspondence makes it possible to interpret any homography
of P G(3, q) as a homography of P G(5, q) leaving H5 invariant. Suppose the
matrix A = (aij) gives a homography θ of P G(3, q). Let x = (x0, x1, x2, x3)
and y = (y0, y1, y2, y3) be distinct points and put ℓ = 〈x, y〉. Then if θ : x ↦→

540 CHAPTER 12. THE KLEIN CORRESPONDENCE
x ′ = xA, and θ : y ↦→ y ′ = yA , put ℓ ′ = 〈x ′ , y ′ 〉. Then
i.e.,
G(ℓ) = (p01, p23, p02, p31, p03, p12)
G(ℓ ′ ) = (p ′ 01 , p′ 23 , p′ 02 , p′ 31 , p′ 03 , p′ 12 )
so p ′ ij =
x′ i x ′ j
y ′ i y ′
j
=
3
r=0
xrari 3 r=0 yrari
=
=
3
r=0
3
xrari 3 s=0 ysasi
3
r=0 s=0
xrari xrarj
ysasi ysasj
p ′ ij
=
xrarj 3 s=0 ysasj
0≤r

12.2. TIME OUT FOR H5 541
Conversely, it is clear that each line of H5 corresponds to the lines of
P G(3, q) lying in a flat pencil. Let ℓ0, . . . , ℓq be the lines of a flat pencil
through the point p. By Theorem 12.1.3 the set of 1+q +q 2 lines of P G(3, q)
through p correspond to the points of a totally singular plane of H5. Call
this plane of H5 a Latin plane. Similarly, the set of 1 + q + q 2 lines lying
in the plane containing ℓ0, . . . , ℓq also correspond to the points of a totally
singular plane of H5. Call this plane a Greek plane. Using G and its inverse
H it is easy to see that two totally singular planes of the same type intersect
in a unique point, and two totally singular planes of different types intersect
in a line or in the empty set. Moreover, each totally singular line of H5
lies in a unique totally singular plane of each type. Now suppose that π
is a totally singular plane of H5 whose points correspond to the 1 + q + q 2
lines M1, . . . , M 1+q+q 2 of P G(3, q). If these lines are contained in a plane,
clearly they are all the lines of that plane, and the original totally singular
plane of H5 must be a Greek plane. Otherwise, some three of the lines, say
M1, M2, M3 are not coplanar but each two meet. The only possibility is
that they all three meet at a point p. Then the only way the remaining lines
can meet all three of these lines is for them all to be incident with p, in which
case the original totally singular plane of H5 must be a Latin plane. This
completes a proof of the following theorem.
Theorem 12.1.4. The totally singular planes of H5 are partitioned into two
classes, with two such planes being in the same class if and only if they are
the same or intersect in a point, i.e., if and only if their intersection has even
projective dimension. Moreover, each totally singular line is contained in a
unique totally singular plane of each type. A Greek plane of H5 corresponds
to the lines of a single plane of P G(3, q); a Latin plane of H5 corresponds to
the set of lines of P G(3, q) through a fixed point of P G(3, q).
12.2 Time out for H5
Since the lines of P G(3, q) are in one-to-one correspondence with the points
of the Klein quadric H5, we pause to discuss this nonsingular quadric in
P G(5, q) just a bit before proceeding to investigate further the Klein corre-

542 CHAPTER 12. THE KLEIN CORRESPONDENCE
spondence. First, let
⎛
⎜
D = ⎜
⎝
0 1 0 0 0 0
0 0 0 0 0 0
0 0 0 1 0 0
0 0 0 0 0 0
0 0 0 0 0 1
0 0 0 0 0 0
⎞
⎟ , and E = D + D
⎟
⎠
T .
In what follows let V denote the 6-dimensional vector space underlying
P G(5, q), i.e., V = {(a0, a1, . . . , a5) : ai ∈ Fq}. Then for (a), (b) ∈ V put
1. ϖ(a) = (a)D(a) T ;
2. ϖ((a), (b)) = (a) · E · (b) T .
So H5 = {(a) ∈ P G(5, q) : ϖ(a) = 0}, and two points (a) and (b) of
H5 are collinear on a line lying in H5 if and only if ϖ((a), (b)) = 0. Let
(a) = (a0, . . . , a5) and (b) = (b0, . . . , b5) be points of H5. Then
ϖ(a) = a0a1 + a2a3 + a4a5, (12.8)
ϖ((a), (b)) = a0b1 + a1b0 + a2b3 + a3b2 + a4b5 + a5b4. (12.9)
For a ∈ V , a ⊥ = {b ∈ V : ϖ(a, b) = 0. Of course these “perps” are
immediately carried over to perps of points and subspaces of P G(5, q). If
W1 and W2 are subspaces of V for which V = W1 ⊕ W2 and W ⊥ 1 = W2 (so
W ⊥ 2 = W1), write V = W1 ⊥ W2.
Let ϖi = ϖ|Wi , i = 1, 2. We have shown in Chapter 5 that if W1 and
W2 have even dimension and ϖ1 and ϖ2 are both nonsingular, then both are
hyperbolic or both are elliptic since H5 is hyperbolic.
Lemma 12.2.1. Here are some basic facts about H5:
(i) The number of points of H5 is the number of lines of P G(3, q), which
is (q 2 + 1)(q 2 + q + 1).
(ii) Let x and y be distinct points of H5 and let ℓ be the line ℓ = 〈x, y〉.
Then the following are equivalent:
• ℓ has an additional point z on H5.
• Each point of ℓ is on H5.
• ϖ(x, y) = 0.

12.2. TIME OUT FOR H5 543
• ℓ ⊂ ℓ ⊥ and ℓ ⊥ ∩ H5 is a pair of planes meeting in the line ℓ.
(iii) With the same hypothesis as in part (ii), the following are equivalent:
• x and y are the only points of H5 on ℓ.
• ϖ(x, y) = 0.
• V = ℓ ⊥ ℓ ⊥ ; ℓ ⊥ ∩ H5 is an H3, i.e., a hyperbolic quadric in P G(3, q).
Lemma 12.2.2. Let ℓ be a line of P G(5, q) disjoint from H5, i.e., ℓ ∩ H5 is
anisotropic, that is, elliptic. Hence ℓ is disjoint from ℓ ⊥ , so V = ℓ ⊥ ℓ ⊥ and
ℓ ⊥ ∩ H5 must be an elliptic quadric E3 in P G(3, q).
Lemma 12.2.3. Let x be a point of H5 and y a point of P G(5, q)\H5. Then
the line ℓ = 〈x, y〉 has a unique point z = y + ax on H5 in addition to x if
y ∈ x ⊥ , and has no point of H5 other than x if y ∈ x ⊥ . In this latter case
ℓ ⊥ is a solid P G(3, q) meeting H5 in a cone over a conic.
Lemma 12.2.4. Let (x) be a point of P G(5, q) not on H5, i.e., ϖ(x) = 0.
Then [x] ∩ H5 is a parabolic quadric (nonsingular quadric) in P G(4, q), so
ϖ is nonsingular on x ⊥ = [xE], and [xE] ∩ H5 is a parabolic quadric in
P G(4, q), i.e., it is a classical generalized quadrangle often denoted by Q(4, q).
Note that (x) is not on H5 if and only if (xE) is not on H5. Hence we
may also say: The point (x) ∈ P G(5, q) \ H5 if and only if the hyperplane
intersection [x] ∩ H5 is a parabolic quadric P4, i.e., a GQ Q(4, q).
We illustrate these results with a collection of examples. For this, define
the following points:
X1 = (0, 1, 0, 0, 0, 0) ∈ H5; X ⊥ 1 = [1, 0, 0, 0, 0, 0]
X ⊥ 1 ∩ H5 = {(x0, 0, x2, x3, x4, x5) : x2x3 + x4x5 = 0}
= cone over H3
with vertex X1.
X2 = (0, 0, 0, 0, 1, 0) ∈ H5 ∩ X⊥ 1 ; X⊥ 2
ℓ12 = 〈X1, X2〉 ⊂ H5;
ℓ ⊥ 12 ∩ H5 = {(0, x1, x2, x3, x4, 0) : x2x3 = 0}
= pair of planes one Latin, one Greek
= [0, 0, 0, 0, 0, 1]
x2 = 0 : Klein corresponds to all lines through (0, 0, 0, 1)
x3 = 0 : Klein corresponds to all lines in plane [0, 1, 0, 0]

544 CHAPTER 12. THE KLEIN CORRESPONDENCE
X3 = (1, 0, 0, 0, 0, 0) ∈ H5 \ X⊥ 1 ; X⊥ 3 = [0, 1, 0, 0, 0, 0]
ℓ13 = 〈X1, X3〉 ℓ13 ∩ H5 = {X1, X3}
ℓ ⊥ 13 ∩ H5 = {(0, 0, x2, x3, x4, x5) : x2x3 + x4x5 = 0} ∼ = H3.
X4 = (0, 0, 0, 0, −1, 1) ∈ X ⊥ 1 \ H5; X ⊥ 4 = [0, 0, 0, 0, 1, −1]
X ⊥ 4 ∩ H5 = {(x0, x1, x2, x3, x4, x4) : x0x1 + x2x3 + x 2 4 = 0}
ℓ14 = 〈X1, X4〉 ℓ14 ∩ H5 = {X1}
ℓ ⊥ 14 ∩ H5 = {(0, x1, x2, x3, x4, x4) : x2x3 + x 2 4 = 0}
= cone with vertex X1 over a conic.
X5 = (1, 0, 0, 0, −1, 1) ∈ P G(5, q) \ (X⊥ 1 ∪ H5); X ⊥ 5
X ⊥ 5 ∩ H5 = {(x0, x1, x2, x3, x4, x1 + x4) :
x0x1 + x2x3 + x4x1 + x 2 4 = 0} ∼ = Q(4, q)
ℓ15 = 〈X1, X5〉 ℓ15 ∩ H5 = {X1, X1 + X5}
= [0, 1, 0, 0, 0, 1, −1]
ℓ ⊥ 15 ∩ H5 = {(0, x1, x2, x3, x4, x1 + x4) : x2x3 + x4(x1 + x4) = 0}
∼ = H3.
For ease of reference we list some of the above material in the following
corollary.
Corollary 12.2.5. The following information is contained in the preceding
paragraphs.
(i) ℓ ⊂ H5 =⇒ ℓ ⊥ ∩ H5 is a pair of planes through ℓ.
(ii) |ℓ ∩ H5| = 2 =⇒ ℓ ⊥ ∩ H5 is a hyperbolic quadric H3.
(iii) |ℓ ∩ H5| = 1 =⇒ ℓ ⊥ ∩ H5 is a cone with vertex ℓ ∩ H5 over a conic.
(iv) |ℓ ∩ H5| = 0 =⇒ ℓ ⊥ ∩ H5 is an elliptic quadric E3.

12.3. RETURN TO THE KLEIN CORRESPONDENCE 545
ℓ(A,0)
✉ ✉ ✉
p = (1, 0, x, y) p + u(0, 0, 0, 1)
✉ ✉
✉
ap + bq
q = (0, 1, 0, z)
ℓ(A,u)
✉
ap + bq + u(0, 0, −b, a)
q + u(0, 0, −1, 0)
Figure 12.1: RA is a regulus
ℓ∞
✉
✉
(0, 0, 0, 1)
(0, 0, −b, a)
(0, 0, −1, 0)
12.3 Return to the Klein Correspondence
Define an equivalence relation on the set of 2 × 2 matrices over Fq
as follows.
0 1
With P =
, let A, B be any 2 × 2 matrices over F . Then A ≡ B
−1 0
iff B = A + uP for some u ∈ F . A line of P G(3, q) will be given as the row
space of a 2 × 4 matrix with rank 2. For example, if O (resp., I) is the 2 × 2
zero (resp., identity) matrix over F , put
ℓ∞ = 〈(O, I)〉 . (12.10)
Then if u ∈ F and A is any 2 × 2 matrix over F , put
ℓ(A,u) = 〈(I, A + uP )〉 . (12.11)
Lemma 12.3.1. RA = {ℓ∞} ∪ {ℓ(A,u) : u ∈ F } is a regulus.
x y
Proof. If A ≡ B, then RA = RB, so we may assume that A = . It
0 z
is clear that RA consists of q + 1 pairwise skew lines. Figure 1. shows the
transversals of RA (the horizontal lines for (0, 0) = (a, b) ∈ F 2 ).
For a line ℓ of P G(3, q) recall that G(ℓ) denotes the point of H5 that
corresponds to ℓ.

546 CHAPTER 12. THE KLEIN CORRESPONDENCE
Lemma 12.3.2. If R is a regulus, then {G(ℓ) : ℓ ∈ R} is a conic in H5.
Proof. Since any two reguli of P G(3, q) are projectively equivalent, without
loss of generality we may assume that R is given as in the preceding lemma.
1 0 x y + u
Then G(ℓ∞) = (0, 1, 0, 0, 0, 0) and G(ℓ(A,u)) = G
0 1 −u z
= (1, xz + yu + u2 , −u, y + u, z, −x). It is now easy to check that the points
of H5 that are the images of the lines in R lie in the plane π1 spanned by
the points (0, 1, 0, 0, 0, 0), (0, 0, −1, +1, 0, 0) and (1, 0, 0, y, z, −x) and form a
conic there. To see this, keep in mind that x, y, z are fixed. Define the
points
P = (1, 0, 0, y, z, −x); Q = (0, 0, −1, 1, 0, 0); R = (0, 1, 0, 0, 0, 0).
Then 1 · P + u · Q + (xz + yu + u 2 ) · R is the general point of H5 in the
image of R, along with the point R. If we take P , Q and R as fundamental
points generating the plane π1, we just need to see that the points in the
image of R satisfy a nonsingular quadratic equation. Let (w0, w1, w2) ↔
w0 · P + w1 · Q + ω2 · R be the (homogeneous) coordinates of a general point
of π1 and consider the quadratic equation
xzw 2 0 + yw0w1 − w0w2 + w 2 1 = 0.
It is easy to check that this is the nondegenerate conic C1 consisting of
the points in the image of R.
We can say a bit more here. First determine the points of π1 ∩ H5:
aP + bQ + cR = (a, c, −b, ay + b, az, −ax). This point is on H5 if and only if
ac = aby + b 2 + a 2 xz.
There are two cases. If a = 0 and this point is on H5 it quickly follows that
the point is R. If a = 0, then WOLG we may assume a = 1, in which case
c = by + b 2 + xz and the point is one of those on the conic C1. So π1 ∩ H5 is a
nondegenerate (parabolic) quadric. First suppose that q is odd. Then since
π1 ∩ H5 is a nondegenerate quadric, i.e., a conic C2, there is no point of π1
in π ⊥ 1 . Let π2 = π ⊥ 1 . Then π2 ∩ H5 is a conic which Klein corresponds to the
lines of the regulus R opp opposite to R. Second, suppose that q = 2 e and
again put π2 = π ⊥ 1 . Then the conic C1 has a nucleus N which is the unique
point of π1 ∩π ⊥ 1 . In this case the regulus R opp opposite R Klein corresponds
to the conic C2 = π2 ∩ H5, and N is also the nucleus of C2. We state this as
a theorem.

12.3. RETURN TO THE KLEIN CORRESPONDENCE 547
Theorem 12.3.3. Let R be a regulus in P G(3, q) and R opp its opposite
regulus. Let
C1 = {G(ℓ) : ℓ ∈ R} = π1 ∩ H5; C2 = {G(ℓ) : ℓ ∈ R opp }.
Put π2 = π ⊥ 1 . Then π2 ∩ H5 = C2.
(i) If q is odd, then π1 ∩ π2 = ∅.
(ii) If q = 2 e , and if N is the nucleus of C1, then π1 ∩ π2 = {N} and N
is also the nucleus of C2.
Let L1, L2, L3 be three pairwise-skew lines of P G(3, q), so G(L1), G(L2)
and G(L3) are three points of H5 such that no two of them are collinear in H5.
This means also that they are not on a line of P G(5, q). Let π be the plane
they span. Then π ∩ H5 must be a conic C whose points Klein-correspond to
a set of q + 1 pairwise disjoint lines down in P G(3, q). As above, there is a
conic C ⊥ contained in H5 such that each point of C ⊥ is collinear in H5 with
each point of C. Hence the points of C ⊥ Klein-correspond to the opposite
regulus of the regulus H(C). Hence:
Corollary 12.3.4. Then Klein correspondence G gives a bijection between
the reguli of P G(3, q) and the conics lying in H5.
Lemma 12.3.5. Let R1 and R2 be two reguli of P G(3, q) which share precisely
one line ℓ and for which all lines of R1 \ {ℓ} are disjoint from all lines
of R2 \ {ℓ}. Using the Klein correspondence, the lines of Ri are mapped onto
the points of a conic Ci, i = 1, 2. Denote by πi the plane containing Ci so
that πi ∩ H5 = Ci, i = 1, 2. If ℓ Klein corresponds to the point V , C1 and C2
have exactly one point V in common and no two points of C1 ∪ C2 are on a
common line contained in H5. Then π1 ∩ π2 is a line through V and tangent
to both C1 and C2. Also, 〈π1, π2〉 is a solid which meets H5 in a quadric which
must be elliptic, since no two points of C1 ∪ C2 are on a common line of H5.
It follows (from part (iv) of Theorem 5.4.5) that 〈π1, π2〉 ⊥ = π⊥ 1 ∩ π⊥ 2 is an
elliptic line, which says that π⊥ 1 ∩ π⊥ 2 is disjoint from H5.
Proof. Suppose that π1 and π2 have only the point V in common, i.e., π1 and
π2 generate a projective 4-space P G(4, q). Then for each point pi of C2 \{V },
1 ≤ i ≤ q, π1 and pi generate a projective 3-space Pi intersecting H5 in an
elliptic quadric. Using the polarity ⊥ w.r.t. H5, we obtain q distinct lines
P ⊥
i through the point P G(4, q) ⊥ and contained in π⊥ 1 . However, these q lines
P ⊥
i must be exterior to the conic π⊥ 1 ∩ H5, clearly an impossibility. Hence it

548 CHAPTER 12. THE KLEIN CORRESPONDENCE
must be that π1 ∩ π2 is a line, and clearly it must be tangent to both C1 and
C2 at V .
Corollary 12.3.6. Let S be a line spread of P G(3, q) for which S is the
union of q reguli Ri, 1 ≤ i ≤ q, which pairwise intersect in exactly the same
line ℓ. Then S corresponds to a flock of a quadratic cone.
Proof. Let S be a spread of P G(3, q) of the indicated type. Under the Klein
correspondence, the lines of Ri correspond to the points of a conic Ci and ℓ
corresponds to the point V . Denote by πi the plane containing Ci. By the
preceding Lemma, each two planes πi and πj, i = j, intersect in the unique
line ℓij that is tangent to Ci (resp., Cj) in πi (resp., πj). This uniqueness
shows that all the ℓij coincide, i.e., the planes πi all contain a common line
m through V which is tangent to H5 at V . This forces their perps π⊥ i to all
lie in a common 3-space Σ = m⊥ , which meets H5 in a cone K with vertex
V . No two points in C1 ∪ · · · ∪ Cq lie on a line contained in H5, so as we saw
in the preceding proof, if C∗ i = π⊥ i ∩ H5, then F = {C∗ 1 , . . . , C∗ q } is a flock of
K.
12.4 Dual Coordinates for Lines
When we write (a) = (a0, a1, . . . , an), we usually mean that (a) is a point of
P G(n, q). When we write [b] = [b0, b1, . . . , bn], we mean that [b] is a hyperplane
of P G(n, q), and (a) is incident with [b] if and only if n i=0 aibi = 0.
Frequently, in what follows the same (n+1)-tuple will be used to coordinatize
both a point and a hyperplane in the same paragraph or even in the same
sentence.
Note: We now change notation. In place of pij write ℓij. (This should
make it easier to compare our results with those of Hirschfeld[Hi85]. )
When we first associated a line of P G(3, q) with a point of P G(5, q), we
used the fact that a line is determined by any two of its points. However, a
line of P G(3, q) is also determined as the intersection of any two of the q + 1
planes of P G(3, q) containing that line.
Consider the general case. If [u] and [v] are two distinct planes through
ℓ, where [u] = [u0, u1, u2, u3] and [v] = [v0, v1, v2, v3], then a dual coordinate
vector of ℓ is
ˆL = ( ˆ ℓ01, ˆ ℓ23, ˆ ℓ02, ˆ ℓ31, ˆ ℓ03, ˆ ℓ12),

12.4. DUAL COORDINATES FOR LINES 549
where ˆ ℓij = uivj − ujvi.
To a line ℓ of P G(3, q), where G(ℓ) = (ℓ01, ℓ23, ℓ02, ℓ31, ℓ03, ℓ12) are associated
the matrices Λ and ˆ Λ, where
⎛
⎞
and
Λ =
⎜
⎝
0 ℓ23 ℓ31 ℓ12
−ℓ23 0 ℓ03 −ℓ02
−ℓ31 −ℓ03 0 ℓ01
−ℓ12 ℓ02 −ℓ01 0
⎟
⎠
⎛
0 ℓ23
ˆ ˆℓ31
ˆℓ12
⎜
ˆΛ
⎜ −
= ⎜
⎝
ˆ ℓ23 0 ℓ03
ˆ −ˆ ℓ02
−ˆ ℓ31 −ˆ ℓ03 0 ℓ01
ˆ
−ˆ ℓ12
ˆℓ02 −ˆ ⎞
⎟
⎠
ℓ01 0
.
Both Λ and ˆ Λ are skew-symmetric and singular, viz., each has rank 2
with the row space of one being the null space of the other. Compare with
Eqs. 12.5 and 12.6 to see that ℓ is the null space of Λ and hence the row
space of ˆ Λ.
If, for example, ℓ01 = 0, the last two rows of the matrix Λ are linearly
independent. Then ℓ = [ℓ12, −ℓ02, ℓ01, 0] ∩ [−ℓ31, −ℓ03, 0, ℓ01].
Then ˆ L can be computed as follows:
ˆℓ01 = ℓ12 −ℓ02
−ℓ31 −ℓ03
= −ℓ12ℓ03 − ℓ02ℓ31 = ℓ01ℓ23 by Eq. 12.1.
−ℓ31 0
ˆℓ02 = ℓ12 ℓ01
Continuing in this fashion we find that
This establishes the following:
= ℓ01ℓ31.
ˆL = ( ˆ ℓ01, ˆ ℓ23, ˆ ℓ02, ˆ ℓ31, ˆ ℓ03, ˆ ℓ12)
= ℓ01(ℓ23, ℓ01, ℓ31, ℓ02, ℓ12, ℓ03).
Lemma 12.4.1. Up to a scalar factor, ˆ L is L with its components permuted
as follows. Let R be the 2 × 2 permutation matrix with entry 1 in the two

550 CHAPTER 12. THE KLEIN CORRESPONDENCE
off-diagonal positions and with zero diagonal. Let ˆ R be the 6 × 6 permutation
matrix that is the direct sum of three copies of R. Then up to a scalar,
and
So in fact we have the following:
Λ =
ˆΛ =
⎛
⎜
⎝
⎛
⎜
⎝
0 ℓ23 ℓ31 ℓ12
−ℓ23 0 ℓ03 −ℓ02
−ℓ31 −ℓ03 0 ℓ01
−ℓ12 ℓ02 −ℓ01 0
0 ℓ01 ℓ02 ℓ03
−ℓ01 0 ℓ12 −ℓ31
−ℓ02 −ℓ12 0 ℓ23
−ℓ03 ℓ31 −ℓ23 0
ˆL = L · ˆ R.
Lemma 12.4.2. With the notation as above,
0.
(i) For any line ℓ, ϖ(ℓ) := ϖ(G(ℓ)) = 0;
⎞
⎟
⎠ =
⎛
0 ℓ01
ˆ ˆℓ02
ˆℓ03
⎜ −
⎜
⎝
ˆ ℓ01 0 ℓ12
ˆ −ˆ ℓ31
−ˆ ℓ02 −ˆ ℓ12 0 ℓ23
ˆ
−ˆ ℓ03
ˆℓ31 −ˆ ℓ23 0
⎞
⎟
⎠ =
⎛
0 ℓ23
ˆ ˆℓ31
ˆℓ12
⎜ −
⎜
⎝
ˆ ℓ23 0 ℓ03
ˆ −ˆ ℓ02
−ˆ ℓ31 −ˆ ℓ03 0 ℓ01
ˆ
−ˆ ℓ12
ˆℓ02 −ˆ ℓ01 0
⎞
⎟
⎠ ,
⎞
⎟
⎠ .
(ii) Two lines ℓ and ℓ ′ intersect if and only if ϖ(ℓ, ℓ ′ ) := ϖ(G(ℓ), G(ℓ ′ )) =
Proof. Eq. 12.1 takes care of part (i). Part (ii) is just Theorem 12.1.2.
Lemma 12.4.3. ℓ = H(L) contains the point (x) if and only if (x)Λ = 0.
Dually, ℓ lies in the plane [u] if and only if ˆ Λu = 0. Moreover, the columns
(resp., rows) of Λ are orthogonal to the columns (resp., rows) of ˆ Λ. So the
null space of one matrix is the row space of the other. This says ℓ contains
the point (x) if and only if (x) is in the row space of ˆ Λ, and dually, ℓ is in
the plane [u] if and only if u is in the row space of Λ.
Proof. This is just a rearrangement of the information contained in Eqs. 12.5
and 12.6.
This association of a matrix ˆ Λ with a line of P G(3, q) by means of the
point in H5 to which it Klein-corresponds is useful in greater generality.

12.4. DUAL COORDINATES FOR LINES 551
For any point (a) = (a0, . . . , a5) of P G(5, q), think of the components as
being renamed so that (a) = (a01, a23, a02, a31, a03, a12). Then associate with
the vector (a) the matrix
⎛
⎞
Ca =
⎜
⎝
0 a01 a02 a03
−a01 0 a12 −a31
−a02 −a12 0 a23
−a03 a31 −a23 0
⎟
⎠ .
where det(Ca) = (ϖ(a)) 2 . When ϖ(a) = 0, the row space of Ca is the line
that Klein-corresponds to the point (a).
The “companion” matrix to Ca is
⎛
⎞
Ĉa =
⎜
⎝
0 a23 a31 a12
−a23 0 a03 −a02
−a31 −a03 0 a01
−a12 a02 −a01 0
A straightforward computation shows that
Ca · Ĉa = −ϖ(a) · I.
⎟
⎠ .
When (a) ∈ H5, each of Ca and Ĉa has rank 2, and the row space of one
is the null space of the other. When (a) ∈ H5, then
Ĉa = −ϖ(a) · C −1
a .
Fix a = (a01, a23, a02, a31, a03, a12) ∈ P G(5, q). A linear complex Aa is the
set of lines ℓ = H(L) satisfying
Aa = {ℓ = H(L) : aL T = 0} = {ℓ ∈ P G(3, q) : aG(ℓ) T = 0.
It follows that
Aa = {ℓ ∈ P G(3, q) : G(ℓ) ∈ [a] ∩ H5}
= {H(b) : b ∈ [a] ∩ H5},
Let ϖ(a) = a01a23 + a02a31 + a03a12. Then A = Aa is special if ϖ(a) = 0,
in which case it consists of all the lines meeting the line H(â), the axis of the
complex. (Here â = aR.) If ϖ(a) = 0, then A is general.

552 CHAPTER 12. THE KLEIN CORRESPONDENCE
Note that H(â) is defined if and only if ϖ(a) = 0. If this is the case, then
a line ℓ = H(L) meets H(â) iff L · Râ T = La T = 0 iff aL T = 0 iff ℓ ∈ Aa.
Another way to view this is that a special linear complex is the set of lines
corresponding to the points of H5 collinear in H5 with a fixed point â of H5.
In this case the lines of A correspond to the points of the Klein quadric H5
lying in the hyperplane [a], which meets H5 in a cone. (This is a cone whose
base is a hyperbolic quadric in some solid Σ with a vertex which is a point
outside Σ.)
On the other hand, if A is general, then A is the set of lines of P G(3, q)
corresponding to points of H5 lying in the hyperplane [a]. In this case the
intersection H5 ∩ [a] is a nonsingular quadric Q(4, q). We see this in the
following way. First, some aij must be nonzero. WOLG we may assume that
a01 = 0, in which case we may assume that a01 = −1. Then (x0, . . . , x5) ∈ [a]
iff
x0 = a23x1 + a02x2 + a31x3 + a03x4 + a12x5. (12.12)
Assuming this, (x0, x1, . . . , x5) ∈ H5 iff
a23x 2 1 + a02x1x2 + a31x1x3 + a03x1x4 + a12x1x5 + x2x3 + x4x5 = 0. (12.13)
The polar form of this quadratic form on V5 is
B(x, y) = 2a23x1y1 + a02(x1y2 + x2y1) + a31(x1y3 + x3y1) + a03(x1y4 + x4y1)
+ a12(x1y5 + x5y1) + (x2y3 + x3y2) + (x4y5 + x5y4). (12.14)
We know that the intersection [a] ∩ H5 is either a cone with one point as
vertex or a nonsingular quadric, in this case parabolic. It will be nonsingular
if and only if there is no vector in V5 whose polar space is all of V5 and
which is in the intersection [a] ∩ H5. Relative to this polar form, if x =
(x1, x2, x3, x4, x5), then x ⊥ = V5 if and only if
(2a23x1 + a02x2 + a31x3 + a03x4 + a12x5, a02x1 + x3,
But this is zero if and only if
a31x1 + x2, a03x1 + x5, a12x1 + x4) = 0. (12.15)
x2 = −a31x1; x3 = −a02x1; x4 = −a12x1; x5 = −a03x1,

12.4. DUAL COORDINATES FOR LINES 553
and (remember a01 = −1)
−2(a01a23 + a02a31 + a03a12)x1 = 0.
Hence we know the following, if A is special, [a] ∩ H5 is a cone over an
H3, and A is the set of lines of P G(3, q) corresponding to points of the cone.
If A is general, then [a] ∩ H5 is a nondegenerate (parabolic) quadric Q(4, q),
a generalized quadrangle. We take another look at this a little later in this
chapter.
If A = Aa is general, then A determines a null polarity with skewsymmetric
matrix
⎛
⎞
CA = Ca =
⎜
⎝
0 a01 a02 a03
−a01 0 a12 −a31
−a02 −a12 0 a23
−a03 a31 −a23 0
⎟
⎠ .
where det(CA) = (ϖ(a)) 2 = 0.
Then (x) · CA · (y) T = (a) · L T , where L = G(ℓ) if ℓ = 〈(x), (y)〉. Hence
the absolute lines of the polarity associated with the linear complex A are
precisely those which belong to the linear complex, and they are the lines
that Klein-correspond to H5 ∩ [a] = Q(4, q).
We collect this information in the following theorem.
Theorem 12.4.4. Let a = (a01, a23, a02, a31, a03, a12) satisfy ϖ(a) = 0, so
the linear complex A = {ℓ ∈ P G(3, q) : a · G(ℓ) T = 0} is general. Associated
with the linear complex A is a null polarity given by the matrix
⎛
⎞
CA =
⎜
⎝
C −1
A
0 a01 a02 a03
−a01 0 a12 −a31
−a02 −a12 0 a23
−a03 a31 −a23 0
= 1
ϖ(a)
⎛
⎜
⎝
⎟
⎠
0 −a23 −a31 −a12
a23 0 −a03 a02
a31 a03 0 −a01
a12 −a02 a01 0
with inverse
The absolute lines of this polarity are the lines of the linear complex A,
and they are the lines that Klein-correspond to the points of H5∩[a] = Q(4, q).
⎞
⎟
⎠ .

554 CHAPTER 12. THE KLEIN CORRESPONDENCE
At this point it is convenient to determine the null polarities that have
(∞) = (1, 0, 0, 0) and π∞ = [0, 1, 0, 0] T as polar point-plane pair. Writing
{(1, 0, 0, 0) · CA} T = [0, a01, a02, a03] T = λ[0, 1, 0, 0] T , we see that a01 = 0
and a02 = a03 = 0. Without loss of generality we may put a01 = 1. Then
ϖ(a) = a23 = 0. This means that there are q choices for each of a31 and a12
and q − 1 choices for a23. Hence we have proved the following:
Corollary 12.4.5. There are exactly q 2 (q − 1) symplectic polarities interchanging
(∞) and π∞, and in the notation of the preceding paragraph we
have
CA =
⎛
⎜
⎝
0 1 0 0
−1 0 a12 −a31
0 −a12 0 a23
0 a31 −1 0
⎞
⎟
⎠
, and C−1
A
12.5 Linear Congruences
= 1
a23
⎛
⎜
⎝
0 −a23 −a31 −a12
a23 0 0 0
a31 0 0 −1
a12 0 1 0
A linear congruence is the set of lines belonging to two linear complexes.
Two linear complexes Aa and Bb are apolar provided
ϖ(a, b) = aRb T = 0 = a01b23 + a23b01 + a02b31 + a31b02 + a03b12 + a12b03.
and
Let (a) and (b) be distinct points of P G(5, q). So
Aa = {ℓ ∈ P G(3, q) : G(ℓ) ∈ [a] ∩ H5},
Ab = {ℓ ∈ P G(3, q) : G(ℓ) ∈ [b] ∩ H5}.
Then, for example, if ϖ(a) = 0, Aa is the special linear complex consisting
of all lines of P G(3, q) concurrent with the axis
H(â) = row space of
⎛
⎜
⎝
0 a23 a31 a12
−a23 0 a03 −a02
−a31 −a03 0 a01
−a12 a02 −a01 0
⎞
⎟
⎠ .
⎞
⎟
⎠ .

12.5. LINEAR CONGRUENCES 555
On the other hand, if ϖ(a) = 0, then
Aa = {ℓ ∈ P G(3, q) : G(ℓ) ∈ [a] ∩ H5}.
We want to examine the various types of linear congruence. The first
lemma is clearly true.
Lemma 12.5.1. If a linear congruence is the intersection of the two linear
complexes Aa and Ab, it is equally well determined to be the intersection of
any two of the linear complexes of the form
As,t = Ac, where c = sa + tb.
Here [a] ∩ [b] is a solid whose intersection with H5 is completely determined
by the intersection of its polar line
([a] ∩ [b]) ⊥ = 〈[a] ⊥ , [b] ⊥ 〉 = 〈(aR), (bR)〉
with H5.
It is easier to see what the intersection of two linear complexes is if one or
both of them is special, so it will be helpful to determine whether or not one
or both of the two linear complexes given in Lemma 12.5.1 is special. There
are a few cases.
Case 1. ([a] ∩ H5) ∩ ([b] ∩ H5) is an elliptic quadric E3 in P G(3, q)
if and only if
〈(aR), (bR)〉 is disjoint from H5
if and only if
0 = ϖ(saR + tbR) = (saR + tbR)D(sRa T + tRb T ) has no nontrivial
solution
if and only if
0 = s 2 aRDRa T + staR(D + D T )Rb T + t 2 bRDRb T has no nontrivial
solution
if and only if
s 2 ϖ(a) + stϖ((a), (b)) + t 2 ϖ(b) = 0 has no nontrivial solution.
From the above computations we see that

556 CHAPTER 12. THE KLEIN CORRESPONDENCE
Lemma 12.5.2. The number of points of 〈(aP ), (bP )〉 on H5 is the number
of inequivalent nontrivial solutions (s, t) to
s 2 ϖ(a) + stϖ((a), (b)) + t 2 ϖ(b) = 0. (12.16)
Here two nontrivial solutions (s, t) and (s ′ , t ′ ) are equivalent provided
there is a nonzero scalar λ for which (s ′ , t ′ ) = λ(s, t).
In Case 1, when Eq. 12.16 has no solutions, so [a] ∩ [b] ∩ H5 is an elliptic
quadric E3 and the linear congruence Aa ∩ Ab is a regular spread, the linear
congruence is called an elliptic congruence.
The remaining cases deal with the situation where there is at least one
solution to Eq. 12.16, so we assume that ϖ(a) = 0 (by Lemma 12.5.1).
Case 2. ϖ(a) = 0, ϖ(b) = 0 and ϖ((a), (b)) = 0.
In this case Eq. 12.16 becomes stϖ((a), (b)) + t 2 ϖ(b) = 0. The solution
(s, t) = (1, 0) corresponds to (a). The solution (s, t) = (ϖ(b), ϖ((a), (b)) is
a second solution. Put
(c) = ϖ(b) · (a) − ϖ((a), (b)) · (b).
Then ϖ(c) = 0 and a short computation shows ϖ((a), (c)) = 0. This implies
that also ϖ((â), (ĉ)) = 0.
This says that the two axes H( ˆ (a) and H( ˆ (c)) are skew, so Aa ∩ Ac is the
set of lines meeting two skew lines. In this case the linear congruence is called
a hyperbolic congruence. The line 〈(aR), (cR)〉 is hyperbolic, so [a] ∩ [c] ∩ H5
is a hyperbolic space H3 in P G(3, q).
Case 3. ϖ(a) = 0, ϖ((a), (b)) = 0, ϖ(b) = 0. In this case the only
solution (up to equivalence) (s, t) to Eq. 12.16 is (s, t) = (1, 0). So the linear
congruence is the set of all lines concurrent with the axis H(â) which are in
the linear complex Ab. But H(â) is contained in Ab if and only if (mapping
by G) (â) is an element in [b] ∩ H5. Clearly (â) in H5, and it is in [b] if and
only if ϖ((a), (b)) = (aR) · [b] T = 0, which holds by hypothesis. So in this
case the linear congruence (called a parabolic congruence) is the set of lines
meeting an axis ℓ which is contained in a general linear complex W (q) that
contains ℓ. Hence a parabolic congruence contains q 2 + q + 1 lines, a line ℓ
and a flat pencil of lines at each point of ℓ containing ℓ, but such that the
pencils at distinct points of ℓ cover no point in common off the axis ℓ.

12.6. THE NULL SYSTEM ν 557
If ϖ(a) = 0 = ϖ(b) and ϖ((a), (b)) = 0, we are back in Case 2. So the
only remaining case is
Case 4. ϖ(a) = ϖ((a), (b)) = ϖ(b) = 0. In this case there are q + 1
solutions to Eq. 12.16 and the two axes meet. So the linear congruence (called
a degenerate congruence) consists of all the lines meeting two intersecting
lines, i.e., the lines lying in a plane π or passing through a point p in the
plane π. This congruence Klein-corresponds to a pair of totally singular
planes of H5, one Greek, one Latin, that have a line in common.
12.6 The null system ν
0
Put C =
−1
1 0
⊕
0 −1
1
.
0
We say that points (xi) and (yi) of
P G(3, q) are conjugate with respect to the null system ν provided B(¯x, ¯y) =
(xi)C(yi) T = | x0 x1
y0 y1
| + | x2 x3
y2 y3
| = 0.
Note that this is saying that points (xi) and (yi) are conjugate w.r.t. ν if
and only if the Plücker coordinates of the line 〈(xi), (yi)〉 satisfy p01 +p23 = 0.
The null system ν is actually the correspondence ν: points ↔ planes of
P G(3, q) defined by: ν : (xi) ↦→ (xi) ⊥ = {(yi) : (xi)C(yi) T = 0}. This
correspondence is a null polarity (also called a symplectic polarity), i.e.,
1. (xi) ∈ (yi) ⊥ iff (yi) ∈ (xi) ⊥ ;
2. (xi + λyi) ⊥ ⊇ (xi) ⊥ ∩ (yi) ⊥ , so the set of points conjugate
to all points on a line is the set of points on a line;
3. (xi) ∈ (xi) ⊥ for all points (xi).
So in general 〈(xi), (yi)〉 ⊥ = (xi) ⊥ ∩ (yi) ⊥ , but not always will 〈(xi), (yi)〉
and (xi) ⊥ ∩ (yi) ⊥ be the same line. This happens iff (xi) ∈ (yi) ⊥ iff (yi) ∈
(xi) ⊥ . So the self-conjugate lines w.r.t. ν through (xi) are precisely those
of the form ℓ = 〈(xi), (yi)〉 where (xi) = (yi) ∈ (xi) ⊥ . This means the
the q + 1 self-conjugate linese through (xi) are in the flat pencil of lines in
(xi) ⊥ through (xi). Also, if ℓ = 〈(xi), (yi)〉 is self-conjugate, and if (zi) ∈
(xi) ⊥ ∩ (yi) ⊥ = 〈(xi), (yi)〉 ⊥ , then (zi) ∈ ℓ. Hence there are no triangles of
self-conjugate lines.
Let W (q) be the incidence geometry whose points are the points of
P G(3, q) and whose lines are the self-conjugate lines w.r.t. ν. Then W (q)
has (1 + q)(1 + q 2 ) = 1 + q + q 2 + q 3 points. Since it has 1 + q points per line

558 CHAPTER 12. THE KLEIN CORRESPONDENCE
and 1 + q lines per point, it also has (1 + q)(1 + q 2 ) lines. The fact that it
has no triangles leads quickly to the following:
Theorem 12.6.1. W (q) is a generalized quadrangle with order (q, q).
Note 12.6.2. If (xi), (yi), (zi) are any three pairwise noncollinear points
in W (q), then (xi) ⊥ , (yi) ⊥ , (zi) ⊥ are three distinct planes (whose pairwise
intersections are non-self-conjugate lines). But any three planes meet in at
least a point, so there is a point (wi) ∈ W (q) collinear in W (q) with all three
points (xi), (yi), (zi). This says each triad of points of W (q) is centric. A
counting argument shows that each point of W (q) is regular (cf. Cor. 9.3.9).
By Theorems 9.8.3 and 9.8.4 (combinatorial arguments only!), the lines of
W (q) are regular or antiregular according as q = 2 e or q is odd.
If ℓ is a line of W (q), then the first two coordinates of G(ℓ) sum to zero.
If q = 2 e , this means they are equal. In fact if G(ℓ) = (a0, a1, a2, a3, a4, a5) =
(a, −a, a2, a3, a4, a5) where a 2 = a2a3 + a4a5, it follows that at least one of the
entries a2, a3, a4, a5 must be nonzero. Hence we may project the point G(ℓ) of
P G(5, q) to a point of P G(3, q) by just dropping the first two coordinates, i.e.,
(a0, a1, a2, a3, a4, a5) ↦→ (a2, a3, a4, a5). If each point of a line L0 ∈ P G(5, q) is
an image of a line of W (q) (i.e., if it is a line of the appropriate Q(4, q)), then
its points project to the points of a line L in P G(3, q). We say L0 projects
to L.
We are going to show that when q = 2 e the lines ℓ of W (q) through a
point (xi) map by G to the points of a line L0 of Q(4, q) ⊂ H5 that projects
to a line L of W (q).
Theorem 12.6.3. Let ℓ and ℓ ′ be lines of W (q) which are distinct but intersecting.
Say ℓ = 〈(xi), (zi)〉, ℓ ′ = 〈(xi), (z ′ i)〉,
with (xi)C(zi) T = 0 = (xi)C(z ′ i )T , 0 = (zi)C(z ′ i )T .
Then G(ℓ) = (p01, p23, p02, p31, p03, p12) and G(ℓ ′ ) = (p ′ 01 , p′ 23 , p′ 02 , p′ 31 , p′ 03 , p′ 12 )
are collinear in the Klein quadric H5, so
p01
p23
+
p02
p31
+
p03
p12
= 0.
−p ′ 01 p′ 23
−p ′ 02 p′ 31
−p ′ 03 p′ 12
If we project G(ℓ) and G(ℓ ′ ) to the points p = (p02, p31, p03, p12) and p ′ =
(p ′ 02, p ′ 31, p ′ 03, p ′ 12) in P G(3, q), the line through p and p ′ has Plücker coordinates
(p ′′
01, p ′′
23, p ′′
02, p ′′
31, p ′′
03, p ′′
12) = (x0x1 + x2x3, x0x1 − x2x3, −x 2 0, x 2 1, −x 2 2, −x 2 3).

12.6. THE NULL SYSTEM ν 559
Proof. Somewhat tedious but straightforward calculations yield:
p ′′
01 =
p02 p31
p ′ 02 p ′
31
= (x0x1 + x2x3) (z ′ i )C(zi) T ;
p ′′
23 =
p03 p12
p ′ 03 p ′
12
= (x0x1 − x2x3) (z ′ i )C(zi) T ;
p ′′
02 =
p02 p03
p ′ 02 p ′
03
= −x2
′
0 (z i )C(zi) T ;
p ′′
31 =
p12 p31
p ′ 12 p ′
31
= x2
′
1 (z i )C(zi) T ;
p ′′
03 =
p02 p12
p ′ 02 p′
12
= −x2
′
2 (z i)C(zi) T ;
p ′′
12 =
= −x2
′
3 (z i)C(zi) T .
p31 p03
p ′ 31 p′ 03
We give the details for two cases. Since ℓ, ℓ ′ ∈ W (q), we have p01 = −p23
and p ′ 01 = −p′ 23 . These equalities say that
and
Then we calculate
p ′′
01 =
p02 p31
p ′ 02 p ′ 31
x0z1 + x2z3 = x1z0 + x3z2, (12.17)
x0z ′ 1 + x2z ′ 3 = x1z ′ 0 + x3z ′ 2. (12.18)
=
x0 x2
z0 z2
x0 x2
z ′ 0 z ′ 2
x3 x1
z3 z1
x3 x1
z ′ 3 z ′ 1
= (x0z2 − x2z0)(x3z ′ 1 − x1z ′ 3 ) − (x0z ′ 2 − x2z ′ 0 )(x3z1 − x1z3)
= x0x1(z ′ 2z3 − z2z ′ 3) + x2x3(z ′ 0z1 − z0z ′ 1)
+x0x3(z ′ 1z2 − z1z ′ 2) + x1x2(z ′ 3z0 − z ′ 0z3)

560 CHAPTER 12. THE KLEIN CORRESPONDENCE
= (x0x1 + x2x3)(z ′ 2 z3 − z2z ′ 3 ) + (x0x1 + x2x3)(z ′ 0 z1 − z0z ′ 1 )
−x2x3(z ′ 2z3 − z ′ 3z2) − (x0x1(z ′ 0z1 − z0z ′ 1)
+x0x3(z ′ 1z2 − z ′ 2z1) + x1x2(z0z ′ 3 − z ′ 0z3)
= (x0x1 + x2x3)(z ′ 0 z1 − z ′ 1 z0 + z ′ 2 z3 − z ′ 3 z2)
+x0z1(−x3z ′ 2 − x1z0) − x1z0(−x2z ′ 3 − x0z ′ 1 )
+x2z3(−x1z ′ 0 − x3z ′ 2 ) − x3z2(−x0z ′ 1 − x2z ′ 3 )
= (x0x1 + x2x3)(z ′ i )C(zi) T + (x0z1 + x2z3)(−x1z ′ 0 − x3z ′ 2 )
+ (x1z0 + x3z2)(x0z ′ 1 + x2z ′ 3 )
= (x0x1 + x2x3)(z ′ i )C(zi) T + (x1z0 + x3z2)(x0z ′ 1 − x1z ′ 0 + x2z ′ 3 − x3z ′ 2 ))
= (x0x1 + x2x3)(z ′ i)C(zi) T + (x1z0 + x3z2)(0)
= (x0x1 + x2x3)(z ′ i)C(zi) T ,
where the third to the last equation follows by Eq. 12.17, and the next to
last equation follows by an application of Eq. 12.18
The calculation of p ′′
23
03.
other. We evaluate p ′′
p ′′
03 =
p02 p12
p ′ 02 p′ 12
=
is similar, and the four others are similar to each
x0 x2
z0 z2
x0 x2
z ′ 0 z ′ 2
x1 x2
z1 z2
x1 x2
z ′ 1 z ′ 2
= (x0z2 − x2z0)(x1z ′ 2 − x2z ′ 1 ) − (x0z ′ 2 − z′ 0 x2)(x1z2 − x2z1)
= x2z2(z ′ 0x1 − z ′ 1x0) + x2z ′ 2(x0z1 − x1z0) + x 2 2(z ′ 1z0 − z ′ 0z1)

12.6. THE NULL SYSTEM ν 561
= x2z2(z ′ 3 x2 − z ′ 2 x3) + x2z ′ 2 (x3z2 − x2z3) + x 2 2 (z′ 1 z0 − z ′ 0 z1)
= z2x ′ 2(0) + (z2z ′ 3 − z3z ′ 2)(x 2 2) + x 2 2(z ′ 1z0 − z ′ 0z1)
= −x 2 2 (z′ i )C(zi) T .
Theorem 12.6.4. In the preceding theorem it is clear that when q = 2 e , p
and p ′ are collinear in W (q), i.e., p and p ′ are conjugate w.r.t. ν. This says
that for q = 2 e if ℓ ∼ ℓ ′ in W (q), then G(ℓ) and G(ℓ ′ ) project to points on a
line of W (q).
The above results make it clear that the point-line dual of W (q) is isomorphic
to some kind of substructure of the Klein quadric H5. We investigate
this a bit more. For convenience later, note that
B(¯x, ¯y) = p01 + p23. (12.19)
H = {(x0, −x0, x2, x3, x4, x5) ∈ P G(5, q) : xi ∈ F } (12.20)
is a hyperplane of P G(5, q). Moreover,
H ∩ H5 = {(x0, −x0, x2, x3, x4, x5) : x 2 0 = x2x3 + x4x5}. (12.21)
⎛
⎜
Put A = ⎜
⎝
−1 0 0 0 0
0 0 1 0 0
0 0 0 0 0
0 0 0 0 1
0 0 0 0 0
⎞
⎟ . Then put
⎠
Q(4, q) = {(x0, x2, x3, x4, x5) ∈ P G(4, q) : (xi)A(xi) T = 0}.

562 CHAPTER 12. THE KLEIN CORRESPONDENCE
Hence A + AT ⎛
−2
⎜ 0
= ⎜ 0
⎝ 0
0
0
1
0
0
1
0
0
0
0
0
0
⎞
0
0 ⎟
0 ⎟
⎟,
which is clearly nonsingular if q
1 ⎠
0 0 0 1 0
is odd. So in this case Q(4, q) is nondegenerate.
When q = 2e , then A + AT ⎛
⎞
0 0 0 0 0
⎜ 0 0 1 0 0 ⎟
= ⎜ 0 1 0 0 0 ⎟
⎟,
and (1, 0, 0, 0, 0) spans
⎝ 0 0 0 0 1 ⎠
0 0 0 1 0
the left null space of A + AT . But (1, 0, 0, 0, 0) is NOT in Q(4, q), so Q(4, q)
is nondegenerate in all cases.
Recapitulation: Two lines of P G(3, q) meet if and only if their images
in H5 are collinear in H5. Moreover, if l1, l2 are concurrent lines in P G(3, q)
and l1, l2, . . . , lq+1 are the q + 1 lines in the flat pencil spanned by l1, l2, then
G(l1), . . . , G(lq+1) are points of a totally isotropic line of H5. Hence the
point where a flat pencil of lines of W (q) meet can be mapped to the line of
H ∩ H5 ∼ Q(4, q) of points that are the images of the lines of the flat pencil.
It follows that G induces a duality from W (q) onto Q(4, q).
Corollary 12.6.5. Q(4, q) is a GQ(q, q) isomorphic to the point-line dual of
W (q).
If ℓ is a line of W (q), then the first two coordinates of G(ℓ) sum to zero.
Since q = 2 e , this means they are equal. In fact if G(ℓ) = (a0, a1, a2, a3, a4, a5) =
(a, −a, a2, a3, a4, a5) where a 2 = a2a3 + a4a5, it follows that at least one of the
entries a2, a3, a4, a5 must be nonzero. Hence we may project the point G(ℓ)
of P G(5, q) to a point of P G(3, q) by just dropping the first two coordinates,
i.e., (a0, a1, a2, a3, a4, a5) ↦→ (a2, a3, a4, a5). This amounts to projecting the
points of G(W (q)) ∼ = Q(4, q) to P G(3, q) from the nucleus N = (1, 1, 0, 0, 0, 0)
of the Q(4, q) which is H5 ∩ Π, where Π is the hyperplane [1, 1, 0, 0, 0, 0] of
P G(5, q). Let T denote this projection map on the points of GQ(4, q). Since
a line through N cannot be incident with more than one point of G(W (q)),
it follows that if each point of a line L0 ∈ P G(5, q) is an image of a line of
W (q) (i.e., if it is a line of the appropriate Q(4, q)), then its points project
to the points of a line L in P G(3, q). We say L0 projects to L.

12.6. THE NULL SYSTEM ν 563
Theorem 12.6.6. Let q = 2 e . Then there is a duality of W (q) defined as
follows. If x = (x0, x1, x2, x3) is any point of W (q), put ax = x0x1 + x2x3.
Then
δ(x) = row space of
⎛
⎜
⎝
0 ax x 2 0, x 2 2
ax 0 x 2 3 x2 1
x 2 0 x2 3 0 ax
x 2 2 x 2 1 ax 0
If ℓ = 〈x, y〉 with p01 = p23, so ℓ ∈ W (q), then
⎞
δ : ℓ ↦→ T (G(ℓ)) = (p02, p31, p03, p12).
⎟
⎠ = H(ax, ax, x 2 0 , x2 1 , x2 2 , x2 3 ).
For the remainder of this section let q = 2 e and recall Theorem 9.4.10. To
see how the duality δ maps a conic of Σ = P G(3, q) we need only consider
one conic in each orbit. To begin, suppose that C is a conic in the first
orbit. So without loss of generality we may assume that C = {(t, 0, t 2 , 1) :
t ∈ Fq} ∪ {(0, 0, 1, 0)} with nucleus P = (1, 0, 0, 0), which is the pole of the
plane [0, 1, 0, 0] containing C. A straightforward application of the definition
of the map δ yields the following.
δ : (0, 0, 1, 0) ↦→ 〈(1, 0, 0, 0), (0, 0, 0, 1)〉;
δ : (t, 0, t 2 , 1) ↦→ 〈(t 2 , 0, 1, 0), (0, 1, 1, t 2 )〉;
δ : (1, 0, 0, 0) : ↦→ 〈(1, 0, 0, 0), (0, 0, 1, 0)〉.
The first two rows above give the q + 1 lines of a regulus of lines of W (q),
and the third row gives image of the nucleus, which is a transversal of the
regulus. The q other transversals are the lines 〈(1, 0, 0, a), (0, a, 1 + a, 0)〉, all
of which are lines not in W (q). This proves the following:
Theorem 12.6.7. If C is a conic whose nucleus is the pole of the plane
containing C, then δ maps C to the a regulus of lines of W (q), and the
nucleus of C maps to the only transversal of this regulus that is a line of
W (q). Moreover, since there are q 3 − q 2 conics in a plane with a given
nucleus, and there are q 2 (q − 1) reguli of lines of W (q) having a fixed line
as the unique transversal contained in W (q), we see that δ yields a bijection
between the conics in orbit 1 and the reguli of W (q) with a unique transversal
that is a line of W (q).

564 CHAPTER 12. THE KLEIN CORRESPONDENCE
If C is a conic in Orbit 2 with nucleus r, then δ maps the hyperconic
C + = C ∪ {r} to regulus of lines of W (q) with a unique transversal in W (q),
but in this case the transversal is the image of one of the points of the conic
and the image of the nucleus is one of the lines of the regulus.
If ℓ is a line of Σ not in W (q), so ℓ = ℓ ⊥ , the images of ℓ and ℓ ⊥ are two
opposite reguli with all their lines in W (q). Clearly every regulus of lines in
W (q) with at least two transversals which are lines of W (q) must have q + 1
transversals which are lines of W (q). Also, in this case it is easy to see that
the two reguli must be the image under δ of the points of two lines ℓ and ℓ ′
with ℓ ′ = ℓ ⊥ = ℓ.
Let R be a regulus of lines of W (q). We have accounted for those with at
least one transversal which is a line of W (q). What about those reguli R with
all lines in W (q) but no transversal in W (q)? In Exercise 9.4.10.1 we showed
that no such regulus of lines of W (q) can exist. This general discussion has
an interesting corollary.
Corollary 12.6.8. Let R be a regulus in P G(3, q), q = 2 e . Suppose that R
is contained in a spread S of P G(3, q), all of whose lines belong to a linear
complex A of lines. Then the spread S is a regular spread, i.e., given any
three lines of S, the regulus of P G(3, q) containing the three lines consists of
lines of A.
Proof. Let W (q) be the symplectic geometry whose lines are the lines of A.
Then S is a spread of W (q). We know that R has either 1 or q+1 transversals
in A. But if R had q + 1 transversals in A (i.e., in W (q)), then R would be
a span of lines in W (q), so by the point-line dual of Theorem 9.10.2 could
have at most two lines of S. Hence R has exactly one line of A.
Let δ be a duality of W (q). Then δ maps S to an ovoid Ω of W (q) (which
is also an ovoid of P G(3, q) by Lemma 7.6.2). Ω defines a null polarity ν
of P G(3, q) such that each point N not on Ω lies on q + 1 lines tangent to
Ω that lie in a plane π. Then N is both the pole of the plane π w.r.t. the
symplectic polarity ν and the nucleus of the oval π ∩Ω. Since R has a unique
transversal in W (q) it is mapped by δ to a conic C of P G(3, q) whose nucleus
is the pole of the plane containing C. As C is a conic contained in an ovoid
Ω of P G(3, q), by Theorem 11.6.2 Ω must be an elliptic quadric. As each
triad of points of Ω must have a unique center, which is then the nucleus of a
conic in Ω, under δ we have that each triple of lines of S must be contained
in a unique regulus of lines of S, i.e., S is a regular spread.

12.7. A POLARITY π OF W (Q) WHEN Q = 2 2M+1 565
Exercise 12.6.8.1. Put p1 = (1, 0, 0, 0), p2 = (1, 0, 1, 0), p3 = (0, 1, 0, 0),
p4 = (0, 1, 1, k) where k = 0, 1. Put L∞ = 〈p1, p2〉, Lc = 〈p3 + cp1, p4 + cp2〉.
Then R = {L∞} ∪ {Lc : c ∈ Fq} is a regulus with all its lines in W (q). It
opposite regulus has the lines M∞ = 〈p1, p3〉 and Ma = 〈p2 + ap1, p4 + ap3〉.
The unique transversal in W (q) is Ma with a 2 = 1+k. Compute the image of
R and of R opp under the Klein map G. If G(R) is the conic C in the plane
π of P G(5, q), then G(R opp ) is a conic C ∗ in the plane π ⊥ . You should find
that N = (0, 0, 1, 0, k, 1) is the nucleus of both C and C ∗ . Of course each
point of C is collinear in H5 with each point of C ∗ . π ∩ π ⊥ = N. Let Π
denote the hyperplane x0 = x1 of P G(5, q). Recall that this hyperplane meets
H5 in quadric Q(4, q) with nucleus P = (1, 1, 0, 0, 0, 0). π ⊥ meets Π in the
line 〈N, P 〉, which is exterior to H5, i.e., it is an elliptic line.
12.7 A polarity π of W (q) when q = 2 2m+1
A map π that maps points of W (q) to lines of W (q) and maps lines of W (q)
to points of W (q) is a polarity of W (q) provided it preserves incidence and
π 2 = id. Preserving incidence means that if ℓ = 〈p, q〉 is a line of W (q), then
π(p) must meet π(q) at π(ℓ), i.e., π(〈p, q〉) = π(p) ∩ π(q).
Also note that if α ∈ Aut(Fq), then replacing each coordinate (of point,
hyperplane, etc.) by its image under α induces an automorphism of the entire
projective space. In the present situation, since the matrix C is equal to its
image under this map, each automorphism of Fq induces an isomorphism of
W (q).
From now on let q = 2 e = 2 2m+1 with σ = 2 m+1 : x ↦→ x σ ∈ Aut(F ),
F = GF (q). So σ 2 = 2. Define a map π from the points of P G(3, q) to the
lines of W (q) by
π : (x0, x1, x2, x3) ↦→ row space of
⎛
⎜
⎝
0 a σ
2 x σ 0 xσ 2
a σ
2 0 x σ 3 xσ 1
x σ 0 xσ 3 0 a σ
2
x σ 2 xσ 1
which is the line with Plücker coordinates
where a = (x0x1 + x2x3).
a σ
2 0
(a σ
2 , a σ
2 , x σ 0 , xσ 1 , xσ 2 , xσ 3 ) = (p01, p23, p02, p31, p03, p12),
⎞
⎟
⎠ , a = x0x1 + x2x3,

566 CHAPTER 12. THE KLEIN CORRESPONDENCE
Also define π from the set of lines of W (q) to the points of P G(3, q) by:
if (xi)C(zi) T
x0, x1, x2, x3
= 0, then π :
↦→ (p
z0, z1, z2, z3
σ
2
02, p σ
2
31, p σ
2
03, p σ
2
12).
So π(xi) is the line whose last four Plücker coordinates are (xσ 0 , xσ1 , xσ2 , xσ3 ).
And if L is a line whose last four Plücker coordinates are (p02, p31, p03, p12),
then π(L) = (p σ
2 2 2 2
02, p σ
31, p σ
03, p σ
12). Since σ2 = 2, clearly π2 = id.
Theorem 12.7.1. π(xi) ∩ π(zi) = π((xi) · (zi)) when (xi)C(zi) T = 0. So π
is a polarity of W (q).
Proof. π(xi) is the line with Plücker coordinates px = (a σ
2
x , a σ
2
x , xσ 0, xσ 1, xσ 2, xσ 3),
ax = x0x1 + x2x3. Similarly, π(zi) is the line with Plücker coordinates pz =
(a σ
2
z , a σ
2
z , zσ 0 , zσ 1 , zσ 2 , zσ 3 ), az = z0z1 +z2z3. The theorem will be proved if we can
σ
x0 x2 2
show that the point π(L) =
,
x3
σ
x1 2
,
x0
σ
x3 2
,
z0 z2
z3 z1
z0 z3
x1 x2
z1 z2
is on the two lines with Plücker coordinates px and pz. By Theorem 12.2.3
the lines of W (q) through π(L) are mapped to points of H5 lying on a line
L0
whose projection to P G(3, q) has Plücker coordinates
−, −,
x0
σ
x2
,
x3
σ
x1
,
x0
σ
x3
,
x1
x2 σ
. But the line of
z0 z2
z3 z1
z0 z3
z1 z2
H5 through the points px and pz, when projected to P G(3, q), also has Plücker
coordinates equal to those just given. Hence the two lines are the same, completing
the proof.
Now that we know π is a polarity of W (q), we can show that the set Ω
of points that are self-conjugate w.r.t. π (i.e., (xi) ∈ π(xi)) forms an ovoid
of W (q), and hence an ovoid of P G(3, q) by 7.6.2 .
Theorem 12.7.2. Let P = P G(3, q), q = 2 e , have a null system ν, i.e., ν
is a bijection from the points of P G(3, q) to the planes of P G(3, q) such that
(i) x ∈ ν(x) for all points x ∈ P.
(ii) x ∈ ν(y) iff y ∈ ν(x) for all x, y ∈ P.
Let B be the set of lines L of P G(3, q) that are self-conjugate w.r.t. ν,
i.e., L = ν(x)∩ν(y) for distinct points x, y of L. Then S = (P, B, ∈) is a GQ
of order q, usually denoted W (q), in which all points are regular. Let π be a
polarity of S. So π : P ↔ B satisfies x ∈ π(y) iff y ∈ π(x) for all x, y ∈ P ;
and π(L) ∈ L ′ iff π(L ′ ) ∈ L for all L, L ′ ∈ B. Put Ω = {x ∈ P : x ∈ π(x)};
Ω ∗ = {L ∈ B : π(L) ∈ L}. Then:
σ
2

12.7. A POLARITY π OF W (Q) WHEN Q = 2 2M+1 567
(a) Ω is an ovoid of S; hence Ω is an ovoid of P G(3, q)
(b) Ω ∗ is a spread of S and hence a spread of P G(3, q);
(c) The lines of B passing through a point x form the plane ν(x).
If x ∈ Ω, ν(x) ∩ Ω = {x}. If x ∈ Ω, ν(x) ∩ Ω is an oval .
Proof. Suppose x and y are distinct points of Ω on a line L of B. If “∼”
denotes “collinear on a line of B”, we have: x ∈ π(x), y ∈ π(y), x ∼ y, so
π(x) meets π(y). Hence, L ∈ {π(x), π(y)}, since otherwise L, π(x), π(y)
are the three sides of a triangle. So suppose L = π(x). Then y ∈ π(x) ⇒
x ∈ π(y). Since y ∈ π(y), clearly π(y) = x · y = L = π(x), forcing x = y, a
contradiction. So we know each line of B is incident with at most one point
of Ω. L ∈ Ω ∗ iff π(L) ∈ L iff π(L) ∈ Ω. Now assume that L is a line of B
for which L ∈ Ω ∗ , i.e., π(L) ∈ L. Since S is a GQ, there is a unique point
u and a unique line M for which π(L) ∈ M ∈ B and u ∈ M ∩ L. Then
L · u · M · π(L) =⇒ π(L) · π(u) · π(M) · L, where we write x · y to indicate
that x and y are incident in S. If π(M) = u this gives a triangle. Hence
π(u) = M and π(M) = u. Consequently u ∈ Ω and M ∈ Ω ∗ . So each line of
B is incident with at least one, and hence exactly one, point of Ω. It follows
that Ω is an ovoid of S, and hence an ovoid of P G(3, q) by 7.6.2. Dually, Ω ∗
is a spread of S. But it is then clear that the lines of Ω ∗ partition the points
of P , i.e., Ω ∗ is a spread of P G(3, q).
Although we officially already know that Ω is an ovoid of P G(3, q), we
give an independent proof of this that gives a little more insight into the
relationship between Ω, π and ν. Since a line of B is now known to contain
a unique point of Ω, part (a) will be proved if we show that each line L ∈
P G(3, q)\B has either 0 or 2 points of Ω. So suppose L is a line of P G(3, q)\
B, x ∈ L∩Ω, y ∈ L, y = x. Put L ′ = ν(x)∩ν(y) = ν(L). Note: L and L ′ are
skew. For v ∈ L ∩ L ′ ⇒ v ∈ ν(x) ∩ ν(y) ⇒ L =< x, y >⊆ ν(v). And since L
would be a line through v in ν(v), L ∈ B, a contradiction. Put z = π(x)∩L ′ .
π(x) ∩ L ′ is a point, since π(x) and L ′ are both in ν(x), but π(x) ∈ B and
L ′ ∈ B. Finally, put w = π(z) ∩ L ′ . Here observe that z ∈ π(x) iff x ∈ π(z)
iff π(z) is a line of B containing x, so π(z) and L ′ are in ν(x), π(z) ∈ B,
L ′ ∈ B. Since L ∩ L ′ = ∅, x = w. We claim that y ∈ Ω iff y is the unique
point y = L ∩ π(w).

568 CHAPTER 12. THE KLEIN CORRESPONDENCE
x ∈ π(x)
p L ∈ B
x ∈ Ω
✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥
❚
❍
❍❍❍❍❍❍❍❍❍❍❍❍❍
❚❚❚❚❚❚
L
✔❚
✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥
✔
❚❚
′ ✉
✔
❡✉
✔
✔
✔
✔
✔ w
= ν(L) ∈ B
✔
✉
z ✔✉
✔
π(z)
π(w) π(x)
First, z ∈ L ′ = ν(x) ∩ ν(y) ⇒ x, y ∈ ν(z) → L ⊆ ν(z). And w ∈ π(z) ⇒
z ∈ π(w) ⇒ π(w) is a line of B through z, so π(w) is in ν(z). Then L ⊆ ν(z),
π(w) ∈ ν(z), L ∈ B, π(w) ∈ B ⇒ L ∩ π(w) is a unique point p. We know
x is the unique point of Ω on π(z), so w ∈ Ω ⇒ w ∈ π(w). So there is a
unique point on π(w) collinear with w in S. But w ∈ L ′ ⊆ ν(y), so y ∈ ν(w).
In fact, w ∈ ν(x) ∩ ν(y) ⇒ L = x · y ⊆ ν(w). So p = π(w) ∩ L is collinear
with w and must be the unique point of π(w) collinear with w. Now p ∼ w,
p ∈ π(w), w ∈ π(p), w ∈ π(w) ⇒ π(p) meets π(w) in S. And no triangles in
S forces p · w = π(p), i.e., p ∈ Ω. So p = L ∩ π(w) ∈ Ω.
Conversely, suppose y ∈ Ω. Put w ′ = π(y) ∩ π(z). So y · z = π(w ′ ), and
y ∈ π(w ′ ) ∩ L. Since y and w ′ are collinear in S, and π(z) ⊆ ν(x), we have
w ′ ∈ ν(y) ∩ π(z) ⊆ ν(y) ∩ ν(x) ∩ π(z) = L ′ ∩ π(z) = w. So w = w ′ , implying
y = π(w) ∩ L, as claimed. Condition (c) is clear.

12.8. THE OVOID OF J. TITS 569
π(y)
x ∈ Ω
❅ ❡✉ ❡✉ L ∈ B
❅❅❅❅❅❅❅❅❅❅❅y
= π(w) ∩ L
❅ ❅
z = π(x) ∩ L
π(w)
π(x)
′
w
✉
✉
π(z)
L ′ = ν(L)
The above proof shows the following: If x ∈ Ω, x ∈ L ∈ B, then the other
point y of L∩Ω is obtained as follows: Put z = π(x)∩ν(L), w = π(z)∩ν(L).
Then y = π(w) ∩ L. Moreover, y · w = π(y).
12.8 The ovoid of J. Tits
Return to the specific polarity π of Theorem 12.7.1, with q = 2 2m+1 , etc. Put
Ω = {(xi) : (xi) ∈ π(xi)}. π(xi) was defined as the row space of a certain
matrix, but it is also easy to confirm that π(xi) is the right null space of

570 CHAPTER 12. THE KLEIN CORRESPONDENCE
⎛
⎜
⎝
0 a σ/2
x x σ 1 x σ 3
a σ/2
x 0 x σ 2 x σ 0
x σ 1 x σ 2 0 a σ/2
x
x σ 3 x σ 0 a σ/2
x
0
⎞
⎟
⎠ , ax = x0x1 + x2x3.
And we know there must be exactly 1 + q 2 points in Ω. But then it is
easy to check directly that p = (0, 1, 0, 0) ∈ Ω, and if f(a, b) = a σ+2 +ab+b σ ,
then (1, f(a, b), a, b) ∈ Ω. Hence we have
Theorem 12.8.1. Ω = {(0, 1, 0, 0)} ∪ {(1, f(a, b), a, b) : a, b ∈ F } is the Tits
ovoid.
Define ηt ∈ P GL(3, q) by: For 0 = t ∈ F , ηt : (xi) ↦→ (xi)[ηt], where
[ηt] =
⎛
⎜
⎝
And put N =< ηt : 0 = t ∈ F >.
1 0 0 0
0 t σ+2 0 0
0 0 t 0
0 0 0 t σ+1
⎞
⎟
⎠ .
Theorem 12.8.2. N fixes (0, 1, 0, 0) and (1, 0, 0, 0) and permutes the other
points of Ω in q + 1 orbits, each of size q − 1.
Proof. Straightforward, given the above description of Ω.
Theorem 12.8.3. f(a, b)+f(x, y)+x(a σ+1 +b)+ya = f(a+x, b+a σ x+y).
Proof. Straightforward.
For a, b ∈ F , define τa,b ∈ P GL(3, q) by τa,b : (xi) ↦→ (xi)[τa,b], where
[τa,b] =
⎛
⎜
⎝
1 f(a, b) a b
0 1 0 0
0 a σ+1 + b 1 a σ
0 a 0 1
⎞
⎟
⎠ .
Theorem 12.8.4. τa,b leaves Ω invariant. In fact,

12.8. THE OVOID OF J. TITS 571
(i) τa,b : (1, f(x, y), x, y) ↦→ (1, f(a + x, b + a σ x + y), a + x, b + a σ x + y).
(ii) [(τa,b) −1 ] = [τ a,a σ+1 +b] =
⎛
⎜
⎝
1 f(a, b) a a σ+1 + b
0 1 0 0
0 b 1 a σ
0 a 0 1
(iii) f(a, b) = f(a, y) iff y = b or y = b + a σ+1 .
(iv) ηt ◦ τa,b ◦ (ηt) −1 = τ a
t ,
b
t
σ+1 .
Proof. The proof of (i) is essentially Theorem 12.8.3, and parts (ii), (iii) and
(iv) are easy to check.
Put T = 〈τa,b : a, b ∈ F 〉. The following is now clear:
Theorem 12.8.5. We have
(i) T is a group.
(ii) T fixes (0, 1, 0, 0) and is sharply transitive on Ω \ {(0, 1, 0, 0)}.
(iii) Gp = 〈N , T 〉 = N · T = T ⋊ N , a semidirect product.
The tangent plane to Ω at (1, f(x, y), x, y) is ν(1, f(x, y), x, y) = [f(x, y), 1, y, x] T .
As it contains the point (1, f(a, b), a, b) iff (a, b) = (x, y), we have
Theorem 12.8.6. The following hold:
(i) f(x, y) + f(a, b) + ay + bx = 0 iff (a, b) = (x, y).
And putting x = y = 0,
(ii) f(a, b) = 0 iff a = b = 0.
Also,
Theorem 12.8.7. The following hold:
(i) f(a, b) σ+1 = b σ+2 + abf(a, b) σ + a σ f(a, b) 2 .
(ii)
1
f(a,b)
= f
b
f(a,b) ,
a . f(a,b)
Proof. Just substitute the value of f(a, b) in (i) and expand. Doing the same
for (ii) will show that it is equivalent to (i).
⎞
⎟
⎠

572 CHAPTER 12. THE KLEIN CORRESPONDENCE
and
A brief computation shows that
π(0, 1, 0, 0) = 〈(0, 1, 0, 0), (0, 0, 0, 1)〉 ⊂
⊂ ν(0, 1, 0, 0) = [1, 0, 0, 0] T ,
π(1, f(x, y), x, y) = 〈(0, x σ+1 + y, 1, x σ ), (1, y σ , 0, x σ+1 + y)〉 ⊂
⊂ ν(1, f(x, y), x, y) = [f(x, y), 1, y, x] T .
It follows that for each point p ∈ Ω, the stabilizer of the point p in Sz(q)
must fix the line π(p).
Define θ ∈ P GL(3, q) by: θ
: (xi) ↦→ (xi)[θ], where
0 1 0 1
[θ] = ⊕ .
1 0 1 0
Theorem 12.8.8. The map θ leaves Ω invariant. In particular,
(i) θ : (0, 1, 0, 0) ↔ (1, 0, 0, 0), and
(ii) θ : (1, f(a, b), a, b) ↔
b 1, f f(a,b) ,
a
f(a,b)
b , f(a,b) ,
a
f(a,b) .
Proof. Writing out the image of (1, f(a, b), a, b) under θ shows that Theorem
12.8.6(ii) proves the result.
Now put Sz(q) = 〈N , T , θ〉, q = 22m+1 . Then it is clear that Sz(q)
is doubly transitive on the points of Ω. The stabilizer of one point (e.g.,
(0, 1, 0, 0)) in Sz(q) is isomorphic to 〈N , T 〉. In fact, J. Tits [Ti62] proves
that Sz(q) is the complete group of all collineations of P G(3, q) leaving Ω
invariant. In the present work we assume this without proof.
Suppose that x ∈ Ω, so ν(x) is the plane tangent to Ω at x. Since Sz(q)
is (doubly) transitive on points, we may assume that p = (0, 1, 0, 0) and then
study the stabilizer Gp = N T of the point p in Sz(q). Each element of Gp
has a matrix of the form
ηt · τa,b =
⎛
⎜
⎝
1 f(a, b) a b
0 t σ+2 0 0
0 t(a σ+1 + b) t ta σ
0 t σ+1 a 0 t σ+1
⎞
⎟
⎠ .

12.8. THE OVOID OF J. TITS 573
As we saw above, π(p) = 〈p, (0, 0, 0, 1)〉 is fixed by these collineations,
and the other lines of ν(p) through p are permuted as follows:
ηt · τα,b : 〈p, (0, 0, 1, c)〉 ↦→ 〈p, (0, 0, 1, a σ + ct σ )〉.
Here a, b, t ∈ Fq with t = 0. So Gp fixes the line π(p) and is transitive on the
q other lines of ν(p) through p.
The planes containing π(p) are the planes [x, 0, z, 0]. τa,b : [x, 0, 1, 0] ↦→
[x + a, 0, 1, 0], so τ0,b fixes each plane through π(p). ν([x, 0, 1, 0]) = (0, x, 0, 1)
is the nucleus of
Ωx = Ω ∩ [x, 0, 1, 0] = {(0, 1, 0, 0)} ∪ {(1, f(x, b), x, b) : b ∈ Fq}.
In particular, τ0,b : (1, f(x, y), x, y) ↦→ (1, f(x, b+y), x, b+y), and τ0,b induces
an elation of [x, 0, 1, 0] with axis π(p) and center (0, xb + b σ , 0, b). It follows
that π(p) is the axis of the translation oval Ωx. We state this as a theorem.
Theorem 12.8.9. If p ∈ Ω, then π(p) is the axis of the oval Ωx = ν(x) ∩ Ω
for each point x ∈ π(p) \ {p}, i.e., π(p) is the axis of each (translation) oval
of Ω determined by a cutting plane through π(p).
Such a family of q ovals containing a common tangent which is an axis of
each of them is called an axial pencil.
Now suppose that x ∈ Ω, so that ν(x)is a plane secant to Ω, and Ωx =
ν(x) ∩ Ω is an oval in ν(x) with nucleus x. Moreover, as we show next, the
point π(x) ∩ ν(x) is a special point of the oval Ωx called the noed of Ωx.
(Sometimes the noed is called the corner of Ωx.)
Of course, if x ∈ Ω, then x ∈ π(x). Suppose this holds. Then the only
lines of ν(x) which are self-conjugate w.r.t. ν pass through x. Hence π(x) ⊆
ν(x), so z = π(x) ∩ ν(x) is a well-defined point. Now z ∈ π(x) ⇒ x ∈ π(z).
And z ∈ ν(x) ⇒ x ∈ ν(z). Suppose for a moment that z ∈ Ω, i.e., z ∈ π(z).
So z is the unique point of π(x) collinear in S with x, and x = π(z) ∩ ν(z) is
the unique point of π(z) collinear with z. But x collinear with z in S implies
π(x) must be concurrent with π(z), giving a triangle. The only possibility is
then that z ∈ π(z). So z = π(x) ∩ ν(x) ∈ Ωx. And the subgroup of Sz(q)
fixing Ωx must fix the noed zx = π(x) ∩ ν(x). We have proved the following:
Theorem 12.8.10. If x ∈ Ω, then zx = π(x) ∩ ν(x) is a point of Ωx (i.e.,
the noed), which must be fixed by the subgroup of Sz(q) fixing x (or leaving

574 CHAPTER 12. THE KLEIN CORRESPONDENCE
invariant the plane Ωx). Also, if p = (0, 1, 0, 0) ∈ Ω, it is easy to check that p
is the noed of each Ωx, i.e., of each plane section of Ω by a plane containing
π(p).
At this point we know that Sz(q) is doubly transitive on the points of Ω,
so in particular it is transitive on the set of secant lines. The q + 1 ovals of
Ω lying in the q + 1 planes through a fixed secant line form a bundle of Ω.
For example, if x = (0, 0, x2, x3), ν(x) = [0, 0, x3, x2] T , and Ω ∩ ν(x) =
Ωx = {(0, 1, 0, 0)} ∪ {(1, f(a, b), a, b) : ax3 + bx2 = 0}. As x2, x3 vary, these
ovals Ωx vary over the q + 1 ovals that contain the two points (0, 1, 0, 0) and
(1, 0, 0, 0) of Ω. Hence this set of q + 1 ovals is a bundle. of ovals. Note that
θ interchanges (0, 0, 1, 0) and (0, 0, 0, 1), and
ηt : (0, 0, 1, y) ↦→ (0, 0, 1, yt σ ).
Also, both N and θ leave the secant line L =< (1, 0, 0, 0), (0, 1, 0, 0) > invariant.
So < N , θ > permutes the planes through L in at most two orbits. But
(0, 0, 0, 1) θ·τ1,1 = (0, 0, 1, 1), so at least they are all in the same orbit under
Sz(q). At this point we see that Sz(q) is transitive on the pairs (ℓ, π), where ℓ
is a secant line contained in a cutting plane π. This establishes the following:
Theorem 12.8.11. Each oval section of Ω is a translation oval equivalent to
Oσ. Moreover, if x is the nucleus of some oval section, then the line joining
the nucleus x and its noed zx must be the axis of the oval Ωx = ν(x) ∩ Ω.
So consider an example of one of the planes of the bundle mentioned
above. Let x = (0, 0, 1, 1), so ν(x) = [0, 0, 1, 1], and ν(x) ∩ Ω = Ωx =
{(0, 1, 0, 0)} ∪ {(1, a σ+2 + a 2 + a σ , a, a) : a ∈ F }.
π(0, 0, 1, 1) =< (1, 0, 1, 0), (0, 1, 0, 1) > .
Here π(0, 0, 1, 1) ∩ Ω = (1, 1, 1, 1), the noed of Ωx. But the plane [0, 0, 1, 1]
contains p = (0, 1, 0, 0) and meets ν(p) in the tangent line m = 〈p, (0, 0, 1, 1)〉,
which is different from π(p). Hence the q planes through the line m different
from [1, 0, 0, 0] meet Ω in an oval for which m is not an axis, i.e., the q ovals
form a non-axial pencil.
[τ −1
1,1 θηtθτ1,1] =
⎛
⎜
⎝
t σ+2 1 + t σ+2 t σ+1 + t σ+2 t σ+2 + t σ+1
0 1 0 0
0 1 + t t σ+1 t + t σ+1
0 1 + t 0 t
⎞
⎟
⎠

12.8. THE OVOID OF J. TITS 575
fixes (1, 1, 1, 1) and (0, 1, 0, 0) and (for a = 1) maps (1, a σ+2 + a 2 + a σ , a, a)
to (1, b σ+2 + b 2 + b σ , b, b), where b = 1 + t −1 + at −1 , i.e., a = 1 + (b + 1)t. So
these collineations, all of which leave invariant the ovoid Ω, act transitively
on the q − 1 points of Ωx other than (1, 1, 1, 1) and (0, 1, 0, 0).
Now consider the homography
φ : (x0, x1, x2, x3) ↦→ (x0, x1, x3, x2).
Clearly φ commutes with ν, so φ preserves W (q), and φ fixes p = (0, 1, 0, 0).
But φ maps the line π(p) to 〈p, (0, 0, 1, 0)〉 = π(p), so φ cannot belong to
Sz(q). Put Ω = Ω1 and Ω φ = Ω2. This gives a very important example as
described in the next thereom.
Theorem 12.8.12. Let p = (0, 1, 0, 0) and ℓ = 〈p, (0, 0, 0, 1)〉, and let Ω1
and Ω2 be the two Tits ovoids described above, both determining the same
symplectic geometry W (q). Then ℓ is a line of W (q) (i.e., tangent to both
Ω1 and Ω2 at p). However, ν(p) = [1, 0, 0, 0] is the tangent to Ωi at p for
both i = 1 and i = 2. Hence each other plane through ℓ is a cutting plane
with respect to Ωi, i = 1, 2, but such a plane meets Ω1 in an oval for which
ℓ is an axis, and meets Ω2 in an oval for which ℓ is not an axis. (Since
ℓ = 〈p, (0, 0, 0, 1)〉 is the axis of the intersection with Ω of the planes through
ℓ, ℓ φ = 〈p, (0, 0, 1, 0)〉 is the axis of the intersection with Ω of the planes
through ℓ φ .)
=
The spread associated with Ω is π(Ω). In detail:
π(0, 1, 0, 0) =
π(1, f(a, b), a, b) =
⎛
⎜
⎝
=
0 1 0 0
0 0 0 1
;
a σ a 2(σ+1) + (ab) σ + b 2 a σ+1 + b 0
1 b σ 0 a σ+1 + b
a σ+1 + b 0 b σ a 2(σ+1) + (ab) σ + b 2
0 a σ+1 + b 1 a σ
1 b σ 0 a σ+1 + b
0 a σ+1 + b 1 a σ
.
⎞
⎟
⎠
Exercise: Write out the conditions for these lines to be a spread and
discover that this holds iff f(a, b) + f(c, d) = ad + bc iff (a, b) = (c, d).

576 CHAPTER 12. THE KLEIN CORRESPONDENCE
12.9 The Hermitian geometry H(3, q 2 )
Over E = GF (q 2 ) ⊇ F = GF (q), consider the Hermitian form on E 4 given
by (where ā = a q for a ∈ E):
U((x0, x1, x2, x3), (y0, y1, y2, y3)) =
3
xi¯y3−i. (12.22)
Put H(3, q 2 ) = {(x0, x1, x2, x3) ∈ P g(3, q 2 ) : x0¯x3 + x1¯x2 + x2¯x1 + x3¯x0 = 0}.
Note: H(3, q 2 ) is a generalized quadrangle GQ(q 2 , q).
For a line l = 〈(xi), (yi)〉 in P G(3, q 2 ), recall the Klein map G (with
inverse H) defined by
where lij =
i=0
G(l) = L = (l01, l23, l02, l31, l03, l12) ∈ Q + (5, q) = H5,
xi xj
yi yj
.
Recall that if a line ℓ of P G(3, q) has coordinates L = (l01, l23, l02, l31, l03, l12),
it has dual coordinates ˆ L = (l23, l01, l31, l02, l12, l03).
There is a unitary polarity µ : (a0, a1, a2, a3) ↔ {(ā3, ā2, ā1, ā0) T } ⊥ =
[ā3, ā2, ā1, ā0] T . Here (a0, a1, a2, a3) belongs to H(3, q 2 ) if and only if (a0, a1, a2, a3) ∈
(a0, a1, a2, a3) µ .
So if l =
a0 a1 a2 a3
b0 b1 b2 b3
, then
l µ = {(ā3, ā2, ā1, ā0) T } ⊥ ∩ {( ¯ b3, ¯ b2, ¯ b1, ¯ b0) T } ⊥ .
So l µ has dual coordinates
ā3 ā2
,
ā1
ā0
,
¯ b3 ¯ b2
¯ b1 ¯ b0
ā3 ā1
¯b3 ¯b1
,
ā0 ā2
¯b0 ¯b2
,
= ( ¯ l32, ¯ l10, ¯ l31, ¯ l02, ¯ l30, ¯ l21).
ā3 ā0
¯b3 ¯b0
,
Hence l µ has regular coordinates ( ¯ l01, ¯ l23, − ¯ l02, − ¯ l31, ¯ l12, ¯ l03).
ā2 ā1
¯b2 ¯b1 Hence l µ ⊆ H(3, q 2 ) if and only if l = l µ if and only if there is a nonzero
ρ ∈ GF (q 2 ) for which
(l01, l23, l02, l31, l03, l12) = ρ( ¯ l01, ¯ l23, − ¯ l02, − ¯ l31, ¯ l12, ¯ l03).

12.9. THE HERMITIAN GEOMETRY H(3, Q 2 ) 577
Apply the involution a ↦→ ā (which preserves H(3, q 2 )) to get
This implies
( ¯ l01, ¯ l23, ¯ l02, ¯ l31, ¯ l03, ¯ l12) = ¯ρ(l01, l23, −l02, −l31, l12, l03).
which in turn implies that ρ¯ρ = 1.
ρ¯ρ(l01, l23, −l02, −l31, l12, l03) =
ρ( ¯ l01, ¯ l23, ¯ l02, ¯ l31, ¯ l03, ¯ l12) =
(l01, l23, −l02, −l31, l12, l03),
Recap: l ⊆ H(3, q 2 ) if and only if there exists a ρ ∈ GF (q 2 ) such that
(i) ρˆρ = 1, and
(ii) (l01, l23, l02, l31, l03, l12) = ρ( ¯ l01, ¯ l23, − ¯ l02, − ¯ l31, ¯ l12, ¯ l03).
For 0 = α ∈ E, replace l with αl, still on H(3, q 2 ), so that ρ is replaced
with some nonzero ρ0 for which ρ0 ¯ρ0 = 1. Then
(αl01, αl23, αl02, αl31, αl03, αl12) =
ρ0(ā ¯ l01, ā ¯ l23, −ā ¯ l02, −ā ¯ l31, ā ¯ l12, ā ¯ l03) =
α(l01, l23, l02, l31, l03, l12) =
αρ( ¯ l01, ¯ l23, − ¯ l02, − ¯ l31, ¯ l12, ¯ l03).
Hence αρ = ρ0 ¯α. Since ρ0 ¯ρ0 = 1, there exists an α0 ∈ E such that ρ = ¯α0
α0
(by Theorem 20.2.2). If we use α0 in place of α, then α0ρ = ρ0 ¯α0, so that
ρ = ρ0 · ¯α0
α0 = ρ0ρ, implying ρ0 = 1. So l ∈ P G(3, q) has a representation
L = (l01, l23, l02, l31, l03, l12) = ( ¯l01, ¯l23, −¯l02, −¯l31, ¯l12, ¯l03). This says that l
on H(3, q2 ) has a representation of the form L = (a, b, cα0, dα0, u, ū) where
a, b, c, d ∈ F , u ∈ E, and α0 ∈ E is fixed with α q
0 = −α0, and ab+cdα2 0 +uū =
0.
Now consider the cases q odd and q even separately.
Case 1. q odd. Project (a, b, cα0, dα0, u, ū) to (a, b, c, d, e, f), where u =
e+fα0. So l on H(3, q 2 ) maps to (a, b, c, d, e, f) where ab+cdα 2 0 +e2 −f 2 α 2 0 =
0. Put

578 CHAPTER 12. THE KLEIN CORRESPONDENCE
so that
⎛
0
⎜ 0
⎜
A = ⎜ 0
⎜
⎝
1
0
0
0
0
0
0
0
α
0
0
0
0
2 0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0 0 0 0 0 −α2 ⎞
⎟ ,
⎟
⎠
0
⎛
A + A T ⎜
= ⎜
⎝
0 1 0 0 0 0
1 0 0 0 0 0
0 0 0 α 2 0 0 0
0 0 α 2 0 0 0 0
0 0 0 0 1 0
0 0 0 2 0 −2α 2 0
So H(3, q2 ) maps to the quadric Q− (5, q) given by A with polar form given
by A+AT . Here Q− (5, q) is the orthogonal sum of two hyperbolic lines and an
1 0
elliptic line. (Exercise: Show that Q((x4, x5)) = (x4, x5)
0 −α2
x4
0 x5
⎞
⎟ .
⎟
⎠
gives an elliptic line.) Hence H(3, q 2 ) maps to (the point-line dual of)
Q − (5, q), which is an elliptic quadric.
Case 2. q = 2 e . Here α q
0 = −α0 = α0 just says that α0 ∈ GF (q). So l on
H(3, q 2 ) maps to a point L = (a, b, c, d, u, ū) with ab + cd + uū = 0. Write
F = GF (q) ⊆ GF (q 2 ) = E = F (ζ), where ζ 2 + ζ + δ = 0, for some δ ∈ F
with tr(δ) = 1.
Here ζ + ¯ ζ = 1; ζ · ¯ ζ = δ. If α = a + bζ, then
(i) α¯α = a 2 + ab + b 2 δ;
(ii) α + ¯α = b.
If u = e + fζ, then ζ(e + f ¯ ζ) + ¯ ζ(e + fζ) = e, and u + ū = f.
Consider the linear map
T : P G(5, q 2 ) → P G(5, q 2 ) : (x0, . . . , x5) ↦→ (x ′ 0 , . . . , x′ 5 ),

12.10. A CLOSER LOOK AT THE KLEIN CORRESPONDENCE 579
where
x ′ i = xi, 0 ≤ i ≤ 3,
x ′ 4 = ζx4 + ¯zx5; x ′ 5 = x4 + x5.
It follows that x4 = x ′ 4 + ¯ ζx ′ 5 ; x5 = x ′ 4 + ζx′ 5 .
For l ∈ H(3, q 2 ), l = H(L), suppose that L = (a, b, c, d, ū, u), a, b, c, d ∈
F , u ∈ E, with ab + cd + ūu = 0. So
T : L ↦→ (a, b, c, d, ζū + ¯ ζu, ū + u) = (a, b, c, d, e, f),
where u = e + fζ. And ab + cd + ūu = 0 becomes
⎛
0
⎜ 0
⎜
(a, b, c, d, e, f) ⎜
⎝
1
0
0
0
1
0
1
⎞ ⎛
a
⎟ ⎜
⎟ ⎜ b
⎟ ⎜
⎟ ⎜ c
⎟ ⎜
⎟ ⎜ d
1 ⎠ ⎝ e
0 δ f
⎞
⎟ = 0.
⎟
⎠
This quadratic form is elliptic. So for q even we also have that H(3, q 2 ) is
isomorphic to the point-line dual of the elliptic orthogonal geometry Q − (5, q).
12.10 A Closer Look at the Klein Correspondence
In this section we collect a variety of observations about the Klein correspondence
between lines of P G(3, q) and points of the Klein quadric H5 in
P G(5, q). This is adapted shamelessly from Table 15.10 (pages 29 – 31 of
Hirschfeld [Hi85]).

580 CHAPTER 12. THE KLEIN CORRESPONDENCE
Table of properties of Klein correspondence
Σ = P G(3, q) q P G(5, q)
(q 2 + 1)(q 2 + q + 1) lines in Σ − (q 2 + 1)(q 2 + q + 1) points on H5
Two skew lines − The two points of H5 on a bisecant
Two intersecting lines − Two points whose join is a line of H5
A (flat) pencil of lines − A line of H5
The three sides of a triangle − The three vertices of a triangle whose
sides lie on H5
A plane (of lines) − A Greek plane of H5
(All lines through) a point − A Latin plane of H5
q + 1 points on a line − q + 1 Latin planes through a point of H5
q + 1 planes on a line − q + 1 Greek planes through a point of H5
A unique line through two points − Two Latin planes meet in a unique point
A unique line on two planes − Two Greek planes meet in a unique point
A pencil of lines lies in a unique plane − A line on H5 lies in just one Greek
and passes through a unique point plane and just one Latin plane
The lines through a point and in a plane − A Latin plane and a Greek plane
form a pencil or the emptyset according meet in a line or in the empty set
as the point is or is not on the plane
The edges of a trihedral angle and the − the vertices and sides of a triangle
pencils generated by pairs of the edges in a Latin plane

12.10. A CLOSER LOOK AT THE KLEIN CORRESPONDENCE 581
Table of Klein properties continued
The sides of a triangle and the pencils − The vertices and sides of a triangle
generated by pairs of the sides in a Greek plane
The three face planes of a trihedral − The three Greek planes through the
angle sides of a triangle in a Latin plane
The three vertices of a triangle − The three Latin planes through the sides
of a triangle in a Greek plane
The 3-dimensional Principle of Duality − The consistent interchange of Latin
and Greek planes
A dual conic − A conic in a Greek plane
The generators of a quadratic cone − A conic in a Latin plane
Two non-coplanar pencils with a − A pair of intersecting lines: the
common line and distinct centers section of H5 by a tangent plane
A regulus: the q + 1 transversals − A conic: the section of H5 by the
to three skew lines plane of a triangle with vertices
but not sides on H5
A hyperbolic congruence: the (q + 1) 2 − A hyperbolic quadric H3: the section
transversals of two skew lines ℓ, ℓ ′ of H5 by a solid Π3whose polar
line meets H5 in two points
Two families of q + 1 pencils each − The two reguli on H3
joining a point on one of ℓ, ℓ ′ to all
the points on the other

582 CHAPTER 12. THE KLEIN CORRESPONDENCE
Table of Klein properties continued
Σ = P G(3, q) q P G(5, q)
An elliptic congruence: q 2 + 1 lines − An elliptic quadric E3 : the section
forming a regular spread of H5 by a solid whose polar line is
skew to H5. No two points of E3
are conjugate with respect to H5.
The regulus containing three lines of an − The conic section of H5 by a plane
elliptic congruence lies in the congruence through 3 points on a solid section
E3 lies on E3
An arbitrary spread of q 2 + 1 lines − q 2 + 1 points of H5, one in each Latin
and in each Greek plane; an ovoid of H5
A packing, i.e., a partition of the lines of − A partition of H5 into ovoids
P G(3, q) into q 2 + q + 1 spreads
A packing of P G(3, q) by regular spreads − A partition of H5 into elliptic
quadrics E3
A parabolic congruence: q 2 + q + 1 lines − A quadratic cone with vertex V : the
meeting the axis ℓ, no two of which meet section of H5 by a tangent solid Σ,
off ℓ whose polar line is therefore a tangent
line at V
The q + 1 pencils of lines in the − The q + 1 generators of the cone
parabolic congruence
The q 3 reguli in the parabolic − the q 3 conic sections by planes
congruence not through the vertex
A special linear complex: the lines − A cone with H3 as base: the section
meeting (or equal to) the axis of H5 by a non-tangent hyperplane

12.10. A CLOSER LOOK AT THE KLEIN CORRESPONDENCE 583
Table of Klein properties continued
Σ = P G(3, q) q P G(5, q)
A general linear complex: (q 2 + 1)(q + 1) − A parabolic quadric P4 : the section
lines of a symplectic geometry of H5 by a non-tangent hyperplane
The (q 2 + 1)(q + 1) pencils in a general − The (q 2 + 1)(q + 1) lines on
linear complex, one through each point P4; one in each Latin plane and
and one in each plane and one in each Greek plane
A pair of apolar linear complexes − A pair of conjugate hyperplanes
The null polarity whose self-polar odd The harmonic homology whose axial
lines form a general linear complex hyperplane and center are the
representing P G(4, q) and its pole
The null polarity whose even The elation whose axial hyperplane and
self-polar lines form a center are the representing P G(4, q) and
general linear complex its pole, which lies in P G(4, q) and is
the nucleus of P4.
The pseudo polarity with even The projectivity (x0, . . . , x5) ↦→
bilinear form (x5, x1, x2, x1 + x3, x2 + x4, x0)
x0y0 + x0y1 + x1y0 + x2y3 + x3y2
with fixed points the plane
x1 = x2 = x0 + x5 = 0
The ordinary polarity with odd The projectivity
self-conjugate quadric (x0, . . . , x5) ↦→ (x5, 2x3, 2x4, 2νx1,
E3 : −νx 2 0 + x 2 1 + x2x3 = 0 2νx2, 4νx0) with no fixed points
The ordinary polarity with even The projectivity (x0, . . . , x5) ↦→
self-conjugate quadric (x5, x1, −x2, −x3, x4, x0)
H3 : x0x1 + x2x3 = 0 with fixed points the planes x0 − x5 =
= x2 = x3 = 0; x0 + x5 = x − 1 = x4

584 CHAPTER 12. THE KLEIN CORRESPONDENCE
Table of Klein properties continued
Σ = P G(3, q) q P G(5, q)
A pair of complementary − A pair of conics on H5 in polar
reguli on H3
planes Π2, Π ′ 2
H3 does not define a null polarity odd Π2 ∩ Π ′ 2 = ∅
H3 does define a null polarity even Π2 ∩ Π ′ 2 = Π0
Four skew lines with 0, 1, or − Four points generating a solid whose
2 transversals polar line is skew, tangent, or
bisecant to H5
12.11 Spreads from Flocks: The Thas-Walker
Construction
In this section we give a construction discovered independently by J. A. Thas
and M. Walker [Wa76]. It is a nice application of the Klein correspondence
(and is a more general converse of Cor. 12.3.5).
Let F = {C1, C2, . . . , Ct} be a flock of a quadric Q of P G(3, q), with
t = q − 1, q + 1, or q according as Q is elliptic, hyperbolic or a cone. Let
πi be the plane for which πi ∩ Q = Ci. Embed Q in the hyperbolic (Klein)
quadric H5. Let π⊥ i be the polar plane of πi with respect to H5 meeting H5
in the conic C∗ i = π⊥ i ∩H5. For i = j, πi ∩πj is a line exterior to H5, implying
that (πi ∩ πj) ⊥ = π⊥ i ∪ π⊥ j lies in a 3-space that intersects H5 in an elliptic
quadric. In particular, no two points of C ∗ i ∪ C∗ j
are on a common line of H5.
Thus
(a) When Q is elliptic and F has carriers x and y, C ∗ 1 ∪ C ∗ 2 ∪ · · · ∪ C ∗ q−1 ∪
{x, y} = O ∗ has size q 2 + 1 and no two points of this set are on a common
line of H5.
(b) When Q is hyperbolic, C ∗ 1 ∪ · · · ∪ C∗ q+1 = O∗ has size q 2 + 1 and no
two points of this set are on a common line of H5.

12.11. SPREADS FROM FLOCKS: THE THAS-WALKER CONSTRUCTION585
(c) when Q is a quadratic cone with vertex x, C∗ 1 ∪ C∗ 2 ∪ · · · ∪ C∗ q = O∗
has size q2 + 1 and no two of its points are on a common line of H5.
So in each of the three cases O∗ is a set of q2 + 1 points, no two on a line
of H5, which also implies that no two are in a same plane of H5. There are
(1 + q)(1 + q2 ) Latin planes and that many Greek planes of H5. Moreover,
there are 1 + q Greek planes on a give point of H5 and 1 + q Latin planes
on that point. Since no two points of O∗ can lie in the same plane, it follows
that each plane of H5 contains exactly one point of O∗ . Such a set O∗ is
called an ovoid of H5.
Under the Klein correspondence the points of O∗ correspond to q2 + 1
mutually skew lines of P G(3, q), i.e., a spread of P G(3, q). This construction
of spreads is referred to as the Thas-Walker construction. It was noted in
Chapter 6 that flocks of elliptic quadrics are all linear, as are the flocks of a
hyperbolic quadric when q is even. But there are many nonlinear flocks of
quadratic cones for all q and a few of hyperbolic quadrics when q is odd.

586 CHAPTER 12. THE KLEIN CORRESPONDENCE

Chapter 13
q-Clans and Their Geometries
13.1 Preliminaries
Let q be an arbitrary prime power, and let Fq = GF (q) be the Galois field
with q elements. Let
x y
A =
and A
w z
′
′ x y
=
′
w ′ z ′
be arbitrary 2 × 2 matrices over Fq.
Definition 13.1.1. We say A and A ′ are equivalent and write A ≡ A ′
provided x = x ′ , z = z ′ , and y + w = y ′ + w ′ .
Theorem 13.1.2. αAα T = αA ′ α T for all α ∈ F 2 q iff A ≡ A′ .
Proof. Write α = (a, b). Then
Let P be the matrix P =
αAα T = a 2 x + ab(y + w) + b 2 z. (13.1)
0 1
−1 0
Theorem 13.1.3. Let A, A ′ , B and M be arbitrary 2 × 2 matrices over F ,
and let σ ∈ Aut(F ). Then
.
1. A ≡ B iff B = A + uP for some u ∈ Fq;
587

588 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
2. A ≡ B iff A σ ≡ B σ ;
3. If A ≡ A ′ , then BAB T + M ≡ BA ′ B T + M.
Proof. Parts 1. and 2. are clear. For Part 3, first compute the product
BABT .
a
c
b x
d w
y a
z b
c
d
= (13.2)
2 2 a x + ab(y + w) + b z
acx + bdz + ady + bcw
.
acx + bdz + adw + bcy c 2 x + cd(y + w) + d 2 z
Now the result follows easily.
Definition 13.1.4. The matrix A is said to be anisotropic provided αAα T =
0 iff α = (0, 0).
Theorem 13.1.5. Let A, A ′ be as above, B ∈ GL(2, q), σ ∈ Aut(F ), M
any 2 × 2 matrix over F . Then
1. A ≡ A ′ implies A is anisotropic iff A ′ is anisotropic;
2. A is anisotropic iff BA σ B T is anisotropic;
3. A−A ′ is anisotropic iff (BA σ B T +M)−(B(A ′ ) σ B T +M) is anisotropic.
Proof. Easy exercise.
x y
From Eq. 13.1 it is clear that A =
is anisotropic iff a
w z
2x +
ab(y + w) + b 2 z is irreducible over F (as a homogeneous polynomial in a and
b). Using the quadratic formula when q is odd and the usual trace condition
when q = 2 e , we obtain the following result.
Theorem 13.1.6. The matrix A =
x y
w z
is anisotropic iff
1. −det(A + A T ) = (y + w) 2 − 4xz is a nonsquare in F , if q is odd;
2. xz/(y + w) 2 has absolute trace equal to 1, if q = 2 e .

13.2. Q-CLANS 589
x y x
When q is odd, ≡ y+w
w z
2
when q = 2e
x y x y + w
,
≡
w z 0 z
y+w
2
Hence in what follows, when we deal with an anisotropic matrix A, for
x y/2
q odd we may write A ≡
, and for q even we may write
y/2 z
x y
A ≡ . Then Eq. 13.2 may be rewritten as follows.
0 z
a b x y/2 a c
= (13.3)
c d y/2 z b d
z
.
;
a 2 x + aby + b 2 z acx + (ad + bc)y/2 + bdz
acx + (ad + bc)y/2 + bdz c 2 x + cdy + d 2 z
a b
c d
x y
0 z
a 2 x + aby + b 2 z (ad + bc)y
0 c 2 x + cdy + d 2 z
13.2 q-Clans
a c
b d
Definition 13.2.1. A q-clan is a set C =
, when q is even.
xt yt
At ≡
, when q is odd, and
≡ (13.4)
0 zt
: t ∈ F
(equivalence classes of) 2 × 2 matrices over F for which all pairwise differences
As − At (s, t ∈ F, s = t) are anisotropic.
For q odd, we assume in general that At ∈ C is symmetric and write
xt yt/2
At =
. For q = 2
yt/2 zt
e we assume in general that At ∈ C is
xt yt
upper triangular and write At =
.
0 zt
So for both q even and q odd we have
xt yt
C = At ≡
: t ∈ F , where X : t ↦→ xt, Y : t ↦→ yt, Z :
0 zt
t ↦→ zt are functions from F to F . It is easy to check that if C is a q-clan,
then X and Z must be one-to-one, and for q = 2e Y is also one-to-one.
of q

590 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
a b
Moreover, by Theorem 13.1.5 if 0 = u ∈ F, σ ∈ Aut(F ), B =
∈
c d
GL(2, q), and if x, y, z are any fixed elements of F , replacing At ∈ C with
A ′ t ≡ µBAσt BT
x y
+ gives a new q-clan C
0 z
′ = {A ′ t : t ∈ F }. Hence
it follows that the definition below of equivalence for q-clans does give a
reasonable notion of equivalence for q-clans.
Definition 13.2.2. Let C = {At : t ∈ F } and C ′ = {A ′ t : t ∈ F } be qclans.
We say C and C ′ are equivalent and write C ∼ C ′ provided there exist
the following: 0 = µ ∈ F, B ∈ GL(2, q), σ ∈ Aut(F ), M a 2 × 2 matrix
over F , and a permutation π : t ↦→ ¯t on F for which the following holds:
A ′ ¯t ≡ µBAσ t BT + M, for all t ∈ F. (13.5)
Theorem 13.1.6 has an obvious interpretation for q-clans.
Theorem 13.2.3. Let X :
t ↦→ xt, Y : t ↦→ yt, Z : t
↦→ zt be three functions
xt yt
from F to F . Put C = At ≡
: t ∈ F . Then C is a q-clan iff
0 zt
the following holds:
(i) −det At + AT
t − As + AT
s = (yt − ys) 2 − 4(xt − xs)(zt − zs) is
a nonsquare in F for distinct t, s ∈ F , if q is odd.
(ii) tr
= 1 for distinct s, t ∈ F , if q = 2e .
(xs+xt)(zs+zt)
(ys+yt) 2
13.3 Flocks of a Quadratic Cone
Let K = {(x0, x1, x2, x3) ∈ P G(3, q) : x 2 1 = x0x2}. So K is a quadratic cone
in P G(3, q) with vertex V = (0, 0, 0, 1).
Definition 13.3.1. A flock of K is a partition F = {Ct : t ∈ F } of K \{V }
into q, pairwise disjoint conics. Each conic Ct ∈ F is a plane intersection
Ct = πt ∩ K, where πt = [xt, yt, zt, 1] T is a plane not containing the vertex
V . We also speak of πt as a plane of the flock F.
The following important connection between flocks and q-clans was first
discovered by J. A. Thas [Th87].

