subregular nilpotent representations of lie algebras in prime ...
subregular nilpotent representations of lie algebras in prime ...
subregular nilpotent representations of lie algebras in prime ...
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
We have clearly ( ) = 0 for all 2 X, hence<br />
(f + )= (f) for all f 2 h , 2 X. (4)<br />
Here (as usual) we write <strong>in</strong>stead <strong>of</strong> d by abuse <strong>of</strong> notation.<br />
For each 2 g set<br />
= f f 2 h j jh = (f) g: (5)<br />
Given f 2 h we can regard Kf as a b + {module with n + act<strong>in</strong>g trivially. Thisis<br />
then a U (b + ){module for all 2 g satisfy<strong>in</strong>g (n + )=0andf 2 . For all<br />
these we get then an <strong>in</strong>duced U (g){module (a \baby Verma module")<br />
Z (f) =U (g) U (b + ) Kf : (6)<br />
Denote its \standard generator" 1 1by vf . As before we usually write Z ( )<br />
<strong>in</strong>stead <strong>of</strong> Z (d ) for 2 X; we should keep <strong>in</strong> m<strong>in</strong>d that Z ( ) depends only on<br />
d , hence on the coset + pX.<br />
If f 0 2 h satis es f 0 (h ) = 0 for all 2 R, then we can extend Kf 0 to a<br />
g{module with all x act<strong>in</strong>g as 0. This is then a U (f 0 )(g){module where we extend<br />
(f 0 ) to a l<strong>in</strong>ear form on g such that (f 0 )(n + + n , ) = 0. A trivial form <strong>of</strong> the<br />
tensor identity yields then<br />
Z (f) Kf 0 ' Z + (f 0 )(f + f 0 ) (7)<br />
for all f 2 h and 2 g with f 2 and (n + )=0.<br />
A.3. Fix a simple root for the rest <strong>of</strong> Section A. Denote by p = b + + g, the<br />
correspond<strong>in</strong>g parabolic subalgebra. For all f 2 h and 2 g with (n + )=0<br />
and f 2 consider the <strong>in</strong>duced U (p ){module<br />
Z ; (f) =U (p ) U (b + ) Kf : (1)<br />
Denote the \standard generator" 1 1nowby v0 f . It satis es x v0<br />
f = 0 for all<br />
2 R + and hv0 f = f(h)v0<br />
f for all h 2 h; thexi , v0<br />
Z ; (f). We have<br />
f with 0 i