If $G$ is a topological group acting effectively on a topological space $X$ and $H$ is a subgroup of $G$ then is it true that $\overline{H}x = \overline{Hx}$, where $\overline{H}$ is the closure of $H$ in $G$ and $Hx$ denotes the $H$ - orbit of $x$.

I thought one way would be easy - proving $\overline{Hx} \subseteq \overline{H}x$. However for that I need to show that $\overline{H}x$ is closed and I am unsure how to do that (or if it is even true). So I am not sure how to prove either one is contained in the other.

I am particularly trying this for the case when $G=\text{ Homeo}(X)$ with compact open topology, $X$ is locally connected, locally compact hausdorff space and $H$ is a subgroup of $G$.

a. Is $\overline{Hx} = \overline{H}x$ for any $G$ and $X$?

b. If not then is it true for my particular case?

  • $\begingroup$ Could you say what an effective action means? $\endgroup$ – Stefan Hamcke Oct 19 '15 at 10:15
  • $\begingroup$ Never mind, I just read that it means faithful. $\endgroup$ – Stefan Hamcke Oct 19 '15 at 10:16
  • $\begingroup$ @StefanHamcke: For whatever reason, effective is the more commonly used word in topology. $\endgroup$ – user98602 Oct 19 '15 at 10:18

This was a fun question. Unfortunately, what you hope for is false.

Let $X = S^1$. In this answer it will frequently be useful to consider $S^1$ as the quotient $[0,1]/(0 \sim 1)$. (Actually, the answer will be even easier if we just consider $X= [0,1]$, but I figured you might like a manifold counterexample.)

Let $f: S^1 \to S^1$ be the homeomorphism given by $f([x]) = [x^2]$, and let $H$ be the $\Bbb Z$ subgroup of $\text{Homeo}(S^1)$ generated by $f$. For any point $x \neq [0]$, $f^n(x) \to [0]$ as $n \to \infty$, so $\overline{Hx}$ includes $[0]$ (and indeed has that as its only accumulation point). However, $\overline{H}x$ cannot include $[0]$; because $H$ fixes $[0]$, so does $\overline{H}$, and in particular there is no $g \in \overline{H}$ with $g[0] = x$. So $\overline{Hx} \not\subset \overline Hx$. You can modify this example so that $f$ is smooth if you want to, or even 'linear' ($SL_2(\Bbb R)$ acts on $S^1$ by first acting on the plane and then projecting radially.)


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