13.3. FLOCKS OF A QUADRATIC CONE 591
Theorem 13.3.2. Let X : t ↦→ xt, Y : t ↦→ yt, Z : t ↦→ zt be three functions
on F . For t ∈ F , put πt = [xt, yt, zt, 1] T , Ct = πt ∩ K, F = {Ct
: t ∈ F }.
xt yt
Also put At ≡
and C = {At : t ∈ F }. Then C is a q-clan iff
0 zt
F is a flock.
Proof. We already know that if C is a q-clan, then X and Z are permutations
on F . Suppose zs = zt for distinct s, t ∈ F . Then (0, 0, 1, −zs) ∈ πs ∩
πt ∩ K. So if F is a flock, Z (and similarly, X) must be one-to-one. So
we may suppose in any case that X and Z are permutations. Then the
point (a, b, c, d) ∈ P G(3, q) belongs to πs ∩ πt ∩ K for distinct s, t ∈ F
iff b 2 = ac and −d = axs + bys + czs = axt + byt + czt. In this case,
a(xs −xt)+b(ys −yt)+c(zs −zt) = 0. If a = 0, then b = 0 and −d = czs = czt
implies c = 0 = d, an impossibility. Hence πs ∩ πt ∩ K = ∅ iff (multiply
through by a = 0 and use ac = b 2 ) a 2 (xs − xt) + ab(ys − yt) + b 2 (zs − zt) = 0
iff (a, b) (As − At) (a, b) T = 0. It follows that F is a flock iff C is a q-clan.
Theorem 13.3.3. From Theorem 4.3.1 it follows that the most general projective
semilinear map T of P G(3, q) defined as a map on the planes of
P G(3, q) not through the vertex V , that leaves invariant the cone K, is given
a b
by the following. There is a matrix B =
∈ GL(2, q), 0 = λ ∈
c d
F, σ ∈ Aut(F ), fixed elements x0, y0, z0 ∈ F , for which
T :
⎡
⎢
⎣
x
y
z
1
⎤
⎥
⎦ ↦−→
⎡
⎢
⎣
x ′
y ′
z ′
1
⎤
⎥
⎦ =
⎛
⎜
⎝
λa 2 λab λb 2 x0
2λac λ(ad + bc) 2λbd y0
λc 2 λcd λd 2 z0
0 0 0 1
⎞ ⎡
⎟ ⎢
⎟ ⎢
⎠ ⎣
x σ
y σ
z σ
1
⎤
⎥
⎦ .
(13.6)
Proof. This is an interesting exercise. See [Hi85].
xt yt
Now let C = At ≡
: t ∈ F be a q-clan. Let F(C) =
0 zt
πt ∩ K : πt = [xt, yt, zt, 1] T , t ∈ F be the corresponding flock. Put λ =
1, B = I, σ = id, M = −A0. Then C ′ = {A ′ t ≡ At − A0 : t ∈ F }
is a q-clan equivalent to C with A ′ 0 =
0 0
. And T : [x, y, z, 1]
0 0
T ↦−→
[x−x0, y −y0, z −z0, 1] T defines a projective semilinear map on P G(3, q) that

592 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
leaves the cone K invariant and maps F(C) to F(C ′ ). And the plane π ′ 0 of
the flock F(C ′ ) is the plane π ′ 0 = [0, 0, 0, 1] T . Hence from now on we assume
without loss of generality that each q-clan C contains the zero matrix and
the corresponding flock F(C) contains a conic in the plane π = [0, 0, 0, 1] T .
(It is possible, however, that the matrices of C are labeled so that As ∈ C is
the zero matrix for some nonzero s ∈ F .)
Theorem 13.3.3 has as an immediate consequence the following very important
theorem.
xt yt
Theorem 13.3.4. Let C = At ≡
: t ∈ F
0 zt
and C ′
= A ′ t ≡
′ x t y ′
t : t ∈ F be two (not necessarily distinct)
0 z ′ t
q-clans, normalized so that A0 = A ′ 0 is the zero matrix, with corresponding
flocks F(C) and F(C ′ ). Then F(C) and F(C ′ ) are projectively equivalent iff
there exist the following:
(i) 0 = λ∈ F ,
a b
(ii) B = ∈ GL(2, q),
c d
(iii) σ ∈ Aut(F ),
(iv) π : F → F : t ↦→ ¯t, a permutation, such that the following
condition is satisfied:
⎡
⎢
⎣
x ′ ¯t
y ′ ¯t
z ′ ¯t
1
⎤
⎥
⎦ =
⎛
⎜
⎝
λa 2 λab λb 2 x ′ ¯0
2λac λ(ad + bc) 2λbd y ′ ¯0
λc 2 λcd λd 2 z ′ ¯0
0 0 0 1
⎞ ⎡
⎟ ⎢
⎟ ⎢
⎠ ⎣
x σ t
y σ t
z σ t
1
⎤
⎥
⎦ , t ∈ F. (13.7)
For q odd, so the matrices of C and C ′ are symmetric, Eq. 13.7 is equivalent
to
A ′ ¯t = λBAσt BT + A ′ ¯0 , t ∈ F. (13.8)
For q = 2e , so the matrices of C and C ′ are upper triangular, Eq. 13.7 is
equivalent to
A ′ ¯t ≡ λBAσt B T + A ′ ¯0 , t ∈ F. (In fact this holds for all q.) (13.9)
Clearly Theorem 13.3.4 is just a very explicit way of saying that C ∼ C ′
iff F(C) and F(C ′ ) are projectively equivalent.

13.4. 4-GONAL FAMILIES FROM Q-CLANS 593
13.4 4-Gonal Families from q-Clans
Throughout this section C is a
given q-clan normalized so that A0 is the zero
0 1
matrix. And P =
.
−1 0
For α = (a, b), β = (c, d) ∈ F 2 , put α ◦ β = ac + bd = α(β) T . Then
we may define a group K = {(α, c, β) ∈ F 2 × F × F 2 : α, β ∈ F 2 , c ∈ F }
with binary operation
(α, c, β) ◦ (α ′ , c ′ , β ′ ) = (α + α ′ , c + c ′ + β ◦ α ′ , β + β ′ ). (13.10)
This makes K into a group of order q 5 with (α, c, β) −1 =
(−α, α ◦ β − c, −β). Starting with the q-clan C put Kt = At + A T t for
all At ∈ C. Also put gt(α) = αAtα T . It is then easy to check that the
following holds.
gt(α + β) = gt(α) + gt(β) + αKtβ T . (13.11)
Then we may define the following subgroups of K.
A(∞) = {(0, 0, β) ∈ K : β ∈ F 2 };
A ∗ (∞) = {(0, c, β) ∈ K : c ∈ F, β ∈ F 2 };
and for all t ∈ F
A(t) = {(α, gt(α), αKt) ∈ K : α ∈ F 2 };
A ∗ (t) = {(α, c, αKt) ∈ K : α ∈ F 2 , c ∈ F }.
Put ˜ F = F ∪ {∞}, J (C) = {A(t) : t ∈ ˜ F }, J ∗ (C) = {A ∗ (t) : t ∈ ˜ F }.
Theorem 13.4.1. The triple (K, J (C), J ∗ (C)) satisfies the conditions K1
and K2 of Section 9.15. The resulting GQ is denoted GQ(C) and has parameters
(q 2 , q).
Proof. The proof is organized into a sequence of steps. The property needed
at several crucial points is that since C is a q-clan, the pairwise difference of
distinct matrices in C is always anisotropic. Then if q is odd, −det(Kt−Ku) is
a nonsquare in F , implying Kt−Ku is nonsingular. And if q = 2e , Kt−Ku =
(yt − yu)P , which is nonsingular since Y : t ↦→ yt is one-to-one.
Step 1. A∗ (∞) A(t) = {e} = A∗
t A(∞) for all t ∈ F .
Step 2. A∗ (t) A(u) = {e} for distiinct t, u ∈ F , precisely because
Kt − Ku is nonsingular.

594 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
Step 3. For distinct t, u ∈ F , the following are equivalent:
(i) A(∞)A(t) A(u) = {e};
(ii) A(t)A(∞) A(u) = {e};
(iii) A(t)A(u) A(∞) = {e};
(iv) At − Au is anisotropic.
Step 4. For distinct t, u, v ∈ F, A(t)A(u) A(v) = e iff B = (Kt −
Kv) −1 )(At−Av)(Kt−Kv) −1 +(Kv−Ku) −1 (Av−Au)(Kv−Ku) −1 is anisotropic.
Step 5. The conclusion of the proof with separate cases for q odd and q
even.
Steps 1 and 2 are quite easy, and together they show that K2 is satisfied.
Step 3 is equally easy, and we leave the details to the reader. Step 4 is rather
more intricate, so we include the details.
For distinct t, u, v ∈ F an arbitrary element of A(t)A(u) A(v) must
have the form (α + β, gt(α) + gu(β) + αKtβ T , αKt + βKu)
= (α + β, gv(α + β), (α + β)Kv). From equality of the third coordinates
we have α(Kt − Kv) = β(Kv − Ku), i.e., we may put γ = α(Kt − Kv) =
β(Kv − Ku), so α = γ(Kt − Kv) −1 and β = γ(Kv − Ku) −1 . Equating middle
coordinates, we now obtain
0 = gt(α) + gu(β) + αKtβ T − gv(α) − gv(β) − αKvβ T
= gt(α) − gv(α) + α(Kt − Kv)β T + gu(β) − gv(β)
= αAtα T + β(Kv − Ku)β T + β(Au − Av)β T
= α(At − Av)α T + β(Av − Au) T β T
= γ(Kt − Kv) −1 (At − Av)(Kt − Kv) −1 γ T +
γ(Kv − Ku) −1 (Av − Au)(Kv − Ku) −1 γ T = γBγ T .
This essentially completes Step 4. Now note that
0 = (Kv − Ku) − (Kt − Ku) + (Kt − Kv)
= (Kt − Kv)[(Kt − Kv) −1 − (Kt − Kv) −1 (Kt − Ku)(Kv − Ku) −1
+(Kv − Ku) −1 ](Kv − Ku) =⇒
0 = (Kt − Kv) −1 − (Kt − Kv) −1 (Kt − Ku)(Kv − Ku) −1 + (Kv − Ku) −1 =⇒
(Kt − Kv) −1 + (Kv − Ku) −1 =
(Kt − Kv) −1 (Kt − Ku)(Kv − Ku) −1 .
And now it follows that B + B T = (Kt − Kv) −1 + (Kv − Ku) −1
= (Kt − Kv) −1 (Kt − Ku)(Kv − Ku) −1 .
First suppose that q is odd. Since C is a q-clan, −det(B + B T ) =
(−det(Kt − Kv) −1 )(−det(Kt − Ku)(−det(Kv − Ku) −1 ) is the product of three
nonsquares in F , i.e., −det(B + B T ) is a nonsquare. Hence B is anisotropic.
By Step 4, the proof is complete in this case.
Now suppose that q = 2 e . If A is upper triangular with upper right hand

13.5. SOME FAMILIES OF Q-CLANS 595
entry equal to y and K = A + A T , then K −1 AK −1 = P (y −2 A)P . Hence the
matrix B of Step 4 appears in this case as B = P ((yt − yv) −2 (At − Av) +
(yv − yu) −2 (Av − Au))P , which is anisotropic iff
xt−xv
(yt−yv) 2 + xv−xu
(yv−yu) 2
yt−yv
(yt−yv) 2 + yv−yu
(yv−yu) 2
X Y
=
0
0 Z
zt−zv
(yt−yv) 2 + zv−zu
(yv−yu) 2
1 = tr(XY/Z 2 ) = tr(N/D), where
N = xtzt(yv + yu) 4 + xuzu(yt + yv) 4 + xvzv(yt + yu) 4
+ (xtzu + xuzt)(yv + yt) 2 (yv + yu) 2
+ (xtzv + xvzt)(yu + yt) 2 (yu + yv) 2
+ (xuzv + xvzu)(yt + yu) 2 (yu + yv) 2 ,
and D = (yt + yu) 2 (yt + yv) 2 (yu + yv) 2 .
But since At + Au, At + Av and Au + Av are anisotropic,
(xt+xu)(zt+zu)
1 = tr (yt+yu) 2 + (xt+xv)(zt+zv)
(yt+yv) 2 + (xu+xv)(zu+zv)
(yu+yv) 2
= tr(N/D).
It follows that B is anisotropic, completing the proof.
is anisotropic iff
Given a q-clan C, the geometries of major interest to us are the generalized
quadrangle GQ(C) of order (q 2 , q) and the associated flock F(C). Before
proceeding with a discussion of other related geometries we pause to give in
some detail several of the known infinite families of q-clans.
13.5 Some Families of q-Clans
Since a general theory without examples is clearly not interesting, we pause
here to give several of the known infinite families that are rather easily shown
to exist. Later in special sections the various known families will be studied
in greater detail after more theory has been developed. For example, the
known flocks will be studied in terms of the intersection patterns of their
planes.
13.5.1 The Classical q-Clans
Let x2 + bx + c be a fixed irreducible polynomial over F . For each t ∈ F put
t tb
At =
. It is easy to see that for distinct s, t ∈ F, As − At =
0 tc
1 b
(s−t) is anisotropic. The planes of the corresponding flock are πt =
0 c
[t, bt, ct, 1] T , t ∈ F . This flock is characterized by the fact that it is linear, i.e.,

596 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
each plane πt contains the line L = {(−by − cz, y, z, 0) ∈ P G(3, q) : y, z, ∈ F }.
In [PT84] it is shown that the corresponding GQ is isomorphic to H(3, q 2 )
arising from a nonsingular hermitian surface in P G(3, q 2 ), and it is the pointline
dual of the GQ Q(5, q) arising from the nonsingular elliptic quadric in
P G(5, q).
13.5.2 W.M. Kantor [Ka80]; J.C. Fisher and J.A. Thas
[FT79]; M. Walker [Wa76]
2 t 3t
Let q ≡ 2 (mod 3) and put At ≡
0 3t3
. For q = pe , p odd, −det((At −
As) + (At − As) T ) = −3(t − s) 4 is a nonsquare in F iff −3 is a nonsquare iff
q ≡ 2 (mod 3). For q = 2e
2 2
t − s t − s
, At −As =
0 t3 − s3
, and (t−s)(t3−s3 )
(t2−s2 ) 2 =
1 +
t t 2
+ has trace 1 iff 1 = tr(1) iff e is odd.
t+s t+s
In this example the flock is linear iff q = 2. It is easy enough to show
that no three of the planes of the flock contain a common line and no four of
the planes contain a common point. J.A. Thas [Th93] has shown that this
property characterizes this flock, with finitely many values of q for which the
proof is not yet complete.
13.5.3 W.M Kantor [Ka86]
Let q be odd, σ ∈ Aut(F ), m a nonsquare in F . For t ∈ F , put
At =
t 0
0 −mt σ
. In this case it is almost trivial to show that C =
{At : t ∈ F } is a q- clan. All planes πt = [t, 0, −mt σ , 1] T of the flock F(C)
contain the point (0, 1, 0, 0). The flock is linear iff σ = id. The planes
πt, πu, πr, for distinct t, u, r ∈ F , contain a common line iff (u − t)/(r − t)
is fixed by σ. J.A. Thas [Th87] has shown that any flock whose planes
all contain a common point must be linear or equivalent to the one in this
example.

13.5. SOME FAMILIES OF Q-CLANS 597
13.5.4 W.M. Kantor [Ka86] for q odd; S.E. Payne [Pa85]
for q even
3 t 5t
For q ≡ ±2(mod 5) put At ≡
0 5t5
, t ∈ F . For distinct s, t ∈ F ,
when q is odd, −det (As − At) + (As − At) T = 5(s − t) 2 (t2 + 3ts + s2 ) 2 is a
nonsquare in F iff q ≡ ±2(mod 5). And when q = 2e , (t − s)(t5 − s5 )/(t3 −
s3 ) 2 =
ts
t2 +ts+s2 2
ts + t2 +ts+s2
+ 1 has trace 1 iff tr(1) = 1 iff e is odd.
The associated flock is linear iff q = 2 or q = 3. The planes πt, πu, πr for
distinct t, u, r ∈ F contain a common line iff tur = 0 and t + u + r = 0. It
follows that for q even, q > 2, no three planes contain a common line. And
it follows that for all q no four planes of the flock contain a common line.
13.5.5 H. Gevaert and N.L. Johnson [GJ88]; q = 5 e
Let q = 5e and let k be a nonsquare of F . Then for t ∈ F put At =
2 t 3t
. Put M = (At−As)+(At−As) T . Then −det(M) =
3t 2 t(1 + kt 2 ) 2 /k
(t − s) 2 (k(t − s) 2 − 1) 2 /k is a nonsquare.
The corresponding flock was originally derived from some “likeable” planes
of W.M. Kantor.
13.5.6 H. Gevaert and N.L. Johnson [GJ88]; q = 3 e
Let q = 3 e , m a nonsquare of F , At ≡
t t 3
0 −t(t 8 + m 2 )/m
, t ∈ F . Put
M = At − As + A T t − AT s . Then −det(M) = (t − s)2 [((t − s) 4 − m] 2 /m is
a nonsquare of F .
The corresponding flock was originally derived from some planes of M.
Ganley.
There are three more infinite families known. For q odd, there is a flock
discovered by J.C. Fisher (see Fisher and Thas [FT79]); for q = 2 e there
is the Subiaco q-clan discovered by W. Cherowitzo, T. Penttila, I. Pinneri
and G. Royle [CPPR96], and also for q = 4 k there is the Adelaide q-clan
discovered by Cherowitzo, O’Keefe and and Penttila [COP01]. For small
odd q some “sporadic” examples are known. For q even there are no known
sporadic examples, i.e., each known q-clan is a member of an infinite family.
For a more complete study of the known examples with q even, see [CP02].

598 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
13.6 q-Clans, Flocks & Spreads of PG(3,q)
Starting with a q-clan C we show how to construct a line spread of PG(3,q)
(similar to a construction of Gevaert and Johnson). Or we can start with
the associated flock F(C) and use the Klein correspondence to give the construction
discovered independently by Thas and Walker (see Section 12.11)
to find a spread of PG(3,q). We show that these two approaches are equivalent,
and extend Theorem 13.3.4 to include the fact that equivalent q-clans
give equivalent spreads. There is a standard (Bose-Barlotti) construction of
a translation plane from a spread of PG(3,q), with isomorphic translation
planes corresponding to projectively equivalent spreads. In this book we ignore
the corresponding planes most the time, since our real interest is in the
generalized quadrangles. However, see Section 13.8
*******************************************************************************
for a brief treatment of these planes as well as the ∆-planes derived from
them.
0 1
With P =
, let A, B be any 2 × 2 matrices over F . Then
−1 0
A ≡ B iff B = A + µP for some µ ∈ F . A line of P G(3, q) will be given as
the row space of a 2 × 4 matrix with rank 2. For example, if O (resp., I) is
the 2 × 2 zero (resp., identity) matrix over F , put
L∞ = 〈(O, I)〉 . (13.12)
Then if u ∈ F and A is any 2 × 2 matrix over F , put
L(A,u) = 〈(I, A + uP )〉 . (13.13)
Lemma 13.6.1. RA = {L∞} ∪ {L(A,u) : u ∈ F } is a regulus.
x y
Proof. If A ≡ B, then RA = RB, so we may assume that A = . It
0 z
is clear that RA consists of q + 1 pairwise skew lines. Figure 1.1 shows the
transversals of RA (the horizontal lines for (0, 0) = (a, b) ∈ F 2 ).
For a line L, L∗ denotes the set of points incident with L. And for a
matrix A, R∗ A denotes the set of points incident with a line of the regulus
RA.

13.6. Q-CLANS, FLOCKS & SPREADS OF PG(3,Q) 599
L(A, 0)
L(A, u)
✉ ✉ ✉
p = (1, 0, x, y) p + µ(0, 0, 0, 1)
✉ ✉
✉
ap + bq
q = (0, 1, 0, z)
✉
ap + bq + µ(0, 0, −b, a)
q + µ(0, 0, −1, 0)
Figure 13.1: RA is a regulus
L∞
Lemma 13.6.2. Let A, B be two 2 × 2 matrices over F .
(i) 〈(I, A)〉 = 〈(I, B)〉 iff A = B.
(ii) 〈(I, A)〉 ∈ RB iff A ≡ B iff RA = RB.
(iii) R ∗ A ∩ R∗ B ⊇ L∗ ∞, with equality iff A − B is anisotropic.
✉
✉
(0, 0, 0, 1)
(0, 0, −b, a)
(0, 0, −1, 0)
Proof. Parts (i) and (ii) are
trivial. For part (iii) we may assume (by part
x1 y1
x2 y2
(ii) ) that A =
, B =
. Then clearly R
0 z1
0 z2
∗ A ∩ R∗B ⊇ L∗∞ .
And the containment is proper iff there exist u, v ∈ F for which L(A,u) =
〈(I, A + uP )〉 and L(B,v)
= 〈(I,
B + vP )〉
have nonempty intersection,
which
I A + uP
I A + uP
holds iff 0 = det
= det
I B + vP
0 B − A + (v − u)P
x2 − x1 y2 − y1 + v − u
= det
= (x2 − x1)(z2 − z1) + (v − u)(y2 − y1) +
u − v z2 − z1
(v − u) 2 = ∆.
If (x2 − x1)(z2 − z1) = 0, we can put u = 0 and v = y1 − y2 to show that
R∗ A ∩ R∗B ⊇ L∗ (A,u) ∩ L∗ (B,v) = ∅. And it also follows easily in this case that
A−B is NOT anisotropic. So from now on assume that (x2−x1)(z2−z1) = 0.
Then
∆ =
1
x2 − x1 y2 − y1
(z2 − z1, v − u)
(z2 − z1)
0 z2 − z1
z2 − z1
v − u
.
(13.14)

600 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
and
1
x2 − x1 y2 − y1 v − u
∆ =
(v − u, x2 − x1)
(x2 − x1)
0 z2 − z1 x2 − x1
(13.15)
So if ∆ = 0, clearly A − B cannot be anisotropic, i.e., if R∗ A ∩ R∗B = L∗∞, then A − B is not anisotropic.
Conversely, suppose R∗ A ∩ R∗B = L∗∞. Hence for all u, v ∈ F , ∆
= 0. We
a
claim that A − B is anisotropic. For suppose (a, b)(B − A) = 0 with
b
(0, 0) = (a, b). If a = 0, put u = 0, v = b
a (z2 − z1) to contradict ∆ = 0 in
Eq. 13.14. If b = 0, put u = 0 and v = a
b (x2 − x1) to contradict ∆ = 0 in
Eq. 13.15.
As an immediate corollary we have the following theorem.
Theorem 13.6.3. Let X : t ↦→ xt; Y : t ↦→ yt; Z : t ↦→ zt be three functions
xt yt
from F to F . For t ∈ F put At =
. Then put C = {At : t ∈ F }.
0 zt
With the notation as above, Rt = RAt is a regulus for each t ∈ F . Put
S(C) = ∪{Rt : t ∈ F }. Then C is a q-clan iff S(C) is a spread of P G(3, q),
in which case X and Z are both one-to-one.
It follows that with each q-clan C there is a corresponding line spread S(C)
of P G(3, q), from which a standard construction provides a translation plane
π(C). One major goal of this section is to show that two q-clans C, C ′ are
equivalent iff their associated spreads S(C), S(C ′ ) are projectively equivalent.
The following Lemma is a big step in this direction.
Lemma 13.6.4. Let C and C ′ be two (not necessarily distinct) q-clans. Then
there is a semilinear transformation mapping S(C) to S(C ′ ) in such a way
that it fixes L∞ and maps the q special reguli of S(C) on L∞ to the q special
reguli of S(C ′ ), if and only if C ∼ C ′ .
Proof. First suppose C = {At : t ∈ F } and C ′ = {A ′ t : t ∈ F } are equivalent.
This means that there exist nonzero λ ∈ F, B ∈ GL(2, q), σ ∈ Aut(F ), a
2 × 2 matrix M over F , and a permutation π : t ↦→ ¯t on the elements of F for
which A ′ ¯t ≡ λBAσ t B T + M for all t ∈ F , i.e., for each t ∈ F there is a ut ∈ F
for which A ′ ¯t = λBAσ t BT + M + utP . Consider the semilinear transformation
defined by

13.6. Q-CLANS, FLOCKS & SPREADS OF PG(3,Q) 601
T : (x0, x1, x2, x3) ↦→ (x σ 0 , xσ1 , xσ2 , xσ3 )
−1 −1 −1 −1 λ B λ B M
O BT Clearly T leaves L∞ invariant, and
T : L(t,u) = 〈(I, At
+ uP )〉 ↦→
−1 −1 −1 −1 σ λ B , λ B M + (At + uσ
T P )B
= I, M + λBAσ t BT + uσ
T λBP B
= I, λBAσ t BT + uσλ(det(B))P + M
= 〈(I, A ′ ¯t − utP + uσλ(det(B))P )〉 .
Hence T : Rt ↦→ R ′ ¯t and thus maps S(C) to S(C ′ ).
For the converse, suppose there is a semi-linear transformation
T : (x0, x1, x2, x3) ↦→ (x σ 0 , xσ1 , xσ 2 , xσ3 )
C1 C2
C3 C4
(here each Ci is a 2 × 2 matrix over F ) that fixes L∞ and maps Rt to R ′ ¯t
for some permutation π : t ↦→ ¯t of
the elements of F . The fact that T fixes
0 0
L∞ implies that C3 = .
0 0
It is an easy exercise to show that if A and B are two 2 × 2 matrices over
F , then AP B is a scalar multiple of P iff A is a scalar multiple of BT . Then
T : 〈(I, At + uP )〉 ∈ Rt ↦→ 〈(C1, C2 + (Aσ t + uσP )C4)〉
= I, C −1
1 C2 + C −1
1 Aσ t C4 + uσC −1
′
1 P C4 ∈ R ¯t, for all u ∈ F . Since this
holds for all u ∈ F , C −1
1 P C4 must be a scalar multiple of P , so C −1
1 = λCT 4
for some nonzero λ ∈ F . Hence
T : 〈(I, At + uP )〉 ↦→
I, λC T 4 A σ t C4 + λC T 4 C2 + λu σ (det(C4))P ∈ R ′ ¯t.
Therefore, A ′ ¯t ≡ λCT 4 Aσ t C4 + λC T 4 C2 for all t ∈ F , implying C ∼ C ′ .
In order to improve the preceding lemma to the theorem we really want, as
well as to obtain related useful information concerning the associated flocks,
we need to use the Klein map from lines of P G(3, q) to points of the Klein
quadric to give the Thas-Walker construction of a spread of P G(3, q) starting
with a flock of a quadratic cone. This construction was given in detail in even
greater generality in Section 12.11.
.

602 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
xt yt
Starting with a q-clan C = {At ≡
: t ∈ F }, we have the flock
0 zt
F(C) = {Ct : t ∈ F }, where Ct = πt ∩ K, πt = [xt, yt, zt, 1] T . We now have
two constuctions of a spread of P G(3, q) starting with C, one going directly,
algebraically, from the q-clan, the other one using the associated conical flock
and using the Klein correspondence.
Lemma 13.6.5. Our two constructions of spreads from a q-clan are equivalent.
(The proof is a variation on a treatment in Gevaert and Johnson
[GJ88].)
Proof. What we wishto establish is that starting with a q-clan
xt yt
C = At =
: t ∈ F , the lines of the regulus Rt (as defined in
0 zt
Theorem 13.6.3) Klein-correspond to to the points of a conic lying in a plane
π⊥ t , where πt is the usual plane [xt, yt, zt, 1] T of the associated flock. For this
proof we use a different (but equivalent) Klein-correspondence. We replace
the Klein map ℓ ↦→ G(ℓ) with the specific map ℓ ↦→ p(ℓ) given below. We
know that the points p(L∞) and p(Lt,u) for u ∈ F form a conic in some plane
π T t . So the plane πt will be the intersection of p(L∞) ⊥ , p(Lt,0) ⊥ , and p(Lt,1) ⊥ .
x0, x1, x2, x3
Given a line L =
y0, y1, y2, y3
specific Klein map
, put pij =
p(L) = (p01, p02, p03, p12, −p13, p23).
So p(L∞) = (0, 0, 0, 0, 0, 1). And L(t,u) =
xi xj
yi yj
1, 0, xt, yt + u
0, 1, −u, zt
, and use the
, so
p(L(t,u)) = (1, −u, zt, −xt, yt + u, u2 + ytu + xtzt). Here H = {(a1, . . . , a6)
∈ P G(5, q) : a1a6 + a2a5 + a3a4 = 0}.
Put (ai) ◦ (bj) = 6 i=1 aib7−i. So (ai) ⊥ (bj) iff (ai) ◦ (bj) = 0. Then
p(L∞) ⊥ = {(b1, . . . , b6) : b1 = 0}
p(L(t,0)) ⊥ = (1, 0, zt, −xt, yt, xtzt) ⊥
= {(bi) : b6 + ztb4 − xtb3 + ytb2 + xtztb1 = 0}.
And p(L(t,1)) ⊥ = (1, −1, zt, −xt, yt + 1, 1 + yt + xtzt) ⊥ =
{(bi) : b6 − b5 + ztb4 − xtb3 + (yt + 1)b2 + (1 + y + t + xtzt)b1 = 0}.
The intersection of these three perps is {(0, b2, b3, b4, b2, b6) : (−b3)xt +
b2yt + b4zt + b6 = 0}. We identify {(0, b2, b3, b4, b2, b6) (which is on H iff
b2 2 = (−b3)b4) with (−b3, b2, b4, b6) = (x0, x1, x2, x3) ∈ Σ. So Σ ∩ H = K :
x2 1 = x0x2. And (−b3)xt + b2yt + b4zt + b6 = 0 becomes

13.6. Q-CLANS, FLOCKS & SPREADS OF PG(3,Q) 603
⎡
xt
⎤
⎢ yt ⎥
(x0, x1, x2, x3) ⎢ ⎥
⎣ zt ⎦ = 0.
1
This says that the regulus Rt Klein - corresponds with C⊥ t where Ct =
πt ∩K, πt = [xt, yt, zt, 1] T . Hence the two spread constructions are essentially
the same.
a b ax ay + bz
Lemma 13.6.6. The transversal L =
to RA
0 0 −b a
x y
where A = that meets L∞ at (0, 0, −b, a) Klein- corresponds to the
0 z
point of Σ given by (−a2 , −ab, −b2 , a2x + aby + b2z). This implies that the q
transversals (of the reguli RA) through the point (0, 0, −b, a) of L∞, which all
lie in a plane [b, −a, 0, 0] T containing L∞, must Klein-correspond to points
on the generator of K through the point (a2 , ab, b2 , 0). Hence the q + 1 planes
through L∞ Klein-correspond to the q + 1 generators of the cone K.
Proof. The proof is a routine application of the Klein correspondence and
subsequent identification with points of Σ as given in the proof of Lemma 13.6.5.
Lemma 13.6.7. Let F = {C1, . . . , Cq} and F ′ = {C ′ 1 , . . . , C′ q } be two flocks
of K. If |F ∩ F ′ | > (q − 1)/2, then F = F ′ .
Proof. Let F ∩ F ′ = {C1, . . . , Cs}. Assume that C ′ i ∈ F. Then for s + 1 ≤
j ≤ q, |C ′ i ∩ Cj| ≤ 2, and clearly C ′ i ⊆ Cs+1 ∪ . . . ∪ Cq. Hence |C ′ i | ≤ 2(q − s),
implying q + 1 ≤ 2(q − s), i.e., s ≤ (q − 1)/2.
The preceding Lemma has an immediate but interesting corollary.
Corollary 13.6.8. If F = {C1, . . . , Cq} is a flock of a quadratic cone K such
that more than (q − 1)/2 of the planes πi (where πi ⊇ Ci) share a common
line, then this flock must be a linear flock. In terms of the ovoid O we obtain:
if more than (q − 1)/2 planes π ⊥ lie in a common projective 3-space, then
O must be an elliptic quadric of that projective 3-space, implying that the
corresponding spread of P G(3, q) is regular.
For small q the preceding Corollary forces all flocks to be linear.
Corollary 13.6.9. For q = 2, 3, or 4, each flock of a quadratic cone in
P G(3, q) must be linear.

604 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
Lemma 13.6.10. Two reguli R and R ′ of P G(3, q) which share precisely
one line L and for which all lines of R \ {L} are disjoint from all lines of
R ′ \ {L}, are contained in exactly one regular spread of P G(3, q).
Proof. Using the Klein correspondence, the lines of R (resp., R ′ ) are mapped
onto the points of a conic C (resp., C ′ ). Denote by π (resp., π ′ ) the plane
containing C (resp., C ′ ). So π ∩ H = C and π ′ ∩ H = C ′ . Moreover, C
and C ′ have exactly one point V in common and no two points of C ∪ C ′
are on a common line of H. First suppose that π and π ′ have only the
point V in common, i.e., π and π ′ generate a projective 4-space P G(4, q).
Then for each point pi of C ′ \ {V }, π and pi generate a projective 3-space
Pi intersecting H in an elliptic quadric, 1 ≤ i ≤ q. Using the polarity ⊥
w.r.t. H, we obtain q distinct lines Pi ⊥ through the point (P G(4, q)) ⊥ and
contained in π ⊥ . However, these q lines Pi ⊥ must be exterior to the conic
π ⊥ ∩ H, clearly an impossibility. Hence it must be that π ∩ π ′ is a line and
that π and π ′ generate a projective 3-space P G(3, q) ′ . Since C and C ′ are
contained in the quadric P G(3, q) ′ ∩ H = Q, and since no line of H contains
points of both C \ {V } and C ′ \ {V }, this quadric Q must be elliptic. Clearly
Q Klein-corresponds to the unique regular spread of P G(3, q) containing R
and R ′ .
Theorem 13.6.11. A flock of a quadratic cone in P G(3, q) is equivalent to
a line spread S of P G(3, q), where S is the union of q reguli Ri, 1 ≤ i ≤ q,
which pairwise intersect in the same line L.
Proof. Starting with a flock of a quadratic cone, we have seen that there
arises a spread of the indicated type.
Conversely, let S be a spread of P G(3, q) of the indicated type. Under
the Klein correspondence, the lines of Ri correspond to the points of a conic
Ci and L corresponds to V . Denote by πi the plane containing Ci. By the
preceding Lemma, each two planes πi and πj, i = j, intersect in a line, say
Lij, which is tangent to H at V . The line Lij is the unique tangent line of
Ci (resp., Cj) at V . This uniqueness shows that all Lij, i = j, coincide. So
all planes πi contain a common tangent line L of H at V , which forces their
perps π ⊥ i to all lie in a common 3-space Σ = L⊥ which meets H in a cone K
with vertex V . And clearly these planes π ⊥ i determine a flock of K.
Theorem 13.6.12. If S is a spread of P G(3, q) which is the union of q
reguli Ri, 1 ≤ i ≤ q, sharing precisely one line L, and if S contains at least

13.6. Q-CLANS, FLOCKS & SPREADS OF PG(3,Q) 605
one regulus R distinct from all the Ri, then S is a regular spread ( i.e., the
regulus determined by any three lines of R is contained in R). This means
that the corresponding translation plane is desarguesian, the associated flock
is linear, and the GQ constructed from the associated q-clan is classical.
Proof. Under the Klein correspondence, the lines of Ri (resp., R) correspond
to the points of a conic Ci (resp., C). Let O be the ovoid C1 ∪ . . . ∪ Cq.
Denote by πi (resp., π) the plane containing Ci (resp., C). Let L correspond
to V ∈ H. By the preceding proof the planes πi share a common line M
which is tangent to Ci at V .
Suppose first that M ∩ π = ∅. If there exists a conic Ci having exactly
one point in common with C, then by the proof of Lemma 13.6.10, π and
πi intersect in a line, contradicting M ∩ π = ∅. So for any Ci, Ci meets C
in exactly zero or two points. Suppose Ci is one of the conics meeting C
in two points. Then πi ∩ π is a line of π, so M ∩ πi ∩ π = ∅, contradicting
M ∩ π = ∅. Consequently M ∩ π = ∅. If M ⊂ π, then C must coincide with
one of the Ci, contrary to hypothesis. Hence |M ∩ π| = 1, implying that
M and π generate a 3-space < M, π >. If V ∈ C, then there are at least
(q + 1)/2 conics Ci having a nontrivial intersection with C. The planes πi of
these conics Ci are contained in < M, π >. Then by Corollary 13.6.8 O is an
elliptic quadric and the corresponding spread is regular. So suppose V ∈ C.
Then each conic Ci has at most one further point in common with C. Since
|C \ {V }| = q = |{C1, . . . , Cq}| and since any point of C \ {V } is on exactly
one Ci, it follows that each conic Ci has exactly one point in common with
C \ {V }. Hence all planes πi are contained in < M, π > and again O is an
elliptic quadric.
The preceding theorem provides just the right finishing touch to
Lemma 13.6.4. Let C and C ′ be two q-clans whose associated spreads S(C)
and S(C ′ ) (with special reguli all containing the line L∞) are projectively
equivalent. If either spread is regular, then so is the other and the q-clan
is classical, etc. So suppose that neither spread is regular and hence each
spread has just the q given reguli. If T is any semi-linear transformation
mapping one spread to the other, it clearly must leave L∞ invariant and
map the q reguli of the one spread to the q reguli of the other. Hence by
Lemma 13.6.4 the two q-clans are equivalent. This completes a proof of the
following theorem.

606 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
Theorem 13.6.13. Two q-clans are equivalent iff their associated spreads
(and flocks!) are projectively equivalent. Moreover, the proof of Lemma 13.6.4
shows how to interpret q-clan equivalence as projective equivalence of the corresponding
spreads.
13.7 The Classical Examples
Before moving on to more interesting topics, we review briefly the classical
situation.
xt yt
Let C = {At =
be a q-clan normalized as usual. Then as a
0 zt
set of planes, the flock F(C) = {πt = [xt, yt, zt, 1] T : t ∈ Fq} contains the
plane π0 = [0, 0, 0, 1] T , and for t ∈ F \ {0}, Lt = πt ∩ π0 = {(a, b, c, 0) :
axt + byt + czt = 0}. It is easy to see that if all planes of the flock contain
the line L1, and if x1 = 1 or y1 = 1 according as q is odd or even, then
(xt, yt, zt) = t(x1, y1, z1) for t ∈ F . It follows that the flock F(C) is linear iff
the matrices of the q-clan have the form At = tA1, for t ∈ F . Such a q-clan
is also called linear. Although the concept of q-clan had not been recognized
explicitly when the monograph [FGQ] was written, section 10.6.1 of [GFQ]
basically contains a proof that when F(C) is linear, the associated GQ(C) is
isomorphic to the classical GQ H(3, q2 ).
If S is a spread of P G(3, q), the corresponding translation plane π(S)
is defined as follows. Embed P G(3, q) as a hyperplane in P G(3, q). The
points of π(S) are of two types: (i) the points of P G(4, q) \ P G(3, q) and (ii)
the lines of S. The lines are also of two types: (a) the planes of P G(4, q)
meeting P G(3, q) in a line of S, and (b) the ”infinite” line L∞ incident with
the points of type (ii). Otherwise the incidence in π(S) is the natural one
induced by incidence in P G(4, q). The spread S is regular provided the
regulus containing any triple (L1, L2, L3) of lines of S is contained in S. It is
by now a ”classical” result that the spread S is regular iff the corresponding
translation plane π(S) is desarguesian (cf., Bruck and Bose [BB66]). For our
present purposes it suffices to accept the following theorem.
Theorem 13.7.1. Let C be a normalized q-clan. The following are equivalent:

13.8. THE BOSE-BARLOTTI ∆-PLANES 607
(i) C is linear.
(ii) GQ(C) is classical, i.e., isomorphic to H(3, q 2 ).
(iii) The flock F(C) is linear.
(iv) The translation plane π(C) is desarguesian.
(v) The ovoid O(C) of the Klein quadric H is classical,
i.e., it is an elliptic quadric.
13.8 The Bose-Barlotti ∆-Planes
It is not our intention to delve very deeply into the theory of finite translation
planes. However, the isomorphism classes of the planes introduced in this section
(each of which is derived from the point-line dual of a translation plane
already associated with a flock of a cone) are in one-to-one correspondence
with the orbits on the generators of the cone under the semi-linear group
preserving the cone K and the given flock. Hence this increases the number
of geometries related to a q-clan and also motivates the determination of the
orbits on the generators of the cone under the appropriate group.
Let P = P G(4, q) and let Σ be a hyperplane of P . If S is a spread of
Σ, we will denote by π = π(P, Σ, S) the translation plane associated with
S, whose points and lines will be called, respectively, π-points and π-lines.
The π-points are the points of P \ Σ and the lines of S. The π-lines are the
planes of P which intersect Σ at a line of S, and Σ. The incidence relation
is inclusion.
Now let π d denote the dual plane of π. Let Σ1 be a hyperplane of P
different from Σ. As Σ ∩ Σ1 is a plane of Σ, there is exactly one line L of S
contained in the plane Σ ∩ Σ1. Following Bose and Barlotti [BB71] we can
construct a projective plane ∆ = ∆(P, Σ, Σ1, S) using P, Σ, Σ1 and S. The
points and the lines of ∆ will be called, respectively, ∆-points and ∆-lines.
The ∆-points are of three types. The ∆-points of type (1) are the planes
of P which intersect Σ at a line of S different from L. The ∆-points of type
(2) are the planes of P which intersect Σ at L and are not contained in Σ1.
The ∆-points of type (3) are the points of L.
The ∆-lines are of three types. The ∆-lines of type (1) are the points of
P which do not belong to Σ or to Σ1. The ∆-lines of type (2) are the planes
of Σ1 which do not contain L. The line L is the unique ∆-line of type (3).
A ∆-line α of type (2) is incident with a ∆-point β of type (2) iff α ∩ β
is a line. In all other cases, a ∆-line and a ∆-point are incident iff they are

608 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
incident in P .
The following comment is for those readers who are familiar with the
concept of derivation of a translation plane. Let R(Σ1) be the set of all πlines
α such that α is either a plane of Σ1 containing L or α = Σ. In π d ,
R(Σ1) defines a derivable net and ∆ is the derived plane of π d obtained by
replacing the derivable net R(Σ1).
The following theorem on ∆-planes is taken without proof from section
8.4 of N.L. Johnson [Jo89].
Theorem 13.8.1. The planes ∆(P, Σ, Σ1, S) and ∆(P, Σ, Σ2, S) are isomorphic
iff there is a collineation σ of Σ such that:
(1) σ : S ↦→ S
(2) if αi = Σ ∩ Σi for i = 1, 2, then σ : α1 ↦→ α2.
We can now put together Lemmas 13.6.4 and 13.6.6 along with Theorem
13.8.1 to obtain the following corollary.
Corollary 13.8.2. Starting with a given q-clan C, the number of pairwise
distinct ∆-planes derived from the plane of the spread S(C) is the same as
the number of orbits on the generators of the cone K under the group of
semilinear collineations preserving the flock F(C).
13.9 Property (G); some projective planes
Throughout this section C = {At : t ∈ F } will be a fixed q-clan normalized
0 0
so that A0 =
and each At is symmetric or upper triangular
0 0
according as q is odd or even. GQ(C) is the GQ of order (q2 , q) constructed
0 1
from C. Also P is always the matrix P =
, α ◦ β = αβ
−1 0
T , so
αP αT = α ◦ αP T = αP ◦ α = 0.
In any GQ of order (q2 , q), each triad of lines (three lines, no two concurrent)
has exactly 1+q transversals (by the point-line dual of Theorem 9.3.6).
Then each triad of those transversals again has 1 + q transversals, but in
general there is no guarantee that this sequence of triads of transversals will
produce a (1 + q) × (1 + q)-grid. We now consider a special situation where
such grids are always present.
Consider the notion of regularity for a line L in the following form: For
any triads (L = L1, L2, L3) and (M1, M2, M3) with Li ∼ Mj whenever 2 ≤

13.9. PROPERTY (G); SOME PROJECTIVE PLANES 609
i + j ≤ 5, then also L3 ∼ M3. Property (G) is a variant that is weaker in two
senses. Not every triad containing L is involved, and the grids that occur do
not imply regularity for the pairs of nonconcurrent lines that are involved.
Also, Property (G) is a variant of the concept of 3-regularity that appears in
?? for GQ of order (s, t) with s = t 2 .
Def. Let L1, M1 be distinct concurrent lines. We say that the pair
(L1, M1) has Property (G) provided whenever (L1, L2, L3, L4) are four distinct
pairwise noncurrent lines and (M1, M2, M3, M4) are four distinct lines for
which Li ∼ Mj whenever 2 ≤ i + j ≤ 7, then also L4 ∼ M4. A GQ S is said
to have Property (G) at a point p provided each pair (L, M) of distinct lines
through p has Property (G).
Theorem 13.9.1. The GQ GQ(C) has Property(G) at the pont (∞). Moreover,
for each t ∈ ˜ F there is a desarguesian projective plane π(t) defined at
the line [A(t)].
Proof. We prove this theorem by exhibiting the (q + 1) × (q + 1) grids that
exist because of Property (G) at (∞) and then coordinatize the plane π(t) as
P G(2, q). For checking these figures it is convenient first to note collinearities
in GQ(C). For t ∈ ˜ F , p t ∼ r means that points p and r lie on a line that is
a coset of A(t), with the appropriate adjustment if r = A ∗ (t)g.
(i) (α, c, β) ∞ ∼ (α, c + γα T , β + γ);
(ii) (α, c, β) ∞ ∼ A ∗ (∞)(α, 0,0);
(iii) (α, c, β) t ∼ (α + γ, c + γAtγ T + γKtα T , β + γKt), t ∈ F, γ ∈ F 2 ;
(iv) (α, c, β) t ∼ A ∗ (t)(0, 0, β − αKt), t ∈ F.
(13.16)
In Figs. 13.2 and 13.3 a and b vary independently over the elements of F ,
and δ = (0, 0) is arbitrary but fixed in F 2 .
The grid liines of Fig. 13.2 that meet [A(∞)] at points other than (∞)
meet [A(∞)] at points of the form A ∗ (∞)(α + bδ, 0,0), b ∈ F . This set
of points is clearly independent of t. Moreover, if we coordinatize the point
A ∗ (∞)(α, 0,0) with (α, 1) ∈ P G(2, q), then the set of points of the form
A ∗ (∞)(α + bδ, 0,0) is coordinatized by the line [δP, −α ◦ δP ] T in P G(2, q).
So the points of [A ∗ (∞)] other than (∞) are the affine points of a projective
plane π(∞) isomorphic to P G(2, q) with infinite line [0, 0, 1] T , and whose

610 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
A
∗ (∞)(α, 0,0) (α, c + aδP αT , β + aδP ) (α, c, β)
A
∗ (∞)(α + bδ, 0,0) p
(α + bδ, c+
b2gt(δ) + bδKtαT [A(∞)]
[A(t)]
,
bδKt + β)
(∞) ∗ A (t)(0, 0, β − αKt + aδP ) A∗ (t)(0, 0, β − αKt)
p = (α + bδ, c + δ(αP + bKt)α T + b 2 δAtα T , β + δ(aP + bKt))
Figure 13.2: Grids on [A(∞)] and [A(t)].
A
∗ (u)(0, 0, β − αKu) (α + aγ, c + a2γAuγ T + aγKuαT , β + aγKu)
(α, c, β)
q1
[A(u)]
p1
❡
p2
q2
[A(t)]
(∞)
A
∗ (t)(0, 0, β − αKt + aγ(Ku − Kt)) A∗ (t)(0, 0, β − αKt)
q1 = A ∗ (u)(0, 0, β − αKu + bγ(Kt − Ku))
q2 = (α + bδ, c + b 2 δAtδ T + bδKtα T , β + bδKt)
Figure 13.3: Grids on [A(u)] and [A(t)].

13.10. THE FUNDAMENTAL LEMMA 611
finite lines are the sets of points incident with the lines of a grid containing
(∞).
Similarly, the grid lines meeting [A(t)] at points different from (∞) meet
it at the points A ∗ (t)(0, 0, β − αKt + aδP ), a ∈ F . And the points of
[A(t)] different from (∞) are the affine points of a plane π(t) isomorphic to
P G(2, q). Coordinatize A ∗ (t)(0, 0, β) as (β, 1) ∈ P G(2, q). Then the affine
line of points of the form A ∗ (t)(0, 0, β − αKt + aδP ), a ∈ F , is coordinatized
by the line [δ, (αKt − β) ◦ δ] T , and the infinite line is [0, 0, 1] T . We claim that
all grids containing (∞) and meeting [A(t)] determine the same affine lines,
and hence exactly the same projective plane π(t). This follows from Fig. 13.3,
which shows that all affine lines of π(t) have the form [δ, (αKt − β) ◦ δ] T for
some δ = (0, 0), δ ∈ F 2 . This is independent of u ∈ ˜ F \ {t}, so a unique
plane π(t) is constructed.
The points p1 = (α+aγ +bδ, c+a 2 γAuγ T +aγKuα T +b 2 δAtδ T +bδKt(α+
aγ) T , β+aγKu+bδKt) and p2 = (α+bδ+aγ, c+b 2 δAtδ T +bδKtα T +a 2 γAuγ T +
aγKu(α + bδ) T , β + bδKt + aγKu) are the same point iff γ(Kt − Ku)δ T = 0.
So if δ is a fixed nonzero element of F 2 , put γ = δ(Kt − Ku)P to obtain a
(q+1)×(q+1) grid. And the set of points on [A(t)] different from (∞) and on
the grid lines, is the set of points of the form A ∗ (t)(0, 0, β−αKt+aγ(Ku−Kt)),
which is coordinatized by the line [γ(Ku −Kt)P, (αKt−β)◦(γ(Ku −Kt)P )] T .
Putting δ = γ(Ku−Kt)P here shows that lines have the general form claimed
above.
13.10 The Fundamental Lemma
The theorem of this section, which we refer to as the Fundamental Lemma,
really contains the heart of the proof of the Fundamental Theorem.
Theorem 13.10.1. Let C = {At : t ∈ F } and C ′ = {A ′ t : t ∈ F } be (not
necessarily distinct) normalized q-clans. Let θ : GQ(C) → GQ(C ′ ) be an isomorphism
for which θ : (∞) ↦→ (∞), [A(∞)] ↦→ [A ′ (∞)], (0, 0,0) ↦→ (0, 0,0).
Then there exist the following:
(i) σ ∈ Aut(F )
(ii) D ∈ GL(2, q)
(iii) 0 = λ ∈ F
(iv) π : F → F : t ↦→ ¯t, a permutation on F for which
(v) A ′ ¯t ≡ λDT Aσ t D + A′ ¯0 ∀t ∈ F.

612 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
Moreover, θ is (induced by) an automorphism of G given by
θ = θ(σ, D, λ, π) : (α, c, β) ↦→ (13.17)
(λ −1 α σ D −T , λ −1 c σ + λ −2 α σ D −T A ′ ¯0 D−1 (α σ ) T , β σ D + λ −1 α σ D −T K ′ ¯0 ).
Conversely, given σ, D, λ, π satisfying the conditions given above, θ defined
by Eq. 13.17 gives such an isomorphism.
Proof. Clearly θ is completely determined as a permutation of the elements
of G. And there must be a permutation π : t ↦→ ¯t on F for which θ : [A(t)] ↦→
[A(¯t)]. Moreover, θ must map grids to grids. Therefore θ must map π(∞)
to π ′ (∞) and π(t) to π ′ (¯t), always mapping infinite points to infinite points
and infinite lines to infinite lines. This means that there is a B ∈ GL(2, q)
and a σ ∈ Aut(F ), and for each t ∈ F a Dt ∈ GL(2, q) and a σt ∈ Aut(F ),
for which
(i) θ : A ∗ (∞)(α, 0,0) ↦→ (A ′ ) ∗ (∞)(α σ B, 0,0),
(ii) θ : A ∗ (t)(0, 0, β) ↦→ (A ′ ) ∗ (¯t)(0, 0, β σt Dt).
(13.18)
Since (α, c, β) ∞ ∼ A ∗ (∞)(α, 0,0), clearly (α, c, β) θ ∞ ∼
(A ′ ) ∗ (∞)(α σ B, 0,0). This implies that as a permutation of the elements of
G, θ must have the form
θ : (α, c, β) ↦→ (α σ B, (α, c, β) θ2 , (α, c, β) θ3 ). (13.19)
for some functions θ2 : F 5 → F and θ3 : F 5 → F 2 that we now proceed to
determine.
Since (α, c, β) t ∼ A ∗ (t)(0, 0, β − αKt) for each t ∈ F ,
(α σ B, (α, c, β) θ2 , (α, c, β) θ3 ) ¯t ∼ (A ′ ) ∗ (¯t)(0, 0, (β − αKt) σt Dt).
This implies (α, c, β) θ3 − α σ BK ′ ¯t = (β − αKt) σt Dt, or
(α, c, β) θ3 = (β − αKt) σt Dt + α σ BK ′ t (independent of c). (13.20)
With α = 0, (0, c, β) θ3 = β σt Dt = β σ0 D0 for all t ∈ F . First put
β = (1, 0), and then β = (0, 1), to force Dt = D0 = D for all t ∈ F . And D
is invertible, so σt = σ0 for all t ∈ F . So now Eq. 13.20 implies
(α, c, β) θ3 = (β − αKt) σ0 D + α σ BK ′ ¯t (∀t ∈ F ; ∀α, β ∈ F 2 ). (13.21)

13.10. THE FUNDAMENTAL LEMMA 613
Then for distinct s, t ∈ F , Eq. 13.21 implies (β − αKt) σ0 D + α σ BK ′ ¯t =
(β − αKs) σ0 D + α σ BK ′ ¯s, which may be rewritten as
[α(Ks − Kt)] σ0 D = α σ B(K ′ ¯s − K ′ ¯t). (13.22)
Put α = (1, 0), then α = (0, 1) to get
Put Eq. 13.23 into Eq. 13.22 to get
(Ks − Kt) σ0 D = B(K ′ ¯s − K ′ ¯t). (13.23)
α σ0 (Ks − Kt) σ0 D = α σ0 B(K ′ ¯s − K ′ ¯t) = α σ B(K ′ ¯s − K ′ ¯t). (13.24)
Since Ks − Kt, D, B, K ′ ¯s − K ′ ¯t are all invertible,
σ0 = σ and D = (Ks − Kt) −σ B(K ′ ¯s − (K ′ ¯t). (13.25)
At this point θ has the following appearance:
θ : (α, c, β) ↦→ (α σ B, (α, c, β) θ2 ,
(β − αKt) σ (Ks − Kt) −σ B(K ′ ¯s − K ′ ¯t) + α σ BK ′ ¯t).
(13.26)
In Fig. 13.10 we have reconstructed Fig. 13.2 specialized so that
(α, c, β) = (0, 0,0), γ = bδ = aδ.
And in Fig. 13.5 we have the image under θ of Fig. 13.10, using especially
Eq. 13.18
Consider the impact of the existence of the center point p. The fact that
p lies on the horizontal line forces p to have the form
p = (γ σ B, γ σ BA ′ ¯tB T (γ σ ) T + δ ◦ γ σ B, γ σ BK ′ ¯t + δ)
for some δ = (0, 0) (by part (i) of Eq. 13.16). Then the fact that p lies on
the vertical line forces p to have the form
p = (γ σ B, γ σ BA ′ ¯tB T (γ σ ) T , γ σ P D + γ σ BK ′ ¯t).
Hence δ = γ σ P D and γ σ P D ◦ γ σ B = 0. This is for all γ = (0, 0). Hence
DB T = µI for some µ with 0 = µ ∈ F . Set λ = µ −1 , so
B = λ −1 D −T for some nonzero λ ∈ F. (13.27)

614 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
A
∗ (∞) (0, 0, γP ) (0, 0,0)
A
∗ (∞)(γ, 0,0) (γ, γAtγ T , γ(P + Kt))
(γ,
γAtγ T ,
γKt)
(∞) A ∗ (t)(0, 0, γP ) A ∗ (t)(0, 0,0)
Figure 13.4:
(A
′ ) ∗ (∞) (0, 0, γσP D)) (0, 0,0)
(A
′ ) ∗ (∞)(γσB, 0,0) p
(γ σ B,
γ σ BA ′ ¯tB T (γ σ ) T ,
γ σ BK ′ ¯t)
(∞) (A ′ ) ∗ (t)(0, 0, γ σ P D) (A ′ ) ∗ (¯t)(0, 0,0)
Figure 13.5:

13.10. THE FUNDAMENTAL LEMMA 615
Using this in Eq. 13.25 we obtain
(Ks − Kt) σ = λ −1 D −T (K ′ ¯s − K ′ ¯t)D −1 for all s, t ∈ F. (13.28)
Then s = 0 in Eq. 13.28 gives
So now Eq. 13.26 may be rewritten as
λD T Kt σ D = K ′ ¯t − K ′ ¯0. (13.29)
θ : (α, c, β) ↦→ (λ −1 α σ D −T , (α, c, β) θ2 , β σ D + λ −1 α σ D −T K ′ ¯0). (13.30)
Putting γ = −α into part (iii) of Eq. 13.16 we obtain p1 = (α, c, β) t ∼
p2 = (0, c − αAtαT , β − αKt), so that pθ ¯t
1 ∼ pθ 2 , where by Eq. 13.30 pθ1 =
(λ−1ασD −T , (α, c, β) θ2 σ −1 σ −T ′ , β D + λ α D K ¯0) and
p2 θ = (0, (0, c − αAtαT , β − αKt) θ2 , (β − αKt) σD). Since (α, c, β) −1 =
(−α, α ◦ β − c, −β), p1 θ ◦ (p2 θ ) −1 = (λ−1ασD −T , (α, c, β) θ2 σ , β D+
λ−1ασD −T K ′ ¯0)◦(0, −(0, c−αAtαT , β−αKt) θ2 , −(β−αKt) σD) = (λ−1ασD −T , (α, c, β) θ2− (0, c−αAtαT , β−αKt) θ2 −1 σ −T ′ , λ α D K ¯0+α σKt σ D) must be in A ′ (¯t). Eq. 13.29
implies that the third coordinate is O.K., but from the middle coordinate we
have
(α, c, β) θ2 = (0, c − αAtα T , β − αKt) θ2 + λ −2 α σ D −T A ′ ¯tD −1 (α σ ) T . (13.31)
Put t = 0 to obtain
(α, c, β) θ2 = (0, c, β) θ2 + λ −2 α σ D −T A ′ ¯0D −1 (α σ ) T . (13.32)
Now put γ = −β in part (i) of Eq. 13.16 to obtain p1 = (α, c, β) ∞ ∼ p2 =
(α, c − βα T ,0), so p1 θ ∞ ∼ p2 θ . Then by Eq. 13.30
and
So
p1 θ = (λ −1 α σ D −T , (α, c, β) θ2 , β σ D + λ −1 α σ D −T K ′ ¯0)
p2 θ = (λ −1 α σ D −T , (α, c − βα T ,0) θ2 , λ −1 α σ D −T K ′ ¯0).
p1 θ ◦ (p2 θ ) −1 = (λ −1 α σ D −T , (α, c, β) θ2 , β σ D + λ −1 α σ D −T K ′ ¯0)◦
(−λ −1 α σ D −T , λ −2 α σ D −T K ′ ¯0D −1 (α σ ) T − (α, c − βα T ,0) θ2 ,

616 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
−λ −1 α σ D −T K ′ ¯0) = (0, (α, c, β) θ2 − (α, c − βα T ,0) θ2 − λ −1 β σ (α σ ) T , β σ D)
must be in A ′ (∞). Hence
(α, c, β) θ2 = (α, c − βα T ,0) θ2 + λ −1 α σ (β σ ) T . (13.33)
Set α = 0, so (0, c, β) θ2 = (0, c,0) θ2 , which is independent of β. Now use
this in Eq. 13.32.
(α, c, β) θ2 = (0, c,0) θ2 + λ −2 α σ D −T A ′ ¯0D −1 (α σ ) T . (13.34)
Then (0, c,0) θ2 −2 σ −T ′ +λ α D A ¯0D −1 (ασ T 13.34
)
= (α, c, β)
θ2 13.33
= (α, c−βα T ,0) θ2 +
λ−1ασ (βσ T 13.34
) = (0, c − βαT ,0) θ2 −2 σ −T ′ + λ α D A ¯0(α σ ) T + λ−1ασ (βσ ) T . Put
c = αβT into these last equations to obtain
(0, αβ T ,0) θ2 = λ −1 α σ (β σ ) T ,
where we used the fact that (0, 0,0) θ = (0, 0,0) implies (0, 0,0) θ2 = 0. Hence
So Eq. 13.34 becomes
(0, c,0) θ2 = λ −1 c σ . (13.35)
(α, c, β) θ2 = λ −1 c σ + λ −2 α σ D −T A ′ ¯0D −1 (α σ ) T . (13.36)
Now use Eq. 13.36 in Eq. 13.30.
θ : (α, c, β) ↦→ (λ −1 α σ D −T ,
λ −1 c σ + λ −2 α σ D −T A ′ ¯0D −1 (α σ ) T , β σ D + λ −1 α σ D −T K ′ ¯0). (13.37)
Note : We have now shown that θ is an automorphism of G.
Conversely, let 0 = λ ∈ F, σ ∈ Aut(F ), D ∈ GL(2, q), and ¯0 ∈ F .
Then it is routine to check that θ defined by Eq. 13.37 is an automorphism
of G which maps A(∞) to A ′ (∞). So θ is an isomorphism from GQ(C)
to GQ(C ′ ) iff θ maps A(t) to A(¯t) for all t ∈ F , for some permutation
π : t ↦→ ¯t. Consider the effect of θ on A(t). Here θ : (α, αAtα T , αKt) ↦→
(λ −1 α σ D −T , λ −1 α σ At σ (α σ ) T +
λ −2 α σ D −T A ′ ¯0D −1 A ′ ¯0D −1 (α σ ) T , α σ Kt σ D +λ −1 α σ D −T K ′ ¯0), which must be in
A ′ (¯t). From the middle coordinates,
λ −2 α σ D −T A ′ ¯tD −1 (α σ ) T = λ −1 α σ At σ (α σ ) T + λ −2 α σ D −T A ′ ¯0D −1 (α σ ) T . This

13.11. THE FUNDAMENTAL THEOREM 617
can be rewritten as (α σ D −T [λD T At σ D + A ′ ¯0 − A ′ ¯t](α σ D −T ) T for all α ∈
F 2 , t ∈ F , which is equivalent to
λD T At σ D ≡ A ′ ¯t − A ′ ¯0 for all t ∈ F. (13.38)
From the third coordinates we obtain
α σ (Kt σ D + λ −1 D −T K ′ ¯0) = λ −1 α σ D −T K ′ ¯t. As this holds for all α ∈ F 2 , t ∈
F , multiplying on the left by λD T , we obtain
λD T Kt σ D = K ′ ¯t − K ′ ¯0. (13.39)
We claim that the condition in Eq. 13.38 implies that of Eq. 13.39. First
suppose that q is odd. If K and M are symmetric matrices, then K ≡ M
iff K = M. So from Eq. 13.38, λDT At σ D = A ′ ¯t − A ′ ¯0. Add this to its
transpose to get Eq. 13.39. Now suppose Eq. 13.38 holds with q = 2e .
Let DT
a b
= with ∆ = det(D). If K and M are upper triangular
c d
matrices, then K ≡ M iff K = M. So from Eq. 13.38
x ′ ¯t − x ′ ¯0 y ′ ¯t − y ′ ¯0
0 z ′ ¯t − z ′ ¯0
= A ′ ¯t − A ′ ¯0 ≡
λD T At σ
2 a xt
D ≡
σ + abyt σ + b2zt σ ∆yt σ
0 c 2 xt σ + cdyt σ + d 2 zt σ
So in particular, y ′ ¯t−y ′ ¯0 = λ∆yt σ . But Eq. 13.39 says (y ′ ¯t−y ′ ¯0)P = K ′ ¯t−
K ′ ¯0 = λD T Kt σ D = λyt σ D T P D = λyt σ ∆P , which holds iff y ′ ¯t − y ′ ¯0 = λ∆yt σ .
So for all q, Eq. 13.38 does imply Eq. 13.39.
This completes the proof (in the hard direction) of the Fundamental
Lemma, i.e., the proof of Theorem 13.10.1 The converse is a straightforward
exercise essentially contained in the steps already given.
13.11 The Fundamental Theorem
Frequently it is convenient to change the parameters in the Fundamental
Lemma. Put µ = λ −1 and B = λ −1 D −T , so D = µB −T . Then collect
the results of Lemma 13.6.4, Theorems 13.6.13 and 13.3.4 along with the
Fundamental Lemma into the following Fundamental Theorem (F.T.).
.

618 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
Theorem 13.11.1. (The Fundamental Theorem of q-Clan Geometry) For
xt yt
any prime power q, let C = {At ≡
: t ∈ F } and C
0 zt
′ = {A ′ t ≡
′ x t y ′
t : t ∈ F } be two (not necessarily distinct) normalized q-clans.
0 z ′ t
Then the following are equivalent:
(i) C ∼ C ′ ;
(ii) The flocks F(C) and F(C ′ ) are projectively equivalent;
(iii) GQ(C) and GQ(C ′ ) are isomorphic by an isomorphism mapping
(∞) to (∞), [A(∞)] to [A ′ (∞)], and (¯0, 0, ¯0) to (¯0, 0, ¯0);
(iv) The associated spreads S(C) and S(C ′ ) are equivalent .
Moreover, these four conditions hold iff there exist σ ∈ Aut(F ), B ∈ GL(2, q), 0 =
u ∈ F , and a permutation π : t ↦→ ¯t on F for which
(v) A ′ ¯t ≡ uB−1 A σ t B−T + A ′ ¯0
for all t ∈ F.
The corresponding isomorphism θ : GQ(C) ↦→ GQ(C ′ ) is given by
(vi) θ = θ(u, B, σ, π) : (α, c, β) ↦→
(α σ B, uc σ + α σ BA ′ ¯0 BT (α σ ) T , uβ σ B −T + α σ BK ′ ¯0 ).
And the corresponding isomorphism between the associated translation
planes is induced by the following projective equivalence of the associated
linespreads of P G(3, q).
(vii) T : (x0, x1, x2, x3) ↦→ (xσ 0 , xσ 1, xσ 2, xσ
−1 u B
3)
O
−1 ′ u BA ¯0
B−T S(C) ↦→ S(C ′ ).
If B−1
a b
= , the corresponding isomorphism Tθ : F(C) ↦→ F(C
c d
′ ) is
defined for planes of the flocks by
:

13.11. THE FUNDAMENTAL THEOREM 619
(viii) Tθ :
⎡
⎢
⎣
x ′ ¯t
y ′ ¯t
z ′ ¯t
1
⎤
⎥
⎦ =
⎛
⎜
⎝
⎡
⎢
⎣
xt
yt
zt
1
⎤
⎥
⎦ ↦→
ua 2 uab ub 2 x ′ ¯0
2uac u(ad + bc) 2ubd y ′ ¯0
uc 2 ucd ud 2 z ′ ¯0
0 0 0 1
⎞ ⎡
⎟ ⎢
⎟ ⎢
⎠ ⎣
For the remainder of this section we consider the F.T. under the additional
hypothesis that C = C ′ , or more specifically that At = A ′ t, t ∈ F . Then for
the normalized q-clan C, we define the following groups of collineations of
GQ(C), where G denotes the group of all collineations of GQ(C) that fix the
point ∞ .
(i) G0 = {θ ∈ G : (¯0, 0, ¯0) θ = (¯0, 0, ¯0)}; (13.40)
(ii) H = {θ ∈ G0 : [A(∞)] θ = [A(∞)};
(iii) M = {θ ∈ H : [A(0)] θ = [A(0)]};
(iv) N = {θ = θa ∈ M : θa : (α, c, β) ↦→ (aα, a 2 c, aβ), 0 = a ∈ F }.
x σ t
y σ t
z σ t
1
⎤
⎥
⎦ .
The following observation is a trivial consequence of the F.T.
Corollary 13.11.2. M always contains the following subgroup:
M∗ = {θ = θ(a2 , aI, σ, π : t ↦→ tσ ) : (α, c, β) ↦→ (aασ , a2cσ , aβσ ) : 0 = a ∈
F, σ ∈ Aut(F ) with Atσ = Aσ t for t ∈ F }.
Now let θi = θ(µi, Bi, σi, πi), i = 1, 2. We want to compute θ = θ1 ◦ θ2,
where the notation means do θ1 first. Then
(α, c, β) θ1◦θ2 =
(ασ1B1, µ1cσ1 σ1 + α B1A0π1 BT 1 (ασ1 T ) , µ1βσ1 −T
B1 + ασ1B1K0 π1 ) θ2
= ((ασ1B1) σ2B2, µ2[µ1cσ1 σ1 + α B1A0π1 BT 1 (ασ1 T σ2 ) ]
+(ασ1B1) σ2B2A0 π2 BT 2 ((ασ1B1) σ2 T ) , µ2(µ1β σ1 −T
B1 +ασ1B1K0 π1 ) σ2 −T
B2 + (ασ1B1) σ2B2K0 π2 )
= (ασ1◦σ2 σ2 B1 B2, µ σ2
1 µ2C σ1◦σ2 + µ2ασ1◦σ2 σ2 B1 A σ2
0π1 (B σ2
1 ) T (ασ1◦σ2 T )
+ασ1◦σ2 σ2 B1 B2A0π2 (B σ2
1 B1) T (ασ1◦σ2 T σ2
−T
) , µ 1 B2)
+µ2ασ1◦σ2 σ2 B1 Kσ2 0π1 B −T
2
1 µ2β σ1◦σ2 (B σ2
+ ασ1◦σ2 B σ2
1 B2K0 π 2 ).

620 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
Since θ = θ1 ◦ θ2 = θ(µ, B, σ, π) for some µ, B, σ, and π according to the
F.T., we can read off µ = µ σ2
1 µ2, B = B σ2
1 B2, σ = σ1 ◦ σ2. So we have at
least
θ(µ1, B1, σ1, π1) ◦ θ(µ2, B2, σ2, π2) = θ(µ σ2
1 µ2, B σ2
1 B2, σ1 ◦ σ2, π), (13.41)
where we say a little more about π. We have now the fact that
θ1 ◦ θ2 : (α, c, β) ↦→
(α σ B, µc σ + µ2α σ B σ2
1 Aσ2 0π1 (B σ2
1 )T (α σ ) T + α σ BA0π2 B T (α σ ) T ,
µβ σ B −T + µ2α σ B σ2
1 Kσ2 0π1 B −T
2 + ασBK0π2 ),
where this image must equal (α σ B, µc σ +α σ BA0 πBT (α σ ) T , µβ σ B −T +α σ BK0 π).
This is equivalent to
(i) α σ [µ2B σ2
1 A σ2
0 π 1 (B σ2
1 ) T + B σ2
1 B2A0 π 2 (B2) T (B σ2
1 ) T ](α σ ) T (13.42)
= α σ (B σ2
1 B2A0 πBT 2 (B σ2
1 ) T )(α σ ) T
(ii) B σ2
σ2
1 B2K0π = µ2B1 Kσ2 0π1 B −T
2
Since [A(0)] θ1
↦→ [A(0π1 θ2
)]
pothesis µ2B −1
2 A σ2
0π1 B −T
2 + A0π2 + A(0π1 ) π ≡
µ2B −1
2 K σ2
0π1 B −T
2
+ Bσ2
0 0
1 B2K0 π 2 .
↦→ [A((0π1 π2 π π1◦π2 ) )], at least 0
= 0 . So by hy-
0 0
, since θ2 ∈ G0, and also
= K0 π 2 +K(0 π 1 ) π 2 . The first relationship verifies Eq. 13.42(i);
the latter equality verifies Eq. 13.42(ii). In fact, this gives a direct proof that
the composition in Eq. 13.41 is correct. But π is essentially controlled by σ,
B, µ, and 0 π according to the F.T. (v), which may be written out as follows.
xtπ = µ(a2x σ t + abyσ t + b2z σ t ) + x0π (13.43)
ytπ = µ(2acxσt + (ad + bc)yσ t + 2bdzσ t ) + y0π zt π = µ(c2 x σ t + cdy σ t + d 2 z σ t ) + z0 π
Since t ↦→ xt and t ↦→ zt must be permutations, t π is determined (though not
conveniently!) by σ, B, µ, and 0 π . This suggests that we make the following
additional normalizing convention: If q is odd, we assume that the members
of C are indexed so that xt = t for all t ∈ F . And if q = 2 e , so that t ↦→ yt
is also a permutation, we assume that the members of C are indexed so that
yt = t for all t ∈ F . This makes it possible to be more explicit with the

13.12. RECOORDINATIZATION: SHIFTS, FLIPS & SCALES 621
description of π.
For θ(σ, B, µ, π) ∈ H with B −1 =
a b
c d
(i) t π = µ(a 2 t σ + aby σ t + b2 z σ t ) + 0π , if q is odd.
(ii) t π = µ(ad + bc)t σ + 0 π , if q is even.
, (13.44)
Read Theorem 13.11.1(v) as equality with q odd, and apply θ1 ◦ θ2 with
π as in Eq. 13.41.
At
π = µ2B −1
= µ σ1
1
2
−1
(µ1B1 Aσ1 t B −T
1 + A0π1 ) σ2 −T
B2 + A0π2 (13.45)
1 B2) −1 A σ1◦σ2
t (B σ2
1 B2) −T + µ2B −1
2 Aσ2 0π1 B −T
2 + A0π2 .
µ2((B σ2
And here µ2B −1
2 A σ2
0 π 1 B −T
2 + A0 π 2 = A0 π 1 ◦π 2 , confirming again (put t = 0
in Eq. 13.45) that 0 π = 0 π1◦π2 .
A relationship similar to Eq. 13.45 holds for q even, and it follows that
(use Theorem 13.11.1(vii)):
Theorem 13.11.3. T : θ ↦→ Tθ is a homomorphism from H onto the subgroup
of P ΓL(4, q) leaving invariant the cone K and the flock F(C). The
kernel of T is N .
We define N to be the kernel of GQ(C) This definition of kernel may not
be as satisfactory as that for translation generalized quadrangles (TGQ) studied
in [FGQ], but we shall see later that for many nonclassical q-clans C the
kernel of GQ(C) plays a role similar to one played by the multiplicative group
of the kernel of a TGQ. Note that the kernel N is a group of collineations of
GQ(C) fixing each line through the point (∞) for every q-clan C.
13.12 Recoordinatization: Shifts, Flips
& Scales
Any automorphism of G replaces one 4-gonal family for G with another. But
in general we cannot expect that a 4-gonal family J (C) arising from a q-clan
C will be replaced with one for which a q-clan is clearly present. In this section
we study three basic automorphisms of G that even replace normalized qclans
with normalized q-clans. One goal is to develop a procedure to use the
F.T. to describe all elements of G0. A second goal is to assign to each line

622 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
through (∞) in some GQ(C) an equivalence class of flocks (or q-clans) in such
a way that two lines through (∞) are in the same G0-orbit if and only if their
flocks (or q-clans) are equivalent. This provides a very concrete algebraic
approach to the concept of derivation first presented in [BLT90] when q is
odd and includes the case for q = 2 e first worked out in [BLP94] for a
representation of G that works only in characteristic 2. That representation
will be considered in a later chapter where q is restricted to be a power of 2.
Throughout this section we assume that C is a normalized q-clan with
the specific indexing given in Section 13.11. Our first observation is just a
special corollary of the F.T.
Corollary 13.12.1. Let B ∈ GL(2, q) and π : F → F : t ↦→ t a permutation.
For each At ∈ C set A ′
t = B−1AtB −T . Then C ′ = {A ′ : t ∈ F } is a q-clan
t
equivalent to C, and θ : GQ(C) → GQ(C ′ ) : (α, c, β) ↦→ (αB, c, βB−T ) is an
isomorphism.
Lemma 13.12.2. Shift by s. Fix s ∈ F . Define τs : G → G : (α, c, β) ↦→
(α, c−αAsα T , β −αKs). For At ∈ C, put A τs
t = At −As, so Aτs x = Ax+s −As.
Then Cτs = {A τs
t : t ∈ F } is a normalized q-clan equivalent to C, and τs =
θ(1, I, id, π : t ↦→ t − s) : GQ(C) → GQ(Cτs ) is an isomorphism.
Proof: This is immediate from the F.T. For q odd, xt = t, so x τs
t = (t + s) −
s = t. For q even, yt = t, so y τs
t = (t + s) − s = t. Hence Cτs is normalized
and 0 = −s.
Lemma 13.12.3. Scale by a. Let 0 = a ∈ F . Define σa : G → G :
(α, c, β) ↦→ (α, ac, aβ). For At ∈ C, put A σa
t = aAt, so A σa
at = aAt, or
A σa
t = aAt/a. Then Cσa = {A σa
t : t ∈ F } is a normalized q-clan equivalent to
C, and σa = θ(a, I, id, π : t ↦→ at) : GQ(C) → GQ(Cσa ) is an isomorphism.
Before considering the third type of automorphism of G, we collect some
rather trivial results that are helpful in doing computations.
0 1
Lemma 13.12.4. Put P =
, and for B ∈ GL(2, q), put ∆ =
−1 0
det(B).
(i) P T = P −1 = −P .
(ii) P T BP = ∆B −T = P BP T ≡ ∆B −1 .

13.12. RECOORDINATIZATION: SHIFTS, FLIPS & SCALES 623
(iii) B T P B = ∆P ; B T P T B = ∆P T .
(iv) If q is odd and A is nonsingular and symmetric with K = A+A T = 2A,
then P T K −1 AKP = (det(K)) −1 A.
(v) If q = 2 e and A is upper triangular with K = A + A T = (det(K)) 1/2 P
nonsingular, then P T K −1 AKP = (det(K)) −1 A.
Lemma 13.12.5. The flip. Define ϕ : G → G : (α, c, β) ↦→ (βP, c −
αβT , αP T ). Then ϕ is an automorphism of G that interchanges A(0) and
A(∞). It maps a normalized q-clan C = {At : t ∈ F } to a q-clan Cϕ =
{A ϕ
t = −P T K −1
t AtK −1
t P = −(det(Kt)) −1At : t ∈ F }. The permutation
π : t ↦→ t is chosen so that Cϕ will have a normalized indexing. If q is even
with yt = t, clearly π : t ↦→ t = t−1 for all t ∈ ˜ F . For q odd with xt = t, the
permutation π : t ↦→ t depends on the functions t ↦→ yt and t ↦→ zt. Explicitly:
π : t ↦→ t = t/(4xtzt − y2 t ).
Proof: It is routine to check that ϕ is an automorphism of G interchanging
A(0) and A(∞). So A ϕ (0) = A(∞), A ϕ (∞) = A(0). And
ϕ : (α, αAtα T , αKt) ↦→ (αKtP, αAtα T − αKtα T , αP T )
= (αKtP, αKtP (−P T K −1
t AtK −1
t P )(αKtP ) T , αKtP (P T K −1
t P T ))
= (γ, γ(−(det(Kt)) −1 At)γ T , −(det(Kt)) −1 γKt),
for γ = αKtP, 0 = t ∈ F.
Shifting by s and scaling by a are automorphisms of G that map normalized
q-clans to equivalent normalized q-clans, and affect on ˜ F (the index set
for the lines through (∞)) the permutations t ↦→ t = t − s and t ↦→ t = at,
respectively. So the group generated by these permutations stabilizes ∞. On
the other hand, the flip ϕ effects a permutation π on ˜ F that interchanges 0
and ∞, and for q = 2 e π : t ↦→ t −1 for all t ∈ ˜ F . In general for q odd, π is
determined by t ↦→ yt and t ↦→ zt, and C and C ϕ are not equivalent!
For s ∈ F define the shift-flip is by
(i) is = τs ◦ ϕ, s ∈ F. (13.46)
(ii) is : (α, c, β) ↦→ ((β − αKs)P, c − αβ T + αAsα T , αP T )
(iii) i −1
s : (α, c, β) ↦→ (βP, c − αβ T + (βP )As(βP ) T , αP T + βP Ks).
Compute the effect of is on an element of A(t). is : (α, αAtα T , αKt) ↦→
(α(Kt − Ks)P, −α(At − As) T , αP T ). Put γ = α(Kt − Ks)P , so for t = s,

624 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
α = γP (Kt − Ks) −1 . Then the image is
(γ, γ(−P T (Kt − Ks) −1 (At − As)(Kt − Ks) −1 P )γ T , −γP T (Kt − Ks) −1 P )
=
(γ, γ(−(det(Kt − Ks)) −1 (At − As))γ T , γ(−(det(Kt − Ks)) −1 )(Kt − Ks)). So
we note that
(i) is : C → Cis : At ↦→ A is
t = (13.47)
−(det(Kt − Ks)) −1 (At − As),
t = −(t − s)(det(Kt − Ks)) −1 .
(ii) If q = 2 e , t = (t + s) −1 .
Also define i∞ = id : G → G. Then to each line of GQ(C) through (∞)
we are going to assign an equivalence class of q-clans and its associated class
of flocks. Start with a normalized q-clan C. For each s ∈ ˜ F , applying is to
G yields a normalized q-clan C is . We assign to the line [A(s)] of GQ(C) the
class of q-clans equivalent to C is , and also the class of flocks projectively equivalent
to F(C is ). The following application of the F.T. makes this assignment
natural and useful.
Theorem 13.12.6. Let C be a normalized q-clan. Then for arbitrary s, t ∈
˜F , there is a collineation (an automorphism) of GQ(C) mapping [A(s)] to
[A(t)] if and only if F(C is ) and F(C it ) are projectively equivalent (i.e., if and
only if C is ∼ C it ).
Proof. Since τs always effects the permutation t ↦→ t − s and ϕ always interchanges
0 and ∞, the shift-flip is always maps s to ∞, i.e., is : GQ(C) →
GQ(C is ) : [A(s)] ↦→ [A is (∞)]. Let θ be an automorphism of GQ(C) map-
ping [A(s)] to [A(t)]. Without loss of generality we may assume that θ fixes
(0, 0, 0) (and of course (∞)). Then θ = i−1 s ◦ θ ◦ it : GQ(Cis ) → GQ(Cit ) :
[Ais (∞)] ↦→ [Ait (∞)]. Since clearly θ : (∞) ↦→ (∞) and (0, 0, 0) ↦→ (0, 0, 0),
by the F.T. F(C is ) and F(C it ) are projectively equivalent. Conversely, given
that F(C is ) and F(C it ) are projectively equivalent, there is an appropriate
θ : GQ(Cis ) → GQ(Cit ) : [Ais (∞)] ↦→ [Ait (∞)]. Hence we may put
θ = is ◦ θ ◦ i −1
t : GQ(C) → GQ(C) : [A(s)] ↦→ [A(t)].
13.13 Additive q-Clans and Dual TGQ
Recall that a symmetry about a line L of a GQ S of order (s, t) is a
collineation of that GQ that fixes each line concurrent with L. The group

13.13. ADDITIVE Q-CLANS AND DUAL TGQ 625
of symmetries about L has order at most s and L is an axis of symmetry
provided there are s symmetries about L. In Chapter 8 of [FGQ] it is shown
that if there is a point p for which each line through p is an axis of symmerty,
then the group T generated by the hypothesized symmetries is a maximal
group of elations about the point p, and T is elementary abelian. T is unique
and is called the group of translations about p and the GQ is said to be a
translation generalized quadrangle (TGQ) with base point p. If even one
axis of symmetry exists, then s ≤ t. Many more results about T GQ are
obtained in [FGQ], but here we concern ourselves only with determining how
to recognize normalized q-clans whose associated GQ have point-line duals
that are TGQ. A much more extensive and up-to-date treatment of TGQ is
contained in the recent book [TTVM06].
First note that starting with any q-clan C, the point (∞) is a center of
symmetry for GQ(C). Specifically, for each d ∈ F , the map (α, c, β) ↦→
(α, c + d, β) is a symmetry about (∞), and there are q = t of them. Also,
the group G of elations about p is transitive on the points of [A(∞)] different
from (∞), so that in order to determine when all points of [A(∞)] are centers
of symmetry, we need only determine when A ∗ (∞) is a center of symmetry.
Theorem 13.13.1. The point A ∗ (∞) is a center of symmetry (i.e., the pointline
dual of GQ(C) is a TGQ with base “point” equal to the line [A(∞)] of
GQ(C)) if and only if (after a suitable ordering of the members of the q-clan)
the map t ↦→ At is additive, i.e., At + As = At+s for all s, t ∈ F .
Proof. Throughout the proof C denotes a normalized q-clan. First suppose
that the map t ↦→ At is additive. Then it is a routine exercise to show that
for each x ∈ F , the map
θx : (α, c, β) :↦→ (α, c + αAxα T , β + αKx)
induces a symmetry about A ∗ (∞) mapping the line [A(t)] to the line [A(t+x)]
for each t ∈ F .
For the converse, let θ be a collineation of GQ(C) that is a symmetry about
A ∗ (∞), i.e., θ fixes each point collinear with A ∗ (∞), i.e., it fixes each element
of A ∗ (∞) and each point on the line [A(∞)]. Hence θ fixes both (¯0, 0, ¯0) and
(∞) and must therefore be in G0. Clearly θ must fix the line [A(∞)], so in
fact θ ∈ H. Then by the F.T. θ = θ(u, B, σ, π) for appropriate u, B, σ and π.
Since θ must fix each element of A ∗ (∞), we have that for all c ∈ F, β ∈ F 2 ,
(¯0, c, β) = (¯0, uc σ , uβ σ B −T ). This quickly forces u = 1, σ = id, B = I. So it

626 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
remains only to determine π : t ↦→ ¯t. Suppose θ : [A(0)] ↦→ [A(x)] for some
x ∈ F , i.e., ¯0 = x. Apply θ to an arbitrary element of A(t), which must give
an element of A(¯t).
θ : (α, αAtα T , αKt) ↦→ (α, αAtα T + αAxα T , αKt + αKx) ∈ A¯t. This
quickly forces α(At+Ax)α T = αA¯tα T for all α ∈ F 2 , t ∈ F . Hence At+Ax ≡
A¯t for all t ∈ F . If q is odd, then At is symmetric with xt = t, and if q is
even, then At is upper triangular with yt = t. In either case it now follows
that t + x = ¯t. Hence in all cases At+x = At + Ax.
13.14 Property (G) at a second point of ∞ ⊥
The following result appears in [PT91].
Theorem 13.14.1. A flock GQ has Property (G) at some point of (∞) ⊥ \
{(∞)} if and only if it is classical, i.e., it is isomorphic to H(3, s 2 ).
Proof. Recall that a grid Γ is said to cover a point p provided p is on two lines
of Γ. And S = GQ(C) has Property (G) at the point p provided each 3 × 3
grid that covers p can be extended (uniquely, of course) to a (q + 1) × (q + 1)
grid.
First suppose that S has Property (G) at some point p not collinear with
(∞). Then under the action of the elation group G, S has Property (G) at
all points not collinear with (∞). Then if z is any point collinear with but
different from (∞), clearly each 3 × 3 grid covering z must also cover some
point not collinear with (∞) and hence be extendable to a (q + 1) × (q + 1)
grid, forcing S to have Property (G) at z. Hence it suffices to show that if
S has Property (G) at some point z ∈ (∞) ⊥ \ {(∞)}, then C is a classical
q-clan. Moreover, since we can use a shift-flip to recoordinatize C, without
loss of generality we may suppose that z is on the line [A(∞)].
First, recall that the classical GQ are those associted with a q-clan of
the form C ′
= A ′
t at
t =
: t ∈ F . Here a, b are constants for which
0 bt
x 2 + ax + b is irreducible over F . Secondly, by the remarks above, under the
action of G we may assume that z = A ∗ (∞)(¯0, 0, ¯0). Finally, by reindexing
the members of C, we may assume that xt = t for all t ∈ F and A0 =
0 0
0 0
.

13.14. PROPERTY (G) AT A SECOND POINT OF ∞ ⊥ 627
Put α = (−1, 0), γ = (0, −1) ∈ F 2 . For each t ∈ F put βt = (t, zt). Then
{A(∞), A(∞)(α, 0, ¯0), A(∞)(γ, 0¯0)} is a triad of lines with q + 1 transversals
as indicated in Figure 13.6.
By Property (G) at z, there must be (q + 1) − 3 = q − 2 values of
δ = (d1, d2) ∈ F 2 \ {(0, 0), α, γ} for which some line through A ∗ (∞)(δ, 0, ¯0)
meets each of the lines A(t)(¯0, 0, βt), t ∈ F . A(∞)(δ, 0, ¯0) meets A(t)(¯0, 0, βt)
at the point (δ, δAtδ T , δKt + βt) on the line
A(∞)(δ, δAtδ T , δKt + βt), which means that for the appropriate δ these lines
are the same for all t ∈ F . A little computation shows that this is the case
if and only if
δ(As − At)δ T + δ(β T t ) = 0 (13.48)
for all s, t ∈ F .
Put t = 0, s = 0, and recall βt = (t, zt), to obtain
(d 2 1 + d1)s + d1d2ys + (d 2 2 + d2)zs = 0 (13.49)
for all s ∈ F . For q ≤ 4 all the GQ are classical by Corollary 13.6.9. So we
assume that q ≥ 5. We have d2 1 + d1 = e2 1 + e1 iff d1 = e1 or d1 = −e1 − 1.
If d2 = 0 in Eq. 13.49, then (d2 1 + d1)s = 0 for all s ∈ F , so d1 = 0 or
d1 = −1. This says δ ∈ {(0, 0), α}, impossible. Similarly, d1 = 0. Since there
must be at least three choices of δ (= (0, 0), α, γ) satisfying Eq. 13.49 for all
s ∈ F , it will follow that ys and zs must each be just a constant times s,
completing the proof. For suppose (d1, d2), (e1, e2), (f1, f2) are three choices
of δ (= (0, 0), α, γ) satisfying Eq. 13.49 for all s ∈ F . Put
⎛
d
M(d, e, f) = ⎝
2 1 + d1 d1d2 d2 ⎞
2 + d2
+ e2 ⎠ .
e2 1 + e1 e1e2 e2 2
f 2 1 + f1 f1f2 f 2 2 + f2
For each s ∈ F , (s, ys, zs) is in the right null space of M(d, e, f). So the rank
rk(M(d, e, f)) is 1 or 2. If rk(M(d, e, f)) = 2, then xt = t, yt = at, zt = bt,
completing the proof.
Now suppose
rk(M(d, e, f)) = 1 for all choices of three values of δ. (13.50)
If d1 = 0, then d2 ∈ {0, −1} from Eq. 13.49. Similarly, if d2 = 0, then
d1 ∈ {0, −1}. If d1 = −1, then d2 = 0 by Eqs. 13.49 and 13.50. From

628 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
Eq. 13.50 it also follows that for the q − 2 elements δ (= (0, 0), α, γ) the d ′ 1s are the q − 2 of F \ {0, −1}.
First let q be odd. then we may choose d1 = 1. But then Eq. 13.50
forces d2 = −2, e1 + e2 = −1, f1 + f2 = −1. Hence always d2 = −d1 − 1.
From Eq. 13.49 it follows that s − ys + zs = 0, so zs = ys − s. Then
y2 s−4xszs = y2 s−4s(ys−s) = (ys−2s) 2 , which is a square in F , a contradiction.
Next let q = 2e . Then we may choose e1 = d1 + 1, with d1 arbitrary in
F \ {0, 1}. But then Eq. 13.50 forces d2 to be d1 + 1. From Eq. 13.49 it
follows that s + ys + zs = 0, so zs = ys + s. Then zsxs/y 2 s = s(ys + s)/y 2 s
. The
equation x 2 + x + s(ys + s)/y 2 s = 0 has the solution x = s/ys, a contradiction.
Hence always rk(M(d, e, f)) = 2 and S is classical.
13.15 An additional regular pair of points
In studying flock GQ with parameters (q 2 , q) that have subquadrangles of
order q, J. A. Thas in [Th01c], and J. A.Thas and K.Thas in [JTKT06] show
that flock GQ having a regular point in (∞) ⊥ \ {(∞)} must be classical if
q is even, and must be the point-line dual of a translation GQ when q is
odd. Their proofs are a bit heavy and apparently only work for flock GQ. In
Section 9.17 of this book we give a general construction of STGQ. In [Pa89]
this model is used to study certain families of GQ and some computations
are given that could possibly lead to a proof that a flock GQ with a regular
point in (∞) ⊥ \ {(∞)} is the point-line dual of a TGQ (so classical when q
is even). This subject is of interest to the author but does not especially fit
within the present book.
13.16 Flocks Whose Planes Contain a Common
Point
The three theorems of this section are all taken from J. A. Thas [Th87].
The first theorem deals with the case q even and depends on Theorem
8.3.6 (D.G. Glynn [Gl84]). (Of course by now we have seen the much
stronger result Theorem 11.6.2 by M. Brown [Br00a] saying that if even one
of the ovals that is a plane section of the ovoid Ω is a conic, then Ω must be
an elliptic quadric.

13.16. FLOCKS WHOSE PLANES CONTAIN A COMMON POINT 629
Theorem 13.16.1. Let q = 2 e and let F = {C1, . . . , Cq} be a flock of the
cone K projecting an irreducible conic of the plane P G(2, q). If the planes
π1, . . . , πq, with πi ⊃ Ci all contain a common point, then F is linear.
Proof. Since all flocks are linear if q ≤ 4, we may assume that q > 4. Let the
cone K be embedded in the Klein quadric H of P G(5, q). If π∗ i is the polar
plane of πi with respect to H5 and if π∗ i ∩ H5 = C∗ i , then C∗ 1 ∪ C∗ 2 ∪ . . . ∪ C∗ q =
O∗ is an ovoid of H5 (see the proof of the Thas-Walker Construction in
Section 12.11). By the proofs of Lemma 13.6.9 and Theorem 13.6.10 the
planes π∗ 1 , . . . , π∗ q contain a common tangent line L of H5 passing through
the vertex V of K. By hypothesis the planes π1, . . . , πq contain a common
point y, so the planes π∗ 1, . . . , π∗ q are contained in the polar space P G(4, q)
of y with respect to H5. Here y is the nucleus of the parabolic quadric
P4 = P G(4, q) ∩ H5 (because y⊥ = P G(4, q) and y ∈ H5), so that no line
through y contains two points of H5. Clearly the points of O∗ all lie in P4
and hence form an ovoid of P4. So each line of P4 is on exactly one point of
O∗ .
Now let z1, z2 be any two points of O∗ and let C∗ = P4 ∩ 〈y, z1, z2〉. C∗ will meet O∗ in at least the two points z1, z2, and, of course, it meets P4
in a conic. Put V ∗ = C∗ ∩ O∗ and v = |V ∗ |. Count in two different ways
the ordered pairs (u, z) with u ∈ O∗ \ V ∗ , z ∈ C∗ \ V ∗ , and uz a line of P4.
There are q2 + 1 − v choices for u, and the perp in P4 of u contains y. Hence
u⊥ ∩ 〈y, z1, z2〉 must be a tangent line to C∗ so contain a unique point of C∗ .
But of course this point of C∗ cannot be in O∗ itself. Hence for each u there
is a unique z. On the other hand there are q + 1 − v choices for z, each on
q +1lines of P4, each of which must contain a point of O∗ but cannot contain
a point of C∗ . This implies
q 2 + 1 − v = (q + 1 − v)(q + 1) =⇒ v = 2.
This says that 〈y, z1, z2〉 contains only the points z1, z2 of O ∗ . Now project
O ∗ from y onto a solid P G(3, q) contained in y ⊥ = P G(4, q). Since no line
through y can contain two points of O ∗ , the image O ∗∗ is a set of q 2 + 1
points of P G(3, q). If there were three points of O ∗∗ on a line m, that would
mean there were three points of O ∗ in the plane 〈m, y〉, something we just
showed was impossible. Hence O ∗∗ is an ovoid of P G(3, q) (since q > 2).
Remember that O ∗ is the union of q conics C ∗ 1, . . . , C ∗ q , meeting pairwise
at just the point V and having a common tangent line L there. Under the
, . . . , C∗∗ having a
projection to O ∗∗ these conics are mapped to conics C ∗∗
1
q

630 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
common tangent line, the image of L. (Since y ∈ πi and πi ∩ π ⊥ i
= ∅, y ∈ L.)
Hence O ∗∗ is an ovoid of P G(3, q) with a pencil of conics, so O ∗∗ must be
an elliptic quadric (by Theorem 8.4.1) But the inverse projection maps these
conics back to conics, i.e., O ∗ is the union of conics. Fix two of the conics
of O ∗∗ , say C ∗∗ and C ′∗∗ , sharing a common secant line. Their preimages
back in P4 span a 3-space Σ that contains the preimage of any conic of O ∗∗
meeting C ∗∗ and C ′∗∗ each in two points. It is possible to show that each
one of these conics of O ∗ is contained in that same 3-space Σ. Hence O ∗ is
an ovoid of a 3-space with conics as plane sections and therefore must be an
elliptic quadric. Then the perp Σ ⊥ of Σ with respect to H5 must be a line
contained in the planes of the original flock.
Clearly the same result holds for any cone projecting a hyperoval O ∗ which
contains an irreducible conic, and also for any cone projecting a pointed conic.
Now let q be odd and recall that Segre’s theorem says that each oval of
P G(2, q) is an irreducible conic. So let C be an irreducible conic of P G(2, q)
and let K be the cone with vertex V ∈ P G(3, q) \ P G(2, q) and base C. Let
F = {π1, . . . , πq} be a flock of K with Ci = πi ∩ K.
Theorem 13.16.2. If the planes π1, . . . , πq contain a common interior point
x of K, then F is linear.
Proof. Let x0x1 = x 2 2 be the equation of the cone K, and let πi have coordinates
πi = [ai, bi, ci, 1]. Recall that the condition for the planes to form a
flock is that
(ci − cj) 2 − 4(ai − aj)(bi − bj) = ❆ ∈ Fq whenever i = j.
Since the automorphism group of P G(3, q) leaving K invariant is transitive
on the interior points of K, we may assume that x = (1, −m, 0, 0), with m a
given nonsquare of Fq. As x is contained in πi, we have ai = mbi, 1 ≤ i ≤ q.
Let Fq2 be the field obtained by adjoining a root ω of x2 = m to Fq. If
α ∈ Fq2, then the conjugate of α with respect to Fq is denoted by α. To the
q planes πi we let correspond the q elements ci + 2biω = αi of Fq
2. We claim
that αi − αj, i = j, is never a square in F q 2. For suppose that with some
distinct fixed i and j αi − αj is a square in F q 2, say αi − αj = (r + sω) 2 with

13.16. FLOCKS WHOSE PLANES CONTAIN A COMMON POINT 631
r, s ∈ Fq. Then we have
(ci − cj) 2 − 4(ai − aj)(bi − bj) = (ci − cj) 2 − 4(bi − bj) 2 m
= (ci − cj) 2 − 4(bi − bj) 2 ω 2 =
(ci − cj + 2(bi − bj)ω) × (ci − cj − 2(bi − bj)ω)
= (αi − αj)(αi − αj) = (r + sω) 2 (r − sω) 2 = (r 2 − s 2 m) 2 .
Hence (ci − cj) 2 − 4(ai − aj)(bi − bj) is a square in Fq, contradicting the
condition for a flock. So αi − αj is never a nonzero square in Fq2. Let v be a given nonsquare of Fq2. The set W consisting of the q elements
v(αi − α1) contains 0 and has the property that δ − γ is a square whenever
δ, γ ∈ W . Thus the theorem (Cor. 20.25.6) of Blokhuis says that W = {dα :
d ∈ Fq} where α is some fixed nonzero square of Fq2. Hence v(αi−α1)/α ∈ Fq
for 1 ≤ i ≤ q. Suppose v/α = u + u ′ ω for u, u ′ ∈ Fq, with (u, u ′ ) = (0, 0).
Then in the product (u + u ′ ω)(αi − α1), the coefficient on ω must equal zero.
It follows that u ′ ci + 2biu = u ′ c1 + 2b1u, 1 ≤ i ≤ q. Consequently each plane
πi = [mbi, bi, ci, 1] contains the point (0, 2u, u ′ , −u ′ c1 − 2b1u). Thus the q
planes πi all contain the line 〈(1, −m, 0, 0), (0, 2u, j ′ , −u ′ c1 − 2b1u)〉, i.e., the
flock is linear.
Theorem 13.16.3. Let F = {π1, π2, . . . , πq} be a flock of the quadratic cone
K and suppose that the planes πi all contain a common exterior point of K.
If m is a given nonsquare of Fq, then the coordinates can be chosen so that K
has equation x0x1 = x 2 2 and the plane πi has equation aix0 − ma σ i x1 + x3 = 0,
{a1, . . . , aq} = Fq and σ ∈Aut(Fq). Conversely, given a nonsquare m of Fq,
the planes πi with equation aix0 − ma σ i x1 + x3 = 0, {a1, . . . , aq} = Fq and
σ any automorphism of Fq, define a flock F of the cone K with equation
x0x1 = x 2 2. Moreover, the planes πi all contain the exterior point (0, 0, 1, 0)
of K. Finally, the flock is linear iff σ = 1.
Proof. Consider the cone K with equation x0x1 = x 2 2 and the q planes πi =
[ai, −ma σ i , 0, 1] where {a1, . . . , aq} = Fq. Let m be a given nonsquare of Fq
and let σ be any automorphism of Fq. All these planes contain the exterior
point (0, 0, 1, 0) of K. Since −4(ai − aj)(−ma σ i + maσ j ) = 4m(ai − aj) σ+1 is
a nonsquare whenever i = j, the planes πi define a flock of K. It is now a
routine exercise to compute the line
πi ∩ πj = 〈(0, 0, 1, 0), (m(a σ j − aσ i ), aj − ai, 0, −m(aia σ j − aσ i aj))〉.

632 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES
WLOG we may assume the planes have been indexed so that a1 = 0 and
a2 = 1. Then let j ∈ {3, 4, . . . , q}. So
π1 ∩ π2 = 〈(0, 0, 1, 0), (m, 1, 0, 0)〉; π1 ∩ πj = 〈(0, 0, 1, 0), (ma σ j , aj, 0, 0)〉.
These are the same line if and only if there is a nonzero λ ∈ Fq for which
λ(m, 1, 0, 0) = (ma σ j , aj, 0, 0) ⇐⇒ a σ j = aj.
Hence the flock is linear if and only if σ = 1.
Conversely, consider any flock F = {π1, . . . , πq} of a quadratic cone K
where the planes of the flock contain a common exterior point x of K. Coordinates
can be chosen in such a way that x = (0, 0, 1, 0), that π1 = [0, 0, 0, 1],
and that K has equation x0x1 = x2 2 . Then the plane πi = [ai, bi, 0, 1], with
a1 = b1 = 0. In the section showing that flocks of a quadratic cone and
q-clans are equivalent objects we showed that i ↦→ ai and i ↦→ bi must both
be permutations, so that the map θ : ai ↦→ bi, 1 ≤ i ≤ q, is a permutation of
the elements of Fq (with 0θ = 0). Also, the condition for F to be a flock is
that −(ai − aj)(aθ i − aθj ) be a nonsquare whenever i = j. Putting aj = 0 and
ai = 1, we see that −1θ is a nonsquare. Hence ((aθ i /1θ ) − (aθ j/1θ ))/(ai − aj)
is a nonzero square whenever i = j. Let σ : ai ↦→ (aθ i /1θ ), 1 ≤ i ≤ q.
Then σ is a permutation of the elements of Fq for which 0σ = 0 and 1σ = 1.
Moreover, (aσ i − aσ j )/(ai − aj) is a nonzero square whenever i = j. By a
theorem of Carlitz [Ca60] (see Theorem 20.26.1) σ is an automorphism of
Fq. Notice that the plane πi = [ai, 1 θ a σ i
, 0, 1]. Now let m be any given
nonsquare of Fq and consider the following transformation of coordinates:
x0 = x ′ 0 ; x1 = (−m/1 θ )x ′ 1 ; x2 = sx ′ 2 ; x3 = x ′ 3 , with −m/1θ = s 2 . Then K
still has equation x0x1 = x 2 2 and πi = [ai, −ma σ i , 0, 1], 1 ≤ i ≤ q, completing
the proof.

