Existence of proper invariant subset in an irreducible action - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T05:59:04Z http://mathoverflow.net/feeds/question/63595 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63595/existence-of-proper-invariant-subset-in-an-irreducible-action Existence of proper invariant subset in an irreducible action Dennis Gulko 2011-05-01T09:56:51Z 2011-05-05T00:31:33Z <p>Let $G&lt;\rm{GL}_n(\mathbb{k})$ be a linear group, where $\mathbb{k}$ is an algebraically closed field. Assume that the linear action of $G$ on $\mathbb{k}^n$ is strongly-irreducible (i.e. there are no $H$-invariant proper subspaces of $\mathbb{k}^n$, except $0$, for any $H&lt; G$ of finite index). Equip $\mathbb{k}^n$ with the Zariski topology. Could there be a proper non-empty open subset $U\subset\mathbb{k}^n$ which is $G$-invariant, and $\mathbb{k}^n\setminus U\neq {0}$?</p> <p>Thanks for any help.</p> http://mathoverflow.net/questions/63595/existence-of-proper-invariant-subset-in-an-irreducible-action/63598#63598 Answer by Alain Valette for Existence of proper invariant subset in an irreducible action Alain Valette 2011-05-01T10:37:38Z 2011-05-01T10:37:38Z <p>Let $Q$ be a non-degenerate quadratic form on $\mathbb{K}^n$, and $G=O(Q)$ its orthogonal group. I think that the set $U$ of vectors $x\in\mathbb{K}^n$ with $Q(x)\neq 0$ does the job.</p> http://mathoverflow.net/questions/63595/existence-of-proper-invariant-subset-in-an-irreducible-action/63961#63961 Answer by George McNinch for Existence of proper invariant subset in an irreducible action George McNinch 2011-05-05T00:31:33Z 2011-05-05T00:31:33Z <p>Here are some more examples in the spirit of that of Alain Valette: let $G$ be a quasi-simple algebraic group and consider its adjoint representation on $\mathfrak{g} = \operatorname{Lie}(G)$. Then most of the time $\mathfrak{g}$ is a simple $G$-module (e.g. it is simple if the characteristic of $k$ is "very good" for $G$, and in particular, if the characteristic of $k$ is 0). And the action of $G$ leaves invariant the (closed) nilpotent subvariety $\mathcal{N}$ of $\mathfrak{g}$, and hence also its (open) complement $U = \mathfrak{g} \setminus \mathcal{N}$.</p> <p>The adjoint action of $G$ on $\mathfrak{g}$ also leaves invariant these proper open sets of $\mathfrak{g}$:</p> <p>$\bullet$ the regular elements $\mathfrak{g}_{\operatorname{reg}}$ (i.e. those elements with minimal dimensional centralizer), and</p> <p>$\bullet$ the regular semisimple elements $\mathfrak{g}_{\operatorname{rs}}$ (those semisimple elements whose connected centralizer is a maximal torus).</p>