subregular nilpotent representations of lie algebras in prime ...

G.5. We have quite generally for all ; 2 C0 and all simple L in C
T QL ' M
(QL0)m(L;L0 )
where L 0 runs over representatives of simple modules in C and
L 0
m(L; L 0 ) = dim Hom(T QL;L 0 ) = dim Hom(QL;T L 0 )=[T L 0 : L]: (2)
Lemma: Let 2 C 0
0 ,let be a short simple root with 2 J( ). Then we have
Proof : Set = $ , .We claim that
63
(1)
n( )Q 2P 0 : (3)
Q ' T Q : (4)
Then the claim follows from G.3(2) and Lemma G.4. In order to get (4), we use
(1) and (2): If L 0 is a simple module in C with T L 0 6=0,then 2 (L 0 ), hence
L 0 ' L or L 0 ' L 0. It remains to recall that T L ' L and T L 0 ' L 0 by
G.1.
Remark: Let 2 C 0 0 ,let be a long simple root with 2 J( ). One gets similarly,
using Remark G.4 instead of Lemma G.4 that n( )Q 2P0 or n( )(Q ;1 +Q ;2 ) 2
P 0 .
G.6. Recall the integers m from D.6(3).
Theorem: Let 2 C 0 we have
0 . Suppose that all roots in R have the same length. Then
[Q0 : L0]=jW j: (1)
and
for all 2 J( ) and
for all ; 2 J( ).
[Q : L 0]=[Q 0 : L ]=jW j m (2)
[Q : L ]=jW j m m (3)
Proof : Wemay assume that (b + ) = 0. Lemma G.5 and G.4(1) imply that we
can apply G.3(5) to all simple modules L and L 0 in C .Wehave
jW ( + pX)j = jW j (4)
by C.1. So the claim follows from [Z ( ):L ]=m (see F.5) and [Z ( ):L 0 ]=1
(see C.2).