13.16. FLOCKS WHOSE PLANES CONTAIN A COMMON POINT 633
[A(∞)]
✉
A∗ (∞)(γ, 0, ¯0)
✉
A∗ (∞)(α, 0, ¯0)
✉
A∗ (∞)(¯0, 0, ¯0)
A(0)
✉
(γ, 0, ¯0)
✉
(α, 0, ¯0)
✉
(¯0, 0, ¯0)
A ∗ (0)
✉
(γ, γAtγ T , γKt + βt)
✉
(α, αAtαT , αKt + βt)
✉
(¯0, 0, βt)
A(∞)(γ, 0, ¯0)
A(∞)(α, 0, ¯0)
A(∞)
[A(0)]
✉❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵
✉
(∞)
✉ [A(t)]
A
❵
∗ (t)(¯0, 0, βt)
A(t)(¯0, 0, βt)
Figure 13.6: Property (G) at A ∗ (∞)(¯0, 0, ¯0)

634 CHAPTER 13. Q-CLANS AND THEIR GEOMETRIES

Chapter 14
Finite Circle Geometries
14.1 The Circle Geometries
In this chapter we are interested only in the finite circle geometries, and then
primarily in the finite circle planes, so we could give axioms for the various
circle geometries in terms of their parameters. However, we prefer to give
the basic axioms and let the reader choose whether to prove the basic results
on the parameters or take them as the starting point.
The study of circle planes, sometimes referred to as Benz planes (see
[Be37]), was originally motivated by the study of quadrics in the projective
space P G(3, q) over a field F . In particular, we will be studying the following
three quadrics: the elliptic quadric in P G(3, q), the quadratic cone formed
by taking a cone with vertex V and base some conic in a plane not containing
V , and the hyperbolic quadric. The points, lines and plane sections of each
of these quadrics yields a geometry that will serve as a prototype for a circle
plane with 0, 1 or 2 parallel classes of lines. First, any three points of the
quadric, no two on a line, span a plane that intersects the quadric in an
irreducible plane quadric. Secondly, given an irreducible plane section C of
a quadric Q and a point P ∈ C, there is a unique line ℓ tangent to C at
P in the plane of C. Given any point Q ∈ Q \ C not collinear with P , the
plane 〈Q, ℓ〉 meets Q in the unique irreducible plane section of Q containing
Q and touching C at P . Thirdly, if the quadric Q contains lines, then every
line meets every irreducible plane section of Q in a unique point. Further,
the lines occur in rulings where the lines in a ruling partition the points of
Q, and any two lines in distinct rulings intersect in a unique point.
635

636 CHAPTER 14. FINITE CIRCLE GEOMETRIES
These geometric properties of the quadrics are formalized in the definition
of a circle plane, where the concept of a circle takes the place of an irreducible
plane section of a quadric.
Definition 14.1.1. A circle plane S = (P, L, C) is an incidence structure
of points, lines and circles, respectively, such that the following axioms are
satisfied:
C1 Three pairwise non-collinear points are incident with a unique circle.
C2 (the “touching axiom”) For any circle C, point P ∈ C and point Q
not collinear with P such that Q is not on C, there is a unique circle D
incident with Q and touching C at P (i.e., C and D are incident with
exactly one common point P ).
C3 The lineset L is partitioned into parallel classes. Each point is incident
with exactly one line in each parallel class, and each line and each circle
have a unique point in common.
C4 Two lines in different parallel classes intersect in a unique point.
C5 Each circle is incident with at least three points, and there exist more
than one circle.
The touching axiom has the following interesting consequence. Suppose
that P is a point of the circle C, and that C1 and C2 are two circles both
touching C at P . If there were a point Q on both C1 and C2 with Q = P ,
then the touching axiom would be violated. Hence the circles touching C
at P pairwise touch each other at P , so effectively partition the points not
collinear with P . More generally, the circles on a fixed point P are partitioned
into sets which touch pairwise at P and partition the points of S not collinear
with P . Such a set of circles is called a pencil, and P is called the base point
of the pencil.
A circle plane with 0, 1 or 2 parallel classes of lines is called an inversive
(or Möbius) plane, a Laguerre plane, or a Minkowski plane, respectively.
In addition to the quadrics in P G(3, q), there are other sets of points that
may be used to define circle planes. The main one of interest to us in the
next section is the ovoid.
In the quadric models for circles planes there is a local projective structure
in the sense that for any point P on the quadric, the quotient geometry

14.1. THE CIRCLE GEOMETRIES 637
P G(3, q)/P is a projective plane. If we also remove the plane tangent to the
quadric at P , then in the quotient space we have an affine plane. In this
context, a tangent plane meets the quadric in just the point P in the elliptic
case, a line in the quadratic cone case and two lines intersecting at P in the
hyperbolic case. For example, in the elliptic quadric case, the “points” of
the quotient space, which are in general the lines of P G(3, q) through P not
in the tangent plane at P , may be identified with the points of the quadric
different from P and not collinear with P . The “lines” of the quotient space,
which are in general the planes of P G(3, q) passing through P but different
from the tangent plane at P , may be identified with the circles through P .
Then it is clear that this quotient structure is an affine plane of order q. This
local or internal structure may be generalized to arbitrary circle planes.
Definition 14.1.2. Let S = (P, L, C) be a circle plane. Then the derived
plane at P ∈ P is the point-line geometry SP with points, lines and incidence
as follows:
• Points: points of S not collinear with P .
• Lines: circles incident with Φ in S and lines of S not incident with P .
• Incidence: inherited from the incidence in S.
Since three pairwise non-collinear points define a unique circle in S we
have that two points define a unique line in SP (whether they be collinear
or non-collinear in S). The touching axiom for S implies that the circles
of S incident with P fall into parallel classes as lines of SP . The lines of
S not incident with P form a parallel class of lines in SP . Note also that
any circle of S not incident with P meets any circle incident with P in at
most two points (by Axiom C1). Hence we have the following important and
well-known result:
Theorem 14.1.3. Let S = (P, C, L) be a circle plane and P ∈ P.
1. SP is an affine plane, and we denote its projective completion by SP .
2. If C is a circle of S not incident with P , then in SP the points of
SP incident with C (and hence not collinear with P ) plus the point of
SP that is the parallel class of SP corresponding to the lines of S not
incident with P , form an oval of SP .

638 CHAPTER 14. FINITE CIRCLE GEOMETRIES
14.2 Finite Inversive Planes
Recall that a circle plane with no line was called an inversive plane. We
repeat the axioms for such a geometry.
An inversive plane or Möbius plane I = (P, C, I) (here “I” denotes the
incidence relation) is a point-block incidence structure on a set P of points
whose blocks, i.e., members of C, are called circles, and satisfying the following
conditions:
I(i) Every three distinct points of P are incident with a unique common
circle C ∈ C.
I(ii) For each triple (C, P, Q) where C is a circle, P is a point incident
with C and Q is a point not incident with C, there is a unique circle D
incident with both P and Q, but with no point incident with C other than
P .
I(iii) There exists a non-incident point-circle pair, and every circle is
incident with at least three points.
Two circles are called disjoint, tangent, or secant according as they have
0, 1 or 2 common points, respectively. Any subset of the points of a circle
is called a concircular set. For any two distinct points P and Q the set of
all circles through both P and Q is called the bundle with carriers P and
Q. A maximal set of mutually tangent circles through a point P is called a
pencil with carrier P . Using (i) and (ii) it is easy to show that any bundle
or pencil covers the points of I.
A flock F of I is a set of pairwise disjoint circles such that, with the
exception of precisely two points, called the carriers of F, every point of I
lies on a (necessarily unique) circle of F. For a finite inversive plane, i.e., one
with a finite set of points, it is easy to verify that there is a positive integer
q called the order of I, such that
1. Each circle contains exactly q + 1 points.
2. |P| = q 2 + 1.
3. |C| = q(q 2 + 1).
4. Each point is on exactly q(q + 1) circles.
5. Each pencil contains exactly q circles.
6. A flock contains exactly q − 1 circles.

14.2. FINITE INVERSIVE PLANES 639
In the finite case a circle can be thought of as a subset of q + 1 points
of I. Then a definition equivalent to that just given comes from the idea
of the derived geometry IP of I at the point P . Here IP is the structure
IP = (P ′ , C ′ , I ′ ) where P ′ = P \ {P }, C ′ = {C ∈ C : P IC}; and I ′ is induced
by I. Then I is an inversive plane of order q if and only if IP is an affine
plane of order q for each point P ∈ P.
The most common way to construct an inversive plane I is as follows.
Construction 14.2.1. (Ovoidal or egglike inversive plane) Let Ω be an ovoid
of P G(3, q). Put P = Ω. Each of the q 3 + q planes of P G(3, q) not tangent
to Ω must meet Ω in an oval which we call a circle. These circles are the
circles of I and incidence I is just that inherited from incidence in P G(3, q).
It is clear that any three points of Ω lie on a unique circle. So suppose
that P is a point of Ω contained in the circle C = π ∩ Ω for some cutting
plane π. Suppose that Q is a point of Ω not contained in C, so not in π.
The tangent plane πP at P intersects π in a line L such that L ∩ O = {P }.
Then the plane π ′ = 〈L, Q〉 is not tangent to O and hence intersects O in a
circle D. Since π ∩ π ′ = L, C and D have no point in common other than
P . If D ′ is any other circle through P and Q, tangent to C, then D ′ lies in a
plane τ, such that τ ∩ π is a line in π intersecting O in {P }, and thus lying
in the tangent plane at P . Hence τ ∩ π = L; τ = π ′ , and D = D ′ . There is
therefore a unique circle D through P and Q and tangent to C, establishing
the second axiom for inversive planes. The third axiom is clearly satisfied,
so we have an inversive plane.
Such an inversive plane is called ovoidal or egglike. A celebrated theorem
of Dembowski says that each finite inversive plane with even order must be
ovoidal. Later we will see a proof of this theorem found by J. A. Thas using
the theory of generalized quadrangles.
There are other models of inversive planes equivalent to the one just
given, but which look quite different. In fact, in this chapter we will develop
properties of I using a model presented by W. F. Orr [Or73].
Construction 14.2.2. (Bruck [Br69]) Let S be a regular line-spread of
P G(3, q). So S is a set of q 2 + 1 pairwise skew lines of P G(3, q) such that
if L0, L1, L2 are any three distinct lines of S, then the lines of the regulus
R(L0, L1, L2) are all contained in S. Construct a point-circle geometry IB
as follows. The points of IB are the lines of S. The circles of IB are the
reguli contained in S. Then it is clear that each three points of IB lie in a

640 CHAPTER 14. FINITE CIRCLE GEOMETRIES
unique circle. For a regulus R contained in S, a line L of R, and a line M
of S not in R, there are q + 1 reguli containing L and M, q of which meet R
in one further line. Hence there is a unique regulus containing L and M and
containing no further line of R. So axioms I(i) and I(ii) are both satisfied.
And I(III) is clearly satisfied. Indeed, the properties 1. – 6. listed above are
easily verified.
Construction 14.2.3. (Projective sublines) For this construction of I =
(P, C, I) let the points be the elements of Fq2 ∪ {∞}, which may be thought
of as the points of the projective line P G(1, q2 ). Then P G(1, q) = Fq ∪ {∞}
is a projective subline of order q. The circles of C may be described as all the
images of P G(1, q) under maps of the form x ↦→ ax+b,
where ad − bc = 0.
cx+d
Construction 14.2.4. This model was developed in detail in the Ph.D. thesis
of W. F. Orr [Or73].
Let F = Fq ⊂ F q 2 = E, q any prime power. Let M ′ be the collection
of all 2 × 2 matrices with entries in E. Define g : M ′ → M ′ : A ↦→ A ′ by
the following: If A = (aij), then A ′ = (aij q ). Let A ∗ be the classical adjoint
a b
matrix of the matrix A. Hence if A =
c d
Note that (A ∗ ) ∗ = A and (AB) ∗ = B ∗ A ∗ .
, then A ∗ =
d −b
−c a
Let M0 be the subset of M ′ consisting of all A in M ′ for which A ∗ = −A ′ ,
i.e., d = −a q , b = b q , c = c q , so b, c ∈ F . It follows easily that M0 is a vector
space over the field F , and
M0 =
x a
b −x q
Fix y ∈ E \ F . Then the set of four matrices
1 0
0 −1
,
0 0
1 0
,
: a, b ∈ F ; x ∈ E .
0 1
0 0
;
y 0
0 −y q
is a basis for the vector space M0 over F . Let M denote the corresponding
projective space isomorphic to P G(3, q). For A = 0 in M0, let 〈A〉 represent
the element of M containing A. The points, lines, and planes of M are,
respectively, the 1-, 2- and 3-dimensional subspaces of M0.
Define a mapping from M0 onto F by
;
A ↦→ A = det(A) = −x q+1 − ab (14.1)
.

14.2. FINITE INVERSIVE PLANES 641
So A ↦→ Q(A) = A is a quadratic form on M0 as a vector space over F .
Hence it determines a quadric surface O in M consisting of all elements 〈A〉
of M for which A = 0. Associated with the quadratic form is the bilinear
form
and the associated scalar matrix,
h(A, B) = A + B − A − B, (14.2)
h(A, B)I = (A + B)(A + B) ∗ − AA ∗ − BB ∗
where I is the 2 × 2 identity matrix.
= AB ∗ + BA ∗ , (14.3)
Lemma 14.2.5. The quadric O is an elliptic quadric in M ∼ = P G(3, q).
Proof. Let R be a nonsingular matrix in M ′ , and consider the map A ↦→
RAR ′∗ for A ∈ M0. Note that A∗ = −A ′ is equivalent to A ′∗ = −A. Then
(RAR ′∗ ) ′∗ = (R ′ A ′ R∗ ) ∗ = RA ′∗R ′∗ = R(−A)R ′∗ = −(RAR ′∗ ), so this map
leaves M0 invariant, is a bijection and preserves addition and scalar multiplication
by elements of F . Hence it is an invertible linear map on M0
and induces an homography φR on M. If A ∈ M0 with A = 0, then
RAR ′∗
0 0
= 0, so φR leaves inivariant the quadric O. Let P (0) =
1 0
0 1
and P (∞) = . Then P (0) and P (∞) are singular matrices and
0 0
x a
any singular matrix
b −xq
with any one of a, b, x being zero must be
a scalar times P (0) or times P (∞). Note that if 0 = x ∈ E, then xq+1 ∈ F .
Then it follows that if A is any singular matrix not a scalar multiple of P (0)
or P (∞) it must be of the form
q+1 x −x
A = a
1 −xq
Consider the mapping φR for R =
for 0 = a ∈ F ; 0 = x ∈ E.
1 x
0 1
.

642 CHAPTER 14. FINITE CIRCLE GEOMETRIES
RP (0)R ′∗ =
RP (∞)R ′∗ =
x −x q+1
1 −x q
0 1
0 0
= a −1 A, (14.4)
= P (∞). (14.5)
Thus φR fixes P (∞) and maps P (0) onto A, the general singular element
of M0.
The homography group of M fixing O and 〈P (∞)〉 is thus transitive on
the points of O \{P (∞)}. Rulings of O (if any were to exist) must be carried
into rulings of O by such collineations. Therefore, since a ruled quadric
contains in P G(3, q) exactly two ruling lines through each of its points, any
homography that fixed a point of O would have to fixe the two ruling lines
through that point. Hence O cannot be ruled.
We now have a new model of the ovoidal inversive plane I. The points
of I are the points of O. The circles of I are the non-empty intersections of
O with nontangent planes. The incidence relation is just containment. As
we saw above for general ovoidal inversive planes, I is indeed an inversive
plane.
Associated with the quadric O is a polarity ⊥ of M. In terms of the polar
form h, for a point 〈A〉 of M,
〈A〉 ⊥ = {〈B〉 ∈ M : h(A, B) = 0}. (14.6)
Since h is a nonsingular bilinear form, 〈A〉 ⊥ is a subspace of M of projective
dimension 2. Then ⊥ is defined on planes of M by Eq. 14.6 and the
condition that A ⊥⊥ = A. To each singular matrix P in M0 corresponds the
element 〈P 〉 of O, and thus a point of I. To each nonsingular matrix C in
M0 corresponds an element 〈C〉 of M \O and thus a plane 〈C〉 ⊥ of M, which
intersects O in a circle of I.
If P and C are elements of M0 with P singular and C nonsingular, then
〈P 〉 and 〈C〉 correspond to a point and a circle, respectively, of I. The point
and circle are incident if and only if h(P, C) = 0.
A collineation of an inversive plane I is a bijection of the set of points of
I onto itself which sends concircular sets onto concircular sets. An involution
is a collineation of order 2. An involution which fixes every point of a circle

14.2. FINITE INVERSIVE PLANES 643
〈A〉 is called an inversion with axis A. If there exists an inversion with a
given axis A, then it is unique and it fixes no points other than those of A.
Lemma 14.2.6. Let R be a nonsingular matrix of M ′ , and consider the two
mappings
φR : A ↦→ RAR ′∗ , and φ ′ R : A ↦→ RA′ R ′∗ ,
respectively on the ponts and circles of I, where A ranges over the elements
of M0. Then
(i) φR and φ ′ R are homographies leaving I invariant.
(ii) For a circle 〈C〉 of I, the mapping φ ′ C is inversion with respect to
〈C〉, and 〈A〉 is fixed by φ ′ C , for A in M0, exactly when either h(A, C) = 0
or 〈A〉 = 〈C〉 or both.
Proof. Since RAR ′∗ = R·A·R ′∗, we see that φR preserves singularity
and nonsingularity. Moreover, φR and φ ′ R are both invertible linear maps
from M0 onto M0. For 〈P 〉 in O and 〈C〉 in M \ O, we know that 〈P 〉 is
incident with 〈C〉 if and only if h(P, C) = 0. Then
h(RP R ′∗ , RCR ′∗ ) = R(P + C)R ′∗ − RP R ′∗ − RCR ′∗
= R · h(P, C) · R ′∗
= 0.
Thus φR sends 〈P 〉 and 〈C〉 into an incident point-circle pair. A similar
argument works for φ ′ R , proving (i).
For C ∈ M0, C ′∗ = −C, so the collineation φ ′ C is induced by the mapping
A ↦→ −CA ′ C on M0. If 〈P 〉 is a point of 〈C〉 ⊥ , then
Thus,
h(P, C)I = P C ∗ + CP ∗
= 0.
P C ∗ = −CP ∗
= CP ′ .
So P ↦→ −CP ′ C = −P C∗C = −CP , implying φ ′ C : 〈P 〉 ↦→ 〈P 〉.
Also, (φ ′ C )2 is induced by A ↦→ C(CA ′ C) ′ C = CC ′ AC ′ C = C2A, so
(φ ′ C )2 is the identity mapping on M. Hence φ ′ C is an involution fixing every
point of the circle 〈C〉, i.e., it is inversion with respect to 〈C〉.

644 CHAPTER 14. FINITE CIRCLE GEOMETRIES
In addition, 〈A〉 ∈ M0 is fixed by φ ′ C precisely when CA′ C = aA for some
a ∈ Fq, in which case C 2 A ′ = a 2 A, a 2 = C 2 . This leads to
CA ∗ = −CA ′
= (−CA ′ CC ∗ )C −1
= −aAC ∗ C −1
= ±AC ∗ .
Thus either h(A, C)I = AC ∗ + CA ∗ = 0 or h(A, C)I = AC ∗ + AC ∗ . In the
latter case, AC ∗ = kI for some k ∈ Fq, and AC ∗ C = kC, i.e., CA = kC,
which implies 〈A〉 = 〈C〉, proving (ii).
It is interesting to note that when q is odd the two cases of the preceding
Lemma are mutually exclusive, the first corresponding to 〈A〉 ∈ 〈C〉 ⊥ , and
the second to 〈A〉 = 〈C〉. On the other hand, when q is even, h(C, C) = 0
for all C, so that only the first case occurs. If, for a circle 〈D〉, h(C, D) = 0,
we say that D is orthogonal to C. Clearly, “orthogonality” is a symmetric
relation, and if q is even, all circles are self-orthogonal. The plane 〈C〉 ⊥ ,
where ⊥ is the polarity with respect to O, is the plane consisting of all
points of M which are fixed by φ ′ C and is called the orthogonal plane to 〈C〉
in M.
Lemma 14.2.7. Let L be a line in M. Then the points of L which are not
on O correspond to a bundle, a pencil, or a flock of circles in I according as
L intersects O in 0, 1 or 2 points, respectively. When L is tangent to O, the
point of tangency is the carrier of the pencil. When L is disjoint from O,
then the line L ⊥ intersects O in two points, the carriers of the corresponding
bundle. When L is secant to O, its points of secancy are interchanged by
inversion with respect to the circles corresponding to any of its other points.
Proof. Under the polarity ⊥, each point of L is mapped onto a plane through
the line L ⊥ . A point 〈P 〉 of L on O is mapped onto a plane through L ⊥
tangent to O at 〈P 〉. Thus L contains 0, 1 or 2 points of O according as L ⊥
is in 0, 1 or 2 planes tangent to O, according as L ⊥ contains 2, 1 or 0 points
of O, respectively. We treat the three cases separately.
If L is disjoint from O, then since “elliptic ⊥ hyperbolic = elliptic,” it
must be that L ⊥ meets O in two points 〈P 〉 and 〈Q〉, and each plane through
L ⊥ contains 〈P 〉 and 〈Q〉. Therefore the points of L correspond to the q + 1
circles of the bundle through 〈P 〉 and 〈Q〉.

14.2. FINITE INVERSIVE PLANES 645
If L is tangent to O at 〈P 〉, then L⊥ must also be tangent to O at 〈P 〉,
and any two planes through L⊥ have only the point 〈P 〉 in common on O.
Thus the q points of L other than 〈P 〉 correspond to a pencil of circles of I
tangent at 〈P 〉.
If L is secant to O at 〈P 〉 and 〈Q〉, then L⊥ is disjoint from O and is
contained in the tangent planes to O at 〈P 〉 and 〈Q〉. Thus any two planes
through L⊥ have no point of O in common, and the q − 1 points of L other
than 〈P 〉 and 〈Q〉 correspond to a flock of I. We shall call such a flock linear.
Consider 〈C〉 in L \ O. Then φ ′ C fixes the points of the orthogonal plane to
〈C〉, which contains L⊥ . Therefore φ ′ C must fix L, and it either fixes 〈P 〉 and
〈Q〉 or it interchanges them. But since 〈P 〉 and 〈Q〉 cannot be incident with
〈C〉, they cannot be fixed by φ ′ C . We conclude that φ′ C interchanges 〈P 〉 and
〈Q〉.
We are now in a good position to indicate the close connection between
this model of Orr and the model of Construction 14.2.3
Each point of O can be represented as 〈P1〉 for some P1 ∈ M0 with
det(P1) = 0, i.e.,
P1 =
y a
b −y q
where −ab = y q+1 . Since P1 = 0, a and b cannot both be zero. If b = 0, let
P = b −1 P1. If b = 0 (forcing y = 0), let P = a −1 P1. Then P has one of the
two forms:
or
q+1 x −x
P (x) =
1 −xq P (∞) =
,
, for some x ∈ Fq 2,
0 1
0 0
Thus we have a one-to-one correspondence between points of O and the
elements of the set Fq2 ∪ {∞}, where ∞ is considered formally as a new
symbol.
a b
Let R =
∈ M
c d
′ be nonsingular. The mappings φR and φ ′ R
defined above induce the following mappings on Fq2 ∪ {∞}.
.

646 CHAPTER 14. FINITE CIRCLE GEOMETRIES
and
φR =⇒ x ↦→
ax + b
, (14.7)
cx + d
φ ′ R =⇒ x ↦→ axq + b
cxq , (14.8)
+ d
where we observe the usual conventions ∞/a = ∞; a/∞ = 0; a/0 = ∞;
and ∞q = ∞.
Thus, in particular inversion φ ′ C with respect to the circle 〈C〉 of the form
x a
b −xq
induces the map
φ ′ C =⇒ y ↦→ xyq + a
byq . (14.9)
− xq Thus the circles in Fq2 ∪ {∞} are the fixed points of the mappings of the
form in Eq. reffcircle2 eq 9. One can show that these are just the images of
Fq ∪ {∞} under the mappings of the form Eq. 14.7. (See [Br69].) The group
of mappings φR is thus transitive on theordered triples of distinct points of
q+1 y −y
I. To see this, show that (with P (y) =
1 yq
∈ 〈C〉)
φ ′ C : y ↦→ y ⇐⇒ a + xyq = by q+1 − yx q
x a
⇐⇒ 0 = h
b −xq
q+1 y −y
,
1 yq
= 0.
14.3 Miquelian Inversive Planes of Odd Order
Let q be any power of an odd prime, and otherwise continue to use the
notation of the preceding section. So I is an inversive plane of order q
derived from an elliptic quadric O in P G(3, q).
Let t be a fixed solution to t q = −t = 0. So we may think of 1, t as
a basis of F q 2 over Fq. Since t 2 = −t q+1 ∈ Fq, it is easy to see that t 2 is a
nonsquare in Fq. Recall the bilinear form h from the preceding section and
define
A.B = 1
2 h(A, B), for A, B ∈ M0. (14.10)

14.3. MIQUELIAN INVERSIVE PLANES OF ODD ORDER 647
For the associated scalar matrix,
(A.B)I = 1
2 (AB∗ + BA ∗ ), (14.11)
where I is the 2 × 2 identity matrix.
In particular, we have A.A = A = det(A). We also define the product
A × B = (A.B) 2 − A · B, (14.12)
and note that both these products have values in Fq.
Lemma 14.3.1. Let 〈C〉 and 〈D〉 be distinct circles of I (as defined in the
preceding section). Then 〈C〉 and 〈D〉 are disjoint, tangent, or secant according
as C × D is a nonzero square, zero, or a nonsquare in Fq, respectively.
Proof. Since the group of mappings φR is transitive on the circles of I, we
assume first that
C =
D =
t 0
= tI; where t
0 t
q = −t = 0;
y a
b −yq
.
Note that P (∞) is on 〈D〉 if and only if
0 =
0
0
1 y
0 b
a
−yq ∗
y
+
b
a
−yq = −bI.
0 1
0 0
Firstassume that 〈P (∞)〉 is not on 〈D〉, so b = 0. Then the points
q+1 x −x
P (x) =
1 −xq
of 〈C〉 are exactly those points for which
0 = (P (x).C)I
= 1
2 (P (x)C∗ + CP (x) ∗ =
)
1
2 (P (x) · tI + tI · P (x)∗ =
)
1
2 t · (P (x) + P (x)∗ =
)
1
2 t
q+1 x −x
1 −xq
q −x
+
−1
q+1 x
x
.
∗

648 CHAPTER 14. FINITE CIRCLE GEOMETRIES
Thus x = x q and x ∈ Fq.
Similarly, 〈P (x)〉 is on 〈D〉 if and only if
0 = (P (x).D)I
q+1 x −x
=
1 −xq y a
b −y q
∗
= 1
2 (bxq+1 − yx q − y q x − a)I.
+
y a
b −y q
−x q x q+1
−1 x
Thus for 〈P (x)〉 to be on both 〈C〉 and 〈D〉 we must have x ∈ Fq and
bx 2 −(y+y q )x−a = 0. The discriminant of this equation is d = (y+y q ) 2 +4ab,
which lies in Fq. Thus when 〈P (∞)〉 is not on 〈D〉, 〈C〉 and 〈D〉 have in
common 0, 1 or 2 points according as d is a nonsquare, zero, or a nonzero
square in Fq. On the other hand, notice that
Hence
(C.D)I = 1
2 t(D + D∗ ),
C.D = 1
2 t(y − yq ).
C × D = 1
4 t2 (y − y q ) 2 − t 2 (−y q+1 − ab)
= 1
4 t2 (y + y q ) 2 + 4ab
= 1
4 t2 d.
Since t2 is a nonsquare in Fq, and thus C×D = kd, where k is a nonsquare
in Fq, we have 〈C〉 and 〈D〉 are disjoint, tangent, or secant exactly as C × D
is a nonzero square, zero, or a nonsquare in Fq, respectively.
Now note that 〈P (∞)〉 is on 〈C〉. It is on 〈D〉 if and only if D =
y a
0 −yq
. In this case 〈P (x)〉 is on 〈C〉 and 〈D〉 if and only if x ∈ Fq and
−a = x(y + yq ).

14.3. MIQUELIAN INVERSIVE PLANES OF ODD ORDER 649
If y + y q = 0, then a = 0 and 〈C〉 = 〈D〉. So,
(i) a = 0 =⇒ 〈C〉 = 〈D〉 or 〈C〉 and 〈D〉 are secant at P (∞) and P (0);
(ii) a = 0; y + yq = 0 =⇒ 〈C〉 and 〈D〉 are secant at P (∞) and P (x),
x = −a
;
y + yq (iii) a = 0; y + yq = 0 =⇒ 〈C〉 and 〈D〉 are tangent atP (∞).
In this case C × D = 1
4t2 (y + yq ) 2 = 0, if y + yq = 0;
❆ , if y + yq = 0.
following:
y + y q = 0 ⇐⇒ C × D = ❆ ⇐⇒ 〈C〉 and 〈D〉 are secant;
So we have the
y + y q = 0 ⇐⇒ C × D = 0 ⇐⇒ 〈C〉 = 〈D〉 or 〈C〉 and 〈D〉
This proves the Lemma for the case C = tI.
are tangent at P (∞).
More generally, suppose we have a linear fractional mapping φR taking
C0 to C and D0 to D. Then C = RC0R ′∗ and D = RD0R ′∗ , so that
Therefore,
(C.D)I = 1
2 (CD∗ + DC ∗ )
= 1
2 (RC0R ′∗ r ′ D ∗ ) R∗ + RD0R ′∗ C ∗ 0 R∗ )
= R ′ R 1
2 (C0D ∗ 0 + D0C ∗ 0 )R∗
= R · R ′ · (C0.D0)I.
C × D = (C.D) 2 − (C.C)(D.D) =
= R 2 R ′ 2 (C0.C0) 2 = R · R ′∗ · R · R ′∗ · C0 · D0
= R 2 R ′ 2 (C0 × D0)
= (R q+1 ) 2 (C0 × D0).

650 CHAPTER 14. FINITE CIRCLE GEOMETRIES
Since (R q+1 ) 2 is the square of a nonzero element of Fq, we see that the
square/nonsquare nature of C × D is preserved under the linear fractional
transformations φR. Since this group of transformations is transitive on
circles of I, it follows that the Lemma holds for all pairs of circles.
For two circles in I with q = 2 e , tangency and orthogonality are equivalent
conditions, so that every two circles have a common tangent circle. However,
in the present case (q odd) things are different. We may define the relation
∼ for circles 〈C〉 and 〈D〉 of I by:
C ∼ D ⇐⇒ 〈C〉 and 〈D〉 have a common tangent circle. (14.13)
Lemma 14.3.2. For each odd prime power q, the relation ∼ satisfies the
following:
(i) For all circles 〈C〉 and 〈D〉 of I, C ∼ D if and only if C · D is a
square in Fq. Suppose that CD = = 0. Then if 〈C〉 and 〈D〉 are
tangent at a point 〈P 〉, they have a pencil of common tangent circles at
〈P 〉 and one common tangent at each point of 〈C〉 \ 〈D〉 (respectively,
〈D〉 \ 〈C〉). If 〈C〉 and 〈D〉 are not tangent, they have two common
tangents at each point of 〈C〉 \ 〈D〉 (respectively, 〈D〉 \ 〈C〉).
(ii) The relation ∼ is an equivalence relation, separating the circles of I
into two equivalence classes.
(iii) For 〈C〉 and 〈D〉 distinct nontangent circles in I and C ∼ D, the
relation ∼ separates the set of circles orthogonal to both 〈C〉 and 〈D〉
into two classes of equal size: (1) Those secant to both 〈C〉 and 〈D〉;
and (2) those disjoint from both 〈C〉 and 〈D〉.
(iv) For 〈C〉 and 〈D〉 distinct nontangent circles in I, and C ∼ D, the
relation ∼ separates the set of circles orthogonal to both 〈C〉 and 〈D〉
into two classes of equal size: (1) those secant to 〈C〉 and disjoint from
〈D〉 ; and (2) those disjoint from 〈C〉 and secant to 〈D〉.
Proof. Suppose C ∼ D, so there is a circle 〈E〉 tangent to both 〈C〉 and 〈D〉.
By Lemma 14.3.1 D × E = C × E = 0, i.e.,
(D.E) 2 = D · E; (C.E) 2 = C · E.

14.3. MIQUELIAN INVERSIVE PLANES OF ODD ORDER 651
Multiplying these two equations gives
(D.E) 2 (C.E) 2 = D · C · E 2 ,
so that D · C is a square in Fq.
Conversely, suppose that C · D = k 2 for some nonzero k ∈ Fq. Pick
any point 〈P 〉 of 〈C〉. Then P = P.C = 0. Consider the circle E = C +xP ,
where x is in Fq. We have
C.E = C.(C + xP ) = C = 0;
D.E = (D.C) + x(D.P );
P.E = 0;
E = C = 0.
Therefore, C × E = C 2 − CC = 0, which implies by Lemma 14.3.1
that 〈C〉 is tangent to 〈E〉. Further
E × D = (C + xP ) × D
= ((C + xP ).D) 2 − C + xP D
= (C.D + x(P.D)) 2 − CD
= (C.D) 2 + 2x(C.D)(P.D) + x 2 (P.D) 2 − CD
= x 2 (P.D) 2 + 2x(P.D)(C.D) + (C.D) 2 − CD.
The discriminant of this quadratic expression in x is equal to
4CD(P.D) 2 . If 〈P 〉 is not a point of 〈D〉, then P.D = 0 and the discriminant
is a nonzero square in Fq, implying that the equation E × D = 0
has two distinct solutions, and thus 〈C〉 and 〈D〉 have a common tangent
circle. In fact, if 〈C〉 and 〈D〉 are not tangent, we have shown that they have
exactly two common tangent circles through each point of 〈C〉 which is not
on 〈D〉. If they are tangent circles, they have a pencil of common tangent
circles through their point of tangency, and the above quadratic expression
reduces to
E × D = x(x(P.D) 2 + 2(P.D)(C.D)) = 0
⇐⇒
x = 0 or x = −2(C.D)
,
(P.D)

652 CHAPTER 14. FINITE CIRCLE GEOMETRIES
and thus one of the two distinct solutions is x = 0, E = C, implying that
〈C〉 and 〈D〉 have a unique common tangent circle through each point of
〈C〉 other than their point of tangency. In any case, the proof of Part (i) is
complete.
For part (ii), the relation ∼ is clearly symmetric and reflexive. Transitivity
follows from the fact that the nonzero squares of Fq form a multiplicative
group. Thus ∼ is an equivalence relation separating the circles of I into
two equivalence classes, those with nonsquare norm and those with nonzero
square norm.
Now suppose 〈C〉 and 〈D〉 are disjoint or secant, i.e., C × D = 0. Let 〈E〉
be a common orthogonal circle to 〈C〉 and 〈D〉, i.e., C.E = D.E = 0. Then
by definition of C × E and D × E,
C × E = −CE
and
D × E = −DE.
If C ∼ D, then (C × E)(D × E) is a nonzero square in Fq, and thus the
quantities C × E and D × E are either both squares or both nonsquares in
Fq. So by Lemma 14.3.1 〈E〉 is either secant to both 〈C〉 and 〈D〉 or disjoint
from them both. If C ∼ D, then (C × E)(D × E) is a nonsquare in Fq,
and thus exactly one of the quantities (C × E), (D × E) is a square, and
the other is a nonsquare in Fq. Again, by Lemma 14.3.1 this means that E
is secant to exactly one of the two circles 〈C〉, 〈D〉, and disjoint from the
other. We can pick two circles from differrent equivalence classes on any
linear flock or bundle. Then each of their noncommon points must be in one
of their common orthogonal circles, which must intersect one of them in two
points. Since the two circles have the same number of points, the two classes
of common orthogonal circles must be equal in number. This completes the
proof of parts (iii) and (iv), and hence of the Lemma.
Recall the mapping φR and φ ′ R defined above. For a circle 〈C〉,
C φR = |RCR ′∗
and
= R q+1 · C
C φ′ R = RC ′ R ′∗
= R q+1 · C.

14.3. MIQUELIAN INVERSIVE PLANES OF ODD ORDER 653
Thus the two equivalence classes are either fixed or interchanged by φR
and φ ′ R , according as Rq+1 is a square or a nonsquare in Fq, respectively.
That they may in fact be interchanged is shown by the matrix
R =
k 0
0 1
where Rq+1 = kq+1 is a nonsquare in Fq. The existence of such a k in Fq2 is clear, since the mapping x ↦→ xq+1 sends Fq2 onto Fq. Also, note that for
a circle 〈C〉 of I, inversion φ ′ C with respect to 〈C〉 fixes the two equivalence
classes, since C is in Fq, and Cq+1 = C2 .
Let {Ai} n i=1 be a set of matrices in M0. Define the n × n symmetric
matrix A = (aij) where aij = Ai.Aj, 1 ≤ i, j ≤ n. Since we may take various
representatives for the circle 〈C〉 or the point 〈P 〉 by multiplying by nonzero
elements of Fq, we consider A to be equivalent to any matrix obtained by
multiplying any row of A and the corresponding column of A by a nonzero
element of Fq. In particular, the diagonal elements of A will be multiplied
by nonzero squares in Fq. In this way we can ”normalize” A in the following
way. If Ai and Aj are nonsingular matrices with Ai ∼ Aj, i.e., 〈Ai〉 and
〈Aj〉 are equivalent circles, so that AiAj = = 0, then aii = Ai
and ajj = Aj are both squares or both nonsquares, so we may assume we
have multiplied the appropriate rows and columns of A by the appropriate
elements of Fq to arrange it so that aii = ajj.
Transforming I by φR, where R q+1 is a nonsquare in Fq produces aij ↦→
R q+1 aij and interchanges the two equivalence classes of circles. We may
therefore assume, for any fixed circle 〈Ai〉 of the set, that aii = Ai is a
square in Fq. In particular, we may assume that aii = Ai = 1. Any circle
〈Aj〉 not equivalent to 〈Ai〉 would then have that ajj = Aj = ❆ , for some
fixed nonsquare ❆ .
Suppose A is normalized as described above. Then Ai × Aj = (Ai.Aj) 2 −
AiAj = (aij) 2 − aiiajj. Thus our results in Lemmas 14.3.1 and 14.3.2
for circles 〈Ai〉 and 〈Aj〉 translate as:
,

654 CHAPTER 14. FINITE CIRCLE GEOMETRIES
(i) Ai ∼ Aj ⇐⇒ aii = ajj;
(ii) Ai and Aj are disjoint, tangent, or secant
according as the determinant
−
aii aij
aij ajj
is a nonzero square, zero, or a nonsquare in Fq respectively.
The following lemma will facilitate the construction of sets of circles and
points in I.
Lemma 14.3.3. Let A be a symmetric, nonsingular, 4 × 4 matrix with
entries aij in Fq, q an odd prime power. Then there exists a quadruple
(A1, A2, A3, A4) of distinct nonzero matrices of M0 such that Ai.Aj = aij for
1 ≤ i, j ≤ 4, if and only if the determinant of A is a nonsquare in Fq.
Proof. As before, let t be a fixed element of F q 2 for which t = −t q = 0, so
that t 2 is a nonsquare in Fq. Then the 4-dimensional vector space M0 over
Fq has a basis consisting of the following matrices:
E1 =
E3 =
t 0
0 t
0 1
1 0
1 0
; E2 =
0 −1
0 −1
; E4 =
1 0
Then the 4 × 4 matrix E with entries eij = Ei.Ej is:
⎛
t
⎜
E = ⎜
⎝
2 0
0
0
−1
0
0
0
−1
⎞
0
0 ⎟
0 ⎠
0 0 0 1
;
;
.
(14.14)
and we see that det(E) = t 2 is a nonsquare in Fq.
Clearly the diagonal of E has either one or three nonsquares, depending
on whether −1 is a square or nonsquare.
Suppose that (A1, A2, A3, A4) exists as described above, so that Ai =
4
j=1 uijEj, for 1 ≤ i ≤ 4, where the uij are elements of Fq. Let U = (uij).

14.3. MIQUELIAN INVERSIVE PLANES OF ODD ORDER 655
Compute:
which is equivalent to
aij = Ai.Aj
=
4
=
k,m=1
4
k,m=1
uikEk.Emujm
uikekmujm,
A = UEU T .
Thus det(A) = (det(U)) 2 det(E) = 0, and hence U is nonsingular and det(A)
is a nonsquare in Fq.
Conversely, suppose we are given a symmetric 4 × 4 matrix A with entries
in Fq having a nonsquare determinant. There exists a nonsingular 4 × 4
matrix S such that SAS T is a diagonal matrix with determinant equal to
(det(S)) 2 det(A), i.e., a nonsquare in Fq. So the principal diagonal of this
matrix must contain either exactly one or exactly three nonsquares in Fq. It
contains the same number of nonsquares as those in E, and we may assume
that they occur in the same order and the matrix can be “normalized”as described
above, by multiplying each row and the corresponding column by the
same nonzero element of Fq, which operation is equivalent to multiplying on
the left by a nonsingular matrix and on the right by its transpose. Therefore,
in this case, there is a nonsingular 4 × 4 matrix V such that
E = VAV T
or
A = UEU T , U = V −1 .
On the other hand, if E and SAS T do not contain the same number of
nonsquare diagonal entries, one must contain exactly as many squares as the
other contains nonsquares. Recall that there is an element R of M with
R q+1 a nonsquare in Fq, so that the mapping φR interchanges the two
equivalence classes of circles under ∼. Specifically, if B and C are in M0,
then

656 CHAPTER 14. FINITE CIRCLE GEOMETRIES
(B φR ).(C φR ) = kB.C,
where k is a fixed nonsquare in Fq. Let
Then
and the e ∗ ij
E ∗ i
= EφR
i
=
e ∗ ij = E∗ i .E∗ j
m,n=1
4
gijEj, 1 ≤ i ≤ 4.
j=1
= keij
4
= gijenmgjm, 1 ≤ i, j ≤ 4,
are the entries of a 4 × 4 matrix
E ∗ = kE = GEG T ,
where G is the (nonsingular) 4 × 4 matrix with entries gij. Since k is a nonsquare,
we see that E ∗ has exactly the same number of nonsquare entries on
its main diagonal as E has squares. Hence E ∗ has exactly as many nonsquare
entries on its main diagonal as SAS T has. We proceed as above to order and
normalize the diagonal entries of E ∗ and SAS T , and we obtain a nonsingular
4 × 4 matrix V such that
Hence
Consider the quadruple
VAV T = E ∗ = GEG T .
A = UEU T , with U = V −1 G.
Ai =
4
uijEj; 1 ≤ i ≤ 4.
j=1
We have Ai.Aj = 4
k,m=1 uikekmujm = aij, and thus this is the required
quadruple in M0, completing the proof of the Lemma.
Note that since G and V are invertible, so is U. Then since the Ei form
a basis for M0, so do the matrices Ai, 1 ≤ i ≤ 4.

14.3. MIQUELIAN INVERSIVE PLANES OF ODD ORDER 657
Corollary 14.3.4. Let A1, A2, A3 be three distinct nonzero elements of M0,
and let A0 be the 3 × 3 symmetric matrix with entries aij = Ai.Aj, 1 ≤ i, j ≤
3. Assume that A0 is nonsingular. If we suppose that A0 can be extended
to a 4 × 4 symmetric matrix A with entries aij, 1 ≤ i, j ≤ 4, and whose
determinant is a nonsquare in Fq, then there exists a nonzero element A4 of
M0 such that aij = Ai.Aj for 1 ≤ i, j ≤ 4.
Proof. By Lemma 14.3.3 there must exist a quadruple (B1, B2, B3, B4) of
elements of M0 such that Bi.Bj = aij for 1 ≤ i, j ≤ 4. This implies that
Bi.Bj = Ai.Aj for 1 ≤ i, j ≤ 3. By the proof of Lemma 14.3.3, since
E1, E2, E3, E4 form a basis for M0, so must B1, B2, B3, B4 also form a basis.
In M, each of 〈B1〉, 〈B2〉, 〈B3〉 has an orthogonal plane, and these three
planes must have at least one point of intersection in projective 3-space.
Suppose 〈D〉 is such a point, where D = 4
i=1 xiBi. Thus 〈D〉 is orthogonal
to 〈Bj〉 for 1 ≤ j ≤ 3, and D.Bj = 0 for j = 1, 2, 3.
If x4 = 0, then D = 3
i=1 xiD.Bi = 0, and thus 〈D〉 is a point of M
incident with the three circles 〈Bj〉, 1 ≤ j ≤ 3. These are three homogeneous
equation in three unknowns, x1, x2, x3, which have a nontrivial solution only
when det(A0) = 0, contrary to the assumption of the Corollary.
x4
We may therefore assume that x4 = 0, and therefore that D = x4D.B4 =
4
j=1 aj4xj is nonzero, because if it were zero we would have 4
j=1 xjaij =
0 for all i = 1, 2, 3, 4. But this cannot happen because A is nonsingular.
Therefore 〈D〉 is a circle in M orthogonal to 〈B1〉, 〈B2〉, and 〈B3〉.
Similarly, we see that we may extend (A1, A2, A3) to a basis of M0 and
apply the same process to obtain a mutual orthogonal circle 〈C〉 such that
D = 0, and C.Ai = D.Bi = 0 for 1 ≤ i ≤ 3. The matrices representing,
respectively, the quadruple (〈A1〉, 〈A2〉, 〈A3〉, 〈C〉) and the quadruple
(〈B1〉, 〈B2〉, 〈B3〉, 〈D〉), are:
⎛
⎜
⎝
0
A0 0
0
0 0 0 C
⎞
⎟
⎠ ,
⎛
⎜
and ⎜
⎝
A0
0
0
0
⎞
⎟
⎠
0 0 0 D
.
Lemma 14.3.3 says that the determinants of these matrices, namely det(A0)C
and det(A0)D, must be nonsquares in Fq. Thus CD is a square, and
we may normalize these matrices so that C = D. Since x4 = 0, we may
write B4 as B4 = 3 i=1 yiBi + zD, where y1, y2, y3, z are in Fq.
Put A4 = 3 i=1 yiAi + zC. Then we have for k = 1, 2, 3:

658 CHAPTER 14. FINITE CIRCLE GEOMETRIES
and
Ak.A4 =
A4.A4 =
=
3
i=1
yiaik
3
yiBi.Bk
i=1
= Bk.B4
= ak4,
3
yiAi.A4 + zA4.C
i=1
= B4.B4
= a44.
Thus for all i, j = 1, 2, 3, 4 we have Ai.Aj = aij, and the Corollary is
proved.
Our goal, of course, is to prove Orr’s theorem that says that each flock of
an elliptic quadric in P G(3, q), q odd, must be linear. The next three lemmas
take us nearly there.
Lemma 14.3.5. Let K be a flock of I, i.e., a flock of O in M0, with carriers
〈P 〉 and 〈Q〉 (q odd as always). In particular, K is a set {〈C1〉, . . . , 〈Cq−1〉}
of q − 1 pairwise disjoint circles. Then;
(i) K contains exactly q−1
circles in each of the equivalence classes under
2
∼.
(ii) No circle of K has a tangent circle through both 〈P 〉 and 〈Q〉.
(iii) No three distinct circles of K have a common tangent circle.
(iv) No two distinct circles of K have a common tangent circle through
either 〈P 〉 or 〈Q〉.

14.3. MIQUELIAN INVERSIVE PLANES OF ODD ORDER 659
Proof. If 〈E〉 is a circle through 〈P 〉 but not through 〈Q〉, then 〈E〉 has
q points other than 〈P 〉, an odd number, and each of these points must
be on a circle of K. Since 〈E〉 passes through 〈P 〉, 〈E〉 is not a circle of
K. Thus it intersects each circle of K in at most two points. It cannot
intersect each in an even number of poitns, hence it is tangent to at least
one circle of K, i.e., it is in the same equivalence class under ∼ as at least
one circle of K. We may pick 〈E〉 in either equivalence class, and thus we
conclude that K contains at least one circle in each class. Suppose that
C1 ∼ C2 ∼ · · · ∼ Cn ∼ Cn+1 ∼ · · · ∼ Cq−1, where 1 ≤ n ≤ q − 2. So there
are n circles in one class and q − 1 − n in the other.
Consider the set of tangent circles to 〈C1〉 at one of its points 〈R〉. These
circles, together with 〈C1〉, form a pencil of q circles covering the points of
I. Each of 〈P 〉 must be on a unique circle in this pencil, distinct from 〈C1〉.
Let 〈D〉 be any one of the circles in this pencil, other than 〈Cq〉, and assume
either that 〈D〉 is disjoint from both 〈P 〉 and 〈Q〉 or that it contains both
〈P 〉 and 〈Q〉. Then 〈D〉 contains an even number, q + 1 of points, an odd
number of them distinct from 〈P 〉, 〈Q〉, and 〈R〉. This odd number of points
must be covered by the circles of K other 〈C1〉 (K being a flock). Since 〈D〉
intersects 〈C1〉, it cannot be in K, and thus it intersects each circle of K in
at most two points. An odd number of points cannot be covered by distinct
pairs of points, and therefore 〈D〉 must be tangent to at least one circle 〈Ck〉
of K distinct from 〈C1〉. Hence C1 ∼ D ∼ Ck, implying 2 ≤ k ≤ n.
If there is a circle through 〈P 〉 and 〈Q〉 and tangent to 〈C1〉 at 〈R〉, then
each of the q − 1 tangent circles to 〈C1〉 at 〈R〉 satisfies the above conditions
and thus is tangent to some 〈Ck〉, 2 ≤ k ≤ n.
If, on the other hand, 〈P 〉 and 〈Q〉 lie on distinct tangent circles to 〈C1〉
through 〈R〉, we can only conclude that each of the remaining q − 3 circles of
the pencil has a tangent circle 〈Ck〉, 2 ≤ k ≤ n. We are guaranteed at least
a set of m ≥ q − 3 circles of the pencil, each of which has a tangent circle
among 〈C2〉, · · · , 〈Cn〉, with m = q − 3 if and only if 〈C1〉 has no tangent
circle through 〈P 〉, 〈Q〉 and 〈R〉.
Recall from the proof of Lemma 14.3.2 that each of the circles 〈C2〉, · · · , 〈Cn〉
has exactly two common tangent circles with 〈C1〉 through 〈R〉. Hence
2(n − 1) ≥ m ≥ q − 3, (14.15)
=⇒ 2n ≥ q − 1. (14.16)
We may pick any one of the q − n − 1 ≥ 1 circles of K in the other
equivalence class and a point on that circle, and carry through the same

660 CHAPTER 14. FINITE CIRCLE GEOMETRIES
procedure, which will give us
2(q − n − 1) ≥ q − 1, (14.17)
=⇒ q − 1 ≥ 2n. (14.18)
It follows that n = q−1
, and thus part (i) of the Lemma is proved.
2
But it also follows that equality holds in Eq. 14.15, which means that
〈C1〉 has no tangent circle through 〈P 〉, 〈Q〉, 〈R〉. Since 〈C1〉 was chosen
arbitrarily from K, and 〈R〉 is an arbitrary point of 〈C1〉, it must be the case
the part (ii) holds.
Also, since equality holds in Eq. 14.15 we see that each tangent circle to
〈C1〉 which misses both 〈P 〉 and 〈Q〉 has exactly one other tangent circle
in K and therefore no circle missing both 〈P 〉 and 〈Q〉 has three distinct
tangent circles in K. Similarly, every common tangent circle of 〈C1〉, 〈Ck〉
for 2 ≤ k ≤ n, must miss both 〈P 〉 and 〈Q〉. Therefore no circle through
〈P 〉 or 〈Q〉 has two distinct tangent circles in K, and no circle of I has three
distinct tangent circles in K. This completes a proof of parts (iii) and (iv),
completing a proof of the lemma.
Lemma 14.3.6. (Common tangent circles)
1. Let 〈R〉 be a point incident with neither of two disjoint circles 〈C〉 and
〈D〉. If 〈C〉 and 〈D〉 have a common tangent circle through 〈R〉, then
they have exactly 4 common tangent circles through 〈R〉.
2. If a circle 〈C〉 of I has a tangent circle through the distinct points 〈P 〉
and 〈Q〉 not on 〈C〉, then it has precisely two such tangent circles.
Proof. First suppose that 〈T 〉 is a circle through 〈R〉 and tangent to both 〈C〉
and 〈D〉. Then C ∼ D ∼ T , and we may assume that C = D = T = 1.
The three conditions on T translate as
T.R = 0; T × C = 0; and T × D = 0.
Hence (T.C) 2 = T C = 1 = (T.D) 2 . Since T.C = ±1, we may assume
that T.C = 1 and T.D = ±1. Letting C, D, R, T equal A1, A2, A3, A4,
respectively, we have the 4 × 4 matrix A with entries aij = Ai.Aj for 1 ≤
i, j ≤ 4, where
A =
⎛
⎜
⎝
1 a c 1
a 1 d e
c d 0 0
1 e 0 1
⎞
⎟
⎠ ;

14.3. MIQUELIAN INVERSIVE PLANES OF ODD ORDER 661
where a, b, c, d, e are given elements of Fq with e = ±1.
We have two matrices, A + and A− , depending on whether e = +1 or
e = −1, respectively. Note: det(A + ) = 2(a − 1)cd and det(A− ) = 2(a + 1)cd.
Since 〈C〉 and 〈D〉 are disjoint, by Lemma 14.3.1 C × D = a2 − 1 = (a +
1)(a − 1) is a nonzero square in Fq. Thus a = ±1, and since 〈R〉 is disjoint
from 〈C〉 and 〈D〉, c = R.C and d = R.D are nonzero. Therefore A + and A− are nonsingular, and Lemma 14.3.3 implies that there exists a quadruple in
M0 representing one of them precisely when its determinant is a nonsquare
in Fq. But (det(A + ))(det(A− )) = 4(a2 − 1)c2d2 , and a2 − 1 is a nonzero
square in Fq. Thus A + corresponds to a quadruple in M0 precisely when A− corresponds to a quadruple in M0.
If 〈R〉 were a carrier of the linear flock containing 〈C〉 and 〈D〉, they
would have no common tangent circle through 〈R〉 by Lemma 14.3.5(iv).
Hence we may assume that R is not a linear combination of C and D in M0,
and that det(A0) = 0, where A0 is the submatrix of A determined by its first
three rows and its first three columns.
Assume then that (C, D, R, T ) is represented by A + . Then det(A− ) is
a nonsquare in Fq, and thus, by Corollary 14.3.4 there exists an element of
S of M0 such that (C, D, R, S) is represented by A− , and 〈S〉 is therefore a
circle through 〈R〉 tangent to both 〈C〉 and 〈D〉.
Let 〈E〉 be the common circle orthogonal to 〈C〉, 〈D〉 and 〈R〉, i.e., 〈E〉
is a circle in the bundle orthogonal to 〈C〉 and 〈D〉, and 〈E〉 contains 〈R〉.
We claim that 〈T 〉 φ′ E and 〈S〉 φ′ E, the inverses with respect to 〈E〉 of 〈T 〉 and
〈S〉, are also tangent to 〈C〉 and 〈D〉 and pass through 〈R〉. Since φ ′ E fixes
〈C〉, 〈D〉 and 〈R〉, it must send circles tangent to 〈C〉 into circles tangent to
〈C〉, circles tangent to 〈D〉 into circles tangent to 〈D〉, and circles through
〈R〉 into circles through 〈R〉.
Now we have four circles satisfying the required conditions, namely 〈T 〉,
〈S〉, 〈T 〉 φ′ E and 〈S〉 φ′ E. That 〈S〉 is distinct from 〈T 〉 and 〈T 〉 φ′ E is clear,
since A + cannot be transformed into A− by multiplying each column and
the corresponding row by the same element of Fq. Similarly, 〈T 〉 is distinct
from 〈S〉 φ′ E. If 〈T 〉 φ′ E = 〈T 〉, this would imply that the points of tangency
of 〈T 〉 with 〈C〉 and 〈D〉 were fixed by φ ′ E , i.e., are on 〈E〉. 〈T 〉 would thus
have three distinct points in common with 〈E〉, so the two would be equal,
and 〈T 〉 could not be tangent to 〈C〉 and 〈D〉. Similarly, we can see that 〈S〉
is distinct from 〈S〉 φ′ E.
We have, therefore, four distinct circles satisfying the given requirements.
Now, any such circle 〈U〉 must produce, together with C, D and R, either

662 CHAPTER 14. FINITE CIRCLE GEOMETRIES
A + or A − . Without loss of generality, we assume that (A1, A2, A3, U) is
represented by A + . Since A1, A2, A3, A4 form a basis for M,
U =
Ak.U =
4
xiAi,
i=1
4
aikxi = ak4, 1 ≤ k ≤ 3.
i=1
If we hold x4 fixed we have three nonhomogeneous equations in the unknowns
x1, x2, x3 with matrix A0, which is nonsingular. Thus there is a
unique solution, and x1, x2, x3 can be written as linear functions of x4. We
have the additional condition
U =
=
4
xiAi.U
i=1
3
xiai4 +
i=1
= (1 + x4)
3
i=1
4
i=1
x4xiai4
ai4xi + x 2 4a44 = a44,
which is a nontrivial quadratic equation in x4. Thus there are at most two
solutions, and either U = T or U = T φ′ E. Applying the same procedure to
A − , we see that 〈T 〉, 〈T 〉 φ′ E, 〈S〉, 〈S〉 φ′ E are the only such circles in I. Hence
1. of Lemma 14.3.6 is proved.
Obs. 14.3.7. We may use this part 1. of Lemma 14.3.6 to count the points
〈R〉 disjoint from 〈C〉 and 〈D〉 through which there are such circles (i.e.,
tangent to 〈C〉 and 〈D〉 and containing 〈R〉). Since 〈C〉 and 〈D〉 have 2(q+1)
common tangent circles, four of which pass through each such point 〈R〉, and
each one of which contains q − 1 such points 〈R〉, we conclude that there
are precisely q2−1 points disjoint from 〈C〉 and 〈D〉 through which there pass
2
common tangents to 〈C〉 and 〈D〉. Thus, since q is odd, two disjoint circles
separate the rest of the plane into two classes of points: q2−1 points “between”
2
the circles, and q2−4q+1 others. (If we follow through the same procedure for
2

14.3. MIQUELIAN INVERSIVE PLANES OF ODD ORDER 663
the two secant circles, we will find that precisely one of the two matrices A +
or A − occurs, and thus through every point disjoint from the given circles
there are precisely two common tangent circles to the given circles.)
To prove part 2. of Lemma 14.3.6 assume that 〈T 〉 is a circle tangent to
the circle 〈C〉 and that 〈T 〉 passes through two points 〈P 〉 and 〈Q〉 not on
〈C〉. Suppose that the quadruple 〈C〉, 〈P 〉, 〈Q〉, 〈T 〉) is represented by the
4 × 4 matrix
B =
⎛
⎜
⎝
1 p q 1
p 0 r 0
q r 0 0
1 0 0 1
⎞
⎟
⎠ ; p, q, r, = 0;
where we let B1, B2, B3, B4 be C, P, Q, T , respectively. This matrix must
have nonsquare determinant equal to 2pqr, which implies that {C, P, Q, T }
is a basis for M0. Suppose there is another circle U = 4 i=1 xiBi through 〈P 〉
and 〈Q〉 and tangent to 〈C〉. Then, by the proof of part 1. we see that there
are at most two solutions U. We know T to be one solution. The bundle
through 〈P 〉 and 〈Q〉 must cover the q + 1 points of 〈C〉, an even number of
points. Thus 〈C〉 must have an even number of tangent circles in this bundle.
It has, therefore, precisely two, completing a proof of Lemma 14.3.6.
Lemma 14.3.8. Let J be a bundle with carriers 〈P 〉, 〈Q〉 in I, q odd, and
assume that 〈C〉 is a circle not in J such that all the circles 〈D〉 of J which
are not disjoint from 〈C〉 are in the same equivalence class under ∼. Then
〈C〉 is in the linear flock K orthogonal to J.
Proof. If 〈C〉 passes through 〈P 〉 or 〈Q〉, then it intersects all the circles of J.
But these circles are not all of the same equivalence class, by Lemma 14.3.2,
and thus 〈C〉 must be disjoint from both 〈P 〉 and 〈Q〉.
〈C〉 has q + 1 points, each of them in one circle of J. Through each
of these points passes a circle of the linear flock K. Let J0 and K0 be the
subsets of circles of J and K, respectively, which are not disjoint from 〈C〉.
Each circle of J0 is then secant to a circle of K0. Since, by hypothesis, the
circles of J0 all lie in the same equivalence class, those of K0 must also lie in
the same equivalence class, by Lemma 14.3.2. If 〈C〉 is not in K, it has at
most two of its q + 1 points in common with each circle of K0. This would
mean that K0 is a set of at least q+1
circles in the linear flock K, all in the
2
same equivalence class. But K contains exactly q − 1 circles, half of them in

664 CHAPTER 14. FINITE CIRCLE GEOMETRIES
each equivalence class, by Lemma 14.3.2. So we see that 〈C〉 must, indeed,
be a circle of K, and the Lemma is proved.
We are now ready for the proof of Orr’s theorem.
Theorem 14.3.9. Let I be an egglike inversive plane of order q, q an odd
prime power, and let K be a flock of I with carriers 〈P 〉 and 〈Q〉. Then K
is the linear flock of circles, inversion with respect to which interchanges 〈P 〉
and 〈Q〉.
Proof. By part (i) of Lemma 14.3.5, we can represent the circles of K as
K1 ∪ K2, where K1 consists of q−1
circles of one equivalence class under ∼,
2
and K2 consists of q−1
circles of the other equivalence class. By part (ii) of
2
Lemma 14.3.5 no circle of K has a tangent circle through 〈P 〉 and 〈Q〉. We
shall derive an equivalent statement of this condition in terms of the bilinear
from on M0.
Consider a quadruple (C, P, Q, T ) of elements of M where 〈P 〉 and 〈Q〉
are points of I disjoint from the circle 〈C〉, and 〈T 〉 is a circle through 〈P 〉
and 〈Q〉 tangent to 〈C〉. This quadruple can then be represented by a 4 × 4
matrix of the form:
⎛
⎜
⎝
c p s c
p 0 r 0
s r 0 0
c 0 0 c
where p, r, s, c are nonzero elements of Fq. This matrix has determinant equal
to 2psrc. By Corollary 14.3.4 such a T exists for a given triple C, P, Q exactly
when this determinant is a nonsquare in Fq. In other words, for 〈C〉 in K,
Lemma 14.3.6 part 2. implies:
⎞
⎟
⎠ ,
2C(P.C)(Q.C)(P.Q) is a nonzero square inFq. (14.19)
By the proof Lemma 14.3.6 we see that the condition that the circles
〈C〉 ∼ 〈D〉 have a common tangent circle through the point 〈R〉, disjoint
from both of them, is equivalent to the condition that the matrix A + have
nonsquare determinant. If we assume that C = D = 1, this determinant
equals 2(C.D − 1)(R.C)(R.D).
Now suppose that the sets K1 and K2 are chosen such that C = 1 for
all 〈C〉 in K1. Then Lemma 14.3.5 part (iv) says that for distinct circles 〈C〉
and 〈D〉 in K1

14.3. MIQUELIAN INVERSIVE PLANES OF ODD ORDER 665
2(C.D − 1)(P.C)(P.D) and 2(C.D − 1)(Q.C)(Q.D) (14.20)
are nonzero squares in Fq.
Now let 〈C〉 ∈ K1 and 〈D〉 ∈ K2, and let 〈R〉 be an arbitrary point of
〈D〉, and let 〈E〉 the unique circle through 〈P 〉, 〈Q〉 and 〈R〉. Consider the
set of circles through 〈R〉 and tangent to 〈C〉. There is precisely one such
circle at each point of 〈C〉, by the definition of inversive plane, and thus there
are exactly q +1 of them. A circle 〈T 〉 through 〈R〉 and tangent to 〈C〉 which
misses both 〈P 〉 and 〈Q〉 has q points disjoint from 〈C〉, and these points
must be covered by the circles of K. Since this is an odd number of points,
and 〈T 〉 is not in K, it must be tangent to at least one circle of K other
than 〈C〉, i.e., it must be tangent to precisely one circle 〈B〉 of K1 other than
〈C〉, since by Lemma 14.3.5 (iii) it could not be tangent to more than two
circles of K. Then by Lemma 14.3.6 〈C〉 and 〈B〉 have precisely 4 common
tangent circles through 〈R〉. Thus the circles through 〈R〉 tangent to 〈C〉
and disjoint from 〈P 〉 and 〈Q〉 are divided into disjoint classes of 4 distinct
circles, and the number of such tangent circles is a multiple of 4, say 4n.
Suppose 〈C〉 has a tangent circle through 〈P 〉 and 〈R〉. Then by Lemma 14.3.6
part 2. it has exactly two such tangent circles. Thus the number np of circles
through 〈P 〉 and 〈R〉 tangent to 〈C〉 must be either 0 or 2. Similarly, the
number nQ of circles through 〈Q〉 and 〈R〉 tangent to 〈C〉 is either 0 or 2.
Counting all the circles through 〈R〉 tangent to 〈C〉, we have
q + 1 = 4n + nP + nQ.
There are two cases depending on the value of q:
If q ≡ −1 (mod 4), then q +1 is a multiple of 4, and thus nP +nQ must be
divisible by 4. This can happen in precisely two ways. Either nP = nQ = 0,
so that 〈C〉 has no tangent circle through 〈P 〉 and 〈R〉 and also no tangent
circle through 〈Q〉 and 〈R〉. Or nP = nQ = 2 and 〈C〉 has a tangent circle
through 〈P 〉 and 〈R〉 and a (necessarily distinct) tangent circle through 〈Q〉
and 〈R〉. By Eq. 14.19 either 2(P.C)(R.C)(P.R) and 2(Q.C)(R.C)(Q.R) are
both nonzero squares in Fq, or they are both nonsquares in Fq. In either case,
their product 4(P.C)(Q.C)(R.C) 2 (P.R)((Q.R) is a nonzero square. Hence we
may conclude that
(P.C)(Q.C)(P.R)(Q.R) is a nonzero square in Fq. (14.21)

666 CHAPTER 14. FINITE CIRCLE GEOMETRIES
If q ≡ 1 (mod 4), then q+1 ≡ 2 (mod 4), and thus nP = nQ ≡ 2 (mod 4),
i.e., precisely one of nP , nQ is equal to 2 and the other is 0. In other words,
〈C〉 has a tangent circle through 〈R〉 and through precisely one of 〈P 〉 and
〈Q〉. Thus one of the quantities 2(P.C)(R.C)(P.R) and 2(Q.C)(R.C)(Q.R)
is a nonsquare and the other is a nonzero square in Fq. Their product must
therefore be a nonsquare, and we conclude that for q ≡ 1 (mod 4) the quantity
(P.C)(Q.C)(P.R)(Q.R) is a nonsquare in Fq. (14.22)
Return to the circle 〈E〉 through 〈P 〉, 〈Q〉 and 〈R〉, and ask : In which
equivalence class does 〈E〉 lie? The quadruple (P, Q, R, E) of elements of M0
is represented by the matrix:
⎛
⎜
⎝
0 P.Q P.R 0
P.Q 0 R.Q 0
P.R R.Q 0 0
0 0 0 E
⎞
⎟
⎠ ,
whose determinant 2E(P.Q)(R.Q)(P.R) must be a nonsquare in Fq.
We have shown in Eq. 14.20 that, since 〈C〉 has no tangent through both
〈P 〉 and 〈Q〉, the quantity 2(P.C)(Q.C)(P.Q) is a nonzero square in Fq.
Multiplying this by the determinant of the above matrix gives us that
and hence
4E(P.R) 2 (Q.R)(P.C)(Q.C) is a nonsquare in Fq,
E(P.Q)(Q.R)(C.P )((Q.C) is a nonsquare in Fq. (14.23)
For q ≡ −1 (mod 4), multiplying Eqs. 14.23 and 14.21 gives us that E
is a nonsquare in Fq. For q ≡ 1 (mod 4), multiplying Eqs. 14.23 and 14.22
gives us that E is a nonzero square in Fq.
We recall that 〈D〉 was chosen as an arbitrary circle of K2 and that 〈R〉
was an arbitrary point of 〈D〉. Thus we have shown that for all points 〈R〉
of 〈D〉 the circles defined by 〈P 〉, 〈Q〉 and 〈R〉 are of the same equivalence
class under ∼. In other words, all circles of the bundle through 〈P 〉 and 〈Q〉
which intersect 〈D〉 are of the same equivalence class. By Lemma 14.3.8 〈D〉
must be in the linear flock with carriers 〈P 〉 and 〈Q〉. Thus every circle of
K2 is in this linear flock. Clearly we may reverse the roles of K1 and K2 by

14.4. DEMBOWSKI’S THEOREM 667
picking 〈C〉 in K2 and 〈D〉 in K1, and thus every circle of K1 is also in this
linear flock. Therefore K itself must be exactly the linear flock with carriers
〈P 〉 and 〈Q〉. This completes a proof of the Theorem.
14.4 Dembowski’s Theorem
The theorem referred to in the title of this section is the following:
Theorem 14.4.1. Any finite inversive plane I of even order is egglike.
The proof we give here is due to J. A. Thas [Th84]. It uses the theory
of generalized quadrangles and is very much shorter than the original proof
due to P. Dembowski.
Proof. Let I = (P, B, I) be an inversive plane of even order n. We may
assume that I is containment ∈. Define a new incidence structure S =
(P ′ , B ′ , I ′ ) as follows. P ′ = P ∪ B. B ′ is the set of pencils of circles. A pencil
is incident with a point x ∈ P (the carrier of the pencil) and a maximal set
of mutually tangent circles at x. I ′ is the natural incidence. We now show
that S is is a GQ of order n.
Each line of S is incident with n+1 points of S, each point of S is incident
with n + 1 lines of S, and two lines are incident with at most one common
point.
Let p ∈ P ′ , L ∈ B ′ , p not incident with L in S. We must show that there
are unique w ∈ P ′ and M ∈ B ′ for which pI ′ MI ′ wI ′ L. Let x be the element
of P in L, i.e., L is a pencil of circles pairwise tangent at x. There are three
cases to consider:
Case (1) p ∈ P ;
Case (2) p ∈ B and x ∈ p;
Case (3) p ∈ B and x ∈ p.
In Case (1), w is the unique circle C of L which contains p, and M is the
element of B ′ defined by p and C.
In Case (2), w = x and M is the element of B ′ defined by p and x.
So suppose we are in Case (3). So p is a circle of I and x is a point
of I not in p. Let πx be the derived structure of I with respect to x. So
πx is an affine plane of order n whose points are the points of P \ {x} and
whose lines are the circles in B containing x. If L = {x, C1, . . . , Cn}, then
{C1 \{x}, . . . , Cn \{x}} = P is a parallel class of πx. Let πx be the projective

668 CHAPTER 14. FINITE CIRCLE GEOMETRIES
completion of πx. In πx, the circle p is an oval. Since n is even, the oval p
has a nucleus k. Clearly k is not on the line at infinity of πx, since otherwise
p would have one of its points at infinity, a contradiction. We are looking for
circles w in L which are tangent to p. So in πx we are looking for lines in P
which are tangent to p. There is just one such line in P, being the line of P
which contains the nucleus k of p. It follows that in S there is just one point
w which is incident with L and collinear with p.
Hence S is a GQ of order n.
Our next goal is to show that the GQ S is classical, i.e., isomorphic
to W (q). To do this we need to show that each point is regular, which by
Cor 9.3.9 is equivalent to showing that each triad of points is centric. Suppose
that {p1, p2, p3} is a triad of points of S. There are four cases to consider.
Case 1 p1, p2, p3 ∈ P . Then the element of B which contains p1, p2, p3 is
a point of S in {p1, p2, p3} ⊥ .
Case 2 p1, p2 ∈ P , p3 ∈ B, {p1, p2} ∩ p3 = ∅. Let πp1 be the derived
structure of I with respect to p1, with πp1 its projective completion. The
circle p3 is an oval in πp1. The kernel k of the oval p3 is a point of πp1 (as
above). With any line T of πp1 which contains p2 and kthere corresponds a
circle C = T ∪ {p1} of I which contains p1, p2 and is tangent to p3. Clearly
C ∈ {p1, p2, p3} ⊥ .
Case 3 p1 ∈ P , p2, p3 ∈ B, p1 ∈ p2, p1 ∈ p3, and p2, p3 nontangent. Let
πp1 be the derived structure of I with respect to p1. In πp1, the kernels of
the ovals p2 and p3 are denoted by k2 and k3, respectively. With any line T
of πp1 which contains k2 and k3 there corresponds a circle C = T ∪ {p1} of I
which contains p1 and is tangent to p2 and p3. Clearly C ∈ {p1, p2, p3} ⊥ .
Case 4 p1, p2, p3 ∈ B and pairwise nontangent. If x ∈ p1 ∩ p2 ∩ p3, then
x ∈ {p1, p2, p3} ⊥ . So assume that p1 ∩ p2 ∩ p3 = ∅. We show that in I there
is a circle that is tangent to all of p1, p2, p3.
Let x ∈ p1. First, suppose that x ∈ p2 or x ∈ p3, say x ∈ p2. In the
tangent pencil (of I) at x containing p2, the unique circle Cx containing
the nucleus of p3 considered as an oval in πx) is tangent to p3. If x ∈ p2
and x ∈ p3, let k2 (respectively, k3) be the nucleus of p2 (respectively, p3)
considered as an oval in πx. Any circle Cx of I containing x, k2 and k3 is
tangent to p2 and p3. If Cx is not uniquely defined by x, it can only be
because k2 = k3. In that case each circle containing x and k2 is tangent to
both p2 and p3. Clearly k2 ∈ p1, because then p1 would also be tangent to p2
and p3, contradicting our main hypothesis. In I there is a circle C containing
k2 and tangent to p1 at x. Then since C is tangent to p1, p2 and p3, it must

14.5. FINITE LAGUERRE PLANES 669
be that C ∈ {p1, p2, p3} ⊥ . So we may assume that Cx is uniquely defined by
x. So we are now in the case where for each x ∈ p1 there is a unique circle
Cx containing x and tangent to both p2 and p3. Then |Cx ∩ p1| = 1 or 2 for
each x ∈ p1. If |Cx ∩ p1| = 2 for all x ∈ p1, then the pairs Cx ∩ p1 define
a partition of p1, an impossibility since |p1| = n + 1 is odd. Hence at least
one of the circles Cx, say C, is tangent to p1. So C is tangent to all three of
p1, p2, p3 and C ∈ {p1, p2, p3} ⊥ .
At this point we know that S is isomorphic to W (n), n = 2 e . So identify
S and W (n) and denote by P G(3, q) the ambient 3-space around W (n).
In P G(3, n), the set P has exactly one point in common with each line of
W (n). (Each line of W (n) is of the form {x, C1, . . . , Cn}, where C1, . . . , Cn
is a pencil of circles pairwise tangent at x.) It follows that P is an ovoid of
the GQ W (n), and hence an ovoid of P G(3, q) by Lemma ??. The circleset
B of I is the set of all points of P G(3, n) \ P . And in I a point x is incident
with a circle C iff the line xC of P G(3, n) belongs to W (n), i.e., is tangent
to P . We can now see an isomorphism θ from the inversive plane I to the
inversive plane defined by the ovoid P : θ maps each point of P to itself, and
if C ∈ B = P G(3, n) \ P , then C θ is the set of all points of P on lines of
W (n) through C. Hence I is egglike.
14.5 Finite Laguerre Planes
A finite circle plane with one parallel class of lines was called a Laguerre
plane. So in detail the definition appears as follows:
A Laguerre plane S = (P, L, C) is an incidence structure of points, lines
and circles, respectively, such that the following axioms are satisfied:
I(i) Three pairwise non-collinear points are incident with a unique circle.
I(ii) For any circle C ∈ C, point P ∈ P and point Q not in C and not
collinear with P there is a unique circle D incident with Q and touching
C at P , i.e., C and D have P as their only common point.
I(iii) Each point P ∈ P is incident with a unique line L ∈ L, and each line
meets each circle in a unique point.
I(iv) Every circle is incident with at least three points and there are at least
two circles.

670 CHAPTER 14. FINITE CIRCLE GEOMETRIES
Let O be an oval in the plane π embedded as a hyperplane in Σ =
P G(3, q), and let V be a point of Σ \ π. Then the cone K with vertex
V and base O yields a Laguerre plane S = (P, C, L) in the following way.
P = K \ {V }; L consists of the lines of Σ joining V to points of O; C is the
set of plane sections of K by planes of Σ not containing V . Clearly each line
of S is incident with q points.
In general, if a Laguerre plane S has a line incident with a finite number
q of points, then it is easy to show that each line is incident with q points,
there are q + 1 lines, each circle has q + 1 points, there are q 2 + q points in P,
and every derived plane has order q. The number q is called the order of S.
Let S = (P, C, L) be a Laguerre plane of order q. Let L0, L1, . . . , Lq be
the generators (i.e., lines) of S, and suppose that P is a point on L0. The
projective closure SP of the derived plane SP may be described as follows (if
L is a line of S, then L ∗ is the set of points incident with L):
Points of SP :
(i) points of S not on L0. (So L ∗ 1, . . . , L ∗ q partition these q 2 points.)
(ii) the tangent pencils of circles at P
(iii) the symbol (∞)
Lines of SP :
(a) generators L1, . . . , Lq of S not incident with P
(b) circles incident with P
(c) the symbol [∞]
A point of type (i) is incident with the generator of S that contains it
and with the q circles that contain it and P . A point of type (ii) is incident
with the q circles in it and with the line [∞]. The point (∞) is incident with
the q generators and with the line [∞].
Let C be a circle of S not incident with P . It is clear that C is an oval
of SP with nucleus (∞).
The following theorem first appeared in [PT75]:
Theorem 14.5.1. A Laguerre plane of odd order s (s > 1) is equivalent to
a GQ of order s with a distinguished antiregular point.

14.5. FINITE LAGUERRE PLANES 671
Proof. Let S = (P, B, I) be a GQ of order s with a fixed antiregular point
x∞, so s must be odd, implying s ≥ 3. Let L0, . . . , Ls be the lines of S
incident with x∞. Let
L ∗ i = {y ∈ P : x∞ = yILi}, and put P ∗ = x ⊥ ∞ \ {x∞}.
For each y ∈ P \ {x∞} ⊥ put
Finally, put
Cy = {x∞, y} ⊥ , and put C = {Cy : y ∈ P \ {x∞}} .
L = {L ∗ i
: 0 ≤ i ≤ s}.
Then we will prove that S ∗ = (P ∗ , C, L) is a Laguerre plane of order s. It
is immediately clear that both I((iii) and I(iv) hold. For I(i), let x1, x2, x3
be pairwise noncollinear points of P ∗ . Hence {x1, x2, x3} is a triad of points
in S with center x∞. Since x∞ is antiregular in S, each triad of points in
P ∗ has a unique second center y ∈ P \ {x∞} ⊥ , so the the triad {x1, x2, x3}
belongs to a unique circle Cy, establishing the validity of I(i). Now note that
|P ∗ | = s 2 + s, |L ∗ i | = s for each i = 0, . . . , s, and |Cy| = 1 + s for all Cy ∈ C.
A simple counting argument now shows that I(ii) must also hold, proving
that S ∗ is a Laguerre plane.
Conversely, suppose that S ∗ = (P ∗ , C, L) is a Laguerre plane of odd order
s. We must construct a GQ S = (P, B, I) with an antiregular point. We
start by describing the points and lines of S.
Points of S are of three types:
(i) a symbol (∞);
(ii) the points of P ∗ ;
(iii) the circles C ∈ C.
Lines of S are of two types:
(a) the lines (called generators) in L;
(b) the pencils of mutually tangent circles of S ∗ .
Incidence I is then defined as follows: the point (∞) is on each line of type
(a); each point x ∈ P∗ is incident with the unique generator L∗ i with x ∈ L
and with each tangent pencil of circles that consists of circles tangent at x.
A point x of type (iii) is incident with each pencil containing x, one centered
at each of the s + 1 points of P∗ in x.

672 CHAPTER 14. FINITE CIRCLE GEOMETRIES
It is now easy to see that S = (P, B, I) is an incidence structure with
1 + s points (respectively, lines) incident with each line (point, respectively),
with 1 + s + s2 + s3 points and also that many lines, and with no two points
ever incident with the same two lines. A simple counting argument shows
that to prove that S is a GQ it suffices to show that if x ∈ P is a point not
on a line L ∈ B, then there is at least one point incident with L and collinear
with x.
There are several cases to consider, but the only one not really trivial is
when x is of type (iii) (i.e., a circle) and L is a pencil of circles tangent at the
point y. If y ∈ x, then y is the only point incident with L that is collinear
in S with the point x (on the line L). So suppose that y ∈ x. Suppose that
y ∈ L∗ i with L∗i ∩x = {z}. An involution θ of the circle x is defined as follows:
θ(z) = z; for u ∈ x \ {z}, θ(u) = u if the circle C of L that contains u is
tangent to x at u; for u ∈ x \ {z}, if the circle C of L that contains u is not
tangent to x, then θ(u) is defined by C ∩ x = {u, θ(u)}. As |x| = 1 + s is
even and θ(z) = z, θ has at least two fixed points. Hence L contains at least
one circle C that is tangent to x. Since C and x are collinear points of S,
the line L contains at least one point collinear with x. Hence S is a GQ of
order s.
It remains only to show that (∞) is an antiregular point of S. By Theorem
9.8.4, part 1, this is equivalent to showing that each triad contained in
(∞) ⊥ has exactly two centers. One center is the point (∞). It is fairly easy
to see that the unique other center is the circle containing that triad.
Of course by the Theorem of Bagchi, Brouwer and Wilbrink we now know
that each point of S is antiregular, i.e., each triad of points of S has exactly
0 or 2 centers.
We now want to interpret for the Laguerre plane S ∗ several properties
of the GQ S. The next theorem interprets what it means for one of the
generators of the Laguerre plane to be a regular line in the associated GQ.
The result was called the ”tilt property” in [PT75].
Theorem 14.5.2. (Tilt Property) Writing out the property of L0 being regular
gives the following. Let L0, L1, . . . , Ls be the generators of the Laguerre
plane S ∗ . Then L0 is a regular line in the associated GQ S if and only if the
following holds: Let C be any circle of S ∗ and suppose that C ∩ L ∗ 0 = {x0},
C ∩ L ∗ i = {xi}, for arbitrary but fixed i, 1 ≤ i ≤ s. Let C1 be tangent to C at
x1 and meet L0 at y0. Let C0 be tangent to C at x0 and meet L1 at y1. Then

14.5. FINITE LAGUERRE PLANES 673
the circle C ′ tangent to C1 at y0 and containing y1 must be the same circle
that is tangent to C2 at y1 and contains y0.
Another condition equivalent to the regularity of L0 is the fact that each
triad of lines containing L0 must have either 1 or 1 + s transversals, which is
equivalent to each triad of lines containing L0 having at least one transversal.
At this point we leave as an exercise the task of interpreting this property in
one (or more) of the associated Laguerre planes.
In view of Theorem 14.5.1 the following statements are almost trivial.
Theorem 14.5.3. The generalized quadrangle S of order s having an antiregular
point is isomorphic to the system consisting of the points and lines
of a nondegenerate hyperquadric in P G(4, s) if and only if the corresponding
Laguerre plane S ∗ is isomorphic to the classical model arising from a cone
in P G(3, s). If S is a generalized quadrangle having an antiregular point
x∞, then the affine plane A(x∞, y), y ∼ x∞, y = x∞, is just the internal
(derived) structure at the point y of the corresponding Laguerre plane S ∗ .
The following theorem appeared in [CK73] and independently in [PT75].
Theorem 14.5.4. Let S ∗ = (P ∗ , C, L) be a finite Laguerre plane of odd order
s. If the internal structure at one of the points of S ∗ is desarguesian, then
S ∗ is isomorphic to the classical model arising from a cone in P G(3, s).
Proof. Suppose that the internal structure S ∗ y = (P ∗ y , B ∗ y, ∈) at the point y
of the Laguerre plane S ∗ = (P ∗ , C, L) is desarguesian. Here P ∗ y = P∗ \ Ly,
where Ly is the generator of S ∗ through y; B ∗ y = (L\{Ly})∪{C ∈ C : y ∈ C}.
If C ∈ C is a circle that does not contain y, then Cy = C ∩ P ∗ is an s-arc of
the projective completion πy of the affine plane S ∗ of order s. As each line
of L \ {Ly} has a unique point of Cy, they are the tangent lines to Cy and
they meet in πy at the ideal point (∞). Hence the set Cy ∪ {(∞)} is an oval
of πy. Moreover, the ideal line [∞]y of πy is tangent to this oval at the point
(∞). As s is odd, the oval Cy ∪ {(∞)} is an irreducible conic. As there are
s 3 − s 2 circles of the type Cy, we obtain in this way s 3 − s 2 irreducible conics
in πy that are tangent to [∞] at the point (∞). Now consider two circles C1
and C2 that have in common a point z ∈ Ly, z = y, and that also have in
common a second point u. The corresponding s-arcs of πy have one point u
in common. Suppose that the corresponding conics O1 and L2 are tangent
at u. Then with their tangent line at u there corresponds a circle of S ∗ that
is tangent to C1 and C2 at u. Consequently C1 and C2 are tangent at u, a

674 CHAPTER 14. FINITE CIRCLE GEOMETRIES
contradiction. So with the s 2 circles through z there correspond s 2 mutually
osculating conics of πy. Hence the representation of S ∗ in πy is a classical
representation of the Laguerre plane arising from a cone in P G(3, q). See
Observation 6.2.2.
The following corollary is just a reinterpretation of Theorem 14.5.4.
Corollary 14.5.5. Let S = (P, B, I) be a generalized quadrangle of order s
havingt an antiregular point x∞, so that s must be odd. If there is a point
y ∼ x∞, y = x∞, for which the plane A(x∞, y) is desarguesian, then S
must be isomorphic to the system consisting of the points and lines of a
nondegenerate hyperquadric in P G(4, s).
14.6 Finite Minkowski Planes
A finite circle plane with two parallel class of lines was called a Minkowski
plane. So in detail the definition appears as follows:
A Minkowski plane S = (P, L, C) is an incidence structure of points, lines
and circles, respectively, such that the following axioms are satisfied:
M(i) Three pairwise non-collinear points are incident with a unique circle.
M(ii) For any circle C ∈ C, point P ∈ P and point Q not in C and
not collinear with P there is a unique circle D incident with Q and
touching C at P , i.e., C and D have P as their only common point.
M(iii) L = M1 ∪ M2, where M1 and M2 are each a parallel class of lines
(i.e., generators). Each point P ∈ P is incident with a unique line
in M1 and with a unique line in M2, and each line meets each circle
in a unique point.
M(iv) Every circle is incident with at least three points and there are at
least two circles.
The classical Minkowski plane S = (P, M1∪M2, C) arises in the following
fashion. Let H3 be a hyperbolic quadric in P G(3, q). Let P be the set of
points on H3 and let M1 and M2 be the two rulings of H3. The circles of S
are the intersections of H3 with nontangent planes. So each circle is a conic
lying in H3, and each three noncollinear points of P lie in a unique circle.

14.6. FINITE MINKOWSKI PLANES 675
Each point P of P lies on one line L1 ∈ M1 and one line L2 ∈ M2. The
tangent plane to H3 at P is the plane containing L1 and L2. Suppose P is
a point of the circle C which is the intersection of the plane π with H3, and
suppose that Q is a a point not collinear with P and not on the circle C. The
plane π intersects the tangent plane at P in a line ℓ, so ℓ is the unique line of
π tangent to C at P . The plane 〈Q, ℓ〉 meets H3 in the unique conic section
of H3 containing Q and touching C at P . It is also clear that the lines in a
ruling partition the points of P and two lines in distinct rulings intersect in
a unique point with the plane spanned by those two lines being the tangent
plane to H3 at that point. So S is a Minkowski plane (of order q).
In 1974 it was discovered independently by W. Heise [He74] and M. Percsy
[Pe74] that any finite Minkowski plane of even order is classical. We give the
proof of this result found by J. A. Thas [Th84].
Theorem 14.6.1. (Heise-Percsy) Any finite Minkowski plane S = (P, L, C)
of even order s is classical.
Proof. Let S = (P, L, C) be a Minkowski plane of even order s. Here
L = M1 ∪ M2 is the family of two parallel classes of generators. We may
assume that incidence in S is containment ∈. We now define a new point-line
incidence structure S ′ = (P ′ , B ′ , I ′ ). Here P ′ = P ∪ C is the pointset. In the
lineset B ′ there are two types of elements: The elements of type (i) consist
of a point x ∈ P together with a maximal set of mutually tangent circles at
x (x is called the carrier of this element of B ′ ), and the elements of type (ii)
are the generators. The incidence I ′ is the natural incidence. The first main
step is to show that S ′ is a GQ of order s.
A routine check shows that each line of S ′ is incident with s + 1 points
of S ′ , each point of S ′ is incident with s + 1 points of S ′ , and two lines are
incident with at most one common point. Let p ∈ P ′ . L ∈ B ′ , p I ′ L. We
must show that there are unique w ∈ P ′ and M ∈ B ′ for which pI ′ MI ′ wI ′ L.
There are several cases to consider.
Case 1. p ∈ P, L ∈ L, p ∈ L. Assume, for example, that L ∈ M1. Then
w is the unique point of L which is collinear with p, and M is the generator
in M2 which contains p.
Case 2. p ∈ P and L is a line of type (i) of S ′ . Let x be the element of
P in L. If x and p are on a generator N , then clearly w = x and N = M.
If x and p are noncollinear in S, then w is the unique circle C of L through
p and M is the line of type (i) of S ′ defined by p and C.

676 CHAPTER 14. FINITE CIRCLE GEOMETRIES
Case 3. p ∈ L, L ∈ L. Then w is the common point of p and L in S, and
M is the line of type (i) of S ′ defined by w and p.
Case 4. p is a circle and L is a pencil with carrier x. If x is in the
circle p, then w = x and M is the line of type (i) of S ′ defined by x and
p, i.e., the tangent pencil with carrier x and containing p. Now assume
that x ∈ p. Let πx be the derived structure of S with respect to x. Then
πx is an affine plane of order s. If C1, . . . , Cs are the circles in L, then
{C1 \ {x}, . . . , Cs \ {x}} = P is a parallel class of the affine plane πx. Also
the generators in M1 (respectively, M2) which do not contain x define a
parallel class P ′ (respectively, P ′′ ) in πx. Note that P, P ′ , P ′′ are distinct.
Let πx be the projective completion of πx. The points of πx defined by P ′
and P ′′ are denoted by z ′ and z ′′ respectively. If R ′ and R ′′ are the lines of S
through x, then (p\(R ′ ∪R ′′ ))∪{z ′ , z ′′ } = O is an oval of πx. Since s is even,
the oval O has a kernel N. Since the line at infinity of πx has two points in
common with O, the kernel N is an affine point of πx. We are looking for
circles w in L which are tangent to p. So in πx we are looking for lines in P
which are tangent to O \ {z ′ , z ′′ }. There is just one such line in P , being the
line of P which contains N. It follows that in S ′ there is just one point w
which is incident with L and collinear with p.
Hence S ′ is a GQ of order s. In S ′ we have M ⊥ 1 = M2, so that any two
lines of Mi form a regular pair of lines, i = 1, 2.
The second main step is to show that the GQ S ′ is classical, i.e., each
point is regular. To do this we show that each triad of points has a center.
Case 1. Let p1, p2, p3 ∈ P be pairwise noncollinear in S. Then the circle
of S that contains p1, p2, p3 belongs to {p1, p2, p3} ⊥ in S ′ .
Case 2. Let p1, p2 ∈ P, p3 ∈ C, with p1 and p2 noncollinear in S and
{p1, p2} ∩ p3 = ∅. Let πp1 be the derived structure of S with respect to p1.
The liines in M1 (resp., M2) which do not contains p1 define a parallel class
P ′ (resp., P ′′ ) in πp1. Let πp1 be the projective completion of πp1. The points
of πp1 defined by P ′ and P ′′ are denoted by z ′ and z ′′ , respectively. If R ′ and
R ′′ are the lines of S through p1, then (p3 \ (R ′ ∪ R ′′ )) ∪ {z ′ , z ′′ } = O is an
oval of πp1. First assume that in πp1 at least one of the lines p2z ′ , p2z ′′ , say
p2z ′ , contains the kernel N of O. Then in S the circle p3 contains a point
x which is collinear with p1 and p2. Clearly x ∈ {p1, p2, p3} ⊥ . Now assume
that N is on no one of the lines p2z ′ , p2z ′′ . Then with the line p2N of πp1
there corresponds a circle C ∈ {p1, p2, p3} ⊥ .
Case 3. Let p1 ∈ P, p2, p3 ∈ C with p1 ∈ p2 ∪ p3 and with p2 and p3

14.6. FINITE MINKOWSKI PLANES 677
not tangent. Let πp1 be the derived structure at p1. The ovals of πp1 which
correspond to p2 and p3 are denoted by O2 and O3, respectively. Let Ni be
the kernel of Oi in πp1, i = 1, 2. First suppose that N2 = N3 and that the
line N2N3 belongs to one of the parallel classes P ′ , P ′′ , say P ′ . Then in S the
points N2 and N3 are collinear and the circles p2 and p3 have a common point
which is collinear with p1. Clearly x ∈ {p1, p2, p3} ⊥ . Suppose now that either
N2 = N3 or that N2 = N3 and the line N2N3 is not in P ′ or P ′′ . Then with
any line T of πp1 which contains N2 and N3 there corresponds a circle C of S
which contains p1 and is tangent to p2 and p3. In this case C ∈ {p1, p2, p3}.
Case 4. Suppose p1, p2, p3 ∈ C and are pairwise not tangent. If x ∈ p1∩p2∩
p3, then clearly x ∈ {p1, p2, p3} ⊥ . So we may assume that {p1, p2, p3} ⊥ = ∅.
We show that in S there is a circle which is tangent to p1, p2, and p3. Let
x ∈ p1. First suppose that x ∈ p2 or x ∈ p3, say x ∈ p2. By Case 4. of
the proof that S ′ is a GQ, there is a unique circle Cx in S which contains
x and is tangent to p2 and p3. So Cx ∈ {p1, p2, p3} ⊥ . Next, suppose that
x ∈ p2 ∪ p3. By Case 3. (the preceding paragraph), there is a circle Cx which
contains x and is tangent to p2 and p3, or there is a point y ∈ p2 ∩ p3 which
is collinear with x in S. If there is a point y ∈ p2 ∩ p3 which is collinear
with x in S, then the union of the lines of S through y is also denoted by Cx
(here |Cx ∩ p1| = 2). Now suppose that there is a point x ∈ p1, x ∈ p2 ∪ p3,
for which the element Cx is not uniquely defined by x. Then in the plane
πx the kernels of the ovals O2 and O3 which correspond to the circles p2 and
p3, respectively, coincide. Let this common kernel be N. Hence each circle
containing x and N is tangent to p2 and p3. If N ∈ p1, then p1 is tangent to
p2 and p3, a contradiction. So N ∈ p1. Since N is a point of πx, we have that
in S, N is not collinear with x. So in S there is a circle C which contains
N and is tangent to p1 at x. Since C is tangent to each of p1, p2 and p3, we
have C ∈ {p1, p2, p3} ⊥ . So we may assume that Cx is uniquely defined by
any x ∈ p1. If |Cx ∩ p1| = 2 for all x ∈ p1, then the pairs Cx ∩ p1 define a
partition of p1, a contradiction since |p1| = s + 1 is odd. Consequently there
is at least one circle Cx, say C, which is tangent to p1. So C ∈ {p1, p2, p3} ⊥ ,
completing the proof that each triad of points of S ′ is centric.
Since each triad of points of S ′ is centric, we know that all points are regular
and S ′ is isomorphic to W (s) with s = 2 e . So identify S ′ and W (s), and
let P G(3, s) denote the projective 3-space that contains W (s). In P G(3, s)
the linesets M1 and M2 are the two reguli of a hyperbolic quadric. Hence
in P G(3, s) the pointset P is a hyperbolic quadric. the circleset of S is the

678 CHAPTER 14. FINITE CIRCLE GEOMETRIES
set of all points of P G(3, s) \ P. And in S a point x is incident with a circle
C if and only if the line xC of P G(3, s) belongs to W (s), i.e., is tangent to
P. Hence S is isomorphic to the Minkowski plane defined by the quadric P.
This isomorphism θ is explicitly described as follows: θ : x ↦→ x for all x ∈ P;
θ : L ↦→ L for all generators L ∈ L; if C ∈ C = P G(3, s) \ P, then C θ is the
set of all points of P on lines of W (s) through C. Hence S is classical.
In the preceding proof a Minkowski plane of even order was used to construct
a generalized quadrangle which was shown to be classical, implying
that the Minkowski plane was classical. In the next proof we start with a
generalized quadrangle having a regular pair of lines with a certain property
and use it to produce a Minkowski plane of even order. Then knowing that
the Minkowski plane must be classical, we show that the GQ must also be
classical.
Theorem 14.6.2. Let S = (P, B, I) be a GQ of order s > 2. Suppose that
S has a regular pair (L0, L1) of nonconcurrent lines with the property that
any triad of points lying on lines of {l0, L1} ⊥ has a center. Then s = 2 e
and S is the generalized quadrangle W (s) arising from a symplectic polarity
π of P G(3, s). Moreover, the points lying on the lines of {L0, L1} ⊥ form a
nonsingular ruled quadric of P G(3, s) for which the tangent lines coincide
with the totally isotropic lines of the polarity π.
Proof. Put {L0, L1} ⊥ = {M0, . . . , Ms} and {L0, L1} ⊥⊥ = {L0, L1, L2, . . . , Ls}.
We may order the point V = {xij : 0 ≤ i, j ≤ s} on the lines of {L0, L1} ⊥ so
that Li and Mj meet at the point xij. Assume that each triad of points of V
is centric. Using these hypotheses we now construct an incidence structure
M ∗ = (V, L, C) that will turn out to be a Minkowski plane of order s.
The points of M ∗ are just the points of V . The elements of L1 =
{L0, . . . , Ls} and of L2 = {M0, . . . , Ms} are the generators of M ∗ : L =
L1 ∪ L2. For each point y ∈ P \ V there is a set Cy of s + 1 points of
V collinear in S with y. This set Cy will be called a circle and we put
C = {Cy : y ∈ P \ V }. So |C| = s 3 − s. Incidence in M ∗ is the one naturally
induced by incidence in S. The following observations are immediate:
• Each point of V lies on a unique generator in L1 and on a unique
generator in L2, and each generator of L1 meets each generator of L2
in a unique point.

14.6. FINITE MINKOWSKI PLANES 679
• Each circle meets each generator in a unique point.
• For each C ∈ C we have |C| >≥ 3, and there is a pair (x, C) ∈ V × C
with x ∈ C.
• Each triad of points of V (i.e., to two lying on a generator) is contained
in a unique circle. This follows easily since by hypothesis each triad
of points of V is contained in at least one circle. However, there are
(s+1) 2 s 2 (s−1) 2 ordered triads, with each circle containing (s+1s(s−1)
ordered triads, and there are s 3 − s circles. It follows that each triad
must be contained in a unique circle.
To complete a proof that M ∗ is a Minkowski plane we must prove the
following: Let x, y be noncollinear points of V and let C be a circle containing
x but not y. Then there is a unique circle C ′ containing y for which C ∩C ′ =
{x}.
So suppose that x and y are noncollinear points of V , and let Cp be a
circle with nucleus p containing x but not y. We must show that there is a
unique circle C ′ containing y and tangent to Cp at x. First, the number of
circles containing x is the number s 2 − s of points of P \ V collinear in S
with x. Second, if z ∈ Cp, z = x, the number of circles (including Cp) that
contain x and z is the number s − 1 of points in {x, z} ⊥ ∩ (P \ V ). Hence the
number of circles containing x and intersecting Cp in two points is s(s − 2),
leaving s − 1 circles C ′ with Cp ∩ C ′ = {x}. We claim that these are precisely
the circles Cq for q on the line tangent to Cp at x, x = q = p. For if q is
such a point, and if Cq contains a point z of Cp, z = x, then (z, q, p) must
be the vertices of a triangle in S. Hence if L is the tangent of Cp at x, those
circles C ′ for which Cp ∩ C ′ = {x} are precisely the circles Cq, with q on L,
x = q = p. Finally, y is collinear with a unique point q of L, x = q = p, and
Cq is clearly the unique circle containing y for which Cq ∩ C ′ = {x}. Hence
M ∗ is a Minkowski plane.
We now prove that s must be even. Let x and y be noncollinear points of
V and Cp a circle through x not containing y. Let C ′ p ′ be the unique circle
containing y which is tangent to Cp at x. Hence p ′ is on the tangent line L
of Cp at x. Let z be any one of the s − 2 points of ”Cp for which z = x and
z is not collinear with y, and let Cp ′′ be the unique circle containing z for
which Cp ′′ is tangent to Cp ′ at y. So p′′ must lie on the tangent line L ′ to Cp ′
at y. Clearly Cp = Cp ′′ and z ∈ Cp ∩ Cp ′′. But p′′ cannot be on the tangent

680 CHAPTER 14. FINITE CIRCLE GEOMETRIES
Lp ′′ to Cp at z without forcing {p, p ′ , p ′′ } to be a triangle in S. Hence Cp ′′
cannot be tangent to Cp at z, i.e., |Cp ∩ Cp ′′| = 2. This shows that the s − 2
points of Cp different from x and not collinear with y are split into pairs by
the circles C ′′ tangent to Cp ′ at y. Hence s − 2 is even.
Since any Minkowski plane M ∗ of even order is ovoiday, s is of the form
s = 2 e . This means there is a nonsingular ruled quadric Q in P G(3, 2 e )
and a bijection σ of Q onto V , such that each generator of Q is mapped
onto a generator of M ∗ . If W (2 e ) = (P ′ , B ′ , ∈) is the GQ arising from the
symplectic polarity π defined by Q, then we define as follows the bijection φ
of P ′ onto P:
(i) φ : x ↦→ x σ for x ∈ Q.
(ii) For x ∈ P ′ \ Q, let H be the polar plane of x with respect to π.
Then φ maps x to the nucleus of the circle (Q ∩ H) σ of M ∗ . It then follows
readily that φ determines an isomorphism of W (2 e ) onto S = (P, B, I) with
Q φ = V , and the proof is essentially complete.
In a GQ of order s a pair (x, y) of noncollinear points is regular if and
only if each triad {x, y, z} is centric. Hence if the hypothesis of the preceding
theorem is altered to say that each pair {x, y} of noncollinear points lying on
lines of the span {L0, L1} ⊥ is regular, then the conclusion remains valid.

Chapter 15
Property G
In Chapter 13 the concept of Property (G) was introduced and it was shown
that flock GQ have this property at the point (∞). In this chapter we study
Property (G) abstractly and eventually show that it characterizes those GQ
arising from flocks of a quadratic cone. In this chapter, however, we study
the point-line dual version.
15.1 Property (G) in GQ(s, s 2 )
Throughout this chapter we let S = (P, B, I) be a GQ with parameters
(s, t) = (s, s 2 ). By Theorem 9.3.6 we know that if {x, y, z} is a triad of
points (i.e., no two of x, y, z are collinear), then |{x, y, z}| ⊥ = s + 1. If
{u, v, w} is a triad of points contained in {x, y, z} ⊥ , then {u, v, w} ⊥ is a set
of 1 + s points containing x, y, z. If the set {u, v, w} ⊥ is independent of the
triad {u, v, w} in {x, y, z} ⊥ , then {x, y, z} ⊥⊥ = {u, v, w} ⊥ and we say that
the triad {x, y, z} is 3-regular.
Theorem 15.1.1. Let {x, y} be a pair of noncollinear points of S for which
each triad of points contained in {x, y} ⊥ is 3-regular. Then the incidence
structure I(x, y) with pointset P = {x, y} ⊥ and as circles the sets {z, z ′ , z ′′ } ⊥⊥ ,
where z, z ′ , z ′′ are any distinct points in {x, y} ⊥ , is an inversive plane of order
s.
Proof. The fact that each triad of points contained in {x, y} ⊥ is 3-regular
means that the circles are well-defined, i.e., each three points lie on a unique
circle. Moreover, an easy counting argument shows that the derived geometry
at any fixed point of P is an affine plane.
681

682 CHAPTER 15. PROPERTY G
Theorem 15.1.2. Let T = {x, y, z} be a 3-regular triple. Then each point
of P \ (T ⊥ ∪ T ⊥⊥ ) is collinear with precisely two points of T ⊥ ∪ T ⊥⊥ .
Proof. Put X = {x, y, z} ⊥⊥ and Y = {x, y, z} ⊥ . Then |X| = |Y | = s + 1,
and the result follows from Theorem 9.7.5.
Theorem 15.1.3. Let T = {x, y, z} be a 3-regular triple. Let P ′ be the set
of points incident with lines of the form uv, u ∈ T ⊥ , v ∈ T ⊥⊥ . If L is a line
through no point of T ⊥ ∪ T ⊥⊥ , and if L is incident with k points of P ′ , then
(i) k ∈ {0, 2} if s is odd;
(ii) k ∈ {1, s + 1}, if s is even.
(iii) If s is even and B ′ is the set of lines incident with at least two points
of P ′ , then S ′ = (P ′ , B ′ , I ′ ) is a subquadrangle of order s. (Clearly I ′ is the
incidence I restricted to P ′ × B ′ .) Moreover, {x, y} is a regular pair of S ′
with {x, y} ⊥′ = {x, y, z} ⊥ and {x, y} ⊥′ ⊥ ′
= {x, yu, z} ⊥⊥ .
Proof. Let L be a line which is incident with no point of T ⊥ ∪T ⊥⊥ . If w ∈ T ⊥
with wIMImIL, and if M is not a line of the form uv with u ∈ T ⊥ , v ∈ T ⊥⊥ ,
there is just one point w ′ ∈ T ⊥ \ {w} which is collinear with m. Hence the
number r of lines uv, u ∈ T ⊥ , v ∈ T ⊥⊥ , which are concurrent with L, has
the parity of |T ⊥ |s + 1. Clearly r is also the number of points in P ′ which
are incident with L.
Let {L1, L2, . . . , } = L be the set of all lines which are incident with no
point of T ⊥ ∪T ⊥⊥ , and let ri be the number of points of P ′ which are incident
with Li. We have |L| = s 3 (s 2 − 1) and |P ′ \ (T ⊥ ∪ T ⊥⊥ | = (s + 1) 2 (s − 1).
Clearly
i ri = (s + 1) 2 (s − 1)s 2 , and
i ri(ri − 1) is the number of ordered
triples (uv, u ′ v ′ , Li), with u, u ′ distinct points of T ⊥ , with v, v ′ distinct points
of T ⊥⊥ , and with uv ∼ Li ∼ u ′ v ′ , where u, v, u ′ , v ′ are not incident with Li.
Hence
i ri(ri − 1) = (s + 1) 2 s 2 (s − 1).
Let s be odd. Then ri is even, so
i ri(ri−2) ≥ 0 with equality if and only
if ri ∈ {0, 2} for all i. Since
i ri(ri−2) = (s+1) 2 s 2 (s−1)−(s+1) 2 (s−1)s 2 =
0, we have indeed ri ∈ {0.2} for all i.
Now let s be even. Then ri is odd, implying
i (ri−1)(ri−(s+1)) ≤ 0 with
equality if and only if ri ∈ {1, s + 1} for all i. Since
i (ri − 1)(ri − (s + 1)) =
(s + 1) 2 s 2 (s − 1) − (s + 1)(s + 1) 2 (s − 1)s 2 + (s + 1)s 3 (s 2 − 1) = 0, we have
indeed ri ∈ {1, s + 1} for all i for all i.
This completes the proof of parts (i) and (ii). For proof of part (iii) let s
be even. We have |P ′ | = (s + 1)(s − 1) + 2(s + 1) = (s + 1)(s 2 + 1).

