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48<br />

E.6. We now drop the assumptions xed throughout E.1{5 and consider a more<br />

general situation.<br />

Let = 0 + 1 2 g with<br />

0(n , + n + )=0 and 1(b + )=0: (1)<br />

So this looks like E.1(1); but we no longer assume that = 0 + 1 is a Jordan<br />

decomposition, i.e., the analogue <strong>of</strong> E.1(2) will not hold <strong>in</strong> general.<br />

On the other hand, we x aga<strong>in</strong> a set I as <strong>in</strong> E.2 and assume the analogue <strong>of</strong><br />

E.2(1):<br />

0(h )=0 for all 2 I and 1(pI) =0: (2)<br />

Let PI be standard parabolic subgroup <strong>of</strong> G with Lie algebra pI. Denote its<br />

unipotent radical by Ru(PI). For each g 2 Ru(PI) and all x 2 pI the di erence<br />

Ad(g ,1 )(x),x belongs to the nilradical <strong>of</strong> pI. S<strong>in</strong>ce vanishes on that nilradical,<br />

we get<br />

(Recall that (g )(x) = (Ad(g ,1 )(x)).<br />

(g ) jpI = jpI : (3)<br />

Lemma: There exist g 2 Ru(PI) and 0 1 2 g with 0 1(pI) =0such that g has<br />

Jordan decomposition g = 0 + 0 1 .<br />

Pro<strong>of</strong> : This follows from the pro<strong>of</strong> <strong>in</strong> Subsection 3.8 <strong>in</strong> [13]. If we apply the<br />

construction there to our (as their l), then we get an element g 2 G such that<br />

g has Jordan decomposition g = 0 + 0 1 where 0 1 = g , 0. Wejusthave<br />

to check that the g used there actually is <strong>in</strong> Ru(PI). (If so, then we get 0 1(pI) =0<br />

from 1(pI) = 0, s<strong>in</strong>ce 0 1 co<strong>in</strong>cides by (3) with 1 on pI.)<br />

Well, <strong>in</strong> [13] one considers the set (denoted by ) <strong>of</strong> all positive roots with<br />

(h ) 6= 0. Then g is constructed as a product <strong>of</strong> elements from the root subgroups<br />

U with 2 . Now our assumption (2) says that R + n ZI. This imp<strong>lie</strong>s that<br />

g 2 Ru(PI) as claimed.<br />

E.7. Keep the assumptions <strong>of</strong> E.6 and choose g as <strong>in</strong> Lemma E.6. We can<br />

apply E.1{5 with s = 0, n = 0 1 and with replaced by g . Set <strong>in</strong> particular<br />

R1 = f 2 R j 0(h )=0g and l = h L 2R1 g .Weget also 1 as <strong>in</strong> E.5(1).<br />

Proposition: There exists an equivalence <strong>of</strong>categories<br />

such that<br />

F : U (g){modules ! Ug (l){modules<br />

F(U (g) U (pI) M) ' Ug (l) U (l\pI) (M 1) (1)<br />

for all nite dimensional U (gI){modules M extended trivially to a pI{module.<br />

Pro<strong>of</strong> :We construct F as a composition <strong>of</strong> two equivalences. The rst one is<br />

U (g){modules ! Ug (g){modules; V 7! g V (2)

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