subregular nilpotent representations of lie algebras in prime ...

Now Lemma C.1 implies that the stabiliser of +pX in W is equal to the stabiliser
of in W . Using (1) and the fact that QL is a summand of Q, we get that
dim Q = p N jW j dim QL = p N jW j[Z ( ):L] p N jW j:
So we have tohave equality everywhere, hence QL = Q and [Z ( ):L] = 1. So
the claim follows.
C.3. Proposition C.2 yields a representation theoretic proof of the following
special case of a recent theorem of Brown and Gordon in [2], 3.18:
Corollary: If 2 C 0 0 , then the subcategory C is a block of the category of all
nite dimensional U (g){modules.
Proof : If C is not a block, then it is a direct sum of two non-trivial subcategories
that are closed under taking subquotients. The indecomposable module Q
considered in C.2 would have to belong to one of them. Then so would be all
subquotients Z (w ) with w 2 W of Q, hence all simple modules in C . Then
the other subcategory will contain no simple modules at all, hence be trivial.
Remark: The theorem in [2] says that all C with 2 X are blocks. That proves
the conjecture by Humphreys in [8], Section 18. (For arbitrary type such a result
had previously been known only for in standard Levi form, cf. [8].) The corollary
here together with H.1 yields that conjecture in case R has no components of
exceptional type (and p 6= 2 if it has components not of type A).
C.4. Let w0 2 W denote the unique element withw0(R + )=,R + .
Proposition: Let 2 C 0 0 ,letLdenote the simple module in C with projective
cover isomorphic to T , Z (, ).
a) Up to isomorphism L is the only simple module in C with T , L 6= 0.
b) The socle of Z ( ) and the head of Z (w0 ) are isomorphic to L. We have
dim Homg(Z (w0 );Z ( )) = 1: (1)
Each non-zero homomorphism from Z (w0 ) to Z ( ) has image equal to the
socle of Z ( ).
Proof : a) It is known that Z (, ) is simple, cf. [7], Thm. 4.2. (The assumption
in that theorem that p is good for R is not needed for this particular result.)
Therefore Z (, ) is the only simple module in C, (up to isomorphism). If M is
amoduleinC with T , M 6= 0, then we get
0 6= Homg(Z (, );T , M) ' Homg(T , Z (, );M):
If we assume additionally that M is simple, then we get that M is a homomorphic
image of the projective cover of L, hence isomorphic to L.
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