15.1. PROPERTY (G) IN GQ(S, S 2 ) 683
Let L be a line of B ′ . If L is incident with some point of T ⊥ ∪ T ⊥⊥ , then
clearly L is of type uv, with u ∈ T ⊥ and v ∈ T ⊥⊥ . Then all points incident
with L are in P ′ . If L is incident with no point of T ⊥ ∪T ⊥⊥ , then by parts (i)
and (ii) L is again incident with s + 1 points of P ′ . Now by Theorem 9.10.3
S ′ = (P ′ , B ′ , I ′ ) is a subquadrangle of order (s, t ′ ). Since |P ′ | = (s+1)(st ′ +1)
we have t ′ = s. Hence S ′ is a subquadrangle of order s. Since T ⊥ ∪T ⊥⊥ ⊂ P ′ ,
and |T ⊥ | = |T ⊥⊥ | = s + 1, and each point of T ⊥ is collinear with each point
of T ⊥⊥ , we have {x, y} ⊥′ = {x, y, z} ⊥ and {x, y} ⊥′ ⊥ ′
= {x, y, z} ⊥⊥ .
This theorem shows that if T = {x, y, z} is a 3-regular triad of points,
then each point of P \ (T ⊥ ∪ T ⊥⊥ ) is collinear with precisely two points of
(T ⊥ ∪ T ⊥⊥ ). This means that if p ∈ (T ⊥ ∪ T ⊥⊥ ) and if p is not on a line from
(T ⊥ to T ⊥⊥ ), then either p is collinear with two points of T ⊥ ( and no point
of T ⊥⊥ , or p is collinear with two points of T ⊥⊥ and no point of T ⊥ .
We want to weaken the notion of 3-regularity to a property that will characterize
the flock GQ. The correct property turns out to be a slightly generalized
point-line dual of the concept of Property (G) introduced in Chapter 13.
In that chapter, we showed (in our present context) that the point-line dual
of a flock GQ has Property (G) at a line that is the dual of the point (∞).
Definition: Property (G) for a GQ S of order (s, s 2 ), s > 1. Let x, y
be distinct points incident with a line L. Then S has property (G) at {x, y}
provided each triad T with x ∈ T ⊆ y ⊥ is 3-regular, which is if and only if
each triad T ′ with y ∈ T ′ ⊆ x ⊥ is 3-regular. If S has property (G) at each
pair {x, y} of distinct points of the line L, we say that S has property (G)
at L. Similarly, S has property (G) at a flag (x, L) provided it has property
(G) at (x, y) whenever x = yIL.
Lemma 15.1.4. Suppose that (x, y) has Prop.(G), x = y ∼ x with L0 = xy.
Suppose u3 ∈ {y, u1, u2} ⊥⊥ \ {y, u1, u2} for a triad T = {y, u1, u2} ∈ x ⊥ . Put
Li = xui, i = 1, 2, 3. Now choose v2 ∈ L2, x = v2 = u2. Then {y, u1, v2} ⊥⊥
contains a point of L3.
Proof. Suppose that this is not the case. Put T0 = {y, u1, v2}.
We know u3 is collinear with x ∈ T ⊥ 0 on a line not joining x to T ⊥⊥
0 . So by
Theorem 15.1.3 there must be a unique second point x ′ ∈ T ⊥ 0 with u3 ∼ x ′ .
Then x ′ ∈ {y, u1, u3} ⊥ = {y, u1, u2} ⊥ , which forces x ′ ∼ u2. So x ′ ∼ v2 and
x ′ ∼ u2 forces v2u2, an impossibility. Hence it must be true that {y, u1, v2} ⊥⊥
contains a point of L3.

684 CHAPTER 15. PROPERTY G
Corollary 15.1.5. With exactly the same setup as in Lemma 15.1.4 suppose
also that v1IL1 with x = v1 = u1. Then {y, v1, v2} ⊥⊥ has a point v3 of L3.
Proof. Apply the lemma first to {y, v1, u2} ⊥⊥ and then to {y, v1, v2} ⊥⊥ .
Corollary 15.1.6. Let L1, . . . , Ls be the lines through x (different from L0 =
xy) containing points of {y, u1, u2} ⊥⊥ . Then the set P ′ of points different
from x on L1, . . . , Ls form an affine plane A ′ of order s. If u, v are distinct
points of P ′ , then the line of A ′ through u, v is the line xu = xv if u ∼ v in
S, and otherwise is the set {y, u, v} ⊥⊥ \ {y}.
We now want to construct an affine space Sx,y = (Pxy, Bxy, Hxy, IXY ) on
the points of Pxy = x ⊥ \ y ⊥ . First note that
1. The points of Sxy are the elements of Pxy, and |Pxy| = s 3 .
2. The lines of Sxy, i.e., the elements of Bxy, come in two types:
(a). T ⊥⊥ \ {y}, where T is a triad with y ∈ T ⊆ x ⊥ .
(b) the lines xw of S, where w ∈ x ⊥ \ y ⊥ .
3. The planes of S, i.e., the elements of Hxy, also come in two types:
(a) {x, z} ⊥ \ {y}, where z ∈ y ⊥ \ x ⊥ .
(b) The union of all “lines” of type (b) containing a point of some “line”
T ⊥⊥ \ {y} of type (a).
Clearly the “lines” of a “plane” of type (a) are the “lines” of type (a) contained
in it. The “planes” of type (b) are those constructed in the Cor. 15.1.6.
Property (G) implies that “planes” of type (a) are affine planes. Hxy contains
s3 “planes” of type (a) and s2 (s2−1) s(s−1) = s2 + s “planes” of type (b).
4. Incidence IXY is just containment.
Theorem 15.1.7. Sxy = (Pxy, Bxy, Hxy) is an affine space AG(3, q), with
q = s a prime power, where Pxy, Bxy, Hxy are the points, lines and planes,
respectively.
Proof. The uniqueness of S for s = 2, 3 essentially allows us to concentrate
on s > 3.
Let u, v, w be three distinct points of Pxy not on a “line” of Bxy and not
in a “plane” of type (b) of Hxy. Then no two of u, v, w are collinear in S
(i.e., no two lie on a common line of S through x, and {y, u, v} ⊥⊥ contains no
point of xw. Since w ∼ x ∈ {y, u, v} ⊥ and w ⊥ ∩ {y, u, v} ⊥⊥ = ∅, w must be
collinear with a unique second point x ′ of {y, u, v} ⊥ . Then {x, x ′ } ⊥ contains

15.2. PROPERTY (G) AND SUBQUADRANGLES 685
y, u, v, w, so u, v, w are contained in a “plane” of type (a) of Hxy. Since s > 3,
we may invoke the theorem of Buekenhout (see Section 1.5).
Corollary 15.1.8. If one pair of distinct collinear points has Property (G),
then s = q is a prime power.
15.2 Property (G) and Subquadrangles
From now on S is a GQ with parameters (q, q 2 ) and with a pair (x, y) of
distinct points on a line L0 having Property (G). For our first lemma q may
be either odd or even.
Lemma 15.2.1. Let Mi = {y, ui, zi} ⊥⊥ , i = 1, 2, be two parallel “lines”
of some “plane” of type (b) of AG(3, q), with u1 ∼ u2, z1 ∼ z2. Clearly
{y, u1, z1} ⊥ ∩ {y, u2, z2} ⊥ = {x}. Let r ∈ {y, u2, z2} ⊥ . Then r must lie on
some line joining a point (i.e., the point y) of M1 to a point of M ⊥ 1 .
Proof. We know that if r is not on such a line, then since r is collinear with
one point y of {y, u1, z1} ⊥⊥ , it must be collinear with a second point y ′ of
{y, u1, z1} ⊥⊥ . Let y ′′ be the point of M2 = {y, u2, z2} ⊥⊥ on xy ′ . Since r
must be collinear with y ′′ , clearly y ′′ = y, and M1, M2 are not parallel, a
contradiction. So the line yr contains a point r ′ of {y, u1, z1} ⊥ .
In the above it is clear that as r runs over the q points of {y, u2, z2} ⊥ \{x},
also r ′ runs over the q points of {y, u1, z1} ⊥ \ {x}. Hence {y, u1, z1} ⊥ \ {x} =
N1 and {y, u2, z2} ⊥ \ {x} = N2 are parallel lines in the affine plane defined
by y and N1.
For the next theorem we need q to be even so we can use part (iii) of
Theorem 15.1.3.
Theorem 15.2.2. Let S = (P, B, I) be a GQ of order (q, q 2 ), q even, for
which at least one pair {x, y} of distinct collinear points has Property (G).
Each “line” {y, u, v} ⊥⊥ of type (a) of Bxy defines a subquadrangle S ′ of order
q of S. In this way there arise q 3 + q 2 subquadrangles of order q of S. Also,
in any such subquadrnalge, each pair {y, u} with x ∈ {y, u} ⊥ (and so each
pair {x, z} with y ∈ {x, z} ⊥ ) is regular.
Proof. Assume that the two distinct “lines” M1, M2 of type (a) define the
same subquadrangle S ′ of order q. Met Mi ∪ {y} = {y, ui, zi} ⊥⊥ , i = 1, 2.

686 CHAPTER 15. PROPERTY G
We know {y, ui} is regular in S ′ , i = 1, 2, and {y, ui} ⊥′ ⊥ ′
= Mi ∪ {}, i = 1, 2.
Since {y, ui} ⊥′ ⊥ ′
= {vi, wi} ⊥⊥ for any two points vi, wi ∈ {y, ui} ⊥′ ⊥ ′
, i = 1, 2,
it is clear that {y, u1} ⊥′ ⊥ ′
∩{y, u2} ⊥′ ⊥ ′
= {y}. Also, as x ∈ {y, ui} ⊥′ , i = 1, 2,
any “line” of type (b) of Bxy containing a point of M1 also contains a point
of M2. Hence M1 and M2 are parallel“lines” in the affine “plane” of type (b)
of AG(3, q)) defined by M1.
Conversely, suppose M1 and M2 are distinct parallel “lines” of type (a)
of an affine “plane” of type (b) of AG(3, q). Just above we have a proof that
if r ∈ M ⊥ 2 , then r is in the subquadrangle S ′ of order q defined by M1. So
M ⊥ 2 ⊆ S′ . Since M2 is a “line” of the affine “plane” of type (b) defined by
M1, also M2 is contained in S ′ . Hence S ′ must also be the subquadrangle
determined by M2. Hence the “lines” of type (a) determine q4 +q3 q = q3 + q2 subquadrangles of order q.
Now consider a point u of the subquadrangle S ′ defined by M1, with
y ∼ u and x ∈ {y, u} ⊥ . As xu has a point in common with M1, the point u
belongs to the plane π defined by M1 and x. Let M2 be the line of π parallel
to M1 and passing through u. Then M2 also defines the subquadrangle S ′ ,
and it follows that {y, u} is regular in S ′ .
For the next theorem let S have Property (G) at the flag (x, L), L = xy,
and suppose that s is even. Let T = {y, z, u} be a triad with y ∈ T ⊆ x ⊥ .
Let S ′ be the subquadrangle with order q containing T ⊥ ∪ T ⊥⊥ .
Theorem 15.2.3. With the above notation, the point x is regular in S ′ . If
L has Property (G), then all points of L are regular in S ′ , L is regular in S ′
and L is regular in S.
Proof. Let w be a point of S ′ with x ∼ x, w ∼ y. Further, let r, r ′ , r ′′ be
distinct points of {x, w} ⊥′ , with rIL. Then {r, r ′ , r ′′ } ⊥ and {r, r ′ , r ′′ } ⊥⊥ are
contained in a subquadrangle S ′′ of order q. Clearly the lines xr, xr ′ , xr ′′ ,
wr, wr ′ , wr ′′ are common line of S ′ and S ′′ . And S ′ ∩ S ′′ is a subquadrangle
with parameters (q, t ′ ), t > 1. By Theorem 9.10.2 we have S ′ = S ′′ . It
follows that {r, r ′ } ⊥′ ⊥ ′
= {r, r ′ , r ′′ } ⊥⊥ , {r, r ′ } ⊥′ = (r, r ′ , r ′′ } ⊥ . So {r, r ′ } and
{x, w} are regular in S ′ . This shows that x is regular in S ′ . Hence if L has
Property (G), each point of L is regular in S ′ . Since q is even, L is regular
in S ′ .
Now consider a line M of S, M ∼ L. Let y ∼ x ′ IM, x ∼ y ′ IM, x ′′ ∈
{y, y ′ } ⊥ \{x, x ′ }. The subquadrangle S of order q defined by x, x ′ , x ′′ contains

15.2. PROPERTY (G) AND SUBQUADRANGLES 687
L and M. Since L is regular in S, the span {L, M} ⊥′ ⊥ ′
= {L, M} ⊥⊥ has size
q + 1. Hence L is regular in S.
Still with the hypothesis that S has Property (G) at the flag (x, L), fix u,
x = uIL. Let R = {u, u1, u2} ⊥⊥ = {u, u1, . . . , uq}, with {u, u1, u2} a triad in
x ⊥ . Put Li = xui, 1 ≤ i ≤ q. Let PxR be the set of all points different from
x and lying on the lines L, L1, . . . , Lq and L. Put BxR = {L, L1, L2, . . . , Lq},
and let CsR be the set of spans {w, w1, w2} ⊥⊥ , wIL, w1IL1, w2IL2, and
x ∈ {w, w1, w2}. Incidence IxR is the natural one.
Theorem 15.2.4. L = (PxR, BxR, CxR, IxR) is a Laguerre plane of order q.
For each point yIL, y = x, the derived or internal affine plane Ly of L at
y is the affine plane AG(2, q). Hence for q odd, L is the classical Laguerre
plane arising from a quadratic cone in P G(3, q).
Proof. Clearly |PxR| = q 2 + q, |BxR| = q + 1, |CxR| = q 3 . Each line of L is
incident with q points of O. (We will soon show that each circle is incident
with q + 1 points of L, one on each line of L.)
Let C = {w, w1, w2} ⊥⊥ , C ′ = {w ′ , w1, w2} ⊥⊥ be circles of CxR. Consider
w ′′ ∈ {w, w1, w2} ⊥⊥ \ {w, w1, w2}. Suppose that xw ′′ has no point of
{w ′ , w1, w2} ⊥⊥ . So w ′′ is collinear with no point of {w ′ , w1, w2} ⊥⊥ . Hence w ′′
is collinear with two points x, r of {w ′ , w1, w2} ⊥ . So r ∈ {w1, w2, w ′′ } ⊥ =
{w, w1, w2} ⊥ , which implies r ∼ w. But r, w, w ′ would be a triangle, impossible.
Hence xw ′′ has a point in common with {w ′ , w1, w2} ⊥⊥
If w = u, then {w, w1, w2} ⊥⊥ } has a point in common with each Li
by an earlier lemma. The preceding argument shows that {w, w1, w2} ⊥⊥
and {u, w1, w2} ⊥⊥ lie on the same lines L, L1, . . . , Lq through x. It follows
that each line of L and each circle of L have exactly one point in common.
Each triad of points of L is on at most one circle, so there are at most
(q 2 +q)(a 2 (q 2 −q)
(q+1)q(q−1) = q3 circles. But there are exactly q 3 circles, implying that
each triad of points of L is on a unique circle of L. Consequently L is a
Laguerre plane of order q. Let x = yIL. The internal affine plane Ly of
L at y is the affine plane of type (b) in AG(3, q), so must be isomorphic to
AG(2, q). So if q is odd, L is classical.
Let O be an oval in some P G(2, q) embedded in P G(3, q), and let V ∈
P G(3, q)\P G(2, q). Then let K be the cone obtained by projecting O from V .
And let L(O) be the corresponding Laguerre plane. Every known Laguerre
plane is some L(O), and if q is odd, each L(O) is classical, i.e., O is a conic.

688 CHAPTER 15. PROPERTY G
Theorem 15.2.5. Assume q is even and the Laguerre plane L obtained by
Property (G) at (x, L) is isomorphic to some L(O). Let C be a circle of L
and let S ′ be the subquadrangle of order q containing C and C ⊥ . The plane
πx from the regularity of the point x is Desarguesian
Proof. The lines of the plane πx not containing x are circles of L. Any two of
these lines have exactly one point in common. So we have q2 circles of L that
are pairwise mutually tangent. Let C1, . . . , Cq2 be the corresponding circles
of L(O). Since q is even, the q + 1 tangent planes of the cone K defining
L(O) contain a common line - the nucleus line N of K.
The nucleus of the ovals Ci is the intersection of N with any tangent
line of Ci. Since any two ovals of {C1, . . . , Cq2} are mutually tangent, they
have a common tangent line, hence have the same nucleus. Consequently
C1, . . . , Cq2 have a common nucleus n on N. Projecting K \ {x} from n, the
incidence structure formed by the points of K \ {x} and circles C1, . . . , Cq2 is isomorphic to a dual affine plane whose points are the lines of P G(3, q)
joining n to points of K \ {x}, and whose lines are the planes of P G(3, q)
containing n but not x. This dual affine plane is Desarguesian, hence the
dual affine plane obtained from πx by deleting x and the lines through x is
Desarguesian. It follows that the projective plane πx is Desarguesian.
Theorem 15.2.6. (J. A. Thas and H. Van Maldeghem [TVM97]) Let S =
(P, B, I) be a GQ of order (q 2 , q), q even, satisfying Property (G) at the point
x. Then x is regular in S and the dual net N ∗ x satisfies the Axiom of Veblen.
Consequently, N ∗ x ∼ = H 3 q .
Proof. By Theorem 15.2.3 the point x is regular. Let y be a point of the dual
net (i.e., y ∈ x⊥ \ {x}). Let A1 and A2 be distinct lines of N ∗ x containing
y, and let B1 and B2 be distinct lines of N ∗ x not containing y. Suppose
Ai ∩ Bj = ∅ for all i, j ∈ {1, 2}. Let z = A1 ∩ B1 and let zIM, x not incident
with M. Let xIL with z not on L; let u = A1 ∩ L; v = B1 ∩ L. Put C
equal to the line of S through u meeting M; D equal to the line of S through
v meeting M. Let N be the line of S through x and z. Since S satisfies
Property (G) at x, the triple {C, D, N} is 3-regular. By Theorem 15.1.3 the
lines through the points on the grid determined by {C, D, N} ⊥∪{C, D, N} ⊥⊥
form the lineset of a subquadrangle S ′ of order (q, q) of S. As x is regular for
S, it is also regular for S ′ . Hence the point x defines a projective plane πx of
order q. Clearly A1, A2, B1, B2 are lines of pix. Hence B1 and B2 intersect in
πx. Consequently N ∗ x satisfies the Axiom of Veblen. By the result of Thas
and De Clerck [TDC77] (Theorem 2.7.4), N ∗ x ∼ = H3 q .

15.3. TETRADIC SETS 689
15.3 Tetradic Sets
Let (∞, π∞) be an incident point-plane pair of Σ = P G(3, q). If X, Y, Z, W
are four distinct points of Σ \ π∞, then we say that {X, Y, Z, W } is a tetrad
with respect to (∞, π∞) provided {∞, X, Y, Z, W } is a cap of Σ such that
there exists a plane of Σ containing ∞ and exactly three of X, Y, Z, W . If
∞ and π∞ are understood, then we will refer to {X, Y, Z, W } simply as a
tetrad.
A tetradic set of ovoids with respect to (∞, π∞) is a set of ovoids of Σ,
each element of which contains ∞, has tangent plane π∞ at ∞ and such that
every tetrad with respect to (∞, π∞) is contained in a unique ovoid of the
set. If all the ovoids are elliptic quadrics, then we call the set a tetradic set of
elliptic quadrics. In this chapter we will be concerned primarily with tetradic
sets of elliptic quadrics.
Lemma 15.3.1. The number of tetrads with respect to (∞, π∞) is
q6 (q − 1) 3 (q − 2)(q + 1)
.
6
Proof. There are q + 1 lines ℓ in π∞ through ∞. Then there are q planes π
through ℓ and different from π∞. Then there are q
(unordered) triples of
3
lines through ∞ in π \ π∞. Once the three lines have been fixed, there are
q2 (q − 1) triples (X, Y, Z) of points, one on each of the three lines, such that
(∞, X, Y, Z) is a cap. Then there are (q3 +q 2 +q +1−2(q2 +q +1)+(q +1) =
q3 − q2 choices for a point W ∈ Σ \ (π ∪ π∞) such that (∞, X, Y, Z, W ) is a
cap. Multiplying these numbers together gives the desired result.
Let Θ be a tetradic set of ovoids with respect to the incident point-plane
pair (∞, π∞).
Lemma 15.3.2. The size of Θ is q 3 (q − 1).
Proof. If π∞ is tangent to the ovoid Ω at the point ∞, there are q2 + q
secant
planes through ∞. In each secant plane π through ∞, there are
q
(unordered) triples {X, Y, Z} of Ω \ {∞}. There are q + 1 ovoid points
3
of π and (q2 + 1) − (q + 1) = q2 − q points W of Ω \ π. Thus there are
(q2 + q) q 2 q
(q − q) = 3
3 (q+1)(q−1) 2 (q−2)
tetrads with respect to (∞, π∞) in Ω.
6
As each tetrad with respect to (∞, π∞) is contained
in a unique ovoid in Θ,
q6 (q−1) 3 (q+1)(q−2) q3 (q−1) 2 (q+1)(q−2)
it must be that there are
/
= q 6
6
3 (q − 1)
ovoids in a tetradic set of ovoids.

690 CHAPTER 15. PROPERTY G
We leave a proof of the following lemma to the reader.
Lemma 15.3.3. The homographies that are central collineations of P G(3, q)
with center ∞ = (1, 0, 0, 0) are those with matrices of the form
⎛
1
⎜ a21
A = ⎜
⎝ a31
0
λ
0
0
0
λ
⎞
0
0 ⎟
0 ⎠
a41 0 0 λ
.
It is easy to check that
⎛
A −1 =
⎜
⎝
1 0 0 0
−a21λ −1 λ −1 0 0
−a31λ −1 0 λ −1 0
−a41λ −1 0 0 λ −1
⎞
⎟
⎠ .
Suppose such a collineation θ fixes each point of a tetrad (w.r.t. (∞, π∞))
where π is the plane different from π∞ containing three of the points of the
tetrad different from ∞. Since θ fixes each point of a 4-arc of π it must fix
all points of π. So θ must be an elation with axis π and center ∞. But
then if it fixes the fourth point of the tetrad, a point not in π, it must be
the identity map. Hence each tetrad belongs to an orbit of size q 3 (q − 1).
Moreover, if we let G∞ be the group of all central collineations with center
∞, for each θ ∈ G∞ and each point R different from ∞, the points ∞, R
and R θ are collinear. Hence if T is a tetrad contained in an ovoid Ω having
π∞ as a tangent plane at ∞, then Ω and Ω θ intersect in just the point ∞,
so that each point of T is in just one of the ovoids Ω θ , θ ∈ G∞. Hence the
q 3 (q + 1)(q − 1) 2 (q − 2)/6 tetrads contained in Ω give rise under the action
of G∞ to q 3 (q − 1) × q 3 (q + 1)(q − 1) 2 (q − 2)/6 = q 6 (q − 1) 3 (q + 1)(q − 2)
tetrads. Since this is the total number of tetrads of the appropriate kind, the
following lemma has been proved.
Lemma 15.3.4. Let Ω be an ovoid having π∞ as its tangent plane at the point
∞, and let G∞ be the group of all homographies that are central collineations
with ∞ as center. Then
is a tetradic set of ovoids.
Θ = {Ω θ : θ ∈ G∞}

15.3. TETRADIC SETS 691
Until further notice let (∞, π∞) be an incident point-plane pair of P G(3, q),
and let Θ be a tetradic set of ovoids with respect to (∞, π∞).
Lemma 15.3.5. Let X, Y, Z ∈ P G(3, q)\π∞ be such that ∞ ∈ 〈X, Y, Z〉 = π
and {X, Y, Z, ∞} is a 4-arc in π. Then there are q elements Ω1, . . . , Ωq of Θ
containing X, Y, Z and |Ωi ∩ Ωj| ≤ q + 1, for i, j ∈ {1, . . . , q}, i = j.
Proof. If W is any point of P G(3, q)\(π ∪π∞), then {X, Y, Z, W } is a tetrad
with respect to (∞, π∞). Hence the ovoids of Θ containing X, Y, Z must
partition the points of P G(3, q) \ (π ∪ π∞) into sets of size q 2 − q. It follows
that there are q such ovoids Ω1, . . . , Ωq, and |Ωi ∩ Ωj| ≤ q + 1 for i = j since
the intersection |Ωi ∩ Ωj| ≤ q + 1 must be contained in π.
Lemma 15.3.6. Two distinct elements of Θ intersect in at most q+3 points.
If q is odd, this is at most q + 2 points.
Proof. Let Ω, Ω ′ ∈ Θ, Ω = Ω ′ . If Ω ∩ Ω ′ contains three points defining a
plane containing ∞, then by the previous lemma |Ω∩Ω ′ | ≤ q +1. So suppose
that each plane of P G(3, q) on ∞ meets Ω ∩ Ω ′ in at most two points. Hence
the lines joining the point ∞ with the points of (Ω ∩ Ω ′ ) \ {∞} form an arc
in the quotient space P G(3, q)/∞. Thus |(Ω ∩ Ω ′ ) \ {∞}| ≤ q + 2, implying
that |(Ω ∩ Ω ′ )| ≤ q + 3. It is clear that this bound can be improved by 1
when q is odd.
Lemma 15.3.7. Let q be even. If Ω, Ω ′ ∈ Θ, Ω = Ω ′ are such that Ω ∩ Ω ′
contains three points X, Y, Z spanning a plane π containing ∞, then Ω∩Ω ′ =
π ∩ Ω, implying |Ω ∩ Ω ′ | = q + 1.
Proof. Let X, Y, Z ∈ P G(3, q) \ π∞ be such that ∞ ∈ π = 〈X, Y, Z〉 and
such that {X, Y, Z, ∞} is a 4-arc in π. By Lemma 15.3.3 there are q ovoids
Ω1, . . . , Ωq of Θ containing X, Y, Z, and for i = j, Ωi ∩ Ωj ⊆ π ∪ π∞, i.e.,
Ωi ∩ Ωj ⊆ Ωi ∩ π, so |Ωi ∩ Ωj| ≤ q + 1. What we want to show is that
Ωi ∩ π = Ωj ∩ π for all i, j. So by way of looking for a contradiction, we
assume that this is not the case. We may label the Ωi so that WOLG we
have O1 = Ω1 ∩ π = Ω2 ∩ π = O2. Then let X ′ ∈ O2 \ O1, i.e., X ′ ∈ Ω2 \ Ω1,
but X ′ ∈ π, hence also π = 〈X ′ , Y, Z〉. Again by Lemma 15.3.3 there are q
elements Ω1, . . . , Ωq of Θ that each contain X ′ , Y, Z (and are tangent to π∞
at ∞). Clearly Ω2 is one of these ovoids, and without loss of generality we
may assume that Ω2 = Ω2. We know that Ω1 \π, Ω2 \π, . . . , Ωq \π partitions
the points of P G(3, q)\(π ∪π∞), and also that Ω1 \π, Ω2 \π = Ω2 \π, Ω3 \π,

692 CHAPTER 15. PROPERTY G
. . . , Ωq \ π partition the points of P G(3, q) \ (π ∪ π∞). Since X ′ ∈ Ωj for all
j and X ′ ∈ Ω1, clearly Ω1 = Ωi for all i. Consequently for some i, Ωi is not
any Ωj. Without loss of generality we may suppose that Ω1 = Ωi for all i.
Now consider Ω1∩Ωi, i ∈ {1, . . . , q}. Since {X, Y, Z} ⊂ Ω1∩Ω2, it follows
that |Ω1∩Ω2| ≤ q+1 and |(Ω1∩Ω2)\π| = 0 since Ω1∩Ω2 = Ω1∩Ω2 ⊆ π. Also
|Ω1 ∩ Ωi| ≤ q + 3 and |(Ω1 ∩ Ωi) \ π| ≤ q for i = 2, since Y, Z, ∞ ∈ Ω1 ∩ Ωi ∩ π
for i = 2. However, Ω1 \ π, Ω3 \ π, . . . , Ωq \ π partitions the q 2 − q points
of Ω1 \ π, from which it follows that |(Ω1 ∩ Ωi) \ π| = q for i = 2. This
implies that |Ω1 ∩ Ωi| = q + 3 for i = 2. This forces Ω1 ∩ Ωi ∩ π = {Y, Z, ∞},
so that X ∈ Ωi for i = 2. Also for i = 2 the lines spanned by ∞ and
the points of (Ωi ∩ Ω1) \ {∞} correspond to an hyperoval in the quotient
space P G(3, q)/∞, implying that each plane on ∞ contains two points of
(Ωi ∩ Ω1 \ {∞} or no points of (Ωi ∩ Ω1) \ {∞}. However, let π ′ be any plane
through the line 〈∞, X〉 different from the plane π (so in particular π ′ = π∞).
Then the ovoids Ωi with i = 2 partition the q −1 points of (Ω1 ∩π ′ )\〈∞, X〉.
This forces q − 1 to be even, which is a contradiction. Hence we know that
Ωi ∩ π = Ωj ∩ π for all i, j, forcing |Ωi ∩ Ωj| = q + 1 for all i = j.
15.4 Tetradic Sets of Elliptic Quadrics
In this section let Θ be a tetradic set of q 3 (q−1) elliptic quadrics with respect
to the incident point-plane pair (∞, π∞).
Let {X, Y, Z, ∞} be a 4-cap of Σ with ∞ ∈ π = 〈X, Y, Z〉 = π∞. There
must be a unique conic C of π containing X, Y, Z, ∞ and with tangent line
ℓ∞ = π ∩ π∞. Hence any elliptic quadric of Θ containing X, Y, Z must also
contain C. Moreover, if W is any point of Σ \ (π ∪ π∞), then {X, Y, Z, W }
is a tetrad. Hence the elliptic quadrics of Θ containing C must partitioin
the points of Σ \ (π ∪ π∞). It follows that there are exactly q such elliptic
quadrics, and they meet pairwise exactly in C. From Section felliptic 5 we
see that these q quadrics are an orbit of the group of elations with center ∞
and axis π∞. More generally, by applying that section to any oval of Θ we
have the following theorem.
Theorem 15.4.1. Let Θ be a tetradic set of q 3 (q − 1) elliptic quadrics with
respect to the pair (∞, π∞). Let G1 be the group of q 3 elations of P G(3, q)
having ∞ as center. G1 acts on Θ with q − 1 orbits, each of size q 3 . This
permits us to define an equivalence relation ˆ⊲⊳ on Θ by O1 ˆ⊲⊳O2 if and only

15.4. TETRADIC SETS OF ELLIPTIC QUADRICS 693
if there is a g ∈ G1 for which O2 = O g
1 . Moreover, if O ∈ Θ, let [O] denote
the equivalence class of O. There is a unique rosette of elliptic quadrics
contained in [O] with base point ∞, base plane π∞ and containing O. The
rosette is generated by the action of the elations with center ∞ and axis π∞.
The remaining q3 − q elliptic quadrics in [O] share a common conic with
O. Hence two elliptic quadrics in [O] either intersect in the point ∞ or in
a conic containing ∞. We also note that the q3 elliptic quadrics in [O] are
partitioned into q2 disjoint rosettes with base point ∞.
Even more is true. Let G2 be the group of q3 elations of Σ with axis π∞.
Then the orbits of G2 on the elements of Θ are the same as those of G1, and
hence the same as those of the group G = 〈G1, G2〉 of order q5 .
Lemma 15.4.2. Let X, Y, Z be three distinct, non-collinear points of P G(3, q)\
π∞ not coplanar with ∞. Then there are exactly q − 1 elliptic quadrics of Θ
containing {X, Y, Z}, one from each equivalence class.
Proof. Let π = 〈X, Y, ∞〉 and let ℓ be a line of π incident with ∞ but not
with X nor Y . Let W = ℓ∩〈X, Y 〉. If A ∈ ℓ\{W, ∞}, then there is a unique
elliptic quadric of Θ containing the tetrad {X, Y, Z, A}. There are q − 1 such
points A and hence q − 1 such elliptic quadrics, since any elliptic quadric of
Θ must contain a point of ℓ \ {∞}. As two equivalent elliptic quadrics must
intersect in either the single point ∞ or in a containing ∞, it follows that the
q − 1 elliptic quadrics on {X, Y, Z, ∞} are in distinct equivalence classes.
Lemma 15.4.3. Let O1 and O2 be two inequivalent elliptic quadrics of Θ.
Then |O1 ∩ O2| ≤ q + 2.
Proof. The elliptic quadrics O1 and O2 may not intersect in a conic containing
∞ and hence no plane on ∞ contains more than two points of
(O1 ∩ O2) \ {∞}. Thus the lines of P G(3, q) spanned by ∞ and points
of (O1 ∩ O2) \ {∞} form an arc in the quotient space P G(3, q)/∞ with tangent
line π∞/∞. We now show that such an arc has size at most q + 1,
forcing |O1 ∩ O2| ≤ q + 2. So suppose that there is such an arc with size
q + 2. This means that each plane (not π∞) through ∞ must contain 1 or
three points of the intersection of the two elliptic quadrics. By Lemma 6.5.1
there is a plane π containing one of the lines in the arc of the quotient space
for which the pole N of π is the same for both O1 and O2, and hence π contains
three points of (O1 ∩ O2), say ∞, X, Y . So π ∩ Oi is a conic Ci, i = 1, 2,
containing ∞, X and Y , and with both conics having the same nucleus, i.e.,

694 CHAPTER 15. PROPERTY G
tangents at these three points. Hence C1 = C2. But the remaining q points of
(O1 ∩O2) outside the plane π mean that there are 2q +1 > q +3 points in the
intersection of the quadrics, an impossibility. Hence |(O1 ∩ O2)| ≤ q + 2.
Lemma 15.4.4. Let O1 and O2 be inequivalent elliptic quadrics of Θ with
|O1 ∩ O2| ≥ 3. Then |O1 ∩ O2| = q + 2.
Proof. Consider fixed X, Y ∈ (O1 ∩ O2) \ {∞}, X = Y , and let [O2] be
the equivalence class of O2. There are q2 − q triples {X, Y, Z} such that
X, Y, Z ⊂ O1 and ∞ ∈ 〈X, Y, Z〉. By Lemma 15.4.2 each such triple is
contained in a unique element of [O2].
We know that X, Y ∈ O2 and that any other elliptic quadric O ′ 2
∈ [O2]
with X, Y ∈ O ′ 2 meets O2 in points contained in a plane on ∞, which must
be 〈X, Y, ∞〉. Further, it must be that O2 ∩ O ′ 2 = O2 ∩ 〈X, Y, ∞〉. There are
exactly q elliptic quadrics of [O2] containing O2 ∩ 〈X, Y, ∞〉.
Now count pairs (O ′ 2, {X, Y, Z}) where O ′ 2 ∈ [O2] and X, Y, Z are distinct
points of O ′ 2 . From above we know that the count is in fact q2 − q. However,
we also have that |O1 ∩O2| ≤ q +2 and so there are at most q −1 such triples
{X, Y, Z} and q such O ′ 2 . Hence the count is bounded above by q(q − 1) =
q2 − q. It follows that |O1 ∩ O ′ 2 | = q + 2 and certainly |O1 ∩ O2| = q + 2.
Corollary 15.4.5. If O1 and O2 are inequivalent elliptic quadrics of Θ, then
|O1 ∩ O2| = 2 or q + 2.
Proof. We show that |O1 ∩ O2| = 1 is impossible. Consider the rosette R
of elliptic quadrics in [O2] containing O2. The elliptic quadric O1 intersects
each of the q elliptic quadrics of the rosette in 1, 2, or q + 2 points. The
elements of R also partition the points of P G(3, q)\π∞ . Hence the q 2 points
of O1 \ {∞} are partitioned into q sets of size 0, 1, or q + 1. This can only
be done with one set of size 1 and q − 1 of size q + 1.
Lemma 15.4.6. Let O be an elliptic quadric of Θ and E an equivalence
class of Θ such that O ∈ E. If X ∈ O \ {∞}, then there is a unique O ′ ∈ E
such that O ∩ O ′ = {X, ∞}.
Proof. For fixed X ∈ O \ {∞} the number of pairs (Y, Z) with Y, Z ∈ O and
Y = Z such that ∞ ∈ 〈X, Y, X〉 is (q 2 − 1)(q 2 − q). By Lemma 15.4.2 each
triple {X, Y, Z} is contained in a unique element of E. Further, each of the
q 2 rosettes of E contains a unique elliptic quadric on the point X. If such an
elliptic quadric O ′ intersects O in q + 2 points, then we have q(q − 1) pairs

15.4. TETRADIC SETS OF ELLIPTIC QUADRICS 695
(Y, Z) with Y, Z ∈ O ∩ O ′ and Y = Z such that ∞ ∈ 〈X, Y, Z〉. If on the
other hand, |O ∩ O ′ | = 2, then there are no such pairs (Y, Z).
Hence it follows that there are q 2 − 1 elliptic quadrics of E containing X
and meeting O in q + 2 points and a unique elliptic quadric of E containing
X and meeting O in 2 points.
Let T be a set of q − 1 ovoids of P G(3, q) meeting pairwise in exactly
two fixed points and sharing the tangent planes at those two fixed points.
Such a set T is called a transversal of ovoids. These two common points are
called the base points of the transversal and the two common tangent planes
are called the base planes of the transversal.
Lemma 15.4.7. Let O be an elliptic quadric of Θ and X ∈ O \ {∞}. The
q − 2 elliptic quadrics meeting O in exactly {X, ∞} also meet pairwise in
exactly {X, ∞} and have a common tangent plane at X, thus together with
O forming a transversal.
Proof. Let πX be the tangent plane to O at X. For Y ∈ P G(3, q) \ (π∞ ∪
πX ∪ 〈∞, X〉 ∪ O) count pairs (Z, O ′ ) with Z ∈ O \ {X, ∞}, O ′ ∈ Θ and
{X, Y, Z, ∞} ⊂ O ′ . Suppose that Z ∈ O \ 〈X, Y, ∞〉. Then there are q 2 − q
chooices for Z and {X, Y, Z} is contained in q − 1 elliptic quadrics, giving
(q 2 − q)(q − 1) pairs.
Now suppose thqt Z ∈ O ∩ 〈X, Y, ∞〉 and let C = O ∩ 〈X, Y, ∞〉. Note
that Y ∈ C. Since Y ∈ πX it follows that 〈X, Y 〉 is not tangent to C and
hence meets C in a point of C \ {X}. Similarly, 〈Y, ∞〉 meets C in a second
poitn, leaving q − 3 possible choices for Z. For each such choice of Z the
points X, Y, Z define a unique conic C ′ in ∠X, Y, ∞〉 containing ∞ and with
tangent π∞ ∩ 〈X, Y, ∞〉. There are q elliptic quadrics of Θ containing C ′
giving q(q − 3) pairs (Z, O ′ ) with Z ∈ O ∩ 〈X, Y, ∞〉. ] So in total we have
(q 2 − q)(q − 1) + q(q − 3) = q(q − 2)(q + 1) pairs (Z, O ′ ).
Counting these pairs in a second way we consider the number of elliptic
quadrics of Θ containing {X, Y }. In 〈∞, X, Y 〉 there are q − 1 conics on
X, Y, ∞ with tangent 〈∞, X, Y 〉∩π∞, which means there are q(q −1) elliptic
quadrics containing {X, Y }, q in each class. If such an elliptic quadric is in
the same class as O, then we know that the possible intersection sizes with
O are 1 and q + 1, so they must all be q + 1, since the intersection is at
least {∞, X}. This gives q(q − 1) pairs (Z, O ′ ). there are q(q − 2) elliptic
quadrics containing {X, ∞} which are inequivalent to O, and by the earlier

696 CHAPTER 15. PROPERTY G
counts there are exactly q(q − 2)(q + 1) − q(q − 1) = q(q 2 − 2q − 1) pairs
(Z, O ′ ) where O ′ is of this type. In this case |O ∩ O ′ | = 2 or q + 2, so each
O ′ gives rise to 0 or q pairs (Z, O ′ ), respectively. From this it follows that
there must be q 2 − 2q − 1 elliptic quadrics intersecting O in q + 2 points and
(q 2 − 2q) − (q 2 − 2q − 1) = 1 intersecting O in the two points X, ∞.
Hence there is a unique elliptic quadric of Θ meeting O in exactly {X, ∞}
and containing the point Y ∈ P G(3, q) \ (π∞ ∪ πX ∪ 〈∞, X〉 ∪ O).
In Lemma 15.4.6 we saw that there are q−2 elliptic quadrics of Θ meeting
O in exactly {X, ∞}. By the above, these q −2 elliptic quadrics plus O cover
the q 3 − a 2 − q + 1 = (q 2 − 1)(q − 1) points of P G(3, q) \ (π∞ ∪ πX ∪ 〈∞, X〉).
It follows that these elliptic quadrics partition the pointset into q − 2 sets
of size q 2 − 1. Hence the elliptic quadrics meet pairwise in exactly {X, ∞}
and have πX as tangent plane at X, thus forming a transversal of elliptic
quadrics.
Lemma 15.4.8. Let (X, π) be an incident point-plane pair of P G(3, q) such
that X ∈ π∞ and ∞ ∈ π. Then there are exactly q − 1 elliptic quadrics of
Θ containing X and with tangent plane π at X. Further, these q − 1 elliptic
quadrics form a transversal with one elliptic quadric from each equivalence
class of Θ.
Proof. Since each rosette of elliptic quadrics contained in Θ is generated by
the action on one elliptic quadric of the elations of P G(3, q) with center ∞
and axis π∞, it follows that each plane not on ∞ is tangent to exactly one
elliptic quadric of a given rosette.
There are (q − 1)q2 rosettes of elliptic quadrics in Θ, so (q − 1)q2 elliptic
quadrics of Θ with π as a tangent plane at one of the poins of π \ π∞. By
Lemma 15.4.7 if π is tangent to one elliptic quadric of Θ at a point, then it
is tangent to the q − 1 elliptic quadrics of a transversal of Θ at that point.
Since two elliptic quadrics in the same equivalence class have intersection
size 1 or q + 1, it follows that the elliptic quadrics of the transversal are one
from each equivalence of class of Θ.
Suppose that for X ∈ π \ π∞ there are two transversals O1, . . . Oq−1 and
O ′ 1 , . . . , O′ q−1
containing X and with tangent plane π. We investigate how
O ′ 1 intersects O1, . . . , Oq−1. In fact O1, . . . , O partition the q 2 − 1points of
O ′ 1 \ {X, ∞} into q − 1 sets of size q or q − 1, which is a contradiction.
Hence there can be only one transversal of elliptic quadrics of Θ with
X as a base point and π as the corresponding base plane. Since there are

15.5. FLOCKS AND TETRADIC SETS 697
q 2 (q − 1) elliptic quadrics with tangent plane π, there are q 2 transversals of
Θ with base plane π and a unique such transversal with base point X for
each X ∈ π \ π∞.
Lemma 15.4.9. Let T = {O1, . . . , Oq−1} be a transversal of elliptic quadrics
of Θ with base point X and base plane π, X ∈ π∞ and ∞ ∈ π. Then every
plane π ′ such that X, ∞, π∞ ∩ π ⊂ π ′ is tangent to a unique element of
T . Further, any two elements of T have only π and π∞ as common tangent
planes.
Proof. Let π ′ be a plane such that X, ∞, π∞ ∩ π ⊂ π ′ . The elements of
T partition the q 2 − q − 1 points of π ′ \ (π∞ ∪ π ∪ 〈∞, X〉), which can only
be into q − 2 conics and one single point. That is, π ′ is tangent to a unique
element of T .
There are (q 2 − 1)(q − 1) such planes, which is the same as the number of
pairs (Oi, π ′′ ) where π ′′ is tangent to Oi ∈ T at a point not ∞ nor X. Hence
no two elements of T have a common tangent not at ∞ or X. Since a plane
of P G(3, q) (distinct from π∞ and π) on π∞ ∩ π or on ∞ cannot be a tangent
plane to an elliptic quadric of T , the lemma is proved.
15.5 Flocks and Tetradic Sets
Before reading the next two sections it would be wise to review Sections 6.5
and 6.6.
Theorem 15.5.1. Let F be a flock of a quadratic cone in P G(3, q). Then
F gives rise to a tetradic set of elliptic quadrics of P G(3, q).
Proof. Let F be a flock of the quadratic cone K of P G(3, q). Let V be the
vertex of K, P a point of K \ {V } and π a plane of P G(3, q) such that
P, V ∈ π. If we project the elements of F from P onto π we obtain a set
{C1, . . . , Cq−1, m} where C1, . . . , Cq−1 are q − 1 conics and m is a line of π.
The conics C1, . . . , Cq−1 meet pairwise in the point ∞ = 〈V, P 〉 ∩ π and with
common tangent ℓ where 〈ℓ, V 〉 is the plane meeting K in the line 〈V, P 〉.
The line m is the intersection of π with the plane containing the element of
F containing P . Hence the points of m are disjoint from all the Ci. (Review
Section 6.2.)
Now let π∞ be a plane of P G(3, q) containing ℓ and distinct from π. Let
R be any point of P G(3, q) \ (π ∪ π∞) and πR the plane 〈m, R〉. Then by

698 CHAPTER 15. PROPERTY G
Theorem 6.6.1 there is a unique set of elliptic quadrics T = {O1, . . . , Oq−1}
such that Oi ∩ π = Ci, 1 ≤ i ≤ q − 1, R is contained in all of the Oi and
π∞ and πR are tangent planes to all of the Oi. Further, O1, . . . , Oq−1 is a
transversal of elliptic quadrics.
Let G be the group of q 3 elations of P G(3, q) with center ∞, and define
Θ = {O g : g ∈ G, O ∈ T }. We show that Θ is a tetradic set of elliptic
quadrics. Since no elliptic quadric of T may be fixed by an element of G,
it follows that |Θ| = q 3 (q − 1), and it suffices to show that each tetrad of
P G(3, q) with respect to (∞, π∞) is contained in at least one elliptic quadric
of Θ.
Now let π ′ be a plane containing ∞ and distinct from from π∞. Let
Oi ∩ π ′ = C ′ i , 1 ≤ i ≤ q − 1, and m′ = πR ∩ π ′ . Any element of G fixes
the plane π ′ and induces an elation with center ∞ in π ′ . No such elation in
π ′ fixes a C ′ i and also cannot map C ′ i to C ′ j for 1 ≤ i = j ≤ q − 1. Hence
{(C ′ i )g : 1 ≤ i ≤ q − 1, g ∈ G} is a set of q 2 (q − 1) conics in π ′ containing
∞ and with π∞ ∩ π ′ as tangent line. Hence every conic in π ′ containing ∞
and with π∞ ∩ π ′ as tangent is contained in an element of Θ. In fact, taking
images under the group of elations with center ∞ and axis π ′ we get a set of
q elliptic quadrics partitioning the points of P G(3, q) \ (π∞ ∩ π ′ . It follows
from this that every tetrad of P G(3, q) with respect to (∞, π∞) is contained
in an element of Θ, implying that Θ is a tetradic set of elliptic quadrics with
respect to (∞, π∞).
Theorem 15.5.2. Every tetradic set of elliptic quadrics of P G(3, q) arises
from a flock of a quadratic cone of P G(3, q).
Proof. Suppose that Θ is a tetradic set of elliptic quadrics of P G(3, q) with
respect to (∞, π∞). Let T = {O1, . . . , Oq−1} be a transversal of elliptic
quadrics of Θ with base point R and base plane πR. Let π be any plane
containing ∞, but not containing R and distinct from π∞. Consider the set
{C1, . . . , Cq−1, m} where Ci = Oi ∩ π, 1 ≤ i ≤ q − 1, and m = π ∩ πR. By
Obs. 6.2.1 and Lemma 15.4.8 there is no pair (Ci, Cj), i, j ∈ {1, . . . , q − 1},
i = j, such that Ci is the image of Cj under an elation of π with center ∞.
Hence {C1, . . . , Cq−1, m} is a flock in the planar model of the quadratic cone
of P G(3, q).
At this point we want to view the planar representation of a flock of a
quadratic cone using coordinates. So first suppose that K is the cone in

15.5. FLOCKS AND TETRADIC SETS 699
P G(3, q) with equation x0x2 = x2 1
xt yt
C =
At =
and vertex V = (0, 0, 0, 1). Let
0 zt
: t ∈ Fq
be a q-clan with associated flock F of K with planes
F = {πt = [xt, yt, zt, 1] : t ∈ Fq}.
Let P = (0, 0, 1, 0) and Pλ = (0, 0, 1, λ). We are going to project the
planes of the flock from P onto the plane [0, 0, 1, 0] through the vertex V .
For λ = 1, the plane πλ meets the cone K in the conic
Cλ = {(1, t, t 2 , −gv(1, t)) : t ∈ Fq} ∪ {(0, 0, 1, λ)},
where gλ(1, t) = xλ + tyλ + t 2 zλ.
Projecting Cλ to [0, 0, 1, 0] (for λ = 1) from P gives the conic
Cλ = {(1, t, 0, −gv(1, t)) : t ∈ Fq} ∪ {(0, 0, 0, 1)}.
The plane π1 maps to a line n disjoint from all the Cλ, λ = 1.
The homography (x0, x1, x2, x3) ↦→ (x0, x1, x3, x2 + x0) maps the conics
Cλ to the conics
C ′ λ = {(1, t, −gv(1, t), 1) : t ∈ Fq} ∪ {(0, 0, 0, 1)},
and maps the line n to the line n ′ = [1, 0, 0, 0] ∩ [0, 0, 0, 1]. These are conics
that pairwise meet in the point (0, 0, 1, 0) with common tangent n ′ at
(0, 0, 1, 0). Since they are projections from the conics of a flock, it also follows
that no two of them are in the same orbit of the elations of [0, 0, 1, 0]
with center (0, 0, 1, 0) and axis n. (See ??) This completes a proof of the
following.
Lemma 15.5.3. Let F ′ be the following set of q − 1 conics and one line n ′
in the plane [1, 0, 0, −1]:
and
F ′ = {C ′ λ = {(1, t, −gv(1, t), 1) : t ∈ Fq} ∪ {(0, 0, 1, 0)} : 1 = λ ∈ Fq}
n ′ = [1, 0, 0, 0] ∩ [0, 0, 0, 1].
This is projectively equivalent to a planar representation of the flock Fof
the cone K : x0x2 = x 2 1 whose planes are of the form
πt = [xt, yt, zt, 1], t ∈ Fq.

700 CHAPTER 15. PROPERTY G
15.6 Coordinates for the AG(3, q) from Flock
GQ
Let G be the group with elements (α, c, β), α, β ∈ F 2 q , c ∈ Fq and binary
operation
(α, c, β) · (α ′ , c ′ , β ′ ) = (α + α ′ , c + c ′ + β ◦ α ′ , β + β ′ ),
where (a1, a2) ◦ (b1, b2) = a1b1 + a2b2. The center of G is the group
Z = {(0, c, 0) : c ∈ Fq}.
Start with a normalized q-clan C = {At : t ∈ Fq} (i.e., A0 is the zero
matrix, and put Kt = At + A T t . Construct subgroups
and
A(t) = {(α, gt(α), αKt) : α ∈ F 2 q }, t ∈ Fq,
A(∞) = {(0, 0, β) : β ∈ F q 2}.
For each t ∈ ˜ Fq put A ∗ (t) = A(t)Z.
Proceed as in Chapter 13 to construct a GQ S with parameters (s, t) =
(q 2 , q) and recall that S has Property (G) at the point (∞). Then consider
the point-line dual ˜ S of S.
So (∞) is a line with points [A(t)]for each t ∈ ˜ Fq. For each t ∈ ˜ Fq and
each g ∈ G, the coset A ∗ (t)g is a line through the point [A(t)] and incident
with the points A(t)hg, h ∈ A ∗ (t). Also, each g ∈ G is a line incident with
points A(t)g. Since ˜ S has Property (G) at the line (∞), in particular it has
Property (G) at the two flags (X = [A(∞)], (∞)) and (Y = [A(0)], (∞)).
Before proceeding we remind the reader that
A(∞)(α, c, β) = A(∞)(α, c − α ◦ β, 0) (15.1)
A(0)(α, c, β) = A(0)(0, c, β). (15.2)
Construct the affine space ˜ SXY ∼ = AG(3, q). The points of AG(3, q) are
the points of ˜ S collinear with X but not on the line XY = (∞). These
are the points that are of the form A(∞)(α, c, β) = A(∞)(α, c − α ◦ β, 0).

15.6. COORDINATES FOR THE AG(3, Q) FROM FLOCK GQ 701
So the points of AG(3, q) have the form A(∞)(α, c, 0). Give the point
A(∞)((a1, a2, c, 0) the coordinates (a1, a2, c).
The planes of type (a) of AG(3, q) are the sets {X, Z} ⊥ \{Y }, where X ∼
Z ∼ Y . Such a point Z looks like A(0)g = A(0)(α, c, β) = A(0)(0, c, b1, b2),
where β = (b1, b2). So if we write Z = A(∞)(0, c, b1, b2), then
{X, Z} ⊥ \ {Y } = {A(∞)(α, c, β) : α ∈ F q 2}
= {(a1, a2, c − a1b1 − a2b2) : a1, a2 ∈ Fq}. (15.3)
Embed the affine space AG(3, q) in P G(3, q) by
(a1, a2, c) ↩→ (a1, a2, c, 1),
which is never on the plane π∞ = [0, 0, 0, 1]. Then the plane {X, Z} ⊥ \ {Y }
of type (a) becomes the plane [b1, b2, 1, −c].
The line A ∗ (∞)g = A ∗ (∞)(α, c, β) = A ∗ (∞)(α, 0, 0) of type (b) is mapped
to the line
[1, 0, 0, −a1] ∩ [0, 1, 0, −a2] = {A(∞)(α, c, 0) : c ∈ Fq}.
Each such line of type (b) contains the point ∞ = (0, 0, 1, 0) and is not
contained in the plane [0, 0, 0, 1].
A line of type (a) is the intersection of two planes of type (a) and never
contains the point (0, 0, 1, 0). We ask: can two planes of type (a) meet in a
line of π∞?
(x, y, z, 0) ∈ [b1, b2, 1, −c] ∩ [b ′ 1 , b′ 2 , 1, −c′ ] iff x(b1 − b ′ 1 ) + y(b2 − b ′ 2 ) = 0.
If even one of b1 − b ′ 1 , b2 − b ′ 2 is nonzero, then we get a single point (x, y, z, 0)
in the intersection of the three planes. So suppose that b1 = b ′ 1 and b2 = b ′ 2 .
If c = c ′ , then [b1, b2, 1, −c] ∩ [b1, b2, 1, −c ′ ] is a line of π∞ = [0, 0, 0, 1].
The line A∗ (0)(0, 0, β)of ˜ S contains the points A(0)(0, c, β) for all c ∈ Fq.
These points represent parallel planes of type (a) in AG(3, q).
If g is any line of ˜ S not meeting the line (∞), and g ′ is any other such
line, there is a collineation of ˜ S fixing all points of the line (∞) and mapping
g ′ to g. This induces a collineation of the affine space AG(3, q) and hence
of the corresponding projective space P G(3, q) (fixing the point (0, 0, 1, 0)
since the lines of type (b) are permuted among themselves). Hence we may

702 CHAPTER 15. PROPERTY G
assume that g = (0, 0, 0), since g ∼ XY = (∞). The point U on g collinear
with Y = [A(0)] is the point A(0)(0, 0, 0) = A(0). This point corresponds
to the plane [0, 0, 1, 0]. The point r = A(∞) is the point on g collinear with
X = [A(∞)], and r corresponds to the point (0, 0, 0, 1). Now for 0 = v ∈ Fq
we want to compute points of the ovoid OZ where Z = A(v). So we need to
determine which points of the form A(∞)(α ∗ , c ∗ , β ∗ ) are collinear with
A(v) = {((a1, a2), a 2 1 xv+a1a2yv+a 2 2 zv, 2a1xv+a2yv, a1yv+2a2zv) : a1, a2 ∈ Fq}.
A(∞)(α ∗ , c ∗ , β ∗ ) ∼ A(v) if and only if
A(∞)(α ∗ , c ∗ − α ∗ ◦ β ∗ , 0) ∩ A(v) = ∅.
It is easy to see that we need α ∗ = (a1, a2). Using the two other equalities
imposed on the system by this condition we find the points of OZ are those
points of the form
A(∞)((a1, a2), −a 2 1 xv − a1a2yv − a 2 2 zv, 0)
↩→ (a1, a2, −)a1, a2, −(a1, a2)Av
a1
a2
, 1).
It is then routine to work out the fact that the upper triangular matrix
giving OZ is the matrix
⎛
⎞
.
Rewriting,
A = AOZ =
⎜
⎝
xv yv 0 0
0 zv 0 0
0 0 0 1
0 0 0 0
⎟
⎠ .
OZv = {(x, t, −gv(s, t), 1) : s, t ∈ Fq} ∪ {(0, 0, 1, 0)}.
Here OZv ∩ π∞ = {(0, 0, 1, 0)} precisely because Av is anisotropic. Here
⎛
⎞
A + A T = ⎝
Kv
0
0
0 1
1 0
For each nonzero v, the tangent plane at ∞ = (0, 0, 1, 0) is π∞ = [0, 0, 0, 1],
and the tangent plane at (0, 0, 0, 1) is [0, 0, 1, 0]. So the points A(t), 0 = t ∈
⎠ .

15.7. PROP.(G) & TETRADIC SETS OF QUADRICS 703
Fq, correspond to ovoids giving a transversal with base point - base plane
pairs ((0, 0, 1, 0), [0, 0, 0, 1]) and ((0, 0, 0, 1), [0, 0, 1, 0]).
Now
OZv ∩ [1, 0, 0, −1] = {(1, t, −gv(1, t), 1) : t ∈ Fq} ∪ {(0, 0, 1, 0)}.
By Lemma 15.5.3 we see that this set of conics is equivalent to the planar
representation of the flock we started with. Moreover, the ovoids OZv
completely determine the GQ, since any tetradic set of elliptic quadrics containing
this transversal must consist precisely of the orbits of G1 (elations
with center ∞ = (0, 0, 1, 0)) containing these q − 1 elliptic quadrics.
15.7 Prop.(G) & Tetradic Sets of Quadrics
The goal of this section is to use a tetradic set of elliptic quadrics of P G(3, q)
to construct a GQ of order (q, q 2 ) that is the point-line dual of a flock GQ.
We follow the treatment of [BBP04] with inspiration from [Th99].
Theorem 15.7.1. Let Θ be a tetradic set of elliptic quadrics of P G(3, q)
with respect to the incident point-plane pair (∞, π∞). Consider the following
incidence structure GQ(Θ):

704 CHAPTER 15. PROPERTY G
Points : (i) ∞
(ii) π∞.
(iii) Equivalence classes of Θ under the action of elations with
center ∞.
(iv) Points of P G(3, q) \ π∞.
(v) Planes of P G(3, q)not incident with ∞.
(vi) Elements of Θ.
Lines : (a) [∞].
(b) Lines of P G(3, q) not in π∞ but incident with ∞.
(c) Lines of π∞ not incident with ∞.
(d) Rosetts of elliptic quadrics in Θ.
(e) Triples (T , π, X), where T is a transversal of elliptic quadrics
with base point-base plane pairs (∞, π∞) and (X, π).
Incidence (i) : The point ∞ is incident with [∞] and all lines of type (b).
(ii) : The point π∞is incident with [∞]and all lines of type (c).
(iii) : An equivalence class E is incident with [∞]and all rosettes
contained in E.
(iv) : The point X ∈ P G(3, q) \ π∞ is incident with the line
〈X, ∞〉 of P G(3, q), and triples (T , π, X) where T
is a transversal of ovoids in Θ with some base plane π = π∞
and corresponding base point X.
(v) : The plane π, not incident with ∞, is incident with π ∩ π∞
and with triples (T , π, X) where T is a transversal of elliptic
quadrics in Θ, and (X, π) = (∞, π∞) is a base point-base plane
pair of T .
(vi) : the elliptic quadric O ∈ Θ is incident with the rosette in Θ
containing it and each triple (T , π, X) with O ∈ T .
Then GQ(Θ) is a GQ of order (q, q 2 ).

15.7. PROP.(G) & TETRADIC SETS OF QUADRICS 705
Proof. First we check the order of the incidence structure of GQ(Θ). First
we check that each line of GQ(Θ) is incident with q + 1 points of GQ(Θ).
The lines are as follows. The line [∞] contains the points ∞, pi∞ and the
q − 1 equivalence classes. A line ℓ of P G(3, q), not in pi∞, incident with ∞
contains the point ∞ and the q affine points incident with ℓ. A line m of π∞
not incident with ∞ contains the points π∞ and the q planes through m. A
line R, a rosette of elliptic quadrics, has points the q ovoids in R, and the
equivalence class E containing R. A line (T , π, X) has points X, π and the
q − 1 ovoids in the transversal T .
The points ∞ and π∞ are both clearly incident with q 2 + 1 lines. Each
equivalence class E of Θ contains q 3 elliptic quadrics partitioned into q 2
rosettes. Together with [∞]this gives q 2 + 1 lines incident with E. If X ∈
P G(3, q) \ π∞, then for each plane π, ∞ ∈ π and X ∈ π, there is a unique
triple (T , π, X), where T is a transversal of elliptic quadrics in T . Together
with the line 〈∞, X〉 this gives q 2 + 1 incident lines. A similar argument
shows that a plane π with ∞ ∈ π is incident with q 2 + 1 lines. If O ∈ Θ,
then for each X ∈ O \ {∞} there is a transversal of elliptic quadrics in Θ
containing O and with base point X. Together with the unique rosette of
elliptic quadrics in Θ containing O this gives q 2 + 1 incident lines.
We now check those cases of the main theorem for finite GQ which are
not straightforward. Suppose that E is an equivalence class of Θ and T a
transversal of Θ. Then there is a unique elliptic quadric of T in E.
Suppose that X is a point of P G(3, q) \ π∞. If R is a rosette of Θ not
incident with X in GQ(Θ), then since the ovfoids of R partition the points of
P G(3, q)\π∞ there is an ovoid O ∈ R such that X ∈ O. We know that there
is a unique transversal T containing O and with base point X, and if πS is
the tangent plane of O at X, then (T , πX, X) is the unique line of GQ(Θ)
incident with X and meeting R. Next suppose that (T , π ′ , X ′ ) is not incident
with X in GQ(Θ). If X ∈ π ′ , then there is a unique transversal with base
point X and base plane π ′ . If X ∈ 〈X ′ , ∞〉, then this is the unique line of
GQ(Θ) incident with X and meeting (T , π ′ , X ′ ). Finally, if X ∈ π ′ ∪ 〈X ′ , ∞〉
we know that the elements of T partition the set of such points, so X is
contained in a unique element of T , and hence must be the base point of a
unique transversal containing an element of T .
Suppose that π is a plane of P G(3, q) not incident with ∞. Let R be a
rosette of ovoids of Θ. Then R is generated by the action of elations with
center ∞ and axis π∞ on any of the elliptic quadrics of R. It follows that
π is tangent to a unique element of R and so is the base plane of a unique

706 CHAPTER 15. PROPERTY G
transversal with an elliptic quadric in R. Next suppose that (T , π ′ , X) is
not incident in GQ(Θ) with π. If X ∈ π, then there is a unique transversal
with base plane π and base point X. If π ∩ π ′ ⊂ π∞, then this is the unique
line of GQ(Θ) incident with π and meeting (T , π ′ , X). Finally, if X ∈ π and
π ∩ π ′ ⊂ π∞, then π is tangent to a unique elliptic quadric of T and hence
there is a unique transversal with base plane π and containing an elliptic
quadric of T .
Now suppose that O ∈ Θ. Let R be a rosette of Θ not containing O.
If O is in the same equivalence class as the elliptic quadrics of R, then the
unique rosette containing O is the unique line of GQ(Θ) incident with O
and meeting R. If O is inequivalent to to ovoids in R, then by the proof
of Cor. 15.4.5 there is a unique elliptic quadric of R meeting O in exactly
two points and hence by Lemma 15.4.7 contained in a transversal with O.
Next suppose (T , π, X) is not incident with O in GQ(Θ). If X ∈ O, then
O is contained in a unique transversal with base point X and base plane
distinct from π. Similarly, if π is a tangent plane to O, then O is contained
in a unique transversal with base point π and base point distinct from X.
Finally, suppose that X ∈ O and that π is not a tangent plane to O. Now
π ∩ O is a conic C not containing X and the line 〈X, ∞〉 of P G(3, q) contains
a unique point Y of O \ {∞} with Y = X. By Cor. 15.4.5 the q − 1 elliptic
quadrics of T partition the q 2 − q − 2 points of O \ (C ∪ {Y, ∞}) into sets
of size 0, 1, q or q + 1. There are only two ways in which this may be done.
First, with one set of size 0 and q − 2 sets of size q + 1, in other words O
is in a rosette with a unique element of T and in a transversal with none.
Secondly, with one set of size 1, one set of size q and q − 3 sets of size q + 1,
in other words O is in a transversal with a unique element of T and in a
rosette with none. In either case there is a unique line of GQ(Θ) incident
with O and meeting (T , π, X).
Corollary 15.7.2. Let Θ be a tetradic set of elliptic quadrics of P G(3, q)
and GQ(Θ) the associated GQ of order (q, q 2 ). Then GQ(Θ) is a dual flock
GQ.
Corollary 15.7.3. Let Θ be a tetradic set of elliptic quadrics of P G(3, q)
with respect to (∞, π∞) and let GQ(Θ) ∗ the corresponding flock GQ. Then
GQ(Θ) ∗ is an EGQ with base point [∞] ∗ and the elation group about [∞] ∗
is induced by the group of P G(3, q) of order q 5 generated by all elations with
center ∞ and all elations with axis π∞.

15.7. PROP.(G) & TETRADIC SETS OF QUADRICS 707
Proof. Any collineation of P G(3, q) fixing the set Θ induces a collineation of
GQ(Θ) and by Lemma 6.5.9 the group G of collineations of P G(3, q) of order
q 5 generated by all elations with center ∞ and all elations with axis π∞ fixes
Θ. Calculation shows that this group fixes the equivalence classes of Θ and
that for O ∈ Θ the group GO acts regularly on the points of O \ {∞}. Hence
the induced collineation group of GQ(Θ) ∗ fixes [∞] ∗ pointwise and fixes no
point not collinear with [∞] ∗ , and is thus an elation group about [∞] ∗ .
Theorem 15.7.4. Let S = (P, B, I) be a GQ of order (s, s 2 ) satisfying
Property (G) at a pair of collinear points (X, Y ). If s is odd, then S is the
dual of a flock GQ. If s is even and all ovoids Oz of SXY for Z ∈ P\(X ⊥ ∪Y ⊥ )
are elliptic quadrics, then we have the same conclusion.
Proof. Let SXY be the projective three-space constructed from the pair (X, Y ).
Hence s is a prime power q. Let Θ be the set of ovoids in SXY ∼ = P G(3, q)
associated with S. If q is odd, then Θ is a set of elliptic quadrics. If q is even,
then this is also the case by hypothesis. WE will show that Θ is a tetradic
set of elliptic quadrics in P G(3, q).
The points of XY \ {X, Y } divide the elliptic quadrics of Θ into q −
1 equivalence classes. Two elliptic quadrics in the same equivalence class
intersect in either 1 or q +1 points, while two elliptic quadrics of Θ in distinct
equivalence classes intersect in either 2 or q+2 points. We will show that two
elliptic quadrics in different equivalence classes cannot have an intersection
containing a conic.
Without loss of generality suppose that the elliptic quadrics of Θ have
common point ∞ = (0, 1, 0, 0) and common tangent plane [1, 0, 0, 0]. If q is
odd, let O ∈ Θ with O = {(1, s 2 − ηt 2 , s, t) : s, t ∈ Fq} ∪ {(0, 1, 0, 0)}, where
η is a fixed non-square in Fq. let C ⊂ O be the conic C = {(1, s 2 , s, 0) :
s ∈ Fq} ∪ {(0, 1, 0, 0)}. Let O ′ be a second elliptic quadric containing C.
By the proof of Lemma (giving bowtie relation) we may assume that O ′ =
{(1, s 2 −ηt 2 +bt, s+ct, dt) : s, t ∈ Fq}∪{∞} for b, c, d ∈ Fq, d = 0. It follows
that |O ∩ O ′ | = k + 1 where k is the number of solution pairs (s, t) to
2cst = ηd 2 t 2 − c 2 t 2 − ηt 2 + bt. (15.4)
Any (s, 0) is a solution pair corresponding to the points C \ {(0, 1, 0, 0)}.
If O ′ ∈ Θ and is inequivalent to O, then it must be that |O ∩ O ′ | = q + 2
and there is a unique solution to Eq. 15.4 with t = 0. Under the assumption
that t = 0, Eq. 15.4 becomes 2cs = ηd 2 t − c 2 t − ηt + b. If c = 0 we have

708 CHAPTER 15. PROPERTY G
ηd 2 t − ηt + b = 0, to which there are either no solutios, or a unique solutioin
for t and q solution pairs (s, t), so we may suppose that c = 0. Hence we
may write
s = ηd2t − c2t − ηt + b
,
2c
which yields a solution in s for each choice of t, that is q solution pairs in
total.
If q is even, let O = {(1, s 2 +st+ρt 2 , s, t) : s, t ∈ Fq}∪{∞} with ρ a fixed
element of Fq such that tr(ρ) = 1. let C ⊂ O be the conic {(1, s 2 , s, 0) : s ∈
Fq} ∪ {∞}. Let O ′ be a second elliptic quadric containing C which, by the
proof of Theorem (two part theorem) we may assume is O ′ = {(1, s 2 + st +
ρt 2 +bt, s+ct, dt) : s, t ∈ Fq}∪{∞} for b, c, d ∈ Fq, d = 0. For |O∩O ′ | = q+2
we need a unique solution to (ρ + c 2 + cd + ρd 2 )t + (d + 1)s + b = 0 with
t = 0. However, the existence of one such solution implies the existence of
at least q such solutions.
Now the group of an elliptic quadric is transitive on pairs (P, C) where C
isa conic section of the elliptic quadric and P ∈ C. Hence we may conclude
that if two elliptic quadrics of Θ contain a common conic, then they are in
the same equivalence class.
Let C be a conic in the plane π containing ∞ such that π = π∞ and
π ∩ π∞ is a tangent to C. Then C is contained in at most q elements of Θ
since in S the set of points C \ {X} contains a triad and so has at most q + 1
centers, one of which is X. Counting the number of such conics and noting
that |Θ| = q 3 (q − 1) we conclude that each such conic is contained in exactly
q elliptic quadrics of Θ.
Now let {A, B, C, ∞} be a 4-arc in a plane π of P G(3, q) such that π =
π∞. The conic C is contained in q elliptic quadrics O1, . . . , Oq of Θ all of
which must be in the same equivalence class. Hence O1, . . . , Oq intersect in
exactly C and partition the points of P G(3, q)4 \ (π∞ ∪ π). thus for any
5-cap {A, B, C, Z, ∞} of P G(3, q) with A, B, C, Z ∈ π∞, ∞ ∈ 〈A, B, C〉 and
Z ∈ 〈A, B, C〉, there is a unique elliptic quadric on {A, B, C, Z}. Hence Θ
is a tetradic set of elliptic quadrics of P G(3, q) with respect to (∞, π∞) and
by Cor. 15.7.2 S is the corresponding dual flock GQ.

15.8. THE ⊲⊳ EQUIVALENCE RELATION 709
15.8 The ⊲⊳ Equivalence Relation
Until further notice we assume that q = 2 e .
Recall the following. If a = (a01, a23, a02, a31, a03, a12) is a point of P G(5, q),
we have ϖ(a) = a01a23 + a02a31 + a03a12. Then if ϖ(a) = 0, (i.e., a ∈ H5) we
know that [a] ∩ H5 = [a01, a23, a02, a31, a03, a12] ∩ H5 is a nonsingular quadric
isomorphic to Q(4, q). Hence it is a GQ of order (q, q). If we put
⎛
⎞
Ca =
so C −1
a
⎜
⎝
= 1
ϖ(a)
0 a01 a02 a03
−a01 0 a12 −a31
−a02 −a12 0 a23
−a03 a31 −a23 0
⎛
⎜
⎝
⎟
⎠ ,
0 −a23 −a31 −a12
a23 0 −a03 a02
a31 a03 0 −a01
a12 −a02 a01 0
⎞
⎟
⎠ ,
then Ca is skew-symmetric and determines a symplectic polarity whose absolute
lines are exactly the lines of P G(3, q) that Klein-correspond to the
points of the GQ [a] ∩ H5. Hence these lines of P G(3, q) are the lines of the
linear complex
Aa = {ℓ ∈ P G(3, q) : [a] · G(ℓ) T = 0}.
Let ((∞), π∞) be an incident point-plane pair. To help us study tetradic
sets of elliptic quadrics (and more generally, tetradic sets of ovoids), we
introduce an equivalence relation ⊲⊳ on the set of symplectic polarities that
interchange (∞) and π∞. The lines through (∞) lying in the plane π∞
Klein-correspond to the points of a line α lying in H5. So defining ⊲⊳ on the
symplectic polarities interchanging (∞) and π∞ is equivalent to defining an
equivalence relation ⊲⊳ on the set of nonsingular hyperplane intersections of
H5 that contain the line α. For two such GQ(4, q) L1 and L2 we say L1 ⊲⊳ L2
provided L1 and L2 contain a quadratic cone with vertex being some point
of α.
Theorem 15.8.1. The relation ⊲⊳ defined on nonsingular hyperplane sections
of H5 that contain the line α is an equivalence relation.
Proof. It is clear that ⊲⊳ is reflexive and symmetric, so the proof amounts to
showing that it is transitive. This is equivalent to the following: If L1 and L2

710 CHAPTER 15. PROPERTY G
share a cone at the point p of α, and L2 and L3 share a cone at the point q of α,
then there is some point r of α for which L1 and L3 share a cone at r. Without
loss of generality we may assume that (∞) = (1, 0, 0, 0) and π∞ = [0, 1, 0, 0].
So the lines of π∞ through (∞) are of the form ℓ = 〈(1, 0, 0, 0), (0, 0, z, w)〉,
and they Klein correspond to the points of the form (0, 0, z, 0, w, 0), which
constitute the line α = 〈(0, 0, 1, 0, 0, 0), (0, 0, 0, 0, 1, 0)〉.
The group of homographies leaving both H5 and α invariant is doubly
transitive on the points of α. Hence we may suppose L1 and L2 share a cone
at p = (0, 0, 0, 0, 1, 0), and L2 and L3 share a cone at ℓλ = (0, 0, 1, 0, −λ, 0).
We need to show that L1 and L3 share a cone at some point ℓµ.
By hypothesis (1, 0, 0, 0) · Ca is a constant times [0, 1, 0, 0], so a01 = 0 and
a02 = a03 = 0. Dually, C −1
a · [0, 1, 0, 0] T = (1, 0, 0, 0) T , so a23 = 0. Without
loss of generality we may put a01 = 1, a23 = a = 0, a31 = b, a12 = c. Hence
a = (1, a, 0, b, 0, c) with a = 0. These a give all the symplectic polarities that
interchange (∞) and π∞.
Next we want to determine the set of all a for which [a] ∩ H5 contains
a specific cone centered at (0, 0, 0, 0, 1, 0) and containing α. All such cones
are projectively equivalent, so we pick a particular one. Actually, we pick
a = (1, −1, 0, 0, 0, 0) and see which cone at (0, 0, 0, 0, 1, 0) it has.
First, what cone does [1, −1, 0, 0, 0, 0]∩H5 have centered at (0, 0, 0, 0, 1, 0)?
This is the set of points in [1, −1, 0, 0, 0, 0] ∩ H5 that are orthogonal to
(0, 0, 0, 0, 1, 0), i.e., in (0, 0, 0, 0, 1, 0) ⊥ = [0, 0, 0, 0, 0, 1]. The point y =
(y0, . . . , y5) ∈ [0, 0, 0, 0, 0, 1] ∩ [1, −1, 0, 0, 0, 0] if and only if y =
(y1, y1, y2, y3, y4, 0). And this y is in H5 if and only if y2 1 +y2y3 = 0. If y3 = 0,
we get the line α, so without loss of generality we may put y3 = −1. Then
we get the cone generators (all but α):
ℓt = 〈(0, 0, 0, 0, 1, 0), (t, t, t 2 , −1, 0, 0)〉, t ∈ Fq.
Next we ask: given q = (0, 0, 1, 0, −λ, 0) ∈ α \ {p = (0, 0, 0, 0, 1, 0)}, what
cone does [1, −1, 0, 0, 0, 0]∩H5 have that is centered at (0, 0, 1, 0, −λ, 0)? This
is the set of points of the form y = (y1, y1, y2, λy5, y4, y5) with y 2 1 + λy2y5 +
y4y5 = 0. If y5 = 0, then y1 = 0 and we get the line α again. So we may
assume that y5 = −1. Then add −y2(0, 0, 1, 0, −λ, 0) to (y1, y1, y2, −λ, y4, −1)
to get the cone generators (all but α):
mλ(t) = 〈(0, 0, 1, 0, −λ, 0), (t, t, 0, −λ, t 2 , −1)〉, t ∈ Fq.
Now we know what cone [1, −1, 0, 0, 0, 0] ∩ H5 has at each point of α.

15.8. THE ⊲⊳ EQUIVALENCE RELATION 711
Next we determine which [1, a, 0, b, 0, c] ∩ H5 contain the cone with generators
ℓt. Then (t, t, t 2 , −1, 0, 0) ∈ [1, a, 0, b, 0, c] for all t ∈ Fq if and only if
a = −1 and b = 0. This says
{[1, −1, 0, 0, 0, c] : c ∈ Fq}
is the complete set of hyperplanes whose intersection with H5 contains the
cone contained by [1, −1, 0, 0, 0, 0] ∩ H5 at the point (0, 0, 0, 0, 1, 0).
The next step is to determine the hyperplanes [1, a, 0, b, 0, c] that contain
the cone with points (t, t, 0, −λ, t 2 , −1), t ∈ Fq. It is easy to check that
{[1, −1, 0, b, 0, −λb] : b ∈ Fq}
is the set of hyperplanes whose intersections contain the cone shared with
[1, −1, 0, 0, 0, 0] ∩ H5 at the point (0, 0, 1, 0, −λ, 0).
Finally, suppose we have L1 = [1, −1, 0, 0, 0, c]∩H5, c = 0, sharing a cone
with [1, −1, 0, 0, 0, 0]∩H5 at (0, 0, 0, 0, 1, 0), and L2 = [1, −1, 0, b, 0, −λb]∩H5
sharing a cone with [1, −1, 0, 0, 0, 0]∩H5 at (0, 0, 1, 0, −λ, 0). Do [1, −1, 0, 0, 0, c]∩
H5 and [1, −1, 0, b, 0, −λb] ∩ H5 share a cone at some point (0, 0, 1, 0, −µ, 0)?
Calculating points of
[1, −1, 0, 0, 0, 0] ∩ [1, −1, 0, b, 0, λb] ∩ [0, 0, 0, 1, 0, −µ] ∩ H5
leads to µ = λ + cb −1 . Now it is easy to check that
[1, −1, 0, 0, 0, 0] ∩ [1, −1, 0, b, 0, −λb] ∩ [0, 0, 0, 1, 0, −µ] ∩ H5
is a conical cone. This completes a proof that ⊲⊳ is an equivalence relation
on the set of q 2 (q − 1) nonsingular hyperplane sections of H5 that contain
the line α. Given one such hyperplane section L1, at each of the q + 1 points
of α it shares its cone there with q − 1 other hyperplane sections. So each
equivalence class contains q 2 members and there are q −1 equivalence classes.
For example, the class containing [1, −1, 0, 0, 0, 0]∩H5 consists of all sections
of the form [1, −1, 0, b, 0, c]. The complete set of all nonsingular hyperplane
sections containing α consists of those of the form [1, a, 0, b, 0, c] ∩ H5 with
a, b, c ∈ Fq, a = 0.
We shall also use ⊲⊳ to denote the corresponding equivalence relation on
the set of symplectic polarities interchanging ∞ and π∞. Consequently the
equivalence relation ⊲⊳ partitions the symplectic polarities of P G(3, q) that

712 CHAPTER 15. PROPERTY G
interchange ∞ and π∞ into q − 1 equivalence classes of q 2 elements each. We
now explore this equivalence relation on the polarities.
With a little more care we can show that
[1, a, 0, b1, 0, c1]∩H5 ⊲⊳ [1, a, 0, b2, 0, c2]∩H5 at the point (0, 0, b1−b2, 0, c1−
c2, 0). Let ℓ = 〈(1, 0, 0, 0), (0, 0, b1 − b2, c1 − c2)〉. Then G(ℓ) = (0, 0, b1 −
b2, 0, c1 −c2, 0). Let φi be the symplectic polarity of P G(3, q) associated with
ai = (1, a, 0, bi, 0, ci), i = 1, 2. Let Ci = Cai be the corresponding skewsymmetric
matrix. Then φ1 ◦ φ2 (do φ1 first) maps a point x to the plane
(xC1) T ≡ C1xT , then maps this to the point xC1 · C −1
2 . Here we have
⎛
⎞
⎛
⎞
0 1 0 0
0 −1 −b2/a −c2/a
⎜
C1 = ⎜ −1 0 c1 −b1 ⎟
⎜
⎟
⎝ 0 −c1 0 a ⎠ , and C−1 2 = ⎜ 1 0 0 0 ⎟
⎝ b2/a 0 0 −1/a ⎠
0 b1 −a 0
c2/a 0 1 0
.
Hence
C1 · C −1
2 =
⎛
⎜
⎝
1 0 0 0
c1b2−c2b1 1 a b2−b1
a
c2−c1
a
c2 − c1 0 1 0
b1 − b2 0 0 1
⎞
⎟
⎠ ,
from which C1 · C −1
−1 2 is obtained by interchanging the subscripts 1 and
2.
It is now easy to check that φ1 ◦ φ2 fixes each point on ℓ (and only those
points) as well as each plane containing ℓ (and only those planes). We say
that φ1 and φ2 are equivalent about ℓ.
Now let ℓ be a line through (∞) but not lying in π∞. Without loss
of generality we have ℓ = 〈(1, 0, 0, 0), (0, 1, z, w)〉. Let φ1 and φ2 be inequivalent
symplectic polarities interchanging (∞) and π∞ associated with
a1 = (1, a1, 0, b1, 0, c1) and a2 = (1, a2, 0, b2, 0, c2), respectively, where a1 = a2.
Since ℓ passes through (∞), ℓφi ⊂ π∞. We want to find ℓ so that ℓφ1 φ2 = ℓ is
some line m ⊂ π∞. It is routine to determine
φi : (0, 1, z, w) ↦→ [−1, −ciz + biw, ci − aiw, −bi + aiz],
from which it is easy to solve for unique z and w, and we find
ℓ = 〈(1, 0, 0, 0), (0, a1 − a2, b1 − b2, c1 − c2)〉.

15.8. THE ⊲⊳ EQUIVALENCE RELATION 713
A routine computation shows that
m = π∞ ∩ ℓ φi = 〈(a1c2 − a2c1, 0, a1 − a2, 0), (a2b1 − a1b2, 0, 0, a1 − a2)〉.
Compute
⎛
Ca1 · C −1
a2 =
⎜
⎝
1 0 0 0
a2
a2
1
0
0
a1
0
a2
0
0
c1b2−b1c2
a1c2−a2c1
b1a2−b2a1
a2
c2−c1
a2
a1
a2
⎞
⎟
⎠ .
It is now easy to check that φ1 ◦ φ2 fixes each point on ℓ, each plane
containing ℓ, each point on m and each plane containing m. Moreover, a
little more work shows that these are the only points and planes fixed by
φ1 ◦ φ2.
This completes a proof of the following lemma.
Lemma 15.8.2. Let φ1, φ2 be two (of the q 2 (q − 1)) distinct symplectic
polarities interchanging (∞) and π∞ as above. If φ1 and φ2 are equivalent,
then φ1 ◦ φ2 fixes the points and planes incident with a unique line ℓ for
which (∞) ∈ ℓ ⊂ π∞. The two polarities are said to be equivalent about
ℓ. Moreover, if ℓ is any line of π∞ through the poing ∞, there are exactly q
distinct symplectic polarities interchanging ∞ and π∞ and pairwise equivalent
about ℓ.
If φ1 and φ2 are not equivalent, then there exist skew lines ℓ and m with
(∞) ∈ ℓ ⊂ π∞ and (∞) ∈ m ⊂ π∞ such that φ1 ◦ φ2 fixes all the points and
planes incident with either ℓ or m, and ℓ φ1 = ℓ φ2 = m.
Lemma 15.8.3. Let Ω, Ω ′ ∈ Θ be such that {X, Y, Z} ⊂ Ω ∩ Ω ′ and ∞ ∈
〈X, Y, Z〉 = π. Then Ω and Ω ′ define distinct polarities equivalent about the
line ℓ = π ∩ π∞. Moreover, Ω ∩ Ω ′ is an oval.
Proof. By Lemma 15.3.7 there are q elements Ω = Ω1, Ω ′ = Ω2, . . . , Ωq of Θ
that contain {X, Y, Z} and meet pairwise in O = Ω ∩ π. We also know that
the Ωi partition the points of P G(3, q) \ (π ∪ π∞).
Consider a plane π ′ such that ∞ ∈ π ′ , m = π ′ ∩ π∞ = ℓ and X, Y ∈ π ′ .
The sets O − i = (π′ ∩ Ωi) \ π partition the q 2 − q points of π ′ \ (π ∪ π∞) into
q sets of size q − 1. Let P be the point 〈X, Y 〉 ∩ π ′ . Let n be any line of
π ′ through P but different from 〈P, ∞〉. Then n contains P , a point of m,
and q − 1 other points of π ′ partitioned by the O −
i into sets of size 0, 1 or 2.
Since q − 1 is odd, the line n is a tangent to at least one of the O −
i . Since

714 CHAPTER 15. PROPERTY G
m must also be a tangent to the oval Ωi ∩ π ′ , the intersection n ∩ µ must be
the nucleus of the oval Ωi ∩ π ′ . Considering the q lines on P different from
π ′ ∩ π, we see that the ovals Ωi ∩ π ′ , i = 1, . . . , q have distinct nuclei on m.
This is true for any line m = ℓ with ∞ ∈ m ⊂ π∞. Fix such an m. Let o
be a line through ∞ in the plane π but different form ℓ. Then o is a secant
to O, but we may suppose that X, Y, Z are labeled so that X, Y are not in o.
Let π ′ be the plane containing o and m. Let P = 〈X, Y 〉 ∩ o. Let M1, . . . , Mq
be the distinct points of π∞ \ {∞} on m labeled so that Mi is the nucleus of
Ωi ∩π ′ . Let φi be the polarity defined by Ωi. It follows that Mi is mapped by
φi to the plane π ′ , which is then mapped by φj to Mj. Hence φi ◦ φj cannot
fix any point of π∞ not on ℓ. It follows by Lemma 15.8.2 that no two of the
polarities φi can be inequivalent and further must be distinct and equivalent
about ℓ.
Lemma 15.8.4. Let X, Y ∈ Σ \ π∞, and let π be the plane 〈∞, X, Y 〉.
Suppose that X, Y ∈ Ω, Ω ′ ∈ Θ. Then either Ω ∩ Ω ′ = π ∩ Ω or Ω ∩ π and
Ω ′ ∩ π have distinct nuclei. In particular this means that Ω and Ω ′ must
define distinct polarities not equivalent about the line ℓ.
Proof. Suppose that Ω∩Ω ′ = Ω∩π. By Lemma 15.3.7 Ω∩Ω ′ must be exactly
{X, Y, ∞}. It follows that there are exactly q − 1 ovoids Ω = Ω1, Ω ′ =
Ω2, . . . , Ωq−1 of Θ meeting π in ovals O1, . . . , Oq−1 which intersect pairwise
in exactly {X, Y, ∞} and partition the points of π not on the lines π∞ ∩ π,
〈∞, X〉, 〈∞, Y 〉 and 〈X, Y 〉. Let Z = 〈X, Y 〉 ∩ π ∩ π∞ and let P be any
point of (π ∩ π∞ \ {∞, Z}. If ℓ is a line through P meeting 〈X, Y 〉 in a point
different from all of X, Y, Z, then the Oi partition the q − 3 (odd) points of
ℓ \ {P, 〈∞, X〉, 〈∞, Y 〉, 〈X, Y 〉}, and hence ℓ is tangent to at least one of the
Oi (since each intersection of ℓ with an oval has 0, 1 or 2 points). But since
π ∩ π∞ is a tangent and ℓ is tangent to Oi, their intersection P must be the
nucleus. It follows that each of the q − 1 points P must be the nucleus of
exactly one of the Oi so they all have different nuclei all on the line ℓ. (So at
least the two associated polarities are not equivalent about the line ℓ.)
Lemma 15.8.5. If Ω, Ω ′ ∈ Θ are such that Ω and Ω ′ define the same polarity
of Σ, then |Ω ∩ Ω ′ | ≤ 2.
Proof. By Lemma 15.8.3, Ω∩Ω ′ may not contain an oval. Then by Lemma 15.8.4,
Ω ∩ Ω ′ may have at most two points, one of which is ∞, of course.
Lemma 15.8.6. If Ω, Ω ′ define equivalent, distinct polarities, then |Ω∩Ω ′ | ≤
q + 1.

15.8. THE ⊲⊳ EQUIVALENCE RELATION 715
Proof. Let ℓ be a line of π∞ incident with ∞ such that the polarities defined
by Ω and Ω ′ are equivalent about ℓ. Then either Ω ∩ Ω ′ is an oval in a plane
about ℓ, in which case by Lemma 15.8.3 we know there is a set of q ovoids
Ω = Ω1, Ω ′ = Ω2, . . . , Ωq meeting pairwise in that oval, or by Lemma 15.8.4
each plane about ℓ, not π∞, contains at most one point of (Ω ∩ Ω ′ ) \ {∞}.
In either case |Ω ∩ Ω ′ | ≤ q + 1.
Lemma 15.8.7. Let Ω ∈ Θ, let π b e a plane incident with ∞ distinct from
π∞, and put O = π ∩ Ω. Let Ω = Ω1, Ω2, . . . , Ωq be the q elements of Θ
containing O, and for π ′ a plane containing π ∩ π∞, distinct from π and π∞
let Oi = Ωi∩π ′ . Let the q elements of Θ containing Oi be Ω1 i = Ωi, Ω2 i , . . . , Ωqi
and let E = {Ω j
i : i, j ∈ {1, 2, . . . , q}}. Then E has the following properties.
(1) The set of polarities defined by elements of E is the set of q distinct
polarities equivalent with the polarity defined by Ω about the line Π∩π∞, and
each polarity is defined by q elements of E.
(2) Let π ′′ be any plane on π ∩ π∞, not π∞, then the elements of E
intersect π ′′ in q distinct ovals which meet pairwise in ∞ and and with q
elements of E containing each of these ovals.
(3) Elements of E defining a fixed polarity meet pairwise in ∞ and hence
define a rosette with base point ∞ and base plane π∞.
Proof. Let Ω ∈ Θ, and let π be any plane containing ∞ but distinct from
π∞. If Ω = Ω1, Ω2, . . . , Ωq are the q elements of Θ containing O = π ∩Ω, then
by Lemma 15.8.3 Ω1, . . . , Ωq define q distinct polarities all equivalent about
the line π ∩ π∞. Now let π ′ be any plane containing π ∩ π∞ but different
from π and π∞, and let Oi = Ωi ∩ π ′ . Let the q elements of Θ containing Oi
be Ω1 i = Ωi, Ω2 i , . . . , Ω q
i and put E = {Ωji
: 1 ≤ i, j ≤ q}. So for each fixed i
the polarities defined by all q Ω j
i are distinct but equivalent about π ∩ π∞.
But the Ω1 i , 1 ≤ i ≤ q are all equivalent about the line π ∩ π∞. Hence the
polarities defined by the Ω j
i in E are all equivalent about the line π ∩ π∞ and
thus must be the q distinct polarities equivalent with the polarity defined by
Ω about the line π ∩ π∞, and each polarity is defined by q elements of E.
This completes a proof of part (1).
Consider a plane π ′′ on π ′ ∩ π∞, not π∞ nor π ′ . for fixed i ∈ {1, . . . , q}
we have Ω j
i ∩ Ωki = Oi ⊂ π ′ for j, k ∈ {1, . . . , q}, j = k, so the ovoids Ω j
i ,
j ∈ {1, . . . , q}, intesect π ′′ in ovals (O j
i )′′ meeting pairwise in ∞. Consider
two such sets {(O j
1) ′′ : j = 1, 2, . . . , q} and {(O j
i )′′ : j = 1, 2, . . . , q}, for i = 1,
whose elements each partition the points of π ′′ \π∞. Without loss of generality
are labelled such that (by Lemma 15.8.4) we have
we may suppose that the O j
i

716 CHAPTER 15. PROPERTY G
that either (a) (O j
1 )′′ = (O j
i )′′ for 1 ≤ j ≤ q, or (b) |((O j
1 )′′ ∩(Ok i )′′ )\{∞}| = 1
for all j, k ∈ {1, . . . , q}. Now suppose that π ′′ = π. In π we have that
(O 1 1 )′′ = (O 1 i
)) = Ω ∩ π for i ∈ {1, . . . , q}. Hence we must have case (a). If
π ′′ = π, then since Ω 1 i ∩ Ω1 j = Ω ∩ π for i = j, it follows that the ovoids Ω1 i ,
k = 1, 2, . . . , q, intersect pairwise on π ′′ in only ∞, so we again have case (a),
completing a proof of part (2).
At this point we know that two elements of E must meet in either 1 or
q + 1 points, so by Lemma 15.8.5 the q elements of E defining the same
polarity must meet pairwise in ∞ and form a rosette with base point ∞ and
base plane π∞.
Lemma 15.8.8. Let φ be a polarity of P G(3, q) such that φ interchanges ∞
and π∞. Then φ is induced by exactly q element of Θ which form a rosette
with base point ∞ and base plane π∞.
Proof. If φ is induced by one element of Θ, then by Lemma 15.8.7 and
Lemma 15.8.5 it is induced by exactly q elements of Θ forming a rosette.
Since |Θ|/q is the number of polarities interchanging ∞ and π∞ every such
polarity is induced by an element of Θ.
Theorem 15.8.9. Let Ω, Ω ′ ∈ Θ and let φ(Ω), φ(Ω ′ ) be the polarities defined
by Ω and Ω ′ , respectively. If, abusing notation, we define Ω ⊲⊳ Ω ′ provided
φ(Ω) ⊲⊳ φ(Ω ′ ), then ⊲⊳ defines an equivalence relation on Θ with the following
properties.
(1) The set Θ is partitioned by ⊲⊳ into q − 1 equivalence classes each of
size q 3 .
(2) Each equivalence class is partitioned into q 2 rosettes.
(3) If Ω ⊲⊳ Ω ′ , then Ω ∩ Ω ′ is either {∞} or an oval.
(4) Let π be a plane, not π∞, containing ∞, and E an equivalence class
of Θ under ⊲⊳. The elements of E intersect π in a set L of q 2 ovals such that
the incidence structure with point set π \ π∞, lines the ovals in L plus the
lines of π incident with ∞, not π ∩ π∞, and natural incidence, is an affine
plane of order q.
Proof. (1) and (2) follow from Lemmas 15.8.7 and 15.8.8.
(3) Suppose that Ω ⊲⊳ Ω ′ and Ω ∩ Ω ′ = {∞}. The polarities φ(Ω) and
φ(Ω ′ ) are equivalent about some line ℓ with ∞ ∈ ℓ ⊂ π∞. Since there are
q polarities equivalent with φ(Ω) and φ(Ω ′ ) about ℓ there are exactly q 2
elements of the equivalence class of Θ defined by Ω defining one of these
polarities, and we apply Lemma 15.8.7.

15.8. THE ⊲⊳ EQUIVALENCE RELATION 717
(4) From Lemma 15.8.7 we know that the elements of an equivlence class
of Θ intersect π in exactly q 2 ovals. (There are q 3 elements of Θ, each of the
oval intersections in at least q elements of Θ - but also at most q of them
from earlier - so exactly q.) Let O = Ω ∩ π and O ′ = Ω ′ ∩ π for Ω, Ω ′ ∈ E
be two such distinct ovals. If φ(Ω) and φ(Ω ′ ) are equivalent about π ∩ π∞,
then by Lemma 15.8.7 it follows that O ∩ O ′ = {∞}. If φ(Ω) and φ(Ω ′ )
are equivalent about some line m = π ∩ π∞, then by Lemma 15.8.7 the set
Ω ∩ Ω ′ is an oval in a plane π ′ on m. It follows that O ∩ O ′ = π ′ ∩ π ∩ O
so |(O ∩ O ′ ) \ {∞}| = 1. Also, each line of π incident with ∞, not π ∩ π∞,
intersects each oval Ω ∩ π, Ω ∈ E, in exactly one point other than ∞. From
the above it follows that the incidence structure in (4) is an affine plane of
order q.
Theorem 15.8.10. Let Ω, Ω ′ ∈ Θ. Then |Ω ∩ Ω ′ | ≤ q + 2.
Proof. In view of Lemma 15.3.6 we need only prove that |Ω ∩ Ω ′ | = q + 3. So
suppose |Ω∩Ω ′ | = q +3. By Lemma 15.3.5 the q +2 lines joining the point ∞
to the q + 2 other points of Ω∩Ω ′ must form a q + 2-arc in the quotient space
P G(3, q)/∞. This means that each plane incident with ∞ contains either 0
or 2 points of (Ω ∩ Ω ′ ) \ {∞}, and Ω and Ω ′ define inequivalent polarities by
Lemma 15.8.6.
Let R be the rosette of ovoids of Θ containing Ω. Let π be a plane
incident with ∞, not π∞. By Theorem 15.8.9 the equivalence class of Ω
defines an affine plane A on the point set π \ π∞ with the intersection of the
elements of R with π forming a parallel class of A. Lemma 15.8.3 implies
that (Ω ′ ∩ π) \ {∞} is a q-arc in A. Moreover, with the addition of the
parallel class of A consisting of the lines of π, not π ∩ π∞, on ∞, we obtain
an oval in A, the projective completion of A. Further, the parallel class of
A corresponding to R is not the nucleus of the oval in A. Hence for each
Ω ∈ R we have that |(π ∩Ω ∩Ω ′ )\{∞}| is either 0 or 2, and (Ω∩Ω ′ )\{∞} is
a hyperoval in P G(3, q)/∞. Hence |Ω ∩ Ω ′ | = q + 3. The elements of R must
partition the points of Ω ′ \ {∞}, forcing |Ω ′ | = 1 + q(q + 2) = q 2 + 2q + 1, a
contradiction.
Lemma 15.8.11. Let X, Y, Z be three distinct, non-collinear points of P G(3, q)\
π∞ not coplanar with ∞. Then there are exactly q −1 ovoids of Θ containing
{X, Y, Z}, one from each equivalence class.
Proof. Let π = 〈X, Y, ∞〉 and let ℓ be a line of π incident with ∞ but not
with X nor Y . Let W = ℓ ∩ 〈X, Y 〉. If A ∈ ℓ \ {W, ∞}, then there is

718 CHAPTER 15. PROPERTY G
a unique element of Θ containing the tetrad {X, Y, Z, A}. Hence there are
q − 1 ovoids of Θ containing X, Y, Z, which, by Theorem 15.8.9 (3), are in
distinct equivalence classes.
Lemma 15.8.12. Let Ω1 and Ω2 be inequivalent ovoids of Θ with |Ω1 ∩Ω2| ≥
3. Then |Ω1 ∩ Ω2| = q + 2.
Proof. Consider fixed X, Y ∈ (Ω1∩Ω2)\{∞}, X = y. There are q2 −q triples
{X, Y, Z} such that X, Y, Z ⊂ Ω1 and ∞ ∈ 〈X, Y, Z〉. By Lemma 15.8.11
each such triple is contained in a unique element of the equivalence class of
Ω2.
We know that X, Y ∈ Ω2 and that any other ovoid Ω ′ 2 ∈ [Ω2] with
X, Y ∈ Ω ′ 2 meets Ω2 in points contained in a plane on ∞ which must be
〈X, Y, ∞〉. Further, it must be that Ω2 ∩ Ω ′ 2 = Ω2 ∩ 〈X, Y, ∞〉. There are
exactly q elements of [Ω2] containing Ω2 ∩ 〈X, Y, ∞〉.
Now count pairs (Ω ′ 2, {X, Y, Z}) where Ω ′ 2 ∈ [Ω2] and X, Y, Z are distinct
points of Ω ′ 2 . From above we know that the count is in act q2 − q. However,
we also have that |Ω1 ∩ Ω ′ 2 | ≤ q + 2, and hence there are at most q − 1
such triples {X, Y, Z} and q such Ω ′ 2. Hence the count is bounded above
by q(q − 1) = q2 − q. It follows that |W1 ∩ Ω ′ 2 | = q + 2 and certainly
|W1 ∩ Ω2| = q + 2.
Corollary 15.8.13. If Ω1 and Ω2 are inequivalent ovoids of Θ, then |Ω1 ∩
Ω2| = 2 or q + 2.
Proof. We show that |Ω1 ∩ Ω2| = 1 is impossible. Consider the rosette R of
ovoids in the equivalence class [Ω2] containing Ω2. The ovoid Ω1 intersects
each of the q ovoids of the rosette R in 1, 2 or q + 2 points. The elements of
R also partition the points of P G(3, q)\π∞. Hence the q 2 points of Ω1 \{∞}
are partitioned into q sets of size 0, 1, or q + 1. Suppose there are a, b and
c, respectively, of these kinds of sets, so a, b, c are non-negative integers for
which a + b + c = q and a · 0 + b · 1 + c · (q + 1) = q 2 . From the second
equation subtract q times the first to get c = a · q + b(q − 1). Since a, b, c ≤ q,
clearly a = 0. Then b = 1 and c = q − 1. This says there are q − 1 elements
of R meeting Ω1 in q + 2 points and a unique one meeting Ω1 in exactly 2
points.
Lemma 15.8.14. Let Ω ∈ Θ and let E be an equivalence class of Θ such
that Ω ∈ E. If X ∈ Ω \ {∞}, then there is a unique Ω ′ ∈ E such that
Ω ∩ Ω ′ = {X, ∞}.

15.8. THE ⊲⊳ EQUIVALENCE RELATION 719
Proof. For fixed X ∈ Ω \ {∞} the number of pairs (Y, Z) with Y, Z ∈ Ω
and Y = Z such that ∞ ∈ 〈X, Y, Z〉 is (q 2 − q)(q 2 − q). By Lemma 15.8.11
each triple {X, Y, Z} is contained in a unique element of E. Further, each
of the q 2 rosettes of E contains a unique ovoid on the point X. If such an
ovoid Ω ′ intersects Ω in q + 2 points, then we have (q(q − 1) pairs (Y, Z) with
Y, Z ∈ Ω ∩ Ω ′ and Y = Z such that ∞ ∈ 〈X, Y, Z〉. If on the other hand,
|Ω ∩ Ω ′ | = 2, then there are no such pairs (Y, Z).
Hence it followsthat there are q 2 −1 ovoids in E containing X and meeting
Ω in q + 2 points and a unique ovoid in E containing X and meeting Ω in 2
points.
Lemma 15.8.15. Let Ω ∈ Θ and X ∈ Ω \ {∞}. The q − 2 ovoids meeting Ω
in exactly {X, ∞} also meet pairwise in exactly {X, ∞} and have a common
tangent plane at X. Thus together with Ω they form a transversal.
Proof. Let πX be the tangent plane to Ω at X. For Y ∈ P G(3, q) \ (π∞ ∪
πX ∪ 〈∞, X〉 ∪ Ω) count pairs (Z, Ω ′ ) with Z ∈ Ω \ {X, ∞},Ω ′ ∈ Θ and
{X, Y, Z, ∞} ⊂ Ω ′ . Suppose that Z ∈ Ω \ 〈X, Y, ∞〉. Then there are q 2 − q
choices for Z and {X, Y, Z} is contained in q −1 ovoids, giving (q 2 −q)(q −1)
pairs.
Now suppose thqt Z ∈ Ω ∩ 〈X, Y, ∞〉 and let O = Ω ∩ 〈X, Y, ∞〉. Note
that Y ∈ O. Since Y ∈ πX it follows that 〈X, Y 〉 is not tangent to O and
hence meets O in a point of O \ {X}. Similarly, 〈Y, ∞〉 meets O in a second
point, leaving q − 3 possible choices for Z. For each such choice of Z the
points X, Y, Z define a unique oval O ′ in ∠X, Y, ∞〉 containing ∞ and with
tangent π∞ ∩〈X, Y, ∞〉. There are q ovoids of Θ containing O ′ giving q(q −3)
pairs (Z, Ω ′ ) withZ ∈ Ω ∩ 〈X, Y, ∞〉.
So in total we have (q 2 − q)(q − 1) + q(q − 3) = q(q − 2)(q + 1) pairs
(Z, O ′ ).
Counting these pairs in a second way we consider the number of ovoids
of Θ containing {X, Y }. In 〈∞, X, Y 〉 there are q − 1 ovals on X, Y, ∞ with
tangent 〈∞, X, Y 〉 ∩ π∞, which means there are q(q − 1) ovoids containing
{X, Y }, q in each class. If such an ovoid is in the same class as Ω, then we
know that the possible intersection sizes with Ω are 1 and q +1, so they must
all be q +1, since the intersection is at least {∞, X}. This gives q(q −1) pairs
(Z, Ω ′ ). There are q(q − 2) ovoids containing {X, ∞} which are inequivalent
to Ω, and by the earlier counts there are exactly q(q − 2)(q + 1) − q(q − 1) =
q(q 2 − 2q − 1) pairs (Z, Ω ′ ) where Ω ′ is of this type. In this case |Ω ∩ Ω ′ | = 2

720 CHAPTER 15. PROPERTY G
or q + 2, so each Ω ′ gives rise to 0 or q pairs (Z, Ω ′ ), respectively. From this
it follows that there must be q 2 − 2q − 1 ovoids intersecting Ω in q + 2 points
and (q 2 − 2q) − (q 2 − 2q − 1) = 1 intersecting Ω in the two points X, ∞.
Hence there is a unique ovoid of Θ meeting Ω in exactly {X, ∞} and
containing the point Y ∈ P G(3, q) \ (π∞ ∪ πX ∪ 〈∞, X〉 ∪ O).
In the proof of Lemma 15.8.14 we saw that there are q − 2 ovoids of Θ
meeting Ω in exactly {X, ∞}. By the above, these q − 2 ovoids plus Ω cover
the q 3 − a 2 − q + 1 = (q 2 − 1)(q − 1) points of P G(3, q) \ (π∞ ∪ πX ∪ 〈∞, X〉).
It follows that these ovoids partition the pointset into q −2 sets of size q 2 −1.
Hence the ovoids meet pairwise in exactly {X, ∞} and have πX as tangent
plane at X, thus forming a transversal of ovoids.
Lemma 15.8.16. Let (X, π) be an incident point-plane pair of P G(3, q)
such that X ∈ π∞ and ∞ ∈ π. Then there are exactly q − 1 ovoids of Θ
containing X and with tangent plane π at X. Further, these q − 1 ovoids
form a transversal with one ovoid from each equivalence class of Θ.
Proof. Since each rosette of ovoids contained in Θ is generated by the action
on one ovoid of the elations of P G(3, q) with center ∞ and axis π∞, it follows
that each plane not on ∞ is tangent to exactly one ovoid of a given rosette.
There are (q − 1)q2 rosettes of ovoids in Θ, so (q − 1)q2 ovoids of Θ with
π as a tangent plane at one of the poins of π \ π∞. By Lemma 15.8.15 if π is
tangent to one ovoid of Θ at a point, then it is tangent to the q −1 ovoids of a
transversal of Θ at that point. Since two ovoids in the same equivalence class
have intersection size 1 or q + 1, it follows that the ovoids of the transversal
are one from each equivalence of class of Θ.
Suppose that for X ∈ π \ π∞ there are two transversals Ω1, . . . Ωq−1 and
Ω ′ 1 , . . . , Ω′ q−1
containing X and with tangent plane π. We investigate how
Ω ′ 1 intersects Ω1, . . . , Ωq−1. In fact Ω1, . . . , Ωq−1 partition the q 2 − 1points of
Ω ′ 1 \ {X, ∞} into q − 1 sets of size q or q − 1, which is a contradiction.
Hence there can be only one transversal of ovoids of Θ with X as a base
point and π as the corresponding base plane. Since there are q 2 (q − 1) ovoids
with tangent plane π, there are q 2 transversals of Θ with base plane π and a
unique such transversal with base point X for each X ∈ π \ π∞.
Lemma 15.8.17. Let T = {Ω1, . . . , Ωq−1} be a transversal of ovoids of Θ
with base point X and base plane π, X ∈ π∞ and ∞ ∈ π. Then every plane
π ′ such that X, ∞, π∞∩π ⊂ π ′ is tangent to a unique element of Θ. Further,
any two elements of T have only π and π∞ as common tangent planes.

