subregular nilpotent representations of lie algebras in prime ...

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So the Hom space in (2) is isomorphic to
Homl(Z 0( , a + 1; ; l);Z 0( + 1; ; l)): (8)
Note that l is the Lie algebra of some reductive group satisfying the same
assumptions as G. Furthermore l is either of type A1 A1 or C2. Sowe can apply
Lemma F.2 or Lemma F.3 and get
Homl(Z 0(s 0 0 ; ; l);Z 0( 0 ; l)) 6= 0 (9)
for all 0 2 X. Here 0 is the dot action for l, de ned as w 0 = w( + 0 ), 0 where
0 is half the sum of the positive roots in R \ (Q + Q ). We have h , 0 ; _ i =
h 1; _ i by Lemma E.4 and the de nition E.5(1). This implies
s 0 ( + 1) = + 1 ,h + 1 + 0 ; _ i = + 1 ,h + ; _ i
=(s )+ 1 , a + 1 (mod pX):
So we can rewrite the rst module in (8) as Z 0(s 0 ( + 1); ; l). Now (9) implies
that the Hom space in (8) is non-zero. This yields the claim, hence Proposition
F.1.
F.5. Wenow return to the situation from Theorem D.6. So we assume in addition
that (D2) holds. Let 2 g be subregular nilpotent. We exclude the case where
R is of type G2; ifR is of type E8 or F4, we assume that p>h+1. (Theresults
here as well as in F.6{11 hold in these cases also for p h+1 provided the socle of
Z ( ) as in D.4 has the expected dimension.) Recall the de nition of the integers
m from D.6(1).
Theorem: Let 2 C 0 0 . Let be a short root with 2 J( ). If L1 and L2
are composition factors of Z ( ) with (L1) = (L2) =f g, then L1 ' L2. If
(b + )=0, then L1 has multiplicity m as a composition factor of Z ( ).
Proof : Lemma B.13 and D.5(3) show: If this theorem holds for one subregular ,
then it holds for all subregular . Assume from now on that we choose and the
simple root as in D.12. We get then from Lemma D.12 elements x1, x2 2 W
with xi = and
L1 ' Z (x1 ; ); L2 ' Z (x2 ; ):
We have x1x ,1
2 = . Since W is a re ection group, there are roots 1,
2;:::; r with x1x ,1
2 = s 1s 2 :::s r and s i = for all i.
Set yi = s is i+1 :::s rx2 for 1 i r +1;we get in particular y1 = x1 and
yr+1 = x2. Wehave y ,1
i = for all i and yi = s iyi+1 for all i r. If we can
show that Z (yi ; ) and Z (yi+1 ; ) are isomorphic to each other for all i r,
then the claim will follow.
This shows that we may assume that there exists a root 0 orthogonal to
with x2 = s 0x1. Since is short and since we exclude type G2, we can apply
Proposition F.1 and get
Homg(Z (x1 ; );Z (x2 ; )) 6= 0:
This implies that these modules are isomorphic to each other, since they are simple.
The claim on [Z ( ):L1] follows now from Theorem D.6.