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Gruber P. Convex and Discrete Geometry

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312 <strong>Convex</strong> Polytopes<br />

Embed E d into E d+1 as usual (first d coordinates) <strong>and</strong> let q1,...,qd+1 ∈ Z d+1<br />

be the vertices of the lattice simplex T = S + (0,...,0, 1) in the hyperplane<br />

{x : xd+1 = 1}. The lattice Z d+1 has determinant d(Z d+1 ) = 1. Let L be the<br />

sublattice of Z d+1 with basis {q1,...,qd+1}.ThesetF ={α1q1 +···+αd+1qd+1 :<br />

0 ≤ αi < 1} is a fundamental parallelotope of L. Thus, for the determinant of L<br />

we have d(L) = Vd+1(F), where Vd+1 st<strong>and</strong>s for the volume in E d+1 . Then the<br />

following hold.<br />

(2) For each u ∈ Z d+1 , there is precisely one point v ∈ F ∩ Z d+1 such that<br />

u ∈ v + L. Further, Vd+1(F) = #(F ∩ Z d+1 ) = h, say.<br />

If u = β1q1 +···+βd+1qd+1 ∈ Z d+1 , then v = (β1 −⌊β1⌋)q1 +···+(βd+1 −<br />

⌊βd+1⌋)qd+1 is the unique point v ∈ F ∩ Z d+1 with u ∈ v + L. The index of the<br />

sublattice (subgroup) L in the lattice (group) Z d+1 is thus #(F ∩ Z d+1 ). Since the<br />

index is also equal to d(L)/d(Z d+1 ) = Vd+1(F), we see that Vd+1(F) = #(F ∩<br />

Z d+1 ), concluding the proof of (2). Compare Sect. 21.3. For each u ∈ nT ∩ Z d+1 ,<br />

according to (2), we have<br />

�d+1<br />

(3) u = v + miqi with suitable integers mi ≥ 0.<br />

i=1<br />

To see this, note that u = β1q1 +···+βd+1qd+1 with βi ≥ 0 <strong>and</strong> put mi =⌊βi⌋.<br />

Considering the last coordinate, (3) shows that<br />

�d+1<br />

(4) n = vd+1 + mi with integers mi ≥ 0.<br />

i=1<br />

Conversely, given v ∈ F ∩ Z d+1 , any solution of (4) in integers mi ≥ 0 gives rise to<br />

a unique point u ∈ nT ∩ Z d+1 with u ∈ v + L. The number of such solutions of (4)<br />

equals the coefficient of t n in the power series<br />

t vd+1 2 d+1 t<br />

(1 + t + t +···) =<br />

vd+1<br />

(1 − t) d+1<br />

= t vd+1<br />

� � � � �<br />

−d − 1 −d − 1<br />

1 + (−t) + (−t)<br />

1<br />

2<br />

2 �<br />

+···<br />

= t vd+1<br />

� � � � �<br />

d + 1 d + 2<br />

1 + t + t<br />

1<br />

2<br />

2 �<br />

+··· ,<br />

where we have applied Newton’s binomial series. Hence<br />

(5) L(nS)(= L(nT)) equals the coefficient of tn in the power series<br />

� � � � �<br />

d + 1 d + 2<br />

q(t) 1 + t + t<br />

1<br />

2<br />

2 �<br />

+··· , where q(t) = �<br />

v∈F∩Z d+1<br />

t vd+1 .<br />

To determine the degree of q(·), note that each v ∈ F ∩ Z d+1 can be represented<br />

in the form v = α1q1 +···αd+1qd+1 where 0 ≤ αi < 1. Each qi has last coordinate

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