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48 E.6. We now drop the assumptions xed throughout E.1{5 and consider a more general situation. Let = 0 + 1 2 g with 0(n , + n + )=0 and 1(b + )=0: (1) So this looks like E.1(1); but we no longer assume that = 0 + 1 is a Jordan decomposition, i.e., the analogue of E.1(2) will not hold in general. On the other hand, we x again a set I as in E.2 and assume the analogue of E.2(1): 0(h )=0 for all 2 I and 1(pI) =0: (2) Let PI be standard parabolic subgroup of G with Lie algebra pI. Denote its unipotent radical by Ru(PI). For each g 2 Ru(PI) and all x 2 pI the di erence Ad(g ,1 )(x),x belongs to the nilradical of pI. Since vanishes on that nilradical, we get (Recall that (g )(x) = (Ad(g ,1 )(x)). (g ) jpI = jpI : (3) Lemma: There exist g 2 Ru(PI) and 0 1 2 g with 0 1(pI) =0such that g has Jordan decomposition g = 0 + 0 1 . Proof : This follows from the proof in Subsection 3.8 in [13]. If we apply the construction there to our (as their l), then we get an element g 2 G such that g has Jordan decomposition g = 0 + 0 1 where 0 1 = g , 0. Wejusthave to check that the g used there actually is in Ru(PI). (If so, then we get 0 1(pI) =0 from 1(pI) = 0, since 0 1 coincides by (3) with 1 on pI.) Well, in [13] one considers the set (denoted by ) of all positive roots with (h ) 6= 0. Then g is constructed as a product of elements from the root subgroups U with 2 . Now our assumption (2) says that R + n ZI. This implies that g 2 Ru(PI) as claimed. E.7. Keep the assumptions of E.6 and choose g as in Lemma E.6. We can apply E.1{5 with s = 0, n = 0 1 and with replaced by g . Set in particular R1 = f 2 R j 0(h )=0g and l = h L 2R1 g .Weget also 1 as in E.5(1). Proposition: There exists an equivalence ofcategories such that F : U (g){modules ! Ug (l){modules F(U (g) U (pI) M) ' Ug (l) U (l\pI) (M 1) (1) for all nite dimensional U (gI){modules M extended trivially to a pI{module. Proof :We construct F as a composition of two equivalences. The rst one is U (g){modules ! Ug (g){modules; V 7! g V (2)
where g V is equal to V as a vector space and where each x 2 g acts on g V as Ad(g ,1 )(x) actsonV . In case V = U (g) U (pI) M with M as above one checks easily that g V ' Ug (g) Ug (Ad(g)pI) g M ' Ug (g) U (pI) g M: For the second equality use that Ad(g)pI = pI since g 2 Ru(PI) and apply E.6(3). For each x 2 pI the di erence Ad(g ,1 )x , x belongs to the nilradical of pI. This nilradical acts trivially on M. Therefore g M is isomorphic to M as a pI{module. We get thus g V ' Ug (g) U (pI) M: (3) Let F be the composition of the functor in (2) with the equivalence of categories V 0 7! (V 0 ) u from [6], with u as in E.1. Then F is an equivalence of categories from U (g){modules to Ug (l){modules with F(V )=( g V ) u . Combining (3) and Proposition E.5 we get (1). F We assume in this section that (B1), (B2), and (D1) hold. From F.5 on we shall also assume (D2). F.1. Let be a simple root. We write (as before) p = b + g, for the corresponding minimal parabolic subalgebra of g. Let . We want to show: be a root orthogonal to Proposition: Suppose that the root system R \ (Q + Q ) has type A1 A1 or B2. In the second case assume that is a short root in that subsystem. Then we have Homg(Z (s ; );Z ( ; )) 6= 0 (1) for all 2 X and all 2 g with (p )=0. The proof will occupy the following subsections until F.4. We shall rst prove the proposition in the rank 2 case (i.e., if R = R \ (Q + Q )) and then use the `old S' to reduce to that case. Remark: This proposition does not generalise to the situation where R\(Q +Q ) has type B2 and where is long. For example, if R is of type B2, then [10], 3.13 shows that there exist simple modules Z ( 1; 1) and Z ( 3; 1) that are not isomorphic to each other, but where 3 = s 1 with the positive root orthogonal to 1. F.2. Lemma: Proposition F.1 holds if R is of type A1 A1. Proof : In this case also is a simple root. Let e be the integer with 0 e
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 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 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

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