subregular nilpotent representations of lie algebras in prime ...

Because u is the nilradical of a parabolic subalgebra with Levi factor l h, the
set of all with g u is stable under the Weyl group of l. Nows belongs to
that Weyl group; this implies s ((,R +
2 ) [ R+ 3 )=(,R+ 2 ) [ R+ 3 , hence
0=
X
2(,R +
2 )[R+ 3
h ; _ i = ,a2 + a3:
So a2 = a3; plugging this into (1) we get our claim.
E.5. Set
1 = X
2R +
2
47
: (1)
Each 2 I belongs to the basis of R with positive system R + , and it belongs to
the basis of R1 with positive system R + \ R1. This implies h ; _ i =1=h 0 ; _ i.
So Lemma E.4 shows that h 1; _ i = 0, hence 1(h ) = 0 for all 2 I. It follows
that 1 vanishes on the intersection of h with the derived Lie algebra of gI. We get
therefore a one dimensional gI{module where each x with 2 R \ ZI acts as 0,
and each h 2 h as 1(h). This is a restricted gI{module. Its tensor product with
a U (gI){module, say N, is again a U (gI){module; we shall denote it be N 1.
Proposition: Let M be a nite dimensional U (gI){module extended trivially to
a U (pI){module. Then we have an isomorphism of U (l){modules
(U (g) U (pI) M) u ' U (l) U (l\pI) (M 1): (2)
Proof : Abbreviate the left hand side in (2) by M 0 . Write x = Q 2R +
2
x p,1
, .
Lemma E.3 implies that the subspace xM of M 0 is stable under all x satisfying
2 R1 \ R + or 2 (,R + ) \ ZI. (Note that M is stable under these x .) On the
other hand, we have for all h 2 h and m 2 M
hxm= xhm , X
2R +
2
(p , 1) (h) xm = xhm + 1(h) xm: (3)
Because l \ pI is spanned by h and the x as above, it follows that xM is a
U (l\pI){submodule of M 0 .Formula (3) and Lemma E.3 show that this submodule
is isomorphic to M 1.
The universal property of an induced module yields a homomorphism of U (l){
modules
U (l) U (l\pI) xM ! M 0 ; u xm 7! uxm:
Lemma E.2 implies that this map is bijective. The claim follows.