subregular nilpotent representations of lie algebras in prime ...

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Remark: Let L be a simple module in C . Suppose that the projective cover QL
of L satis es the assumption of our lemma. Then (3) and B.12(2) imply
hence
for all simple L 0 in C .
[QL] =jW ( + pX)j [Z ( ):L][Z ( )]; (4)
[QL : L 0 ]=jW ( + pX)j [Z ( ):L][Z ( ):L 0 ] (5)
G.4. We return to the assumptions and conventions from G.1. In particular, we
have 2 g subregular nilpotent. Given 2 C 0 0 we denote by Q 0 (resp. Q or
Q ;i ) the projective cover in C of the simple module L 0 (resp. L or L ;i ).
Proposition C.2 says that Q 0 ' T , Z (, ). So we get
for all 2 C 0 0.
Q 0 2P 0
Lemma: Let be ashortsimpleroot. Then there exists an integer n( ) > 0 such
that
n( )Q $ , 2P 0
$ , : (2)
Proof : Set = $ , . There are in C (up to isomorphism) only two simple
modules: L and L 0 , hence only two indecomposable projective modules: Q
and Q 0 . Lemma G.2 yields a G{module E such thatQ is a direct summand of
E Z (, ). Then there exist integers n( ) > 0 and m 0 with
n( )
pr (E Z (, )) ' (Q )
Now the de nition of P 0 and (1) yield the claim.
(Q 0 ) m :
Remark: Let be a long simple root. Set = $ , . Should (against expectations)
there be only one simple module in C with invariant f g, thenwegetas
above aninteger n( ) > 0withn( )Q 2P 0 .Ifhowever, as expected, there are
two such modules, then we get instead
(1)
n( )(Q ;1 + Q ;2 ) 2P 0 : (3)
Indeed, we have in this situation by Proposition F.7 an element g 2 G with
g = and g L ;1 ' L ;2 , hence g Q ;1 ' Q ;2 . On the other hand, we have
above g E ' E [since this is a G{module] and g Z (, ) ' Z (, ) [since this is the
only simple module in C, ]. Therefore the multiplicities of Q ;1 and of g Q ;1 as
direct summands of E Z (, )have to be equal. (The same argumentshow that
g Q ' Q for all Q in P 0 .)