15.9. PROP.(G) & TETRADIC SETS OF OVOIDS 721
Proof. Let π ′ be a plane such that X, ∞, π∞ ∩ π ⊂ π ′ . The elements of
Θ partition the q 2 − q − 1 points of π ′ \ (π∞ ∪ π ∪ 〈∞, X〉), which can only
be into q − 2 conics and one single point. That is, π ′ is tangent to a unique
element of Θ.
There are (q 2 − 1)(q − 1) such planes, which is the same as the number of
pairs (Ωi, π ′′ ) where π ′′ is tangent to Ωi ∈ Θ at a point not ∞ nor X. Hence
no two elements of Θ have a common tangent not at ∞ or X. Since a plane
of P G(3, q) (distinct from π∞ and π) on π∞ ∩ π or on ∞ cannot be a tangent
plane to an elliptic quadric of Θ, the lemma is proved.
15.9 Prop.(G) & Tetradic Sets of Ovoids
In this section we use the properties of a tetradic set of ovoids of P G(3, q)
to prove the existence of a GQ of order (q, q 2 ) with Property (G) at a flag.
The construction of the GQ follows Barwick, Brown and Penttila [BBP04]
and J. A. Thas [Th99] to prove the existence of a GQ of order (q, q 2 ) with
Property (G) at a flag. The construction of the GQ follows Barwick, Brown
and Penttila [BBP04] and J. A. Thas [Th99].
Theorem 15.9.1. Let Θ be a tetradic set of ovoids of P G(3, q) with respect
to the incident point-plane pair (∞, π∞). Then let GQ(Θ) be the following
incidence structure.

722 CHAPTER 15. PROPERTY G
Points : (i) ∞
(ii) π∞.
(iii) Equivalence classes of Θ
(iv) Points of P G(3, q) \ π∞.
(v) Planes of P G(3, q)not incident with ∞.
(vi) Elements of Θ.
Lines : (a) [∞].
(b) Lines of P G(3, q) not in π∞ but incident with ∞.
(c) Lines of π∞ not incident with ∞.
(d) Rosettes of Θ.
(e) Transversals of Θ.
Incidence (i) : The point ∞ is incident with [∞] and all lines of type (b).
(ii) : The point π∞is incident with [∞]and all lines of type (c).
(iii) : An equivalence class E is incident with [∞]and all rosettes
contained in E.
(iv) : The point X ∈ P G(3, q) \ π∞ is incident with the line
〈X, ∞〉 of P G(3, q), and transversals T for which X
is a base point.
(v) : The plane π, not incident with ∞, is incident with π ∩ π∞
and with transversals T for which π is a base plane.
(vi) An ovoid of T is incident with each rosette and
transversal containing it.
Then GQ(Θ) is a GQ of order (q, q 2 ).

15.9. PROP.(G) & TETRADIC SETS OF OVOIDS 723
T
[∞]
X π O ∈ T
〈X, ∞〉 π ∩ π∞ Rosette in E
containing O
(∞) π∞ E = [O]
Incidence diagram for GQ(Θ)
Proof. We check the order of the incidence structure of GQ(Θ). First we
check that each line of GQ(Θ) is incident with q + 1 points of GQ(Θ). The
lines are as follows. The line [∞] contains the points ∞, π∞ and the q − 1
equivalence classes. A line ℓ of P G(3, q), not in pi∞, incident with ∞ contains
the point ∞ and the q affine points incident with ℓ. A line m of π∞ not
incident with ∞ contains the points π∞ and the q planes through m. A
line R, a rosette of ovoids in Θ, has as points the q ovoids in R, and the
equivalence class E containing R. A line T has as points its base point X
and base plane π as well as the the q − 1 ovoids in the transversal T .
The points ∞ and π∞ are both clearly incident with q 2 + 1 lines. Each
equivalence class E of Θ contains q 3 ovoids partitioned into q 2 rosettes. Together
with [∞]this gives q 2 + 1 lines incident with E. If X ∈ P G(3, q) \ π∞,
then for each plane π, ∞ ∈ π and X ∈ π, there is a unique triple (T , π, X),
where T is a transversal of ovoids in T . Together with the line 〈∞, X〉 these
q 2 transversals give q 2 + 1 incident lines. A similar argument shows that a
plane π with ∞ ∈ π is incident with q 2 + 1 lines. If Ω ∈ Θ, then for each
X ∈ Ω \ {∞} there is a transversal of ovoids in Θ containing Ω and with
base point X. Together with the unique rosette of ovoids in Θ containing Ω
this gives q 2 + 1 incident lines.
We now check those cases of the main theorem for finite GQ which are
not straightforward. Suppose that E is an equivalence class of Θ and T a
transversal of Θ. Then there is a unique ovoid of T in E. Suppose that
X is a point of P G(3, q) \ π∞. If R is a rosette of Θ not incident with X
in GQ(Θ), then since the ovoids of R partition the points of P G(3, q) \ π∞

724 CHAPTER 15. PROPERTY G
there is an ovoid Ω ∈ R such that X ∈ Ω. We know that there is a unique
transversal T containing Ω and with base point X, and if πX is the tangent
plane of Ω at X, then T is the unique line of GQ(Θ) incident with X and
meeting R. Next suppose that T is not incident with X in GQ(Θ). If X ∈ π ′ ,
then there is a unique transversal with base point X and base plane π ′ . If
X ∈ 〈X ′ , ∞〉, then this is the unique line of GQ(Θ) incident with X and
meeting T . Finally, if X ∈ π ′ ∪ 〈X ′ , ∞〉 we know that the elements of T
partition the set of such points, so X is contained in a unique element of
T , and hence must be the base point of a unique transversal containing an
element of T .
Suppose that π is a plane of P G(3, q) not incident with ∞. Let R be a
rosette of ovoids of Θ. Then R is generated by the action of elations with
center ∞ and axis π∞ on any of the elliptic quadrics of R. It follows that
π is tangent to a unique element of R and so is the base plane of a unique
transversal with an ovoid in R. Next suppose that T is not incident in
GQ(Θ) with π. If X ∈ π, then there is a unique transversal with base plane
π and base point X. If π ∩ π ′ ⊂ π∞, then this is the unique line of GQ(Θ)
incident with π and meeting T . Finally, if X ∈ π and π ∩ π ′ ⊂ π∞, then π is
tangent to a unique ovoid of T and hence there is a unique transversal with
base plane π and containing an ovoid of T .
Now suppose that Ω ∈ Θ. Let R be a rosette of Θ not containing Ω. If Ω
is in the same equivalence class as the ovoids of R, then the unique rosette
containing Ω is the unique line of GQ(Θ) incident with Ω and meeting R.
If Ω is inequivalent to to ovoids in R, then by the proof of Cor. 15.8.13
there is a unique ovoid of R meeting Ω in exactly two points and hence
by Lemma 15.8.15 contained in a transversal with Ω. Next suppose T , is
not incident with Ω in GQ(Θ). If X ∈ Ω, then Ω is contained in a unique
transversal with base point X and base plane distinct from π. Similarly, if π
is a tangent plane to Ω, then Ω is contained in a unique transversal with base
point π and base point distinct from X. Finally, suppose that X ∈ Ω and
that π is not a tangent plane to Ω. Now π ∩ Ω is an oval O not containing
X and the line 〈X, ∞〉 of P G(3, q) contains a unique point Y of Ω \ {∞}
with Y = X. By Cor. 15.8.13 the q − 1 ovoids of T partition the q 2 − q − 2
points of Ω \ (O ∪ {Y, ∞}) into sets of size 0, 1, q or q + 1. There are only
two ways in which this may be done. First, with one set of size 0 and q − 2
sets of size q + 1, in other words Ω is in a rosette with a unique element of
T and in a transversal with none. Secondly, with one set of size 1, one set
of size q and q − 3 sets of size q + 1, in other words Ω is in a transversal with

15.9. PROP.(G) & TETRADIC SETS OF OVOIDS 725
a unique element of T and in a rosette with none. In either case there is a
unique line of GQ(Θ) incident with Ω and meeting T .
Theorem 15.9.2. Let Θ be a tetradic set of ovoids of P G(3, q). Then GQ(Θ)
satisfies Property (G) at the flag (∞, [∞]).
Proof. First show that GQ(Θ) satisfies Property (G) at the pair (∞, π∞).
Let E be an equivalence class of Θ and let {∞, Ω1, Ω2} be a triad of
GQ(Θ) contained in E ⊥ . Then Ω1, Ω2 ∈ Θ are equivalent ovoids such that
Ω1 ∩ Ω2 is an oval O containing ∞. In fact, by Lemma 15.8.7 there are
q pairwise equivalent ovoids Ω1, . . . , Ωq ∈ Θ such that O ⊂ Ωi, 1 ≤ i ≤
q. Hence in GQ(Θ) we have that {∞, Ω1, Ω2} ⊥⊥ = {∞, Ω1, Ω2, . . . , Σq}.
Thus GQ(Θ) has Property (G) at the pairs {∞, E} and also at the flag
(∞, [∞]).
There is also a converse.
Theorem 15.9.3. Let S = (P, B, I) be a GQ of order (q, q 2 ) satisfying
Property (G) at the flag (X, ℓ). Let Y ∈ ℓ \ {X}, SXY the AG(3, q) defined
from X and Y and SXY = P G(3, q) the projective completion of SXY with
plane at infinity π∞ and the q 2 parallel lines of SXY corresponding to the lines
of S on X, meeting in the point ∞ of π∞. Then Θ = {({X, Z} ⊥ \ ℓ) ∪ {∞} :
Z ∈ ℓ and Z ∼ X, Y } is a tetradic set of ovoids with respect to (∞, π∞).
Proof. Consider A1, A2, A3 ∈ X⊥ \ ℓ such that {∞, A1, A2, A3} is a cap
in P G(3, q), so {A1, A2, A3} has at most one center in Y ⊥ . The triad
{A1, A2, A3} has q + 1 centers in S: X and q others V1, . . . , Vq with Vi ∼
Ui ∈ ℓ, i = 1, . . . , q. Supppose that the Ui, 1 ≤ i ≤ q are distinct. Then
without loss of generality we may suppose that V1 ∈ Y ⊥ and A1, A2, A3
span a plane of P G(3, q) not containing ∞. So if A1, A2, A3 span a plane
π of P G(3, q) containing ∞, then the Ui are not all distinct, so we may
suppose that U1 = U2 = Y . Since S satisfies Property (G) at the pair
{X, U1} it follows that {A1, A2, A3} are collinear in SXU1 and {A1, A2, A3} ⊥ =
{V1, V2, ∞} ⊥⊥ ⊂ U ⊥ 1 . If {A1, A2, A3} ⊥⊥ = {A1, A2, A3, . . . , Aq, U1}, then
{A1, A2, Ai} ⊥ = {A1, A2, A3} ⊥ has no point of Y ⊥ for i = 4, . . . , q, implying
that A1, A2, Ai span a plane on ∞ which must be 〈A1, A2, ∞〉 = π. Hence
{Vi, Vj} ⊥ ⊂ π and V1, . . . , Vq must partition the points of P G(3, q)\(π ∪π∞).
Thus any tetrad with respect to (∞, π∞) containing A1, A2, A3 is contained
in a unique V ⊥
i and hence a unique ovoid of Θ.

726 CHAPTER 15. PROPERTY G
15.10 Using Tetradic Sets to Characterize GQ
If S is a GQ(s, s 2 ), then a 3-regular point X of S is a point with the property
that for each triad T containing X, |T ⊥⊥ | = s + 1. The existence of such a
point characterizes the GQs T3(Ω).
Theorem 15.10.1. Let Θ be a tetradic set of ovoids of P G(3, q) with respect
to (P, πP ) and let GQ(Θ) be the associated GQ of order (q, q 2 ). Then the
following are equivalent:
(i) GQ(Θ) is isomorphic to T3(Ω) with 3-regular point ∞.
(ii) Ω ∈ Θ and the group of collineations of P G(3, q) with center P acts
regularly on Θ.
(iii) Ω ∈ Θ and Θ is a tetradic set with respect to (P, πP ) with the property
that if Ω1, Ω2 ∈ Θ and |Ω1 ∩ Ω2| = q + 2, then Ω1 ∩ Ω2 is P plus an
oval in a plane not containing P .
Proof. Let H be a hyperplane of P G(4, q) containing the ovoid Ω, and let
T3(Ω) be constructed from Ω ⊂ H ⊂ P G(4, q). The GQ T3(Ω) satisfies
Property (G) at any flag ((∞), P ), where (∞) is the 3-regular point of type
(iii) (in the usual description of T3(Ω)) and P ∈ Ω. In T3(Ω) the point P of
Ω plays the role of a line. Let Σ ∈ P be a point of T3(Ω) incident with the
line P different from the point (∞), i.e., Σ is a P G(3, q) meeting H in the
tangent plane πP to Ω at P . The affine three-space SΣ(∞) is Σ \ πP . The
group of collineations of P G(4, q) with axis H and center in Σ induces the
required group of T3(Ω). Suppose that Ω is an element of the tetradic set Θ
with respect to (P, πP ) and that the collineations with center P act regularly
on Θ. This tetradic set is then the same as associated with T3(Ω) and any
pair {(∞), Σ} incident with the lien P of T3(Ω). This shows that (i) and (ii)
are equivalent.
Now let Θ be a tetradic set of ovoids with respect to (P, πp) such that if
Ω1, Ω2 ∈ Θ and |Ω1 ∩ Ω2| = q + 2, then Ω1 ∩ Ω2 is P plus an oval O in a
plane not containing P . Let X, Y, Z be three distinct, non-collinear points
of P G(3, q) \ πp with P ∈ 〈X, Y, Z〉 = π. By Lemma 15.8.11 there are q − 1
elements Ω1, . . . , Ωq−1 of Θ containing {X, Y, Z}. By our hypothesis Ω1 ∩ Ωi,
2 ≤ i ≤ q − 1, is the point P plus an oval Oi. Since X, Y, Z ∈ Oi for
2 ≤ i ≤ q − 1, it follows that Oi = π ∩ Ω1, for 2 ≤ i ≤ q − 1, and in GQ(Θ we
have that {X, Y, Z} ⊥ = {Ω1, . . . , Ωq−1, P, π} and |{S, Y, Z} ⊥⊥ | = |O| = q +1.

15.11. DUAL TETRADIC SETS WITH Q = 2 E 727
Now if {X, Y, U} ⊂ P ⊥ is a triad of GQ(Θ) with UI[P ], then by Property
(G) at {P, U} it follows that |{X, Y, U} ⊥⊥ = q + 1, forcing P to be 3-regular.
By a theorem of J. A. Thas (??) GQ(Θ) must be T3(Ω).
Conversely, if T3(Ω) has tetradic set Θ with respect to (p, πP ), then the
group action we saw earlier in this chapter gives the result.
15.11 Dual Tetradic Sets with q = 2 e
Let Θ be a tetradic set of ovoids of P G(3, q) with respect to (∞, π∞). For
Ω ∈ Θ, let φ(Ω) be the polarity defined by Ω. Let ∗ be a fixed duality
of P G(3, q) interchanging ∞ and π∞. for Ω ∈ Θ let ˆ Ω = Ω φ(Ω)∗ and let
ˆΘ = { ˆ Ω : Ω ∈ Θ}. If ˆ Θ is a tetradic set of ovoids with respect to (∞, π∞),
then we say that the set Θ is dual tetradic and ˆ Θ is the dual tetradic set.
Our present goal is to prove that if a tetradic set is also dual tetradic, then
every element of the tetradic set is an elliptic quadric.
Let Θ be a set of ovoids that is both tetradic and dual tetradic with
respect to (∞, π∞).
Lemma 15.11.1. The equivalence classes of Θ defined by beiing a tetradic
set are precisely those defined by Θ being a dual tetradic set.
Proof. This follows from the fact that if R = {Ω1, . . . , Ωq} is a rosette of Θ
and T = {Ω ′ 1, . . . , Ω ′ q−1} is a transversal of Θ, then ˆ R = { ˆ Ω1, . . . , ˆ Ωq} is a
rosette of Θ (should it be ˆ Θ?) and ˆ T = { ˆ Ω ′ 1, . . . , ˆ Ω ′ q−1} is a transversal of
Θ (or ˆ Θ?).
Lemma 15.11.2. Let Ω and Ω ′ be two equivalent ovoids of Θ. Then there
are two possibilities:
(1) Ω and Ω ′ lie in a common rosette; or
(2) There exists a point-plane pair (P, π) such that 〈∞, P 〉 = π ∩ π∞,
Ω ∩ Ω ′ = π ∩ Ω, and also Ω and Ω ′ have the same tangent planes incident
with P .
Proof. Suppose that Ω and Ω ′ are not in a common rosette. then Ω ∩ Ω ′ =
O = π∩Ω for some plane π containing ∞. Let {W1 = Ω, Ω2 = Ω ′ , Ω3, . . . , Ωq}
be the set of equivalent ovoids of Θ meeting pairwise in O. Since ˆ Θ is a
tetradic set and has the same equivalence classes as Θl, it follows that the
ovoids Ωi and Ωj, i = j, share q + 1 common tangent planes with a common
point, Pij, say. It also follows that Pij ∈ π∞.

728 CHAPTER 15. PROPERTY G
Now consider a plane π such that O has empty intersection with π. The
ovoids Ω1, . . . , Ωq partition the q 2 − q points of π \ (π∞ ∪ π) into q sets of
size 1 or q + 1. Hence π must be tangent to exactly two ovoids of the set
{Ω1, . . . , Ωq}. So each tangent plane to Ω1, not on the nucleus N of O, is
tangent to exactly one other ovoid Ωi, i = 1.
Consider the set {P1i : i = 2, . . . , q}. If ℓ is a line of π∞ not incident with
∞ nor with N, then there is a tangent plane π ′ = π∞ to Ω1 containing ℓ.
Therefore there exists an i ∈ {2, . . . , q} such that P1i ∈ ℓ. It follows that the
set {P1i : i = 2, . . . , q} blocks the lines of π∞ not incident with ∞ nor with
N and hence {P1i : i = 2, . . . , q} = 〈∞, N〉 \ {∞, N}.
Lemma 15.11.3. Let Ω, Ω ′ ∈ Θ be equivalent ovoids intersecting in an oval
O with nucleus N contained in the plane π and such that Ω, Ω ′ share tangent
planes on the point P ∈ (π ∩ π∞) \ {∞, N}. If φ is the polarity defined by
W , then the oval P φ ∩ Ω is equivalent to P φ ∩ Ω ′ under a collineation of P φ
that is an elation with center P and axis P φ ∩ π∞ = π ∩ π∞.
Proof. ****************************************************************************
By Lemma 15.8.14 we know that the polarities φ and φ ′ defined by Ω and
Ω ′ , respectively, are equivalent about the line π ∩ π∞. Let T be the set of
tangent planes common to Ω and Ω ′ on P . Then T φ and T φ′ are equivalent
ovals in P φ = P φ′ and (T φ′ ) φ′ φ φ ′ = T where φ φ is a collineatioin of P G(3, q).
Both φ ′ and φ fix the lines of P φ on P , so φ ′ φ induces in P φ an elation with
axis 〈∞, P 〉 and center P .
Lemma 15.11.4. Let Ω, Ω ′ ∈ Θ be such that Ω and Ω ′ are contained in a
common rosette R. Let φ be the common polarity defined by the elements of
R. If π is a plane such that ∞ ∈ P , π = π∞, then π ∩ Ω is equivalent to
π ∩ Ω ′ under an elation of π with axis π ∩ π∞ and center π φ .
Proof. Consider a line ℓ ⊂ π∞, ∞ ∈ ℓ. We know that by the proof of
Lemma 15.8.1 there are q 2 elements of Θ whose polarity shares with φ (at
least) the point-polar plane pairs (P, P φ ) for P ∈ ℓ. In each plane π on ℓ,
π = π∞, these q 2 ovoids intersect in q ovals intersecting pairwise in ∞ and
partitioning the points of π \ π∞. By Lemma ?? and Lemma 6.5.9 this set
of ovals is generated by the action on one element of the set by the group of
elations with center π φ and axis π ∩ π∞. In particular, this is true for the
ovals π ∩ Ω and π ∩ Ω ′ . Since Ω and Ω ′ define the same polarity, the above
is true for any line ℓ of π∞ incident with ∞.

15.11. DUAL TETRADIC SETS WITH Q = 2 E 729
Lemma 15.11.5. Let R = {Ω1, Ω2, . . . , Ωq} be a rosette of ovoids in Θ
defining the polarity φ, and let π be a plane, not π∞, incident with ∞, Oi =
Ωi ∩ π for i = 1, 2, . . . , q and N = πφ . If Ai P , i = 1, . . . , q, denotes the lines
incident with the point P ∈ (π ∩ π∞) \ {N, ∞} that are secant to Oi, then
for i, j ∈ {1, 2, . . . , q}, i = j, either Ai P = Aj
P or AiP ∩ Aj
P = ∅.
Proof. Let P ∈ π∞ ∩ π \ {∞, N} and let E be the set of q 2 ovoids of Θ whose
polarities are equivalent with φ about ℓ = π ∩ π∞. By Lemma ?? applied
to ˆ Θ there exist q elements of Ω ′ 1, . . . , Ω ′ q of E that share the same q + 1
tangent planes incident with P . These q + 1 tangent planes intersect π in
the line ℓ and q/2 lines which are external to the ovals Ω ′ i ∩ π, i = 1, 2, . . . , q,
leaving a set of q/2 common secants to the ovals incident with P , which we
denote by AP . By the proof of Lemma ?? three elements of Θ which share a
common oval on ∞ may not share tangent planes at a common point. Hence
{Ωi ∩ π : i = 1, 2, . . . , q}, has size q/2 and is a subset of {O1, . . . , Oq} with
the same secants incident with P . The result follows.
In the preceding Lemma (5.6) let π = P G(2, q) and {O1, O2, . . . , Oq}
have common point (0, 0, 1), common nucleus (0, 1, 0) acted on regular by
the group of elations with center (0, 1, 0) and axis ℓ : x0 = 0. Consider
the point (0, s, l1), s = 0, of ℓ \ {(0, 0, 1), (0, 1, 0)}. The lines incident with
(0, s, 1), not ℓ, are (using line coordinates) [a, 1, s], a ∈ Fq. Let As = {a ∈
Fq : [a, 1, s] is a secant to O1} (As is called the local secant parameter set of
O1 at (0, s, 1)). By the action of the elations (x0, x1, x2) ↦→ (x0, x1 + bx0, x2)
we have that As + b = As or the complement of As in Fq, for all b ∈ Fq.
Hence As is an additive group.
Theorem 15.11.6. (SEE IF WE PROVED THIS ELSEWHERE) Let O be
an oval of P G(2, q), q even, containing the point (0, 0, 1) and with nucleus
(0, 1, 0). Let As be the local secant parameter set of O at (0, s, 1) and suppose
that for s ∈ Fq \ {0} the set As is either a subgroup of Fq of order q/2 or
the other coset of such a subgroup. Then O is a translation oval with axis
x0 = 0.
Proof. By the action of P GL(3, q) we may assume that (1, 0, 0), (1, 1, 1) ∈ O.
Thus 0 ∈ As and As is a group for all s ∈ Fq\{0}. Also, 〈(1, 1, 1), (0, s, 1)〉 is a
secant, from which it follows that s+1 ∈ As for all s ∈ Fq \{0}. Consider the
map ψ : (x0, x1, x2) ↦→ (x0, x0 + x1, x0 + x2), so ψ : [a, , 1, s] ↦→ [a + 1 + s, 1, s].
Since As is a group and 1+s ∈ As, it follows that ψ fixes the set of secants to

730 CHAPTER 15. PROPERTY G
O incident with (0, s, 1). Hence ψ fixes O, that is mO is fixed by an elation
with axis x0 = 0 interchanging (1, 0, 0) and (1, 1, 1).
It now follows, from the above discussion, that if π is any plane, not
π∞, incident with ∞, and if Ω ∈ Θ, then π ∩ Ω is a translation oval with
axis π ∩ π∞. By Penttila and Praeger [PP97] (see ???) any ovoid with an
axial pencil of translation ovals is either a Tits ovoid or an elliptic quadric.
However, if Ω is a Tits ovoid and P ∈ Ω, then there is a unique line ℓ on
P tangent to Ω such that the pencil about ℓ is an axial pencil of translation
ovals (this is the element of the associated Lüneburg spread that is incident
with P ). Hence if Ω ∈ Θ it follows that Ω must be an elliptic quadric ovoid.
The above discussion established the following result.
Theorem 15.11.7. Let Θ be a tetradic set of ovoids of P G(3, q) with respect
to (∞, π∞) that is also dual tetradic. Then Θ is comprised entirely of elliptic
quadrics.
There is an interesting corollary about GQ and flocks.
Theorem 15.11.8. Let S = (P, B, I) be a GQ of order (q, q 2 ) and assume
that S satisfies Property (G) at two distinct flags (X, ℓ) and (Y, ℓ) for some
line ℓ. Then S is the dual of a flock GQ.
Proof. The set of ovoids of SXY associated with S is a tetradic set by Theorem
??. However, S also satisfies Property (G) at the flag (Y, ℓ). So the
associated set of ovoids in SY X is tetradic. Hence we have a set of ovoids that
is both tetradic and dual tetradic, and hence comprised of elliptic quadrics.
By the main theorem of Thas [Th99] we now have that S is the dual of a
flock GQ. (THIS SHOULD BE PROVED IN THIS BOOK ALSO.)
Theorem 15.11.9. Let Ω be an ovoid of P G(3, q). Then the GQ T3(Ω)
satisfies Property (G) at some line if and only if Ω is an elliptic quadric and
T3(Ω) is isomorphic to the classical GQ Q(5, q).
Proof. Suppose that T3(Ω) ∼ = (P, B, I) satisfies Property (G) at a line ℓ.
by Theorem ??, for any pair (X, Y ) of points incident with ℓ we have
that the ovoids of SXY induced by points of P \ (X ⊥ ∪ Y ⊥ ) are elliptic
quadrics. If (∞)Iℓ, then by Theorem ?? Ω is an elliptic quadric and hence
T3(Ω) ∼ = Q(5, q) (SEE FGQ, 3.2.4). If (∞ is not on ℓ, then ℓ is a regular line
(somewhere in chapter 13 - from Thas III). Hence by FGQ, 5.3.3(ii)(b), we
know T3(Ω) ∼ = Q(5, q).

Chapter 16
Laguerre Geometries and GQ
16.1 Laguerre points
This chapter contains material from [Br06] that extends in a significant way
the material in Chapter 15. Most likely it would be possible to rewrite the two
chapters as one with some of the computations omitted, but we deliberately
chose to include it all. We think that the notion of Laguerre geometry might
lead to some interesting new research.
Let Q = (P, B, I) be a GQ with parameters (s, t), s > 1, t > 1. Suppose
that Q has a point p ∈ P such that each triad of points in p ⊥ has a constant
number of centers. Such a point will be called a Laguerre point. By Theorem
9.6.6 this constant number of centers is 1 + t/s, and each triad of points
which contains p has 0 or 1 + t/s centers. Put s = t/s.
We are aware of the following GQ that admit Laguerre points:
Case 1. If s = t, i.e., s = 1, then p is antiregular and s is odd. The only
known example is Q(4, q) with q = s odd.
Case 2. If s 2 = t, then each triad of points has 1 + s centers. Point-line
duals of flock GQ have this property.
Case 3. A particular kind of elation GQ (those with abelian elation
group) are called translation GQ. Either they have s = t = q, a prime power,
or they have parameters of the form (s, t) = (q a , q a+1 ) where q is a prime
power and a is odd. The only known examples with s < t have a = 1, and
when q = 2 e it is known that this must be the case. The known examples
include T2(O) (for O an oval in P G(2, q), T3(Ω) (for Ω an ovoid in P G(3, q)),
731

732 CHAPTER 16. LAGUERRE GEOMETRIES AND GQ
and the point-line duals of certain flock GQ, as well as their translation
duals. (See [PT84] for the concept of translation dual and [Pa89] for an
infinite family of examples known as the Roman GQ. One additional pair of
examples is known, but, sadly, we have have had to ignore this topic in this
book.)
Note: It seems to be a very difficult problem to determine whether or
not there is a translation GQ with parameters (s, t) where 1 < s < t < s 2 .
Consider the incidence geometry S = (P, L, C) constructed starting with
Q (and Laguerre point p) as follows.
P = p ⊥ \ {p} is the pointset of S. The lineset L of S consists of the lines
of Q through p. Finally, C = {p, y} ⊥ : y ∈ P \ {p} ⊥ , and the elements of
C are called circles.
The following counts are completely elementary:
Lemma 16.1.1. (i) |P| = s(1 + t).
(ii) |L| = 1 + t and for ℓ ∈ L, |ℓ| = s.
(iii) |C| = s 2 t and for C ∈ C, |C| = 1 + t.
Put s = t/s.
Let w1 and w2 be distinct points of P \ p ⊥ and suppose that the circles
Ci = {p, wi} ⊥ , i = 1, 2, have a point x in common. There are two possibilities.
If {p, w1, w2} is a triad, then since it has one center x it must have 1 + t/s
centers. So |C1 ∩ C2| = 1 + s. On the other hand, if {p, w1, w2} is not a
triad, it must be that x, w1, and w2 lie on a line ℓ. In this case the s circles
{p, w} ⊥ with x = w ∈ ℓ are pairwise tangent at x. This set of circles will be
called a tangent pencil of circles tangent at x. In any case, different points of
P \ p ⊥ determine different circles, so there really are s 2 t circles. Because p is
a Laguerre point, it is clear that each triple of pairwise noncollinear points of
P is contained in exactly s circles. We notice that each point is on a unique
line, each line meets each circle in exactly one point, each circle contains at
least three points, and there is more than one circle. Finally, suppose that
C = {p, w} ⊥ is a circle containing the point x but not the point y. The
circles tangent to C at x are determined by the points of x ⊥ \ p ⊥ on the line
xw. The unique one that contains y is {p, w ′ } ⊥ where w ′ is the point of xw
collinear with y.
These observations show that S = (P, L, C) satisfies the following axioms
given for a finite Laguerre geometry.

16.1. LAGUERRE POINTS 733
A finite Laguerre geometry S = (P, L, C) is an incidence structure of
points, lines and circles, respectively, that satisfies the following properties:
L1. Three pairwise noncollinear points are incident with a (finite) constant
number s > 0 of circles.
L2. (The touching axiom.) If x and y are noncollinear points and C is
a circle containing x but not containing y, then there is a unique circle D
containing y and tangent to C at x, (i.e., C ∩ D = {x}).
L3. Each point is incident with a unique line, each line meets each circle
in a unique point, and |P| < ∞.
L4. Every circle is incident with at least three points and there is more
than one circle.
So let S = (P, L, C) be a finite Laguerre geometry. The parameters of
S are defined as follows. If some line of S is incident with n points, then
all lines are incident with n points. We let s be the number of circles of
S on three pairwise noncollinear points. Finally, let ℓ = |L|. Given these
parameters (n, s, ℓ), it is routine to make the following counts.
Lemma 16.1.2. .
(i) |P| = nℓ.
(ii) |C| = ℓ for C ∈ C, and |C| = n 3 s.
(iii) There are ns circles incident with two given noncollinear points.
(iv) There are n 2 s circles incident with a given point.
A Laguerre geometry S with parameters (n, s, ℓ) where s = 1 is exactly a
Laguerre plane, which implies that ℓ = n + 1. Moreover, in this case there is
a derived plane SP at P ∈ P whose points are the points of S not collinear
with P , whose lines are the circles of S which are incident with P and the
lines of S not incident with P . Incidence is just that inherited from S. Since
three pairwise noncollinear points define a unique circle in S, it must be that
two points in SP define a unique line in SP (whether they are collinear or
noncollinear in S). The touching axiom for S implies that the circles of S
incident with P fall into parallel classes as lines of SP (corresponding to the
tangent pencils). The lines of S not incident with P form another parallel

734 CHAPTER 16. LAGUERRE GEOMETRIES AND GQ
class of lines. Also, note that any circle of S not incident with P meets any
circle incident with P in at most two points. All this leads to the following
well known result.
Theorem 16.1.3. Let S = (P, L, C) be a Laguerre plane and fix P ∈ P.
(i) SP is an affine plane, and we denote its projective completion by SP .
(ii) If C is a circle of S not incident with P , then in SP the points of SP
incident with C plus the point of SP that is the parallel class of SP
corresponding to the lines of S not incident with P form an oval of SP .
It was proved by Chen and Kaerlein in [CK73] (and independently by S.
E. Payne and J. A. Thas in [PT75]) that if even one of the derived planes
is Desarguesian then the Laguerre plane must be the classical one arising
from a cone over a conic (and the associated GQ is classical) (see Theorem
14.5.4). Payne and Thas showed that a finite GQ with a distinguished
antiregular point is equivalent to a finite Laguerre plane of the same order
(see Theorem 14.5.1).
16.2 Ovoids and Laguerre Geometries
The material in this section is taken from M. R. Brown [Br06]. The standard
model of a Laguerre plane arises from a cone over an oval. We want to see if
we can construct a Laguerre geometry from a cone over an ovoid in P G(3, q).
Let Ω be an ovoid in P G(3, q) embedded as a hyperplane in P G(4, q).
So |Ω| = 1 + q 2 and no three points of Ω are collinear. Let V be a point
of P G(4, q) \ P G(3, q), and let K be the cone over Ω with vertex V . Each
line through V and through a point x of Ω cannot be incident with a second
point of Ω, since that would force it to be contained in P G(3, q). If Π3 is a
hyperplane of P G(4, q) not containing V , then K ∩ Π3 is an ovoid of Π3. To
see this, first note that each of the 1 + q 2 lines (generators) of K (through
V ) must meet Π3 in a unique point. Suppose that points x, y and z are on
distinct generators and lie in K ∩ Π3. If x, y and z are collinear, the three
generators V x, V y and V z lie in a plane π, and π ∩ P G(3, q) must be a
line which contains the three points at which these three generators meet Ω.
Since this is impossible, it is clear that K ∩ Π3 is an ovoid.
It is clear that there is a unique line of K on each point of K different
from V . Secondly, let x, y and z be three pairwise noncollinear points of K,

16.3. INTERNAL STRUCTURE AT A POINT 735
i.e., no two on the same generator. Then 〈x, y, z〉 is a plane π of P G(4, q)
which must intersect K in an oval. Let ℓ be a line through V disjoint from
π. Let V = V0, V1, . . . , Vq be the points of ℓ. Then 〈π, Vi〉, for 0 ≤ i ≤ q
are the q + 1 hyperplane solids intersecting pairwise in the plane π. So
there are q hyperplanes containing π and not containing V . So the three
points x, y and z are contained in s = q ovoidal hyperplane sections, and
there are ℓ = 1 + q 2 generators. Finally, any ovoidal hyperplane section has
a unique tangent plane at each point. The hyperplanes about this tangent
plane, not containing V , give a set of q ovoidal hyperplane sections that meet
pairwise in the common point of tangency. This says that if we let the ovoidal
hyperplane sections be the circles of C, the points of K \ {V } be the points
of P, and the generators of K be the lines of L, this tangent plane property
of the ovoids means that the geometry of K satisfies the touching axiom for
Laguerre geometries. So we have a Laguerre geometry S = (P, L, C) with
parameters (n, s, ℓ) = (q, q, 1 + q 2 ).
16.3 Internal Structure at a Point
Let S = (P, L, C) be a Laguerre geometry and fix a point P ∈ P. Define the
internal structure S P of S at P to be the incidence geometry with
Points: points of S not collinear with P ;
Hyperplanes: circles of S incident with P ;
and
Incidence: inherited from S.
In the case of a Laguerre plane S we compare this structure with the
derived plane SP and its projective closure SP . Let (∞) be the “infinite”
point determined by the parallel class consisting of the original generators of
the Laguerre plane. If [∞] denotes the line of “infinite” points, then we have
the following. The line [∞] and its points are removed from SP to obtain
SP , and then the lines through the (former) point (∞) are removed. So to
get S P we start with a projective plane SP and an incident point-line pair
((∞), [∞]). Then we remove the point-line pair, all the points on the line
and all the lines through the point, to obtain S P .
Before discussing the internal structure of a Laguerre geometry S arising
from a cone whose base is an ovoid in P G(3, q) and whose vertex is a point
V ∈ P G(4, q) \ P G(3, q) we review some combinatorics of the geometry
P G(4, q).

736 CHAPTER 16. LAGUERRE GEOMETRIES AND GQ
Lemma 16.3.1. There are
(i) 1 + q + 2q 2 + 2q 3 + 2q 4 + q 5 + q 6 lines of P G(4, q);
(ii) 1 + q + 2q 2 + 2q 3 + q 4 lines in P G(4, q) equal to or meeting a fixed line
ℓ;
(iii) q 4 + q 5 + q 6 lines skew to a fixed line of P G(4, q);
(iv) Two skew lines span a solid (i.e., a 3-dimensional subspace).
(v) In a solid Σ containing a line ℓ there are 1 + q + 2q 2 + q 3 + q 4 lines, q 4
of them skew to ℓ.
(vi) A line ℓ in P G(4, q) is contained in (q 4 + q 5 + q 6 )/q 4 = 1 + q + q 2 solids.
(vii) A plane is contained in 1 + q solids.
(viii) A line is on (q 2 + q 3 + q 4 )/q 2 = 1 + q + q 2 planes.
(ix) A point is on 1 + q + q 2 + q 3 solids.
Now consider the cone K with vertex V as above. If P V is a generator of
K, it is on q2 planes containing a second generator and 1 + q planes meeting
K in just the line P V . Those q + 1 planes must form the unique solid on P V
meeting K in exactly the line P V . We see this as follows. Given any two
generators QV and RV different from P V , the three generators are contained
in a unique solid meeting K in 1 + q generators including P V . So there are
q2 q 2 / = q + q solids on P V meeting K in 1 + q generators. This leaves a
2 2
unique solid Σ on P V meeting K in precisely P V , i.e., tangent to K at P V .
We are ready to consider the internal structure at a point of the Laguerre
geometry S arising from K. The points of S are the points of P G(4, q)
different from V lying on the lines joining V to points of the ovoid Ω of
P G(3, q). Let P be a point of S. So P is on 1 + q + q 2 + q 3 solids while
the line P V is on 1 + q + q 2 solids, leaving q 3 solids containing P but not V
and meeting K in q 2 points, one point on each generator different from P V .
These sets of q 2 points are the circles through P and are the hyperplanes of
S P .
We want to interpret the internal structure S P in the 3-dimensional quotient
geometry P G(4, q)/P . It is basically the geometry of P G(4, q)/P with

16.3. INTERNAL STRUCTURE AT A POINT 737
the point P V/P and the plane Σ/P removed. (Here Σ is the solid tangent
to K at P V .)
Each point of S not collinear with P spans a distinct line with P (in
P G(4, q), giving q 3 lines through P each meeting K in one point. There are
1+q+q 2 lines through P in the tangent solid Σ. So basically the lines through
P in Σ, considered as points of P G(4, q)/P , are removed from P G(4, q)/P
to form S P , i.e., the points of Σ/P are removed. This means that the points
of S P are identified with the lines of P G(4, q) through P not lying in Σ.
The lines of P G(4, q)/P correspond to planes of P G(4, q) on P . So a
line on the point P V/P is a plane of P G(4, q) containing the line P V , i.
e., a plane lying in Σ or a plane meeting K in the union of two generators.
In both cases it plays no role in the internal structure S P . On the other
hand, the other planes containing P meet K in the q + 1 points of an oval.
If π is a plane of P G(4, q) containing P and meeting K in an oval Ω, then
Ω \ {P } is the pairwise intersection of q solids each meeting K in a circle
minus the point P . We can now view the internal structure S P as that
obtained as follows. In P G(4, q)/P we start with the incident point-plane
pair (P V/P, Σ/P ), delete all points in the plane Σ/P , and delete all planes
through P V/P . The resulting structure is an affine space (described as a
point-plane incident structure) with some additional planes missing.
The preceding examples suggest the following definition.
A Laguerre geometry S = (P, L, C) has classical internal structure at
P ∈ P if S P is a projective space with an incident point-hyperplane pair
removed.
We now seek to determine what relationship might exist between the
parameters n, s and ℓ. So let S = (P, L, C) be a Laguerre geometry with
n points on each line, s circles on three pairwise noncollinear points and
having ℓ lines. It is easy to see that there are nℓ points in total, ℓ points
incident with a circle, n 3 s circles in total, ns circles incident with two given
noncollinear points and n 2 s circles incident with a given point.
If s = 1 the Laguerre geometry is exactly a Laguerre plane. This implies
that ℓ = n + 1 and the geometry SP is an affine plane. But what if s > 1?
We first show that n and s do not necessarily determine ℓ.
Let Ω be an ovoid of P G(3, q) and let Ω ⊂ Ω be such that there is a unique
tangent plane to Ω at each point of Ω. For example, Ω could be Ω with any
point removed. So at any point P of Ω there is the plane πP tangent to Ω
at P , which clearly meets Ω only at P . If π ′ is any other plane containing

738 CHAPTER 16. LAGUERRE GEOMETRIES AND GQ
P , it contains q points of Ω, so is not tangent to Ω. It is easy to see how to
obtain such Ω with fewer points. If we now form the cone in P G(4, q) with
a point V not in P G(3, q) and base Ω, then the geometry of (non-vertex)
points, lines and hyperplane sections of this cone not containing V , we get a
Laguerre geometry with n = s = q and ℓ = |Ω|. The unique tangent plane
at every point of Ω ensures that the touching axiom is still valid. So we see
that there are Laguerre geometries with fixed n and s and varying values for
ℓ. In fact, if Ω is any set of points in P G(n, q), n ≥ 3 such that no three are
collinear and there is a unique tangent plane at each point, the cone over Ω
will give such a Laguerre geometry. However, the next theorem shows that
there is a relationship between n, s and ℓ.
Theorem 16.3.2. Let S be a finite Laguerre geometry with parameters n, s
and ℓ. Then n + 1 ≤ ℓ ≤ ns + 1. If ℓ = n + 1, then s = 1, ℓ = n + 1 and S is
a Laguerre plane, which is equivalent to S having intersection sizes between
two circles being 0,1,2. In the case ℓ = ns + 1 we have (s + 1)|(n 3 − n), n ≥ s
and the possible intersection sizes for two circles are 0, 1 and s + 1. Further,
if ℓ = ns + 1, there exist disjoint circles if and only if n > s.
Proof. It is easy to see that the total number of points is nℓ, the total number
of circles is n 3 s, the number of circles on a pair of non-collinear points is ns
and the number of circles on a given point is n 2 s.
Let C be a fixed circle of S and P a fixed point of C. there are exactly
n − 1 circles meeting C in exactly P and hence n 2 s − n circles distinct from
C meeting C in P and at least one other point. For any point Q ∈ C \ {P }
there are ns−1 circles distinct from C meeting C in at least P and Q. Hence
there are at most (ns−1)(ℓ−1) circles meeting C in P and at least one other
point. Hence, n 2 s − n ≤ (ns − 1)(ℓ − 1), which is equivalent to ℓ ≥ n + 1.
Equality holds if and only if the circles just counted are distinct, that is if
the intersection size of two circles is at most 2. Hence ℓ = n + 1 if and only
if s = 1, in which case S is a Laguerre plane. This proves the first part of
the theorem.
Now let the set of circles meeting C in P and at least one other point
be {C1, . . . , Cn 2 s−n}. For 1 ≤ i ≤ n 2 s − n, put ti = |(Ci ∩ C) \ {P }|. Then
n2s−n i=1
ti = (ℓ − 1)(ns − 1), and the average value of the ti is
t = (ℓ − 1)(ns − 1)/(n 2 s − n) =
ℓ − 1
n .

16.3. INTERNAL STRUCTURE AT A POINT 739
Count the ordered triples (Q, R, C ′ ) such that P ,Q, R are distinct, pairwise
noncollinear points, C ′ = C and {P, Q, R} ⊂ C ∩ C ′ . On the one hand this
number is n2s−n i=1 ti(ti − 1). On the other hand (choosing first the points
Q and R in C) this number is (ℓ − 1)(ℓ − 2)(s − 1). So n2s−n i=1 ti(ti − 1) =
(ℓ − 1)(ℓ − 2)(s − 1). Adding the sums on ti and on ti(ti − 1) we obtain
2
ti = (ℓ − 1)((ℓ − 1)(s − 1) + (ns − 1)).
We now have
0 ≤ (ti − t) 2 = t 2 i − 2t ti + t 2 (n 2 s − n)
= (ℓ − 1)[(ℓ − 1)(s − 1) + ns − 1] − (ℓ − 1)2 (ns − 1)
n
= (ℓ − 1)
(n − 1)(ns + 1 − ℓ).
n
From this it follows that ℓ ≤ ns + 1, with equality if and only if each
ti = t =
ℓ − 1
n
= s.
So if ℓ = ns+1 then any two circles that intersect in at least two points must
actually intersect in exactly t+1 = s+1 points. Moreover, if for a fixed circle
C we count the ordered triples (P, Q, C ′ ) with P = Q and P, Q ∈ C ∩ C ′ ,
then we see that the number of circles distinct from C, meeting C in at least
two (and hence in exactly s + 1) points is (ns + 1)ns(ns − 1)/(s 2 + s). Since
this must be an integer it follows easily that (s + 1)|(n 3 − n). The total
number of circles is n 3 s and the number tangent to C is ℓ(n − 1). This leaves
n 3 s − ℓ(n − 1) − 1 circles available to meet C in s + 1 points. Hence
(ns + 1)(ns − 1)n
s + 1
After some simplification this leads to
≤ n 3 s − ℓ(n − 1) − 1 = n 3 s − (ns + 1)(n − 1).
(n − 1)ns(n − s) ≥ 0, which is equivalent to s ≤ n.
If n = s there are no circles disjoint from C, and the possible intersection
sizes between circles are 1 and s + 1.

740 CHAPTER 16. LAGUERRE GEOMETRIES AND GQ
Corollary 16.3.3. Let S = (P, L, C) be a finite Laguerre geometry with
parameters n, s, ℓ. If S has a classical internal structure at some point P ,
then either
(i) s = 1, ℓ = n+1 and S P arises from a projective plane with an incident
point-line pair removed; or
(ii) s = n, ℓ = n 2 + 1 and S P arises from a projective 3-space with an
incident point-plane pair removed.
Proof. Since S P is assumed to be classical, it follows that any two distinct
circles of S on P touch at P or intersect in a constant number of points.
By the proof of Theorem 16.3.2 we know that ti = t for all i if and only if
ℓ = ns + 1 and then t = (ℓ − 1)/n = s. The projective space giving rise
to S P has order n, implying that s = n k for some integer k. We also know
that n ≥ s, so the only possibilities are s = 1 (and k = 0), or s = n (and
k = 1).
16.4 Laguerre Geometries and GQ
Lemma 16.4.1. Let S = (P, B, ∈) be a generalized quadrangle with parameters
(s, t). Suppose that S has a point (∞) with the property that each triad
of points of S having (∞) as a center has exactly 1 + s centers (so we know
s = t/s). Then construct a circle geometry S ′ = (P ′ , L, C) as follows:
(i) P ′ = (∞) ⊥ \ {(∞)}.
(ii) L = {m ∈ B : (∞) ∈ m}.
(iii) C = {X ⊥ ∩ (∞) ⊥ : X ∈ P \ (∞) ⊥ }.
Incidence is that induced by incidence in S. Then S ′ is a Laguerre geometry
with parameters (n, s, ℓ) = (s, t/s, t + 1).
Proof. It is clear that each line of S ′ has n = s points, there are ℓ = t + 1
lines (generators), and each triple of pairwise noncollinear points of S ′ lies in
s = t/s circles. (If s = t then (∞) is antiregular and s must be odd.) This
essentially completes the proof.
Since the circles of S ′ are points of the GQ S and no triangles are allowed,
we see that the circles of S ′ must also satisfy the following:

16.4. LAGUERRE GEOMETRIES AND GQ 741
L5 = [GQ]: Three pairwise tangent circles are incident with a common
point (of contact).
We now want to start with a Laguerre geometry satisfying [GQ] and see
just when we can construct a GQ from it.
Theorem 16.4.2. Let S = (P, L, C) be a finite Laguerre geometry with
parameters n, s and ℓ and satisfying axiom [GQ]. Construct the following
incidence structure GQ(S) = (P ′ , B ′ , I).
Points of P ′ are of three types:
(i) Points of S.
(ii) Circles of S.
(iii) A point (∞).
Lines of B ′ are of two types:
(a) Lines of S.
(b) Pencils of circles.
Incidence is that inherited from S, plus (∞) is incident with all lines
of type (a) and a point of S is incident with all pencils for which it is the
common point of contact. This incident structure, denoted GQ(S), is a GQ
if and only if S satisfies [GQ] and ℓ = ns + 1, in which case the GQ has
order (n, ns).
Proof. If m is a pencil of circles touching at the point Y , so it is a line not
incident with (∞), then the generator V Y is the unique line through (∞)
meeting m at the point Y . If Y is a point of S not on a generator m, then
(∞) is the unique point on m collinear with Y . Suppose that X is a point
on a generator m ′ and m is a pencil of circles centered at the point Y on a
generator different from m ′ . Let C be the circle of m containing the point
X. Then the unique pencil of circles centered at X and containing C is the
unique line through X meeting m at the point C. If X and Y are distinct
points of the generator m ′ and m is a pencil of circles centered at Y , then
m ′ is the unique line through X meeting m at Y . Suppose that C is a circle
and m is a generator. Let Y be the point of m on C. Then the pencil of
circles centered at Y and containing C is the unique line through C meeting

742 CHAPTER 16. LAGUERRE GEOMETRIES AND GQ
m at the point Y . So finally, let C be a circle not on a pencil m of circles
centered at the point Y . First suppose that the point Y is contained in C.
Let m ′ be the pencil of circles centered at Y and containing C. Then m ′ is
the unique line through C meeting m (at the point Y ).
Now suppose that C is a circle not on a pencil m of circles centered at
the point Y , where Y is not on C. What we want is to show that there is a
pencil of circles containing C and centered at Y .
There are n − 1 circles tangent to C at each of its ℓ points, so (n − 1)ℓ
circles tangent to C. Since a point of S is on n 2 s circles, and each pencil has
n circles, there are ns circles on a given point, so there are (n − 1)ℓns pencils
centered at points not on C. Also, each circle touching C is contained in
ℓ − 1 pencils whose base point is not on C, accounting for ℓ(n − 1)(ℓ − 1)
pencils centered at points not on C and containing circles tangent to C. To
get a GQ we need these to be all of the pencils centered at points not on C,
i.e.,
ℓ(n − 1)(ℓ − 1) = ℓ(n − 1)ns, which is equivalent to ℓ = ns + 1.
Theorem 16.4.3. Let S be a Laguerre geometry with parameters n, s and
ℓ such that ℓ = ns + 1.
(i) If S satisfies [GQ], then (s + 1)|(n + 1); hence if s = 1, then n is odd.
(ii) If n = s, then S satisfies [GQ].
Proof. By the proof of Theorem 16.3.2 we see that since ℓ = ns + 1 the
possible intersection numbers for circles are 0, 1, s + 1. Let C and C ′ be two
circles of S touching at the point P . Let Q be any point of C ′ \ {P }, let
R be the point of C collinear with Q, and consider the circles touching C ′
at Q. Each of these n − 1 circles meets C \ {P, R} in either 0, 1 or s + 1
points, and each of the ℓ − 2 points of C \ {P, R} is contained in a unique
such circle, i.e., a unique circle tangent to C ′ at Q (by axiom L2). Let a, b, c
(respectively) be the number of circles tangent to C ′ at Q and meeting C in
1 + s, 1, 0 points (respectively).
Then
a + b + c = n − 1.
a(s + 1) + b · 1 + c · 0 = (ℓ − 2) · 1 = ns − 1.

16.5. PROPERTY (G) 743
Here a, b and c are nonnegative integers. Moreover, if there is a circle C ′′
tangent to C ′ at Q and tangent to C at some point, then the three circles
C, C ′ , C ′′ violate [GQ]. So if S satisfies [GQ], then b = 0. In this case
a(s + 1) = ns − 1 = n(s + 1) − n − 1, implying that (s + 1)|(n + 1).
If n = s, then a(s + 1) + b = s 2 − 1, i.e., (s + 1)(s − 1 − a) = b ≤ s − 1,
and this forces s − 1 − a = 0 = b. But b = 0 says that S satisfies [GQ].
So, we see that a Laguerre geometry with n = s and ℓ = ns + 1 always
gives rise to a GQ of order (s, s 2 ). Conversely, every GQ of order (s, s 2 ) has
the property that every triad of points has s + 1 centers, so from every point
of such a GQ there arises a Laguerre geometry with parameters n = s = s
and ℓ = s 2 + 1. So we have many constructions of Laguerre geometries, most
of which will not be classical at any point.
16.5 Property (G)
Let S be a GQ with parameters (s 2 , s) for s ≥ 2. Let L0, L1, L2 be three
pairwise noncurrent lines. Then we know that there are exactly s+1 transversals
M0, M1, . . . , Ms, i.e., lines that meet all three of the Li, i = 0, 1, 2. If
each three of the transversals Mj, Mk, Mℓ, 0 ≤ j < k < ℓ ≤ s meet the same
S +1 lines L0, L1, L2, L3, . . . , Ls, we say the original triad of lines is 3-regular.
In the Essay [Pa89], Payne proved the following (see Theorem 13.9.1). If S
is a flock GQ viewed as an elation GQ with the usual base point (∞), and
if {L0, L1, L2, L3} and {M0, M1, M2, M3} are two sets of pairwise nonconcurrent
lines such that L0 meets M0 at the point (∞), and such that Li meets
Mj if 0 ≤ i + j ≤ 5, then also L3 meets M3. This is the same as saying
that if {L0, L1, L2} is a triad of lines with (∞) on L0 and some transversal
M0 of the triad meets L0 at (∞), then the original triad must be 3-regular.
This means that the original triad of lines is contained in a (1 + q) × (1 + q)
grid. In the Essay we described this property by saying that the GQ S has
Property (G) (for grid) at the point (∞). It turned out later that it was
more helpful to discuss an even more localized version of this property and
from the point-line dual point of view.
Let S be a GQ with parameters (q, q 2 ), q > 1. Let X and Y be distinct
collinear points. We say that the GQ S has Property (G) at the (unordered)
pair {X, Y } (or that the pair {X, Y } has Property (G)) if each triad of points
including X and contained in Y ⊥ is 3-regular. This is equivalent to saying

744 CHAPTER 16. LAGUERRE GEOMETRIES AND GQ
that each triad of points including Y and contained in X ⊥ is 3-regular. This
notion of is extended to the following. If (X, m) is a flag (i.e., an incident
point-line pair), then (X, m) has Property (G) (or S has Property (G) at
the flag (X, m)) provided {X, Y } has Property (G) for each point Y on m
with X = Y . Similarly, S has Property (G) at the line m provided it has
Property (G) at the flag (X, m) for each point X on m.
Start with a GQ S = (P, B, I) of order (q, q 2 ) satisfying Property (G) at
the pair of distinct collinear points X and Y . We now introduce a pointline
incidence geometry SXY = (PXY , BXY , IXY ) that will turn out to be
isomorphic to the affine space AG(3, q).
(i) PXY = X ⊥ \ {X, Y } ⊥ .
(ii) Elements of BXY are of two types:
(a) the sets {Y, Z, U} ⊥⊥ , where {Y, Z, U} is a triad with X ∈ {Y, Z, U} ⊥ ,
and
(b) the sets {X, W } ⊥ \ {X}, with X ∼ W ∼ Y . .
(iii) Incidence is given by containment.
By the chapter on Property (G) in this book, we know that the incidence
structure SXY is the deisgn of points and lines of the affine space AG(3, q).
In particular, q is a prime power. (See Theorem 15.1.7.)
The planes of the affine space SXY = AG(3, q) are of two types:
(a) The sets {X, Z} ⊥ \ {Y }, with X ∼ Z and Y ∈ {X, Z} ⊥ , and
(b) Each set which is the union of all elements of type (b) of BXY containing
a point of some line m of type (a) of BXY .
Let SXY be the projective completion of SXY with plane at infinity π∞.
The GQ S has the following interpretation of the GQ S in SXY . The q 2 lines
of type (b) of SXY are parallel, so they define a point ∞ of SXY . Let Z ∈ P
be a point with X ∼ Z ∼ Y and let U be the point of ℓ = 〈XY 〉 such that
Z sin U. Then V = {X, Z} ⊥ \ {U} is a set of q 2 points. Clearly each line of
SXY on ∞ meets V in exactly one point. Further, if U1, U2, U3 are points
of V collinear in SXY , then it must be that Y ∈ {U1, U2, U3} ⊥⊥ , implying

16.5. PROPERTY (G) 745
that Z ∼ Y = U, an impossibility. Hence ΩZ = V ∪ {∞} is an ovoid of SXY
with tangent plane π∞ at ∞. Here π∞ is the “plane at infinity.” If we start
with SXY and delete the point ∞ and all planes on it (i.e., π∞ and all planes
of type (b)) as well as π∞ and all the points on it, we obtain a point-plane
geometry with points X ⊥ \ 〈X, Y 〉 and planes Y ⊥ \ 〈X, Y 〉.
This construction leads to an equivalent formulation of Property(G) at a
pair of collinear points.
Theorem 16.5.1. Let S = (P, B, I) be a GQ with parameters (s, s 2 ), s > 1,
and let X, Y ∈ P with X ∼ Y , X = Y . Then S has Property (G) at
{X, Y } provided the point-plane incidence structure SXY = (PXY , HXY , IXY )
described below is the point-plane incidence structure of P G(3, s) with an
incident point-plane pair removed.
• Point-set PXY = X ⊥ \ 〈X, Y 〉;
• Plane-set HXY = Y ⊥ \ 〈X, Y 〉;
• Incidence IXY is just that inherited from incidence in S.
Corollary 16.5.2. Let S = (P, B, I) be a GQ with parameters (s, s 2 ), x > 1,
and let X, Y ∈ P with X ∼ Y , X = Y . Then S has Property (G) at the pair
{X, Y } provided the Laguerre geometry SX has classical internal structure at
the point Y .
The intersections of the ovoids ΩZ are determined as follows.
First suppose that Z1 ∼ Z2 on a line that meets 〈X, Y 〉 at a point U with
X = U = Y . Then ΩZ1 ∩ ΩZ2 = {∞}, since any larger intersection would
result in a triangle in S.
Second, let m be a line of the GQ S as above not concurrent with the line
〈X, Y 〉. There will be points R and W on m with R ∼ X and W ∼ Y . If
Z1 and Z2 are any distinct points of m \ {R, W }, then ΩZ1 ∩ ΩZ2 = {∞, R}.
Further, the point W corresponds to a plane in SXY which is tangent at R
to both ΩZ1 and ΩZ2. This gives rise to a set T of q − 1 ovoids that intersect
pairwise at the two points R and ∞ and have the same tangent plane at
R (i.e., W ⊥ ∩ X ⊥ ) and the same tangent plane at ∞ (i.e., π∞). Such a
set of ovoids is called a transversal of ovoids. The two common points are
called base points and the two common planes are called base planes of the
transversal. In the Laguerre geometry SX the ovoids in the transversal T of

746 CHAPTER 16. LAGUERRE GEOMETRIES AND GQ
correspond to the circles not incident with Y in a pencil of ovoids, and the
base plane of T at the point R corresponds to the circle in the pencil which
is incident with Y .
Next, suppose that Z1 ∼ Z2 but that there is a point U on 〈X, Y 〉 with
X = U = Y , Z1 ∼ U ∼ Z2. In this case ΩZ1 ∩ ΩZ2 = ({X, Z1, Z2} ⊥ \ {U}) ∪
{∞}, an intersection of size q + 1.
For the last case, suppose that Z1 ∼ U1 = U2 ∼ Z2 with U1, U2 ∈ 〈X, Y 〉
but different from X and Y . In this case ΩZ1 ∩ ΩZ2 = {X, Z1, Z2} ⊥ ∪ {∞},
an intersection of size q + 2.
We have previously seen in Theorem 15.7.4 that if S is a GQ of order
(s, s 2 ), s > 1, s odd, satisfying Property (G) at a pair of collinear points,
then S is the dual of a flock GQ. Consequently, in the s odd case we know
that a Laguerre geometry with a classical internal structure at a point arises
from a dual flock GQ. Hence from now on we turn our attention the the case
s = 2 e .
16.6 Laguerre Sets of Ovoids with q even
A 4-cap is a set of four points, no three of which are collinear. Let (∞, π∞)
be an incident point-plane pair of P G(3, q). Then a Laguerre set Θ of ovoids
of P G(3, q) with respect to (∞, π∞) is a set of ovoids each of which contains
∞, has tangent plane π∞ at ∞ and is such that if X, Y , Z are three distinct
points of P G(3, q) \ π∞ with {∞, X, Y, Z} a 4-cap in P G(3, q), then:
A1 If ∞ ∈ 〈X, Y, Z〉, then |{Ω ∈ Θ : {X, Y, Z} ⊂ Ω}| = q; and
A2 If ∞ ∈ 〈X, Y, Z〉, then |{Ω ∈ Θ : {X, Y, Z} ⊂ Ω}| = q − 1.
Recall (Lemma 15.3.5) that if Θ is a tetradic set of ovoids, then Θ satisfies
A1, and (Lemma 15.4.2) if Θ is a tetradic set of elliptic quadrics then Θ
satisfies A2. Hence any tetradic set of elliptic quadrics is automatically a
Laguerre set of elliptic quadrics.
Our goal is to show that starting with a Laguerre set Θ we can construct
a Laguerre geometry with classical internal structure at a point.
So until further notice we suppose we are given a Laguerre set Θ of ovoids
of P G(3, q), q = 2 e .

16.6. LAGUERRE SETS OF OVOIDS WITH Q EVEN 747
Lemma 16.6.1. |Θ| = q 3 (q − 1).
Proof. There are q 2 +q planes π through ∞ different from π∞, then q 2 points
X of π \ π∞, followed by q 2 − q points Y of π \ π∞ not on the line 〈∞, X〉,
and finally (q 2 − 31 + 2 points Z ∈ π \ π∞ for which ∞ ∈ 〈X, Y, Z〉 = π and
{∞, X, Y, Z} is a 4-cap in π. This gives (q 2 +q)q 2 (q 2 −q)(q 2 −3q +2) ordered
triples (X, Y, Z) of points in P G(3, q) \ π∞ of the type just described. On
the other hand, if Ω is an ovoid with tangent plane π∞ at ∞, then there are
q 2 · (q 2 − 1)(q − 2) = (q 2 + q)q(q − 1)(q − 2) such ordered triples (X, Y, Z)
contained in Ω. Hence |Θ| = q 3 (q − 1).
Lemma 16.6.2. Let X ∈ P G(3, q) \ π∞, then X is contained in q 2 (q − 1)
elements of Θ.
Proof. Let π be a plane on the line m = 〈∞, X〉. There are (q 2 −q)(q 2 −3q+2)
triples (X, Y, Z) of points in π \ π∞ such that {∞, X, Y, Z} is a 4-arc. Thus
there are (q + 1)(q 2 − q)(q 2 − 3q + 2) triples, considering all planes on m.
By A1 there are q(q + 1)(q 2 − q)(q 2 − 3q + 2) elements of Θ on these triples.
However, each element of Θ containing X also contains (q 2 − q)(q − 2) =
(q + 1)(q − 1)(q − 2) triples (X, Y, Z) with ∞ ∈ 〈X, Y, Z〉. Hence the number
of elements of Θ containing X is q 2 (q − 1).
Lemma 16.6.3. Let X, Y ∈ P G(3, q) \ π∞ be such that ∞ ∈ 〈X, Y 〉. Then
there are exactly q 2 − q elements of Θ containing X and Y .
Proof. Any element of Θ containing X and Y meets π = 〈∞, X, Y 〉 in an
oval. Hence there are q −2 triples (X, Y, Z), with Z ∈ π, which are contained
in such an ovoid. In total there are q 2 − 3q + 2 triples (X, Y, Z) in π such
that {∞, X, Y, Z} is a 4-arc and thus contained in q elements of Θ. Hence
there are q(q 2 − 3q + 2)/(q − 2) = q 2 − q elements of Θ containing X and
Y .
Lemma 16.6.4. Let X ∈ P G(3, q) \ π∞ and π a plane such that X ∈ π but
∞ ∈ π. Then there are exactly q − 1 elements of Θ containing X and with
tangent plane π at X.
Proof. We count how many elements of Θ contain X with π not a tangent
plane, that is, meet π in an oval. There are q 2 − 1 choices for a point Y ∈
π\(π∞∪{X}), and hence (q 2 −q)(q 2 −1) elements of Θ on X and Y . For each
element Θ containing X and with π not a tangent plane there are q choices for

748 CHAPTER 16. LAGUERRE GEOMETRIES AND GQ
Y . Hence the number of elements of Θ containing X and with π not tangent
is (q − 1)(q 2 − 1). It follows that the number of elements of Θ containing X
and with π a tangent plane is q 2 (q − 1) − (q − 1)(q 2 − 1) = q − 1.
Recall that when q = 2 e each ovoid Ω determines a symplectic polarity
⊥ of P G(3, q) as follows. Each plane π of P G(3, q) tangent to Ω at a point
X is paired with X by ⊥. Each plane π not tangent to Ω meets Ω in an oval
with nucleus N; π and N are paired by ⊥.
Lemma 16.6.5. Let X ∈ P G(3, q)\π∞ and let π be a plane such that X ∈ π
but ∞ ∈ π. Then there are exactly (q −1) 2 elements Ω of Θ whose associated
polarity interchanges X and π and such that X ∈ Ω, i.e., X is the nucleus
of π ∩ Ω.
Proof. If Ω ∈ Θ, then ⊥Ω denotes the symplectic polarity defined by Ω. So
let X be a fixed point not in π∞ and let π be a fixed plane containing X but
not ∞.
We start by counting the ovoids Ω ∈ Θ for which π is not tangent to Ω.
Each such Ω meets π in an oval for which q/2 lines of π through X meet
Ω in two points. Let N be the set of ordered triples (Y, Z, Ω) such that
X, Y, Z are collinear in π and Y, Z ∈ Ω ∈ Θ. There are q + 1 lines of π
through X, each with q − 1 points different from X and not on π∞. This
gives (q+1)(q−1)(q−2) ordered pairs (Y, Z) with X, Y, Z collinear in π\π∞.
By Lemm ?? there are q 2 − q elements of Θ containing Y and Z (and hence
not containing X). So |N | = q(q + 1)(q − 1) 2 (q − 2). Of course, since X, Y, Z
is a secant to Ω lying in π if (Y, Z, Ω) ∈ N , clearly for such an Ω we have
X ⊥Ω = π. On the other hand, suppose that Ω ∈ Θ with X ∈ Ω and with
X ⊥Ω = π, Let πX be the plane through X with X ⊥Ω = πX. Then π ∩ πX
is a line meeting Ω in just one point. The q other lines of π through X are
split into two sets of size q/2, those in one set having no point of Ω, those
in the other set each meeting Ω in two points. Those that meet Ω in two
points give exactly q ordered pairs (Y, Z) such that (Y, Z, Ω) ∈ N . Hence the
number of Ω ∈ Θ with X ∈ Ω, X ⊥Ω = π, and π is not tangent to Ω equals
|N |/q = (q + 1)(q − 1) 2 (q − 2).
Clearly π has q 2 −1 points different from X and not on π∞. By Lemma 16.6.4
there are q − 1 elements Ω ∈ Θ tangent to π at each of these points. Hence
there are (q 2 − 1)(q − 1) ovoids Ω ∈ Θ such that X ∈ Ω, X ⊥Ω = π and π is
a tangent plane to Ω.
Hence the number of Ω ∈ Θ with X ∈ Ω and for which X ⊥Ω = π is

16.6. LAGUERRE SETS OF OVOIDS WITH Q EVEN 749
(q 3 (q − 1) − q 2 (q − 1)) the total number of Ω ∈ Θ with X ∈ Ω
−(q + 1)(q − 1) 2 (q − 2) minus the number of Ω with X ∈ Ω
π not tangent to Ω and X ⊥Ω = π
−(q 2 − 1)(q − 1) minus the number of Ω with X ∈ Ω
π is a tangent plane to Ω and X ⊥Ω = π
= (q − 1) 2 . equals the number of Ω with X ∈ Ω
and X ⊥Ω = π.
Adding the two results from Lemmas 16.6.4 and 16.6.5 we obtain
Corollary 16.6.6. Let X ∈ P G(3, q) \ π∞ and let π be a plane such that
X ∈ π but ∞ ∈ π. There are exactly q(q − 1) elements of Θ with associated
polarity that interchanges X and π.
Lemma 16.6.7. Let ∞ = X ∈ π∞ and let π be a plane distinct from π∞
such that X, ∞ ∈ π. There are exactly q 2 (q−1) elements of Θ with associated
polarity interchanging X and π.
Proof. There are q 2 (q − 1) ordered pairs of points in π \ π∞ whose span
contains X, and each such pair is contained in q 2 − q elements of Θ by
Lemma ??.Each such element of Θ contains q such ordered pairs (on the
q/2 lines of π through X secant to the oval π ∩ Ω). Hence there are q 2 (q −
1)(q 2 − q)/q = q 2 (q − 1) 2 elements of Θ meeting a line through X in two
points (different from X), i.e., whose polarity does not interchange X and π.
Hence this leaves q 3 (q − 1) − q 2 (q − 1) 2 = q 2 (q − 1) elements Ω of Θ whose
polarity does interchange X and π, i.e., for which X is the nucleus of the
oval Ω ∩ π.
In Section 15.8 we have established an equivalence relation ⊲⊳ on the
set of q 2 (q − 1) symplectic polarities that interchange ∞ and π∞. Two
such polarities ⊥1 and ⊥2 are equivalent if and only if ⊥1 ◦ ⊥2 acts as the
identity on some line m incident with ∞ and contained in π∞. The two
polarities are said to be equivalent about m. It was established in the proof
of Theorem 15.8.1 that there are q − 1 equivalence classes with q 2 polarities
in each class. Morever, (see Lemma 15.8.2) for each line ℓ of π∞ through

750 CHAPTER 16. LAGUERRE GEOMETRIES AND GQ
∞ there are q symplectic polarities interchanging ∞ and π∞ and equivalent
about ℓ. Let ⊥ be such a fixed polarity. For each of the q + 1 lines of π∞
there are q − 1 other such polarities equivalent to ⊥ about that line, giving
at least 1 + (q + 1)(q − 1) = q 2 polarities in each class. Since there are q − 1
classes and q 2 (q − 1) such polarities, we have exactly q 2 polarities in each
class.
If φ1 and φ2 are inequivalent polarities, then there exist skew lines m1
and m2 with ∞ ∈ m1 ⊂ π∞ and ∞ ∈ m2 ⊂ π∞ such that φ1 ◦ φ2 fixes the
points and planes incident with m1 or m2, and both φ1 and φ2 interchange
m1 and m2. The polarities are said to be inequivalent about {m1, m2}.
We extend this equivalence relation to the elements of Θ and speak of
ovoids being equivalent (about a line) or inequivalent (about a pair of skew
lines). (See Theorem 15.8.9.)
Lemma 16.6.8. Among the symplectic polarities of P G(3, q) interchanging
∞ and π∞ two equivalent polarities are induced by the same number of elements
of Θ. Furthermore, the number of elements of Θ with induced polarity
in a fixed class is a multiple of q 2 .
Proof. Let ⊥ be a fixed symplectic polarity interchanging ∞ and π∞. For
this polarity define the following three integers:
a = number of elements of Θ inducing ⊥;
b = number of elements of Θ inducing a polarity distinct but equivalent to ⊥;
c = number of elements of Θ inducing a polarity inequivalent to ⊥ .
We have the following:
a + b + c = q 3 (q − 1). (16.1)
Now we count pairs (X, Ω) where X ∈ P G(3, q) \ π∞, Ω ∈ Θ, and X ⊥ =
X ⊥Ω . There are q 3 points X not in π∞. Fix one and Let π = X ⊥ , so π is a
plane containing X and different from π∞. By Cor 16.6.6 there are q(q − 1)
elements Ω of Θ with associated polarity that interchanges X and π, i.e.,
with X ⊥ = X ⊥Ω . This gives q 4 (q − 1) ordered pairs (X, Ω) of the type we
are counting.
On the other hand, if X ⊥ = X ⊥Ω , then since X ∈ π∞, either ⊥Ω=⊥,
or ⊥ and ⊥Ω are inequivalent. To see this, suppose that ⊥=⊥Ω and put

16.6. LAGUERRE SETS OF OVOIDS WITH Q EVEN 751
m = ∞ ⊥ ∩ X ⊥ = π ∩ π∞ = 〈∞, X〉 ⊥ = 〈∞, X〉 ⊥Ω . Clearly both ⊥ and
⊥Ω interchange m and 〈∞, X〉. If R is any point of m, then R ⊥ contains
R, ∞ and X, so each point of m is interchanged with the plane containing
it and the points ∞ and X. So ⊥ ◦ ⊥Ω fixes each point on m and each
plane through 〈∞, X〉. Interchanging the roles of m and 〈∞, X〉 we see that,
similarly, ⊥ ◦ ⊥Ω fixes each point of 〈∞, X〉 and each plane through m. So
⊥ and ⊥Ω are inequivalent about the pair (m, 〈∞, X〉).
If ⊥=⊥Ω, then X ⊥ = X ⊥Ω for all X ∈ P G(3, q) \ π∞, adding q 3 a to the
count of the pairs (X, Ω). On the other hand, if ⊥ and ⊥Ω are inequivalent,
say about the pair (m, ℓ) with ∞ ∈ m ⊂ π∞ and ∞ ∈ ℓ ⊂ π∞, there are
exactly q points X (of ℓ\{∞}) such that X ⊥ = X ⊥Ω , adding cq to the count.
Hence q 4 (q − 1) = q 3 a + cq, or
q 2 a + c = q 3 (q − 1). (16.2)
From Eqs. 16.1 and 16.2 we have b = (q 2 − 1)a. Since a (and hence
also b) is determined by c, the number of elements of Θ inducing a polarity
inequivalent to ⊥, it follows that a is independent of which polarity in the
equivalence class of ⊥ we use.
Also, the number of elements of Θ inducing a polarity in the class of ⊥
is a + b = q 2 a.
Note: Up to this point we have used only condition A1 of the definition
of Θ.
Theorem 16.6.9. Let X ∈ P G(3, q)\π∞ and let π be a plane such that X ∈
π and ∞ ∈ π. The q−1 elements of Θ (guaranteed to exist by Lemma 16.6.4)
containing X and with tangent plane π at X form a transversal, that is,
intersect pairwise in {∞, X} and (hence) partition the points of P G(3, q) \
(π∞ ∪ π ∪ 〈∞, X〉). Furthermore, the q − 1 ovoids are pairwise inequivalent
about the lines π ∩ π∞ and 〈∞, X〉. This means that each equivalence class
is represented.
Proof. We show that if Y ∈ P G(3, q) \ (π∞ ∪ π ∪ 〈∞, X〉), then there is a
unique element of Θ containing X with tangent plane π that also contains
Y .
Let π ′ = 〈∞, X, Y 〉 and m = π∩π ′ . First count the number of elements of
Θ containing X, Y with m a tangent line. In total there are q(q −1) elements
of Θ containing X and Y (by Lemma ??), and to have m not a tangent the

752 CHAPTER 16. LAGUERRE GEOMETRIES AND GQ
ovoid must contain some point Z ∈ m \ (X ∪ π∞ ∪ 〈∞, Y 〉). Since 〈∞, Y 〉
and m both lie in π ′ they meet in a point of m different from X and not on
π∞, so there are q − 2 choices for Z. Then since ∞ ∈ 〈X, Y, Z〉, by A1 we
have q(q −2) ovoids in Θ with m not tangent (and containing X and Y ), and
hence q(q − 1) − q(q − 2) = q elements of Θ containing X, Y with m tangent.
So we have q(q − 2) ovoids of Θ containing X and Y with m not tangent,
each of which meets π in an oval with q − 1 points not on m and not in
π∞. This gives (q − 1)q(q − 2) pairs (Z ′ , Ω) where Z ′ ∈ π \ (π∞ ∪ m),
X, Y, Z ′ ∈ Ω ∈ Θ and m is not tangent to Ω. By A2 the number of pairs
(Z ′ , Ω) with Z ′ ∈ π \ (π∞ ∪ m) and X, Y, Z ′ ∈ Ω ∈ Θ is (q 2 − q)(q − 1).
It follows that the number of pairs (Z ′ , Ω) with Z ′ ∈ π \ (π∞ ∪ m), X, Y ,
Z ′ ∈ Ω ∈ Θ and m a tangent to Ω is (q 2 −q)(q−1)−(q−1)q(q−2) = q(q−1).
Each Ω gives q of these pairs, so there must be q − 1 Ω ∈ Θ containing X, Y
with π not tangent (X, Z ′ ∈ π means π is not tangent). This leaves one
Ω ∈ Θ containing X and Y with π tangent to Ω at X. We restate this for
emphasis:
For each Y ∈ P G(3, q) \ (π∞ ∪ π ∪ 〈∞, X〉) there is a unique ovoid Ω ∈ Θ
containing X and Y with π tangent to Ω at X.
There are (q 3 +q 2 +q+1)−(2q 2 +q+1+q−1) = q 3 −q 2 −q+1 = (q−1)(q 2 −1)
points of P G(3, q)\(π∞ ∪π ∪〈∞, X〉), exactly partitioned by the q −1 ovoids
Ω ∈ Θ containing X and with tangent plane π. Hence these q − 1 ovoids
form a transversal with base points ∞ and X and base planes π∞ and π.
Let the transversal be T = {Ω1, Ω2, . . . , Ωq−1}. Since all ovoids in the
transversal have tangent plane π at X and tangent plane π∞ at ∞, it must
be that either any two ovoids in the transversal are inequivalent about the
skew lines π ∩ π∞ and 〈∞, X〉, or define the same polarity. We show that the
latter possibility cannot occur. Consider a plane π ′ such that ∞ ∈ π ′ and
X ∈ π ′ . Let {P } = π ′ ∩ π ∩ π∞. Then Ωi ∩ π ′ = Ci is an oval. Now consider
any line m of π ′ such that P, ∞ ∈ m. The q − 1 points of m \ (π ∪ π∞)
are partitioned by the Ci. In particular, since q − 1 is odd, it must be that
m is tangent to at least one of the Ci and the point m ∩ π∞ is the nucleus
of this oval. Hence the nuclei of the Ci are the q − 1 distinct points of
(π ′ ∩ π∞) \ {∞, P }. Hence the polarities of the Ωi are pairwise distinct.
Corollary 16.6.10. Each symplectic polarity of P G(3, q) interchanging ∞
and π∞ is induced by q elements of Θ. Further, each equivalence class of Θ
contains exactly q 3 elements.

16.6. LAGUERRE SETS OF OVOIDS WITH Q EVEN 753
Proof. There are q 3 points X ∈ P G(3, q) \ π∞ and then q 2 planes π incident
with X but not with ∞. There is one ovoid in each equivalence class
interchanging X and π. Each such ovoid gets counted q 2 times - once for
each of its points other than ∞. Then the number of elements of Θ in a
given equivalence class is q 3 q 2 /q 2 = q 3 . Using the equations in the proof of
Lemma 16.6.8 we have that a + b = q 3 , and hence a = q.
We now have enough information to construct a Laguerre geometry from
a Laguerre set.
Theorem 16.6.11. Let Θ be a Laguerre set of ovoids of P G(3, q), q even,
with respect to the incident point-plane pair (∞, π∞). Let S be the incident
structure with
Points of three types:
(i) points of P G(3, q) \ π∞;
(ii) equivalence classes of Θ; and,
(iii) π∞.
Lines of two types:
(a) the lines of P G(3, q) \ π∞ incident with ∞; and
(b) a formal line [∞].
Circles are of two types:
(α) the elements of Θ; and
(β) the planes of P G(3, q) not incident with ∞.
Incidence is natural, plus the line [∞] is incident with all points of type
(ii) and (iii), while π∞ is incident with all circles of type (β).
Then S is a Laguerre geometry with parameters n = s = q, ℓ = q 2 + 1 and
with classical internal structure at the point π∞.

754 CHAPTER 16. LAGUERRE GEOMETRIES AND GQ
Proof. Axioms A1 and A2 of a Laguerre set ensure that three non-collinear
points of S of type (i) are incident with q common circles. Two non-collinear
points of type (i) and π∞ are incident with q common circles of type (β).
If X, Y are two, distinct non-collinear points of type (i), then let π be
any plane with X, Y ∈ π and ∞ ∈ π and let P = 〈X, Y 〉 ∩ π∞. for
N ∈ (π ∩ π∞) \ {P }, choosing Z ∈ 〈X, N〉 \ {X, N} we have q − 1 elements of
Θ containing {X, Y, Z} by A2. Hence in total there are (q − 1) 2 elements of
Θ whose intersection with π does not have nucleus N. Thus by Lemma ??
there are exactly q − 1 elements of Θ whose intersection with π does have
nucleus N. These ovoids are hence pairwise inequivalent. Fixing any one of
these ovoids we see that for each such plane π the ovoid is inequivalent to
q − 2 distinct elements of Θ that contain X and Y , and hence is inequivalent
to at most q(q − 1) − q(q − 2) = q. Since there are q(q − 1) elements of
Θ containing X, Y and q − 1 equivalence classes, this means that there are
exactly q in each equivalence class. Hence in S the points X and Y and a
point of type (ii) are incident with exactly q circles. Hence s = q for S.
Theorem 16.6.9 demonstrates the touching axiom for S except in the case
of touching at a point of type (ii) or (iii). (The case for (iii) is supposed to
be easy. So we consider the case for (ii).) First note that by Cor. 16.6.10 a
fixed polarity interchanging ∞ and π∞ is induced by q elements of Θ, say
Ω1, . . . , Ωq. WOLG suppose that X is an element of (Ω1 ∩ Ω2) \ {∞}. Then
Ω1 and Ω2 have the same tangent plane at X and so define inequivalent
polarities by Theorem 16.6.9, a contradiction. Hence Ω1, . . . , Ωq intersect
pairwise in ∞ and as circles in S intersect pairwise in the equivalence class
of their polarity. Now we show that no other element of Θ defining a polarity
equivalent to that of Ω1 intersects Ω1 in exactly ∞. For the q 2 − q elements
of Θ equivalent to Ω1 and defining a different polarity from that of Ω1, let
t1, . . . , t q 3 −q be the intersection sizes of these ovoids with Ω1 \ {∞}. Then we
have
q3−q
i=1
ti = q 2 (q 2 − 1); ti(ti − 1) = q 2 (q 2 − 1)(q − 1);
and the average value of the ti is t = q. Calculating (ti − t) 2 = 0, we see
ti = t for all i. So the only elements of Θ equivalent to Ω1 that intersect Ω1
in exactly ∞ are Ω2, . . . , Ωq. Hence we have the touching axiom for S.
Finally, the internal structure of S at π∞ consists of the points and hyperplanes
of P G(3, q) with (∞, π∞) removed.

16.6. LAGUERRE SETS OF OVOIDS WITH Q EVEN 755
As an exercise verify that the converse of this theorem also holds.

756 CHAPTER 16. LAGUERRE GEOMETRIES AND GQ

Chapter 17
Translation Oval Cones
The material in this chapter is adapted primarily from the work of W.E.
Cherowitzo [Ch98] Throughout the chapter we assume q = 2 e .
17.1 Representations of Flocks
Let π = P G(2, q) be embedded in P G(3, q) as the hyperplane [0, 0, 0, 1]. Let
A be any set of points of π and put
KA = {(x, y, z, w) : (x, y, z, 0) ∈ A} ∪ {(0, 0, 0, 1)}.
Then KA is the cone over A with vertex V = (0, 0, 0, 1). In this chapter we
primarily treat the case where the base A is an oval in π and is called the
base oval or carrier. If ℓ is a line all of whose points lie in the cone, then ℓ
must be a line joining the vertex with some point of the carrier. Such a line
is called a generator of the cone.
The main results deal with the case where A is a monomial oval. The
special case where A is a translation oval plays an especially important role
in the theory of spreads of the GQ T2(O), so these cones will be studied
in detail. The so-called flock quadrangles, a major topic of investigation in
other chapters, all arise from quadratic cones, i.e., KA where A is a conic.
Let k be an integer satisfying
(k, q − 1) = (k − 1, q − 1) = 1. (17.1)
Recall from Chapter 4 that M denotes the set of such integers and that I is
the set of such integers k for which x ↦→ x k is an o-polynomial.
757

758 CHAPTER 17. TRANSLATION OVAL CONES
Define a monomial oval cone in PG(3,q) to be the set of points given by
Kk = {(x, y, z, w) ∈ P G(3, q) : y k = xz k−1 }, for k ∈ I.
So Kk is a cone with vertex V = (0, 0, 0, 1) and base oval Ok = {(x, y, z, 0) ∈
[0, 0, 0, 1] : y k = xz k−1 }. A flock of Kk is a set of q planes of P G(3, q) not
passing through the vertex which do not intersect each other at any point
of Kk. So the plane sections of Kk by the q planes of a flock partition the
points of Kk \ {V } into projectively equivalent ovals. The point (0, 1, 0, 0) of
π is the nucleus of the base oval. Suppose that π1 and π2 are two planes
not through V that contain the same point Y on the nuclear generator
〈(0, 1, 0, 0), (0, 0, 0, 1)〉. Let ℓ = π1 ∩ π2. So ℓ is a line of πi through the
nucleus of the oval Oi = πi ∩Kk, hence must contain a unique point Pi of Oi,
i = 1, 2. It is clear that P1 = P2, which implies that π1 and π2 must meet at
a point of Kk. This proves the following
Lemma 17.1.1. A flock of the cone Kk remains a flock of the extended cone
defined by
K +
k = Kk ∪ {(0, 1, 0, w) : w ∈ Fq}.
Let ℓ1 = 〈(1, 0, 0, 0), (0, 0, 0, 1)〉, ℓ2 = 〈(0, 1, 0, 0), (0, 0, 0, 1)〉, and ℓ3 =
〈(0, 0, 1, 0), (0, 0, 0, 1)〉. Note that the plane [a, b, c, 1] meets the lines ℓ1, ℓ2, ℓ3,
in the points (1, 0, 0, a), (0, 1, 0, b), (0, 0, 1, c), respectively. This proves the
following
Lemma 17.1.2. If F = {πt = [at, bt, ct, 1] : t ∈ Fq} is a flock of the monomial
oval cone Kk where the planes of the flock are indexed by the elements
of Fq, then the three functions
t ↦→ at, t ↦→ bt, t ↦→ ct,
are permutations of the elements of Fq.
Suppose that for some 0 = t ∈ Fq and distinct r, s ∈ Fq the point
(t k , t, 1, w) ∈ πr ∩ πs = [ar, br, cr, 1] ∩ [as, bs, cs, 1]. It follows that if we put
a = ar + as = 0, b = br + bs = 0, c = cr + cs = 0, then
Multiply by
1
a k−1
b k
k−1
to obtain
T k + T +
at k + bt + c = 0.
a 1
k−1 c
b k
k−1
, where T =
a 1
k−1
b 1
k−1
t.

17.2. NORMALIZED α-CONES 759
This proves the following:
Theorem 17.1.3. F = {πt = [at, bt, ct, 1] : t ∈ Fq} is a flock of the cone Kk
if and only if
(ar + as) 1
k−1 (cr + cs)
(br + bs) k
k−1
where D 0 k = {tk + t : t ∈ Fq} and D 1 k = Fq \ D 0 k .
∈ D 1 k , ∀ r = s, (17.2)
Let ℓ be a line disjoint from the cone Kk. The set of q planes through ℓ
but not through V form a linear flock of Kk. If the oval Ok is a translation
oval (for example, a conic), any two linear flocks are projectively equivalent,
but for a general monomial oval cone this will not be the case.
We should make it clear what we mean by saying that two flocks are
projectively equivalent. Two flocks F1 and F2 of a cone K are projectively
equivalent provided there is a semi-linear transformation of P G(3, q) leaving
the cone invariant and mapping F1 to F2. For a given representation of a
flock, a re-ordering of the planes may be necessary.
17.2 Normalized α-Cones
Let α = 2 i , with (i, e) = 1, so α is an automorphism of Fq of maximal order,
i.e., a generator of Aut(Fq). Put k = α above, and call the resulting cone Kα
an α-cone. . The point V = (0, 0, 0, 1) is the vertex; each line of Kα is called
a generator, and the line 〈(0, 0, 0, 1), (0, 1, 0, 0)〉 is the nuclear generator. If
π is any plane of P G(3, q) not containing the vertex, it meets Kα in an oval
O isomorphic to Oα, and with the property that the line of π tangent to O
at its point on 〈(0, 0, 0, 1), (1, 0, 0, 0)〉 is an axis of the translation oval O, the
unique axis if α = 2. Hence we call 〈(0, 0, 0, 1), (1, 0, 0, 0)〉 an axial generator.
Starting with a knowledge of the collineation group leaving a translation
oval invariant, it is a worthwhile exercise to prove the following:
Theorem 17.2.1. The subgroup of P ΓL(4, q) leaving invariant the cone Kα
consists of the following collineations:
θ : (x0, x1, x2, x3) ↦→ (x σ 0 , xσ1 , xσ2 , xσ3 )M, M =
⎛
⎜
⎝
a ασ 0 0 x
0 a σ 0 y
(as) ασ (as) σ 1 z
0 0 0 w
⎞
⎟
⎠ .

760 CHAPTER 17. TRANSLATION OVAL CONES
Here a, s, x, y, z, w are elements of Fq with a and w not zero, and σ any
automorphisms of Fq.
For convenience in computing the images of planes we give the inverse of
M.
M −1 =
⎛
⎜
⎝
a −ασ 0 0 xa ασ w −1
0 a −σ 0 ya −σ w −1
s ασ s σ 1 xs ασ w −1 + ys σ w −1 + zw −1
0 0 0 w −1
⎞
⎟
⎠ . (17.3)
Suppose that a and s have been fixed with a = 0, and that [r, v, t, 1] is
any plane not on the vertex V (0, 0, 0, 1). Let w be any non-zero element of
Fq and put x = rw, y = vw, z = tw. Then θ maps the plane [r, v, t, 1] to
[0, 0, 0, 1]. Hence we may move any plane not on the vertex V (0, 0, 0, 1) to the
plane [0, 0, 0, 1] without moving the cone or its axial generator or its nuclear
generator. And the collineations fixing the cone and the plane [0, 0, 0, 1] are
given by σ ∈ Aut(Fq), a, w, s ∈ Fq, a = 0 = w, with
M =
⎛
⎜
⎝
a ασ 0 0 0
0 a σ 0 0
(as) ασ (as) σ 1 0
0 0 0 w
⎞
⎟
⎠ , and M −1 =
⎛
⎜
⎝
a ασ 0 0 0
0 a σ 0 0
(as) ασ (as) σ 1 0
0 0 0 w −1
⎞
⎟
⎠ .
(17.4)
If α = 2, then the unique axial generator of Kα is the line
〈V (0, 0, 0, 1), Y (1, 0, 0, 0)〉. The general plane through the axial generator is
[0, x, y, 0]. We name the plane [0, 0, 1, 0] as P G(2, q) = [0, 0, 1, 0]. It is fixed
by the collineations indicated in Eq. 17.4. The plane [0, 1, ay, 0], y ∈ Fq,
is mapped to [0, 1, (s + y) σ , 0]. If we pick s = y, then the plane [0, 1, ay, 0]
is mapped to [0, 1, 0, 0] without moving the cone, its axial generator, or the
plane P G(2, q).
At this point we are still free to pick nonzero a and w and an automorphism
σ and leave invariant the cone and its vertex and both its axial
and nuclear generators, the planes ζ = [0, 1, 0, 0], P G(2, q) = [0, 0, 1, 0] and
π = [0, 0, 0, 1]. Suppose a is fixed and 0 = λ ∈ Fq. If (1, 0, 0, λ α ), λ = 0,
is an arbitrary point of the axial generator different from V (0, 0, 0, 1) and
Y (1, 0, 0, 0), put w = (aλ −1 ) σα . Then (1, 0, 0, λ σ ) is mapped to (1, 0, 0, 1)
without moving any of the structures so carefully arranged above. At this

17.3. α-FLOCKS 761
point we can still choose nonzero a. The cone meets the plane π : x3 = 0 in
the oval xα 1 + x0x α−1
2 = x3 = 0 with nucleus (0, 1, 0, 0), containing the point
Y (1, 0, 0, 0) and with the line 〈Y (1, 0, 0, 0), (0, 1, 0, 0)〉 as an axis (unique if
α = 2). For arbitrary nonzero a ∈ Fq, the collineation
(x0, x1, x2, x3) ↦→ (x0, x1, x2, x3) σ
⎛
⎜
⎝
a σα 0 0 0
0 a σ 0 0
0 0 1 0
0 0 0 a σα
fixes the structure set up above and maps (1, 1, 0, 0) to (a σα , a σ , 0, 0) ≡
(1, (a σ ) 1−α , 0, 0). Since a ↦→ a 1−α is a permutation of the nonzero elements of
Fq, we may map (1, 1, 0, 0) to any point of the axis 〈Y (1, 0, 0, 0), (0, 1, 0, 0)〉
other than Y (1, 0, 0, 0) or (0, 1, 0, 0). So far we have not used the field automorphism
σ. Hence we have proved the following theorem.
Theorem 17.2.2. Let Kα be an α-cone in P G(3, q) with vertex V , with axial
generator L1 and nuclear generator L2. (This means that for any plane π
not containing the vertex, the oval O = π ∩ Kα has nucleus N = L2 ∩ π,
contains the point Q = L1 ∩π, and the line 〈, Q, N〉 is an axis of O. Let π(3)
be an arbitrary but fixed plane not containing the vertex V . Let π(2) be the
plane containing the axial generator and the nuclear generator of Kα. Let
π(1) be any other plane containing the axial generator. Let Y be the point
of π(3) on the axial generator, and let U be any point of the axial generator
different from V and from Y (i.e., not in π(3)). Finally, let P be any point
of π(2) ∩ π(3) not on the axial generator or nuclear generator of Kα. Then
coordinates for P G(3, q) may be chosen so that the following hold:
V = (0, 0, 0, 1), π(3) = [0, 0, 0, 1], π(2) = [0, 0, 1, 0], π(1) = [0, 1, 0, 0],
Y = (1, 0, 0, 0), U = (1, 0, 0, 1), P = (1, 1, 0, 0), L2 ∩ π(3) = (0, 1, 0, 0).
17.3 α-Flocks
A flock of Kα will be called an α-flock. By part 10. of Theorem 4.13.1, we
have the following:
⎞
⎟
⎠

762 CHAPTER 17. TRANSLATION OVAL CONES
Corollary 17.3.1. F = {πt = [at, bt, ct, 1] : t ∈ Fq} is an α-flock if and only
if
tr
(at + as) 1
α−1 (ct + cs)
(bt + bs) α
= 1 ∀ t = s.
α−1
Corollary 17.3.2. F = {πt = [at, bt, ct, 1] : t ∈ Fq} is an α-flock if and only
if F ′ = {π ′ t = [bt, at, ct, 1] : t ∈ Fq} is a 1
α -flock.
Proof. The corollary follows immediately from the following observation:
b
1
α−1−1 a α−1
α −1 −1
= a 1
α−1
b α
α−1
The geometric explanation for this last result is that it is equivalent to
the fact that if we take the α-cone Kα, remove the line 〈(0, 0, 0, 1), (1, 0, 0, 0)〉
and add the nuclear line 〈(0, 0, 0, 1), (0, 1, 0, 0)〉, we obtain a cone equivalent
to K1/α, and that any flock of Kα is also a flock of this new cone. Note: This
last observation really means that there is essentially no difference between
an α-flock and a 1
α -flock.
We now want to find a convenient canonical form for α-flocks. Suppose
F = {πt : t ∈ Fq} is such an α-flock. Then π0 (0 ∈ Fq) may play the role of
π(3) in the preceding theorem, i.e., we may assume that π0 = [0, 0, 0, 1], so
a0 = b0 = c0 = 0. Also, we may reindex the planes of the α-flock so that any
one of the three functions is any permutation fixing 0 that we choose. Hence
we choose to have bt = t 1/α . At this time we also adopt a standard notation:
write at = f(t) and ct = g(t). Here we know f(t) and g(t) may be assumed
to be permutations of the elements of Fq for which f(0) = g(0) = 0. Then
Corollary 17.3.1 reads
Corollary 17.3.3. Let f(t) and g(t) be permutations for which f(0) =
g(0) = 0. Then F = {πt = [f(t), t 1
and only if
α , g(t), 1] : t ∈ Fq} is an α-flock if
f(t)
1
α−1 + f(s)
(g(t) + g(s)) = 1 ∀ t = s.
tr
t + s
And each α-flock may be described in this form.
.

17.3. α-FLOCKS 763
Lemma 17.3.4. Let F = {πt = [f(t), t 1
α , g(t), 1] : t ∈ Fq} be a flock as
above. Let d be any nonzero element of Fq. Write
κ =
f(d)
g(d)
d
¯f(t) = f(dt)
f(d) ;
¯g(t) = g(dt)
g(d) .
1
α−1
;
Then ¯ F = {¯πt = [ ¯ f(t), t 1
α , κ¯g(t), 1] : t ∈ Fq} is an α-flock equivalent to F
and with ¯ f and ¯g permutations of the elements of Fq for which
(i) ¯ f(0) = ¯g(0) = 0;
(ii) ¯ f(1) = ¯g(1) = 1;
(iii) tr(κ) = 1.
Moreover, each α-flock is projectively equivalent to one in this form. An
α-flock given in this form is said to be normalized.
Proof. To start with f and g are permutations with f(0) = g(0) = 0. Hence
for d = 0, both f(d) = 0 and g(d) = 0. And for all t = s we have (replacing
t with dt and s with ds in Cor. 17.3.3):
1 = tr
= tr
f(dt) + f(ds)
dt + ds
¯f(t) + ¯ f(s)
t + s
1
α−1
κ(g(dt) + g(ds))
1
α−1
κ(¯g(t) + ¯g(s)) .
Putting t = d and s = 0 in Corollary 17.3.3 yields tr(κ) = 1. Parts (i)
and (ii) of the Lemma are clearly true.
To see that F and ¯ F are projectively equivalent, put
⎛
⎜
M = ⎜
⎝
1
f(d)
1
d 1 α
1
f(d) α−1
d
1
⎞
⎟
⎠ .

764 CHAPTER 17. TRANSLATION OVAL CONES
Now let θ be the homography of P G(3, q) defined as a permutation of planes
by
⎡ ⎤ ⎡ ⎤
a
a
⎢
θ : ⎢ b ⎥ ⎢
⎥
⎣ c ⎦ ↦→ M · ⎢ b ⎥
⎣ c ⎦
d
d
.
When θ is followed by the permutation t ↦→ dt, the plane [f(t), t 1
α , g(t), 1]
is mapped to the plane [ ¯ f(t), t 1
α , κ¯g(t), 1], where
κ =
f(d)
d
1
α−1
g(d) has trace 1.
Each plane of an α-flock meets the associated α-cone in a translation oval
isomorphic to Oα. The next result connects the theory of α-flocks with that
of ovals in a different way.
Theorem 17.3.5. If F = {πt = [f(t), t 1
α , κg(t), 1] : t ∈ Fq} is a normalized
α-flock, then f(t) is an o-polynomial.
Proof. We already know that f(t) is a permutation polynomial with f(0) = 0
and f(1) = 1. Hence we have only to verify the slope condition. So suppose
that for three distinct values of s, t, u ∈ Fq we have
f(u) + f(t)
u + t
= f(s) + f(t)
s + t
By the additivity of the trace function we have
0 = 1 +
1
f(s) + f(t)
= tr κ
s + t
f(t) + f(u)
+tr κ
t + u
f(s) + f(t)
= tr κ
s + t
f(s) + f(u)
= tr κ
s + u
= 1.
= f(s) + f(u)
1
α−1
(g(s) + g(t)) +
1
α−1
(g(t) + g(u))
. (17.5)
s + u
1
α−1
(g(s) + g(t) + g(t) + g(u))
1
α−1
(g(s) + g(u))

17.4. α-CLANS 765
This contradiction gives the desired result.
An interesting special case is when α = 2.
Corollary 17.3.6. Let α = 2 and κ be an element of Fq with tr(κ) = 1. If
F = {πt = [f(t), t 1
2 , κg(t), 1] : t ∈ Fq} is a normalized 2-flock, then g(t) is
also an o-polynomial.
Proof. Suppose F = {πt = [f(t), t 1
2 , κg(t), 1] : t ∈ Fq} is a normalized 2-flock,
so
(f(s) + f(t))κ(g(s) + g(t))
(g(s) + g(t))κ(f(s) + f(t))
tr
= 1 = tr
,
s + t
s + t
which implies that F = {πt = [g(t), t 1
2 , κf(t), 1] : t ∈ Fq} is a normalized
2-flock, which implies that g(t) is an o-polynomial.
17.4 α-Clans
The notion of α-clan turns out to be a useful notational device for general
α, and we have already seen that it is especially helpful when α = 2.
An α-clan is a set C of q upper triangular matrices
at bt
C = {At =
: t ∈ Fq},
0 ct
with the property that there is some constant κ ∈ Fq for which tr(κ) = 1
and
tr
κ (as + at) 1
α−1 (cs + ct)
(bs + bt) α
α−1
= 1, ∀ s = t.
It is then clear from the definition and Corollary 17.3.1 that
at bt
Obs. 17.4.1.
0 ct t∈Fq
0, t ∈ Fq, is an α-flock.
is an α-clan if and only if atx+bty+κctz+w =
Obs. 17.4.2.
at
0
bt
ct
is an α-clan if and only if
bt
0
at
ct
is a
1
α-clan. t∈Fq
t∈Fq

766 CHAPTER 17. TRANSLATION OVAL CONES
It is also immediate that
Obs. 17.4.3. Each α-clan is equivalent to one in which each matrix of the
clan has the form
At =
f(t) t 1
α
0 g(t)
, t ∈ Fq.
Here f(t) and g(t) are permutation polynomials fixing 0 and 1 and there is
some constant κ with tr(κ) = 1 for which the equation of Corollary 17.3.1
holds. Moreover, by Theorem 17.3.5 f(t) is an o-polynomial.
Similarly,
Obs. 17.4.4.
If
1
f(t) t α
0 g(t)
t∈Fq
f(dt)
f(d)
0
is an α − clan, then for any nonzero d ∈ Fq,
t 1
α
g(dt)
g(d)
t∈Fq
is also an α − clan.
17.5 Some Families of α-Clans
Theorem 17.5.1.
Fq with maximal order.
t 1
α t 1
α
0 t 1
α
t∈Fq
is an α-clan for each automorphism α of
Proof. Let κ be any element of Fq with tr(κ) = 1. Then we have
⎡
⎤
tr ⎣κ
t 1
α + s 1 1
α−1
α
(t
t + s
1
α + s 1
α ) ⎦ = tr(κ) = 1.
This α-clan corresponds to a linear α-flock, so-called because all of the
planes of the flock meet in a common line 〈(1, 1, 0, 0), (0, κ, 1, 0)〉.
Theorem 17.5.2. If q = 2 e with e odd, then
α-clan for any automorphism α of maximal order.
t 1
α2 t 1
α
0 t 1 1
+ α α2 t∈Fq
is an

17.5. SOME FAMILIES OF α-CLANS 767
Proof. Since e is odd we may take κ = 1, because tr(1) = 1 in these fields.
Then
t 1
α 2 + s 1
α 2
t + s
1
α−1
t 1
α
1
+
α2 + s 1 1
+ α α2
= t 1
and we have by Lemma 4.13.4(b) that
⎡
tr ⎣
1
+ α
α2 + s 1 1
+ α α2 (t + s) 1 1
+ α α2 t 1
α2 + s 1
α2 1
α−1
t
t + s
1 1
+ α α2 + s 1 1
+ α α2 ⎤
⎦ = 1.
=
t 1
α +1 + s 1
α +1
(t + s) 1
α +1
1
α
,
Theorem 17.5.3. When q = 2 2h−1 the automorphism σ = 2 h satisfies σ 2 = 2
and t 1− 1
σ t 1
σ
1
1+ 0 t σ
t∈Fq
is a σ − clan.
Proof. As 2h − 1 is odd, we may take κ = 1. It will be convenient to re-
σ−1 x x
normalize this form to
0 xσ+1
by putting t = xσ . Now consider,
(x σ−1 + y σ−1 ) 1
σ−1 (x σ+1 + y σ+1 )
(x σ + y σ ) 1
σ−1
= (xσ−1 + y σ−1 ) σ+1 (x σ+1 + y σ+1 )
(x + y) σ+2
= (x + y + xσ−1 y 2−σ + x 2−σ y σ−1 )(x σ+1 + y σ+1 )
(x + y) σ+2
= (x + y)((x + y)σ+1 + x σ y + xy σ ) + (x σ+1 + y σ+1 )(x σ−1 y 2−σ + x 2−σ y σ−1 )
(x + y) σ+2
= 1 + (xσy + xyσ )
(x + y) σ+1 + (xσ+1 + yσ+1 )((xσ−1 + yσ−1 )(x2−σ + y2−σ ) + x + y)
(x + y) σ+2
= 1 + (xσy + xyσ )
+
(x + y) σ+1

768 CHAPTER 17. TRANSLATION OVAL CONES
+ (xσ−1 + y σ−1 )((x 3 + y 3 ) + x σ+1 y 2−σ + x 2−σ y σ+1 ) + (x + y)(x σ+1 + y σ+1 )
(x + y) σ+2
= 1 + xy(xσ−1 + y σ−1 )
(x + y) σ+1
+ (x3 + y 3 )(x σ−1 + y σ−1 ) + (x 2σ + y 2σ )(x 2−σ + y 2−σ )
(x + y) σ+2
= 1 + xy(xσ−1 + y σ−1 )
(x + y) σ+1
+ (x2 + y 2 )(x σ−1 + y σ−1 )
(x + y) σ+1
= 1 + (x2 + y 2 )(x σ−1 + y σ−1 )
(x + y) σ+1
+ xy(xσ−1 + yσ−1 )
(x + y) σ+1 +
+ (x2σ + y 2σ )(x 2−σ + y 2−σ )
(x + y) σ+2
2 2 σ−1 σ−1 (x + y )(x + y )
+
(x + y) σ+1
As the trace of the sum of the last two terms is 0, we see that the trace of
our expression is equal to tr(1) = 1.
Theorem 17.5.4. If q = 2 2h−1 , then for the automorphism σ = 2 h we have
that t 3
σ −2 t 1
σ
is a σ-clan.
0 t 3
σ
Proof. We again take κ = 1 put put t = xσ to renormalize the form to
3−2σ x x
0 x3
. Now,
t∈Fq
t∈Fq
(x 3−σ + y 3−2σ ) σ+1 (x 3 + y 3 )
(x + y) σ+2
= (xσ−1 + y σ−1 + x 3σ−4 y 3−2σ + x 3−2σ y 3σ−4 )(x 3 + y 3 )
(x + y) σ+2
= xσ+2 + y σ+2 + x 3 y σ−1 + x σ−1 y 3 + x 3σ−1 y 3−2σ + y 3σ−1 x 3−2σ
(x + y) σ+2
+ x3σ−4 y 6−2σ + y 3σ−4 x 6−2σ
(x + y) σ+2
= xσ+2 + y σ+2 + x 3 y σ−1 + x σ−1 y 3 + x 3σ−1 y 3−2σ + y 3σ−1 x 3−2σ
(x + y) σ+2
3−2σ 3σ−2 3−2σ 3σ−2
x y + y x
+
(x + y) σ+1
σ
σ
.

