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?
1
-
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. $\endgroup$– AsafCommented Aug 28, 2016 at 7:51
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
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 $G=U(n)\subseteq SL(2n,\mathbb R)$.
answered Aug 30, 2016 at 1:07