subregular nilpotent representations of lie algebras in prime ...

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0 +
(b ) = 0 and 0
(x, 0)=0. Since there exists only one subregular nilpotent
orbit in g , there exists g 2 G with g = 0 .
Let us use notations from B.13 like C0 for g = 0 . Applying the results so
far to 0 and 0 (instead of and )wesee that there is a simple U 0(g){module
L1 in C0 of dimension equal to (p ,h + ; _ 0 i)pN ,1 and satisfying (T , ) 0L1 6= 0.
Twisting L1 with g ,1 we get a simple module L2 in C of the same dimension with
T , L2 6= 0(byB.13(2)). Now C.4.a implies that L2 ' L; the claim follows.
Remarks: 1) The proof in a2) shows that the claim holds also in many cases where
R of type E8, F4, and G2 and p h +1. Onemay hope that it holds always in
good characteristic.
2) We have here deduced b) from a). Note conversely: If we know that b) holds
(for a speci c ), then also a) holds (for that ): We have T , Z (w ; ) '
Z (, )forw as in a) [by B.11]. Therefore (and by C.4) the socle of Z ( )is
a composition factor of Z (w ; ). Now a dimension comparison shows that
Z (w ; ) is isomorphic to that socle, hence simple. The remainder of a) follows
now asabove.
D.5. Given 2 C 0 0 write J( ) for the set of simple roots with 2 F (J( )). Note
that each , + $ with 2 J( ) is in the closure of the facet of .Let 2 g be
nilpotent. We associate to each module M in C an invariant by setting
(M) =f 2 J( ) [f0g jT $ , (M) 6= 0g (1)
where we use the convention $0 =0.
This invariant will turn out to be useful in the case where is subregular. In
the general case one should have a ner invariant that keeps track of the behaviour
under all translations to the boundary of a given facet.
There may exist weights ; 0 2 C 0 0 with 6= 0 and C = C 0. If we replace
by 0 in (1), then we will in general get di erent results. So depends not just
on the category C , but also on the choice of . It might therefore be better to
denote this map by . However, usually we x and then no problems should
arise.
Suppose that (b + ) = 0. Then C.9(1) and C.10(1) show that
sbm(w; )=J( ) [f0g for all w 2 W . (2)
In particular, the socle L = sbm(w0; )ofZ ( ) satis es (L) = J( ) [f0g.
Proposition C.4 says that this is the only simple module in C with 0 2 (L).
Let g 2 G. Given M in C we have g M in C(g ) , see B.13. We get now
from B.13(2).
( g M)= (M) (3)