subregular nilpotent representations of lie algebras in prime ...
subregular nilpotent representations of lie algebras in prime ...
subregular nilpotent representations of lie algebras in prime ...
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
C.12. Here and <strong>in</strong> the next two subsections we x a simple root with (x, )=0<br />
and consider I as <strong>in</strong> C.11(1). Note that<br />
WI ( + ZI) \ R R + : (1)<br />
Pick for each 2 WI an element w 2 WI with w ,1 = and set<br />
Lemma C.11.b imp<strong>lie</strong>s that<br />
and that<br />
27<br />
M =sbm(s w ; ): (2)<br />
dim M =(p ,h + ; _ N ,1<br />
i)p<br />
(3)<br />
M ' Z (s w ; ): (4)<br />
Claim: The submodule M <strong>of</strong> Z ( ) depends only on , not the choice <strong>of</strong>w .<br />
Pro<strong>of</strong> :Ifw 0 is a second element <strong>in</strong>WI with (w 0 ) ,1 = , then w = w 0 w ,1 2 WI<br />
satis es w = , hence ws = s w. It follows that s w 0 = ws w 2 WIs w ;<br />
now apply Lemma C.11.a.<br />
C.13. Lemma: Let be a simple root, let w 2 W with w 2 WI . Then<br />
sbm(w; )= sbm (ws ; ) is a homomorphic image <strong>of</strong> Z (ww w ; ).<br />
Pro<strong>of</strong> : The element x = ww w satis es x = , hence xs = s x. S<strong>in</strong>ce ww 2<br />
WI, Lemma C.11.a imp<strong>lie</strong>s that<br />
and<br />
sbm(w; )=sbm(x; )<br />
sbm(ws ; )=sbm(xs ; )=sbm(s x; ):<br />
Now the claim follows from Lemma C.7.c.<br />
C.14. Proposition: Let 2 WI and 2 I with h ; _ i < 0.<br />
a) We have<br />
and<br />
dim(M =M s<br />
M M s<br />
(1)<br />
)=jh ; _ ij h + ; _ ip N ,1 : (2)<br />
b) Assume that h ; _ i = h ; _ i = ,1. Then 0 = w ( + ) is a root and<br />
belongs to WI ; the element x = w 0s w 2 W satis es x ,1 = and<br />
M =M + ' Z (x ; ) if h + ; _ i > 0. (3)