17.5. SOME FAMILIES OF α-CLANS 769
= xσ+2 + y σ+2 + x 3 y σ−1 + x σ−1 y 3 + x 3σ−1 y 3−2σ + y 3σ−1 x 3−2σ
(x + y) σ+2
+ x3−2σ y 3σ−2 + y 3−2σ x 3σ−2
(x + y) σ+1
+ Z.
Here, Z, being of the form w + w σ is an element whose trace is 0. The rest
of the computation consists of several repetitions of the last two steps. We
will recombine terms, find a new term which is a σ power and replace it with
its σ root plus a term whose trace is 0 and accumulate the trace 0 terms in
the Z expressions. First put the two fractions over a common denominator:
= xσ+2 + y σ+2 + x 3 y σ−1 + x σ−1 y 3 + x 4−2σ y 3σ−2 + x 3σ−2 y 4−2σ
(x + y) σ+2
= xσ+2 + y σ+2 + x 3 y σ−1 + x σ−1 y 3
(x + y) σ+2
+ x2σ−2 y 3−σ + x 3−σ y 2σ−2
(x + y) σ+1
= xσ+2 + y σ+2 + x 3 y σ−1 + x σ−1 y 3 + x 2σ−1 y 3−σ + x 3−σ y 2σ−1
(x + y) σ+2
+ x2−σ y 2σ−1 + x 2σ−1 y 2−σ
(x + y) σ+1
= xσ+2 + y σ+2 + x 3 y σ−1 + x σ−1 y 3
(x + y) σ+2
= xσ+2 + y σ+2 + x σ y 2 + x 2 y σ
(x + y) σ+2
= (x + y)σ+2
+ Z ′′
+ xσ−1 y 2 + x σ−1 y 2
(x + y) σ+1
+ Z ′′′
(x + y) σ+2 + Z′′′ = 1 + Z ′′′ .
+ Z ′′′
+ Z
+ Z ′
As the terms Z, Z ′ , Z ′′ and Z ′′′ all have trace 0, the proof is complete.
We now have an easy proof that one of Glynn’s families of hyperovals
exists.
Corollary 17.5.5. (Glynn [Gl83]) For the automorphism σ as above, f(t) =
t 3σ+4 is an o-polynomial.

770 CHAPTER 17. TRANSLATION OVAL CONES
Proof. Apply Obs. 17.4.2 to the σ-clan of Theorem 17.5.3 and put the resulting
1
3σ+4 t t
-clan in standard form, thus obtaining
σ σ
0 t9σ+12
. Hence by
t∈Fq
Theorem 17.3.5 and Obs. 17.4.1 we have that over the field Fq with q = 22h−1 ,
f(t) = t3σ+4 is an o-polynomial.
Theorem 17.5.6. If q = 22h−1 , then for the automorphism σ = 2h we have
that
is a σ-clan.
t 3(σ−1)
σ t 1
σ
0 t σ−1
σ
t∈Fq
Proof. We again take κ = 1 and renormalize the form to
Compute
(x 3σ−3 + y 3σ−3 ) σ+1 (x σ−1 + y σ−1 )
(x + y) σ+2
= (x3 + y 3 + x 6−3σ y 3σ−3 + x 3σ−3 y 6−3σ )(x σ−1 + y σ−1 )
(x + y) σ+2
= xσ+2 + yσ+2 +
(x + y) σ+2
x 3(σ−1) x
0 x σ−1
+ x3 y σ−1 + x σ−1 y 3 + x 3σ−3 y 5−2σ + x 5−2σ y 3σ−3 + x 4σ−4 y 6−3σ + x 6−3σ y 4σ−4
(x + y) σ+2
= xσ+2 + yσ+2 (x + y) σ+2 + x3yσ−1 + xσ−1y3 + x3σ−2y4−2σ + x4−2σy 3σ−2
(x + y) σ+2
+ Z
= xσ+2 + yσ+2 +
(x + y) σ+2
+ x3 y σ−1 + x σ−1 y 3 + x 2σ−1 y 3−σ + x 3−σ y 2σ−1 + x 2σ−2 y 4−σ + x 4−σ y 2σ−2
(x + y) σ+2
= xσ+2 + yσ+2 (x + y) σ+2 + x3yσ−1 + xσ−1y 3 + x2σy 2σ−2 + x2−σy2σ (x + y) σ+2
= xσ+2 + yσ+2 (x + y) σ+2 + x2yσ + xσy2 + Z′′′
(x + y) σ+2
= 1 + Z ′′′ .
As Z, Z ′ , Z ′′ and Z ′′′ all have trace 0, the proof is complete.
+ Z ′′
+ Z ′
.

17.6. HERDS 771
17.6 Herds
The definition of herd has varied a bit in the literature, but the definition we
choose here is that of [Ch98] and is a natural generalization of the original
definition given in [CPPR96]. For an in-depth study of the general concept,
see [CP02].
Let f(t) and g(t) be permutation polynomials over Fq with f(0) = g(0) =
0 and f(1) = g(1) = 1. If there exists an automorphism α of Fq and an
element κ ∈ Fq with trace 1 such that for each s ∈ Fq the function
is a permutation, then the set
t ↦→ hs(t) = κg(t) + sf(t) + s 1
α t 1
α
H(f, g, α, κ) = {f(t)} ∪ {hs(t)}
is called a herd.
If each of the functions in a herd is divided by its value at 1, the resulting
set of functions will be called a normalized herd. If all of the functions in a
herd represent ovals in the plane, i.e., if {(hs(t), t, 1) : t ∈ Fq} ∪ {(1, 0, 0)}
is an oval for each s ∈ Fq, then the herd will be called a herd of ovals or an
oval herd. In a normalized oval herd all the functions are o-polynomials.
Our interest in herds arises from the fact that herds and α-clans are
equivalent objects.
Theorem 17.6.1. H(f, g, α, κ) is a herd if and only if
an α-clan with respect to κ.
Proof. First suppose that
f(t) t 1
α
0 g(t)
t∈Fq
f(t) t 1
α
0 g(t)
t∈Fq
is
is an α-clan with respect to κ
and fix s ∈ Fq. Put κ ′ = κ + s + s 1
α , so clearly tr(κ ′ ) = 1. We are going to
show that for each s ∈ Fq
1
f(t) t α
0
κg(t)+sf(t)+s 1/α t 1/α
κ+s+s 1/α
is an α-clan (with respect to the trace 1 element κ ′ ).
So compute
t∈Fq

772 CHAPTER 17. TRANSLATION OVAL CONES
tr κ ′
f(t) + f(u)
t + u
1
α−1 κ(g(t) + g(u)) + s(f(t) + f(u)) + s 1
α (t 1
α + u 1
α
f(t) + f(u)
= tr κ
t + u
+tr
⎡
= 1 + tr ⎣s
s
(f(t) + f(u)) α
α−1
(t + u) 1
α−1
(f(t) + f(u)) α
α−1
(t + u) 1
α−1
This shows that for each s ∈ Fq,
f(t) t 1
α
0
+
κ ′
1
α−1
(g(t) + g(u)) +
) 1
α + s 1
α
s
hs(t)
hs(1)
(f(t) + f(u)
(t + u)
1
α(α−1)
(f(t) + f(u)) α
α−1
(t + u) 1
α−1
1
α
⎤
⎦ = 1.
t∈Fq
is an α-clan, which in particular forces hs(t) to be a permutation, i.e.,
H(f, g, α, κ) is a herd.
Conversely, suppose that H(f, g, α, κ) is a herd. Fix s ∈ Fq. By hypothesis
hs is a permutation, so that if t = u, then hs(t) = hs(u), i.e.,
hs(x) + hs(y) = 0 ∀ x = y. Then
hs(x) + hs(y) = κ(g(x) + g(y)) + s(f(x) + f(y)) + s 1
α (x 1
α + y 1
α ) = 0
for all x = y. By permitting s to vary and writing Z = s 1
α , we see that
Z α (f(x) + f(y)) + Z(x 1
α + y 1
α ) + κ(g(x) + g(y)) = 0 ∀ x = y.
From this it follows that
f(x) + f(y)
tr κ
x + y
Hence
is an α-clan, completing the proof.
f(t) t 1
α
0 g(t)
t∈Fq
1
α−1
(g(x) + g(y)) = 1, ∀ x = y.
Corollary 17.6.2. If H(f, g, α, κ) is a herd, then f is an o-polynomial.

17.7. FLIPPABLE α-CLANS 773
17.7 Flippable α-Clans
Let
f(t t 1
α
0 g(t)
t∈Fq
be an α-clan with respect κ, and let ρ(t)be a permu-
tation polynomial with ρ(0) = 0. This α-clan is said to be ρ(t)-flippable
provided
0
0
⎧
⎨ f(t)
0
∪ ρ(t)
0 ⎩
t 1 0
α
ρ(t)
⎫
⎬
⎭
g(t)
ρ(t)
0=t∈Fq
is also an α-clan with respect to the same trace 1 element κ. This new αclan
is called the ρ(t)-flip of the original. The first known instance of this
phenomenon is the following:
Theorem 17.7.1. ([BLP94], [PT91]) Each 2-clan
flippable.
f(t) t 1
2
0 g(t)
Proof. By hypothesis there is a κ ∈ Fq with tr(κ) = 1 such that
t∈Fq
is t-
(f(t) + f(s))(g(t) + g(s))
tr κ = 1, ∀ t = s. (17.6)
t + s
Replace x with x −1 in the t-flip to obtain
1 xf x
x 1
2
0 xg 1
x
x∈Fq
. Then,
using the same κ, consider
(xf(1/x) + yf(1/y))(xg(1/x) + yg(1/y))
κ
x + y
2 2 x f(1/x)g(1/x) + y f(1/y)g(1/y) + xy(f(1/x)g(1/y) + f(1/y)g(1/x))
= κ
x + y
⎡
= κ ⎣ (x2 + xy)f ⎤
1 1 2 1 1
g + (y + xy)f g
x x
y y
⎦ +
x + y
+κ
⎡
⎣ (f
1 + f x
1
y
)(g 1
x
1/x + 1/y
+ g
⎤
1 ) y
⎦

774 CHAPTER 17. TRANSLATION OVAL CONES
= κ f(1/x)g(1/x)
1/x
+κ f(1/y)g(1/y)
+κ
1/y
(f(1/x) + f(1/y))(g(1/x) + g(1/y))
1/x + 1/y
Each of these three terms has trace 1, being a special case of Eq. 17.6, so the
sum has trace 1.
Theorem 17.7.2. If the herd H(f, g, α, κ) is an oval herd, then the corresponding
α-clan is t-flippable.
Proof. Since H(f, g, α, κ) is an oval herd, the q + 1 functions of this herd are
all o-polynomials. Specifically, for each s ∈ Fq,
hs(x) =
κg(x) + sf(x) + s 1
α x 1
α
κ + s + s 1
α
is an o-polynomial.
We have seen (chapter on ovals) that if k(x) is an o-polynomial, then k(x)/x
is a permutation polynomial. Hence we may put f ∗ (x) =
g(x 1
1−α )
x 1
1−α
.
1
f(x 1−α )
x 1 and g
1−α
∗ (x) =
. It follows that f ∗ and g ∗ are permutation polynomials fixing 0. Then
Hence we have
h ∗ s (x) = κg∗ (x) + sf ∗ (x) + s 1
α x 1
α
κ + s + s 1 .
α
h ∗ s(x) =
κg(x 1
1−α )
x 1 +
1−α
sf(x 1
1−α )
x 1 + s
1−α
1
α x 1
α
κ + s + s 1
α
This gives a herd whose corresponding α-clan is
⎛
⎞
⎜
⎝
f(x 1
1−α )
x 1
1−α
0
x 1
α
g(x 1
1−α )
x 1
1−α
⎟
⎠ ,
which is the x-flip of the original α-clan. To see this start with:
f(x) x 1
α
0 g(x)
x−flip
↦→
f(x)
x
0
(x1−α ) 1
α
g(x)
x
=
⎛
⎜
⎝
.
f(t 1
1−α )
t 1
1−α
0
t 1
α
g(t 1
1−α )
t 1
1−α
⎞
⎟
⎠ .

17.7. FLIPPABLE α-CLANS 775
There is a converse to the preceding theorem, but before we can even
state it we need to introduce a new concept. Suppose that F = {[xt, yt, zt, 1] :
t ∈ Fq} is an α-flock of the α-cone. Then the shift by s, denoted by τs, s ∈ Fq,
of F is the α-flock F τs = {[x ′ t , y′ t , z′ t , 1] : t ∈ Fq}, where
x ′ t = xt + xs
y ′ t = yt + ys
z ′ t = zt + zs.
Now suppose F = {[f(t), t 1/α , κg(t), 1] : t ∈ Fq}, tr(κ) = 1, is a normalized
α-flock. Let φ denote the t-flip on F, so
F φ = {[f(t)/t, t 1
α −1 , κg(t)/t, 1] : t ∈ Fq}.
For s ∈ Fq, the α-flock F is called (t; s)-shift flippable if and only if F τsφ
is still an α-flock with respect to the same trace 1 element κ. The converse
to Theorem 17.7.2 is included in the following theorem, which also contains
Theorem 17.7.2 itself.
Theorem 17.7.3. The α-flock F = {[f(t), t 1
α , κg(t), 1] : t ∈ Fq} of the αcone
Kα : y α = xz α−1 corresponds to an oval herd H(f, g, α, κ) if and only if
F is (t; s)-shift flippable for all s ∈ Fq.
Proof. First, H(f, g, α, κ) defines an oval herd if and only if f(t) and the
polynomials hu(t) = κg(t) + uf(t) + (ut) 1
α , u ∈ Fq, define q + 1 oval polynomials.
This is if and only if (see Theorem 4.5.2 and its proof),for all s ∈ Fq,
the polynomials fs(t) = (f(t) + f(s))/(s + t) and
hu,s(t) = hu(t) + hu(s)
t + s
= κ(g(t) + g(s)) + u(f(t) + f(s)) + (u(t + s)) 1
α
t + s
are all permutation polynomials satisfying fs(0) = hu,s(0) = 0. KEEP THIS
IN MIND!
If we make the substitution t ′′ = (t + s) 1−α , we obtain
hu,s(t) = κ(g(t′′1/(1−α) + s) + g(s))
t ′′1/(1−α)
Also, fs(t), ∀ s ∈ Fq may be rewritten similarly.
+ u f(t′′1/(1−α) + s) + f(s)
t ′′1/(1−α)
+ (ut ′′ ) 1/α .

776 CHAPTER 17. TRANSLATION OVAL CONES
Now apply the shift-flip (τsφ) to F. First applying the shift by s maps
F onto the α-flock {π ′ t = [(f(t) + f(s)), (t + s) 1/α , κ(g(t) + g(s)), 1] : t ∈ Fq}.
By introducing the parameter t ′ = t + s, this is the set of planes
{π ′ t ′ = [(f(t′ + s) + f(s)), t ′1/α , κ(g(t ′ + s) + g(s)), 1] : t ′ ∈ Fq}.
Note that with t ′ = 0 we have the plane π ′ 0 = [0, 0, 0, 1]. Applying the
t ′ -flip gives the set of planes
{π ′′
{π ′′
t ′ = [f(t′ + s) + f(s)
t ′
1
′
, t α −1 , κ (g(t′ + s) + g(s)
t ′
, 1] : t ′ ∈ Fq}.
Since t ′ = t ′′1/(1−α) , this set of planes is given by
t ′′ = [f(t′′1/(1−α) + s) + f(s)
t ′′1/(1−α)
, t ′′1/α , κ g(t′′1/(1−α) + s) + g(s)
t ′′1/(1−α)
, 1] : t ′′ ∈ Fq},
where π ′′
0 = [0, 0, 0, 1].
On the one hand these planes form an α-flock if and only if F is shiftflippable.
On the other hand these planes form an α-flock if and only if (by
Obs. 17.4.1 α-flocks are equivalent to α-clans; by Theorem 17.6.1 α-clans are
equivalent to herds) the polynomials fs(t), hu,s(t), with
hu,s(t) = k(g(t′′1/(1−α) + s) + g(s))
t ′′1/(1−α)
fs(t) = f(t′′1/(1−α) + s) + f(s)
t ′′1/(1−α)
,
+ u f(t′′1/(1−α) + s) + f(s)
t ′′1/(1−α)
+ (ut) 1/α ,
are permutation polynomials for all s ∈ Fq, satisfying fs(0) = hu,s(0) = 0.
So by the first paragraph of this proof, we see that the herd corresponding
to an α-flock is an oval herd if and only if the α-flock is (t; s)-shift flippable
for all s ∈ Fq.
17.8 A Special α-Clan
In this section we study the α-clan of Theorem 17.5.2.

17.8. A SPECIAL α-CLAN 777
Theorem 17.8.1. Let q = 2 e with e odd, and let α be any automorphism of
Fq of maximal order (to force x ↦→ x 1−α to be a permutation). Then
At =
t 1
α2 t 1
α
0 t 1 1
+ α α2
t∈Fq
is a t-flippable α-clan if and only if t 1 1
+ α α2 is an o-polynomial.
Proof. By Theorem 17.5.2 At is an α-clan with respect to κ = 1. First
suppose that At is t-flippable. Then its t-flip (after renormalizing) is
1 1
+
x α α2 x 1
α
0 x 1+α−α2
α 2 (1−α)
By Theorem 17.3.5 and Obs. 17.4.1, x 1 1
+ α α2 is an o-polynomial.
Conversely, suppose that f(t) = t 1
α
image of Of under the homography θ : x ↦→ xA where
⎛
1 1
⎞
1
A = ⎝ 0 1 0 ⎠ .
0 0 1
.
+ 1
α 2 is an o-polynomial. Consider the
Clearly θ fixes (0, 1, 0) and (0, 0, 1) and interchanges (1, 0, 0) and (1, 1, 1).
And it maps (1, t, f(t)) ↦→ (1, 1 + t, 1 + f(t), i.e., k(t) = f(1 + t) + 1 is an
o-polynomial. But
k(t) = (1 + t) 1 1
+ α α2 + 1 = (1 + t 1
α )(1 + t 1
α2 ) + 1 = t 1
α + t 1
α2 + t 1 1
+ α α2 ,
which must also be an o-polynomial.
Similarly,
⎛
if 0 = d ∈ Fq, then k(dt)/k(d) is also an o-polynomial (use
1 0 0
A = ⎝ 0 1
⎞
0 ⎠). Now put d = s−α and note that
d
1 0 0 k(d)
hs(t) = k(s−α t)
k(s −α ) =
t 1 α
s
1
s
+ t 1
α 2
s 1 α
+ 1
s 1 α
+ t 1 α
+
1
α2 s 1+ 1 α
+ 1
s 1+ 1 α
= s 1
α t 1
α + st 1
α2 + t 1 1
+ α α2 s 1
α + s + 1
which must be an o-polynomial for each s ∈ Fq. Thus H(t 1
an oval herd. Hence by Theorem 17.7.2, At is t-flippable.
α2 , t 1 1
+ α α2 , α, 1) is
,

778 CHAPTER 17. TRANSLATION OVAL CONES
Example 17.8.1.1. Each linear α-flock is t-flippable. The proof of this is
almost trivial.
Example 17.8.1.2. The α-clan with α = 1/σ in Theorem 17.5.2 is tflippable.
With the help of Obs. 17.4.2, its t-flip gives the 1/σ-clan of Theorem
17.5.3. In fact, we can say more: Theorem 17.7.3 provides a new proof
of Theorem 17.5.3.
Proof. Let α = σ−1
2 t
in Theorem 17.8.1, so At =
tσ 0 tσ+2
. Since σ + 2 =
σ(σ + 1) = σ is known to be an o-polynomial, we see that the t-flip (put
σ−1
t = x σ
σ−1 = xσ+2 )
is a 1
σ
t t σ−1
0 t σ+1
σ+2 x x
=
σ
0 x σ+2
σ−1
-clan. Finally, by Obs. 17.4.2 we see that
σ σ+2 x x
0 x σ+2
x−1
is a σ-clan. So we may renormalize and get that
t 1− 1
σ t 1
σ
1
1+ 0 t σ
is a σ − clan.
This provides another proof of Theorem 17.5.3
1
2− Example 17.8.1.3. The α-clan with α = 1/σ in Theorem 17.5.2 is t σ -
flippable, leading to the 1/σ-clan of Theorem 17.5.4. It is also tσ+3-flippable, leading to the 1/σ-clan of Theorem 17.5.6.
Theorem 17.8.1 makes it quite interesting to know for which automor-
phisms α of Fq with maximal order the polynomial t 1
α
+ 1
α 2 is an o-polynomial.
Fortunately, Cherowitzo and Storme [CS98] have provided the answer. They
prove a more general result, but we content ourselves here with the following.
Theorem 17.8.2. Let q = 2e with e odd and let α be an automorphism of Fq
with maximal order, i.e., α = 2i with (i, e) = 1. Then tk is an o-polynomial
where k = 1 1 + α α2 if and only if k is one of the following:
(i) k = 2 + 22 = 6;
(ii) k = σ + 2 with σ2 = 2;
(iii) k = γ + γ2 = γ + σ, where γ2 = σ and σ2 = 2;
(iv) k = q/4 + q/2 = 3q/4.

17.8. A SPECIAL α-CLAN 779
Proof. In the following we use α in place of 1
α . So suppose k = α+α2 = 2 i +2 2i
defines a monomial hyperoval Ok in P G(2, q). Since (i, h) = 1, there is a
smallest positive power r for which α r = 2. The proof is divided into four
cases depending on the value of r. Also, the proof in each case amounts to
an application of Glynn’s criterion Theorem 4.5.15. For the convenience of
the reader we recall this theorem.
Theorem 17.8.3. For nonzero k in the integers modulo q − 1, assume that
(k, q − 1) = (k − 1, q − 1) = 1. Then D(x k ) is a hyperoval of P G(2, q) if and
only if for each nonzero d in the integers modulo q − 1 it is NOT the case
that d ≪ kd.
This theorem of Glynn can be stated more generally. Let q = 2 e and
let α = 2 r , (r, e) = 1. Then it is possible to write each integer d with
0 ≤ d < q − 1 uniquely as d ≡ e−1
i=0 aiα i (mod q − 1) with 0 ≤ di ≤ 1,
i = 0, . . . , e − 1. Then ≪ can be extended to this α-ary expansion. Namely
a ≪ b if and only if ai ≤ bi, i = 0, . . . , e − 1, with a ≡ e−1
i=0 aiα i (mod q − 1),
b ≡ e−1
i=0 biα i (mod q − 1), ai, bi ∈ {0, 1}.
Then Glynn’s theorem is also valid for all of these α-ary expansions be-
cause writing an integer n in binary expansion n = e−1
i=0 bi2 i , 0 ≤ bi ≤ 1,
i = 0, . . . , e − 1, or in α-ary expansion n ≡ e−1
i=0 diα i (mod q − 1), with
0 ≤ di ≤ 1, i = 0, . . . , e − 1, simply amounts to a permutation of the binary
digits bi.
We are now ready for the cases:
Case 1. r odd and 3 ≤ r ≤ e − 2.
(e−r−4)
2
t=0
Let d = α2t + αe−r−2 + αe−r−1 if r ≤ e − 4, and d = 1 + α when
r = e − 2. Keep in mind k = α + α2 .
Then for 3 ≤ r ≤ e − 4, dk = e−r−1 t=1
αt + 2αe−r + αe−r+1 . Since 2 = αr ,
this says dk ≡ e−r−1 t=0
αt + αe−r+1 (mod q − 1). So d ≪ dk.
For r = e−2, dk = α+2α 2 +α3 ≡ 1+α+α 3 (mod q −1). Again d ≪ dk.
So in Case 1 no examples arise.
Case 2. r even and 6 ≤ r ≤ e − 7.
(e−r−5)
2
t=0
Let d = α2t + αe−r−4 + αe−r−2 + αe−r−1 . Then dk = e−r−5 t=1
αt +
αe−r−4 + 2αe−r−3 + αe−r−2 + αe−r−1 + 2αe−r + αe−r+1 .
Using αr = 2 we can simplify this to dk ≡ e−r−4 t=0
αt + αe−r−2 + αe−r−1 +
αe−r+1 + αe−3 (mod q − 1).

780 CHAPTER 17. TRANSLATION OVAL CONES
Again d ≪ dk, and no example arises.
Case 3. r = e − 5 (r ≥ 6).
Let d = 1 + α + α 3 + α 4 . Then dk = α + 2α 2 + α 3 + α 4 + 2α 5 + α 6 ≡
1 + α + a 3 + α 4 + α 6 + α e−3 (mod q − 1), since 2 = α e−5 . Again d ≪ dk, and
no example arises.
Case 4. r = e − 3 (r ≥ 6).
Let d = 1 + α 2 + α 4 + α 5 . Then dk = α + α 2 + α 3 + α 4 + α 5 + 2α 6 + α 7 ≡
α + α 2 + 2α 3 + α 4 + α 5 + α 7 ≡ 1 + α + α 2 + α 4 + α 5 + α 7 (mod q − 1), since
2 = α e−3 . Once more, d ≪ dk and no examples arise.
The above four cases show that if x k is an o-polynomial with k = α + α 2 ,
then r ∈ {1, 2, 4, e − 1}. Let α = 2 i .
For r = 1, k = 2 + 2 2 = 6, which defines the Segre hyperoval. For r = 2,
2i ≡ 1 (mod e). Equivalently, i = (e + 1)/2, and k = σ + 2, where D(k) is a
translation hyperoval, since σ ∗ = σ/(σ − 1) = σ(σ + 1) = σ + 2.
For r = 4, then 4i ≡ 1 (mod e). So i = (e + 1)/4 if e ≡ 3 (mod 4), and
i = (3e + 1)/4 when e ≡ 1 )mod 4). Hence k = σ + γ, with σ 2 = 2 and
γ 2 = σ. As we shall see in the next section, this defines one of the Glynn
hyperovals.
Finally, for r = e − 1, also i = e − 1. So k = 2 −1 + 2 −2 ≡ 3q/4, which
defines a translation hyperoval snce k ≡ 1 − 1
4 .
Corollary 17.8.4. Let q = 2 e , e odd, and let α be an automorphism of Fq
with maximal order e. The q planes πt = [t 1
α 2 , t 1
α , t 1 1
+ α α2 , 1], t ∈ Fq, define an
α-flock F of the α-cone y α = xz α−1 . The α-flock F is t-flippable if and only
if the corresponding herd is an oval herd if and only if one of the following
holds:
(i) α = 1/2;
(ii) α = 1/σ with σ 2 = 2;
(iii) α = 1/γ with γ 4 = 2;
(iv) α = 2.
17.9 The ovals σ + γ of Glynn
The next lemma gives a result of Glynn [Gl83] with a rather different proof
found by Cherowitzo.

17.9. THE OVALS σ + γ OF GLYNN 781
Lemma 17.9.1. Let q = 2e , e odd, and let γ be the automorphism of Fq for
which γ2 = σ where σ2
γ+σ t t
= 2. Then
γ
is a 1
γ -clan.
0 t σγ+σ−γ
γ−1
Proof. The original proof by Glynn[Gl83], although perhaps amazing, is
rather unpleasantly technical. The proof we give here is from Cherowitzo
[Ch98]. It is probably just as technical, but may be a little clearer
conceptually.
Since e is odd we may put κ = 1. Then the lemma is equivalent to the
trace of the following expression being equal to 1:
Note that
W := (tγ+σ + sγ+σ 1
) γ−1−1 (t σγ+σ−γ
γ−1 + s σγ+σ−γ
γ−1 )
1
γ−1
(t γ + s γ ) γ−1
γ −1 −1
= (tγ + sγ ) 1
γ−1 (t σγ+σ−γ
γ−1 + s σγ+σ−γ
γ−1 )
(tγ+1 + sγ+1 ) σ
γ−1
= (γ + 1)(σ + 1) and use this to compute
w :=
σγ + σ − γ
γ − 1
= σγ + 2σ + 3γ + 2.
Also, put T = (t γ + s γ ) 1
γ−1 = (t γ + s γ ) σγ+σ+γ+1 , so
W = T (tw + sw )
(tγ+1 + sγ+1 ) σ .
γ−1
Put X = {σγ, σ, γ, 1}. For each subset U ⊆ X write Ū = X \ U and ΣU
to denote the sum of the elements of U. Then
T =
t ΣU Σ
s Ū .
U⊆X
Then with a little persistence we can write the numerator as T (t w +s w ) =
.

782 CHAPTER 17. TRANSLATION OVAL CONES
A + B + C + D + E + F + G where
A = t w (t σγ+σ+γ+2 + t γ+2 s σγ+σ + t γ+σ s σγ+2 )
+ s w (s σγ+σ+γ+2 + s γ+2 t σγ+σ + s γ+σ t σγ+2 );
B = t w (t σγ+γ+2 s σ + t σγ+σ+γ s 2 + t σγ+γ s σ+2 )
+ s w (s σγ+γ+2 t σ + s σγ+σ+γ t 2 + s σγ+γ t σ+2 );
C = t w (t σγ+σ+2 s γ + t σ+γ+2 s σγ + t 2 s σγ+σ+γ + t σ+2 s σγ+γ )
+ s w (s σγ+σ+2 t γ + s σ+γ+2 t σγ + s 2 t σγ+σ+γ + s σ+2 t σγ+γ );
D = t w (t σγ+σ s γ+2 + t σγ s σ+γ+2 ) + s w (s σγ+σ t γ+2 + s σγ t σ+γ+2 );
E = t w (t σ s σγ+γ+2 ) + s w (s σ t σγ+γ+2 );
F = t w (t γ s σγ+σ+2 + s σγ+σ+γ+2 ) + s w (s γ t σγ+σ+2 + t σγ+σ+γ+2 );
G = t w (t σγ+2 s σ+γ ) + s w (s σγ+2 t σ+γ ).
To show that tr(W ) = 1 whenever s = t, we again want to eliminate
from W terms with trace 0. We want to view this as replacing certain terms
from the numerator with something simpler, but the ‘trick’ appears more
complicated than before because the denominator is more complicated.
, and
From γ
γ−1
+ γ = σ
γ−1 we see 1
x γ
γ−1
y 1
γ
x γ
γ−1
γ
+ y 1
= xγ
x σ
γ−1
γ
x γ
γ−1
= y + xγy 1
γ
x σ ,
γ−1
which must have trace 0. If we think of x being in the form x = t γ+1 + s γ+1 ,
then x γ = t σ+γ + s σ+γ . Choose any two terms u and y of the numerator. Put
H = u + (t σ+γ + s σ+γ )y 1
γ and Z = y + (t σ+γ + s σ+γ )y 1
γ .
Then u + y = H + Z, and Z may be discarded,i.e., u + y may be replaced
with H.
First choose C = u and y = B, so
H = C + (t σ+γ + s σ+γ )B 1
γ
= t 2σγ+2σ+2γ+4 s σ+2γ + t σγ+2σ+2γ+4 s σγ+σ+2γ
+ s 2σγ+2σ+2γ+4 t σ+2γ + s σγ+2σ+2γ+4 t σγ+σ+2γ .

17.9. THE OVALS σ + γ OF GLYNN 783
So replace B + C with H. Next choose y = H and u = D. Put
I = D + (t σ+γ + s σ+γ )H 1
γ
= t 2σγ+2σ+2γ+2 s σ+2γ+2 + t 2σγ+σ+2γ+2 s 2σ+2γ+2
+ s 2σγ+2σ+2γ+2 t σ+2γ+2 + s 2σγ+σ+2γ+2 t 2σ+2γ+2 .
So H +D is replaced by I and the numerator of T (tw +sw ) has been replaced
by A + I + E + F + G.
Now put u = E and y = I, and define J and K by J + K = E + (tσ+γ +
sσ+γ )I 1
γ where
J := E + K + (t σ+γ + s σ+γ )I 1
γ
= t σγ+2σ+2γ+2 s σγ+σ+2γ+2 + t σγ+3σ+2γ+2 s σγ+2γ+2
+ s σγ+2σ+2γ+2 t σγ+σ+2γ+2 + s σγ+3σ+2γ+2 t σγ+2γ+2 .
It follows that K = t σγ+σ+2γ+2 s σγ+2σ+2γ+2 +s σγ+2σ+2γ+2 . So replace I +E
with J + K.
Next, put u = F + K and y = J, and
L := F + K + (t σ+γ + s σ+γ )J 1
γ
= t σγ+σ+2γ+2 s σγ+2σ+2γ+2 + s σγ+σ+2γ+2 t σγ+2σ+2γ+2 .
Replace F + K + J with L, so that the numerator of T (t w + s w ) has now
been replaced by A + L + G. Now notice that
G = (t σ+γ +s σ+γ )(t 2σγ+3σ+4γ+4 +s 2σγ+3σ+4γ+4 ) 1
γ +t 2σγ+3σ+4γ+4 +s 2σγ+3σ+4γ+4 ,
so G can be removed. The remaining complete expression has numerator
A + L where (again use “ · · · ” to indicate the undisplayed mates)
and
A = t 2σγ+3σ+4γ+4 + t σγ+2σ+4γ+4 s σγ+σ + t σγ+3σ+4γ+2 s σγ+2 + · · ·
L = t σγ+σ+2θ+2 s σγ+2σ+2γ+2 + · · · .

784 CHAPTER 17. TRANSLATION OVAL CONES
But now it is straightforward to show that
where
and
A + L
(t γ+1 + s γ+1 ) σ
γ−1
= 1 +
M + N
(tγ+1 + sγ+1 ) σ ,
γ−1
M = t 2σγ+σ+2γ+4 s 2σ+2γ + t σγ+3σ+2γ s σγ+2γ+4 + · · · ,
N = t 2σγ+3σ+2γ+2 s 2γ+2 + t 2σ+4γ+2 s 2σγ+σ+2 + · · · .
But now it is easy to check that N = (t σ+γ + s σ+γ )M 1
γ , completing the
proof.
Example 17.9.1.1. Put α = 1 in Theorem 17.7.3 to obtain the α-clan
γ
σ t tγ 0 tγ+σ
σ−1 t t
. The t-flip of this clan is
γ−1
0 tγ+σ−1
, which after renormalization
gives
γ+σ x xγ
.
0 x σ+γσ−γ
γ−1
Since this is an α-clan by Lemma 17.9.1, by Theorem 17.3.5 and Obs. 17.4.1
we see that x γ+σ is an o-polynomial.
This establishes the existence of this infinite family of ovals found by
Glynn [Gl83] by means of Cherowitzo’s theory of α-clans, a proof very different
from that of Glynn.
Example 17.9.1.2. It is an easy exercise to show that if the α-clan
is ρ(t)-flippable, then so is the 1
α-clan f(t) t 1
α
0 g(t)
t∈Fq
t 1
α f(t)
0 g(t)
In particular, Theorem 17.7.1 implies that all 1-clans
are t-flippable. Hence
2
we may put α = 1
2 into Theorem 17.7.3 to see that the Segre polynomial t6 is
an o-polynomial.
The concept of ρ(t)-flippability of an α-clan has a natural geometric in-
at bt
terpretation. If the α-clan
with associated α-flock atx + bty +
0 ct t∈Fq
κctz + w = 0 is ρ(t)-flippable, then the α-flock associated with the ρ(t)-flip
is given by atx + bty + κctz + ρ(t)w = 0. For each t = 0, the corresponding
planes of the two α-flocks meet in the line atx + bty + κctz = 0 in the
t∈Fq
.

17.10. CHEROWITZO HYPEROVALS 785
plane w = 0. It is now a trivial matter to see that any linear α-flock is
ρ(t)-flippable for any permutation ρ(t), and its ρ(t)-flip is itself. However, in
general a ρ(t)-flippable α-clan need not be equivalent to its ρ(t)-flip.
Lemma 17.9.2. An α-clan
1 + f(t + 1) t 1
α
0 1 + g(t + 1)
f(t) t 1
α
.
0 g(t)
is equivalent to the α-clan
Proof. Re-order the matrices of the α-clan with the permutation t ↦→ t + 1,
1 1
and then add the constant matrix to each matrix of the α-clan.
0 1
17.10 Cherowitzo Hyperovals
Let q = 2 e , e odd, and let σ be the automorphism of Fq for which σ 2 = 2.
Define the polynomial function
ch(t) = x σ + x σ+2 + x 3σ+4 .
W. E. Cherowitzo [Ch88] discovered with the help of a computer that for
several values of q, ch(t) is an o-polynomial. He conjectured at that time that
this formula always gives an o-polynomial, but the proof was quite elusive.
Finally in [Ch98] he was able to show that indeed these “Cherowitzo ovals”
always exist. We devote this section to the surprisingly technical proof.
Lemma 17.10.1. If f(x) = x + x σ−1 + x σ+1 , then
(i) f −1 (x) = xσ+1
1+x 2 +x σ ,
and
(ii) f σ+2 (x)+xf 3 (x)+x σ f 2σ (x)
x
Proof. Put g(x) = xσ+1
1+x 2 +x σ . Then
= 1+(1+x)3σ+4
. x
f(g(x)) = f
x σ+1
1 + x 2 + x σ
=

786 CHAPTER 17. TRANSLATION OVAL CONES
=
xσ+1 1 + x2
+
+ xσ x σ+1
1 + x 2 + x σ
σ−1
+
x σ+1
1 + x 2 + x σ
= xσ+1 (1 + x 2 + x σ ) σ + x(1 + x 2 + x σ ) 2 + (x σ+1 ) σ+1
(1 + x 2 + x σ ) σ+1
= x [xσ (1 + x 2 + x σ ) σ + 1 + x 4 + x 2σ + x 2σ+2 ]
(1 + x 2 + x σ ) σ+1
σ+1
= x [xσ (1 + x 2 + x σ ) σ + 1 + x 2 + x 2σ + x 2 + x 4 + x 2σ+2 ]
(1 + x 2 + x σ ) σ+1
= x [xσ (1 + x 2 + x σ ) σ + (1 + x 2 + x σ ) σ + x 2 (1 + x 2 + x σ ) σ ]
(1 + x 2 + x σ ) σ+1
= x((1 + x2 + x σ ) σ ((1 + x 2 + x σ )
(1 + x 2 + x σ ) σ+1
= x.
On the other hand,
g(f(x)) =
(x + x σ−1 + x σ+1 ) σ+1
1 + (x + x σ−1 + x σ+1 ) 2 + (x + x σ−1 + x σ+1 ) σ
= x + x3 + x 2σ−1 + x 2σ+3 + x σ+1 + x 3−σ + x σ+3
1 + x 2 + x 2σ−2 + x 2σ+2 + x σ + x 2−σ + x σ+2
= x.
This proves part (i). For (ii),
f σ+2 (x) + xf 3 (x) + x σ f 2σ (x)
x
= (x + xσ−1 + x σ+1 ) σ+2 + x(x + x σ−1 + x σ+1 ) 3 + x σ (x + x σ−1 + x σ+1 ) 2σ
x
= (x2 + x 2σ−2 + x 2σ+2 )(x 2 + x 2−σ ) + x 3σ + x 4−σ + x 3σ+4
x
= xσ + x 2σ + x 3σ + x 4 + x σ+4 + x 2σ+4 + x 3σ+4
x
= 1 + (1 + x)4 (1 + x + x 2 + x 3 ) σ
x
= 1 + (1 + x)4 (1 + x) 3σ
x
= 1 + (1 + x)3σ+4
.
x

17.10. CHEROWITZO HYPEROVALS 787
Theorem 17.10.2. With σ2
σ+2 1 + (1 + t)
= 2 as usual,
tσ 0 1 + (1 + t) 3σ+4
is a t-flippable 1/σ-clan.
t 1− 1
σ t 1
σ
Proof. By Theorem 17.5.3,
1 is a σ-clan. Then by Obs. 17.4.2,
1+ 0 t σ
1 1
1− t σ t σ
1 is a 1+ 0 t σ
1
1−σ
-clan, and it is equivalent to (replace t σ
σ 2
σ+2 t t
with t)
σ
0 t3σ+4
,
which by Lemma 17.9.2 is equivalent to the 1
σ-clan
σ+2 1 + (1 + t) tσ 0 1 + (1 + t) 3σ+4
=
2 σ σ+2 t + t + t tσ 0 1 + (1 + t) 3σ+4
. This clan is t-flippable if and only if (by
σ t t
Example 17.9.1.2) the σ-clan
2 + tσ + tσ+2 0 1 + (1 + t) 3σ+4
is t-flippable. So the
σ−1 t t + t
proof of the theorem amounts to showing that
σ−1 + tσ+1
0
1+(1+t) 3σ+4
is a σ-clan.
t
Now put x = f(t) = t + tσ−1 + tσ+1 , so by Lemma 17.10.1
t = h(x) = xσ+1
1+x2 +xσ , and what we must show is that
x(1 + x 2 + x σ ) 1−σ x
0 x(1 + x 2 + x σ ) + x 3 + x 2σ+1 (1 + x 2 + x σ ) 1−σ
is a σ-clan. To do this, we must show that for distinct x and y the following
expression must have trace 1, where we have put k(x) = 1 + x2 + xσ and use
1 = σ + 1:
σ−1
(xk(x) 1−σ + yh(y) 1−σ ) σ+1 (xk(x) + x3 + x2σ+1k(x) 1−σ + yk(y) + y3 + y2σ+1k(y) 1−σ )
(x + y) σ+2
.
At this point things really begin to get messy. In order to keep the
account somewhat readable, we adopt (following Cherowitzo) the following
two conventions. First, only the terms of the numerator will be displayed,
and second, we shall exploit the symmetry in x and y. If one term appears, it
is to be understood that the corresponding term with x and y interchanged
(the mate of the first mentioned term) is also present but not written. We
use the ellipsis to remind the reader of this.
t∈Fq

788 CHAPTER 17. TRANSLATION OVAL CONES
We these conventions in mind, expand the numerator. First expand
x
k(x) σ−1 σ+1 + · · · into the sum of four terms
σ+1 x
k(x) + xσyk(x) σ−2
k(y) σ−1
+ · · · .
Then multiply the four terms on the left times the six terms on the right and
add them together.There are two terms of the form xσ+1yA for some A that
k(x)
combine to give the second term below, but the others are the simple result
of multiplication term by term. The result is
x σ+2 + xσ+1y(1 + yσ )
k(x)
+ xσ+1yk(x) σ−1
k(y) σ−1
+ x3σ+2
k(x) σ + yσx2k(y) σ−2
k(x) σ−2
+
+ yσ x 4 k(y) σ−2
k(x) σ−1
+ xσ+1y 2σ+1
xσ+4
+
k(x)k(y) σ−1 k(x) +
+ xσ+3yk(x) σ−2
k(y) σ−1 + x3σ+1y +
k(x)k(y) σ−1
+ yσ x 2σ+2 k(y) σ−2
k(x) 2σ−2
+ · · ·
These terms, in the order in which they appear will be rererred to as
terms a, b, . . . , k.
The following observation will be referred to as the trick:
A
=
(x + y) σ+2
A 1
σ
(x + y)
σ+1 + Z = (x + y)A 1
α
+ Z,
(x + y) σ+2
where tr(Z ) = 0. Specifically, since we are only interested in showing that
the entire sum has trace 1, we may replace each term A of the numerator
with the term (x + y)A 1
σ , plus a term with trace 0. Apply the trick to the
last four displayed terms and their undisplayed mates. The result (after some
simplification) is (combining terms i + e, d + h, f + k, j + g, and reworking
part of b)
x σ+2 + xσ=1 y(1 + y σ )
k(x)
+ xσ+1 y 2σ=1
k(x)k(y) σ−1
+ xσ+3y k(x) + xσy2k(x) σ−1
k(y( σ−1
+ xσ+2y 2k(x) σ−2
k(y) σ−1

17.10. CHEROWITZO HYPEROVALS 789
+ yx2σ+1 (1 + x) + y 2 x 2σ (1 + x + x σ )
k(x)k(y) σ−1
+ · · · + Z,
where Z (not in the numerator) is an element of trace 0.
At this point by putting every term over a common denominator k(x)k(y) σ ,
it is possible to show that this expression is equal to the following:
x σ+2 + x σ y 2 + x σ+1 y + xσ+1 y σ+1 + x 2σ+1 y
k(x)
+ xσ y 2 (1 + y σ )
k(y) σ
+ xσ+2 y 2 k(x) σ−2
k(y) σ−1
+ xσ+1 y 2σ+1 + yx 2σ+1 (1 + x) + y 2 x 2σ+1
k(x)k(y) σ−1
+ · · · + Z.
Now apply the ‘trick’ to the second term of the second line and then swap
the result with its undisplayed mate and simplify to get
x σ+2 + x σ y 2 + x σ+1 y + xσ+1 y σ+1 + x 2σ+1 y
k(x)
+ xσyσ+4 k(y) σ + xσ+1y σ+2 (1 + x) + xσ+1y σ+1 (1 + x + xσ )
k(x)k(y) σ−1
+ xσ+1 y 2σ+1 + yx 2σ+1 (1 + x) + y 2 x 2σ+1
k(x)k(y) σ−1
+ · · · + Z ′ ,
with Z ′ of trace 0.
Putting the rational terms over a common denominator and simplifying
the numerator gives
x σ+2 + x σ y 2 + x σ+1 y + xσ+1 yk(y)(x + y) σ+1
k(x)k(y) σ
+ xσ y σ+2 (x 2 + y 2 + x σ y 2 + x 2 y σ + x σ+1 y + xy σ+1
k(x)k(y) σ
+ · · · + Z ′ .
Start with the first of the rational expressions and include the denominator
(x + y) σ+2 . First cancel one factor of (x + y) from both the numerator
and denominator to get
xσ+1y(x + y) σ
k(x)k(y) σ−1 .
(x + y) σ+1

790 CHAPTER 17. TRANSLATION OVAL CONES
Then replace this term with its image under σ plus a term Z ∗ of trace 0, to
get
Swap this term with its mate to get
x σ+2 y σ (x + y) 2
k(x) σ k(y) 2−σ (x + y) σ+2 + Z∗ .
x σ+2 + x σ y 2 + x σ+1 y + xσ y σ+2 k(x) σ (x + y) 2
k(x) 2 k(y) σ
+ xσ y σ+2 (x 2 + y 2 + x σ y 2 + x 2 y σ + x σ+1 y + xy σ+1
k(x)k(y) σ
+ · · · + Z ′′ .
Straightforward (but tedious) algebraic manipulation shows that this is
the same as
x σ+2 + x σ y 2 + x σ+1 y + x2σ+2 y σ+2 (x + y) 2
k(x) 2 k(y) σ
+ xσ+1 y σ+2 (x + y) σ+1
k(x)k(y) σ
+ · · · + Z ′′ .
(Here we put all the rational expressions over the common denominator
k(x) 2 k(y) σ and multiply out all the terms to see that equality holds.)
Now replace the second rational term with its image under σ to get
x σ+2 + x σ y 2 + x σ+1 y + x2σ+2 y σ+2 (x + y) 2
k(x) 2 k(y) σ
+ y2σ+2 x σ+2 (x + y) 2
k(y) 2 k(x) σ
+ · · · + Z ′′′ .
At this point it is clear that the two rational terms are mates and hence
cancel with their undisplayed counterparts. Now our expression written out
in full (i.e., with mates included and the denominator (x + y) σ+2 included)
(and reorganized a bit) is
x σ+2 + x σ y 2 + x 2 y σ + y σ+2 + x σ+1 y + xy σ+1
(x + y) σ+2
= 1 + xσ+1 y + xy σ+1
(x + y) σ+2
+ Z′′′ .
+ Z ′′′

17.10. CHEROWITZO HYPEROVALS 791
The rational expression here has trace 0 by Lemma 4.12.4(c). Hence the
full expression has trace equal to tr(1) = 1for all x = y, completing the
proof.
Corollary 17.10.3. f(x) = x σ + x σ+2 + x 3σ+4 is an o-polynomial over Fq
for all q = 2 e with e odd.
Proof. The 1/σ-clan of the Theorem is
t 2 + t σ + t σ+2 t σ
0 t 4 + t σ + t σ+4 + t 2σ + t 2σ+4 + t 3σ + t 3σ+4
So its t-flip is
t + t σ−1 + t σ+1 t σ−1
0 t 3 + t σ−1 + t σ+3 + t 2σ−1 + t 2σ+3 + t 3σ−1 + t 3σ+3
Now put t σ−1 = x σ , i.e., t = x σ
σ−1 = x σ+2 , to get the 1/σ-clan
x σ+2 + x σ + x 3σ+4 x σ
0 x 3σ+6 + x σ + x 5σ+8 + x 3σ+3 + x 7σ+10 + x 5σ+4 + x 9σ+12
By Obs. 17.4.3, f(x) is an o-polynomial.
.
.
.

792 CHAPTER 17. TRANSLATION OVAL CONES

Chapter 18
Generalized Fans and Spreads
of T2(O))
Throughout this chapter q = 2 e . The material in this chapter is adapted
primarily from the two papers [BOPPR04] and [BOPPR??].
In Section 7.6 we introduced the notion of fan of ovals in P G(2, q) to
prove the Plane Representation Theorem. In this chapter we generalize the
notion of fan to that of generalized f-fan of ovals in order to study spreads
of T2(O) where f is an o-polynomial and
O = Of = {((1, x, f(x)) : x ∈ Fq} ∪ {(0, 0, 1)} with nucleus N = (0, 1, 0).
18.1 Generalized f-Fans
We first review the notions of compatibility and matching for ovals. Let O1
and O2 be ovals of P G(2, q) with nuclei N1 and N2, respectively, and having
a point Q in common such that QN1 = QN2. Let P be a point of QNi not
on either of the ovals and distinct from their nuclei. We say that O1 and
O2 are compatible at the point P provided that each secant to O1 through
P is external to O2 (so that each line through P external to O1 is secant to
O2. We need to generalized this a bit. Let O1 and O2 be ovals with nuclei
N1 and N2, respectively, and let P1 and P2 be points of P G(2, q) such that
P1 ∈ O1 ∪ {N1} and P2 ∈ O2 ∪ {N2}. We say that (P1, O1) matches with
(P2, O2) provided there is a collineation g such that g(P1) = P2 and g(O1)
and O2 are compatible at P2.
793

794 CHAPTER 18. GENERALIZED FANS AND SPREADS OF T2(O))
Let f be an o-polynomial over Fq. A generalized f-fan in P G(2, q) is a
family F = {Os : s ∈ Fq} of ovals such that
Os has nucleus (0, 1, 0) for all s ∈ Fq; (18.1)
Os ∩ Ot = (0, 0, 1) for all s = t ∈ Fq; (18.2)
f(s) + f(t)
Os and Ot are compatible at Pst = 0, 1, for all s = t.
s + t
(18.3)
The line with coordinates [1, 0, 0] T is called the common tangent of F.
If F is a generalized f-fan for some o-polynomial f where f is not stated
explicitly, then we shall refer to F as a generalized fan.
In the case where the o-polynomial f is f(x) = x2 (so the corresponding
oval is a conic), a generalized f-fan is called merely a fan (as in Section 7.6).
Let α be a generator of the automorphism group of Fq. Define Kα to be
the set of points
α−1
= x0x2 }
= {(t α , t, 1, x3) : t, x3 ∈ Fq} ∪ {(1, 0, 0, x3) : x3 ∈ Fq}.
Kα = {(x0, x1, x2, x3) : x α 1
Then Kα is the cone studied in Chapter 17 with nuclear generator
〈(0, 0, 0, 1), (0, 1, 0, 0)〉.
18.2 Spreads of T2(Of)
Let f be an o-polynomial over Fq, and use the following notation for certain
planes in P G(3, q):
π∞ = [1, 0, 0, 0] T ;
π = [0, 1, 0, 0] T .
Of = {(0, 1, t, f(t)) : t ∈ Fq} ∪ {(0, 0, 0, 1)} is an oval in π∞ with nucleus
N = (0, 0, 1, 0). Note that the line ℓ = π∞ ∩ π = {(0, 0, z, w) ∈ P G(3, q) :
z, w ∈ Fq} is a tangent to Of at the point Q = (0, 0, 0, 1). Recall the
construction of the GQ T2(Of) in P G(3, q).
The points of T2(Of) are of two types:
(i) The points of Σ = P G(3, q) \ π∞;

18.2. SPREADS OF T2(OF ) 795
(ii) The planes of Σ containing the nucleus N = (0, 0, 1, 0).
The lines of T2(Of) are the lines of Σ meeting Of in a unique point.
Incidence in T2(Of) is just that inherited from Σ.
The following theorem is a main reason that we are so interested in generalized
f-fans.
Theorem 18.2.1. Let Γ be a spread of T2(Of) containing the line ℓ =
〈Q, N〉. For s ∈ Fq let Ks be the set of lines of Γ containing (0, 1, s, f(s)),
and let
Os = {π ∩ m : m ∈ Ks} ∪ {Q}.
Then {Os : s ∈ Fq} is a generalized f-fan F of ovals in π and having ℓ as
common tangent. Conversely, if {Os : s ∈ Fq} is a generalized f-fan F of
ovals of π having ℓ as common tangent, then
Γ(F) = {〈Q, N〉} ∪ {〈X, (0, 1, s, f(s))〉 : s ∈ Fq and X ∈ Os \ {Q}}
is a spread of T2(Of) containing the line ℓ.
Proof. Let Γ be a spread of T2(Of) containing the line ℓ = 〈Q, N〉. So
ℓ is the unique line of Γ incident with the “point” π∞. Consider a point
P ∈ Of \ {Q}. Each plane of P G(3, q) meeting π∞ in the line 〈P, Q〉, the
tangent to Of at P , is a point of T2(Of) and is incident with a unique line
of Γ. Since 〈P, N〉 ∈ Γ, the line of Γ incident with this plane must be a line
of P G(3, q) meeting π∞ in P . Hence each point of Of \ {Q} has q lines of Γ
through it.
Let P and R be distinct points of Of \ {Q} and π ′ a plane of P G(3, q)
meeting π∞ in the line 〈P, R〉. Each of the q − 2 points of Of \ {P, Q, R}
has q lines of Γ on it. These q 2 − 2q lines meet π ′ \ 〈P, R〉 in distinct points.
This leaves 2q points of π ′ \ 〈P, R〉, each of which must be incident with an
element of Γ that is either a line of P G(3, q) on P or a line of P G(3, q) on
R. Since these 2q points are in the same plane π ′ of P G(3, q) they must be
the points of two lines on P or two lines on R.
Now suppose that P ∈ Of \{Q} and that π ′ is a plane of P G(3, q) meeting
π∞ in the line 〈P, Q〉. The q 2 − q lines of Γ that are lines of P G(3, q) meeting
π∞ in a point of Of \ {P, Q} meet π ′ \ π∞ in distinct points, leaving q points
that must be the points of a line on P that is an element of Γ.
Having considered the planes of P G(3, q) on a point P of Of \ {Q} we
see that Ks, the set of lines of Γ on P and the line ℓ form an oval cone in

796 CHAPTER 18. GENERALIZED FANS AND SPREADS OF T2(O))
P G(3, q) with nuclear line 〈P, N〉. Thus Os = {π ∩ m : m ∈ Ks} is an oval
of π with nucleus N.
To prove compatibility of the ovals Os and Ot, s = t, at the point
(0, 0, 1, f(s)+f(t)
), consider a plane different form π∞ and containing the points
s+t
(0, 1, s, f(s)) and (0, 1, t, f(t)). By the above we know that such a plane
contains two lines of Γ, either both through (0, 1, s, f(s)) or both through
(0, 1, t, f(t)). Thus if we intersect this plane with π we obtain a line of π
on (0, 0, 1, f(s)+f(t)
) with is either secant to Os and external to Ot, or vice
s+t
versa. By considering all planes of P G(3, q) not π∞ and containing the line
〈(0, 1, s, f(s)), (0, 1, t, f(t))〉, we see that the ovals Os and Ot are comatible
at the point (0, 0, 1, f(s)+f(t)
). Hence {Os : s ∈ Fq} is a generalized f-fan of
s+t
ovals of P G(2, q).
Now suppose that F = {Os : s ∈ Fq} is a generalized f-fan of π. Let
Γ(F) be the set {ℓ}∪{〈X, (0, 1, s, f(s)〉 : s ∈ Fq and X ∈ Os \{Q} of lines of
T2(Of). Suppose, aiming for a contradiction, that two lines m and n of Γ(F)
are concurrent. Without loss of generality we may assume that neither m or
n is ℓ, and that as lines of P G(3, q), m and n are incident with the points
(0, 1, s, f(s)) and (0, 1, t, f(t)), respectively, with s = t. Since m and n are
concurrent, they span a plane 〈m, n〉 of P G(3, q). The intersection of 〈m, n〉
and π is a line containing (0, 0, 1, f(s)+f(t)
). this line also contains a point Pm
s+t
of m and a point Pn of n. Now Φm ∈ Os and Pn ∈ Ot, and since Os ∩ Ot =
{Q} it follows that Pm = Pn. However, (0, 0, 1, f(s)+f(t)
) ∈ 〈Pm, Pn〉, which
s+t
contradicts the compatibility of Os and Ot at (0, 0, 1, f(s)+f(t)
). Hence no two
s+t
liones in Γ(F) intersect, so Γ(F) is a spread of T2(Of).
The benefit of using the canonical forms in Theorem 18.2.1 is that a
generalized f-fan F corresponds to a unique spread of T2(Of) containing the
line ℓ, and conversely. From this point on Γ(F) will denote the spread of
T2(Of) constructed from a generalized f-fan as in Theorem 18.2.1.

Chapter 19
Appendix 1: Some Elementary
Number Theory
The most basic results of elementary number theory are frequently helpful in
finite geometry. In this Appendix we have given really more than is necessary
for most applications. Indeed, Sections 19.1 and 19.2 contain all that will
be needed most the time, except for the very important Law of Quadratic
Reciprocity which appears in Appendix 2.
19.1 Basic Facts about Congruences
If n is a positive integer and a, b ∈ Z are any integers, we write a ≡ b (mod n)
provided n|(a − b), i.e., provided n divides a − b. It is easy to show that “≡”
is an equivalence relation on the set Z of integers. We assume that the
reader is quite familiar with the most basic facts of arithmetic modulo n, so
we merely recall several of them without proof. However, proofs are given
for some of the results that might not have been proved in a first course in
abstract algebra.
Result 19.1.1. If a ≡ b (mod n) and c ≡ d (mod n), then
(i) a + c ≡ b + d (mod n), and
(ii) ac ≡ bd (mod n).
This makes it possible to define addition and multiplication modulo n and
to obtain a commutative ring with 1. This ring will sometimes be denoted
by Zn(+n, ·n).
797

798CHAPTER 19. APPENDIX 1: SOME ELEMENTARY NUMBER THEORY
Let a, m ∈ Z, m > 0, and put d = (a, m), the greatest common divisor
of a and m. Also let φ(m) be the number of integers i for which 1 ≤ i ≤ m
and (i, m) = 1.
Result 19.1.2. For b ∈ Z, ax ≡ b (mod m) has a solution if and only if d|b.
If d|b and x0 is one solution, there are d distinct solutions mod m given by
x0, x0 +
m , x0 + 2 d
m , . . . , x0 + (d − 1) d
m . d
Result 19.1.3. a φ(m) ≡ 1 (mod m). (Euler’s Theorem).
Result 19.1.4. If (m, n) = 1, then φ(m · n) = φ(m) · φ(n).
Result 19.1.5. If p is prime, φ(p a ) = p a − p a−1 .
The order of a modulo m is defined by:
|a|m = min t : t > 0 and a t ≡ 1 ( mod m) .
This is defined only when d = (a, m) = 1.
From here on suppose that (a, m) = 1, and put h = |a|m.
Result 19.1.6. a i ≡ a j (mod m) iff i ≡ j (mod h). Hence {a, a 2 , . . . , a h ≡
1} is a complete listing of all the distinct powers of a modulo m.
Result 19.1.7. |a j |m = h
(j,h) . Hence |aj |m = h iff (j, h) = 1, and
φ(h) = |{distinct a j mod m : |a j |m = h}|.
Result 19.1.8. |a|m divides φ(m).
Result 19.1.9. If |a|m = h, |b|m = k and (h, k) = 1, then |ab|m = hk.
Proof. Let r = |ab|m. Clearly r divides hk. But b rh ≡ (ab) rh ≡ 1 (mod m),
implying that k divides rh. Similarly, considering a rk modulo m leads to the
fact that h divides rh. Since h and k are relatively prime, kh divides r, so
r = kh.

19.2. A THEOREM OF LUCAS 799
Definition: If |a|m = φ(m), then a is said to be a primitive root modulo
m.
IMPORTANT: If |a|m = φ(m), then {a, a 2 , a 3 , . . . , a φ(m) } is a reduced
residue system (r.r.s) modulo m. Hence, if (b, m) = 1, then b ≡ a j (mod m)
for a unique j in the range 1 ≤ j ≤ φ(m).
Problem: For which positive integers m does m have a primitive root?
(We will solve this problem.)
Result 19.1.10. If |a|m = h and d|h, there are φ(d) elements b ∈ {a, a 2 , . . . , a h }
with |b|m = d.
This can be generalized to cyclic groups.
Result 19.1.11. Let G be a cyclic group (written multiplicatively) of order
m generated by the element g. Then for each d that divides m, G contains
exactly φ(d) elements of order d. This shows that
φ(d) = m.
d|m
19.2 A Theorem of Lucas
Result 19.2.1. (Theorem of Lucas) Let p be a prime, and let a = a0 + a1p +
a2p2 + · · · and b = b0 + b1p + b2p2 + · · · be the expansions of integers a and
b to the base p with b ≥ 0, a > 0. The binomial coefficient a
satisfies the
b
following:
a
≡
b
a0 a1 a2
b0
b1
b2
· · · ( mod p). (19.1)
Proof. The coefficient of xb in the expansion of (1 + x) a is a
, and modulo
b
p we have
(1 + x) a ≡ (1 + x) a0+a1p+a2p 2 +··· ≡ (1 + x) a0 (1 + x p ) a1 (1 + x p 2
) a2 · · · (mod p).
Now just compute the coefficient of x b on both sides of the above equation.

800CHAPTER 19. APPENDIX 1: SOME ELEMENTARY NUMBER THEORY
The theorem of Lucas has an interesting corollary that will be useful in
proving a theorem of Carlitz in the next chapter.
Corollary 19.2.2. Let q = p n where p is an odd prime and q = 2m + 1. Put
M = {a0 + a1p + · · · + an−1p n−1 ; 0 ≤ aj ≤ (p − 1)/2.
Then for any integer t, 0 ≤ t ≤ m, the binomial coefficient m
is prime to
t
p if and only if t ∈ M.
Proof. In the theorem of Lucas put a = p−1
2
= m. Then suppose t = n−1 i=0 bipi . So
p−1
2
Since
p n −1
p−1
p−1
2
bi
m
≡
t
p−1
2
b0
p−1
2
b1
p−1 p−1
+ p + 2 2 p2 + · · ·+ p−1
2 pn−1 =
p−1
· · · 2
bn−1
(mod p).
≡ 0 (mod p) if and only if bi > p−1
, the proof is complete.
2
Essentially the same proof gives a slight generalization that is also often
useful.
Corollary 19.2.3. Let λ = n−1
i=0 αip i , where 0 ≤ αi < p, and
n−1
Mλ = {r = βip i : 0 ≤ βi ≤ αi}. (19.2)
i=0
Then λ
is relatively prime to p exactly when r ∈ Mλ by the Theorem of
r
Lucas, i.e., λ
= 0 in GF (q) if and only if r ∈ Mλ.
r
19.3 The Chinese Remainder Theorem
We begin by stating the Chinese Remainder Theorem (CRT) in a little more
generality than is often given in beginning courses.
Theorem 19.3.1. The CRT for Z: Let a1, . . . , an be any integers, and
let m1, . . . , mn be positive integers, n ≥ 2. Then there is an integer x
for which x ≡ ai (mod mi) for all i, 1 ≤ i ≤ n, if and only if ai ≡
aj (mod gcd(mi, mj)) for all i, j with 1 ≤ i < j ≤ n. Moreover, x is unique
modulo lcm[m1, . . . , mn].

19.3. THE CHINESE REMAINDER THEOREM 801
In order to prove this theorem we first establish the following lemma
relating gcd’s and lcm’s.
Lemma 19.3.2. Let m1, . . . , mk+1 be positive integers, k + 1 ≥ 3. Then
gcd(lcm[m1, . . . , mk], mk+1) = lcm[gcd(m1, mk+1), . . . , gcd(mk, mk+1)].
Proof. We want to invoke a general principle. Suppose we have two integers
m and n and we want to show that they are equal. The idea is to show that
for any prime p, p r ||m iff p r ||n. Because unique factorization holds in the
ring Z, this shows that m = n. So let p be any prime. Note that if p r ||m
and p s ||n, then p min(r,s) ||gcd(m, n) and p max(r,s) ||lcm[m, n].
The proof will be by induction on k, so we start with the case k = 2.
Here we would like to show that
gcd(lcm[m1, m2], m3) = lcm[gcd(m1, m3), gcd(m2, m3)]. (19.3)
Note that m1 and m2 play symmetric roles. So if p a ||m1 and p b ||m2,
without loss of generality we may assume that a ≤ b. Also let p c ||m3. Then
p b ||lcm[m1, m2], and on the left hand side we have
p min(b,c) ||gcd(lcm[m1, m2], m3).
On the right hand side, since min(a, c) ≤ min(b, c), we have
p min(a,c) ||gcd(m1, m3) and p min(b,c) ||gcd(m2, m3) imply that
p min(b,c) ||lcm[gcd(m1, m3), gcd(m2, m3)].
This establishes Eq. 19.3. To handle the inductive step, it will be convenient
to use the standard notation: lcm(A, B) = [A, B], and gcd(A, B) =
(A, B). It is easy to see that
[[m1, . . . , mn], mn+1] = [m1, . . . , mn, mn+1]. (19.4)
Let k ≥ 3. The inductive hypothesis may be stated as:
([m1, . . . , mt], mt+1) = [(m1, mt+1), . . . , (mt, mt+1)] (19.5)
for all t with 2 ≤ t < k. Now we use the inductive hypothesis to show that
Eq. 19.5 holds with t = k. Start with

802CHAPTER 19. APPENDIX 1: SOME ELEMENTARY NUMBER THEORY
([m1, . . . , mk], mk+1) = ([[m1, . . . , mk−1], mk], mk+1) by Eq. 19.4
= [([m1, . . . , mk−1], mk+1), (mk, mk+1)] by Eq. 19.5 with t = 2
= [[(m1, mk+1), . . . , (mk−1, mk+1)], (mk, mk+1)]Eq. 19.5, t = k − 1
= [(m1, mk+1), . . . , (mk, mk+1)]. (19.6)
We are now ready for the proof of the CRT.
Proof. Let m1, m2 be positive integers and a1, a2 any integers. We show that
there is a solution x to x ≡ ai (mod mi) if and only if a1 ≡ a2 (mod gcd(m1, m2)).
First suppose that there is such a solution x. Let d = gcd(m1, m2). Then
x ≡ ai (mod mi) implies that x ≡ ai (mod d), and we can subtract one of
these congruences from the other to obtain: a1 ≡ a2 (mod d).
For the converse, suppose that x ≡ ai (mod d), where d = gcd(m1, m2).
This means that we can write a1 − a2 = td for some integer t, and d =
ym1 + zm2 for some integers y and z. So a1 − a2 = t(ym1 + zm2), implying
that we may put
x = a1 − tym1 = a2 + tzm2.
Clearly x ≡ ai (mod mi) for this choice of x, i = 1, 2.
The idea is to use induction on n where we have already done the case
for n = 2. Suppose that x has been found so that x ≡ ai (mod mi) for
all i, 1 ≤ i ≤ n, as in the statement of the CRT. Fix two indices i, j,
with 1 ≤ i < j ≤ n, and put dij = gcd(mi, mj). Then x ≡ ai (mod mi)
implies x ≡ ai (mod dij). Similarly, x ≡ aj (mod dij). Hence subtracting
one congruence from the other, we see that ai ≡ aj (mod dij). So the
condition of the CRT is satisfied. And of course, if x ≡ ai (mod mi) for
some i, then the same congruence holds modulo any divisor of mi. Hence
x ≡ ai (mod gcd(mi, mk+1)).
Conversely, suppose that n ≥ 2 and that ai ≡ aj (mod gcd(mi, mj)) for
all i, j with 1 ≤ i < j ≤ n. Suppose the CRT has been established for
2 ≤ k ≤ n − 1. We want to establish it for k + 1 = n. To get started,
put M = [m1, . . . , mk]. By the induction hypothesis, we may suppose that
x = x0 has been found such that x0 ≡ ai (mod mi) for all i, 1 ≤ i ≤ k.
Hence we also know that x0 ≡ ai (mod gcd(mi, mk+1)), for 1 ≤ i ≤ k.

19.4. SOLVING CONGRUENCES: THE CASE OF A PRIME MODULUS803
We also know that the congruence x = x0 +tM ≡ ak+1 (mod mk+1) has a
solution t if and only if x0 ≡ ak+1 (mod gcd(M, mk+1)). But we have proved
above that
(M, mk+1) = ([m1, . . . , mk], mk+1) = [(m1, mk+1), . . . , (mk, mk+1)].
Since x0 ≡ ai (mod gcd(mi, mk+1)), for 1 ≤ i ≤ k, clearly
x0 ≡ ai (mod [(m1, mk+1), . . . , (mk, mk+1)]). This shows that there is a solution
t, and we now put x = x0 +tM. Hence x ≡ ak+1 (mod mk+1). And since
M ≡ 0 (mod mi), for 1 ≤ i ≤ k, we have x ≡ x0 + tM ≡ x0 ≡ ai (mod mi)
for all i, 1 ≤ i ≤ k. So x ≡ x0 + tM solves the system of congruences for all
i, 1 ≤ i ≤ k + 1 = n.
If x0 and x1 are both solutions to the system of congruences, then for each
i with 1 ≤ i ≤ n, x0 ≡ x1 (mod mi). Hence x0 ≡ x1 (mod lcm[m1, . . . , mn]),
which is what was meant by the claim that the solution x is unique modulo
lcm[m1, . . . , mn]. This completes the proof of the theorem.
19.4 Solving Congruences: The Case of a Prime
Modulus
Note: From now on until further notice m = p is a prime.
Result 19.4.1. If f(x) ∈ Z[x] has degree n when its coefficients are reduced
modulo p, then f(x) ≡ 0 (mod p) has at most n solutions.
Proof. This is easily proved by considering the division algorithm to obtain
f(x) = (x − a)q(x) + f(a). It follows that x − a divides f(x) modulo p if
and only if f(a) ≡ 0 (mod p). So a1, . . . , ar are roots of f(x) ≡ 0 (mod p)
if and only if (x − a1) · · · (x − ar) divides f(x) modulo p. A degree argument
shows that r ≤ n.
This allows us to prove the next result.
Result 19.4.2. Let d|(p − 1). Then x d − 1 ≡ 0 (mod p) has d solutions mod
p. If |b|p = d, then {b, b 2 , . . . , b d } = {a mod p : a d ≡ 1 (mod p)}.
Proof. Since d|(p − 1), write p − 1 = de. Then x d − 1 divides x p−1 − 1,
so we may write x p − x = x(x p−1 − 1) = x(x d − 1)(x d(e−1) + · · · + x d + 1).

804CHAPTER 19. APPENDIX 1: SOME ELEMENTARY NUMBER THEORY
By Fermat’s little theorem, x p − x ≡ 0 (mod p) has p solutions modulo p.
Hence the right side also has p solutions mod p. It is easy to see that each
of the three factors must have a number of zeros equal to its degree, so that
x d − 1 ≡ 0 (mod p) must have d solutions.
Result 19.4.3. Let q be a prime with q α ||(p − 1), α ≥ 1. Then there are
φ(q α ) = q α − q α−1 residues a mod p with |a|p = q α .
Proof. Put S = {a : aqα ≡ 1 (mod p)} = {a : |a|p divides qα }. Clearly
d|qα and d < qα if and only if d = qβ with 0 ≤ β ≤ α − 1. Put S ′ =
{a mod p : aqα−1 ≡ 1 (mod p)} ⊆ S. Then |S| = qα and |S ′ | = qα−1 imply
that |S \ S ′ | = qα − qα−1 . Clearly |a|p = qα if and only if a ∈ S \ S ′ , so there
are qα − qα−1 elements with order qα mod p.
Result 19.4.4. If p − 1 = q α1
1 · · · q αr
r , qi prime, αi ≥ 1, let ai satisfy
|ai|p = q αi
i , 1 ≤ i ≤ r. (By Result 19.4.3 there are such ai.) Then by a
straightforward generalization of Result 19.1.9, |a1 · a2 · a3 · · · ar|p = p − 1.
Hence there is a primitive root modulo p, which implies there are φ(p − 1)
primitive roots modulo p. And for each d such that d|(p − 1), there are φ(d)
elements b mod p with |b|p = d.
Result 19.4.5. If (a, p) = 1 and d = (n, p − 1), then xn ≡ a (mod p) has d
solutioins or no solution according as a p−1
d ≡ 1 (mod p) or not.
Proof. Let g be a primitive root mod p, so there is an i with g i ≡ a (mod p).
Then x ≡ g j (mod p) satisfies x n ≡ a (mod p) iff g jn ≡ g i (mod p) iff
jn ≡ i (mod p − 1). This has d solutions if d|i and no solution if d does not
divide i. But a p−1
d ≡ g
x n ≡ a (mod p) has d solutions.
i( p−1
d ) ≡ 1 (mod p) iff i(p−1)
d
≡ 0 (mod p − 1) iff d|i iff
As an interesting special case we have the following: x 2 ≡ a (mod p) has
2 or 0 solutions according as a p−1
2 ≡ 1 (mod p) or a p−1
2 ≡ −1 (mod p).
Example: How many solutions does x5 ≡ 6 (mod 101) have?
First note that p = 101 is prime, and d = (5, 100) = 5, so p−1
= 20. So
d
x5 ≡ 6 (mod 101) has 5 solutions if 620 ≡ 1 (mod 101) and no solution if

19.5. THE CASE OF A PRIME POWER MODULUS, ETC. 805
6 20 ≡ 1 (mod 101). Routine computations show that 6 20 ≡ 1 (mod 101), so
that x 5 ≡ 6 (mod 101) has 5 solutions.
Exercise: How many solutions does x 45 ≡ 2 (mod 101) have?
Result 19.4.6. Euler’s Criterion: If p is an odd prime and (a, p) = 1, then
x 2 ≡ a (mod p) has 2 solutions or no solution according as a p−1
2 ≡ 1 or
≡ −1 (mod p).
Proof. d = (n, p − 1) = (2, p − 1) = 2. Consider b = a p−1
d = a p−1
2 . Since
b 2 ≡ a p−1 ≡ 1 (mod p), 0 ≡ b 2 −1 ≡ (b+1)(b−1), clearly b ≡ ±1 (mod p). By
Result ?? there are 2 solutions if b ≡ 1 and no solution if b ≡ −1 (mod p).
Result 19.4.7. x 2 ≡ −1 (mod p) has 1 solution if p = 2, two solutions if
p ≡ 1 (mod 4), and no solution if p ≡ 3 (mod 4).
Proof. If p = 2, x ≡ 1 is the unique solution. If p is odd, use Result 19.4.4:
x 2 ≡ −1 (mod p) has 2 solutions if and only if (−1) p−1
2 ≡ 1 (mod p) iff p−1
2
is even iff 4|(p − 1) iff p = 1 + 4k.
19.5 The Case of a Prime Power Modulus,
etc.
Result 19.5.1. If p is any prime, there exist φ(φ(p 2 )) = φ(p 2 −p) = φ(p(p−
1)) = (p − 1) · φ(p − 1) primitive roots modulo p 2 .
Proof. The first goal is to find one primitive root modulo p 2 . Choose a
primitive root g modulo p, i.e., |g|p = p − 1. We ask: When is g + tp
(0 ≤ t ≤ p − 1) a primitive root mod p 2 ? Put h = |g + tp| p 2. Then
g + tp) h ≡ 1 (mod p 2 ), so that 1 ≡ (g + tp) h ≡ g h (mod p), implying that
(p − 1)|h. Also, h|φ(p 2 ), hence h|(p 2 − p). So we have (p − 1)|h|p(p − 1),
and p is prime, so h = p − 1 or h = p(p − 1). If h = p − 1, then g is
not a primitive root modulo p 2 . If h = p(p − 1), then g is a primitive
root modulo p 2 . We ask: For how many t is |g + tp| p 2 = h = p − 1?
This is the same as: For how many t is (g + tp) p−1 ≡ 1(mod p 2 )? Put
f(x) = x p−1 − 1. So f(g) = g p−1 − 1 ≡ 0 (mod p) and f ′ (x) = (p − 1)x p−2 ,
and f ′ (g) = (p − 1)g p−2 ≡ −g p−2 ≡ 0 (mod p). So the question becomes: For
which t is

806CHAPTER 19. APPENDIX 1: SOME ELEMENTARY NUMBER THEORY
which is the same as
0 ≡ f(g + tp) ≡ f(g) + f ′ (g)tp (mod p 2 ),
0 ≡ f(g)
p + f ′ (g)t (mod p).
Hence there is a unique t mod p for which f(g + tp) ≡ 0 (mod p 2 ). This
implies that there are p − 1 values of t modulo p for which (g + tp) p−1 ≡
1 (mod p 2 ), i.e., |g + tp| p 2 = p(p − 1) = φ(p 2 ), i.e., g + tp is a primitive root
mod p 2 .
Now let b be one of the primitive roots mod p 2 . Then {b, b 2 , . . . , b φ(p2 ) } is
a r. r. s. modulo p 2 . And |b j | p 2 = φ(p 2 ) iff 1 = (j, φ(p 2 )). Hence the number
of primitive roots modulo p 2 is φ(φ(p 2 )) = φ(p(p − 1)) = (p − 1)φ(p − 1).
Result 19.5.2. If p is an odd prime and g is a primitive root mod p 2 , then
g is a primitive root mod p α for all α ≥ 2.
Proof. Let g be a primitive root mod p 2 , and suppose that we have shown
that g is also a primitive root modulo p 3 , p 4 , · · · , p α for some α ≥ 2. We want
to show that g is a primitive root modulo p α+1 . So we know the following:
Continue this down to:
g p−1 ≡ 1 (mod p); (by Fermat)
g p(p−1) ≡ 1 (mod p 2 ), g p−1 ≡ 1(mod p 2 ),
g p2 (p−1) ≡ 1 (mod p 3 ), g p(p−1) ≡ 1 (mod p 3 );
g pα−2 (p−1) ≡ 1 (mod p α−1 ), g p α−3 (p−1) ≡ 1 (mod p α−1 );
g pα−1 (p−1) ≡ 1 (mod p α ), g p α−2 (p−1) ≡ 1 (mod p α ).
Hence gpα−2 (p−1) α−1 p = 1 + tp , where p does not divide t. So g α−1 (p−1) =
(1 + tpα−1 ) p = 1 + p · tpα−1 + p(p−1)
(tp 2
α−1 ) 2 + · · · ≡ 1 + tpα (mod pα+1 ),
since 1 + 2(α − 1) = α + 1 + α − 2 ≥ α + 1. Since p does not divide t,
gpα−1 (p−1) α+1 ≡ 1 (mod p ).
Suppose |g| pα+1 = h. We know h|φ(pα+1 ) = pα (p − 1). And gh ≡
1 (mod pα+1 ), implying gh ≡ 1 (mod pα ), forcing pα−1 (p − 1)|h, since g

19.6. POWER RESIDUES 807
is a primitive root mod p α . Hence |g| p α+1 = p α−1 (p − 1) or p α (p − 1). But
we just showed that g pα−1 (p−1) ≡ 1 (mod p α+1 ), hence |g|p α+1 = p α (p − 1), as
desired.
Result 19.5.3. If p is an odd prime and g is a primitive root modulo p α with
g odd (one of g, g + p α must be odd), then g is a primitive root modulo 2p α .
Result 19.5.4. For α ≥ 3, 2α |(a2α−2 − 1), but φ(2α ) = 2α−1 . So there is no
primitive root modulo 2α .
Proof. We have noticed earlier that 8|(a2−1) for all odd a. Then 2α |(a2α−2−1) and 2|a2a−2 + 1 imply that 2α+1 |(a2α−1 − 1). But φ(2α+1 ) = 2α > 2α−1 .
Result 19.5.5. Suppose m is not a prime power or twice a prime power.
Then there is no primitive root modulo m.
Proof. Under the given hypothesis, m = m1 · m2, with m1 > 2, m2 > 2,
(m1, m2) = 1. So φ(m) = φ(m1)φ(m2), and φ(m1) ≡ φ(m2) ≡ 0 (mod 2),
implying that a φ(m)
2 ≡ 1 (mod m) for all a with (a, m) = 1.
Result 19.5.6. Recapitulation: There is a primitive root modulo m iff m =
1, 2, 4, p α , or 2p α , p an odd prime.
19.6 Power Residues
If (a, m) = 1, we ask how many solutions does x n ≡ a (mod m) have?
Result 19.4.5 gave an answer when m is prime. What about general m?
Result 19.6.1. Suppose m = 1, 2, 4, p α or 2p α , p an odd prime. If (a, m) = 1,
then x n ≡ a (mod m) has
d = (n, φ(m)) solutions if a φ(m)
d ≡ 1 (mod m),
NO solution if a φ(m)
d ≡ 1(mod m).

808CHAPTER 19. APPENDIX 1: SOME ELEMENTARY NUMBER THEORY
Proof. The hypothesis implies that there is a primitive root g modulo m.
So if (a, m) = 1, we have a ≡ g i (mod m) for some i. Then there is an
x ≡ g j (mod m) satisfying x n ≡ a (mod m) if and only if g jn ≡ g i (mod m)
iff jn ≡ i (mod φ(m)) iff d = (n, φ(m)) divides i. If d divides i, there are d
solutions.
Also, a φ(m)
d ≡ g iφ(m)
d ≡ 1 (mod m) iff φ(m)| iφ(m)
d
a (mod m) has d solutions.
iff d|i, which is iff x n ≡
So when m = 2 α , α ≥ 3, the multiplicative group of integers prime to m
is not cyclic. However, it is as close to being cyclic as it could be without
actually being cyclic, as is shown by the next result.
Result 19.6.2. Suppose α ≥ 3. Then |5|2 α = 2α−2 and
{±5, ±5 2 , . . . , ±5 2α−2
} is a r.r.s (mod 2 α ).
Proof. First note that 2α ||(52α−2 − 1) for α = 2, 3. Also, if a ≡ 1 (mod 4),
then 2||(a + 1), and we claim that 2||(5j + 1) for all j ≥ 1. To see this, use
induction, starting with 2α ||(52α−2 − 1) for α ≥ 2, and 2||(52α−2 + 1),
Then 2α+1 ||(52α−2 − 1)(52α−2 + 1) = (52α−1 + 1). Now suppose α ≥ 3.
Let h = |5|2α. φ(2α ) = 2α−1 . And 52α−2 ≡ 1 (mod 2α ) implies that
h|2α−2 . If h < 2α−2 , then 52α−3 ≡ 1 (mod 2α ). But 2α−1 ||(52α−3 − 1), so
52α−3 ≡ 1 (mod 2α implies h = 2α−2 = |5|2α. At least this says that the elements
of {5, 52 , 53 , . . . , 52α−2} are pairwise not congruent modulo 2α . Then
clearly {−5, −52 , . . . , −52α−2} are pairwise not congruent modulo 2α . If 5i ≡
−5j (mod 2α ) with α ≥ 3, then 5i ≡ −5j (mod 4), so 1 ≡ −1 (mod 4), an
impossibility. So {5, −5, 52 , −52 , . . . , 52α−2, −52α−2} is a set of φ(2α ) = 2α−1 pairwise incongruent elements modulo 2 α , each of which is relatively prime
to 2 α .
Result 19.6.3. Suppose α ≥ 3 and a is odd.
(i) If n is odd, then x n ≡ a (mod 2 α ) has a unique solution.
(ii) If n is even with (n, 2 α−2 ) = 2 β , then x n ≡ a (mod 2 α ) has
2 β+1 solutions if a ≡ 1 (mod 2 β+2 , and
NO solution if a ≡ 1 (mod 2 β+2 ).
Example: x 2 ≡ a (mod 2 α ), α ≥ 3, has four solutions iff a ≡ 1 (mod 8).

19.7. THE 4-SQUARE THEOREM OF LAGRANGE 809
Proof. Since a is odd, a ≡ (−1) i 5 j (mod 2 α ) for some i, j. Then x n ≡ a
implies x is odd, so x ≡ (−1) u 5 v (mod 2 α ). So x n ≡ a becomes (−1) nu 5 nv ≡
(−1) i 5 j (mod 2 α ), which is equivalent to nu ≡ i (mod 2) and nv ≡ j (mod 2 α−2 ).
Case 1. n is odd. So u ≡ i (mod 1) and there is a unique v modulot 2 α−2
with nv ≡ j (mod 2 α−2 ). Hence x n ≡ a (mod 2 α ) has a unique solution.
Case 2. n is even. Here nu ≡ i (mod 2) has 2 solutions if i is even,
none if i is odd. Since 2 β = (n, 2 α−2 ), nv ≡ j (mod 2 α−2 ) has 2 β solutions
if j ≡ 0 (mod 2 β ) and NO solution otherwise. So x n ≡ a (mod 2 α ) has
a solution iff i ≡ 0 (mod 2) and j ≡ 0 (mod 2 β ), in which case it has
2 · 2 β = 2 β+1 solutions. This says a ≡ 5 j (mod 2 α ) with j ≡ 0 (mod 2 β ).
But |5| 2 β+2 = 2 β . So 5 j ≡ 1 (mod 2 β+2 ) if and only if 2 β |j which is iff
a ≡ 1 (mod 2 β+2 ).
19.7 The 4-Square Theorem of Lagrange
The famous theorem of Lagrange referred to here says that each positive
integer is the sum of four squares.
Theorem 19.7.1. (Lagrange) Let m be a positive integer. Then there are
integers a1, a2, a3, a4 such that m = 4
1 a2 i .
For example, 2 = 1 2 + 1 2 + 0 2 + 0 2 . Basically the proof preceeds by
establishing that each odd prime is the sum of four squares, and the product
of any two sums of four squares is again a sum of four squares.
Lemma 19.7.2. Let p be an odd prime. then there is an m ∈ Z for which
1 ≤ m < p and mp = x 2 1 + x 2 2 + x 2 3 + x 2 4 for some integers x1, x2, x3, x4 ∈ Z.
Proof. Put S1 = {i2 : 0 ≤ i ≤ p−1
2 }. Put S2 = {−i2 − 1 : 0 ≤ i ≤ p−1
}. Note:
2
|S1| = |S2| = p+1
2 . Since |S1|+|S2| > p, there must be i, j with 0 ≤ i, j ≤ p−1
2
and i2 = −j2 − 1 (mod p), i.e., i2 + j2 + 1 = mp for some m ∈ Z, where
1 ≤ m = 1
p (i2 + j2 + 1) ≤ 1
p−1 2
p−1 2
+ + 1 = p 2
2
1
p2−2p+1 + 1 <
p 2
p2 + 1 < p.
1
p
2
Lemma 19.7.3. The smallest integer m satisfying Lemma 19.7.2 is m = 1.

810CHAPTER 19. APPENDIX 1: SOME ELEMENTARY NUMBER THEORY
Proof. Let m be the least integer satisfying the conditions of Lemma 1. First
suppose m is ev en. The number of x ′ is which are even is 0, 2 or 4, since
mp is even. Hence we may order the xi so that x1 ≡ x2 (mod 2) and x3 ≡
x4 (mod 2). Then
x2 + x2
2
2
x1 − x2
+
2
2
x3 + x4
+
2
= x2 1 + 2x1x2 + x 2 2 + x2 1 − 2x1x2 + x 2 2
4
2
x3 − x4
+
2
+ 2x2 3 + 2x2 4
4
= mp
2 .
So m ′ = m/2 satisfies Lemma 19.7.2, contradicting the minimality of m.
Hence we know that m must be odd, and we assume that 1 < m < p.
Then {i : −
m−1
m−1
≤ i ≤ } is a complete residue system modulo
2
2
m. Define yi by: yi ≡ xi (mod m), −
m−1 ≤ yi ≤ 2
m−1
2 . Then 4 1 y2 i ≡
4 1 x2i = mp ≡ 0 (mod m). Hence mn = 4 1 y2 i for some n ∈ Z. Here
0 ≤ n = 1
4 m 1 y2
4 m−1 2 m
i ≤ = m 2
2−2m+1 < m, i.e, 0 ≤ n < m. If
m
n = 0, then yi = 0, implying xi ≡ 0 (mod m); so 4 1 x2i ≡ 0 (mod m2 ),
and mp ≡ 0 (mod m2 ), forcing p ≡ 0 (mod m). This is impossible since p is
prime and 1 < m < p. So in fact 0 < n < m. Now
m 2
4
2 2
np = mp · mn = xi yi =
x1
+
y1
x2
y2
+
x1
+
y1
x4
y4
x3 x4
y3 y4
+
2
x2 x3
y2 y3
x1
+
y1
x3
y3
+
2
1
xiyi
2
x4 x2
y4 y2
= A 2 1 + A2 2 + A2 3 + A2 4 .
Since yi ≡ xi (mod m), each of A2, A3, A4 is clearly congruent to 0 modulo
m, and A1 ≡ 4 1 x2
2
i = (mp) ≡ 0 (mod m). Dividing by m we have
np = 4
Ai 2,
1 with 1 ≤ n < m. So the proof is easily by the method of
m
“infinite descent.”
We are now in a good position to complete the proof of the Four-Square
Theorem. Lemma 19.7.3 shows that any odd prime is the sum of four squares;
2
2

19.8. CONGRUENCE OF SYMMETRIC MATRICES 811
clearly 2 is the sum of four squares; and the proof of Lemma 19.7.3 includes
a proof that the product of two sums of four squares is again a sum of four
squares. This completes the proof.
19.8 Congruence of Symmetric Matrices
The only known theorem restricting the orders of finite projective planes
without any additional hypotheses is the famous Bruck-Ryser Theorem. The
proof of this theorem uses the 4-square theorem of Lagrange and the theory
of congruent symmetric matrices.
Def. Let S, S ′ be two symmetric matrices over a field F . We say S is
congruent to S ′ (over F ) provided there is an invertible matrix P over F
with S ′ = P T SP . (Note: P T SP is symmetric iff S is symmetric.) It is
rather trivial to show that Congruence over F (denoted c =) is an equivalence
relation.
If x1, . . . , xn are variables over F , the quadratic form over F associated
with a symmetric matrix S is:
⎛ ⎞
f(x1, . . . , xn) = (x1, . . . , xn)S
⎜
⎝
x1
.
xn
⎟
⎠ =
n
i,j=1
xisijxj.
If S ′ c = S, say S ′ = P T SP , P invertible, define new variables y1, . . . , yn
by ¯y T = P −1 ¯x T , so ¯x T = P ¯y T . Then f(x1, . . . , xn) = ¯xS¯x T = ¯yP T ·S ·P ¯y T =
¯yS ′ ¯y T = g(y1, . . . , yn) is the quadratic form associated with S ′ . The point
here is that if S c = S ′ , then {¯xS¯x T : ¯x ∈ F n } = {¯yS ′ ¯y T : ¯y ∈ F }. (This
seems more obvious when written as ¯xS¯x T = ¯yS ′ ¯y T when S ′ = P T SP .)
Lemma 19.8.1. Let S1
S1
·
+ S2
c
= S ′ 1
·
+ S ′ 2.
·
+ S2 =
Proof. Easy exercise for the reader.
S1 0
0 S2
, S1
c
= S ′ 1 , and S2
Let m be any integer. Write m = a2 1 + a2 2 + a2 3 + a2 4, ai ⎛
⎞
∈ Q.
Put H =
⎜
⎝
a1 a2 a3 a4
a2 −a1 a4 −a3
a3 −a4 −a1 a2
a4 a3 −a2 −a1
⎟
⎠ , so that
c
= S ′ 2 . Then

812CHAPTER 19. APPENDIX 1: SOME ELEMENTARY NUMBER THEORY
c
HH T = a 2 i I4 = mI4,
i.e. I4 = mI4. So it is clear (by Lemma 19.8.1 that if n ≡ 0 ( mod4) and
m = 4 1 a2i , then
This proves:
In = I4
·
+ + · · · + ·
I4
c
= mI4
·
+ · · · ·
+ mI4 = mIn.
Lemma 19.8.2. For all ingeters m > 0, and for all n = 4w, 0 < w ∈ Z, we
have
c
mIn = In over Q.
19.9 The Bruck-Ryser Theorem
The only known theorem restricting the orders of finite projective planes
without any additional hypotheses is the famous Bruck-Ryser Theorem. The
proof of this theorem uses the 4-square theorem of Lagrange and the theory
of congruent symmetric matrices.
Let π be a projective plane of order n with pointset P = {x1, . . . , xv} and
lineset B = {l1, . . . , lv}, v = 1+n+n 2 . The incidence matrix of π (associated
with this ordering of points and lines) is A = (ai,j) 1≤i,j≤v , where
ai,j =
0, xj ∈ li; (columns labeled by points; )
1, xj ∈ li, (rows labeled by lines.)
Lemma 19.9.1. Since π has n + 1 points on each line and each two lines
meet in a unique point (resp., n + 1 lines on each point and two points are
on a unique line),
This says that Iv
B = AA T = nI + J = A T A, where J is the all 1’s matrix.
c
= B = nI + J over Q.
The next lemma is a simple exercise.
Lemma 19.9.2. Let n ≡ 1, 2 (mod 4). Then v = 1 + n + n 2 ≡ 3 (mod 4).

19.9. THE BRUCK-RYSER THEOREM 813
We have B ·
+ I1
c
= Iv
·
+ I1 = Iv+1
c
= nIv+1. Hence there is a matrix P
invertible over Q for which nIv+1 = P T (B ·
+ I1)P . So with x and y related
by x T = Py T , we have
x(B ·
+ I1)x T = x(nI +J ·
+ I1)x T = n(x2 1 +· · · x2 v)+(x1 +· · ·+xv) 2 +x2 v+1
and y · nIv+1y T = n(y2 1 + · · · + y2 v+1 ). Hence for any x and y related by
x T = Py T , we have
n(x 2 1 + · · · + x2v ) + (x1 + · · · + xv) 2 + x 2 v+1 = n(y2 1 + · · · + y2 v+1 ). (19.7)
The main technical details remaining are designed to produce specific
rational vectors x and y with x T = Py T and satisfying x2 i = y2 i for 1 ≤ i ≤ v.
Suppose first that this has been accomplished. Then (x1 +· · ·+xv) 2 +x2 v+1 =
ny2 v+1 . From this it follows that n is the sum of two rational squares, say
n =
a 2
c 2,
+ where a, b, c, d are integers, nonnegative with bd = 0, and
b d
we may assume that 1 = (a, b) = (c, d).
If a = 0 or c = 0, then n is a square. Suppose ac = 0, say a, b, c, d are all
positive integers. Then nb2d2 = a2d2 + b2c2 . Considering this modulo b2 , we
have that b divides d. Considering this modulo d2 , we have that d divides b.
So b = d, i.e., nb2 = a2 + c2 .
Let p be an odd prime dividing n but for which p2 does not divide n. If
p divides c, then nb2 = a2 + c2 considered modulo p and p2 implies that p2 divides nb2 , so p divides a and p divides b, an impossibility. Hence p does not
divide c. So a2 + c2 = nb2 ≡ 0 (mod p) pimplies (ac−1 ) 2 ≡ −1 (mod p). But
it is well known that −1 is a square mod p if and only if p ≡ 1 (mod 4). (See
Section ??.) So this shows that each prime dividing the square-free part of
n (when n ≡ 1, 2 (mod 4)) must be congruent to 1 modulo 4. (This assumes
we can find the x and y.)
The theorem we will end up with is the following:
Theorem 19.9.3. If there is a projective plane π of order n with n ≡ 1 or
2 modulo 4, then no prime dividing the square-free part of n is congruent to
3 modulo 4.
Proof. Recall that the matrix P = (pij) is given and we seek x and y with
x T = Py T . If p11 = 1, in x1 = v+1 j=1 p1jyj, set x1 = y1, and if p11 = 1, then
set x1 = −y1.
If p11 = 1, we have x1 = y1 = p11y1 + p12y2 + · · · p1,v+1yv+1 and y1 =
e2y2 + · · · ev+1yv+1, ei = p1i
1−p11 ∈ Q, i = 2, 3, . . . , v + 1. If p11 = 1, then
x1 = −y1 = y1 + p12y2 + · · · + p1vyv + p1,v+1yv+1, which implies that

814CHAPTER 19. APPENDIX 1: SOME ELEMENTARY NUMBER THEORY
y1 = e2y2 + e3y3 + · · · + ev+1yv+1, where ei = P1i ∈ Q, i = 2, 3, . . . , v + 1.
−2
Now x2 = p21y1 + · · · + p2,v+1yv+1. Substituting for y1 (as given above),
x2 = p2y2 + · · · + pv+yv+1, where pi is rational for i = 2, . . . , v + 1.
Continue this process, putting xi = ±yi and obtaining y1, . . . , yi as rational
linear combinations of yi+1, . . . , yv+1. Eventually, x1 = ±y1, . . . , xv−1 =
±yv−1, yv−1 = gvyv + gv+1yv+1, gv, gv+1 ∈ Q; and Exv = qvyv + qv+1yv+1,
qv, qv+1 ∈ Q. Set xv = yv or −yv to obtain yv = hv+1yv+1 and xv+1 =
sv+1yv+1, where hv+1, sv+1 ∈ Q. Now put yv+1 equal to any nonzero rational
number. Then y1, . . . , yv and x1, . . . , xv+1 are uniquely determined so that
x2 i = y2 i , i = 1, 2, . . . , v and x T = Py T . This is just what we needed to
complete the proof.
RECAP: We redo the above proof in matrix notation. First write the
matrix P in block form:
,
or
so x T = Py T has the form
Put xi = ±yi, so
⎛⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎝⎝
⎛
⎜
⎝
⎛
⎜
⎝
x1
.
xv
xv+1
±y1
.
±yv
⎞
P =
⎟
⎠ =
⎞
P1
P1
⎟ ⎜
⎠ = P1 ⎝
±1 0 0
0 ±1 0
·
·
0 ±1
¯R T
¯S t
¯R T
¯S t
⎛
⎞
⎟
⎠
y1
.
yv
− P1
⎛
⎜
⎝
⎞
y1
.
yv
vv+1
⎞
⎟
⎠ .
⎟
⎠ + ¯ R T yv+1,
⎞
⎛
⎟ ⎜
⎟ ⎝
⎠
y1
.
yv
⎞
⎟
⎠ = ¯ R T yv+1.

19.10. NUMBER THEORETIC FUNCTIONS 815
Now pick ±1’s so that
⎛
⎜
⎝
±1
. ..
±1
solve for y1, . . . , yv, and x1, . . . , xv, then xv+1 = ¯ S
⎞
⎟
⎠−P1 is invertible. Put yv+1 = 1,
19.10 Number Theoretic Functions
⎛
⎜
⎝
y1
.
yv
⎞
⎟
⎠ + t · yv+1.
An arithmetic function (sometimes called a number theoretic function) is a
function whose domain is the set P of positive integers and whose range is a
subset of the complex numbers C. Hence C P is just the set of all arithmetic
functions. If f is an arithmetic function not the zero function, f is said to
be multiplicative provided f(mn) = f(m)f(n) whenever (m, n) = 1, and to
be totally multiplicative provided f(mn) = f(m)f(n) for all m, n ∈ P. The
following examples will be of special interest to us here.
Example 19.10.0.1. I(1) = 1 and I(n) = 0 if n > 1.
Example 19.10.0.2. U(n) = 1 for all n ∈ P.
Example 19.10.0.3. E(n) = n for all n ∈ P.
Example 19.10.0.4. The omega function: ω(n) is the number of distinct
primes dividing n.
Example 19.10.0.5. The mu function: µ(n) = (−1) ω(n) , if n is square-free,
and µ(n) = 0 otherwise.
Example 19.10.0.6. Euler’s phi-function: φ(n) is the number of integers
k, 1 ≤ k ≤ n, with (k, n) = 1.
The following additional examples often arise in practice.
Example 19.10.0.7. The Omega function: Ω(n) is the number of primes
dividing n counting multiplicity. So ω(n) = Ω(n) iff n is square-free.
Example 19.10.0.8. The tau function: τ(n) is the number of positive divisors
of n.

