subregular nilpotent representations of lie algebras in prime ...

Assume now that i = 1. The approach above does not work because x7 1 =
$1 =2 WI 1. Set J = f 1; 2; 3g and use the abbreviation
Z ;J( ; 1) =U (pJ ) U (p 1 ) L ; 1( )
for all 2 X. Recall the notation Z ;J( ) from B.6(1).
The nilradical of pJ acts trivially on each Z ;J( ), the centre of the standard
Levi factor gJ acts via scalars. Now gJ is the direct sum of its centre and its
derived Lie algebra DgJ [since p 6= 2.] So a composition series of Z ;J( ) is the
same as one as a DgJ {module.
The Lie algebra DgJ has type B3. Assume for the moment that the restriction
of to DgJ has the standard Levi form considered in [10]. The results in [10] show
then that Z ;I( ) has a composition series with factors (among others)
Z ;J( ; 1) and Z ;J(s 1+2 2+2 3 ; 1):
[If we take 1 = in [10], then we can take 5 = s 1+2 2+2 3 .] Induction to g
yields a chain of submodules in Z ( ) with
Z ( ; 1) and Z (s 1+2 2+2 3 ; 1)
among the factors. Since these two modules are simple with invariant f 1g and
since each composition series of Z ( ) has only two factors with this invariant,
our L has to be isomorphic to one of them.
It remains to be shown that we can choose such that its restriction of to
DgJ has the desired standard Levi form. Well, consider with (p )=0and
(x, ) 6= 0 if and only if 2f 2; 3; 4; 1 + 2 + 3 + 4g. It is not di cult
to check that the centraliser of in g has dimension 6. This implies that is
subregular. (See [20], p. 38, on the connection between centralisers in G and g.)
On the other hand, the restriction of to DgJ has clearly the required form.
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