subregular nilpotent representations of lie algebras in prime ...

Since we assume p to be good, h and h are linearly independent inh.
Therefore X1 is an a ne subspace of h of codimension 2.
If isarootinQ + Q , then h 2 Kh + Kh in h (since p is good).
Each f 2 X1 coincides with (or rather d )onh and h , hence on all h with
2 R \ (Q + Q ). We have then f(h ) 2 Fp and (f)(h ) = 0 (recall A.2) for
these .
On the other hand, if 2 R + with =2 Q + Q , then h is linearly independent
ofh and h (since p is good). It follows that we can nd f 2 X1 with
f(h ) =2 Fp. Therefore the set
X reg
1 = f f 2 X1 j f(h ) =2 Fp for all 2 R + , =2 Q + Q g (3)
is open and dense in X1. By our remarks above, Proposition F.1 will follow from:
Claim: Each f 2 X reg
1
satis es (2).
Proof (of Claim): Let f 2 X reg
1 . The de nition of Xreg 1
implies (cf. A.2)
f 2 R j (f)(h )=0g = R \ (Q + Q ): (4)
We want to apply E.6/7 with 0 = (f) and 1 = and I = f g. [So the in
E.6 is our present (f)+ .] The Lie subalgebra l as in E.7 is by (4) now equal to
l = h
M
2R\(Q +Q )
53
g : (5)
We get from E.6 a linear form, say 0 ,ong [denoted by 0 1 in E.6] with 0 (p )=0
such that (f) + 0 is the Jordan decomposition of some conjugate of (f) + .
Furthermore E.7 yields an equivalence of categories F. Its description in E.7(1)
involves a certain element 1 2 X satisfying 1(h ) = 0. The last property implies
by A.3(4) for all 2 X
L (f); (f + ) 1 ' L (f); (f + + 1):
Therefore E.7(1) applied to M = L (f); (f + )yields
F Z(f + ; ; ) ' Z (f)+ 0(f + + 1; ; l) (6)
using the notation from A.5.
Note that d ,f is a linear form on h that vanishes on Kh +Kh . So A.4(5)
applied to l instead of g yields
Z (f)+ 0(f + ; ; l) Kd ,f ' Z 0( + ; ; l) (7)
for all 2 X. Set now F 0 equal to the composition of F with the functor N 7!
N Kd ,f . Then F 0 is again an equivalence of (appropriate) categories; it satis es
by (6) and (7)
F 0 Z(f + ; ; ) ' Z 0( + + 1; ; l):