subregular nilpotent representations of lie algebras in prime ...

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Remarks: 1) The lemma can be extended to the case where L = L 0 and where
is a simple root with ( ; 0) =0.
2) We see as in H.9 that Ext 1 (L; L) 6= 0 in almost all cases. The only additional
exceptions can occur in types C2 = B2 and C3.
H.11. Lemma: Let 2 [f0g, letL be a simple module in C with L 6= 0.
a) The head and the socle of L are both isomorphic to L.
b) If L 0 is a simple module in C with T ( ) L 0 ' T ( ) L, then L 0 ' L.
Proof :IfL 0 is a simple module in C , then H.9(4),(5) imply that L 0 occurs with
multiplicity 1 both in head and in the socle of L if T ( ) L 0 ' T ( ) L. Otherwise
it does not occur at all. In particular L itself occurs with multiplicity 1. This shows
that a) will follow from b).
Suppose that L 0 is a simple module in C with T ( ) L 0 ' T ( ) L. Lemma
H.7 implies that (L) = (L 0 ). If L 0 is not isomorphic to L, then there has to be a
long simple root with (L) =f g. Lemma H.8 shows that both L and L 0 occur
with multiplicity 1 as composition factors in L. Since they occur both in the
head and in the socle they have to be direct summands. The discussion above of
the socle implies that there cannot be any other contributions to the socle, hence
that L ' L L 0 . Then Lemma H.8 implies that ( ; ) = 0 in case 2 , resp.
( 0; ) = 0 in case =0. Now Lemma H.10 yields a contradiction.
H.12. Proposition: a) If and are simple roots with ( ; )=0, then we
have Ext 1 (L; L 0 )=0for all simple modules L and L 0 in C with (L) =f g and
(L 0 )=f g.
b) Let , be simple roots with ( ; ) < 0, letL be asimplemodule L in C with
(L) =f g. Then there exists a simple module L 0 in C with (L 0 )=f g and
Ext 1 (L; L 0 ) ' K ' Ext 1 (L 0 ;L): (1)
If and have the same length, then the condition Ext 1 (L; L 0 ) 6= 0determines
L 0 up to isomorphism. If and have di erent lengths, then (1) holds for all
possible L and L 0 .
c) If be a simple root. If R is not of type A1, then
Ext 1 (L; L 0 ) ' Ext 1 (L 0 ;L) ' K; if ( 0; ) > 0,
0; if ( 0; )=0
for all simple modules L in C with (L) =f g.
Proof : a & b) Consider simple modules L and L 0 in C with (L) =f g and
(L 0 )=f g. Arguing as in H.9 one gets
Ext 1 (L; L 0 ) ' Homg(rad L; L 0 ): (3)
(2)