subregular nilpotent representations of lie algebras in prime ...

extend Theorem D.6 to type G2, wewould have to show thatM1=M2 has length 3
with all factors of dimension h + ; _ 2 ip N ,1 and invariant f 2g. Theway that
we handle such problems in types Cn and F4 was to look at another subregular 0
and to get simple modules of the right type by twisting. If we dothishere,we get
one simple module L that has to occur as a composition factor of M1=M2 having
the expected dimension and the expected invariant. But one such factor is not
enough when the expected length is 3.
D.12. Return to the situation from Theorem D.6. So we exclude the case where
R is of type G2; ifR is of type E8 or F4, assume that p>h+1. IfR is of type
Bn set = n; otherwise let be the same simple root that was used in D.8{10.
So is a short simple root in all cases. Choose subregular with (p ) = 0. We
claim under these assumptions:
Lemma: Let 2 C 0 0 . Let be a short simple root with 2 J( ). IfLis a simple
module in C with (L) =f g, then there exists an element x 2 W with x =
and L ' Z (x ; ).
Proof : If the claim holds for one as above, then it holds also for all g with
g 2 P ; this follows easily from B.14(3) and D.5(3). This means by Lemma D.1
that it su ces to prove the claim for one special .We can therefore assume that
(x, ) 6= 0 for all simple roots 6= .
Suppose at rst that 6= . (Note that this case does not occur for R of type
Bn.) Since L is isomorphic to one of the L ;i from Theorem D.6, the proofs in
D.8{10 show that L is isomorphic to one of the factors Mj=Mj+1 with 1