4

Let $G$ be a polish group, $H$ an open subgroup of $G$. Now assume that $H$ acts by isometries (For all $h\in H$, the map $X\ni x\longmapsto X$ is an isometry) and continously on a metric space $(X,\delta)$. We define $$F=\{f:G\longrightarrow X|\,\forall h\in H,\,\,\forall g\in G,\,\,f(gh)=h^{-1}f(g)\}$$

$G$ act on $F$ by left translation $$G\times F\ni (g,f)\longmapsto \tau_{g}f \in F$$ where $\tau_{g}f(x)=f(g^{-1}x)$ for all $x\in G$.

My goal is to show that the previous action of $G$ on $F$ is continuous

We assume that we can equipped $F$ with the following metric: $$d:F\times F\ni(f,h)\longmapsto d(f,h)=\underset{g\in G}{\overset{}{\sup}}\,\delta(f(g),h(g))\in \mathbb{R}_{+}$$

If $F$ is equiped with the previuous action, then the action of $G$ on $F$ is by isometries

Since $H$ is open in $G$, we only need to show that the action of $H$ on $F$ is continuous. How to show that this fact? where $F$ is equipped by the toplogy induced by the previous metric.

Thank for any help

Serges
  • 479

2 Answers2

0

This question is my own research. It is not homework.

Thank for your previous answer. For this question, i have an idea. But i am not sure that i am right:

In fact, since the action is by isometries, the map $F\in f\longmapsto \tau_{g}f\in F$ is continuous for all $g\in G$. Now fix $f\in F$ and suppose that $g_{n}\longrightarrow g$ in $G$, then $g^{-1}_{n}a\longrightarrow g^{-1}a$ for all $a\in G$. Therefore, $f(g^{-1}_{n}a)\longrightarrow f(g^{-1}a)$ since $f$ is continuous. So we have $\tau_{g_{n}}f\longrightarrow \tau_{g}f$ pointwise. Now since $G$ is separable and $H$ is open, then the quotient space is countable(separable+discrete). Therefore the following $\underset{a\in G}{\overset{}{\sup}}\,\delta(f(g_{n}^{-1}a),f(g^{-1}a))$ is in fact a max. Therefore, We have: $$d(\tau_{g_{n}}f,\tau_{g}f)=\underset{a\in G}{\overset{}{\max}}\,\delta(f(g_{n}^{-1}a),f(g^{-1}a))\longrightarrow 0$$

Thank for givin your opinion about this idea.

Serges
  • 479
  • What I'm missing here is why the $H$-action on $F$ is continuous with respect to the topology induced by $d$. We know that every $f \in F$ is continuous, but why does this imply $\tau_{h_n}f(g) \to \tau_{h}f(g)$ uniformly in $g \in G$, i.e, with respect to the metric $d$? I agree with pointwise convergence. – Martin Jan 05 '13 at 11:44
  • I guess what really escapes me is how you know $d(f,h) \lt \infty$ for $f,h \in F$. Maybe it would be good to see a specific example of $G$, $H$, $X$ from which you are abstracting. – Martin Jan 05 '13 at 11:53
0

More precisely, since $\underset{g\in G}{\overset{}{\sup}}\,\delta(f(g),h(g))$ can be infinite this map failed two be a metric on $F$ in general.\ In this case i can consider the following equivalence relation on $F$: $$f\sim g \,\Longleftrightarrow d(f,g)< \infty$$ If $F_0$ is any class for the relation $\sim$, then $F_0$ is invariant under the action of $G$. In fact, my goal is to prove that the action on any class $F_0$ is continuous.

Thank one more for your help.

Serges
  • 479
  • I suppose you could do that. I think you would still need some equicontinuity condition on $F_0$ in order to conclude that $H$ acts continuously on it. Maybe this follows in your setting ($H$ open in $G$), but I don't see it right now. I suggest that you look at some explicit examples to see whether this is a reasonable assumption to make if it doesn't follow automatically. – Martin Jan 05 '13 at 13:47
  • Thank for all your Kindly help. I will continue thinking about this question – Serges Jan 05 '13 at 13:49
  • You are welcome. I'll be away for a few days, but I'll check back whether there is some progress. Good luck! – Martin Jan 05 '13 at 13:50
  • Since $H$ is open in $G$ est $G$ is separable, the quotien space $G/ H$ is discrete and separable, thus countable. Since $G=\underset{\xi\in \mathcal{E}}{\overset{}{\bigcup}}\xi H$ where $\mathcal{E}$ is a complete system of representant of left coset. Therefore $\mathcal{E}$ is countable and i guess that in this case, $\underset{a\in G}{\overset{}{\sup}},\delta(f(g_{n}^{-1}a),f(g^{-1}a))$ is in fact a max (by countability) and goes to $0$. – Serges Jan 05 '13 at 14:07