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4 A A.1. Let K be an algebraically closed eld of characteristic p>0. Let G be a reductive algebraic group over K and denote by g the Lie algebra of G. This is a restricted Lie algebra as the Lie algebra of an algebraic group; we denote the p{th-power map by x 7! x [p] . Let T be a maximal torus in G and set h =Lie(T ). Let X = X(T ) the (additive) group of all characters of T and let R X be the root system of G. For each 2 R let g denote the corresponding root subspace of g. Wechoose a system R + of positive roots. Set n + equal to the sum of all g with >0andn , equal to the sum of all g with
We have clearly ( ) = 0 for all 2 X, hence (f + )= (f) for all f 2 h , 2 X. (4) Here (as usual) we write instead of d by abuse of notation. For each 2 g set = f f 2 h j jh = (f) g: (5) Given f 2 h we can regard Kf as a b + {module with n + acting trivially. Thisis then a U (b + ){module for all 2 g satisfying (n + )=0andf 2 . For all these we get then an induced U (g){module (a \baby Verma module") Z (f) =U (g) U (b + ) Kf : (6) Denote its \standard generator" 1 1by vf . As before we usually write Z ( ) instead of Z (d ) for 2 X; we should keep in mind that Z ( ) depends only on d , hence on the coset + pX. If f 0 2 h satis es f 0 (h ) = 0 for all 2 R, then we can extend Kf 0 to a g{module with all x acting as 0. This is then a U (f 0 )(g){module where we extend (f 0 ) to a linear form on g such that (f 0 )(n + + n , ) = 0. A trivial form of the tensor identity yields then Z (f) Kf 0 ' Z + (f 0 )(f + f 0 ) (7) for all f 2 h and 2 g with f 2 and (n + )=0. A.3. Fix a simple root for the rest of Section A. Denote by p = b + + g, the corresponding parabolic subalgebra. For all f 2 h and 2 g with (n + )=0 and f 2 consider the induced U (p ){module Z ; (f) =U (p ) U (b + ) Kf : (1) Denote the \standard generator" 1 1nowby v0 f . It satis es x v0 f = 0 for all 2 R + and hv0 f = f(h)v0 f for all h 2 h; thexi , v0 Z ; (f). We have f with 0 i
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 8 and 9: 6 given by '(v 0 f ,d )=xd , v 0 f
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
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58 with j as above and j 0 = h 1; i
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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
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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
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