subregular nilpotent representations of lie algebras in prime ...

C.12. Here and in the next two subsections we x a simple root with (x, )=0
and consider I as in C.11(1). Note that
WI ( + ZI) \ R R + : (1)
Pick for each 2 WI an element w 2 WI with w ,1 = and set
Lemma C.11.b implies that
and that
27
M =sbm(s w ; ): (2)
dim M =(p ,h + ; _ N ,1
i)p
(3)
M ' Z (s w ; ): (4)
Claim: The submodule M of Z ( ) depends only on , not the choice ofw .
Proof :Ifw 0 is a second element inWI with (w 0 ) ,1 = , then w = w 0 w ,1 2 WI
satis es w = , hence ws = s w. It follows that s w 0 = ws w 2 WIs w ;
now apply Lemma C.11.a.
C.13. Lemma: Let be a simple root, let w 2 W with w 2 WI . Then
sbm(w; )= sbm (ws ; ) is a homomorphic image of Z (ww w ; ).
Proof : The element x = ww w satis es x = , hence xs = s x. Since ww 2
WI, Lemma C.11.a implies that
and
sbm(w; )=sbm(x; )
sbm(ws ; )=sbm(xs ; )=sbm(s x; ):
Now the claim follows from Lemma C.7.c.
C.14. Proposition: Let 2 WI and 2 I with h ; _ i < 0.
a) We have
and
dim(M =M s
M M s
(1)
)=jh ; _ ij h + ; _ ip N ,1 : (2)
b) Assume that h ; _ i = h ; _ i = ,1. Then 0 = w ( + ) is a root and
belongs to WI ; the element x = w 0s w 2 W satis es x ,1 = and
M =M + ' Z (x ; ) if h + ; _ i > 0. (3)