816CHAPTER 19. APPENDIX 1: SOME ELEMENTARY NUMBER THEORY
Example 19.10.0.9. The sigma function: σ(n) is the sum of the positive
divisors of n.
Example 19.10.0.10. A generalization of the sigma function: σk(n) is the
sum of the kth powers of the positive divisors of n.
Dirichlet (convolution) Product of Arithmetic Functions.
Def. If f and g are arithmetic functions, define the Dirichlet product
f ∗ g by:
(f ∗ g)(n) =
f(d)g(n/d) =
d|n
Obs. 19.10.1. f ∗ g = g ∗ f.
d1d2=n
f(d1)g(d2).
Obs. 19.10.2. If f, g, h are arithmetic functions, (f ∗ g) ∗ h = f ∗ (g ∗ h),
and [(f ∗ g) ∗ h)](n) =
d1d2d3=n f(d1)g(d2)h(d3).
Obs. 19.10.3. I ∗f = f ∗I = f for all f. And I is the unique multiplicative
identity.
Obs. 19.10.4. An arithmetic function f has a (necessarily unique) multiplicative
inverse f −1 iff f(1) = 0.
Proof. If f ∗f −1 = I, then f(1)f −1 (1) = (f ∗f −1 )(1) = I(1) = 1, so f(1) = 0.
Conversely, if f(1) = 0, then f −1 (1) = (f(1)) −1 . Use induction on n. For
n > 1, if f −1 (1), f −1 (2), . . . , f −1 (n − 1) are known, f −1 (n) may be obtained
from 0 = I(n) = (f ∗ f −1 )(n) =
d|n f(d)f −1 (n/d) for n > 1.
The following theorem has essentially been proved.
Theorem 19.10.5. The set of all arithmetic functions f with f(1) = 0 forms
a group under Dirichlet multiplication.
Theorem 19.10.6. The set of all multiplicative functions is a subgroup.
Proof. f(1) = 0 = g(1) implies (f ∗ g)(1) = 0. Associativity holds by
Obs. 19.10.2. The identity I is clearly multiplicative. So suppose f, g are
multiplicative. Let (m, n) = 1. Then

19.10. NUMBER THEORETIC FUNCTIONS 817
(f ∗ g)(mn) =
d|mn f(d)g
mn
d
=
d1|m d2|n f(d1d2)g
mn
d1d2
=
d1|m d2|n f(d1)f(d2)g(m/d1)g(n/d2)
=
d1|m f(d1)g(m/d1) ·
d2|n f(d2)g(n/d2)
= (f ∗ g)(m)(f ∗ g)(n).
Finally, we need to show that if f is multiplicative, in which case f −1
exists, then also f −1 is multiplicative. Define g as follows. Put g(1) = 1,
and for every prime p and every j > 0 put g(pj ) = f −1 (pj ). Then extend
g multiplicatively for all n ∈ P. Since f and g are both multiplicative, so
is f ∗ g. Then for any prime power p k , (f ∗ g)(p k ) =
d1d2=p k f(d1)g(d2) =
d1d2=p k f(d1)f −1 (d2) = (f ∗ f −1 )(p k ) = I(p k ). So f ∗ g and I coincide on
prime powers and are multiplicative. Hence f ∗ g = I, implying g = f −1 , i.e.,
f −1 is multiplicative.
Clearly µ is multiplicative, and
d|n µ(d) = 1 if n = 1. For n = pe ,
d|n µ(d) = e
j=0 µ(pj ) = 1 + (−1) + 0 · · · + 0 = 0. Hence µ ∗ U = I and we
have proved the following:
Obs. 19.10.7. µ −1 = U; U −1 = µ.
Theorem 19.10.8. Möbius Inversion: F = U ∗ f iff f = µ ∗ F .
This follows from µ −1 = U and associativity. In its more usual form it
appears as:
F (n) =
f(d) ∀n ∈ P iff f(n) =
µ(d)F (n/d) ∀n ∈ P.
d|n
NOTE: Here we sometimes say F is the sum function of f. When F
and f are related this way it is interesting to note that F is multiplicative if
and only if f is multiplicative. For if f is multiplicative, then F = U ∗ f is
multiplicative. Conversely, if F = U ∗ f is multiplicative, then µ ∗ F = f is
also multiplicative.
Exercise 19.10.8.1. 1. τ = U∗U is multiplicative, and τ(n) =
pα ||n (α+
1).
2. φ = µ ∗ E is multiplicative. (First show E = φ ∗ U.)
d|n

818CHAPTER 19. APPENDIX 1: SOME ELEMENTARY NUMBER THEORY
3. σ = U ∗ E is multiplicative and σ(n) =
p α ||n
4. φ ∗ τ = σ.
5. σ ∗ φ = E ∗ E.
6. E −1 (n) = nµ(n).
pα+1−1 . p−1
Sometimes it is useful to have even more structure on C P . For f, g ∈ C P ,
define the sum of f and g as follows:
(f + g)(n) = f(n) + g(n)
Then a large part of the following theorem has already been proved and
the remainder is left as an exercise.
Theorem 19.10.9. With the above definitions of addition and convolution
product, (C P , +, ∗) is a commutative ring with unity I, and f ∈ C P is a unit
iff f(1) = 0.
Exercise 19.10.9.1. For g ∈ CP , define ˆg ∈ CP by ˆg(n) = ng(n). Show that
g ↦→ ˆg is a ring automorphism. In particular, ˆ g−1 = (ˆg) −1 .
At this point we have completed our basic review of elementary number
theory except for the Law of Quadratic Reciprocity. This famous result will
be dealt with in the next chapter.

Chapter 20
Appendix 2: Algebra in Fq[x]
The projective planes that play a major role in this work are the finite desarguesian
planes isomorphic to P G(2, q) coordinatized by a finite field, i.e.,
a Galois field GF (q) = Fq. We assume that the reader is somewhat familiar
with finite fields, so we review only the most basic concepts and notations.
20.1 The Galois Field GF (q) = Fq
If q = p e , e ≥ 1, with p a prime, there is a Galois field Fq = GF (q) with q
elements. Moreover, each finite field is isomorphic to some Galois field. We
know that Fq contains Fp = GF (p) ∼ = Z/pZ as its prime subfield. If f(x) is
an irreducible polynomial of degree e over Fp, then
Fq ∼ = Fp[x]/(f(x)) = {a0+a1t+· · ·+ae−1t e−1 : ai ∈ Fp and f(t) = 0}. (20.1)
If Fr is a Galois field for some prime power r, then Fr contains an isomorphic
copy of Fq as a subfield if and only if r = p h where e divides h. If
Fq also denotes this subfield, then
Fq = {a ∈ Fr : a q = a}. (20.2)
We know that F ∗ q = Fq \ {0} is a cyclic group whose binary operation
is just multiplication of Fq restricted to F ∗ q . A generator of F ∗ q is called a
primitive element or primitive root for Fq.
The group Aut(Fq) of automorphisms of Fq is cyclic of order e and generated
by the automorphism
819

820 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
It follows that φi : x ↦→ xpi. p : Fq → Fq : x ↦→ x p . (20.3)
If p is a prime and n is a positive integer, then considering f(x) = xpn −x
as a polynomial in Fp[x], the splitting field of f(x) is Fpn. Moreover Fpn consists of precisely the roots of xpn = 0. More generally, if q = pn and m is
a positive integer, then
has splitting field Fqm.
g(x) = x qm
− x = x pmn
− x ∈ Fq[x]
Let N (m, q) be the set of monic irreducible polynomials of degree m over
Fq, q = pn , and put N(m, q) = |N (m, q)|.
Suppose that f(x) ∈ N (m, q). the roots of f(x) = 0 each belong to Fqm. If f(α) = 0, then Fq(α) = Fqm. In particular, αqm −α = 0, so that αqm−1 = 1.
Then f(x)|(xqm −1 q − 1), since f(x) and x m−1 − 1 each have no repeated roots
and each root of f(x) = 0 is a root of xqm−1 − 1 = 0. The exponent e of f(x)
is the smallest positive integer k for which f(x)|(xk − 1). So each α with
f(α) = 0 satisfies αe = 1. Since the roots α1, . . . , αm of f(x) = 0 all have
f(x) as their minimum polynomial over Fq, αe i = αe j = 1 iff f(x)|(xe − 1),
and we see that if e is the minimal positive integer for which f(x)|(xe − 1),
then each root α of f(x) = 0 has multiplicative order e. So e|(qm − 1). If
e = qm −1, then f(x) is a primitive polynomial and each root α is a primitive
element of Fqm. Put
θ(n) = (q n+1 − 1)/(q − 1) = q n + q n−1 + · · · + q + 1.
Let α be a primitive element of Fq m, so α has multiplicative order qm − 1.
Then β = α θ(m−1) has multiplicative order q − 1, so is a primitive element of
Fq.
Suppose that r|m, from which we have (q r −1)|(q m −1), which implies that
θ(r − 1)|θ(m − 1). Thus δ = α θ(m−1)
θ(r−1) has multiplicative order qm −1)·θ(r−1)
θ(m−1)
=
(q − 1)θ(r − 1) = qr − 1, so is a primitive element of Fqr, a subfield of Fqm. For each e dividing qm − 1 put
N e (m, q) = {f(x) ∈ N (m, q) : f(x) has exponent e}.

20.1. THE GALOIS FIELD GF (Q) = FQ 821
Then
N (m, q) = ∪ {N e (m, q) : e|(q m − 1) but e | (q r − 1) for 0 < r < m} .
For an e as in the set described just above, if f(x) ∈ N e (m, q), then f(x) = 0
has m roots, each of whose elements has multiplicative order e and generates
Fq m. In general, if d|(qm − 1), the number of elements of order d is φ(d), and
then |N e (m, q)| = φ(e)/m.
The map t ↦→ t θ(m−1) is a multiplicative homomorphism from F ∗ q m into F ∗ q
with kernel K = {t ∈ F ∗ q m : tθ(m−1) = 1} of size θ(m − 1) and image of size
(q m − 1)/(θ(m − 1) = q − 1, i.e., it is onto F ∗ q . Hence each element c ∈ F ∗ q
is the image of θ(m − 1) elements t ∈ F ∗ q m, and no element of F ∗ q m \ F ∗ q is
such an image. But for some t ∈ F ∗ qm it may be that there is some smaller
exponent e < θ(m − 1) for which te = c ∈ Fq. Note that for f(x) ∈ N (m, q),
if f(t) = 0, then te = c ∈ Fq if and only if f(x)|(xe − c).
Let f(x) ∈ N (m, q). We say that f(x) has subexponent e provided e is the
smallest positive integer k for which f(x) divides xk − c for some c ∈ Fq. Let
α be a primitive element for Fqm and suppose t = αr is a root of f(x) = 0.
Put d = gcd(θ(m − 1), r). Then we have
Lemma 20.1.1. e = θ(m−1)
d
is the subexponent of f(x).
Proof. Clearly t e = α re ∈ Fq if and only if (α re ) q−1 = 1 if and only if
(q m −1)|(re(q−1)) if and only if θ(m−1)|re. Write r = du and θ(m−1) = dv,
where gcd(u, v) = 1. Then dv|due if and only if v|ue if and only if v|e. Hence
the smallest positive integer e for which t e ∈ Fq is e = v = θ(m − 1)/d.
If f(x) ∈ N (m, q) has subexponent e = θ(m−1), we say f(x) is subprimitive,
and its roots are subprimitive for Fqm (or m-subprimitive).
If α is a root of a subprimitive polynomial f(x), then θ(m−1) is the smallest
positive integer e such that αe ∈ Fq. Hence if a subprimitive polynomial
has a primitive root, it must be a primitive polynomial for Fqm. For e = qm − 1,
P (m, q) = |N e (m, q)| = φ(e)
= number of primitive polynomials for Fqm; m
R(m, q) = the number of subprimitive polynomials in N (m, q);
N(m, q) = |N (m, q)|.

822 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
Lemma 20.1.2. The following equalities hold:
(i) P (m, q) = φ(qm−1) m .
(ii) N(m, q) = 1
m
m d|m µ(d)q d .
(iii) R(m, q) = (q−1)φ(θ(m−1))
. m
Proof. Part (i) should be clear. For part (ii) we have that the product of
all monic irreducible polynomials of degree dividing m is xqm − x. Hence,
computing the degree of this product we find
Let m = p k1
1
inversion gives
· · · pks
s
q m =
dN(d, q).
d|m
be the prime decomposition of m. Then Möbius
m · N(m, q) =
d|m
µ(d)q m
d ,
which proves part (ii)
Since µ(d) = 0 if the square of some prime divides ”d, we have
m · N(m, q) = q m −
s
i=1
q m
p i +
1≤i

20.2. RELATIVE TRACE AND NORM 823
TF/K(a) = a + a q + a q2
+ · · · + a qn−1
.
Clearly TF/K(a) ∈ K; TF/K(a + b) = TF/K(a) + TF/K(b) for all a, b ∈ F ;
and for c ∈ K, a ∈ F , TF/K(ca) = cTF/K(a). If e = 1, so q is prime, then
TF/K is the absolute trace function. If a ∈ K, then TF/K(a) = na = 0 if
n ≡ 0 (mod p). Note that TF/K is a K-linear transformation of the field F
thought of as an n-dimensional vector space over K.
For the rest of this section let us write tr(a) = TF/K(a) for a ∈ F . And
for σ ∈ Gal(F/K), define Tσ ∈ HomK(F, F ) by Tσ(a) = a σ − a for all a ∈ F .
It is easy to check that tr(Tσ(a)) = 0 for all a ∈ F . So Im(Tσ) ⊆ ker(tr).
First suppose that σ generates Gal(F/K), so that K is the subfield fixed
by σ. In this case K is exactly the kernel of Tσ, so that the image of Tσ
has q n /q = q n−1 elements. This is a vector subspace of F of codimension 1
over K. Since tr is not the zero map (a polynomial of degree q n−1 cannot
have more than q n−1 roots in F ), it must be that tr(b) = 0 if and only if
b = Tσ(a) for some a ∈ F . This holds for each σ that generates the Galois
group Gal(F/K). Now suppose that σ ∈ Gal(F/K) has fixed field larger
than K, so that the kernel of Tσ is larger than K, say it is GF (q j ) with
1 < j ≤ n. Then the image of Tσ has only q n−j elements in it, so that it is
a proper subset of the kernel of tr = TF/K. We have proved the following
result.
Theorem 20.2.1. Let σ be a generator of the Galois group Gal(F/K). Then
tr(b) = TF/K(b) = 0 if and only if b = Tσ(a) = a σ − a.
The relative norm NF/K of an element a ∈ F is defined by
NF/K(a) = a 1+q+q2 +···+q n−1
= a qn −1
q−1 .
Clearly NF/K(a) ∈ K, and if a = b q−1 then NF/K(a) = 1. The kernel of
x ↦→ x q−1 in F ∗ is K ∗ , so the size of the image I of the map x ↦→ x q−1 is
qn−1 q−1 , and I is a subset of the kernel of the map x ↦→ x qn−1 qn−1 . Hence we have proved
q−1
q−1 which has size
Theorem 20.2.2. NF/K(a) = 1 if and only if a = b q−1 for some b ∈ F .

824 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
20.3 Polynomials over Fq
If q = 2, then x q + x = x(x + 1) = x 2 + (0 + 1)x, and
a = 1. For the
a∈F2
remainder of this section let q > 2, so that xq , xq−1 , x are distinct monomials.
from which it follows that
x q−1 − 1 =
(x − a) = x q−1 − (
a∈F ∗ q
x q − x =
(x − a), (20.4)
a∈F ∗ q
a∈Fq
a)x (q−2)
q−1
+ · · · + (−1)
a∈F ∗ q
a. (20.5)
From Eq 20.5 we see immediately that
a =
a = 0;
a = (−1) q = −1. (20.6)
a∈F ∗ q
a∈Fq
a∈F ∗ q
Note: For fixed a ∈ Fq,
(a − b) =
b = (−1) q = −1. (20.7)
b∈Fq
b=a
b∈F ∗ q
Put Γ[x] = Fq[x]/(x q −x) and G(x) = {f(x) ∈ Fq[x] : deg(f(x)) ≤ q −1}.
Then for each g(x) ∈ Fq[x] there is a unique g(x) ∈ G(x) for which f(x) ≡
g(x) (mod x q − x). Moreover, if g(x) ∈ Fq[x] and f(x) ∈ G(x), then
Put
f(a) = g(a) for all a ∈ Fq iff f(x) ≡ g(x) (mod x q − x). (20.8)
δa,b =
1, if a = b;
0, if a = b.
Recall Lagrange interpolation. For a ∈ Fq, put
x
fa(x) =
q
b∈Fq(x
− b)
− x
b=a
=
. (20.9)
(x − a)(−1) q (a − b)
b∈Fq
b=a
Then fa(b) = δa,b for all a, b ∈ Fq. Let h : Fq → Fq be any function. Put

20.3. POLYNOMIALS OVER FQ 825
Then
f(x) =
h(a)fa(x). (20.10)
a∈Fq
f(b) =
h(a)fa(b) = h(b), for all b ∈ Fq. (20.11)
a∈Fq
This shows that each function h : Fq → Fq is a polynomial function
f. But we want to write f(x) = q−1 r=0 arxr where ar is given in terms of
functional values.
Theorem 20.3.1. Let h : Fq → Fq be any function. Then there is a polynomial
f(x) = q−1
r=0 arx r for which f(b) = h(b) for all b ∈ Fq. Moreover,
(i) a0 = f(0) = h(0);
(ii) aq−1 = −
a∈Fq h(a);
(iii) For 0 < r < q − 1, ar = −
a∈F ∗ q h(a)a−r .
Proof. Put x = 0 in Eq. 20.10 to see that h(0) = f(0) = a0. Next, the
coefficient on xq−1 in fa(x) is (−1) q = −1 (for q odd or even). Hence from
Eq. 20.10, aq−1 = −
h(a). Now suppose 0 < r < q − 1. If a = 0, the
a∈Fq
coefficient on xr in xq−x x−a = xq−1 − 1 is zero. If a = 0, put b(x) = xq−x x−a =
b0 + b1x + · · · + bq−1x q−1 = 0 + b1x + · · · + bq−2x q−2 + x q−1 . So
x q − x = (x − a)(b1x + · · · bq−2x q−2 + x q−1 )
= (−ab1)x + (b1 − ab2)x 2 + (b2 − ab3)x 3 + · · · + (bq−2 − abq−1)x q−1 + x q .
This implies ab1 = 1, or b1 = a −1 . And 1 < r < q−1 forces br−1 −abr = 0,
or br = a −r . So b1 = a −1 ; b2 = b1a −1 = a −2 ; b3 = b2a −1 = a −3 , . . . , and
br = a −r . So the coefficient of x r in fa(x) is 0 if a = 0 and −a −r if a = 0. By
Eq. 20.10, the coefficient on x r is ar = −
a∈F ∗ q h(a)a−r .
Lemma 20.3.2. Assume q > 2 and suppose f(x) ∈ G(x). If a ↦→ f(a) is a
bijection, then deg(f)x) ≤ q − 2.
Proof. By assumption,
f(a) = a = 0, since q > 2. So the
a∈Fq a∈Fq
coefficient on xq−1 is −
f(a), which is 0.
a∈Fq

826 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
Lemma 20.3.3. Let q > 2 and let f(x) permute the nonzero elements of Fq.
If deg(f(x)) ≤ q − 2, it must be that f(0) = 0 and a ↦→ f(a) is a bijection on
Fq.
Proof. From Eq. 20.10 we have f(x) = −
a∈Fa f(a) xq
−x
q−1 = −f(0)(x −
x−a
1) −
a∈F ∗ q f(a) xq
−x
q−1 . The coefficient of x , which must be 0, is −f(0) −
x−a
a∈F ∗
f(a) = −f(0) −
q a∈F ∗ a = −f(0). Hence f(0) = 0 and a ↦→ f(a) is
q
a bijection.
20.4 Symmetric Polynomials
Let R be a commutative ring, and let R[x1, . . . , xn] be the ring of polynomials
over R in n indeterminates. If π : i ↦→ i ′ is a permutation of
[n] = {1, . . . , n}, then π determines an automorphism ζ(π) of R[x1, . . . , xn]
such that ζ(π) : a ↦→ a, a ∈ R, xi ↦→ xi ′, 1 ≤ i ≤ n. A polynomial
f(x1, . . . , xn) ∈ R[x1, . . . , xn] is said to be symmetric in the x’s if
f(x1, . . . , xn) is fixed under ζ(π) for every π in the symmetric group Sn.
The set Σ of symmetric polynomials is a subring of R[x1, . . . , xn] containing
R. The coefficients of z in the polynomial
are symmetric. That is,
g(z) = (z − x1)(z − x2) · · · (z − xn) (20.12)
g(z) = z n − σ1z n−1 + σ2z n−2 + · · · + (−1) n σn, (20.13)
where σi = x1 ′ · · · xi ′ (i.e., σi is the sum over all products of i distinct
xi’s) is homogeneous of degree i and is called the i th elementary symmetric
polynomial in x1, . . . , xn. Also put σ0 = 1.
When the general polynomial
P (z) = anz n + an−1z n−1 + · · · + a0
is written in factored form, on comparing the two expressions we find that
σk = (−1) k an−k/an. (20.14)
Theorem 20.4.1. (The Fundamental Theorem of Symmetric Polynomials)
Σ = R[σ1, . . . , σn]. We make this more specific as follows. Let f(x1, . . . , xn)

20.4. SYMMETRIC POLYNOMIALS 827
be a symmetric polynomial in the indeterminates x1, . . . , xn. Then there is a
polynomial P (x1, . . . , xn) such that
f(x1, . . . , xn) = P (σ1, . . . , σn).
The coefficients of P can be expressed as linear combinations, with integral
coefficients, of the coefficients of f. The degree of P is equal to the
highest power of x1 occurring in f. The elementary symmetric polynomials
σ1, . . . , σn in the indeterminates x1, . . . , xn are algebraically independent over
R. Finally, each xi is algebraic over R[σ1, . . . , σn].
Proof. For the equality Σ = R[σ1, . . . , σn] we need to show that any symmetric
polynomial in x1, . . . , xn can be expressed as a polynomial in the
elementary symmetric polynomials. It suffices to show this for homogeneous
symmetric polynomials, i.e., those in which all of the terms ax k1
1 · · · x kn
n which
occur have the same (total) degree k1 + k2 + · · · + kn. Any polynomial can
be written in precisely one way as a sum of homogeneous polynomials of different
degrees. Moreover, as permutations preserve degrees, if f(x1, . . . , xn)
is symmetric, so are its homogeneous components.
Suppose that f(x1, . . . , xn) is homogeneous, symmetric, and of degree m.
Introduce the lexicographic ordering in the set of monomials of degree m:
x k1
1 · · · xkn n > xl1 1 · · · xln n provided that if k1 = l1, k2 = l2, . . . , ks = ls, but
ks+1 > ls+1, s ≥ 0. For example, x2 1x2x3 > x1x3 2 > x1x2 2x3. Let x k1
1 · · · xkn n
be the highest monomial occurring in f (with nonzero coefficient). Since
f is symmetric, it contains all the monomials obtained from x k1
1
· · · xkn
n by
permuting the xi’s. Hence k1 ≥ k2 ≥ · · · ≥ kn.
Note that if M1 and M2 are monomials of degree r, and N is a monomial
of degree k, then M1 > M2 implies NM1 > NM2. And if also N1 > N2, then
M1N1 > M2N2. It is also clear that x1 · · · xi is the highest monomial in σi.
It follows that the highest monomial in σ d1
1
· · · σdn
n is
x d1
1 (x1x2) d2 (x1x2x3) d3 · · · (x1 · · · xn) dn
= x d1+···dn
1 x d2+···dn
2 · · · x dn
n .
Hence the highest monomial in σ k1−k2
1
σ k2−k3
2
· · · σ kn−1−kn
n−1
σ kn
n
is xk1
1 x k2
2 · · · x kn
n ,
the same as the highest monomial in f. If this monomial has coefficient a
in f, then the highest monomial in f1 = f − aσ k1−k2
1
σ k2−k3
2
· · · σ kn−1−kn
n−1
is less than that of f. Repeat the process with f1, etc. A finite number of
applications yields a representation of f as a polynomial in σ1, . . . , σn with
σ kn
n

828 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
each monomial having degree m in the x’s and weight (k1 − k2) + 2(k2 − k3) +
· · · + (n − 1)(kn−1 − kn) + nkn = k1 + k2 + · · · + kn = m in σ1, . . . , σn. As f
is being decreased by subtracting polynomials in the elementary symmetric
functions, the remaining monomials are strictly smaller in the lexicographic
ordering, but always of the same degree m. So clearly the process terminates
in a finite number of steps.
Note that if ax k1
1 · · · x kn
n is the highest monomial in f, then f1 = f −
aσ k1−k2
1 σ k2−k3
2 · · · σ kn−1−kn
n−1 σkn n where aσk1−k2 1 σ k2−k3
2 · · · σ kn−1−kn
n−1 σkn n has degree
k1 in the elementary symmetric functions.
We next show that the σi
are algebraically independent. Suppose
d1
(d)
ad1···dnσ1 · · · σdn n = 0 where this is summed over a finite set of distinct
(d) = (d1, . . . , dn), di ∈ Z + . If the relation is nontrivial, we have ad1···dn = 0
for some (d). For any (d) define (k) = (k1, . . . , kn) by ki = di +di+1 +· · ·+dn.
then the degree of σ d1
1 · · · σdn n in the x’s is m = n 1 ki = n 1 idi, and the
highest monomial of this degree occurring in σ d1
1 · · · σdn n is x k1
1 x k2
2 · · · xkn n . If
(d ′ ) = d ′ 1 , . . . , d′ n ) and k′ i = d′ i + · · · d′ n for 1 ≤ i ≤ n, then d′ i = di, 1 ≤ i ≤ n.
thus distinct monomials σ d1
1 · · · σdn n in the σ’s have distinct highest monomials
in the x’s occurring in them. Now among those (d) with ad1···dn = 0, choose
the one for which m = idi is maximal and the highest monomial x k1
1 · · · xkn n
is maximal. Then expressing our relation in the σ’s in terms of the x’s, we
get the term x k1
1 · · · x kn
n only once and with nonzero coefficient ad1···dn. This
contradicts the algebraic independence of the xi’s.
x n i
For the last statement of the theorem, 0 = g(xi) = (xi−x1) · · · (xi−xn) =
n−1
− σ1xi + · · · + (−1)nσn. 20.5 Newton’s Formulas
Let sk = n j=1 xkj . So s1 = σ1, s2 = x2 1 + · · · + x2n , etc. Note that we do not
define s0.
Start with the equation
g(z) = z n − σ1z n−1 + σ2z n−2 + · · · + (−1) n σn. (20.15)
Put z = xi and sum over i = 1, . . . , n to obtain
0 = sn − σ1sn−1 + σ2sn−2 + · · · + (−1) n−1 σn−1s1 + (−1) n nσn.

20.5. NEWTON’S FORMULAS 829
Note that this is Eq 20.20 below with m = n. Differentiate to obtain
Here we have
g(z)
(z − xi)
Hence
+ (x m i
g ′ (z) = g(z) g(z)
g(z)
+ + · · · + . (20.16)
(z − x1) (z − x2) (z − xn)
= z n−1 + (xi − σ1)z n−2 + (x 2 i − σ1xi + σ2)z n−3 + · · ·
− σ1x m−1
i
+σ2x m−2
i
+ · · · + (−1) m σm)z n−m−1 + · · · . (20.17)
g ′ (z) = nz n−1 + (s1 − nσ1)z n−2 + (s2 − σ1s1 + nσ2)z n−3 + · · ·
+(sm −σ1sm−1 +σ2sm−2 + · · · + (−1) m nσm)z n−m−1 + · · · (20.18)
Also,
g ′ (z) = nz n−1 + · · · + (−1) m (n − m)σmz n−m−1 + · · · . (20.19)
Equating coefficients on like powers of z, we have
s1 − σ1 = 0
s2 − σ1s1 + 2σ2 = 0
s3 − σ1s2 + σ2s1 − 3σ3 = 0
.
sm − σ1sm−1 + σ2sm−2 + · · · +(−1) m−1 σm−1s1 + (−1) m mσm = 0.
Hence for 1 ≤ m ≤ n, and writing s0 = m, σ0 = 1, we have
m
i=0
This determines s1, . . . , sn.
s1 = σ1
(−1) i σism−i = 0. (20.20)
s2 = σ1s1 − 2σ2 = σ2 1 − 2σ2
s3 = σ1s2 − σ2s1 + 3σ3 = σ 3 1 − 3σ1σ2 + 3σ3
.
sn = σ1sn−1 − σ2sn−2+ · · · (−1) n+1 nσn
(20.21)

830 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
If m > n, multiply g(z) by z m−n .
z m−n g(z) = z m − σ1z m−1 + · · · + (−1) n σnz m−n = 0.
Replace z by xi, 1 ≤ i ≤ n, and add the resulting equations to obtain
sm − σ1sm−1 + σ2sm−2 − · · · + (−1) n σnsm−n = 0,
i.e., n
i=0 (−1)i σism−i = 0.
So in general, Newton’s Formulas may be written as:
min(m,n)
(−1) i σism−i = 0, where s0 = m, σ0 = 1. (20.22)
i=0
Alternatively, we have
sm =
min(m,n)
i=1
(−1) i+1 σism−i, where s0 = m, σ0 = 1, m ≥ 1. (20.23)
Exercise: s1 = σ1; s2 = σ 2 1 − 2σ2; s3 = σ 3 1 − 3σ1σ2 + 3σ3.
20.6 The Resultant of Two Polynomials
Let F be an arbitrary field, and let f, g be two nonzero polynomials ∈ F [x].
Suppose K contains a splitting field for fg over F , so that in K[x],
and
f(x) = a(x − α1)(x − α2) · · · (x − αn),
g(x) = b(x − β1)(x − β2) · · · (x − βm)
for some α1, . . . , αn, β1, . . . , βm ∈ K and some nonzero a, b ∈ F . We define
the resultant R(f, g) of f and g by
R(f, g) = a m b n
n
i=1 j=1
m
(αi − βj) (n = deg f, m = deg g).
We start with some basic facts concerning resultants.

20.6. THE RESULTANT OF TWO POLYNOMIALS 831
Lemma 20.6.1. R(g, f) = (−1) mn R(f, g).
Proof. Obtaining R(g, f) from R(f, g) involves making mn changes of sign:
(root of g - root of f) = -(root of f - root of g).
Lemma 20.6.2. R(f, g) = 0 iff f and g have any common factors (over F ).
Proof. R(f, g) = 0 iff in K f and g have a common root iff over F they have
a common factor.
Lemma 20.6.3. R(f, g) = adeg g n i=1 g(αi), where a is the leading coefficient
of f and αi are the various roots of f. Similarly, R(f, g) = (−1) mnbdeg f m j=1 f(βj),
where b is the leading coefficient of g and the βj are the various roots of g.
Proof. Since g(x) = b m j=1 (x−βj), g(αi) = b m j=1 (αi−βj) and n i=1 g(αi) =
bn n m i=1 j=1 (αi − βj). Multiplying both sides by am gives the result.
Lemma 20.6.4. If g = fq + r, then R(f, g) = a (deg g−deg r) R(f, r), where a
is the leading coefficient of f.
Proof. By Lemma 20.6.3, R(f, g) = a deg g n
i=1 g(αi) = a deg g n
i=1 [f(αi)q(αi)+
r(αi)]. Since the αi are the roots of f, we get R(f, g) = a deg g n
i=1 r(αi). On
the other hand, R(f, r) = a deg r n
i=1 r(αi), also by Lemma 20.6.3, completing
the proof.
Lemma 20.6.5. R(f, b) = b deg f if b is a scalar.
Proof. This follows directly from Lemma 20.6.3 with g(x) = b.
Lemma 20.6.6. If f = f1f2, then R(f, g) = R(f1, g) · R(f2, g). Similarly, if
g = g1g2, then R(f, g) = R(f, g1) · R(f, g2).
Proof. Immediate from the definition.
Lemma 20.6.7. For 0 = c, d, R(cf, dg) = c deg g d deg f R(f, g).
Note: We interpret an empty product to be equal to 1, so that when g(x)
is the nonzero constant b, i.e., g has no roots, the general resultant R(f, g)
is consistently defined. If a and b are both nonzero, clearly R(a, b) = 1. We
interpret R(a, b) to be 1 if only one of a, b is zero, and to be 0 if they both
are zero.

832 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
Lemmas 20.6.1, 20.6.4, and 20.6.5 permit the computation of the resultant
of any two polynomials by using the Euclidean algorithm. It is then easy
to show that the resultant R(f, g) is an element of the field F , even though it
is defined in terms of elements of the bigger field K, since the resultant of f
and g can be expressed in terms of products, sums, differences and quotients
of their coefficients. In particular, if f(x) and g(x) are polynomials over a
field F , then R(f, g) ∈ F , even if the roots of f(x) and g(x) are not in F .
Exercise 20.6.7.1. Prove that R(f, g) is in F .
20.7 The Discriminant of a Polynomial
Let f ∈ F [x] have degree n and leading coefficient a, and let f ′ denote the
formal derivative of f. Then the discriminant of f is defined by aD(f) =
(−1) n(n−1)/2 R(f, f ′ ). By Lemma 20.6.3 this is equivalent to
Lemma 20.7.1. If f(x) = a n i=1 (x − αi), then:
D(f) = a 2n−2
(αi − αj) 2 .
1≤i

20.7. THE DISCRIMINANT OF A POLYNOMIAL 833
= a 2n−1
1≤i

834 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
There is an immediate corollary.
Corollary 20.7.5. Let f and g be monic polynomials in F [x]. Then D(fg) =
D(f)D(g)(R(f, g)) 2 , where R(f, g) ∈ F . An easy induction argument extends
this to the following. Let f1, . . . , fr be monic polynomials in F [x]. Then
D(f1f2 . . . fr) = D(f1)D(f2) · · · D(fr) · R 2 , for some R ∈ F .
Note that the discriminant D(f) is a symmetric polynomial in the roots
α1, . . . , αn. Hence by the Fundamental Theorem of Symmetric Functions,
there is a polynomial P (z1, . . . , zn) such that D(f) = P (σ1, . . . , σn), where
σ1, . . . , σn are the elementary symmetric functions in the αi. It follows that
when f is monic, so the coefficents of f are the elementary symmetric functions,
the discriminant D(f) can be expressed as a linear combination, with
ordinary integral coefficients, of the coefficients of f. Also the coefficients of
P can be expressed as linear combinations, with integral coefficients, of the
coefficients of D(f). The degree of P is equal to the highest power of α1
occurring in D(f).
20.8 Reverse Polynomials
Recall that if f(x) = anxn + an−1xn−1 + · · · a1x + a0 is a polynomial over the
field F , its reverse polynomial is given by ˆ f(x) = xnf( 1
x ) = a0xn + a1xn−1 +
· · · an−1x + an. If an = 0 = a0, then f(x) has zeros α1, . . . , αn which are all
nonzero, and ˆ f(x) has zeros α −1
1 , . . . , α −1
n .
Let f(x) = a(x − α1)(x − α2) · · · (x − αn) in some splitting field of f(x)
over F . Then
ˆf(x) = x n a( 1 1
− α1)(
x x − α2) · · · ( 1
− αn)
x
= a(1 − α1x) · · · (1 − αnx)
= a(−1) n α1α2 · · · αn(x − 1
) · · · (x − 1
).
This implies that the discriminant
D( ˆ f(x)) = [a(−1) n α1α2 · · · αn] 2n−2
α1
1≤i

20.9. ROOTS OF IRREDUCIBLE POLYNOMIALS IN ZP [X] 835
= a 2n−2 (α1α2 · · · αn) 2n−2
= a 2n−2
1≤i

836 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
20.10 Stickelberger’s theorem
We compute D(f) by using the Euclidean algorithm on f and f ′ . Here are
some examples.
Example 1. Let f(x) = x−a. Then f ′ (x) = 1, so D(f) = (−1) (1·0)/2 R(f, 1) =
R(f, 1) = 1 deg f = 1.
Example 2. Let f(x) = x 2 + ax + b. Then D(f) = −R(f, f ′ ). Now
f ′ (x) = 2x + a, and
x 2
x
+ ax + b = (2x + a)
2
Put r = b − a2 . Then
4
D(f) = (−1) 2(1)/2R(f, f ′ ) = (−1)R(f, f ′ )
= −R(f ′ , f)
= 2deg f−deg r (−1)R(f ′ , r)
= −2 (2−0) R(f ′ , b − (a2 /4))
= a2 − 4b.
a
+ + b −
4
a2
.
4
Note: If the characteristic of F is different from 2, the roots of f are
− a
2
1√
± a2 − 4b.
2
So f has two irreducible factors iff a 2 − 4b = D(f) is a square in F .
The main goal of this section is to prove a generalization of this result
that is due to Stickelberger.
Theorem 20.10.1. Stickelberger’s Theorem Let p be an odd prime, f(x)
a monic polynomial of degree m with coefficients in Zp[x], without repeated
roots in any splitting field, i.e., without multiple factors (so that its discriminant
D(f) = 0). Let r be the number of irreducible factors of f(x) in Zp[x].
Then r ≡ m (mod 2) iff the discriminant D(f) is a square in Zp.
The proof of this theorem will come later. For now we want to try it out
on a few cubic polynomials.
Example 3. Let f(x) = x 3 + qx + r. Then f ′ (x) = 3x 2 + q and, doing
Euclid’s algorithm we find:

20.10. STICKELBERGER’S THEOREM 837
Then
x 3 + qx + r = (3x 2 + q)( x
2
) + qx + r ,
3 3
3x 2
2qx 9x 27r
+ q = + r −
3 2q 4q2
+ q + 27r2
4q2
.
D(f) = (−1) 3·2
2 R(f, f ′ ) = −R(x3 + qx + r, 3x2 + q)
= (−1)(−1) 3·2R(3x2 + q, x3 + qx + r)
= (−1)33−1R(3x2 + q, 2qx
+ r)
3
= −9(−1) 2·1R( 2
3qx + r, 3x2 + q)
= −9
2q 2−0 2
27r2
R qx + r, q + 3
3 4q2
= −4q2
2
27r2
R qx + r, q + 3 4q2
= −4q2
q + 27r2
4q2 1 = −4q3 − 27r2 .
We now try out Stickelberger’s theorem on a few cubic polynomials of
the form in Example 3, with F = Z5.
Exercise 20.10.1.1. Verify the entries in the following table.
Information from Actual factorization
f D(f) Stickelberger’s theorem of f ∈ Z5[x]
x 3 + 3x + 1 0 repeated roots (x − 1)(xr + 3) 2
x 3 + 3x + 2 4 1 or 3 factors irreducible (no roots)
x 3 + x 2 2 factors (x + 2)(x 2 + 3x + 3)
x 3 − x 1 1 or three factors x(x + 1)(x − 1)
x 3 + 2x −2 2 factors x(x 2 + 2)
We are now ready to prove Stickelberger’s theorem. We prove it first for
r = 1. The result for general r will then follow fairly easily.
Proof. Case r = 1. We assume f(x) is irreducible of degree m. Ket K be
a field containing Zp in which f(x) splits into linear factors. Let α be a
root of f. Then the roots of f(x) are α, αp , αp2, . . . , αpm−1. Then D(f) =
(δ(α, αp , αp2, . . . , αpm−1)) 2 ∈ F , so D(f) is a square in Zp iff
δ(α, αp , αp2, . . . , αpm−1) is in Zp.
We test to see whether δ(α, αp , αp2, . . . , αpm−1) is in Zp by examining
φp(δ(α, αp , αp2, . . . , αpm−1)).

838 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
φp(δ(α, αp , αp2, . . . , αpm−1)) =
0≤i

20.12. QUADRATIC RECIPROCITY 839
If α is one root of f1 in K, then α, αp , . . . , αpn−1 are the distinct roots
of f1 in K, and the roots of f(x) are 1, α, α2 , . . . , αq−1 . Now we may use
Lemma 20.6.3 to calculate the discriminant D(xq − 1).
Lemma 20.11.1. D(x q − 1) = (−1) (q−1)/2 q q .
Proof. By Lemma 20.6.3 D(x q − 1) = (−1) q(q−1)/2 q
i=1 q(αi ) q−1 =
(−1) (q−1)/2 q q (1·α·α 2 · · · α q−1 ) q−1 = (−1) q(q−1)/2 q q (α q(q−1)/2 ) q−1 = (−1) (q−1)/2 q q .
We now apply Stickelberger’s theorem to x q − 1 in Zp[x]. It reads: r =
1 + q−1
n ≡ q (mod 2) iff (−1)(q−1)/2 q q is a square mod p. Since q is odd, this
yields:
q − 1
2 divides iff (−1)
n
(q−1)/2 q q is a square mod p.
We analyze the left side. 2 divides q−1
q−1
iff n divides . Since n is the
n 2
order of p modulo q, we have established the following:
p (q−1)/2 ≡ 1 (mod q) iff (−1) (q−1)/2 q q is a square mod p.
This completes the difficult part of a proof of the Law of Quadratic Reciprocity,
but we still have a little work to do to be able to state it properly.
20.12 Quadratic Reciprocity
Let p be a prime and a an integer. We say a is a quadratic residue mod p
provided there is an integer x for which x2 ≡ a (mod p). Otherwise we say
a is a quadratic nonresidue mod p. If gcd(a, p) = 1, we define the Legendre
a symbol by p
a 1 if a is a quadratic residue mod p
= p −1 if a is a quadratic nonresidue mod p
Often we are able to determine whether or not a is a quadratic residue
mod p just by manipulating Legendre symbols according to the following
rules.
Theorem 20.12.1. Let a and b be integers and let p and q be distinct odd
primes with gcd(ab, p) = 1. Then
a2 1. = 1.
p

840 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
2. If a ≡ b (mod p), then
3.
4.
5.
6.
ab = p
−1
p
2
p
p
q
a
p
b . p
= (−1) p−1
2 .
= (−1) p2 −1
8 .
q
p
= (−1) p−1 q−1
2 2 .
a = p
b . p
Rule 6. is known as the Law of Quadratic Reciprocity.
Proof. Put F = Zp. Clearly the map σ : F ∗ → F ∗ : a ↦→ a 2 is two-toone,
so exactly half the nonzero residues mod p are quadratic residues, and
they form a multiplicative subgroup of order (p − 1)/2. Moreover, if b is a
primitive element mod p, then b j is a quadratic residue if and only if j is
even. b p−1
2 = −1, so −1 is a quadratic residue if and only if p−1
2 is even.
These observations establish parts 1. through 4. of the theorem.
Now suppose br ≡ a (mod p). Then a p−1
2
p−1
r ≡ b 2 ≡ c (mod p). Here
c2 ≡ 1 (mod p), so c ≡ ±1 (mod p). And here the plus sign holds iff p − 1
divides r p−1
a
iff r is even iff a is a square mod p iff = 1. It follows
2
p
a
that = a p
p−1
2 , a fact that contains result 4. and is usually called Euler’s
lemma.
2
By Euler’s lemma, = 2 p
p−1
2 , so to prove 5. it suffices to prove the
following:
2 p−1
2 ≡
1
iff p ≡
−1
1 or 7
(mod 8).
3 or 5
Start with the identity
2 p−1
p − 1
2 · 1 · 2 · 3 · · ·
= 2 · 4 · 6 · · · (p − 1).
2
We break up the right hand side.
Case 1. p ≡ 1 or 5 (mod 8). Here p−1
is an integer. So the right hand
4
side equals
p − 1 p + 3 p − 1
2 · 4 · · · 2 · 2 · · · 2
4 4
2

20.13. APPLICATION 841
p − 1
≡ 2 · 4 · · ·
2
p − 3
· (−1) · (−3) · · · −
2
p − 1
= 1 · 2 · 3 · · · · (−1)
2
p−1
4 .
Equating left and right sides and cancelling gives
2 p−1
2 ≡ (−1) p−1
1 if p ≡ 1 (mod 8);
4 =
−1 if p ≡ 5 (mod 8).
(mod p)
Case 2. p ≡ 3 or 7 (mod 8). Here p−3
4 is an integer. Then manipulations
similar to those above lead to
2 p−1
p − 1
p − 1
2 · 1 · 2 · 3 · · · ≡ 1 · 2 · 3 · · · · (−1)
2 2
p+1
4 (mod p).
From this one gets
2 p−1
2 ≡ (−1) p+1
4 =
1 if p ≡ 3 (mod 8);
−1 if p ≡ 7 (mod 8).
Putting these two cases together gives 5.
For the proof of 6., recall the last identity of the preceding section but
restated with the help of Euler’s lemma and using the Legendre symbol:
p
(−1)
= 1 iff
q
q−1
2 q
= 1
p
which says
p
=
q
This is clearly equivalent to 6.
20.13 Application
(−1) q−1
2 q
q−1
(
= (−1) 2
p
)
p−1
2
q
.
p
Exercise: Let F be the finite field GF (p) for an arbitrary odd prime p, and
put K = spl(f(x), F ), where f(x) = x 3 −x 2 +1. Determine for which primes
p, [K : F ] = 2.
Solution: The reverse of f(x) is ˆ f(x) = x 3 −x+1, which has discriminant
−4(−1) 3 − 27(1) 2 = −23. There are three possibilities for the factorization

842 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
of f(x) over F , but the number of factors is even (different parity from the
degree of f(x)) precisely when it is 2, which by Stickelberger’s theorem must
be exactly when the discriminant D(f) is a nonsquare. So we need to know
for which odd primes p we have
−23
−1 =
p
= (−1) p−1
2
23
= (−1)
p
p−1
2 (−1) p−1
2
23−1
2
p
p
= ,
23 23
by quadratic reciprocity. So [K : GF (p)] = 2 precisely when p is a nonsquare
mod 23, which is if and only if p is congruent mod 23 to one of the following:
5, 7, 10, 11, 14, 15, 17, 19, 20, 21, 22.
20.14 Lifting Roots Modulo Powers of p
In this section we give an application of the discriminant of a polynomial in
Z[x] to the solution of a congruence modulo a power of a prime p. Throughout
this section f(x) ∈ Z[x].
The basic idea of this section is to start with a solution a to the polynomial
congruence f(a) ≡ 0 (mod p j ) and construct a solution b to f(b) ≡
0 (mod p j+1 ). If b = a + tb j , i.e., b ≡ a (mod p j ), we say a lifts to b or that
b lies over a. But as we shall see, sometimes b is congruent to a modulo a
somewhat lower power of p.
First recall Taylor’s formula.
Theorem 20.14.1.
f(a + x) =
n
i=0
f (i) (a)
i!
· x i .
If a is an integer, then clearly f(a + x) ∈ Z[x] also.
Note: Since f is a polynomial with integer coefficients, and a is an
integer, f (i) (a)
is an integer. The main key to proving this is the fact that the
i!
product of k consecutive positive integers is divisible by k!. So if f(x) = cxm ,
then f (i) (x) = m(m−1)···(m−i+1)
i! cx m−i . Hence the coefficient on x m−i is an
integer. Since the derivative operator is additive, this result is easily extended
to sums of monomials, i.e., to all polynomials.
From this we see that for any nonnegative integer k,

20.14. LIFTING ROOTS MODULO POWERS OF P 843
f(a + tp k ) =
∞
i=0
(tp k ) i · f (i) (a)
i! ≡ f(a) + tpk f ′ (a) (mod p 2k ). (20.24)
This is a very useful congruence! First suppose k = j ≥ 1 and that
f(a) ≡ 0 (mod p j ). Then Eq. 20.24 implies
f(a + tp j ) ≡ f(a) + tp j f ′ (a) (mod p j+1 ). (20.25)
Each term is divisible by p j , and since a ≡ b (mod p j ) implies f(b) ≡
0 (mod p j ), this congruence is equivalent to
f(a + tp j )
p j
≡ f(a)
p j + tf ′ (a) (mod p). (20.26)
Hence the congruence f(b) ≡ 0 (mod p j+1 ) with b = a + tp j is equivalent
to the congruence
tf ′ (a) ≡ − f(a)
p j
(mod p).
So if f ′ (a) ≡ 0 (mod p) we may put t ≡ − f(a)
pjf ′ (mod p) and conclude after a
(a)
small calculation that a+tpj = a+f(a) ¯ f ′ (a), where f ¯ ′ (a)·f ′ (a) ≡ 1 (mod p).
Change notation slightly. Suppose f(a) ≡ 0 (mod p) and suppose that
f ′ (a) ≡ 0 (mod p). In this case we say a = a1 is a nonsingular solution to
the congruence f(a1) ≡ 0 (mod p). We have seen that there is a unique t
modulo p for which a2 = a1 + tp is a solution to f(a2) ≡ 0 (mod p). Note
that since a1 ≡ a2 (mod p), we also have f ′ (a2) ≡ f ′ (a1) ≡ 0 (mod p). Here
a2 = a1 − f(a1) ¯ f ′ (a), where f ¯ ′ (a) · f ′ (a) ≡ 1 (mod p). )
In general, if a1, a2, . . . , aj have been determined in this way, so that
f(aj) ≡ 0 (mod pj ), and f ′ (a1) ≡ 0 (mod p), then aj+1 = aj − f(aj) · f ¯ ′ (a1)
satisfies f(aj+1) ≡ 0 (mod pj+1 ). For each i, 1 ≤ i ≤ j, we also have
ai+1 ≡ ai (mod pi ). This establishes what is usually called Hensel’s Lemma,
but we actually have a bit more:
Theorem 20.14.2. Let a be a solution of the polynomial congruence f(a) ≡
0 (mod p j ).
(i) If f ′ (a) ≡ 0 (mod p) and f(a) ≡ 0 (mod p j+1 ), then a does not lift to
a solution modulo p j+1 .

844 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
(ii) If f ′ (a) ≡ 0 (mod p) and f(a) ≡ 0 (mod p j+1 ), then a lifts to p
solutions b = a + tp j , t modulo p.
(iii) If f ′ (a) ≡ 0 (mod p), then there is a unique value of t modulo p for
which b = a + tp j is a solution of f(b) ≡ 0 (mod p j+1 ).
Part (iii) of the preceding theorem is usualy called Hensel’s Lemma. We
now give a generalization of it that says that if the power of p dividing f(a)
is sufficiently large compared with the power of p dividing f ′ (a), then the
solution can be “lifted” indefinitely. In fact, it is not actually “lifted,” since
in each case the new solution b modulo p j+1 produced from the solution a
modulo p j is congruent to a modulo a somewhat smaller power of p than p j .
Theorem 20.14.3. Let f(x) ∈ Z[x]. Suppose that
(i) f(a) ≡ 0 (mod p j );
(ii) p τ ||f ′ (a);
(iii) j ≥ 2τ + 1.
Then:
(a) If b ≡ a (mod p j−τ ), then f(b) ≡ f(a) (mod p j ) and p τ ||f ′ (b).
(b) There is a unique t modulo p for which f(a + tp j−τ ) ≡ 0 (mod p j+1 ).
Moreover, by part (a), this process may be continued indefinitely.
Proof. Write b = a + t · p j−τ , so b ≡ a (mod p j−τ ). By Taylor’s For-
mula, f(b) = f(a + t · p j−τ ) =
i=0
f (i) (a)
i! (t · p j−τ ) i ≡ f(a) + t · p j−τ ·
f ′ (a) (mod p 2(j−τ) ).
Note: 2(j − τ) = j + j − 2τ ≥ j + 1 by (iii), so p j+1 |p 2(j−τ) . Hence
f(b) = f(a + t · p j−τ ) ≡ f(a) + t · p j−τ · f ′ (a) (mod p j+1 ). But p j |f(a) and
p j |[p j−τ · f ′ (a)], so that p j |f(b). This implies that
Since f ′ (a)
p τ
f(b)
pj = f(a + t · pj−τ )
pj ≡ f(a)
pj + t · f ′ (a)
pτ (mod p).
≡ 0 (mod p), there exists a unique t modulo p for which
f(b) = f(a + t · pj−τ ) ≡ 0 (mod pj+1 ).
Also, f ′ (x) ∈ Z[x], so f ′ (a+t·pj−τ ) = f
i=0
(i+1) (a)
·t i!
i (pj−τ ) i ≡ f ′ (a) (mod pj−τ )
for any integer t. But j − τ ≥ τ + 1, so f ′ (a + t · pj−τ ) ≡ f ′ (a) (mod pτ+1 ).
Since pτ ||f ′ (a), it must be that pτ ||f ′ (a + t · pj−τ ), i.e., pτ ||f ′ (b).
So there is a unique t modulo p for which b = a + t · pj−τ solves f(b) ≡
0 (mod pj+1 ). Since pτ ||f ′ (a) implies pτ ||f ′ (b), the process may be continued
indefinitely.

20.14. LIFTING ROOTS MODULO POWERS OF P 845
Example: Consider the congruence x 2 + x + 7 ≡ 0 (mod 81). (Note
81 = 3 4 .)
First consider the congruence modulo 3: x2 + x + 7 ≡ x2 − 2x + 1 ≡ (x −
1) 2 (mod 3). So a1 = 1 is the unique solution modulo 3. Also, f ′ (x) = 2x+1,
so f ′ (1) ≡ 0 (mod 3) and f(1) = 9 ≡ 0 (mod 32 ). So a2 = 1 + t · 3 for all t
modulo 3, i.e., a2 ∈ {1, 4, 7} modulo 9.
Consider a2 ≡ 1 modulo 9. f ′ (1) ≡ 0 but f(1)
9 = 1 ≡ 0 (mod 3), so a2 = 1
does not lift to a solution modulo 27.
Consider a2 ≡ 4 (mod 9). f ′ (4) ≡ f ′ (1) ≡ 0 (mod 3). Then f(4)
9
= 3 ≡
0 (mod 3), so a2 = 4 lifts to 3 roots a3 = 4 + t · 3 2 , for all t modulo 3. Hence
a3 ∈ {4, 13, 22} modulo 27.
Consider a2 ≡ 7 (mod 9). f(7)
9
= 7 ≡ 0 (mod 9), so 7 does not lift to a
solution modulo 27.
At this point we know that 4, 13 and 22 are the only solutions modulo
27. And f ′ (4) ≡ f ′ (13) ≡ f ′ (22) ≡ f ′ (1) ≡ 0 (mod 3). A routine check now
shows that for each of the values a = 4, 13, 22, f(a)
27 is not congruent to 0
modulo 3. Hence none of these solutions modulo 27 lift to a solution modulo
81, i.e., there is no solution modulo 81.
Finally we are ready to invoke the discriminant.
Theorem 20.14.4. Suppose that p δ ||D(f) and p τ ||f ′ (a) where a satisfies
f(a) ≡ 0 (mod p j ) with j > δ. Then j ≥ 2τ + 1.
From this theorem we see that if the prime p does not divide the discriminant
D(f) and f(a) ≡ 0 (mod p), so δ = 0 and we may take j = 1, then
f ′ (a) ≡ 0 (mod p).
Proof. Suppose f(a) ≡ 0 (mod p j ). If we divide f(x) by x−a, the remainder
is f(a). Then we may write f(x) = (x − a)g(x) + p j r for some g(x) ∈ Z[x]
and some r ∈ Z. We know that the discriminant D(f) is a polynomial
in the coefficients of f. And applying Taylor’s formula to D(f), we find
D(f(x)) = D((x−a)g(x)+p j r) ≡ D((x−a)g(x)) (mod p j ). By Lemma 20.7.3
we know that D((x−a)g(x)) = D(g)g(a) 2 . As f ′ (x) = g(x)+(x−a)g ′ (x), we
find that f ′ (a) = g(a). Hence D(f) ≡ D(g)f ′ (a) 2 (mod p j ). The inequality
j > δ is equivalent to the assertion that D(f) ≡ 0 (mod p j ). This implies
that f ′ (a) 2 ≡ 0 (mod p j ), which says that j > 2τ.

846 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
The two preceding theorems have an immediate corollary.
Corollary 20.14.5. Let f(x) be a polynomial with integral coefficients. Suppose
that p δ ||D(f). Let a be an integer for which f(a) ≡ 0 (mod p j ), that
p τ ||f ′ (a), and suppose that j > δ, so by the preceding theorem j > 2τ. If
b ≡ a (mod p j−τ ), so b = a + tp j−τ , then f(b) ≡ f(a) (mod p j ) and p τ ||f ′ (b).
Moreover, there is a unique t (mod p) such that f(a + tp j−τ ) ≡ 0 (mod p j+1 ).
In this situation, a collection of p τ solutions (mod p j ) gives rise to p τ
solutions (mod p j+1 ), while the power of p dividing f ′ remains constant.
Since the hypotheses of the theorem apply with a replaced by b = a + tp j−τ
and (mod p j ) replaced with (mod p j+1 ) but with τ unchanged, the lifting
may be repeated and continues indefinitely.
20.15 Resultants as Determinants
Let K be an integral domain and let
f(x) = n n−i
i=0 aix
(1)
g(x) = n m−i
i=0 bix
We want necessary and sufficient conditions that f and g have a nonconstant
factor φ(x). Assume a0 = 0, so the degree of f is n.
Lemma 20.15.1. f(x) and g(x) have a nonconstant common divisor φ(x)
if and only if an equation of the form
(2) h(x)f(x) = k(x)g(x) holds where deg(h(x)) ≤ m − 1, deg(k(x)) ≤
n − 1, and both h and k are nonzero.
Proof. Assume (2) holds. Factor both sides of (2) over K. All prime factors
of f(x) divide k(x)g(x), but some prime factor of f(x) divides g(x), since
deg(f(x)) > deg(k(x)).
Conversely, if φ(x) is a common factor, put h(x) = g(x)/φ(x) and k(x) =
f(x)/φ(x).
In the above context, put
(3)
h(x) = m−1
i=0 cix m−1−i ,
k(x) = n−1
i=0 dix n−1−i .

20.15. RESULTANTS AS DETERMINANTS 847
Then the coefficient of x n+m−1−t on the L.H.S. of (2) is t
i=0 aict−i and
on the R.H.S. it is t
i=0 bidt−i. For t = 0, 1, . . . , m + n − 1, the equations
t
aict−i =
i=0
t
i=0
bidt−i
give m+n homogeneous linear equations in the m+n unknowns c0, . . . , cm−1, d0,
. . . , dn−1. (Here put ai, ci, di, bi = 0 for values of i larger than n, m − 1, n −
1, m, respectively.) There is a nontrivial solution for the coefficients ci, di if
and only if the determinant of the system equals zero. Using −di in place of
di, this determinant ⎛ (with unknowns ordered c0, . . . , cm−1, −d1, ⎞.
. . , −dn−1)
a0
⎜ a1 a0 b1 b0
⎜ a2 a1 a0 b2 b1 b0
⎜
.
.
⎜
appears as det ⎜ . bm bm−1 . . . b0
⎜ . bm
⎜ an an−1 . . . bm
⎜
⎝ an an−1 bm
It is standard to take the transpose as the definition of the resultant of f
and g.
⎛
a0 a1 . . . an
⎜ a0 . . . an
⎜ a0 ⎜
. . . . . . an
⎜
.
⎜
..
R(f, g) = det ⎜ b0 b1 . . . bm
⎜ b0 . . . bm
⎜
⎝ b0 . . . . . . bm
. ..
b0
⎞
.. .
⎟ .
⎟
⎠
There are m rows in the top part of this matrix and n rows in the bottom
part, so that the matrix is square of size n + m.
Note: R(f, g) is homogeneous of degree m in the ai and homogeneous of
degree n in the bi and contains the principal term a m 0 bn m .
⎟
⎟.
⎟
⎠

848 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
Note: R(f, g) is zero if and only if either both a0 = b0 = 0 or f and g
have a common factor.
Note: R(f, g) is a rational integral form of degree m+n in the coefficients
of f and g.
Exercise. Write out the discriminant D(f) of f as (−1) n(n−1)/2 R(f, f ′ ),
i.e., plus or minus the determinant of a 2n − 1 by 2n − 1 matrix. Multiply
each of the first n − 1 rows by 4. For 1 ≤ i ≤ n − 1, subtract row n − 1 + i
from row i. Then the first column has only one nonzero entry. It is nan and
occurs in row n. Expanding the determinant along the first column yields
D(f) = an(−1) (n−2)(n−1)/2 det(∆)
n n−2
where ∆ = (di,j) is a (2n − 2) × (2n − 2) matrix whose entries are as follows:
for 1 ≤ i ≤ n − 1 and i ≤ j ≤ i + n − 1,
di,j = (j + 1 − i)an+i−j−1 and dn−1+i,j = (n + i − j)an+i−j.
All other entries are zero. Clearly D(f) is a form of degree 2n − 2.
We now place this theory in a more general setting. Let A be a commutative
ring with identity, and let v0, . . . , vn, w0, . . . , wm be algebraically
independent over A. We form two polynomials:
fv(x) = v0x n + · · · + vn,
gw(x) = w0x m + · · · + wm.
Write (v) = (v0, . . . , vn), and similarly define (w). The resultant of
((v), (w)) or of fv, gw is defined to be the determinant of the following
square matrix of size m + n.
⎛
⎞
v0 v1 . . . vn
⎜ v0 ⎜
. . . vn ⎟
⎜ v0 ⎜
. . . . . . vn ⎟
⎜
.
⎜
.. ⎟
⎜
⎟ .
⎜ w0 w1 . . . wm ⎟
⎜
⎟
⎜ w0 . . . wm ⎟
⎜
⎟
⎝ w0 . . . . . . wm ⎠
. ..

20.15. RESULTANTS AS DETERMINANTS 849
The top part of the matrix has m rows, and the bottom part has n rows.
The blank spaces are supposed to be filled with zeros.
If we substitute elements (a) = (a0, . . . , an) and (b) = (b0, . . . , bm) in A
for (v) and (w), respectively, in the coefficients of fv and gw, then we obtain
polynomials fa and gb with coefficients in A, and we define their resultnt to
be the determinant obtained by substituting (a) for (v) and (b) for (w) in the
determinant. the resultant of fv, gw is denoted R(fv, gw) or R(v, w). Then
R(fa, gb) is obtained by substituting (a) for (v) and (b) for (w).
Note: R(v, w) is a polynomial with “integer” coefficients, i.e., we could
assume A = Z.
If z is a variable, R(zv, w) = zmR(v, w) and R(v, zw) = znR(v, w). Hence
R is homogeneous of degree m in its first set of variables and homogeneous
of degree n in its second set of variables. Furthermore, R(∗v, w) contains the
monnomial vm 0 wn m with coefficient 1, when expressed as a sum of monomials.
Note: Substituting v0 = w0 = 0 yields R = 0 since the first column
vanishes.
Consider the following system of linear equations over Z[v, w] with “unknowns”
1, x, . . ., x n+m−1 .
x m−1 fv(x) = v0x n+m−1 +v1x n+m−2 + · · · +vnx m−1
x m−2 fv(x) = v0x n+m−2 + · · · +vnx m−2
.
fv(x) = v0x n + · · · vn
x n−1 gw(x) = w0x n+m−1 +w1x n+m−2 + · · · +wmx n−1
x n−2 gw(x) = w0x n+m−2 + · · · +wmx n−2
.
gw(x) = w0x m + · · · wm
Let C be the column vector on the L.H.S. and let Ci be the column
vector of coefficients of xm+n−1−i on the R.H.S. ⎛ So the⎞ above equations may
⎜
be expressed as C = (C0, C1, . . . , Cm+n−1) ⎜
⎝
x m+n−1
x m+n−2
By Cramer’s Rule applied to the last variable, which is 1, we have
1 · R(v, w) = det(C0, . . . , Cm+n−1) = det(C0, . . . , Cm+n−2, C). Expanding by
.
x
1
⎟
⎟.
⎟
⎠

850 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
the last column, we see that there are polynomials φv,w and ψv,w in Z[v, w][x]
such that
φv,wfv + ψv,wgw = R(v, w).
Note that R(v, w) ∈ Z[v, w], but φ and ψ involve x.
If λ : Z[v, w] → A is a homomorphism into a commutative ring A and
λ(v) = (a), λ(w) = (b), then
(∗) φa,bfa + ψa,bgb = R(a, b) = R(fa, fb).
Theorem 20.15.2. Let K be a subfield of a field L, and let fa, gb be polynomials
in K[x] having a common root in L. Then R(a, b) = 0.
Proof. Use (∗) to obtain a new proof.
Lemma 20.15.3. Let h(x1, . . . , xn) be a polynomial in n variables over Z.
If h has value 0 when x1 is substituted for x2 and the other xi are fixed, then
h(x1, . . . , xn) is divisible by x1 − x2 in Z[x1, . . . , xn].
Let v0, t1, . . . , tn, w0, u1, . . . , um be algebraically independent over Z. Form
the polynomials
fv = v0(x − t1) · · · (x − tn) = v0x n + · · · + vn,
gw = w0(x − u1) · · · (x − um) = w0x m + · · · wm.
Here vi = (−1) i v0si(t) and wj = (−1) j w0sj(u), where si denotes the i th
elementary symmetric function: si(t) is the sum of all monomials which are
products of the t ′ s, i at a time.
Exercise: Show that v0, v1, . . . , vn, w0, . . . , wm are algebraically independent
over Z.
Theorem 20.15.4. With the above notation, R(fv, gw) = vm 0 wn 0
uj).
n m i=1 j=1 (ti−
Proof. Let S be the R.H.S. Since R(v, w) is homogeneous of degree m in
its first variables and of degree n in its second variables, it follows that
R = vm 0 wn 0 h(t, u), where h(t, u) ∈ Z[t, u]. By the preceding theorem and
lemma (R = 0 if ti is substituted for uj), viewing R as an element of Z[t, u],

20.15. RESULTANTS AS DETERMINANTS 851
it follows that R is divisible by ti − uj for each pair (i, j). Hence S divides
R in Z[t, u], because ti − uj is clearly prime in that ring and different pairs
(i, j) give rise to different primes.
S = vm 0 wn n 0 i=1 ( m
j=1 (ti − uj)) = vm n 0 i=1 g(ti)
= wn 0 (−1) mn m j=1 (v0
n i=1 (uj − ti)) = (−1) mnwn m 0 j=1 f(uj).
This shows that S is first homogeneous of degree n in (w) and secondly,
homogeneous of degree m in (v). For example, g(ti) = w0tm m−1
i + w1ti + · · ·+
wm is clearly homogeneous of degree 1 in (w). Since R has exactly the same
homogeneity properties and is divisible by W , it follows that R = cS for
some integer c. As both R and S have the monomial vm 0 wn m with coefficient
1, it follows that c = 1, as desired.
Corollary 20.15.5. Let fa, gb be polynomials with coefficients in a field K
such that a0b0 = 0, and such that fa, gb split into factors of degree 1 in K[x].
Then R(fa, fb) = 0 if and only if fa and gb have a root in common.
Proof. The “if” has already been established. So assume R = 0 and let
and
fa = a0(x − α1) · · · (x − αn)
gb = b0(x − β1) · · · (x − βm).
Then there is a homomorphism Z[v0, t, w0, u]⊤K such that v0 ↦→ a0, w0 ↦→
b0, ti ↦→ αi, uj ↦→ βj, for all i, j. Then 0 = R(fa, fb) = am 0 bn
0 i j (αi − βj),
implying fa and fb have a root in common. (Here we need R(fv, fw) ↦→
R(fa, fb) to show that R(fa, fb) = am 0 bn
0 i j (αi −βj), from which it is clear
that 0 = R(fa, fb). )
We deduce one more special result. As above, let
fv(x) = v0x n + · · · + vn = v0(x − t1) · · · (x − tn).
From R(fv, gw) = vm n 0 i=1 g(ti), we know that if f ′ v is the derivative of
fv, then R(fv, f ′ v ) = vn−1 0
we have
and
f ′ v
(x) =
n
i=1 f ′ (ti). Using the product rule for derivation
i
v0(x − t1) · · · ˆ
(x − ti) · · · (x − tn)

852 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
f ′ v(ti) = v0(ti − t1) · · · (ti −ˆ ti) · · · (ti − tn).
Define the discriminant of fv to be
D(fv) = D(v) = (−1) n(n−1)/2 v 2n−2
0
(ti − tj).
Theorem 20.15.6. (−1) n(n−1)/2R(fv, f ′ v ) = (−1)n(n−1)/2v 2n−1
0 i=j (ti−tj) =
v0D(fv).
Proof. Put f ′ v (ti) = v0
nj=1
j=i
(ti − tj) into R(fv, f ′ v
i=j
n ) = vn−1 0 i=1 f ′ (ti).
Finally, homogenize the polynomials fv(x) and gw(x), i.e., put
and
Fv(x, y) = y n fv(x/y) = v0x n + v1x n−1 y + · · · + vny n
Gw(x, y) = y n gw(x/y) = w0x m + w1x m−1 y + · · · + wmy m .
If v0 = w0 = 0, then Fv and Gw have the common solution (x, y) = (1, 0).
If either v0 = 0 or w0 = 0, then Fv and Gw have a common solution = (0, 0)
if and only if fv, gw have a common solution if and only if R(fv, gw) =
R(v, w) = R(Fv, Gw) = 0. Hence in any case, Fv and Gw have a common
solution = (0, 0) if and only if R(Fv, Gw) = 0.
20.16 Bezout’s Theorem
Continue with the notation of the previous section. Define the weight of vi to
be i (vi is homogeneous of degree i in the t’s), and define the weight of wj to be
j (wj is homogeneous of degree j in the u’s). A nonzero constant has weight
0, and 0 has every weight. The weight of a monomial cvi1 . . . viswj1 . . . wjt is
by definition i1 + · · · is + j1 + · · · jt. the weight of a sum of (unlike) terms
different from zero is the maximum of the weights of the terms. Such a sum
is isobaric of weight m if all the terms have the same weight m.
Theorem 20.16.1. R(v, w) is isobaric of weight mn.

20.16. BEZOUT’S THEOREM 853
Proof. This follows from R(v, w) = vm 0 wn n m 0 i=1 j=1 (ti−uj), since the weight
of a monomial in v and w is the degree of that monomial as a polynomial
in the t’s and u’s and since R(v, w) is homogeneous of degree mn in the t’s
and u’s. Using a modified version of the proof of the theorem on Symmetric
functions, the isobaricity of R follows. A more direct proof is as follows.
Let vij be the element in the ith row and jth column of R(v, w), i =
1, . . . , m; vij has weight j − i. Let wst be the element in the (m + s) th row
and tth column of R(v, w), s = 1, . . . , n; wst has weight t − s. Any term in
the expansion of the determinant will have weight m i=1 (j − i) + n s=1 (t − s),
where j, t together cover the set 1, 2, . . . , m + n. Hence the term has weight
= mn.
(m+n)(m+n+1)
2
+ −(m+1)m
2
+ −n(n+1)
2
Theorem 20.16.2. (Bezout’s Theorem) If F = 0 and G = 0 are two
projective curves over an algebraically closed field K that intersect in only a
finite number of points, then they intersect in at least one point (in exactly
mn points, counting multiplicities).
Proof. Our proof will be complete only in establishing the existence of at
least one point of intersection.
If both F = 0 and G = 0 pass through (0, 0, 1), or if some pair of intersections
of F = 0 and G = 0 are collinear with (0, 0, 1), certainly F = 0 and
G = 0 have a point of intersection. Hence we suppose that neither of these
conditions holds. Then write
F (X, Y, Z) = v0Z n + v1(X, Y )Z n−1 + · · ·+ vn−1(X, Y )Z + vn(X, Y ). (20.27)
G(X, Y, Z) = w0Z m + w1(X, Y )Z m−1 + · · · + wm−1(X, Y )Z + wm(X, Y ).
(20.28)
The resultant of F and G with respect to Z is
RZ(F, G) = R(v, w) = P (v0, . . . , vn, w0, . . . , wm).
As at least one of the curves does not pass through (0, 0, 1), v0 = 0 or
w0 = 0. P is isobaric of weight mn, and vi is homogeneous of degree i in X
and Y , wj is homogeneous of degree j in X and Y . Hence P is homogeneous
of degree mn in X and Y . We claim that P = H(X, Y ) is not zero as a
polynomial in X and Y . For if this were the case, then R(v, w) = 0 =

854 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
RZ(F, G) would imply the existence of a Z0 ∈ K for each choice of X0,
Y0 ∈ K such that F (X0, Y0, Z0) = 0 = G(X0, Y0, Z0). This would imply that
F (X, Y, Z) = 0 and G(X, Y, Z) = 0 had infinitely many points in common,
contradicting the hypothesis. So H(X, Y ) = 0. But each solution X0 : X0
to H(X0, X0) = 0 gives rise to an intersection (X0, Y0, Z0) of F = 0 and
G = 0, and to only one such solution, since two solutions (X0, Y0, Z0) would
be collinear with (0, 0, 1). Since 0 = H(X, Y ) = Y mn h(X/Y ) has exactly mn
solutions, counting multiplicities, the theorem follows.
The gap in the proof of this theorem concerns the number of intersections
when the curves F = 0 and G = 0 do not satisfy the assumptions made at
the beginning of the proof. The “usual” way around this is to show that
there is a projective transformation simultaneously transforming F = 0 and
G = 0 to curves in an acceptable position, and noting that intersections are
preserved by projective transformations.
Theorem 20.16.3. In the notation of the beginning of this section, R(v, w)
is absolutely irreducible over Z[v0, . . . , vn, w0, . . . , wm], i.e., irreducible in
F [v0, . . . , wm] for F any field.
Proof. If R were factored into two factors A and B (polynomials in v0, . . . , wm),
then A and B could be written as symmetric functions of the roots t1, . . . , tn
and u1, . . . , um. Since R is divisible by t1 − u1, A or B, say A, has to be
divisible by t1 − u1 as well. But being symmetric in the ti and in the uj, A
must be divisible by all other ti − uj, and therefore by their product
Since R = v m 0 wn 0
other factor B, namely B = v p
(ti − uj).
i
j
i j (ti − uj), there remains only one possibility for the
0w q
0. But R as a polynomial in the v’s and
w’s is divisible neither by v0 nor by w0. Hence B = 1. So R is absolutely
irreducible.
Now let R = Fq with q = p e , let n = q − 1, and replace x1, . . . , xn with
the nonzero elements of Fq. Apply the above to the polynomial equation
g(x) = x q−1 − 1 =
(x − a) = x q−1 − σ1x q−2 + · · ·
a∈F ∗ q

20.17. THE HERMITE-DICKSON CRITERION 855
to find for the elementary symmetric functions σ1, σ2, . . . , σq−1 in the nonzero
elements of Fq that
σ1 = σ2 = · · · = σq−2 = 0; σq−1 = −1, (20.29)
the first and last of which we saw earlier. If we now put the results of Eq. 20.29
into Eq. 20.21 (Newton’s formulas), we obtain the following lemma.
Lemma 20.16.4. We have
a t =
a t = 0; for 1 ≤ t ≤ q − 2,
a∈Fq
a∈F ∗ q
and
a q−1 =
a q−1 = −1.
a∈Fq
a∈F ∗ q
20.17 The Hermite-Dickson Criterion
The next result is a famous set of necessary and sufficient conditions for a
polynomial over Fq to be a permutation polynomial.
Theorem 20.17.1. (The Hermite-Dickson Criterion) If f(a) = 0 has a
unique solution a ∈ Fq, then f(t) is a permutation polynomial of Fq if and
only if [f(t)] r , reduced modulo t q − t, has zero coefficient on t q−1 , for all r
with 1 ≤ r ≤ q − 2 and gcd(r, p) = 1.
Proof. Without loss of generality we first assume that f(t) has degree at
most q − 1. Then for any r with 1 ≤ r ≤ q − 2, after reducing modulo t q − t
suppose that
f(t) r q−1
≡
i=0
a (r)
i ti .
Replace t in the above equation with each element of Fq and add the
resulting q equalities and apply Lemma 20.16.4 to obtain
f(b) r = q−1
b∈Fq
b∈Fq
i=0
a (r)
i bi q−1
=
i=0
a (r)
i
⎧
⎨
b
⎩
i
⎫
⎬
⎭
b∈Fq
= −a(r) q−1. (20.30)

856 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
If f(t) is really a permutation polynomial, then we have
f(b) r =
b r = 0 for 1 ≤ r ≤ q − 2.
b∈Fq
b∈Fq
Hence a necessary condition for f(t) to be a permutation polynomial is
that a (r)
q−1 = 0 for 1 ≤ r ≤ q − 2, which says that modulo tq − t, f(t) r has
degree at most q−2 for such r. Clearly f(a) = 0 must have a unique solution.
Conversely, suppose that f(t) satisfies the Hermite-Dickson Criterion.
Note that this means that we are assuming that mod tq − t the polynomial
f(t) r has degree at most q − 2 only for those r that are relatively prime to
p. Put
g(T ) =
q
(T − f(b)) = diT q−i .
b∈Fq
Note that di is the ith elementary symmetric function in the f(b), and
in particular d0 = 1. Also, since there is a (unique!) value of b for which
f(b) = 0, the constant term of g(t) must be zero, i.e., dq = 0. Clearly
g(f(b)) = 0 for each b ∈ Fq. For 1 ≤ r put sr =
b∈Fq f(b)r . For r ≤ q − 2
we see that sr = −a (r)
q−1. By hypothesis,
sr = −a (r)
q−1 = 0 for 1 ≤ r ≤ q − 2 and (r, p) = 1. (20.31)
It is also clear that
i=0
sq−1 =
(f(b)) q−1 = q − 1 ≡ −1,
b∈Fq
since by hypothesis all but one of the f(b) are nonzero and raised to the
power q − 1 must equal 1.
If (r, p) = 1 and rp j ≤ q − 2, then
which implies that
(f(b)) rpj
⎛
= ⎝
f(b) r
⎞
⎠
b∈Fq
b∈Fq
p j
, (20.32)
sr =
f(b) r = 0 for 1 ≤ r ≤ q − 2. (20.33)
b∈Fq

20.18. GREATEST COMMON DIVISORS OF A M ± 1 AND A N ± 1 857
Use Newton’s formulas, say in the form of Eq. 20.23, with n = q and with
di replacing σi, to obtain
sm = sm−1d1 − sm−2d2 + · · · + (−1) q+1 sm−qdq, for m ≥ q, s0 = m, (20.34)
and for 1 ≤ m < q we have
sm = sm−1d1 − sm−2d2 + · · · + (−1) m+1 s0dm. (20.35)
But we know that sm = 0 for 1 ≤ m ≤ q − 2 and sq−1 = −1. So for
1 ≤ m ≤ q − 2, Eq. 20.35 becomes
implying dm = 0 for 1 ≤ m ≤ q − 2.
Put m = q − 1 < q in Eq. 20.35 to obtain
0 = sm = (−1) m+1 s0dm, (20.36)
−1 = sq−1 = sq−2d1 − sq−3d2 + · · · + (−1) q s0dq−1
= (−1) q (q − 1)dq−1 = dq−1. (20.37)
.
At this point we know that d0 = 1, dq−1 = −1, dq = 0 and di = 0 for
1 ≤ i ≤ q − 2. Hence
g(T ) =
(T − f(b)) = T q − T.
b∈Fq
from which it follows that b ↦→ f(b) must be a permutation of the elements
of Fq.
20.18 Greatest Common Divisors of a m ± 1
and a n ± 1
20.18.1 Basic Identities
Let m and n be arbitrary positive integers. By the Division Algorithm we
may write
m = qn + r, 0 ≤ r < n, for unique integers q and r.

858 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
Also, let d = gcd(m, n), and write m = dm ′ , n = dn ′ . We know that d is the
smallest positive integer that can be written in the form d = um + vn for
arbitrary u, v ∈ Z. Suppose that d = u0m + v0n. Then for each integer k we
have
d = u0m + v0n = (u0 + kn ′ )m + (v0 − km ′ )n.
By choosing k positive and large enough, we may assure that v0 − km ′ is less
than zero. Hence we may find integers u and v for which
d = um − vn, u, v > 0, and um > vn.
We start with the following four basic identities, each of which can be
readily verified. (The symbol x is an indeterminate.)
x m ± 1 = (x n − 1)(x m−n + x m−2n + · · · + x m−qn ) + x r ± 1. (20.38)
x m ±1 = (x n +1)(x m−n −x m−2n +· · ·+(−1) q−1 x m−qn )+(−1) q x r ±1. (20.39)
First choose the minus sign in Eq. 20.38 to obtain
x m − 1 = (x n − 1)(x m−n + x m−2n + · · · + x m−qn ) + x r − 1. (20.40)
It is clear that if n|m, i.e., if r = 0, then (x n − 1)|(x m − 1) in Z[x].
Conversely, suppose that (x n −1)|(x m −1). Then (x n −1)|(x r −1). But since
0 ≤ r < n, this is possible if and only if r = 0.
Lemma 20.18.2.
(x n − 1)|(x m − 1) in Z[x] iff n|m. (20.41)
If x is replaced by any integer a, 1 < a, we obtain
a m − 1 = (a n − 1)(a m−n + a m−2n + · · · + a m−qn ) + a r − 1, 0 ≤ a r − 1 < a n − 1.
Here it is clear that (a n − 1)|(a m − 1) if and only if (a n − 1)|(a r − 1) which
holds if and only if a r − 1 = 0 if and only if r = 0 if and only if n divides m.
So we have lemma 20.18.2 with x replaced by a.

20.18. GREATEST COMMON DIVISORS OF A M ± 1 AND A N ± 1 859
Lemma 20.18.3. If 1 < a ∈ Z, then (a n − 1)|(a m − 1) iff n|m.
To see what happens when a < 0 we must work a bit harder. It will be
left to the reader to interpret the later results in this case. At present it is
clear that the following is true.
Lemma 20.18.4. We have half a theorem:
(i) (x d − 1)|gcd(x m − 1, x n − 1),
and if 1 < a ∈ Z,
(ii) (a d − 1)|gcd(a m − 1, a n − 1).
Now consider
(x n −1)(−x um−vn −x um−(v−1)n −· · ·−x um−n )+(x m −1)(x m(u−1) +· · · x m +1)
= −x um−(v−1)n − x um−(v−2)n − · · · − x um−n − x um
+x um−vn + x um−(v−1)n + · · · + x um−n + (x um − 1)
= x um−vn − 1 = x d − 1.
So x d − 1 = (x n − 1)A(x) + (x m − 1)B(x), for some A(x), B(x) ∈ Z[x].
Hence gcd(x n − 1, x m − 1)|(x d − 1). We have completed a proof of the
following theorem.
Theorem 20.18.5. Our first main result says:
(i) gcd(x m − 1, x n − 1) = x gcd(m,n) − 1.
Similarly, replacing x by a, 1 < a ∈ Z, we have
(ii) gcd(a m − 1, a n − 1) = a gcd(m,n) − 1.
20.18.6 gcd(a m + 1, a n + 1), a > 1
Now choose the “+” in Eq. 20.39
x m + 1 = (x n + 1)(x m−n − · · · + (−1) q−1 x m−qn ) + (20.42)
+(−1) q (x r + (−1) q ) (20.43)
= (x n + 1)A(x) + (−1) q (x r + (−1) q ), for some A(x) ∈ Z[x].
From this equation it is easy to deduce the following
(Use |(−1) q (a r + (−1) q )| < a n + 1):

860 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
Lemma 20.18.7. First consider the indeterminate x:
(i) (x n + 1)|(x m + 1) if and only if n|m and m
n
For 1 < a ∈ Z,
(ii) (a n + 1)|(a m + 1) if and only if n|m and m
n
= q is odd.
= q is odd.
On the one hand, (x2m − 1, xn − 1) = x (2m, n) − 1 = xd(2m′ , n ′ )
− 1 =
d ′ n
x − 1, if n = is odd,
d
x2d − 1, if n ′ = n
is even.
d
On the other hand,
(x 2m − 1, x n − 1) = (x d m x − 1
− 1)
xd − 1 · (xm + 1), xn − 1
xd
− 1
Hence it follows that
Lemma 20.18.8.
gcd
= (x d
− 1) x m + 1, xn − 1
xd
.
− 1
x m + 1, xn − 1
xd
1,
=
− 1 xd + 1,
n
d odd,
n
d even.
The same results holds if x is replaced by an integer a > 1.
From now on we consider only the case where x is replaced by
an integer a > 1.
Since (am +1, ad−1) = (am−1+2, ad−1) = (2, ad
2, a odd,
−1) =
1, a even,
dividing an − 1 by ad − 1 removes at most a factor 2 of gcd(am + 1, an − 1).
Suppose that n
d
is odd. When a is even,
a n + 1
a d + 1 = (ad ) n
d −1 − (a d ) n
d −2 + · · · − a d + 1
is always odd. When a is odd, an +1
a d +1
≡ n
d
This allows us to prove the following:
(mod 2).

20.18. GREATEST COMMON DIVISORS OF A M ± 1 AND A N ± 1 861
Lemma 20.18.9. Let n
d be odd. Then gcd(am +1, an
1, if a is even,
−1) =
2, if a is odd.
Proof. Since n
d
is odd, gcd
a m + 1, an −1
a d −1
= 1, and dividing by a d − 1 re-
moves at most a factor of 2. Clearly if a is even, then a m + 1 and a n − 1 are
both odd, so gcd(a m + 1, a n − 1) = 1. And when a is odd, a m + 1 and a n − 1
clearly do have a factor of 2, finishing the proof.
Now suppose that n
d
m
is even, so necessarily d is odd, so am +1
ad is odd. Since
+1
(am + 1, ad − 1) = (am − 1 + 2, ad − 1) = (2, ad − 1), by Lemma 20.18.3
(a m + 1, a n − 1) = (a d m a + 1
+ 1)
ad + 1 , an − 1
a2d − 1 · (ad
− 1)
= (a d m a + 1
+ 1)
ad + 1 , an − 1
a2d
= (a
− 1
d + 1).
This holds for a odd or even. We collect these results in the following general
theorem.
Theorem 20.18.10. Let 1 < a, m, n ∈ Z with d = gcd(m, n). Then
(am + 1, an ⎧
⎨ 1, if
− 1) =
⎩
n is odd and a is even,
d
2, if n
is odd and a is odd,
d
ad + 1, if n is even.
d
Corollary 20.18.11.
At most one of n
d
may assume that n
d
(a m + 1)|(a n − 1) if and only if m|n and n
m
is even.
m and can be even, so without loss of generality we
d
is odd.
Start with
a 2d −1 = (a 2m −1, a 2n −1) = (a d m a − 1
−1)
ad − 1 · (am + 1), an − 1
ad − 1 · (an
+ 1)
= (a d m a − 1
−1)
ad − 1 · (am + 1), (a n
+ 1) by Theorem 20.18.5(ii) and Lemma 20.18.2.

862 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
Case 1. m
d odd. Divide the preceding equation by ad − 1, then multiply
and divide by ad + 1 to get:
a d + 1 = (a d m a − 1
+ 1)
ad − 1 · am + 1
ad + 1 , an + 1
ad
.
+ 1
This implies that
which implies that
1 =
(We need m
d odd to have am +1
ad +1
m a + 1
ad + 1 , an + 1
ad
,
+ 1
a d + 1 = (a m + 1, a n + 1).
∈ Z.)
Case 2. m
d even. Divide the equation preceding Case 1 by ad − 1. Then
2d|m, so (a2d−1)|(am−1), and n
d is odd, so by Lemma 20.18.2 (ad +1)|(an +1).
So we have
a d m a − 1
+ 1 =
ad − 1 · (am + 1), a n
+ 1 .
This implies that
which implies that
1 =
m a − 1
a2d − 1 · (am + 1), an + 1
ad
,
+ 1
1 =
a m + 1, an + 1
ad
.
+ 1
In Eq. 20.42 put x = a, n = d, so r = 0, and q = m is even by assumption,
d
to get:
Then
a m + 1 = (a d + 1)A + (−1) q (a 0 + (−1) q ) = (a d + 1)A + 2.
(a m + 1, a d + 1) = ((a d + 1)A + 2, a d + 1) = (2, a d + 1).
This means that dividing a d + 1 into a n + 1 removes at most a factor of 2 in
gcd(a m + 1, a n + 1), from which we see that (a m + 1, a n + 1) divides 2, i.e., it
equals 1 or 2 according as a is even or odd. Putting Cases 1. and 2. together
we have proved the following theorem.

20.18. GREATEST COMMON DIVISORS OF A M ± 1 AND A N ± 1 863
Theorem 20.18.12. At most one of m n , d d
(a m + 1, a n + 1) =
is even. Then
⎧
⎨ a
⎩
d + 1, if both m n
and are odd;
d d
1, if one of m n , is even and a is even;
d d
2, if one of m n , is even and a is odd.
d d
Theorems 20.18.5, 20.18.10 and 20.18.12 tell the whole story.
20.18.13 An Application to Finite Fields of Characteristic
2
Let q = 2 e , F = GF (q) ⊂ E = GF (q 2 ). Let λ be a primitive root for E,
β = λ k(q−1) , 1 ≤ k ≤ q. So 1 < |β| and |β| divides q +1. Write ¯ β = β q = β −1 .
Put δ = β + ¯ β. So δ ∈ F , but since gcd(q − 1, q + 1) = 1, it follows that
β ∈ F .
Problem: How big is the field GF (2)(δ)?
1. Put r = gcd(k, q + 1), so 1 ≤ r < q + 1. Then |β| = q+1
r .
2. GF (2)(β) is the smallest field GF (2 f ) for which q+1
r divides
2 f − 1.
3. Recall from above: If d = gcd(e, f), then
(2 e + 1, 2 f ⎧
⎨
− 1) =
⎩
1, if f
d
2 d + 1, if f
d
is odd
is even.
4. Since 1 < q+1
r and this must divide 2f − 1, it must be that f
d
so e
d is odd, implying that 2d + 1 divides 2e + 1.
5. 2e +1
r |(2 f − 1) iff 2e +1
r |(2 e + 1, 2 f − 1) iff 2e +1
r |(2 d + 1) with f
d
is equivalent to 2e +1
2 d +1
This implies
|r with e
d
odd and f
d even.
6. f = 2d where d is the smallest integer for which
(i) d|e,
(ii) e
d
is odd,
is even,
even, which

864 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
(iii) 2e +1
2 d +1 |r.
So we now know how big is the field GF (2)(β). Since β ∈ GF (2e ),
which contains δ, clearly β ∈ GF (2)(δ). The minimal polynomial for β over
GF (2)(δ) is x2 + δx + 1 = (x + β)(x + ¯ β), so GF (2)(δ) is the subfield of index
2 in GF (2)(β). This solves our problem:
Solution: The field GF (2)(δ) is the field GF (2d ) where d is the smallest
divisor of e for which e
d is odd and 2e +1
2d |r. In particular, if k and q + 1 are
+1
relatively prime, then GF (2)(δ) = GF (q).
Example: Let e = 6, so q = 64 and q + 1 = 65 = 5 · 13, q − 1 = 63 = 9 · 7,
q 2 − 1 = 4095.
e
(i) k = 15. In this case r = 5. odd implies that d = 2 or d = 6. If
d
d = 2, then 26 +1
22 = 13 does not divide r = 5, so it must be that d = 6. Hence
+1
GF (2)(δ) = GF (q).
(ii) k = 39. In this case r = (39, 65) = 13. Then with d = 2 we have
= 13 does divide r = 13. Hence GF (2)(δ) = GF (4).
2 d + 1 = 5 and 65
5
20.18.14 Application to Finite Fields of Odd Characteristic
Let p be an odd prime, q = p e , F = GF (q) ⊆ E = GF (q 2 ). Let λ be a
primitive root for E, β = λ k(q−1) , where k is determined modulo q + 1. Put
δ = β + ¯ β = β + β q = β + β −1 , since β q+1 = 1. Put r = (k, q + 1), so
|γ| = q+1
r .
1. β ∈ F iff |β| divides q − 1 iff q+1
divides q − 1 iff (q + 1)|r(q − 1) iff
r
q+1
2 |r q−1 q+1
iff 2 2 |r, since q+1 q−1
q+1
, = 1. So either r = and β = −1, or
2 2
2
r = q + 1 and β = 1. In either case F = F (δ) = F (β).
Now assume 1 ≤ k ≤ q with k = q+1
2 .
2. Since δ = β + ¯ β is unchanged if k is replaced with q + 1 − k, if β ∈ F
we may assume WLOG that 1 ≤ r < q+1
q+1
, i.e., 2 < ≤ q + 1. Since δ ∈ F
2 r
but β ∈ F , x2 = δx + 1 = (ξ − β)(x − ¯ β) is the minimal polynomial for β
over GF (p)(δ).
3. β ∈ GF (p f ) iff q+1
r |(pf − 1). Also,

20.19. LINEAR MAPS 865
(p e − 1, p f ⎧
⎨
− 1) =
⎩
2, if f
d
p d + 1, if f
d
is odd;
is even.
We want q+1
r |(pf − 1) with q+1
f
e
> 2, so must be even and must then
r d d
be odd. So pe +1
|(p r f − 1) iff (pe + 1)|r(pf − 1) iff pe +1
pd +1 |r
pf −1
pd
iff +1
pe +1
pd |r. All
+1
this proves the following theorem:
Theorem 20.18.15. GF (p)(δ) = GF (pd ) ⊆ GF (p2d ) = GF (p)(β) iff d is
the smallest positive divisor of e such that e
d is odd (so pd + 1 divides pe + 1),
and
e p + 1
pd
+ 1
divides r = (k, q + 1).
Note: If (k, q + 1) = 1, then GF (p)(δ) = GF (q).
20.19 Linear Maps
Until further notice F = GF (q n ) and K = GF (q), where q = p e is any
power of a prime and n is any positive integer. We may view F as a vector
space over K with dimension n. Then V = HomK(F, F ), the vector space
of all K-linear maps from F to F , is a vector space over K with dimension
n2 , so that |HomK(F, F )| = qn2. On the other hand, we can construct
elements of V = HomK(F, F ) in the following way. Let σ : x ↦→ xq , for
x ∈ F . So σ generates the Galois group Gal(F/K), i.e., σi : x ↦→ xqi, i = 0, 1, 2, . . . , n − 1, gives the distinct automorphisms of F that are the
identity on K. For each choice of ai ∈ F , 0 ≤ i ≤ n − 1, consider the map f :
x ↦→ f(x) = n−1 i=0 aiσi (x) = n−1 qi
i=0 aix . Clearly f ∈ HomK(F, F ). Because
each such polynomial f(x) has degree at most q n−1 , no two of them can
represent the same function. Hence there are (q n ) n such functions, showing
that each K-linear function from F to F must be of this form.
This proves the following theorem:
Theorem 20.19.1. Each f ∈ HomK(F, F ) is given by a linearized q-polynomial,
i.e.,a polynomial of the form

866 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
n−1
f(x) =
i=0
aix qi
, ai ∈ F. (20.44)
Consequently, if f(x) ∈ F [x] has degree at most q n − 1 and has the property
that f(a + b) = f(a) + f(b) for all a, b ∈ F , and f(ca) = cf(a) for all c ∈ K
and all a ∈ F , then f(x) is a linearized q-polynomial.
For ai ∈ F , 0 ≤ i ≤ n − 1, write
¯α = (a0, a1, . . . , an−1), and [¯α] =
For x ∈ F , put
[x] = (x, x q , x q2
, . . . , x qn−1
) T .
a qi
[j−i] . (20.45)
(0≤i,j≤n−1)
Then ¯α uniquely determines an element α of HomK(F, F ) by
It is easy to check that for each j,
It follows that
α : x ↦→ x α n−1
= aix qi
= ¯α[x].
(x α ) qj
n−1
=
i=0
i=0
a qj
i xqi+j
n−1
=
k=0
a qj
[k−j] xqk.
[x α ] = [¯α[x]] = [¯α][x]. (20.46)
Lemma 20.19.2. Let x1, x2, . . . , xr be elements of F . Then {x1, . . . , xr} is
linearly independent over K if and only if {[x1], . . . , [xr]} is linearly independent
over F .
Proof.
First suppose that {x1, . . . , xr} is linearly dependent over K, say
r
i=1 cixi = 0, with ci ∈ K and at least one of the ci not zero, 1 ≤ i ≤ r. Then
0 = r i=1 cqj i xqj i = r qj
i=1 cixi , from which it follows that (0, 0, . . . , 0)T
=
r
i=1 ci[xi], so that {[x1], . . . , [xr]} is linearly dependent over K, and hence
over F .

20.19. LINEAR MAPS 867
Now suppose that {[x1], . . . , [x]r} is linearly dependent over F , say
(0, 0, . . . , 0) T = r
i=1 ci[xi] with ci ∈ F . The goal is to show that each ci ∈ K,
so that the top row of the equation of the hypothesis says that 0 = r
i=1 cixi
with ci ∈ K. The proof proceeds by induction on r. First note that for
r = 1 the result is clear, and suppose that r = 2. Here we may assume that
[x1] = c[x2] with c ∈ F . Then the first row says that x1 = cx2 and its image
under x ↦→ x q says that x1 q = c q x2 q . The second row says that x1 q = cx2 q .
Hence c = x1/x2 = x1 q /x2 q = c q , which forces c to be in K, i.e., {x1, x2} is
linearly dependent over K. This shows that the desired result holds when
r = 2.
Suppose that for 1 ≤ r − 1 whenever {x1, . . . , xr−1} is linearly independent
over K, then {[x1], [x2], . . . , [xr−1]} is linearly independent over F , and
suppose that {x1, . . . , xr−1, xr} is linearly independent over K. In particular
we know that {[x1], [x2], . . . , [xr−1]} is linearly independent over F , so that if
some vector in F n is an F -linear combination of the [xi], 1 ≤ i ≤ r − 1, the
coefficients in the linear combination are unique.
Suppose r
i=1 ci[xi] = [0]. If cr = 0, then all the ci’s equal zero. Without
loss of generality we may assume that cr = 1, and that [xr] = r−1
i=1 di[xi],
di ∈ F . Looking at the second row of this equality we see xr q = r−1
i=1 dixi q .
But taking the qth power of the first row we find xr q = r−1
i=1 di q xi q . Since the
coefficients are unique when writing a vector in F n as an F -linear combination
of {[x1], . . . , [xr−1]}, it must be that di = di q , i.e., di ∈ K, for 1 ≤ i ≤ r − 1.
Hence the first row says that x1, . . . , xr are linearly dependent over K.
Theorem 20.19.3. Let ¯α = (a0, . . . , an−1) ∈ F n be given. The kernel
U = x ∈ F = GF (q n ) : x α n−1
= aix qi
= 0; ai ∈ F
is a subspace over K = GF (q). If U has dimension r (i.e., is a projective
(r−1)-dimensional subspace of F ), then the F -rank of the matrix [¯α] = (a qi
j−i ),
where the indices are taken modulo n, is equal to n − r.
Proof. It is easy to see that the set U of zeros of n−1 i=0
and any element of U is also a zero of
n−1
j=0
a qi
j xqi+j
n−1
=
j=0
i=0
a qi
[j−i] xqi,
qi aix forms a subspace

868 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
where the indices are taken modulo n. For all x ∈ U, the vector [x] in
V (n, q n ) is in the right null space of [¯α]. There are r elements of U linearly
independent over K, and therefore by Lemma 20.19.2 there are r vectors
of the form [x] linearly independent over F and in the right null space of
[¯α]. Hence the nullspace of [¯α] is a subspace over F of dimension at least r,
implying that the F -rank of [¯α] is at most n − r.
As before, r is the K-dimension of the kernel of α, so there are n − r
elements x1 α , . . . , xn−r α in the image of α that are K-independent, so that
by Lemma 20.19.2
[x1 α ] = [¯α][x1], . . . , [xn−r α ] = [¯α][xn−r]
are F -linearly independent. This says the F -rank of B is at least n − r.
Hence the F -rank of B must be exactly n − r.
Corollary 20.19.4. The additive map α is one-to-one and onto if and only
if [¯α] is nonsingular, and in general det[¯α] ∈ K.
Proof. The first part of the Corollary is contained in the preceding theorem.
Now put ∆ = det[¯α]. Then x ↦→ x q is an automorphism of F fixing K
pointwise, so ∆ q = det([¯α] q ) = det(B), where B is obtained from [¯α] by
shifting rows one place up and columns one place left. Hence detB = det[¯α],
i.e., ∆ = ∆ q , implying that ∆ ∈ K. This completes the proof.
Note:
[α ◦ µ][x] = [x α◦µ ] = [(x α ) µ ] = [¯µ · [x α ]] = [¯µ · [¯α][x]] = [¯µ][¯α][x].
This says ([α ◦ µ] − [¯µ][¯α])[x] = 0 for all x ∈ Fq. Let ¯ρ = (r0, . . . , re−1)
be the ith row of [α ◦ µ] − [¯µ][¯α]. So e−1 pi
i=0 rix = 0 for all x, i.e., the map
ρ : x ↦→ ¯ρ[x] is trivial. This implies ¯ρ = ¯0. Hence it follows that
[α ◦ µ] = [¯µ][¯α]. (20.47)
At this point we give three lemmas of B. F. Sherman.
Lemma 20.19.5. If f(x) and g(x) are linearized q-polynomials and f(x)
divides g(x), then there is a linearized q-polynomial r(x) such that g(x) =
r(f(x)).

20.19. LINEAR MAPS 869
Proof. Use induction on the degree of g(x). Suppose that
and
Then
f(x) = plx ql
+ pl−1x ql−1
+ . . . p1x q + p0x
g(x) = gmx qm
+ gm−1x qm−1
+ · · · g1x q + g0x.
t(x) = g(x) − (gm/(pl) qm−l
)(f(x)) qm−l
is a linearized q-polynomial divisible by f(x), and of degree less than the
degree of g(x) (or t(x) is the zero polynomial). Hence by the induction
hypothesis, we can assume that t(x) = w(f(x)), for some q-linearized polynomial
w(x), so that
Put
to give
g(x) = (gm/(pl) qm−l
)(f(x)) qm−l
+ w(f(x)).
r(x) = gm/(pl) qm−l
)x qm−l
+ w(x)
g(x) = r(f(x)).
If, on the other hand, t(x) = 0, then g(x) = (gm/(pl) qm−l)(f(x))
qm−l
r(f(x)) with r(x) = (gm/(pl) qm−l).
Lemma 20.19.6. If a linearized q-polynomial p(x) is fully reducible (into
linear factors) with no repeated roots, then there is a fully reducible linearized
q-polynomial s(x) such that
x qn
− x = s(p(x)) = p(s(x)).
Clearly s(x) is also fully reducible with no repeated roots.
Proof. If p(x) is fully reducible with no repeated roots, then it divides xqn −x,
so Lemma 20.19.5 says there is a linearized q-polynomial s(x) such that
s(p(x)) = xqn − x. If p(x) − a = 0 (a = 0) has one root, then it has the same
number of roots as p(x) = 0, so it is also fully reducible with no repeated
roots. Let {a0 = 0, a1, . . . , ak−1} be the image of the K-linear map a ↦→ p(a).
Then
x qn
− x = p(x)(p(x) − a1)(p(x) − a2) · · · (p(x) − ak−1).
=

870 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
Put s(x) = x(x−a1)(x−a2) · · · (x−ak−1). So xqn −x = s(p(x)), and s(x)
is fully reducible with no repeated roots and thus divides xqn −x. Hence there
is a linearized q-polynomial g(x) for which xqn − x = g(s(x)). Then g(x) qn −
g(x) = g(xqn − x) = g(s(p(x)) = p(x) qn − p(x). Put r(x) = g(x) − p(x). The
previous computation shows that r(x) qn = r(x), which forces r(x) = 0, i.e.,
g(x) = p(x). So s(g(x)) = s(p(x)) = xqn − x = g(s(x)) = p(s(x)).
Lemma 20.19.7. If the linearized q-polynomial p(x) is fully reducible with
no repeated roots, and of degree q m , then p(x) − ax, a = 0, has at most q n−m
roots.
Proof. As p(x) is of degree qm , by Lemma 20.19.6 there is a fully reducible
linearized q-polynomial s(x) such that xqn − x = s(p(x)). So s(x) has degree
qn /qm , and being fully reducible has exactly qn−m roots. But s(p(x) − ax) =
s(p(x)) − s(ax) = xqn − x − s(ax). Here s(p(x) − ax) = 0 has all the roots of
p(x) − ax = 0 (and possibly others) as roots, the right hand side has exactly
the qm−n roots of s(ax) as its roots. So p(x) − ax = 0 could not have more
than qn−m roots.
We are going to determine those linearized polynomials that give points,
lines and hyperplanes of P G(n − 1, q).
20.20 K-Subspaces of F as Linearized Polynomials
Given a subspace U of F of dimension r over K, the polynomial
(x − u)
u∈U
is a monic polynomial of degree q r of the form
x qr
r−1
+
j=0
ajx qj
.
We are in a position to calculate necessary and sufficient conditions on the
coefficients to determine when such a (linearized) polynomial determines a
subspace of dimension r. Note: They are precisely the linearized polynomials
which are factors of the polynomial

20.20. K-SUBSPACES OF F AS LINEARIZED POLYNOMIALS 871
x qn
− x =
(x − u).
u∈F
As a first example we consider the case r = 1.
Example 1. Points of P G(n − 1, q). Here the linearized polynomial has
the form xq − ax, but xq − ax = 0 might have no solution x = 0. If there is
such an x, clearly all solutions are given by cx for c ∈ K. Moreover, xq−1 = a.
By Theorem 20.2.2 we know that for nonzero a ∈ F , xq−1 = a has a solution
x if and only if NF/K(a) = 1, in which case it has q − 1 solutions as desired.
Moreover, there are exactly qn−1 solutions a, accounting for exactly all the
q−1
points of P G(n − 1, q).
Example 2. Hyperplanes of P G(n − 1, q). Here the linearized polynomial
has the form
x qn−1
+ an−2x qn−2
+ · · · + a1x q + a0x.
By Theorem 20.19.3, if ¯α = (a0, a1, . . . , an−2, 1), we want the matrix
⎛
⎜
[¯α] = ⎜
⎝ .
a qn−1
1
a0 a1 a2 · · · an−2 1
1 a q
0
a q
1 · · · a q
n−3 a q
n−2
a q2
n−2 1 a q2
0 · · · a q2
n−4 a q2
n−3
.
a qn−1
2
.
.
a qn−1
3 · · · 1 a qn−1
0
to have rank 1. It is clear that each row is nonzero, so each row spans the row
space. In particular the first row must be a0 times the second row. Writing
; 1 =
this out leads to a1 = a 1+q
0 ; a2 = a 1+q+q2
0 ; . . . an−2 = a 1+q+...qn−2
0
= a0
. This implies that a0 = bq−1 for some b ∈ F ∗ .
The number of distinct a0’s of this form is qn−1 , which is also the number of
q−1
hyperplanes. Hence each hyperplane arises from a polynomial of the form
a q
n−2a0 = a 1+q+···+qn−1
0
f(x) = a0x + a 1+q
q n −1
q−1
0 xq + a 1+q+q2
0
x q2
.
.
⎞
⎟
⎠
+ · · · + a 1+q+···+qn−2
0 x qn−2
+ x qn−1
,
where NF/K(a) = 1.
Example 3. Lines of P G(n − 1, q). Here the linearized polynomial has
the form

872 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
x q2
+ a1x q + a0x.
By Theorem 20.19.3, if ¯α = (a0, a1, 1, 0, 0, · · · , 0), then the matrix [¯α]
should have rank n − 2 (i.e., nullity 2), where
⎛
⎞
⎜
[¯α] = ⎜
⎝
a0 a1 1 0 0 · · · 0
0 a q
0
0 0 a q2
0
.
.
a q
1 1 0 · · · 0
.
0 · · · 0 0 a qn−3
0
a q2
1 1 · · · 0
.
.
.
a qn−3
1
1 0 · · · 0 0 a qn−2
0
.
1
a qn−2
⎟ .
⎟
1 ⎠
a qn−1
1 1 0 · · · 0 0 a qn−1
0
When n = 3, a line is just a hyperplane, so Example 2 shows that the
general line is given by the polynomial f(x) = a0x + a 1+q
0 xq + xq2 where
1 = NF/K(a0) = a 1+q+q2
0 . For this example we just work out the case n = 4.
So to avoid subscripts we write f(x) = ex+cx q +xq2 and we want the matrix
[¯α] to have rank 2, where
⎛
e c 1 0
⎜
[¯α] = ⎜
⎝
0 eq cq 1 0 e
1
q2 cq2 cq3 1 0 eq3 ⎞
⎟
⎠ .
It is clear that the first two rows are independent, so we just need rows 3
and 4 to be linear combinations of the first two rows. Write
(1, 0, e q2
, c q2
) = λ1(e, c, 1, 0) + λ2(0, e q , c q , 1) = (λ1e, λ1c + λ2e q , λ1 + λ2c q , λ2),
and
(c q3
, 1, 0, e q3
) = λ3(e, c, 1, 0) + λ4(0, e q , c q , 1) = (λ3e, λ3c + λ4e q , λ3 + λ4c q , λ4).
Using the first equation we find that λ1 = e −1 and λ2 = c q2
have 0 = e−1c + cq2eq , which is equivalent to
. Then we
e 1+q = −c 1−q2
. (20.48)

20.20. K-SUBSPACES OF F AS LINEARIZED POLYNOMIALS 873
Also, e q2
= e −1 + c q2 +q , which is equivalent to
Using the second equation we find λ4 = eq3 + eq3 +q , which is equivalent to
have 1 = e −1 c 1+q3
Similarly, 0 = e −1 c q3
e(e q2
− c q2 +q
) = 1. (20.49)
and λ3 = e−1cq3. Then we
e = c 1+q3
+ e q3 +q+1
. (20.50)
+ eq3cq , which is equivalent to
c q3 −q = −e q 3 +1 . (20.51)
It is routine to check that Eq. 20.51 is equivalent to Eq. 20.48, so we need
only Eqs. 20.48, 20.49 and 20.50.
Eq. 20.48 is equivalent to
Eq. 20.49 is equivalent to
And, Eq. 20.50 is equivalent to
e 1+q = −c 1+q)(1−q) .
c q+1 = e q − e −q3
.
c q+1 = e q − e 1+q+q2
.
From the latter two equations we find that 1 = e q3 +q 2 +q+1 = NF/K(e).
The steps we used to arrive at the final form are reversible, so that we have
proved the following:
Theorem 20.20.1. The lines of P G(3, q) are given by the q-linearized polynomials
of the form f(x) = xq2 + cxq + ex, where eq3 +q2 c
+q+1 = 1 and
q+1 = eq − eq2 +q+1 .
We note that there are (1+q 2 )(1+q +q 2 ) lines of P G(3, q). If c = 0, then
eq = eq2 +q+1 q is equivalent to 1 = e 2 +1 2 , which has q +1 solutions, accounting
for q2 + 1 lines. There are (q3 + q2 + q + 1) − (q2 + 1) = q3 + q values of
e for which NF/K(e) = 1 but eq2 +1 = 0. Consider any e of this class. A
routine computation shows that (eq − e1+q+q2) (q2 +1)(q−1) 2 = (−1) = 1. This
means that e q − e 1+q+q2
is really a (q + 1)-st power, in q + 1 ways. This gives
(q 3 + q)(q + 1) lines, so that we have all q 2 + 1 + (q 2 + 1)(q 2 + q) lines of
P G(3, q).

