subregular nilpotent representations of lie algebras in prime ...
subregular nilpotent representations of lie algebras in prime ...
subregular nilpotent representations of lie algebras in prime ...
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Lemma: Let 2 g with (b + )=0.<br />
a) Let be a simple root and w 2 W with w > 0. Then<br />
(sbm(w; )= sbm(ws ; )) f g:<br />
b) If L is a composition factor <strong>of</strong> Z ( ) not isomorphic to soc Z ( ), then (L)<br />
is either empty or consists <strong>of</strong> just one simple root.<br />
c) Suppose that is a simple root with (p )=0. Let w 2 W with w ,1 2 J( ).<br />
Then Z (w ; )=fw ,1 g.<br />
Pro<strong>of</strong> :a)Wehave for all 2 J( ) [f0g by C.9(1)<br />
T $ , sbm(w; )= sbm(ws ; ) ' sbm(w; $ , )= sbm(ws ;$ , ): (4)<br />
If 6= , then we have s ($ , )=$ , , hence sbm (w; $ , )=sbm(ws ;<br />
$ , ). So the right hand side <strong>in</strong> (4) is 0 <strong>in</strong> that case; the claim follows.<br />
b) Let w0 = s1s2 :::sN be a reduced decomposition <strong>of</strong> w0. (So si = s i for some<br />
simple root i.) This leads to a cha<strong>in</strong> <strong>of</strong> submodules<br />
Z ( )=sbm(1; ) sbm(s1; ) sbm(s1s2; ) sbm(w0; ):<br />
It follows that L is a composition factor <strong>of</strong> some sbm(w; )= sbm(ws ; ) with<br />
w 2 W and a simple root with w > 0. Now the claim follows from a) s<strong>in</strong>ce<br />
quite generally (M 0 ) (M) forany subquotient M 0 <strong>of</strong> a module M <strong>in</strong> C .<br />
c) Set = w ,1 . The de nition <strong>of</strong> J( ) and <strong>of</strong> C 0 0 imp<strong>lie</strong>s that = satis es<br />
0 < hw + ; _ i