subregular nilpotent representations of lie algebras in prime ...

More documents

Recommendations

Info

6 given by '(v 0 f ,d )=xd , v 0 f . [Note that Z ; (f , d )makes sense because (f) = (f , d )by A.2(4).] If (x, ) 6= 0, then 'f is an isomorphism since x p , acts as the scalar (x, ) p on these modules. If (x, ) = 0, then the image of 'f is spanned over K by all xi , v0 f with d i
A.5. Let R 0 be a root subsystem of R. Then g 0 = h L 2R 0 g the Lie algebra of a reductive closed subgroup of G. We can then apply A.2 also to g 0 working with b + \ g 0 instead of b + .We shall write then Z (f; g 0 )=U (g 0 ) U (b + \g 0 ) Kf (1) for the analogue of Z (f). In case 2 R 0 we can also extend A.3/4 and get Z (f; ; g 0 )=U (g 0 ) U (p \g 0 ) L ; (f) (2) as an analogue of Z (f; ). We do not introduce extra notations for the analogues of Z ; (f) and L ; (f) because the analogous U (p \ g 0 ){modules are just the restrictions of the corresponding U (p ){modules. A.6. In some cases we shall have to consider for a given f 2 h at the same time all such that Z (f; ) is de ned. In that situation it will be convenient towork with a slightly modi ed notation. Set X = f f 2 h j f(h ) 2 Fp g: (1) This is a union of p a ne subspaces of h of codimension 1 unless h =0(inwhich case X = h ). Set =f 2 g j (p )=0g: (2) Extend each (f) tog such that (f)(n + +n , ) = 0. Then Z (f)+ (f; ) is de ned for all f 2 X and 2 . We write Z(f; ; )=Z (f)+ (f; ): (3) Note that then also all Z(f + ; ; )with 2 X are de ned because f + 2 X. A.7. Proposition: Let 1; 2 2 X. For each integer m the set f(f; ) 2 X j dim Homg(Z(f + 1; ; );Z(f + 2; ; )) m g (1) is closed inX . Proof : The point istoshow that the Hom space in (1) can be described as a space of solutions to a system of linear equations where the matrix of the system has size independent of(f; ) and entries that are polynomial functions of f and .If we have, say, l unknowns, then the Hom space has dimension m if and only if the rank of the matrix is l , m if and only if all (l , m +1) (l , m + 1) minors of the matrix are 0. This condition clearly de nes a closed subset of X . In order to show that we are in a situation as described above,letmeintroduce some notation. Let R be the set of all R + {tuples of non-negative integers. To each r =(r( )) 2 R we associate elements x , r = Y >0 r( x, ) and x + Y r( r = x >0 ) 7
Page 1: U N I V E R S I T Y O F A A R H U S
Page 4 and 5: 2 positive roots . (We denote by _
Page 6 and 7: 4 A A.1. Let K be an algebraically
Page 10 and 11: 8 in U(g) where these products are
Page 12 and 13: 10 B Keep the assumptions and notat
Page 14 and 15: 12 B.5. Proposition: Suppose that a
Page 16 and 17: 14 are simple GI{modules. It follow
Page 18 and 19: 16 The assumption on the facet impl
Page 20 and 21: 18 Remark: Our assumptions (B1) and
Page 22 and 23: 20 C Keep all assumptions and notat
Page 24 and 25: 22 b) Let E be a G{module; suppose
Page 26 and 27: 24 C.8. We nowwant to generalise Le
Page 28 and 29: 26 Remark: Let denote the usual Che
Page 30 and 31: 28 Proof :a)Wehave clearly s 2 WI a
Page 32 and 33: 30 Lemma: Let J be asabove and let
Page 34 and 35: 32 0 + (b ) = 0 and 0 (x, 0)=0. Sin
Page 36 and 37: 34 Theorem: Assume that is subregul
Page 38 and 39: 36 D.9. Consider now R of type Bn o
Page 40 and 41: 38 and, if 1 2 J( ), dim T $1, (M3=
Page 42 and 43: 40 Claim: The element w1 = s wIs 0
Page 44 and 45: 42 hence that L is isomorphic to Z
Page 46 and 47: 44 E We return to the more general
Page 48 and 49: 46 for all 2 R1 and m 2 M. Proof :
Page 50 and 51: 48 E.6. We now drop the assumptions
Page 52 and 53: 50 F.3. Lemma: Proposition F.1 hold
Page 54 and 55: 52 factors shows that Z (s ; ) has
Page 56 and 57: 54 So the Hom space in (2) is isomo
Page 58 and 59:
56 Consider nally R of type F4. Wem
Page 60 and 61:
58 with j as above and j 0 = h 1; i
Page 62 and 63:
60 if h + ; _ 2 i > 0. Proof : Note
Page 64 and 65:
62 Remark: Let L be a simple module
Page 66 and 67:
64 Remarks: 1) This result con rms
Page 68 and 69:
66 H.2. Let 2 g be subregular nilpo
Page 70 and 71:
68 Assume that R is not of type A1.
Page 72 and 73:
70 (Here L runs over representative
Page 74 and 75:
72 Let 0 2 J( ) with 0 6= .Ifwewrit
Page 76 and 77:
74 and Homg(M 0 ; M) ' Homg(T ( ) M
Page 78 and 79:
76 Remarks: 1) The lemma can be ext
Page 80:
78 References [1] N. Bourbaki, Grou
show all

subregular nilpotent representations of lie algebras in prime ...

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?