874 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
20.21 Additive Maps on Fq
Theorem 20.21.1. Let α and β be permutations of Fq such that 0 α = 0 = 0 β
and such that x ↦→ x α /x β permutes the nonzero elements of Fq. Then p = 2.
Proof. Let ζ be a generator of the multiplicative group F ∗ q , i.e., a primitive
i α
root for Fq. Then α and β are given by ζ ↦→ ζ i′ i β
, ζ ↦→ ζ i′′ , where i ↦→ i ′ and
i ↦→ i ′′ are permutations of the integers 1, 2, . . . , pe − 1. If pe − 1 = |F ∗ q | = 2k,
then 1 + 2 + · · ·+ 2k = k(2k + 1) ≡ k (mod 2k). Thus 0 =
1≤i≤2k (i′ − i ′′ ) ≡
i ≡ k (mod 2k), an impossibility.
1≤i≤2k
Corollary 20.21.2. Let α and β be additive permutations of the elements of
Fq for which x ↦→ x α /x β permutes the nonzero elements of Fq. Then p = 2.
For the remainder of this section Fq = GF (2 e ).
Theorem 20.21.3. Let α and β be additive permutations of the elements of
Fq = GF (2 e ) that fix 1 and for which xmapstox α /x β permutes the nonzero
elements of Fq. Then α −1 β is an automorphism of Fq of maximal order e.
Proof. Let α and β satisfy the hypotheses of the theorem. Then there are
scalars ai, bi in Fq, 0 ≤ i ≤ e − 1, for which
e−1
α : x ↦→
i=0
aix 2i
; β : x ↦→
e−1
i=0
bix 2i
.
Let A = [¯α] = (aij), so aij = a2i−1 [j−i] , and B = [ ¯ β] = (bij, so bij = b2i−1 [j−i] ,
1 ≤ i, j ≤ e.
Since α and β are permutations, by Lemma 20.19.4 both A and B are
nonsingular. Since x ↦→ xα /xβ is a permutation of the elements of F ∗ q , for
each c ∈ F ∗ q there must a be a (unique) nonzero solution x to xα − cxβ = 0.
Hence e−1 2i
i=0 (ai−λbi)x = 0 has a unique nonzero solution x for each λ ∈ F ∗ q .
By Cor. 20.19.4, the matrix
Cλ = ((a[j−i] − λb[j−i]) 2i−1
), 1 ≤ i, j ≤ e,
has zero determinant for each λ ∈ F ∗ q . And 0 = detA · detB, so detA =
detB = 1. So det Cλ is a polynomial in λ of degree 2e − 1 with constant term
1, leading term 1, and having each nonzero element of Fq as a simple root.
This implies

20.21. ADDITIVE MAPS ON FQ 875
detCλ = λ 2e −1 + 1. (20.52)
For 1 ≤ t ≤ 2e − 2, we now calculate the coefficient of λt in detCλ and
set it equal to 0.
Let ti1, ti2, . . . , tir be the nonzero coefficients in the binary expansion of
t = e−1 i=0 ti2i . Then the coefficient of λt in detCλ is easily seen to be the determinant
of the matrix obtained by replacing rows ti1, . . . , tir of A with rows
ti1, . . . , tir of B Hence we know the following: the rows of A are independent,
the rows of B are independent, and any set of rows formed by taking some
r rows of A and the complementary e − r rows of B is a linearly dependent
set, 1 ≤ r ≤ e − 1. In particular, the ith row of B is a linear combination of
rows 1, . . . , i − 1, i + 1, . . . , e of A. Let βi be the ith row of B. Then
βi = (b 2i−1
[1−i], . . . , b 2i−1
[e−i]).
Then there are scalars d1, . . . , di−1, di+1, . . . , de such that
βi = (d1, . . . , di−1, 0, di+1, . . . , de)A. (20.53)
Apply the automorphism x ↦→ x 2 to Eq. 20.53.
(b 2i
[1−i], . . . , b 2i
[e−i]) = (d 2 1, . . . , d 2 i−1, 0d 2 i+1, . . . d 2
e) a 2k
j−k] . (20.54)
1≤k,j≤e
In Eq. 20.54 first permute the columns cyclically, moving column j (on
the left-hand side) to position j + 1, then permute rows on the far right
moving row k to position k + 1. The result is an equation that says:
βi+1 = (d 2 e, d 2 1, . . . , d 2 i−1, 0, d 2 i+1, . . . , d 2 e−1)A. (20.55)
Repeating this k times we obtain
βi+k = (d 2k
1−k, d 2k
2−k, . . . , d 2k
e−k)A, (20.56)
where the d ′ j s are unique and di = 0. By permuting rows cyclically we may
adopt a new notation:
β1 = (d1, . . . , de)A with d1 = 0, so βi+1 = (d 2i
1−i , . . . , d2i e−i )A. (20.57)

876 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
Let αi denote the i th row of A. For some λ1, λ2 in Fq, not both zero, we
have
λ1β1 + λ2β2 =
e
j=3
fjαj = (λ2d 2 e , λ1d2, λ1d3 + λ2d 2 2 , . . . , λ1dj + λ2d 2 j−1 , . . .)A.
(20.58)
Hence as the rows of A are independent, λ2d 2 e = 0 = λ1d2. If λ1 = 0, then
d2 = 0. If λ2 = 0, then de = 0. Since d1 = 0, we have either both de and d1
are zero, or both d1 and d21 are zero, i.e., there is a string of length two of
consecutive d ′ is (including d1) equal to zero. By changing notation we may
suppose d1 = d2 = 0 and complete the rpoof by a finnite induction.
Suppose there is a string of t zeros, say d1 = · · · = dt = 0, 1 ≤ t ≤ e − 2.
We then wish to show that there is a astring of t + 1 zeros. This yields the
following.
β1 = (0, . . . , 0, dt+1, . . . , de)A;
β2 = (d 2 e , 0, . . . , 0, . . . , d2 e−1 )A;
.
βt = (d2t−1 e+2−t , . . . , d2t−1 e
, 0, . . . , 0, d 2t−1
↑
t
↑
2t − 1
t+1
βt+1 = (d 2t
e+1−t, . . . , d 2t
e , 0, . . . , 0, . . . , d 2t
↑
t + 1
↑
2t
, . . . , d2t−1
e+1−t )A;
e−t)A.
(20.59)
Moreover, there are scalars λ1, . . . , λt+1, not all zero, for which t+1 i=1 λiβi
is some linear combination of αt+2, . . . , αe, since 1 < t + 1 < e. Use Eq. 20.53
to calculate the coefficients on α1, . . . , αt+1 (which must be zero) in t+1 i=1 λiβi,
starting with column t + 1 of Eq. 20.53 and working backwards to column 1.
0 = λ1dt+1
0 = λt+1d2t 0 =
e
λtd2t−1 0 =
2t
e + λt+1de−1 λt−1d2t−2 .
2t−1
2t
e + λtde−1 + λt+1de−2 0 = λ2d2 22
2t
e + λ3de−1 + · · · + λt+1de−(t−1) (20.60)

20.21. ADDITIVE MAPS ON FQ 877
If λt+1 = 0, then de = 0, giving a string of zeros of length t + 1.
If λt+1 = 0, λt = 0, then again de = 0.
If λt+1 = λt = 0, λt−1 = 0, then again de = 0.
Continue in this fashion. . . .
If λt+1 = λt = · · · = λ3 = 0, λ2 = 0, then again de = 0.
If λt+1 = · · · = λ2 = 0, λ1 = 0, then dt+1 = 0, giving a string of zeros of
length t + 1.
It follows that only one di can be nonzero, say d = dt = 0, t = 1. This
says:
b[j] = da 2t−1
[j−(t−1)] , 0 ≤ j ≤ e − 1. (20.61)
Our assumption that 1 = 1 α = 1 β implies that d = 1. Put u = t − 1, so
Eq. 20.55 becomes:
b[j] = a 2u
[j−u] . (20.62)
If x ↦→ x2u is an automorphism of F of order n < e, it is routine to find
n rows of B which together with the complementary e − n rows of A form
a linearly independent set. Hence it must be that gcd(u, e) = 1. Moreover,
Eq. 20.56 is equivalent to
This completes the proof.
x β = (x α ) 2u
⇒ β = α · 2 u . (20.63)
Corollary 20.21.4. If β is an additive permutation of the elements of Fq
for which x ↦→ x/xβ permutes the nonzero elements of Fq. Then β has the
form xβ = dx2u for fixed d ∈ F ∗ q , (u, e) = 1.
Let α and β b e additive permutations of Fq, q = 2 e . The pair (α, β) is
said to be admissible provided the following condition holds:
0 =
3
vi =
i=1
3
vizi =
i=1
3
i=1
v α i zβ
i for distinct, nonzero vi ∈ Fq implies that
z1 = z2 = z3 = 0.
(20.64)
Theorem 20.21.5. Let (α, β) be a pair of additive permutations of the elements
of Fq fixing 1. Then the following are equivalent:

878 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
(i) The pair (α, β) is admissible.
(ii) 0 = 2
1 vizi = 2
1 vα i zβ
i for distinct, nonzero v1, v2 implies z1 = z2 = 0.
(iii) For each c ∈ F ∗ q , the map µc : v ↦→ v α (c/v) β is a permutation of the
elements of F ∗ q .
(iv) For each c ∈ F ∗ q , the map λc : w ↦→ (cw) α /w β is a permutation of the
elements of F ∗ q .
(v) α and β are automorphisms of Fq for which α◦β −1 is an automorphism
of maximal order.
Proof. First let (α, β) be admissible. Add 0 = ( 3 1 vi)z3 to the second summation
of Eq. 20.64 and 0 = ( 3 1 vα i )zβ 3 to the third, so that Eq. 20.64
becomes
0 =
2
vi(zi+z3) =
1
2
1
v α i (zβ
i +zβ
3 ) for distinct nonzero v1, v2 ⇒ z1 = z2 = z3 = 0.
Now put wi = zi + z3, i = 1, 2, so that Eq. 20.65 becomes
0 = v1w1 + v2w2 = v α 1 w β
1 + v α 2 w β
2
(20.65)
(20.66)
for distinct nonzero v1, v2 implies that w1 = w2 = 0.
This is just part (ii) of the theorem. Put wi = c/vi in Eq. 20.66 to obtain
(iii) of the theorem. Put v1 = cw2, v2 = cw1 in Eq. 20.66 to obtain (iv) of
the theorem. It follows that (i) – (iv) are equivalent. The crux of the proof
is to show that (iv) implies (v).
So let (α, β) be a pair of additive permutations of the elements of Fq
fixing 1 and satisfying (iv). For 0 = c ∈ Fq, let αc denote the additive
permutation αc : x ↦→ (cx) α for all x ∈ Fq. Then β = αcγc = δcαc for
unique additive permutations γc and δc. For λc as in (iv), λc = αc ◦ (1 − γc),
implying that 1 − γc : w ↦→ w/wγc is also a permutation of the elements of
F ∗ q . By Cor. 20.21.4 it follows that γc : x ↦→ dcxβc for some nonzero scalar
dc and some automorphism βc : x ↦→ x2tc , (tc, e) = 1, 1 ≤ tc ≤ e − 1. As
1 = 1β = 1αcγc , dc is easily calculated so that
x β = ((cx) α /c α ) 2tc
for x, c ∈ Fq, , c = 0. (20.67)

20.21. ADDITIVE MAPS ON FQ 879
In particular, let t = t1, so Eq. 20.67 implies the following:
β = α · 2 t . (20.68)
It is easy to check that (α, β) is an admissible pair of additive permuta-
tions if and only if (α −1 , β −1 ) is. Hence β −1 = αc −1 δc −1 implies that δc −1 (and
hence δc) has the same form as γc, i.e., δc : x ↦→ ¯ dcx2gc for some nonzero scalar
¯dc, and (gc, e) = 1, 1 ≤ gc ≤ e − 1. Then 1 = 1β = 1δcαc = ( ¯ dc · 1) αc = (c ¯ dc) α
implies ¯ dc = c−1 , from which it follows that xβ = xδcαc = (c−1x2gc ) αc = x2gc α ,
i.e., βα−1 = 2gc = 2g for all c.
Hence
and
Then
x β = ((cx) α /c α ) 2tc
β = 2 g · α = α · 2 t , (20.69)
α = 2 g · α · 2 −t . (20.70)
= ((c2gx2g ) α·2−t/(c2g
) α·2−t)2tc
= ((dx2g) α /dα ) 2td · 2−t+tc−td, where d = c2g = (x 2g ·β ) 2 −t+tc−t d .
β = 2 g β2 −t+tc−td (20.71)
Now use Eq. 20.68 in Eq. 20.71 along with Eq. 20.69 to get
tc = td if d = c 2g
. (20.72)
Since x ↦→ x2g is an automorphism of maximal order, it follows that if c
and d are nonzero conjugates then tc = td.
Now suppose that c and d are distinct, nonzero elements of Fq for which
= ((dx) α /dα ) 2tc implies
tc = td. We claim tc+d = tc. Then x β = ((cx) α /c α ) 2tc
Then
x β = [((c + d)x) α /(c + d) α ] 2t c+d
(dx) α = d α (cx) α /c α , if tc = td. (20.73)
= [((cx) α + (dx) α /(c + d) α ] 2tc+d =

880 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
for all x ∈ Fq.
= [((cx) α + d α (cx) α /c α )/(c + d) α ] 2t c+d
Comparing this last form of x β with Eq. 20.67 we obtain
tc+d = tc when tc = td. (20.74)
By the Normal Basis Theorem for cyclic extensions (or for finite fields)
there is an element c ∈ Fq for which c, c2 , c4 , . . . , c2e−1 form a linear basis
over the prime subfield {0, 1}. As tc = td for d ∈ {c2 , c22, . . . , c2e−1} and for
d equal to any nonzero sum of these elements, it follows that there is only
t : t = tc for all c = 0. Put x = c−1 in Eq. 20.73 to obtain
d α d
= c
α
cα for
all nonzero c, d ∈ Fq, implying that α preserves multiplication. Hence a is
an automorphism of Fq. This completes the proof that (iv) implies (v). The
converse is easy.
20.22 The Number of Roots of a Special Polynomial
Let e = dr, i = dj, (r, j) = 1, F = GF (p d ), E = GF (p e ). For b, c ∈ E our
problem is to determine how many solutions there are to the equation
f(x) = x α − bx − c = 0, where α : x ↦→ x pi
.
Note that α r is the identity on E.
As a first step recall the following:
gcd(p m − 1, p n − 1) = p gcd(m,n) − 1. (20.75)
Secondly, put q = p n and let N be a positive integer. Then the number
of solutions of x N − 1 = 0 in GF (q) is gcd(N, q − 1). If N divides q − 1 and
s is a primitive root of GF (q), then
The solutions of X N = 1 are powers of s k , where k =
q − 1
. (20.76)
N
For Step 3, consider HomF (E, E). Clearly E is an r-dimensional vector
space over F . Then Lα : E → E : x ↦→ xα − x defines an F -linear transformation
of E whose kernel is F . If β : x ↦→ xpd, then β generates Gal(E/F )
and 〈β〉 = 〈α〉.

20.22. THE NUMBER OF ROOTS OF A SPECIAL POLYNOMIAL 881
Also Dα : E → E : x ↦→ x + xα + xα2 + · · · + xαr−1 defines an element
of HomF (E, E), and Dα(xα ) = Dα(x) = trE/F (x). Hence the image of Lα is
contained in the kernel of Dα. Since 1 = dim ker(Lα), r − 1 = dim Im(Lα),
so either Dα = 0 (which is impossible since 1, α, . . . , αr−1 are linearly independent)
or Im(Lα) = ker(Dα). Since [Dα(x)] α = Dα(xα ) = Dα(x), Dα(x)
is in the fixed field of α, i.e., Im(Dα) ⊆ F . Hence dim Im(Dα) = 1 implies
Im(Dα) = F . If xα − x − δ = 0 = yα − y − δ, then (x − y) α = x − y, so
y = x + t, t ∈ F . Hence we have proved the following:
x α − x − δ = 0 has a solution in E iff Dα(δ) = 0, (20.77)
in which case it has exactly p d = |F | solutions.
For step 4, consider the general equation
and
Suppose f(t) = 0, so t α = bt + c. Then
t α2
f(x) = x α − bx − c = 0. (20.78)
= b α t α + c α = b α (bt + c) + c α = tb 1+α + cb α + c α ,
t α3
= t α b α+α2
+ c α b α2
+ c α2
=
= tb 1+α+α2
+ cb α+α2
+ c α b α2
Continuing in this fashion we find t = t αr
= tb 1+α+···+αr−1
+ cb α+···+αr−1
+ c α b α2 +···+αr−1 =
+ c α2
.
+ · · · + c αr−2
b αr−1
+ c αr−1
.
Put tαk where
= bkt+ck, c = c1, b = b1. So t = tbr+cr implies t(1−br)−cr = 0,
br = b 1+α+···+αr−1
= NE/F (b),
and
cr = cb α+α2 +···αr−1 + c α b α2 +···αr−1 We can now state the general result:
+ · · · + c αr−2
b αr−1
+ c αr−1
.
(i) If NE/F (b) = 1, then t = cr
1−br
is the unique root of f(x) = 0.
(ii) If NE/F (b) = 1 and cr = 0, there is no solution.
(iii) If NE/F (b) = 1 and cr = 0, then f(x) = 0 has p d distinct roots.

882 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
It is routine to show that in case (i) t = cr
1−br
clearly (i) and (ii) hold.
is indeed a root, and then
Now suppose that NE/F (b) = 1 and cr = 0. Then 1 = b αr −1
α−1 = b N , where
(since α = p i and i = dj)
N = pdjr − 1
p dj − 1 .
Since gcd(pdjr − 1, pdr − 1) = pdr − 1 and gcd(pdj − 1, pdr − 1) = pd − 1, so
gcd(N, pdr − 1) = pdr−1 pd . In the context of the paragraph preceding Eq. 20.76
−1
k =
(p dr − 1)
(p dr − 1)/p d − 1) = pd − 1.
So by Eq. 20.76 b = spd−1 for some s ∈ E. Since
map
is a permutation and s = t
„
p
dj
−1
pd «
−1
x ↦→ x
„
p
dj
−1
pd «
−1
implies that
b = t pdj −1 = t α−1 = 0.
pdj −1
pd−1 , pdr
− 1 = 1, the
Solve for g such that c = gt α and put x = ty. Then f(x) = x α − bx − c =
t α y α − t α y − gt α = t α (y α − y − g) = 0 iff G(y) = y α − y − g = 0.
Use c = gt α , b = t α−1 in the formula for cr:
cr = cb α+···+αr−1
+ c α b α2 +···+αr−1 + · · · + c αr−2
b αr−1
+ c αr−1
= gt α t αr −α + g α t α 2
t (α−1)α2 (1+α+···+α r−3
+ · · ·
= gt αr
+ g α t αr
+ · · · + g αr−1
t αr
= tDα(g).
Hence cr = 0 iff Dα(g) = 0 iff G(y) = 0 has solutions (and then p d of
them) iff f(x) = 0 has solutions (and then p d of them).
20.23 Equations Involving Squares
In this section we assume that q = p e with p an odd prime. We adapt some
of the material in Dickson [Di58].

20.23. EQUATIONS INVOLVING SQUARES 883
Theorem 20.23.1. Let a, b, k be nonzero elements of Fq.
The number of solutions (x, y) to ax 2 + by 2 = k is
q − 1, if −ab = ;
q + 1, if −ab = ❆ .
Proof. Case 1. −ab = c 2 . Put ax + cy = ρ; ax − cy = σ. So
For σ ∈ F ∗ q
1
(ρ + σ) = ax;
2
1
(ρ − σ) = y.
2c
ρ · σ = a 2 x 2 − c 2 y 2 = a(ax 2 + by 2 ) = ak = 0.
ak put ρ = . Then
σ
x = 1 ak + σ2 1 ak − σ2
(ρ + σ) = ; y = (ρ − σ) =
2a 2aσ 2c 2cσ .
This gives all q − 1 solutions.
Case 2. −ab = ❆ . Choose i ∈ F q 2 such that i 2 = −ab. The roots of
x 2 + ab = 0 are i and −i = i q . Then
ak = (ax) 2 + aby 2 = (ax + iy)(ax − iy) = (ax + iy) q+1 .
Put z = ax + iy. Let ζ be a primitive root of Fq2. Then ak = ζ tinFq, so
t ≡ 0 (mod q + 1). Say t = ℓ(q + 1), and z = ax + iy = ζ s for some s. Then
zq+1 = ak becomes ζ s(q+1) = ζ ℓ(q+1) , which is equivalent to s ≡ ℓ (mod q −1).
So s = ℓ + r(q − 1), 0 ≤ r ≤ q. This gives q + 1 solutions z to zq+1 = ak.
Since {a, i} for an Fq-basis for Fq2, each solution z determines x and y such
that z = ax + iy.
Now suppose that a, b, c ∈ F ∗ q and △ = b 2 − 4ac = 0. We determine how
many x there are for which ax 2 + bx + c = (= 0).
First, how many solutions (x, y) are there to ax 2 + bx + c = y 2 ? This
equation is equivalent to
4a 2 x 2 + 4abx + 4ac = 4ay 2
⇐⇒
(2ax + b) 2 − a(2y) 2 = △ = 0.
Put A = 2ax + b and B = 2y. By the preceding theorem we have the
following:

884 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
Corollary 20.23.2. If a = , there are q−1 solutions (A, B) to A 2 −aB 2 =
△.
If △ = , two of these solutions have B = 0 (i.e., y = 0). But q − 3 of
them have B = 0. In this case we find q−3
2 values of x with ax 2 + bx + c =
= 0.
If △ = ❆ , then y = 0 and for each ±y, 2ax + b = A gives q−1
2 solutions
x with ax 2 + bx + c = , leaving q−1
2 values of x with ax 2 + bx + c = ❆ .
If a = ❆ , there are q + 1 solutions (A, B) to A2 − aB2 = △.
If △ = , two solutions have B = 0, (accounting for 2 values of x),
leaving q − 1 solutions with B = 0. Hence we have q−1
2 values of x with
ax 2 + bx + c = = 0, and q+1
2 values of x with ax2 + bx + c = ❆ .
If △ = ❆ , then B = 0, so q+1
2 values of x have ax 2 + bx + c = = 0,
leaving q−1
2 values of x with ax2 + bx + c = ❆ .
Now put S = |{σ 2 = 0 : σ 2 + 1 = = 0}|. Note: If −1 = , say
−1 = c 2 , then σ 2 = c 2 = −1 is not counted in S.
Put N = |{τ 2 : τ 2 + 1 = ❆ }|.
If −1 = , so q ≡ 1 (mod 4), then S + N = q−1
2
− 1 = q−3
2 .
If −1 = ❆ , so q ≡ 3 )mod 4), then S + N = q−1
2 .
The number of solutions (x, y) to x 2 − y 2 = 1 is q − 1 by Case 1 of
Theorem 20.23.1. The solutions are of three types:
(i) y = 0, x = ±1. (2 solutions)
(ii) y 2 = −1, x = 0. (2 solutions when −1 = ; 0 solutions when
−1 = ❆ .
(iii) y 2 = α, x 2 = α + 1, 4S solutions.
It follows that q − 1 = 2 + 2 + 4S when −1 = ; q − 1 = 2 + 4S when
−1 = ❆ . Putting this together we obtain the following corollary.
Corollary 20.23.3. When q ≡ 1 (mod4), S = q−5
4
When q ≡ 3 (mod 4), S = q−3
4
and N = q+1
4 .
20.24 Vandermonde Determinant
and N = q−1
4 .
Let t1, . . . , tn be commuting indeterminants over K, and let A be the n × n
matrix whose entries are from the commutative ring K[t1, . . . , tn] defined by

20.25. A THEOREM OF BLOKHUIS 885
Then
⎛
⎜
A = ⎜
⎝ .
1 t1 t 2 1 · · · t n−1
1
1 t2 t 2 2 · · · t n−1
2
.
.
1 tn t 2 n · · · t n−1
n
det A =
1≤j

886 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
Hence the lines are partitioned into two classes S and N (for square and
nonsquare). Through 0 there are q+1
lines of S and 2 q+1
lines of N. Now let
2
ℓ be a line of S not through 0, so the line ℓ ′ through 0 parallel to ℓ is also in
S. Thus the q+1
lines through 0 in N meet ℓ in nonsquares, and the other
2
q−1
lines through 0 meeting ℓ are in S and meet ℓ in squares. So ℓ has 2 q+1
2
nonsquares and q−1
squares. Analogously, if ℓ is a line of N not through 0 it
2
has q+1
squares and 2 q−1
2 nonsquares.
A set X ⊆ K is called special provided all differences of elements of X
are squares. So aX is also special if a is a square and “anti-special” if a
is a nonsquare. Similarly, X + a is special for all a ∈ K. If 0 ∈ X, put
X0 = X \ {0}.
A set A of q+1
non-squares such that a − b is a square for all a, b ∈ A,
2
a = b, is called extra-special. For example, we observed above that the set of
nonsquares on a line of S not throgh 0 is extra-special.
Lemma 20.25.1. Let X be an arbitrary special q-set containing 0, and let
A be any extra special set. Then A · X0 is the set of all nonsquares of K.
Proof. Clearly A·X0 contains only nonsquares and contains q+1
2
products. It remains only to show that these products are all distinct. So
suppose ax = by, a, b ∈ A x, y ∈ X0. Then (a−b)x = b(y −x), where (a−b)x
is a square, while b(y − x) is a nonsquare unless x = y, forcing a = b.
2 ·(q−1) = q2 −1
Let σk(X) denote the k th elementary symmetric function of the (finite)
set X, i.e.,
(1 + xt) =
x∈X
|X|
k=0
σk(X)t k , or
|X|
(t + xb) = σk(X)b k t |X|−k .
Let X be a special q-set through 0 and suppose A is extra special. For
a ∈ A, put fa(t) =
x∈X0 (t − ax) = q−1 i=0 (−1)iaiσi(X0)t q−i−1 .
x∈X
Lemma 20.25.2.
a∈A fa(t) = t q2 −1
2 + 1.
Proof.
a∈A fa(t) = a∈A
x∈X0
(t − ax) = n∈K
n= ❆
(t − n) = t q2 −1
2 + 1.
Lemma 20.25.3. Let X be a special q-set through 0. Then σk(X0) = 0 if
0 < k ≤ q−1
2 .
k=0

20.25. A THEOREM OF BLOKHUIS 887
Proof. Suppose otherwise and let m ≤ q−1
be the smallest positive integer
2
with the property that σm(X0) = 0. And let A be any extra special set.
Then
fa(t) =
(t−ax) = t q−1 +(−1) m a m σm(X0)t q−m−1 + terms of lower degree in t
x∈X0
which implies that
a∈A
fa(t) = t q2 −1
2 + (−1) m
a m
a∈A
σm(X0)t q2 −1
2 −m + terms of lower degree.
(To see this, when expanding the product
a∈A fa(t), choose the term tq−1 exactly q−1
times and (−1) 2 mamσm(X0)t q−m−1 once for some a ∈ A, then add
these terms up over all a ∈ A.)
2 +1 and σm(X0) = 0, it follows that
Since
a∈A fa(t) = t q2 −1
a∈A am = 0
for all extra special sets A. Let A [S] = {a s : a ∈ A}. Then it is easy
to check that both A [−1] and A [q] are extra special since A is. Hence also
a∈A a−qm = 0. Since a q2 −1
2 = −1 for any nonsquare a, we finally have
a∈A
a q2 −1
2 −qm = 0 for any extra special A.
Now let t ∈ K \ F and let A be the set of nonsquares on the line through
t parallel to the line through 0 and 1, i.e., A = {t + i : i ∈ F, t + 1 = ❆ }.
Then
0 = 2
i∈F
t+i= ❆
= −
−
i∈F
t+i= ❆
i∈F
(t + i) q2 −1
2 −qm
(t + i) −qm +
i∈F
t+i=
(t + i) −qm −
i∈F
t+i= ❆
t+i=
(t + i) q2−1 2 −qm −
=
i∈F
i∈F
(t + i) −qm
(t + i) −qm
(t + i) q2 −1−qm = F (t).

888 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
The function F (t) vanishes for all t ∈ K \F . But for t ∈ F , t+i is always
a square, so (t + i) q2 −1
2 = (t + i) q2 −1 = 1, implying F (t) = 0. Hence F (t) is
identically zero.
Consider the coefficient of tq2−qm−q in F (t). (Note:
qm − q ⇐⇒ q > 1.) The coefficient is
−
2 q − 1 − am
q
i∈F
2
i
− qm − q
(q2−1−qm)−(q 2−qm−q) 2 q − 1 − qm
=
i
q − 1
q−1
2 q − 1 − qm
=
q − 1
i∈F
q 2 −1
2
= (q2 − qm − 1)(q2 − qm − 2) · · · (q2 − qm − (q − 1))
.
(q − 1)!
− qm < q2 −
But pj ||q − i ⇐⇒ pj ||q2 − qm − i, 1 ≤ i ≤ q − 1, for 1 ≤ j ≤ e. Hence
q2 −1−qm
≡ 0 (mod p), contradicting the fact that F (t) ≡ 0, and completing
q−1
the proof of the lemma.
Lemma 20.25.4. With X as in the previous lemma, σk(X0) = 0 for 0 <
k < q − 1.
Proof. It is easy to check that X [−1]
0 ∪ {0} is also special, so
σq−1−k(X0) = x · σk X0) [−1] = 0, if 0 < k <
x∈X0
Hence σk(X0) = 0 for 0 < k < q − 1.
q − 1
2 .
Theorem 20.25.5. Let X be a special q-set. Then X is a line in S.
Proof. We may assume that 0 ∈ X, since otherwise we take a translation of
X.
f(t) =
q−1
(t − x) = (−1) k σk(X0)t q−1−k = t q−1 +
x.
x∈X0
k=0
x∈X0
So for x, y ∈ X0, 0 = f(x) = f(y) implies x q−1 = y q−1 =⇒ y = ix, i ∈ F .
Hence for fixed x ∈ X0, X = {ix : i ∈ F }, i.e., X is a line of S.
Corollary 20.25.6. If X is a q-subset of K with the property that the difference
of any two elements of X is always a nonsquare, then X is a line of
N.
Proof. Apply the theorem to the set aX where a is any nonsquare of K.

20.26. THEOREMS OF BRUEN, LEVINGER AND CARLITZ 889
20.26 Theorems of Bruen, Levinger and Carlitz
Let q = pn where p is an odd prime, and put m = (q − 1)/2. It is easy to see
that for a ∈ Fq,
a m ⎧
⎨ = 1, if a = ∈ Fq;
= = −1, if a =
⎩
❆ ∈ Fq;
= 0, if a = 0 ∈ Fq.
In 1960 L. Carlitz [Ca60] proved the following theorem.
Theorem 20.26.1. (Theorem of Carlitz [Ca60])
Let f(x) be a permutation polynomial over Fq such that f(0) = 0 and
f(1) = 1. Suppose that
Then f ∈ Aut(Fq), i.e., f(x) = x pj
(f(a) − f(b)) m = (a − b) m ∀ a, b ∈ Fq.
for for some j with 0 ≤ j ≤ n − 1.
More than a decade later Bruen and Levinger [BL73] proved a more
general theorem from which the theorem of Carlitz is an easy corollary. In
this section we recall a few elements of the theory of group actions on a set,
prove the theorem of Bruen and Levinger, and then deduce the theorem of
Carlitz. (See [BL73] for more on the history of this theorem.)
Let X be a nonempty set and SX the symmetric group on X, i.e., SX is
the group of all permutations of the elements of X with the group operation
being the composition of functions. Let G be an arbitrary group. An action
of G on X is a homomorphism µ : G → SX. In other words, µ is a function
from G to SX satisfying
for all g1, g2 ∈ G.
µ(g1) ◦ µ(g2) = µ(g1 · g2) (20.80)
Often (µ(g))(x) is written as g(x) if only one action is being considered.
The only difference between thinking of G as acting on X and thinking of G
as a group of permutations of the elements of X is that for some g1, g2 ∈ G,
g1 = g2, it might be that µ(g1) and µ(g2) are actually the same permutation,
i.e., g1(x) = g2(x) for all x ∈ X. Also, sometimes there are several different
actions of G on X which may be considered in the same context.

890 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
Lemma 20.26.2. Let µ be an action of G on X and let e be the identity of
G. Then the following hold:
(i) µ(e)is the identity permutation on X.
(ii) µ(g −1 ) = [µ(g)] −1 , for each g ∈ G.
(iii) More generally, for each n ∈ Z, µ(g n ) = (µ(g)) n .
Proof. These results are special cases of results usually proved for homomorphisms
in general. We leave the easy proofs to the reader.
Let G act on X. For x, y ∈ X, define x ∼ y iff there is some g ∈ G for
which g(x) = y.
Lemma 20.26.3. The relation “∼” is an equivalence relation on X.
Proof. This is an easy exercise.
The “∼” equivalence classes are called G-orbits in X. The orbit containing
x is denoted x G or sometimes just [x] if there is no likelihood of
confusion.
For g ∈ G, put Xg = {x ∈ X : g(x) = x}, so Xg is the set of elements of
X fixed by g. For x ∈ X, put Gx = {g ∈ G : g(x) = x}. Gx is the stabilizer
of x in G.
Lemma 20.26.4. For x ∈ X, Gx is a subgroup of G (written Gx ≤ G). If
G is finite, then |G| = |[x]| · |Gx|.
Proof. It is an easy exercise to show that Gx is a subgroup of G. Having
done that, define a function f from the set of left cosets of Gx in G to [x] by:
f(gGx) = g(x).
First we show that f is well-defined. If g1Gx = g2Gx, then g −1
2 g1 ∈ Gx,
so that (g −1
2 · g1)(x) = x, which implies g1(x) = g2(x). Hence f(g1Gx) =
f(g2Gx). So f is well-defined. Now we claim f is a bijection. Suppose
f(g1Gx) = f(g2Gx), so by definition g1(x) = g2(x) and

20.26. THEOREMS OF BRUEN, LEVINGER AND CARLITZ 891
(g −1
2 · g1)(x) = x. Hence g −1
2 · g1 ∈ Gx, implying g1Gx = g2Gx, so f is one-toone.
If y ∈ [x], then there is a g ∈ G with g(x) = y. So f(gGx) = g(x) = y,
implying f is onto [x].
Hence f is a bijection from the set of left cosets of Gx in G to [x], i.e.,
|G|/|Gx| = |[x]| as claimed.
Lemma 20.26.5. For some x, y ∈ X and g ∈ G, suppose that g(x) = y.
Then (i) H ≤ Gx iff gHg −1 ≤ Gy; in particular,
(ii) Gg(x) = gGxg −1 .
Proof. Easy exercise.
Lemma 20.26.6. The Orbit-Counting Lemma (Not Burnside’s Lemma) Let
k be the number of G-orbits in X. Then
k = 1
|Xg| .
|G|
g∈G
Proof. Put S = {(x, g) ∈ X × G : g(x) = x}. We determine |S| in two
ways: |S| =
x∈X |Gx| =
g∈G |Xg|. Since x ∼ y iff [x] = [y], in which
case |[x]| = |[y]|, it must be that |G|/|Gx| = |G|/|Gy| whenever x ∼ y. So
y∈[x] |Gy| =
y∈[x] |Gx| = |[x]| · |Gx| = |G|. Hence
x∈X |Gx| = k · |G| =
g∈G |Xg|.
And hence k = g∈G |Xg|
/|G|.
The following situation often arises. There is some given action ν of G
on some set X. F is the set of all functions from X into some set Y . Then
there is a natural action µ of G on F defined by: For each g ∈ G and each
f ∈ F = Y X ,
µ(g)(f) = f ◦ ν(g −1 ).
Sometimes this is written as: f g (x) = f (g −1 (x)).
Theorem 20.26.7. µ : G → SF is an action of G on F.

892 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
Proof. First note that µ(g1 · g2)(f) = f ◦ ν((g1 · g2) −1 ) = f ◦ ν(g −1
2 · g −1
1 ) = f ◦
[ν(g −1
2 ) ◦ ν(g −1
1 )] = [f ◦ ν(g −1
2 )]◦ν(g −1
1 ) = µ(g1)(f◦ν(g −1
2 ) = µ(g1)(µ(g2)(f)) =
(µ(g1) ◦ µ(g2))(f). So µ is an action of G on F provided each µ(g) is actually
a permutation of the elements of F.
So suppose µ(g)(f1) = µ(g)(f2), i.e., f1 ◦ ν(g−1 ) = f2 ◦ ν(g−1 ). But since
ν(g−1 ) is a permutation of the elements of X, it must be that f1 = f2, so
µ(g) is one-to-one on F. For each g ∈ G and f : X → Y , f ◦ ν(g) ∈ Y X and
µ(g)(f ◦ ν(g)) = (f ◦ ν(g)) ◦ ν(g−1 ) = f, implying that µ(g) is onto.
To use Not Burnside’s Lemma to count G-orbits in F, we need to compute
|Fµ(g)| for each g ∈ G.
Lemma 20.26.8. For g ∈ G, let c be the number of cycles of ν(g) as a
permutation on X. Then |Fµ(g)| = |Y | c .
Proof. For f : X → Y , g ∈ G, we want to know when is f ◦ ν(g −1 ) = f,
i.e., (f ◦ ν(g −1 ))(x) = f(x). This is iff f(ν(g −1 )(x)) = f(x). So f must have
the same value at x, ν(g −1 )(x), ν(g −2 )(x), . . ., etc. This just says that f is
constant on the orbits of ν(g) in X. So if c is the number of cycles of ν(g) as
a permutation on X, then |Y | c is the number of functions f : X → Y which
are constant on the orbits of ν(g).
Applying the orbit-counting Lemma to the action µ of G on F = Y X , we
have:
Theorem 20.26.9. The number of G-orbits in F is
1
|G|
g∈G
|Fµ(g)| = 1
|G|
|Y | c(g) ,
where c(g) is the number of cycles of ν(g) as a permutation on X.
An action ν of a group G on a set X is called faithful if ν is one-to one,
so ν(g) = id if and only if g is the identity element of G. It is transitive
on X provided for any points x, y ∈ X there is at least one g ∈ G with
ν(g) : x ↦→ y. Further, ν is said to be k-ply transitive on X provided that if
(x1, . . . , xk) and (y1, . . . , yk) are any two ordered k-tuples of distinct elements
of X, there is some g ∈ G for which νg : xi ↦→ yi for all i = 1, 2, . . . , k.
g∈G

20.26. THEOREMS OF BRUEN, LEVINGER AND CARLITZ 893
We now suppose that Y is replaced by the elements of the Galois field
Fq, q = p n , p a prime. Put
V = {f : X → Fq}
.
If f, g ∈ V , define f + g by (f + g)(a) = f(a) + g(a) for all a ∈ X. Also,
for f ∈ V and a ∈ Fq define af by (af)(b) = a(f(b)) for all b ∈ X. Then
it is easy to show that V is a vector space over Fq with dimension |X|. One
natural basis for V is given by
B = {fa : a ∈ X} where fa(b) = δa,b = 1 ∈ Fq if a = b and
fa(b) = 0 ∈ Fq if a = b.
It follows that if f ∈ V , then for each b ∈ X
This just says that
f(b) =
f(a) · fa(b). (20.81)
a∈X
For f ∈ V, f =
f(a)fa. (20.82)
It is easy to check that B is also an independent set over Fq, so B is indeed
a basis for V .
Note that the set of all constant functions a : b ↦→ a for all b ∈ X (and
fixed a ∈ Fq) is a 1-dimensional subspace of V , and a ↦→ a is an isomorphism
from Fq to this subspace, which we also denote by Fq. So the constant
function a is identified with the element a ∈ Fq.
At this point let ν be a faithful, transitive action of the group G on the
set X. In place of ν(g)(a) we write g(a). So g1(a) = g2(a) for all a ∈ X
implies g1 = g2. And for each pair (a, b) ∈ X × X there is a g ∈ G for which
b = g(a).
Then for f ∈ V , i.e., f : X → Fq, in place of µ(g)(f) = f ◦ g −1 , we write
g(f) = f ◦ g −1 . For a ∈ X, Ga denotes the stabilizer in G of the element a
under the action ν.
The action (under µ) of g ∈ G on the basis element fa is especially nice:
(g(fa))(b) = fa(g −1 (b)) = δa,g −1 (b) = 1 iff g −1 (b) = a iff g(a) = b, so
a∈X
g(fa) = fg(a). (20.83)

894 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
A map φ : V → V is a G-homomorphism provided φ is an Fq-linear map
from V into V satisfying φ(g(f)) = g(φ(f)) for all f ∈ V and all g ∈ G. Put
HomG(V, V ) = {φ : V → V : φ is a G − homomorphism}.
We make HomG(V, V ) into a vector space over Fq as follows. For φ1, φ2 ∈
HomG(V, V ) and a ∈ Fq, define φ1 + φ2 by (φ1 + φ2)(f) = φ1(f) + φ2(f)
for all f ∈ V , and define aφ1 by (aφ1)(f) = a · φ1(f). These are just the
standard vector addition and scalar multiplication that make HomFq(V, V )
into a vector space over Fq. But we show that its subset HomG(V, V ) is
closed under this vector addition and scalar multiplication.
Let φ1, φ2 ∈ HomG(V, V ), i.e., φi(g(f)) = g(φi(f)), i = 1, 2. Then we
must show that (φ1 + φ2)(g(f)) = g([(φ1 + φ2)(f)]) for all f ∈ V and all
g ∈ G.
(φ1 + φ2)(g(f)) = φ1(g(f)) + φ2(g(f)) (Def. of addition in HomG(V, V ))
Then
= g(φ1(f)) + g(φ2(f)) (φ1, φ2 ∈ HomG(V, V ))
(g(φ1(f)) + g(φ2(f)))(x) = g(φ1(f))(x) + g(φ2(f))(x) (addition in V )
= φ1(f)(g −1 (x)) + φ2(f)(g −1 (x)) (G acts on V )
= (φ1(f) + φ2(f))(g −1 (x)) (addition in V )
= (g(φ1(f) + φ2(f)))(x) (G acts on V )
completing the proof that φ1 + φ2 ∈ HomG(V, V ).
Now start with φ ∈ HomG(V, V ) and a ∈ Fq. Then
(g[(aφ)(f)])(x)
= [(aφ)(f)](g −1 (x)) ( Def. of action of G on V )
= [a · φ(f)](g −1 (x)) (Def of scalar multiplication in Hom G(V, V ))
= a[φ(f)(g −1 (x))] (Def of scalar multiplication in V )
= a[g(φ(f))(x)] (Def. of action of G on V )
= a[φ(g(f))(x)] (Hypothesis that φ ∈ HomG(V, V )
= [a · φ(g −1 (f))](x) (Def. of scalar multiplication in V )
= [(aφ)(g −1 (f))](x) (Def. of scalar multiplication in Hom G(V, V )

20.26. THEOREMS OF BRUEN, LEVINGER AND CARLITZ 895
This shows that g[(aφ)(f)] = (aφ)(g(f)), so that aφ ∈ HomG(V, V ).
Hence HomG(V, V ) is a vector space over Fq.
Lemma 20.26.10. Let G be a transitive permutation group on the set X.
Fix a ∈ X and suppose that the stabilizer Ga has s orbits on X. Then
dimFq(HomG(V, V )) = s.
Hence G is doubly transitive on X if and only if dimFq(HomG(V, V )) = 2.
Proof. Recall that V has a basis consisting of the functions fa, a ∈ X. And
by Eq. 20.83 we know that g(fa) = fg(a). Then, since G is transitive on X,
and since φ(fg(a)) = φ(g(fa)) = g(φ(fa)) for φ ∈ HomG(V, V ), it follows that
an element φ of HomG(V, V ) is uniquely determined by the image under φ
of fa for some fixed a. (We will show that this forces the dimension over Fq
of HomG(V, V ) to be equal to the number of linearly independent choices for
the image of fa.)
Suppose that g ∈ Ga, so (g(fa))(b) = fa(g−1 (b)) = 1 iff g−1 (b) = a iff
b = a, so that Ga fixes fa. Then for φ ∈ HomG(V, V ) and g ∈ Ga, we have
g(φ(fa)) = φ(g(fa)) = φ(fa), so Ga
fixes φ(fa). Suppose that φ(fa) = f =
αbfb with αb ∈ Fq. If the definition of φ on the other basis elements fb is
to permit φ ∈ homG(V, V ), then for g ∈ Ga we must have g(φ(fa)) = φ(fa),
so
φ(fa) =
αbfb = g
=
αbg(fb) =
b∈X
b∈X
αbfb
=
αbfg(b) =
b∈X
b∈X
b∈X
α g −1 (b)fb.
So Ga fixes φ(fa) if and only if αb = αc whenever b and c are in the same
orbit of Ga on X.
Suppose the Ga-orbits on X are O1 = {a}, O2, . . . , Os. Then if φ ∈
HomG(V, V ), we know that φ(fa) = s j=1 αj
b∈Oj fb
if and only if Ga
fixes φ(fa). So choose α1, . . . , αs arbitrarily in Fq and put
φ(fa) =
s
j=1
αj
⎛
⎝
b∈Oj
fb
⎞
⎠ .

896 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
Then we need to define φ(fb) for all b ∈ X \ {a} in such a way that if
b = g(a), then
φ(fb)(= φ(fg(a)) = φ(g(fa)) = g(φ(fa)).
But we need this to be well-defined. So suppose that b = g1(a) = g2(a),
i.e., g −1
2 g1 ∈ Ga. Then φ(fb) = g1(φ(fa)) = g2(φ(fa)) if and only if g −1
2 g1 ∈
Ga. Then φ(fb) = g1(φ(fa)) = g2(φ(fa)) if and only (g −1
2 g1)(φ(fa)) = φ(fa).
But we saw above that g −1
2 g1 ∈ Ga implies that (g −1
2 g1)(φ(fa)) = φ(fa). So
φ(fb) is well-defined.
This gives q s distinct φ ∈ HomG(V, V ) that clearly form a vector subspace
of dinemsion s of HomFq(V, V ) and gives all the elements of HomG(V, V ).
Now we specialize further. Suppose that in the notation used earlier in
this section X = Y = Fq. Then we can view V = {f : Fq → Fq} as a
vector space over Fq with dimension equal to q as above. Moreover, the basis
functions fa have the following appearance:
fa(x) = 1 − (x − a) q−1 .
If S is any subspace of V , then S is a G-module provided that g(f) ∈ S
for all f ∈ S and g ∈ G. Suppose that S is a non-trivial G-module (i.e.,
Fq = S = V ), and suppose that ɛ ∈ S, where ɛ : a ↦→ a for all a ∈ Fq.
Then (g(ɛ))(a) = ɛ(g −1 (a)) = g −1 (a). This connects the structure of S to
information about the action of G.
Suppose that q − 1 = de, 1 < d < q − 1. Put K = {a ∈ Fq : a d = 1}. So
K is a subgroup of F ∗ q = Fq \ {0} having order d.
Define two subsets of V as follows:
and
G = {f ∈ V :
f(x) − f(y)
x − y
∈ K whenever x = y}; (20.84)
H = {f ∈ V : f(x) = a + bx; a ∈ Fq, b ∈ K}. (20.85)
Let G0 be the stabilizer of 0 in G, and H0 the stabilizer of 0 in H. Clearly
H0 = {f ∈ V : f(x) = bx for some b ∈ K}.
Lemma 20.26.11. G is a group of permutations of Fq and H ≤ G. The
nontrivial orbits of G0 and of H0 are the e = (q − 1)/d cosets of K in F ∗ q .

20.26. THEOREMS OF BRUEN, LEVINGER AND CARLITZ 897
Proof. If a+bx ∈ H, then (x−y) −1 [(a+bx)−(a+by)] = b ∈ K, so H ≤ G. If
g ∈ G, then g is a permutation of Fq, since, for x = y, (x−y) −1 [g(x)−g(y)] ∈
K implies that g(x) − g(y) = 0. Further, if f, g ∈ G, the quantity
(x − y) −1 [fg(x) − fg(y)] = (x − y) −1 [g(x) − g(y)] ×
× [g(x) − g(y)] −1 [f(g(x)) − f(g(y))]
is a product of two elements of K and is thus itself in K. Hence fg ∈ G and
G is a group.
Let G0 and H0 be the stabilizers of 0 in G, H, respectively. If f ∈ G0, it
follows from the definition of G that f(x) − f(0) = f(x) = kx ∈ Kx. Hence
G0x ⊂ Kx for each x ∈ Fq. However, Kx = {bx : b ∈ K} = {f(x) : f ∈
H0} ⊂ {f(x) : f ∈ G0} = G0x. Thus G0x = H0x = Kx, and the non-trivial
orbits of G0 and H0 are the cosets of K in F ∗ q , each of which has length d,
implying that the number of non-trivial orbits is e = (q − 1)/d.
Define φk ∈ V by φk(x) = x kd , 0 ≤ k ≤ e, and for a ∈ Fq define Ta ∈ H
by Ta(x) = x + a. Then Ta acts on f0 by
(Ta(f0))(x) = f0((Ta) −1 (x)) = f0(x − a) = 1 − (x − a) q−1 = fa(x),
i.e., Ta : f0 ↦→ fa.
Then for f ∈ V we have
f =
f(a)fa =
f(a)Ta(f0). (20.86)
a∈Fq
a∈Fq
Suppose φ ∈ HomG(V, V ), i.e., φ(g(f)) = g(φ(f)) for all g ∈ G. Then
⎛
φ(f) = φ ⎝
⎞
f(a)Ta(f0) ⎠ =
f(a)φ(Ta(f0)) =
f(a)Ta(φ(f0)).
a∈Fq
a∈Fq
a∈Fq
This says that φ is completely determined by the image of f0 under φ.
Suppose that φ(f0)(x) = q−1
i=0 λix i . Let f ∈ H0, so f(x) = b −1 x for
some b ∈ K. Then f(f0) = f0(bx) = 1 − (bx) q−1 = 1 − x q−1 = f0(x), i.e.,

898 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
f(f0) = f0, and for this f
q−1
λix i = φ(f0)(x) = φ(f(f0))(x) = [f(φ(f0))](x)
i=0
This implies that
and hence
= φ(f0) ◦ f −1 q−1
(x) = φ(f0)(bx) = λib i x i .
q−1
λi(1 − b i )x i = 0 ∀x ∈ Fq
i=0
λi(1 − b i ) = 0 ∀i = 0, 1, . . . , q − 1.
Suppose that K = 〈b〉, so b i = 1 iff i ≡ 0 (mod d). It follows that λi = 0
if i ≡ 0 (mod d), which implies
so
φ(f0)(x) =
φ(f0) =
e
k=0
λkdx kd ,
i=0
e
λkdφk, (20.87)
k=0
where φk(x) = x kd , 0 ≤ k ≤ e. We have completed most the proof of the
following theorem.
Theorem 20.26.12. If φ ∈ HomG(V, V ), then φ is completely determined
by the image of f0 under φ, and φ(f0) must be a linear combination (over
Fq) of the e + 1 monomials φk(x). Conversely, given any such linear combination
e k=0 λkxkd , there exists a unique element φ ∈ HomG(V, V ) such
that φ(f0) = λkxkd .
Proof. The first part of the theorem is proved. In the converse part, the
uniqueness is clear, but the existence of φ in HomG(V, V ) is probably not. It
would be clear if we knew that the dimension of HomG(V, V ) were equal to
e + 1. This follows from the fact that the dimension of HomG(V, V ) is equal
to the number of G0-orbits in Fq (by Lemma 20.26.10), and this is e + 1 by
Lemma 20.26.11.

20.26. THEOREMS OF BRUEN, LEVINGER AND CARLITZ 899
Corollary 20.26.13. Fix k with 0 ≤ k ≤ e. The subspace of V spanned by
the polynomial functions (x − a) dk , a ∈ Fq, is a G-module.
Proof. If φ ∈ HomG(V, V ), then φ(V ) is a G-module. By the previous theorem,
for fixed k there exists a φ in HomG(V, V ) with φ(f0) = x kd . Further,
φ(fa)(x) = φ(Ta(f0))(x) = Ta(φ(f0))(x)
= (φ(f0) ◦ T −1
a )(x) = φ(f0)(x − a) = (x − a) kd .
Hence φ(V ) is the module described in the corollary.
Lemma 20.26.14. Let λ = n−1
i=0 αip i , where 0 ≤ αi < p, and
n−1
Mλ = {r = βip i : 0 ≤ βi ≤ αi}. (20.88)
i=0
If Sλ is the subspace of V spanned by the polynomials (x − a) λ , a ∈ Fq, then
Sλ has a basis
{ɛr(x) = x r : r ∈ Mλ}. (20.89)
Proof. Since (x − a) λ = λ λ r λ−r
r=0 x (−a) , Sλ is contained in the subspace
r
of V generated by the ɛr with λ
= 0 (in Fq). This is just the set in Eq. 20.88
r
since λ
is relatively prime to p exactly when r ∈ Mλ by the Theorem of
r
Lucas.
For the converse, note that
a q−1+r−λ (x − a) λ =
a∈Fq
=
λ
a∈F k=0
λ
k=0
λ
k
= (−1) λ−r−1
λ
x
k
k (−1) λ−k a q−1+r−k
x k
λ−k
(−1)
λ
x
r
r ,
a∈Fq
a q−1+r−k
unless λ = q − 1 and r = 0 or q − 1. This follows from the fact that
a∈Fq as = −1 if s ≡ 0 (mod q − 1), and
a∈Fq as = 0 otherwise. Thus
ɛr ∈ Sλ whenever r ∈ Mλ, except possibly in the exceptional case where
λ = q − 1. But ɛq−1 ∈ Sq−1 by definition, and

900 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
where we used Eq. 20.81
ɛ0(x) = 1 =
fa(x) =
−(x − a) q−1 ∈ Sq−1,
a∈Fq
Since d|(q − 1), clearly (d, p) = 1. Write d = n−1
i=0 αip i , where 0 ≤ αi ≤
p − 1, for 0 ≤ i ≤ n − 1 and we know that α0 = 0. Then 1 ∈ Md and
d − 1 ∈ Md. Recall that φr(x) = x dr , so φ1(x) = x d . Then φ1(V ) = Sd
is a G-module. Hence for g ∈ G and r ∈ Md, ɛr ∈ Sd and in particular
ɛ g r (x) = ɛr(gx) = (gx) r ∈ Sd.
Lemma 20.26.15. Let f ∈ G. Then f(x) = u + vx t , where u = f(0),
v = f(1) − f(0), and td ≡ d (mod q − 1). Also v d = 1, i.e., v ∈ K.
Proof. Let u = f(0). Since each translation Ta is in G, (put a = −u)
ψ(x) := T(−u)(f(x)) = f(x) − u ∈ G and ψ(0) = f(0) − u = 0.
a∈Fq
Since 1, d − 1 ∈ Md, both ɛ ψ
1 = ψ and ɛ ψ
d−1 = ψd−1 are in Sd. (Here we
are using the fact that k ∈ Md and ψ ∈ G imply ɛ ψ
k
∈ Sd.)
Also, ψ ∈ G0 and G0x = Kx imply x −1 ψ(x) ∈ K, so x d = ψ(x) · ψ(x) d−1 .
But x d = ψ(x) · ψ(x) d−1 can be a product of two polynomials of degree at
most d only if ψ(x) = vx t and ψ(x) d−1 = v ′ x d−t where vv ′ = 1 and 0 < t < d.
Then ψ d (x) = v d x td = x d , if and only if td ≡ d (mod q − 1) and v d = 1.
Further, v = ψ(1) = f(1) − u = f(1) − f(0).
Theorem 20.26.16. (Bruen and Levinger [BL73]) Let q = p n , p a prime,
and K = {x ∈ Fq : x d = 1} for some proper divisor d of q − 1. Then a
mapping f of Fq into itself satisfies
for x = y in Fq, if and only if f(x) is given by
(x − y) −1 (f(x) − f(y)) ∈ K (20.90)
where a ∈ Fq, b ∈ K, and (q − 1) divides d(p j − 1).
f(x) = a + bx pj
, (20.91)
Proof. It will suffice merely to show that in Lemma 20.26.15 the only choices
for t are t = p j .
Without loss of generality we may assume that f ∈ F has the form
f(x) = x t . Since G is a group containing the translations Ta, for any fixed

20.27. WEDDERBURN’S THEOREM 901
α = 0, h(x) := f(Tα(x)) = (x−α) t is a function in G. But by Lemma 20.26.15
h(x) = u + vxt′ = (x − α) t (where u = h(0) = (−α) t and v = h(1) − h(0)).
The equation (x − α) t − (u + vxt′ ) = 0 of degree at most max(t, t ′ ) < q − 1 is
satisfied by each x ∈ Fq. Thus all coefficients in (x − α) t − (u + vxt′ ) are 0.
This is only possible if t ′ = t, v = 1, and (x − α) t = xt + (−α) t′ . Since α was
arbitrary, this shows that f(x + α) = (x + α) t = xt + αt for all α ∈ Fq, which
implies (since f(x) is multiplicative and additive) that f is an automorphism
of F , i.e., t = pj . By Lemma 20.26.15 d(pj − 1) ≡ 0 mod q − 1).
To obtain the theorem of Carlitz, suppose that p is an odd prime and put
d = (q−1)/2. So a ∈ F ∗ q is a square if and only if a ∈ K = {α ∈ Fq : αd = 1}.
Then if f is a permutation of the elements of Fq fixing 0 and 1 and such that
(x − y) −1 (f(x) − f(y)) ∈ K for all x, y ∈ Fq with x = y, it must be that
f(x) = xpj. Here j can be any integer, since 2d = q − 1 and pj − 1 is even.
20.27 Wedderburn’s Theorem
A skew-field satisfies all the properties of a field except that multiplication is
not commutative. Such an algebraic structure is also called a division ring.
Theorem 20.27.1. The famous theorem of Wedderburn says that a finite
skew-field (i.e., division ring) R is a field. Hence it is a finite Galois field
GF (p r ) for some prime p.
Proof. The following proof is due to Witt [Wi31]. So let R be a finite skewfield.
The unit of R generates the characteristic subfield of R, and this must
be a finite field Fp for some prime p. R is naturally viewed as a vector
space over Fp, so must have some finite dimension r. Then R has exactly
p r elements. The center Z of R consistws of all elements z of R such that
zx = xz for every x ∈ R. Z is a commutative subring of R, and hence is a
finite field containing Fp. Let Z have q = p s elements. We wish to show that
Z is all of R. In any case R is a vector space over Z, say with dimension t.
Then R has q t = p st = p r elements in all. Here t = 1 if and only if Z = R.
If x is any element of R, the normalizer Nx of x (i.e., the elements of R that
commute with x) is easily seen to be a subring containing Z, i.e., a skew-field
over which R may be viewed as a vector space. If Nx had dimension d over
Z, Nx must have q d elements. If R has dimension u over Nx, then R has
(q d ) u = q t elements where d|t. Hence in the multiplicative group R ∗ of the

902 CHAPTER 20. APPENDIX 2: ALGEBRA IN FQ[X]
p r − 1 = q t − 1 nonzero elements of R, an element x not in Z has normalizer
of order q d − 1, where d is a divisor of t and d < t. Counting the elements of
R ∗ we have the class equation for R ∗ :
q t − 1 = q − 1 + qt − 1
qd
, (20.92)
− 1
where q − 1 enumerates the elements of Z and each remaining summand
counts the elements in some class of size (q t − 1)/(q d − 1), where d|t, d < t.
Let w be a fixed, complex primitive dth root of 1. Then
Φd(x) =
1≤j≤d
(d,j)=1
(x − w j )
is the dth cyclotomic polynomial whose zeros are the primitive dth roots of 1
and all of whose coefficients are integers. Recall that for each positive integer
t, x t − 1 =
d|t Φd(x). In particular, if d|t, d < t, then
x t − 1 = Φt(x)Φd(x)G(x)
is the product of three monic polynomials with integer coefficients. Hence
Φt(q) =
1≤j≤t
(t,j)=1
(xt − 1)
(xd = Φt(x)G(x).
− 1)
(q − w j ), w a primitive tth root of 1.
Since w j = 1 is a coomplex root of unity, we have |q − w j | > q − 1. Hence
Φt(q) is a rational integer greater in absolute value than q − 1. If t > 1, then
Φt(q) divides each term of Eq. 20.92 except q −1. But then Φt(q) also divides
q − 1, which is impossible since |Φt(q)| > q − 1. Hence the only possibility
is that t = 1. This means that Z = R, implying that R is commutative and
must be a finite field.

Chapter 21
Appendix 3: Review of
Sesquilinear Forms
It is expected that many readers of this book will not need to study this
chapter, but we have included it as an appendix for easy reference.
21.1 Sesquilinear and Quadratic Forms
Let q = pe , p a prime; F = GF (q). Let V be an n-dimensional vector space
over F (i.e., V has rank n). Let B = {v1, . . . , vn} be an ordered basis for V
over F . So for w ∈ V , w = n i=1 civi for unique scalars c1, . . . , cn ⎛ ⎞
∈ F . Write
[w]B = (c1, . . . , cn). And write B =
then
⎜
⎝
v1
.
vn
⎛
⎜
w = [w]B · B = (c1, . . . , cn) ⎝
⎟
⎠ as a column “super”-vector.
v1
.
vn
⎞
⎟
⎠ =
n
civi.
For example, if V = F n = {(a1, . . . , an) : ai ∈ F } and if B = {e1 =
(1, 0, . . . , 0), e2 = (0, 1, 0, . . . , 0), . . . , en = (0, . . . , 0, 1)}, then for a =
(a1, . . . , an) = a1e1 + · · · + anen, we have [a]B = a.
Return to the general case with ordered basis B. Let A ∈ Mn(q) (i.e., A
is an n × n matrix over F ), and let σ ∈Aut(F ). Define f : V × V :→ F by
903
i=1

904CHAPTER 21. APPENDIX 3: REVIEW OF SESQUILINEAR FORMS
f : (v, w) ↦→ [v]BA([w] T B) σ .
So if [v]B = (c1, . . . , cn), and [w]B = (d1, . . . , dn), then
f(v, w) = (c1, . . . , cn) · A ·
⎛
⎜
⎝
d σ 1
.
d σ n
⎞
⎟
⎠ .
Result 21.1.1. f(av + bv ′ , w) = af(v, w) + bf(v ′ , w), so f is linear in its
first variable.
Result 21.1.2. f(v, aw + bw ′ ) = a σ f(v, w) + b σ f(v, w ′ ), so f is semilinear
in its second variable.
We say f is sesquilinear when it is linear it its first variable and semilinear
in its second variable. All sesquilinear forms on V × V to F arise this way
when dim(V ) < ∞.
Case 1. f is symmetric provided f(v, w) = f(w, v) ∀ w, v ∈ V . This
occurs when σ = id and A is symmetric (A = A T ). Then
f(v, w) = [v]BA[w] T B = [w]BA[w] T B
T
= [w]B · A T · [v] T B = [w]B · A · [v] T B
= f(w, v).
Case 2. f is skew-symmetric provided f(v, w) = −f(w, v) ∀ v, w ∈ V . f
is alternating provided f(v, v) = 0 ∀ v ∈ V .
Note: If f is alternating, then necessarily f is skew-symmetric. If p = 2
and f is skew-symmetric, then necessarily f is alternating. (What happens
when p = 2?)
Result 21.1.3. f is alternating (and hence skew-symmetric) if and only if
σ = id and A T = −A and Aii = 0 for all i. For this reason it is usual to
define the matrix A to be skew-symmetric if and only if A T = −A and (when
p = 2) Aii = 0 for all i.

21.1. SESQUILINEAR AND QUADRATIC FORMS 905
Case 3. f is Hermitian provided σ = id and f(w, v) = f(v, w) σ ∀v, w ∈ F .
Then f(w, v) = f(v, w) σ = f(w, v) σ2,
which implies that σ2 = id = σ. Hence
σ has order 2 in Aut(F ), so q = pe with e even. In that case σ : x ↦→ xpe/2 = ¯x.
Thus we have three types of sesquilinear forms:
symmetric alternating Hermitian
σ = 1 σ = 1 σ 2 = 1 = σ.
A function Q : V → F is a quadratic form on V provided
(i) Q(λv) = λ 2 Q(v) ∀λ ∈ F, ∀v ∈ V ;
(ii) The function fQ : V × V → F defined by
fQ(v1, v2) = Q(v1 + v2) − Q(v1) − Q(v2)
is bilinear.
Here the function fQ is the polar form of Q.
Note: fQ is a symmetric, bilinear form. If Q(v) = [v]B · A · [w] T B , then
fQ(v, w) = [v + w]B · A · [v + w] T B − [v]BA[v] T B − [w]BA[w] T B
= [v]BA[w] T B + [w]BA[v] T B
= [v]B(A + A T )[w] T B ,
which implies that fQ(v, v) = 2 · Q(v). So fQ is the symmetric bilinear form
associated with the symmetric matrix A + A T .
Suppose p = 2 and f is a symmetric bilinear form. Put Q(v) = 1f(v,
v).
2
This gives a quadratic form Q with fQ = f. On the other hand, if p = 2,
there are symmetric forms which are not the
polar form of any quadratic
1 0
form. (Example: If p = 2 put A = .)
0 0
A vector space equipped with a symmetric, alternating, hermitian or
quadratic form is called a polar space. In this book we consider three types
of polar spaces:
Symplectic Polar Space V with an alternating form
Unitary Polar Space V with an hermitian form
Orthogonal Polar Space V with a quadratic form and the associated
symmetric bilinear polar form.

906CHAPTER 21. APPENDIX 3: REVIEW OF SESQUILINEAR FORMS
Let V1, V2 be equipped with the forms β1, β2, respectively, both of which
are symplectic or both of which are unitary forms. An invertible semilinear
map T : V1 → V2 with companion automorphism σ is a semisimilarity from
(V1, β1) to (V2, β2) provided there is a nonzero c ∈ F such that
β2(T (v), T (w)) = c · (β1(v, w)) σ ∀ v, w ∈ V1.
T is a similarity iff σ = 1;
T is an isometry iff σ = id and c = 1. .
Let V1, V2 be equipped with quadratic forms Q1, Q2, respectively. Let
T : V1 → V2 be an invertible semilinear map with companion automorphism
σ. Then T is a semisimilarity from (V1, Q1) to (V2, Q2) provided there is a
nonzero c ∈ F such that
Q2(T (v)) = c · (Q1(v)) σ ∀ v ∈ V1.
T is a similarity iff σ = 1;
T is an isometry iff σ = 1 and c = 1. .
NOTE: If g : V1 → V2 is a semisimilarity/similarity/isometry then g
induces a map ¯g : P V1 → P V2 which is called a projective semisimilarity/similarity/isometry.
Let S = (V, β) be a polar space with β an alternating, hermitian, or
quadratic form (with polar form fβ).
ΓS = group of semisimilarities of V ;
GS = group of similarities of V ;
S = group of isometries of V.
P ΓS, P GS, P S are the corresponding projective groups.
Let Z(S) be the group of form-preserving scalar maps. Then
P ΓS ∼ = Γ/Z(S);
P GS ∼ = GS/Z(S);
P S ∼ = S/Z(S).
Let f be a sesquilinear form on V .
.

21.1. SESQUILINEAR AND QUADRATIC FORMS 907
v ∈ V is isotropic iff f(v, v) = 0.
W ≤ V is totally isotropic iff f(v, w) = 0 ∀ (v, w) ∈ W × W.
NOTE: In a symplectic space each vector is isotropic.
Let Q be a quadratic form on V .
v ∈ V is singular iff Q(v) = 0.
W ≤ V is totally singular iff Q(w) = 0 ∀ w ∈ W.
A Hermitian variety is the set of isotropic points of a unitary projective
space.
A quadric is the set of singular points of an orthogonal projective space.
Two isotropic points of a symplectic or unitary space are collinear provided
they span a totally isotropic line.
Two singular points of an orthogonal space are collinear provided they
span a totally singular line.
Lemma 21.1.4. Let (V, β) be a polar space over F . Suppose the points
P1 = 〈v1〉 and P2 = 〈v2〉 are isotropic if β is alternating or hermitian, and
singular if β is quadratic with polar form fβ.
(a) If β is alternating or hermitian, then P1 and P2 are collinear if and
only if β(v1, v2) = 0.
(b) If β is a quadratic form, then P1 and P2 are collinear if and only if
fβ(v1, v2) = 0.
Proof. (a) Let σ be the companion automorphism of β, and L = 〈v1, v2〉.
Any element of L × L may be written as (av1 + bv2, cv1 + dv2) for some
a, b, c, d ∈ F . Then
β(av1 + bv2, cv1 + dv2) = ac σ β(v1, v1) + ad σ β(v1, v2)
+ bc σ β(v2, v1) + bd σ β(v2, v2) (21.1)
= ad σ β(v1, v2) + bc σ β(v2, v1).
So if L is totally isotropic, we have ad σ β(v1, v2)+bc σ β(v2, v1) = 0 ∀ a, b, c, d ∈
F . In particular, setting a, d = 0 and c = 0 yields β(v1, v2) = 0. Conversely,

908CHAPTER 21. APPENDIX 3: REVIEW OF SESQUILINEAR FORMS
putting β(v1, v2) = 0 and hence β(v2, v1) = 0 (using that an alternating
form is skew-symmetric, and that β(v1, v2) = β(v2, v1) σ if β is hermitian) in
Eq. 21.1 implies that L is totally isotropic.
(b) Let L = 〈v1, v2〉. Vectors on L are of the form av1 + bv2 for a, b ∈ F ,
and
β(av1 + bv2) = a 2 β(v1) + b 2 β(v2) + abfβ(v1, v2)
from which the result follows.
= abfβ(v1, v2),
Corollary 21.1.5. Each two points of a totally isotropic subspace of a symplectic
or hermitian polar space are collinear.
Corollary 21.1.6. Each two points of a totally singular subspace of an orthogonal
space are collinear.
Corollary 21.1.7. If three points of a projective line lie on a quadric, the
line is totally singular.
Proof. 0 = Q(v1) = Q(v2) = Q(v1 + av2) = Q(v1) + a 2 Q(v2) + afβ(v1, v2) =
afQ(v1, v2), implying that fQ(v1, v2) = 0.
21.2 Perps and Radicals
Let f be an alternating, hermitian or polar form on V . For U ≤ V , define
the perp of U to be
U ⊥ = {v ∈ V : f(u, v) = 0 ∀ u ∈ U}.
The radical of f is: rad(f) = V ⊥ . If f is alternating (symplectic) or
hermitian, rad(V ) =rad(f), and f and (V, f) are nondegenerate provided
rad(f) = {0}; degenerate otherwise.
If Q is a quadratic form on V with polar form fQ, the singular radical of
Q (or of V ), denoted rad(Q) (or rad(V )) is
rad(Q) = rad(V ) = {v ∈ radfQ : Q(v) = 0},
with the quadratic form (and space) nondegenerate provided rad(Q) = {0};
degenerate otherwise.
For a subspace U ≤ V , restrict to U in the natural way. If f is alternating
or hermitian, rad(U) = U ∩ U ⊥ . If Q is a quadratic form, rad(U) = {x ∈
U ∩ U ⊥ : Q|U(x) = 0}, where fQ is used to define U ⊥ .

21.3. QUOTIENT SPACES 909
21.3 Quotient spaces
If U ≤ V , it is an additive subgroup of V , so the quotient group V/U =
{v + U : v ∈ V } is a group under the (commutative) addition of cosets:
(v + U) + (w + U) := (v + w) + U.
Scalar multiplicatioin of cosets is given by
a(v + U) := av + U.
This makes V/U into a vector space over the same field F .
Theorem 21.3.1. Let V be a vector space over F .
(a) Suppose f is an alternating (symplectic), hermitian (unitary) or polar
(orthogonal) form on V . Let U = rad(f). Define f ′ on V/U by
f ′ (v1 + U, v2 + U) := f(v1, v2).
Then f ′ is a well-defined nondegenerate form of the same type as f.
(b) Suppose Q is a quadratic form on V with polar form fQ; U = rad(Q).
Define Q ′ on V/U by Q/(v + U) := Q(v). Then Q ′ is a well-defined nondegenerate
quadratic form.
Proof. We give the proof of (b). After reading through it, the reader should
write out the proof of (a). U = rad(Q) means u ∈ U if and only if
1. Q(u) = 0 and 2. fQ(u, v) = 0 ∀ v ∈ V . Recall that v + U = w + U
if and only if v − w ∈ U if and only if w = v + u for some u ∈ U. So we
want Q ′ (v + U) to be equal to Q ′ (v + u + U) for all v ∈ V , u ∈ U. But
Q ′ (v + u + U) = Q(v + u) = Q(v) + Q(u) + fQ(v, u) = Q(v). This shows that
Q ′ is well-defined.
Q ′ (λ(v + U)) = Q ′ (λv + U) = Q(λv) = λ 2 Q(v) = λ 2 (Q ′ (v + U)). Hence
Q ′ satisfies the first property of being a quadratic form on V/U.
fQ ′(w + U, v + U) = Q′ ((w + U) + (v + U)) − Q ′ (w + U) − Q ′ (v + U)
= Q ′ (w + v + U) − Q ′ (w + U) − Q ′ (v + U)
= Q(w + v) − Q(w) − Q(v) = fQ(w, v).
Since fQ is (symmetric) bilinear, it is easy to check that fQ ′ is (symmetric)
bilinear, so Q ′ is a quadratic form on V/U. The claim is that Q ′ is nondegenerate,
i.e., rad(Q ′ ) is just the zero subspace of V/U. (Recall 0 + U = U
is the zero element of V/U.)

910CHAPTER 21. APPENDIX 3: REVIEW OF SESQUILINEAR FORMS
w + U ∈ rad(Q ′ )
iff Q ′ (w + U) = 0 and fQ ′(w + U, v + U) = 0 ∀ v + U ∈ V/U
iff Q(w) = 0 and fQ(w, v) = 0 ∀ v ∈ V
iff w ∈ rad(Q) = U iff w + U = U.
This implies that rad(Q ′ ) = {0 + U} is the zero subspace of V/U.
21.4 Polar Spaces
Theorem 21.4.1. In a nondegenerate polar space the following properties
hold:
PS1. Every two isotropic/singular points lie on at most one totally
isotropic/singular line.
PS2. If an isotropic/singular point P is not on a totally isotropic/singular
line L, then either P is collinear with a unique point of L or P is collinear
with all points of L.
PS3. No isotropic/singular point is collinear with all isotropic/singular
points.
Proof. PS1 is clear because a polar space lives in a projective space. PS3
follows because the polar space is nondegenerate. We give a proof of PS2 for
the quadratic case. The reader should supply a proof for the symplectic and
unitary spaces.
Let Q be a nondegenerate quadratic form on V with polar form fQ. Let
P = 〈u〉 be a singular point, i.e., Q(u) = 0. Let L be a totally singular line
(so Q(w) = 0 for all w ∈ L) not containing P . Recall that for R1 = 〈v〉 on
L, P is collinear with R1 if and only if fQ(u, v) = 0.
If not every point of L is collinear with P , there is some point R1 = 〈v〉 of
L with fQ(u, v) = 0. Let R2 = 〈w〉 be any other point of L, so points 〈w+av〉
as a varies over the elements of F are all the points of L except R1. P is
collinear with 〈w+av〉 if and only if 0 = fQ(u, w+av) = fQ(u, w)+fQ(u, v)·a
if and only if
a = −fQ(u, w)
fQ(u, v) ,
so L has a unique point collinear with P .

21.4. POLAR SPACES 911
Note: If f is a sesquilinear form on V (with companion automorphism σ)
given by f(u, v) = [u]BD([v] σ B )T for some ordered basis
then
B =
⎛
⎜
⎝
v1
.
vn
⎞
⎟
⎠ ,
⎛
⎜
f(vi, vj) = (0 · · · 1i · · · 0)D ⎜
⎝
0.
1 σ j
.
0
⎞
⎟ = Dij.
⎟
⎠
In general, if f is sesquilinear and B = {v1, . . . , vn} is any ordered basis
of V , the gram matrix D opf f with respect to B is given by Dij = f(vi, vj).
Lemma 21.4.2. With the notation as above, f is nondegenerate if and only
if det(D) = 0.
Proof. the vector y is in the radical of f if and only if f(x, y) = 0 for all
x ∈ V if and only if [x]BD([y] σ B )T = 0 ∀ x ∈ V if and only if D([y] σ B )T = 0.
So rad(f) has a nonzero vector if and only if det(D) = 0.
This includes the case f = fQ and q is odd. But what about characteristic
2? Even then rad(f) = {0} if and only if det(D) = 0.
Let f be an alternating, hermitian, or symmetric bilinear polar form of a
quadratic form, on a finite dimensional vector space V . Let U be a subspace
of V , dim(U) = m, dim(V ) = n. Let β ′ = {u1, . . . , um} be an ordered basis
of U. Complete β ′ to a basis β = {u1, . . . , um, um+1, . . . , un} of V . Put
D = (Dij), where Dij = f(ui, uj), 1 ≤ i, j ≤ n. Then f(u, v) = [u]βD([v] σ β )T .
f alternating ⇐⇒ D T = −D (with 0’s on diagonal and σ = id)
f hermitian ⇐⇒ D T = D σ (σ ∈ Aut(F ), σ = id = σ 2 )
f polar =⇒ D T = D (σ = id).
U ⊥ = {v ∈ V : f(v, u) = 0 ∀u ∈ U} = {v ∈ V : f(u, v) = 0 ∀u ∈ U}. It
follows from f(v, m i=1 ciui) = m i=1 cσi f(v, ui) that v ∈ U ⊥ ⇐⇒ f(v, ui) =
0 for all i = 1, . . . , m.

912CHAPTER 21. APPENDIX 3: REVIEW OF SESQUILINEAR FORMS
Theorem 21.4.3. Let f be a nondegenerate alternating, hermitian or polar
form (with companion automorphism σ on an n-dimensional vector space V
over V .
(a) U ≤ W =⇒ W ⊥ ≤ U ⊥ ; and the map g : P V → P V : U ↦→ U ⊥ is a
polarity.
(b) If rad(f) = {0}, then dim(U) + dim(U ⊥ ) = dim(V ) + dim(V ⊥ ).
(Recall: If the space is nondegenerate, the only time we could have
rad(f) = {0} is when f = fQ, p = 2, and Q(v) = 0 for some v ∈ rad(f).)
Proof. First we give a proof of (a). Here we need the assumption that the
⎛ ⎞
0.
⎜ ⎟
⎜ ⎟
⎜ ⎟
form f is nondegenerate, i.e., rad(f) = {0}. Then f(v, ui) = [v]BD ⎜ ⎟ =
⎜ ⎟
⎝ ⎠
[v]B · ith
column of D , which implies f(v, u1) = · · · = f(v, um) = 0 iff [v]B
is in the left null space of the matrix D ′ consisting of the first m columns of
D. We assume that det(D) = 0, so D ′ has rank m and dim(U ⊥ ) = n − m.
This implies dim(U) + dim(U ⊥ ) = dim(V ). As (U ⊥ ) ⊥ ⊇ U and dim(U ⊥ ) ⊥ =
n−dim(U ⊥ ) = m, clearly U ⊥⊥ = U. Hence g = g−1 and g is a bijection and
thus a polarity.
For the proof of part (b) we may drop the assumption that rad(f) = {0}.
We may define a nondegenerate form f ′ on V/V ⊥ of the same type as f. By
part (a),
dim(U/V ⊥ ) + dim((U/V ⊥ ) ⊥ ) =
= dim(U/V ⊥ ) + dim(U ⊥ /V ⊥ ) = dim(V/V ⊥ ).
=⇒ dim(U) − dim(V ⊥ ) + dim(U ⊥ ) − dim(V ⊥ ) = dim(V ) − dim(V ⊥ )
=⇒ dim(U) + dim(U ⊥ ) = dim(V ) + dim(V ⊥ ).
The next theorem is a famous one due to Witt. We state the general
result but refer the reader elsewhere (e.g., Taylor - pp. 57-58) for a complete
proof. In the next chapter where we concentrate on quadratic forms we do
give a proof in the orthogonal case.
1i
0.

21.4. POLAR SPACES 913
Theorem 21.4.4. (Witt) Let f be an alternating, hermitian or polar form
on V , with U, W ≤ V , and h : U → W an isometry. Then there is an
isometry g : V → V such that g|U = h if and only if
h(U ∩ rad(f)) = W ∩ rad(f).
Corollary 21.4.5. If rad(f) = {0}, then the isometry h : U → W extends
to g : V → V . Hence if U and W are isometric, then U ⊥ and W ⊥ are
isometric.
Proof. By Witt’s theorem, there is an isometry g : V → V with g|U(U) = W .
Then
g(U ⊥ ) = g ({v ∈ V : f(u, v) = 0 ∀ u ∈ U})
= {g(v) ∈ F : f(gU, gv) = 0 ∀ gu ∈ W }
= W ⊥ .
A totally isotropic/singular subspace of maximum dimension will be referred
to as a maximal.
Corollary 21.4.6. All maximals of a finite dimensional polar space have the
same dimension (i.e., rank).
Proof. Any two totally isotropic/singular subspaces of the same dimension
are isometric. Use Witt’s theorem to get an element of P ΓS taking one to the
other. Suppose U1, U2 are maximals and dim(U) < dim(U2). Let U ′ 1
≤ U2
with dim(U1) = dim(U ′ 1). There must be a g ∈ P ΓS for which g(U1) = U ′ 1.
Then g −1 (U2) properly contains U1, a contradiction.
The rank (i.e., vector space dimension) of a maximal is called the Witt
index of the form f.
Suppose U is a totally isotropic/singular subspace of a polar space S =
(V, β). U ⊆ U ⊥ implies that dim(U) ≤ dim(U ⊥ ) =⇒ 2dim(U) ≤ dim(V ) +
dim(V ⊥ ). Hence
Corollary
21.4.7. The
Witt index of S is less than or equal to
⊥ dim(V ) + dim(V ) .
1
2

914CHAPTER 21. APPENDIX 3: REVIEW OF SESQUILINEAR FORMS
21.5 Hyperbolic Pairs
Let f be an alternating or hermitian form on V . A pair (v1, v2) of isotropic
vectors of V that have f(v1, v2) = 1 is said to be a hyperbolic pair, and the
line they span is a hyperbolic line.
If Q is a quadratic form on V with polar form fQ, a pair (v1, v2) of singular
vectors of V that have fQ(v1, v2) = 1 is a hyperbolic pair, and the line they
span is a hyperbolic line.
Lemma 21.5.1. Any two hyperbolic lines of the same type are isometric.
(a) Any two symplectic hyperbolic lines are isometric.
(b) Any two unitary hyperbolic lines are isometric.
(c) Any two orthogonal hyperbolic lines are isometric.
Proof. Let (V, f) be a symplectic/unitary hyperbolic line. Similarly, let
(V ′ , f ′ ) be a symplectic/unitary hyperbolic line. Let f and f ′ have companion
automorphism σ. Here V = 〈u, v〉 with f(u, u) = 0 = f(v, v), f(u, v) = 1.
Then f(v, u) = −1 in the symplectic case and f(v, u) = 1 σ = 1 in the
unitary case. Similarly, V ′ = 〈u ′ , v ′ 〉 with f ′ (u ′ , u ′ ) = 0, . . . , etc. Define
T ∈ HomF (V, V ′ ) by T (u) = u ′ , T (v) = v ′ , and extend linearly. It is an easy
exercise to show that
f ′ (T (au + bv), T (cu + dv)) = f(au + bv, cu + dv),
implying that T : V → V ′ is an isometry. This proves (a) and (b).
Let (V, Q) be an orthogonal hyperbolic line. So V = 〈u, v〉, Q(u) =
Q(v) = 0 and fQ(u, v) = fQ(v, u) = 1. Similarly, V ′ = 〈u ′ , v ′ 〉, Q ′ (u ′ ) = 0,
. . . , etc. Define T : V → V ′ by T (u) = u ′ , T (v) = v ′ . Then
Q(au + bv) = Q(au) + Q(bv) + fQ(au, bv)
= a 2 Q(u) + b 2 Q(v) + ab · fQ(u, v)
= 0 + 0 + ab.
Similarly, Q ′ (T (au + bv)) = Q ′ (au ′ + bv ′ ) = ab = Q(au + bv), implying that
T is an isometry.
A subspace of a polar space is anisotropic provided it contains NO nonzero
isotropic/singular vectors. A line L of a polar space is tangent (at P ) if it
contains exactly one isotropic/singular point P . (Note: Sometimes a line is
said to be tangent to the polar space if it is contained entirely in it. However,
for the remainder of this chapter we will use the definition just given.)

21.5. HYPERBOLIC PAIRS 915
Lemma 21.5.2. A line of a polar space is either
• anisotropic
• tangent
• hyperbolic
• totally isotropic/singular.
Proof. Suppose that L is a line of a polar space (V, β) over a field F , where
L contains at least one isotropic/singular point P = 〈v1〉. Note that rad(L)
is either {0}, a point of L, or all of L. We show that this characterizes L as
being (respectively) hyperbolic, tangent or totally isotropic/singular.
We start with the alternating and hermitian cases. First suppose that
rad(L) is {0}. If β is alternating, there is a w ∈ L such that β(v1, w) = a = 0.
Then v2 = a −1 w is an isotropic vector with β(v1, v2) = 1. Similarly, if β is
an hermitian form with companion automorphism σ, there must be a w ∈ L
such that β(v1, w) = a = 0. Choose any d ∈ F with d + d σ = 0. Let
c = (d + d σ ) −1 d σ β(w, w). Then v2 = −cu(a · a σ ) −1 + w(a σ ) −1 is an isotropic
vector with f(v1, v2) = 1. In both cases L is an hyperbolic line. Conversely,
let β be an alternating or hermitian form and suppose L contains a hyperbolic
pair (v1, v2) with w a nonzero vector of rad(L). Writing v2 = aw+bv1 for some
a, b ∈ F , we have 1 = β(v1, v2) = aβ(v1, w) + bβ(v1, v1) = 0, a contradiction.
This shows that if L is hyperbolic, then rad(L) must be {0} in the alternating
and hermitian cases.
If rad(L) = L, then L ⊥ = L, that is, L is a totally isotropic line, and
conversely. If rad(L) is a point P = 〈v1〉 of L, then L must be tangent.
Otherwise there is some isotropic point 〈u〉 of L distinct from P so β(v1, u) =
0 (since rad(L) = L ∩ L ⊥ ). This would force L to be a totally isotropic line,
in which case rad(L) = L.
Finally, let (L, Q) be an orthogonal polar space with L a line, so Q is
a quadratic form with polar form fQ. If L has no singular point, then L is
anisotropic. So suppose 〈v1〉 is a singular point of L. Recall that rad(Q) =
rad(L) = {w ∈ L : Q(w) = 0 and fQ(u, w) = 0 ∀ u ∈ L}.
If L has three points 〈v1〉, 〈v2〉, 〈v3〉 with Q(vi) = 0, we know all points 〈v〉
on L satisfy Q(v) = 0 and fQ(u, v) = 0 for all 〈u〉, 〈v〉 on L. So rad(L) = L
and L is totally singular.
Suppose that 〈v1〉, 〈v2〉 are the only distinct points of L with Q(v1) =
Q(v2) = 0. Then fQ(v1, v2) = 0, say fQ(v1, v2) = a = 0. So fQ(v1, a −1 v2) = 1

916CHAPTER 21. APPENDIX 3: REVIEW OF SESQUILINEAR FORMS
and (v1, a −1 v2) forms a hyperbolic pair, so L is a hyperbolic line. The only
possible points in rad(L) are 〈v1〉 and 〈v2〉. But v1 ∈ rad(L) if and only
if fQ(v1, w) = 0 for all 〈w〉 ∈ L iff fQ(v1, v2) = 0 iff L is totally singular.
So rad(L) = {0} when L is hyperbolic, and rad(L) = L when L is totally
singular.
Suppose 〈v〉 is the unique point of L for which Q(v) = 0, i.e., 〈v〉 is the
unique singular point of L. Let 〈w〉 be any other point of L, so Q(w) = 0 and
Q(w+av) = 0 for all a ∈ F . Hence Q(w+av) = Q(w)+a 2 Q(v)+afQ(w, v) =
Q(w) + afQ(w, v) = 0 for all a ∈ F . This forces fQ(w, v) = 0 (otherwise put
), so rad(L) = 〈v〉.
a = −Q(w)
fQ(w,v)
Recap: L is anisotropic (implying rad(L) = {0})
or
L has at least one singular point and
L = hyperbolic line ⇐⇒ rad(L) = {0}
L = tangent line ⇐⇒ rad(L) is a point
L = totally singular ⇐⇒ rad(L) = L.
For symplectic and unitary spaces, the same is true with “singular” replaced
by “isotropic.”

Chapter 22
Appendix 4: Interlacing of
Eigenvalues
The general theory of interlacing of eigenvalues has been applied with great
success to produce restrictions on the values of parameters of strongly regular
graphs and other finite geometries including generalized quadrangles. When
this technique was first applied to GQ in the early 1970’s, the present author
referred to it as the “Higman-Sims” technique because of the introduction
by these two authors (D. G. Higman and C. Sims) of the general method in
algebraic combinatorics. Later the method was generalized by W. Haemers
and is now referred to simply as the method of interlacing of eigenvalues. We
start by giving the more elementary aspects of the theory of interlacing of
eigenvalues and then applying this theory to obtain the results on GQ that
have been known for at least two decades.
22.1 Real Symmetric Matrices: A Review
If x = (x1, . . . , xn) T and y = (y1, . . . , yn) T are column vectors of real numbers,
then x · y = xiyi denotes the usual dot product. If A is a real,
symmetric, n × n matrix, for each x = 0, define the Rayleigh Quotient R(x)
for A by
R(x) =
917
x · Ax
. (22.1)
x · x

918 CHAPTER 22. APPENDIX 4: INTERLACING OF EIGENVALUES
Hence
Put O = {x ∈ R n : x · x = 1}, and note that for k = 0, x = 0,
R(kx) = R(x). (22.2)
{R(x) : x = 0} = {x · Ax : x ∈ O}. (22.3)
The set W (A) = {R(x) : x ∈ O} is called the numerical range of A. It
is the continuous image of a compact set, and hence is a closed, bounded
interval with a maximum M and a minimum m. Let µ1 ≥ µ2 ≥ · · · µn be
the eigenvalues of A associated with the eigenvectors x1, . . . , xn chosen so
that xi ∈ O, and so that T = (x1, . . . , xn) is a real orthogonal matrix with
T T AT = U = diag(µ1, . . . , µn).
Observe that if x = T y, then x ∈ O if and only if y ∈ O. Since T is
nonsingular, T maps O onto O in a one-to-one manner. Hence
Similarly,
M = maxx∈O{x · Ax} = maxy∈O{T y · AT y}
= maxy∈O{y T T T AT y} = maxy∈O{y T Uy}
= maxy∈O{
n
j=1
minx∈OR(x) = miny∈O{
µjy 2 j }.
n
j=1
µjy 2 j } = m.
By the ordering of the eigenvalues it follows that
for y ∈ O. Similarly,
n
j=1
µjy 2 j
≤ µ1
y 2 j = µ1(y · y) = µ1,
n
µjy 2 j ≥ µn, if y ∈ O.
j=1
Furthermore, y = (1, 0, . . . , 0) T and y = (0, . . . , 0, 1) T yield µjy 2 j = µ1 and
µjy 2 j = µn, respectively. Since (1, 0, . . . , o) T and (0, . . . , 0, 1) T belong to
O, we have M = µ1 and m = µn. This proves the following theorem.

22.1. REAL SYMMETRIC MATRICES: A REVIEW 919
Theorem 22.1.1. Let A be an n×n real symmetric matrix, with eigenvalues
µ1 ≥ · · · ≥ µn. Then for any nonzero x ∈ R n ,
µn ≤ R(x) ≤ µ1, (22.4)
µ1 = maxx∈OR(x), µn = minx∈OR(x). (22.5)
Using the same notation as above, we give an additional useful result.
Theorem 22.1.2. If 0 = x ∈ R n satisfies R(x) = µi, for either i = 1 or
i = n, then x is an eigenvector of A belonging to the eigenvalue µi.
Proof. Without loss of generality we may assume x ∈ O, and we first consider
the case R(x) = xT Ax = µn. Let x = cixi, so that xT Ax = c2 i µi
≥
2
µn ci = µn, with equality holding if and only if µi = µn whenever ci = 0.
Hence R(x) = µn if and only if x belongs to the eigenspace associated with
µn. The argument for µ1 is similar.
We continue to let A = (aij) denote an n × n real symmetric matrix. Let
∆ = ∆1 + · · · + ∆r and Γ = Γ1 + · · · Γs be partitions of {1, . . . , n}. Suppose
that Γ is a refinement of ∆, and write i ⊆ j whenever Γi ⊆ ∆j, 1 ≤ i ≤ s.
Set
and
δij = aµν; µ ∈ ∆i; ν ∈ ∆j;
γij = aµν; µ ∈ Γi; ν ∈ Γj.
So δij = δji and γij = γji by the symmetry of A. Define the following
matrices:
A ∆ =
δij
δi
1≤i,j≤r
; A Γ =
γij
γi
;
1≤i,j≤s
Ā∆ = diag( δ1, . . . , δr); ĀΓ = diag( √ γ1, . . . , √ γs);
Â∆ = Ā∆A ∆ ( Ā∆) −1 =
ÂΓ = ĀΓA Γ ( ĀΓ) −1 =
δij
δiδj
1≤i,j≤r
γij
√ .
γiγj
1≤i,j≤s
;

920 CHAPTER 22. APPENDIX 4: INTERLACING OF EIGENVALUES
Hence Â∆ and ÂΓ are real symmetric matrices with real eigenvalues equal
to those of A ∆ and A Γ , respectively. The smallest and largest eigenvalues
of Â∆ and ÂΓ are the minimum and maximum, respectively, of y T Â∆y/y T y
and x T ÂΓx/x T x, 0 = y ∈ R r , 0 = x ∈ R s .
Let y = (y1, . . . , yr) T be any nonzero vector in Rr . Then put x =
(. . . , xα, . . .) T
, where xα = yi γα/δi, whenever α ⊆ i (i.e., Γα ⊆ ∆i). Then
s
α=1
=
implying x T x = y T y. And
=
x 2 α =
r
i=1
r
i=1
y 2 i
δi
x T ÂΓx =
α⊆i
α∈i
s
α,β=1
γα
yi
r
P
α ∈ i β ∈ j γαβ
√
γαγβ
i,j=1
=
=
r
i,j=1
r
i,j=1
yi
yi
P
γα/δi
=
r
i=1
2 y 2 i ,
γαβ
xα √ xβ
γαγβ
· yi
√
γα
√
δi
α ⊆ i β ⊆ j γαβ
δiδj
δij
δiδj
This implies that any value of
is a corollary of Theorems 22.1.1 and 22.1.2
· yj
√
γβ
δj
yj
yj = y T Â∆y.
yT Â∆y
yT is also a value of y
xT ÂΓx
xT , so the following
x
Theorem 22.1.3. If µ1 ≤ µ2 ≤ · · · ≤ µr are the eigenvalues of A ∆ and
λ1 ≤ · · · ≤ λs are the eigenvalues of A Γ , then λ1 ≤ µ1 ≤ µr ≤ λs. If
y = (y1, . . . , yr) T satisfies A ∆ y = λ1y (so λ1 = µ1), then A Γ x = λ1x, where
x = (· · · xα · · · ) T is defined by xα = yi whenever α ⊆ i. A similar result
holds in case λs = µr.

22.1. REAL SYMMETRIC MATRICES: A REVIEW 921
Proof. The first part of the theorem is evident. So let 0 = y = (y1, . . . , yr)
satisfy A∆y = λ1y = µ1y. Then Ā∆y
√ √
= (y1 δ1, . . . , yr δr) T is an eigenvector
of Â∆ belonging to λ1 = µ1. This goes both ways:
(y1, . . . , yr) T is an eigenvector of A ∆ belonging to µ if and only if
(y1 δ1, . . . , yr δr) T is an eigenvector of Â∆ belonging to µ.
Similarly,
(x1, . . . , xs) T is an eigenvector of A Γ belonging to λ if and only if
√ √
(x1 γ1, . . . , xs γs) T is an eigenvector of ÂΓ belonging to λ.
Since in this case Ā∆y
√ √
= (y1 δ1, . . . , yr δr) T is an eigenvector of Â∆
belonging to λ1 = µ1, by the proof of Theorem 22.1.2 z = (· · · zα · · · ) T ,
√
zα = yi γα for α ⊆ i, is an eigenvector of ÂΓ belonging to λ1. It follows that
x = (· · · xα · · · ) T , xα = yα for α ⊆ i, is an eigenvector of AΓ associated with
λ1. A similar proof holds in case λs = µr.

922 CHAPTER 22. APPENDIX 4: INTERLACING OF EIGENVALUES

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Index
(0, 2)-set, 157
(p, L)-elation, 425
(p, L)-transitive, 426
(x, L, y)-elation, 426
(x, L, y)-symmetry, 426
G-module, 896
T2(O), 447
Y × Z, 320
Y · Z, 320
α-clan, 765
α-cone, 759
α-derivation, 232
α-flock, 761
α-flock, normalized, 763
⊲⊳ equivalence relation, 709
ˆ⊲⊳, 693
P(V ), 64
D(f), 179
ρ(t)-flip, 773
k-arc, 157, 379
k-arc in a GQ, 379
k-cap, 330
k-ply transitive, 892
q-binomial theorem, 87
q-isomorphism, 266
[GQ], 741
3-regular, 681
4-cap, 746
absolute point (line), 417
admissible pair, 877
941
affine coordinate, 105
affine plane, 16, 27
alternating, 904
anisotropic, 246, 914
anisotropic matrix, 588
antiregular, 396, 409
apolar linear complexes, 554
arithmetic function, 815
axial generator, 759
axial pencil, 573
Axiom of Veblen, 94
axis, 81, 122
axis of collineation, 207
axis of oval, 208
axis of symmetry, 625
base point, 636
basis of a projective space, 50
Bezout’s Theorem, 853
bowtie equivalence, 749
bundle, 297, 574, 638
cap, 295
carrier (of a cone), 757
carrier of a pencil, 367
carriers of a flock, 295
center, 81, 122
center of collineation, 207
central collineation, 122, 207
Cherowitzo ovals, 785
Chinese Remainder Thm (CRT), 800

942 INDEX
circle, 279
circle plane, 636
circles, 732
classical internal structure, 737
collineation, 121
compatible ovals, 347
complete quadrilateral, 110
cone, 242, 757
congruent, 251
conic, 160
conjugate, 240
constricting about a regular spread,
464
coordinate matrix, 249
coregular, 409
corner, 573
correlation, 103
cyclotomic coset, 354
degenerate, 240, 251, 908
dependent set of points, 50
derived plane, 734
Desargues’ configuration, 68
Desargues, theorem of, 67
dilatations, 133
dimension of projective space, 54
Dirichlet convolution (product), 816
discriminant, 262, 832
division ring, 901
doily, 380
dual coordinate vector of line, 548
dual tetradic set, 727
duality (of GQ), 417
egglike inversive plane, 639
elation, 123, 207, 425
elementary symmetric function, 826
elliptic congruence, 556
elliptic quadric ovoid, 342
elliptic space, 257
equivalent matrices, 587
Euler’s Criterion, 805
Euler’s phi-function, 815
Eulerian generating function, 90
expansion about a regular point,
445
exponent of a polynomial, 820
exterior line, 157
exterior point of oval, 158
external lines lemma, 222
faithful, 892
fan of ovals, 347
flag, 18, 54
flippable, 773
flock, 293
flock of elliptic quadric, 295
flock of hyperbolic quadric, 279
flock of quadratic cone, 291
flock, linear, of I, 645
frame, 149
Gaussian coefficient, 83
generalized f-fan, 794
generalized quadrangle, 377
generator of a cone, 757
generator of cone, 289
germ, 246
Glynn’s Criterion for ovals, 187
Glynn’s hyperovals, 217
Glynn’s ovals 3σ + 4, 769
Glynn’s ovals γ + σ, 784
gram matrix, 911
Greek plane, 541
group action, 889
harmonic conjugate, 111

INDEX 943
Hensel’s Lemma, 844
herd, 771
Hermite-Dickson Criterion, 855
Hermitian, 905
hermitian polarity, 103, 104
Higman’s inequality, 382
homogeneous coordinates, 76
homography, 80, 105, 154, 207
homology, 207
hull, 16, 18
hyperbolic congruence, 556
hyperbolic line, 243, 379, 914
hyperbolic space, 243, 257
hyperboloid, 118
hyperoval, 159
hyperoval cone, 293
hyperoval flock, 293
hyperplane, 238
in perspective from a point (line),
68
independent set of points, 50
index of quadratic form, 268
index of singular orthogonal transformation,
273
interior point of oval, 158
internal structure, 735
inversive plane, 638
isometry, 242, 262, 906
isotropic, 243, 907
Klein correspondence, 539
Lagrange interpolation, 824
Laguerre geometry, 733
Laguerre plane, 636
Laguerre point, 731
larc, 417
Latin plane, 541
lattice, 82
Lemma of Tangents, 175
linear complex, 551
linear congruence, 554
linear flock, 279, 759
linear space, 15
linearized q-polynomial, 199, 865
local secant parameter set, 729
Lucas, 799
Möbius inversion, 817
Möbius plane, 636, 638
maximal flag, 54
Minkowski plane, 636, 674
monomial o-polynomial, 180
monomial oval cone, 758
Moufang panel, 426
net, 412
Newton’s Formulas, 830
noed, 573
non-axial pencil, 574
non-homogeneous coordinate, 105
nondegenerate, 240, 908
norm, 823
nuclear generator, 289, 293, 759
nucleus of oval, 159
null polarity, 102, 336, 557
null system, 557
o-permutation, 180
o-polynomials, 180
opposite regulus, 117
orbit-counting lemma, 891
order of P GL(3, q), 155
order of finite GQ, 378
order of inversive plane, 638
ordered frame, 149
Orr, 639, 664

944 INDEX
Orr’s Theorem, 328
orthogonal, 240
orthogonal linear map, 266
orthogonal polar space, 906
orthogonal polarity, 103
oval, 158
oval derivation, 229
ovals compatible at a point, 793
ovoid in P G(3, q), 329
ovoid of H5, 582
ovoid of GQ, 416
ovoid of Klein quadric, 585
ovoidal inversive plane, 639
packing, 463
packing (of spreads), 461
packing of P G(3, q), 582
panel, 426
Pappus’ configuration, 71
Pappus, Theorem of, 71
parabolic congruence, 556
parabolic space, 257
partial flock, 293
partial geometry P aG(α, s, t), 93
pencil, 367, 638
pencil of circles, 636
perp, 379, 908
perp perp, 379
pivotal spread, 463
Plücker coordinates, 537
plane representation theorem, 346
pointed conic, 179, 211
polar, 103, 240, 336
polar form, 238
polar form fQ, 905
polar space, 905
polarity, 102, 336, 338
polarity of GQ, 417
pole, 103, 336
primitive element, 819
primitive root, 799, 807
principle of duality, 48
projection of a flock, 291
projective collineation, 148
projective geometry, 54
projective ovoid of T2(O), 509
projective plane, 16
projective semilinear group, 80
projective space; synthetic, 17
Property (G), 743
property (G), 609, 683
pseudo polarity, 103
pseudoline, 497
quadratic cone, 289
quadratic form, 238, 905
quadratic reciprocity, 840
quadratic residue, 839
quadric, 160, 238
quiver, 491
quotient geometry, 56
quotient space, 909
radical, 908
reciprocity, 101
reduced representative set, 354
regular, 409
regular hyperoval, 179, 211
regular oval, 179
regular point or line, 379
regular spread, 463
regular spread of P G(3, q), 564
regulus, 93, 117
resultant, 830, 848
reverse polynomial, 834
rosette of ovoids, 310

INDEX 945
secant line, 157
Segre’s oval, 183
Segre’s Theorem, 174
self-polar, 104
semi-oval, 350
semi-ovoid, 350
semilinear, 904
semilinear map, 80
semisimilarity, 242, 906
sesquilinear, 904
similarity, 242, 906
singular element of G, 273
singular point, 240
singular vector, 240, 907
skew-field, 901
skew-symmetric, 904
skew-translation GQ (STGQ), 433
span, 16, 379
span of pointsets, 18
spread of a GQ, 417
Stickelberger’s Theorem, 836
subprimitive, 821
subspace of linear space, 15
sum function, 817
symmetric polynomials, 826
symmetry, 426
symmetry with respect to hyperplane,
271
symplectic GQ W (q), 385
symplectic polar space, 905
symplectic polarity, 102, 557
syntheme-duad geometry, 376
tangent line, 157
tangent line to an ovoid, 329
tetrad, 689
tetradic set, 689
Thas-Walker construction, 585
Tits ovoid, 342
totally isotropic, 243, 907
totally singular, 907
trace, 379, 822
trace (absolute), 218, 259
transitive, 892
translation, 123
translation dual, 732
translation oval, 208
transversal of ovoids, 695, 745
triad, 379
trivial subspace, 54
unitary polar space, 905
unitary polarity, 103, 104
Vandermonde determinant, 884
Wedderburn’s Theorem, 901
whorl, 425
Witt index, 246, 913