Consider a topological group $G$ (or smooth Lie group) and a topological space $M$ (or smooth manifold) and a group homomorphism $\phi:G\rightarrow Sym(M)$, where $Sym(M)$ is the symmetry group of M, i.e. the group of all homeomorphisms (or diffeomorphisms).

Now consider the map

$$\phi':G\times M\rightarrow M, (g,x)\mapsto \left(\phi(g)\right)(x).$$

I am wondering, if the map $\phi'$ is continuous (or smooth)?

I have recognized that if $G$ is a discrete group, then $\phi'$ is continuous. But I would like to know what happens in the more general case?!

Maybe one needs to impose more assumptions on $G,M$ etc. in order to make the map $\phi'$ continuous (smooth) like the case of discrete topology. But I'm not sure about that.

Best wishes

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    $\begingroup$ If $G$ has a topology or a Lie group structure, you need to invoke it. The action is continuous/smooth iff the map $G \to \text{Diff}(M)$ is continuous/smooth. Of course, you need to correctly interpret what a smooth map to $\text{Diff}(M)$ is. For this, look up "Frechet Lie group". I'm only considering here the case of $M$ a closed manifold. There are probably point-set topological problems if $M$ can be a messy space. $\endgroup$ – user98602 Aug 19 '15 at 14:20
  • $\begingroup$ Thanks for your comment! Do you know a reference where I can find a proof? $\endgroup$ – asd Aug 19 '15 at 14:28
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    $\begingroup$ Not off the top of my head. Try proving for $\text{Homeo}(M)$ in the continuous case. The relevant topology is the subset of $\text{Cont}(M,M)$ in the compact-open topology. This topology is essentially built for theorems like the above to be true, so you should be able to prove it yourself. $\endgroup$ – user98602 Aug 19 '15 at 14:34

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