Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We consider a Hausdorff topological group $G$ acting on a topological space $X$ [action simply means a continuous map $G\times X\rightarrow X$ verifying $(gh)(x)=g(h(x))$, and $1(x)=x$].

The set $H=\{g\in G : \forall x\in X, g(x)=x\}$ is a normal subgroup of G. For example, it is closed if singletons are closed in $X$ .

I would like to find a Hausdorff topological group $\mathbf{G}$ and a topological space $\mathbf{X}$ such that $\mathbf{H}$ is not closed.

Note: If we do not require $G$ to be Hausdorff there is a plethora of counterexamples.

share|improve this question
    
I am assuming that the space $X$ is $T1$ (i.e. points are closed). Consider the orbit map $G\rightarrow X$ taking $g\mapsto gx$. If the action is continuous then this maps is continuous. Then $H$ is the preimage of $x$ which is closed. Am I missing something? –  DBS Jul 29 '13 at 3:35
    
Of course, if singletons are closed in $X$ then $H$ is closed. But that is not what I am asking is it? In any case I clarified the post above. –  John Jul 29 '13 at 3:41
    
Okay thanks for the clarification. If you take $G= \mathbb{R}/\mathbb{Z}$ and $H = \mathbb{Q}/\mathbb{Z}$ inside $G$. The $H$ is a normal subgroup of $G$. Take $X= G/H$ with left $G$ action. Then $H$ is not a closed subgroup of $G$ but it is the stabilizer of a point $e.H$. –  DBS Jul 29 '13 at 3:49
    
Isn't that an answer? –  Kevin Carlson Jul 29 '13 at 4:07
add comment

1 Answer 1

up vote 3 down vote accepted

Let $G$ be a Hausdorff topological group. Let $N$ be a non-closed normal subgroup of $G$. Let $X = G/N$. $G$ acts on $X$. Let $g$ be an element of $G$. We denote by $G_g$ the set $\{h \in G |\ hgN = gN\}$. Then $G_g = gNg^{-1} = N$. Hence $H = \{g\in G : \forall x\in X, g(x)=x\} = \bigcap \{G_g |\ \forall g \in G\} = N$. Hence $H$ is not closed.

For example, let $G = \mathbb{R}$ and $N = \mathbb{Q}$.

share|improve this answer
    
Thanks, very nice. –  John Jul 29 '13 at 13:44
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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