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An exercise in Armstrong says if the topological group $G$ acts on the topological space $X$ by homeomorphisms, then the stabilizer $\text{st}(x)=\{g\in G\mid g(x)=x\}$ is a closed subset of $G$.

If $X$ were Hausdorff this would be easy. But he doesn't assume that. I can't manage to prove it or find a counter-example.

Note that Armstrong assumes $G$ is Hausdorff in his definition of topological group. And he assumes the identity in $G$ gives the trivial homeomorphism, and he assumes the map $G\times X\rightarrow X$ is continuous.

I've searched this site and the web in general and can't find a resolution of this question either way.

Gregory Grant
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  • Since $X$ is a topological space, are we assuming that the action is continuous? (I expect so, but I thought it safer to check.) – Brian M. Scott Apr 01 '15 at 11:44
  • @BrianM.Scott Yes, he requires that the action is by homeomorphisms, that the map $G\times X\rightarrow X$ is continuous, and that $e\in G$ maps to the trivial homeomorphism of $X$ to itself. Thank you for your help! – Gregory Grant Apr 01 '15 at 11:49
  • Does anybody have the 1997 edition of Armstrong? I have the 1983 version, maybe this is a mistake that was fixed in the '97? – Gregory Grant Apr 01 '15 at 11:58

1 Answers1

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Take $G:=GL_2(\mathbb{R})$, $H:=GL_2(\mathbb{Q})$ and $X:=G/H$. Because $G$ is a topological group, it induces a topology on $X$ (the quotient topology), open sets are $U.GL_2(\mathbb{Q})$ where $U$ is an open set of $G$.

The action $g.(zGL_2(\mathbb{Q})):=(gz)GL_2(\mathbb{Q})$ must be continuous for the quotient topology.

Nevertheless $Stab_G(GL_2(\mathbb{Q}))=GL_2(\mathbb{Q})$ which is clearly non closed.

Clément Guérin
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  • Thank you, I thought about this very carefully and I think I believe it works. But couldn't the same thing be done simpler with just $\mathbb R$ acting on $\mathbb R/\mathbb Q$ by translation? Anyway thanks again I'm going to accept this answer. – Gregory Grant Apr 01 '15 at 15:36
  • I don't know why I wanted to do it with matrices. But it is indeed simpler with your setting. – Clément Guérin Apr 01 '15 at 17:55
  • So that settles it, Armstrong 1983 is wrong. I will have to get the 1997 version and see if it was fixed. – Gregory Grant Apr 01 '15 at 18:11