subregular nilpotent representations of lie algebras in prime ...

Remark: IfR has type A1 and if we denote the unique simple root by , then one
has
[ 0L ]=[ L 0]=2[L ]+2[L 0]
It is well known that in this case
(Note that = 0 and look at [19].)
Ext 1 (L ;L 0) ' Ext 1 (L 0;L ) ' K 2 :
H.10. We nowwant to extend some results from H.9 to the cases with two root
lengths. Recall that we exclude type G2. Note rst that H.9(4) and H.9(5) hold
without restriction. We next look at H.9(7).
Lemma: Let be a simple root and let L be a simple module in C with (L) =
f g. If is a simple root with ( ; )=0or if =0and ( 0; )=0,thenwe
have a short exact sequence
that does not split.
0 ! L ,! L ,! L ! 0 (1)
Proof : Lemma H.8 shows that L has length 2 with both composition factors, say
L1 and L2, satisfying (L1) = (L2) =f g. Wewant toshow that L1 ' L2 ' L.
If so, then we get a short exact sequence as in (1). If that sequence splits, then we
get dim Homg(L; L) = 2 contradicting H.9(5).
If is short, then L1 ' L2 ' L follows from Theorem F.5. So consider
the case where is long. We may assume that there exists a simple root with
(p )=0andw 2 W with w = and L ' Z (w ; ). The proof of Lemma
H.8 shows that our claim will follow ifwe can show that Z (ws ; ) is isomorphic
to L where s = s in case 2 and s = s 0 in case =0. We distinguish two
cases:
The type ofR is Bn. We may assume that we are in the situation of [10], Section
3. In particular, we have = 1 and = j with 1 j1). So we get indeed s"j = "j and the claim follows in type Bn.
The type of R is Cn or F4. We may assume that we are in the situation of
F.6(1),(2) or F.6(3){(5). We have to show that ws 2 W2w in the notations from
F.6(1) or F.6(3). Since ws = sw 0w with 0 = or 0 = 0 this means that
sw 0 2 W2. Now( 0 ; ) = 0 and w = imply that w 0 is orthogonal to . So
it su ces to show that w 0 is a short root. (Recall the de nition of W2.) That
is obvious in case = 0. Otherwise note that in type Cn or F4 all simple roots
orthogonal to a long simple root are short.
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