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56 Consider nally R of type F4. Wemay assume that (p 1) = 0. The set of roots orthogonal to 1 is a root system of type C3. (Note that 1 is conjugate to the largest root, i.e., to $1.) A basis of this root system is 4, 3, "2 + "3 = 1 +2 2 +2 3. The stabiliser W1 of 1 is the Weyl group of this subsystem. The short roots orthogonal to 1 formarootsystemoftype D3 = A3. The corresponding Weyl group W2 is a subgroup of index 2 in W1; wehave W1 = W2 [ W2s"2+"3: (3) We have 2f 1; 2g. The simple modules in C with invariant f g are (by D.3.a, D.5.c and D.14) the Z (x ; 1) withx 2 W and x = 1. The x with this property are a coset for W1, in fact equal to W1 in case = 1, equal to W1s2s1 in case = 2. Proposition F.1 implies for all w 2 W2 (and x as above) that Z (wx ; 1) ' Z (x ; 1). This yields for all x 2 W1 in case = 1 and in case = 2 Z (x ; 1) ' Z ( ; 1); if x 2 W2, Z (s"2+"3 ; 1); if x=2 W2, Z (xs2s1 ; 1) ' Z (s2s1 ; 1); if x 2 W2, Z (s"2+"3s2s1 ; 1); if x=2 W2. Remark: Lusztig's conjectures predict that we should have two classes also for types Cn and F4. F.7. Given 2 g each g 2 CG( ) (the stabiliser of under the coadjoint action) permutes the isomorphism classes of simple U (g){modules via L 7! g L. Since there are only nitely many classes, one sees easily that CG( ) 0 , the connected component of the identity inCG( ), acts trivially. So does the centre Z(G). This means that we are really looking at an action of the \component group" A( )=CG( )=(Z(G)CG( ) 0 ). (It is really the component group for the adjoint group.) Note that this action is \the same" for all ina xedG{orbit: Given h 2 G we have CG(h )=hCG( )h ,1 ; therefore conjugation with h induces an isomorphism A( ) ,! A(h ). Furthermore, the map L 7! h L induces a bijection from the set of isomorphism classes of simple U (g){modules to the corresponding set for Uh (g). This bijection is compatible with the actions of A( ) and A(h )identi ed as above. We have used before that U(g) G acts on g L as it does on L. Therefore the action of A( ) permutes (for nilpotent) the simple modules in each C . (For general one would have tointroduce a new notation.) By D.5(3) the action of A( ) preserves the invariant (when de ned). Suppose now that is subregular and nilpotent. If all roots have the same length, then F.5 and the remarks above show that A( ) acts trivially. Actually, we have in these cases A( ) = 1 at least for large p, see [20], 7.5. In the next subsections we are going to prove: (4) (5)
Proposition: Suppose that R has type Bn, Cn, orF4; ifR has type F4, assume that p>h+1. Let 2 C 0 0 , let be a long root with 2 J( ). If is subregular, then A( ) permutes transitively the isomorphism classes of simple modules in C with invariant f g. Remarks: 1)Wehave A( ) ' Z=2Z for the groups considered in the proposition, at least if p is large, see [20], 7.5. (In types Bn and Cn it su ces to assume that p>2, see [21], IV.2.26.) 2) The claim is of course trivial in the (unexpected) case that there is only one isomorphism class of such modules. If however there are two and if L and L 0 are representatives for these classes, then the proposition implies [by B.13(3)] that L and L 0 occur with the same multiplicity inZ ( ) in case (b + )=0. That multiplicity has then to be equal to m =2 in the notation from D.6(1). F.8. We look rstattype Bn. I shall now use freely the notations and assumptions from [10], Section 3. In particular, we have subregular with (x, ) 6= 0if and only if 2f 2; 3;:::; ng. Wework with an arbitrary 2 C0. We can nd in NG(T ) a representative g of s"1 with g = . (Each representative g will satisfy Ad(g)x, i 2 Kx, i for all i>1 since s"1 i = i. Multiplying g with a suitable element inT we can get Ad(g)x, i = x, i for all i>1. Then g = holds.) Claim: Suppose that 1 i 2n with bi
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U N I V E R S I T Y O F A A R H U S
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2 positive roots . (We denote by _
Page 6 and 7:
4 A A.1. Let K be an algebraically
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 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
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