0
$\begingroup$

Let $G$ be an algebraic closed subgroup of $SL(n,\mathbb{R})$ whose action on $\mathbb{R}^n$ is strongly irreducible, i.e. there is no finite union of proper nonzero linear subspaces of $\mathbb{R^n}$ which is invariant under $G$. Is it true that $G$ is semisimple?

$\endgroup$
  • 1
    $\begingroup$ One needs to be careful here regarding algebraic group definitions, I'll take the question as real linear Lie group. In this case, $SO(2)(\mathbb{R})$ would give you a counterexample in the plane. $\endgroup$ – Asaf Aug 28 '16 at 7:51
3
$\begingroup$

First of all, $G$ has to be redective since the fiexed point set of the unipotent radical would be a non-trivial $G$-invarinant subspace.

Secondly, as noticed by the commentators, one cannot conclude that $G$ is semisimple since the center of $G^0$ my act via a non-trivial character to $SO(2)$. This problem occurs in all even dimensions. Counterexamples are the unitary groups $G=U(n)\subseteq SL(2n,\mathbb R)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.