Re: Invariant bilinear forms on semi-direct products

From: Ignat Soroko (ignat.soroko_at_gmail.com)
Date: 03/26/05 

Date: Sat, 26 Mar 2005 23:00:08 +0000 (UTC)

José Carlos Santos wrote:
> On 24-03-2005 3:00, deitmar@uni-tuebingen.de wrote:
>
> > No. Consider the following example: g is spanned by one element
> > H and V is spanned by one vector X. Assume H.X=X.
> > Then the semi-direct product is spanned by H and X and one has
> >
> > [H,X]=X
> >
> > Let b be an invariant bilinear form, then
> >
> > b(X,X) = b(X,[H,X]) = b([X,H],X) = -b(X,X),
> >
> > so b(X,X)=0. Next
> >
> > b(H,X) = b(H,[H,X]) = b([H,H],X) = b(0,X) = 0
> >
> > and finally,
> >
> > b(X,H) = b([H,X],H) = b(H,[X,H]) = -b(H,X) = 0
> >
> > So b is degenerate.
>
> Thanks a lot. As a matter of fact, I forgot to add the hypothesis
that
> in g one can define an invariant non-degenerate bilinear product, but
> your answer covers that case too.
>
> What if one alse assumes that g is a simple Lie algebra?
>
> Best regards,
>
> Jose Carlos Santos

Dear Jose,

Under this condition it turns out that the only nontrivial irreducible
module V over a simple complex Lie algebra L such that L sd V allows a
nondegenerate invariant bilinear form, is the adjoint module V.

First, note that b(u,v) = 0 for any u, v from V, since u = [w,x] for
some w in V, x in L. Hence b(u,v) = b([w,x],v) = -b(x,[w,v]) = 0.

Second, we can choose a basis in L sd V consisting of root and weight
vectors. If x is from L_\alpha, v is from V_\lambda, h from the Cartan
subalgebra then

\alpha(h)*b(x,v) = b([h,x],v) = - b(x,h.v) = -\lambda(h) b(x,v)

Hence, if \alpha + \lambda <> 0, we can choose h so that
(\alpha+\lambda)(h)<>0, which yields b(x,v)=0.

The only possibility to have b(x,v)<>0 remains when x in L_\alpha, v in
V_{-\alpha}. In fact, if V is the adjoint module, such a form exists:
one can assign b(x,v) to the value of the Killing form K(x,v).

But if V is not adjoint, a nondegenerate invariant form does not exist.
Since the set of weights of the module V is different from the set of
roots, one can either find a weight in weights(V) not belonging to
roots(L), or a root in roots(L) not belonging to weights(V). In any
case, there is a sequence of root elements x1,...,xk such that either

xk. ... .x1.x = 0 and
x1. ... .xk.w = v for some w<>0

or vice versa

x1. ... .xk.y = x for some y<>0 and
xk. ... .x1.v = 0

In the first case we have

b(x,v) = b( x, x1.x2. ... .xk.w ) = +- b( xk. ... .x1.x, w ) = 0,

and in the second:

b(x,v) = b( x1.x2. ... .xk.y, v ) = +- b( y, xk. ... .x1.v ) = 0.

This proves that V lies in the kernel of b, if V is not adjoint.

Ignat Soroko

Quantcast