subregular nilpotent representations of lie algebras in prime ...

G.2. Let now be an arbitrary linear form on g with (b + )=0.
Lemma: Each projective U (g){module is a direct summand of some E Z (, )
with E a G{module.
Proof : It su ces to look at the projective cover QL of a simple U (g){module L.
There exists a weight 2 X such that L is a homomorphic image of Z ( ). In
fact, we can nd a dominant weight such thatL is a homomorphic image of
Z (, + w0 )sinceZ ( ) depends only on + pX and since each coset in X=pX
has a representative of the form , + w0 as above. Now take E as the simple
module with highest weight . Then E Z (, ) has a ltration with factors
Z (, + 0 ) with 0 running over the weights of L. The factor Z (, + w0 )
occurs thus as a homomorphic image of E Z (, ). So does then L. Because
E Z (, ) is projective, this implies that QL is a direct summand of E Z (, ).
G.3. Let be as in G.2. We saythataU (g){module M has a Z{ ltration if
there exists a chain of submodules in M, beginning with 0 and ending with M,
such that all subsequent factors are isomorphic to some Z ( )with 2 X. If so,
then also each E M with E a G{module, and each pr (M) with 2 X has a
Z{ ltration, cf. the proof in B.3. Therefore all modules of the form
pr r (Er pr r,1 (Er,1 pr 2 (E12 pr 1 (E1 Z (, ))) :::)) (1)
have aZ{ ltration. All these modules are projective, since Z (, ) is projective,
see [7], Thm. 4.1.
Set P equal to the Grothendieck group of all projective U (g){modules. So
this is a free Abelian group with the projective covers of the simple U (g){modules
as a basis. We shall usually write Q also for the class in P of a projective U (g){
module Q.
Let P 0 denote the subgroup of P generated by all modules as in (1). For each
2 X let P be the Grothendieck group of all projective modules in C , and set
P 0 = P 0 \P . Then P is the direct sum of all P (and P 0 that of all P 0 ) with
running over a suitable set of representatives. It is clear that
for all ; 2 X.
T P 0
Lemma: Let 2 X, let Q be aprojective module in C . If there exists an integer
m>0 with Q m 2P 0 , then we have
[Q] =
in the Grothendieck groupofallmodules in C .
P 0
61
(2)
dim Q
[Z ( )] (3)
pN Proof : Since this Grothendieck group is free over Z, it su ces to prove the claim
for Q m .Sowemay assume that Q 2P 0 .SoQ is a Z{linear combination (in P 0 )
of modules as in (1) [with r = ]. Since the claim in (3) is additive inQ, wemay
assume that Q is a module as in (1), hence that Q has a Z{ ltration. But then
the claim follows from the fact that [Z (w )]=[Z ( )] for all w 2 W .