subregular nilpotent representations of lie algebras in prime ...

We have clearly ( ) = 0 for all 2 X, hence
(f + )= (f) for all f 2 h , 2 X. (4)
Here (as usual) we write instead of d by abuse of notation.
For each 2 g set
= f f 2 h j jh = (f) g: (5)
Given f 2 h we can regard Kf as a b + {module with n + acting trivially. Thisis
then a U (b + ){module for all 2 g satisfying (n + )=0andf 2 . For all
these we get then an induced U (g){module (a \baby Verma module")
Z (f) =U (g) U (b + ) Kf : (6)
Denote its \standard generator" 1 1by vf . As before we usually write Z ( )
instead of Z (d ) for 2 X; we should keep in mind that Z ( ) depends only on
d , hence on the coset + pX.
If f 0 2 h satis es f 0 (h ) = 0 for all 2 R, then we can extend Kf 0 to a
g{module with all x acting as 0. This is then a U (f 0 )(g){module where we extend
(f 0 ) to a linear form on g such that (f 0 )(n + + n , ) = 0. A trivial form of the
tensor identity yields then
Z (f) Kf 0 ' Z + (f 0 )(f + f 0 ) (7)
for all f 2 h and 2 g with f 2 and (n + )=0.
A.3. Fix a simple root for the rest of Section A. Denote by p = b + + g, the
corresponding parabolic subalgebra. For all f 2 h and 2 g with (n + )=0
and f 2 consider the induced U (p ){module
Z ; (f) =U (p ) U (b + ) Kf : (1)
Denote the \standard generator" 1 1nowby v0 f . It satis es x v0
f = 0 for all
2 R + and hv0 f = f(h)v0
f for all h 2 h; thexi , v0
Z ; (f). We have
f with 0 i