2 Fixed a mistake in a reference

I got interested in a similar problem (not quite the same thing, though) a few years back but could not find anything really interesting. Here is what I found in the litterature;

• In case $X$ is compact and $G$ is abelian then there is an invariant metric (compatible with the topology, of course) under the action of $G$ iff $G$ is relatively compact in the homeomorphism group of $X$ (endowed with the compact-open convergence topology) iff the action of $G$ is topologically equicontinuous. This is a result of Marjanovic," On topological isometries", Indag. Math. 31(1960)31(1969), 184–189

"Topologically equicontinuous" means the following: for any $x,y \in X$, any open set $V$ containing $y$, there is an open set $U$ containing $x$ and an open subset $W$ containing $y$ and contained in $V$ such that for all $g \in G$ $g(U) \cap W \neq \emptyset \Rightarrow g(U) \subseteq V$

• In case $X$ is locally compact then some extensions of this result are known, but as far as I know they are obtained in the case when the action of $G$ extends to a continuous action on the one-point compactification. I know of two papers on this:

C. Borges, How to recognize homeomorphisms and isometries, Pacific Journal of Mathematics 37(3) (1971), 625–633.

M. Tak Kiang, On some semigroups of mappings, Indag. Math. 33(1972), 18–22

I have not found anything in the litterature more recent than that, but I probably haven't looked hard enough - this was more curiosity than serious research on my part.

I did obtain the following: assume $G$ is abelian (this is the only case I thought about, as I was mostly interested in the case when $G$ is generated by $1$ element), that $X$ is Polish and that the action of $G$ on $X$ is topologically transitive. Then there is a compatible $G$-invariant metric if, and only if, $G$ is a topologically equicontinuous group of homeomorphisms of $X$. Under the same assumptions, there is a complete invariant metric iff any invariant metric is complete.

In case you can read French there are some notes on this on my webpage - I wrote them for myelf so probably you should not take anything that's written on faith...

1

I got interested in a similar problem (not quite the same thing, though) a few years back but could not find anything really interesting. Here is what I found in the litterature;

• In case $X$ is compact and $G$ is abelian then there is an invariant metric (compatible with the topology, of course) under the action of $G$ iff $G$ is relatively compact in the homeomorphism group of $X$ (endowed with the compact-open convergence topology) iff the action of $G$ is topologically equicontinuous. This is a result of Marjanovic," On topological isometries", Indag. Math. 31(1960), 184–189

"Topologically equicontinuous" means the following: for any $x,y \in X$, any open set $V$ containing $y$, there is an open set $U$ containing $x$ and an open subset $W$ containing $y$ and contained in $V$ such that for all $g \in G$ $g(U) \cap W \neq \emptyset \Rightarrow g(U) \subseteq V$

• In case $X$ is locally compact then some extensions of this result are known, but as far as I know they are obtained in the case when the action of $G$ extends to a continuous action on the one-point compactification. I know of two papers on this:

C. Borges, How to recognize homeomorphisms and isometries, Pacific Journal of Mathematics 37(3) (1971), 625–633.

M. Tak Kiang, On some semigroups of mappings, Indag. Math. 33(1972), 18–22

I have not found anything in the litterature more recent than that, but I probably haven't looked hard enough - this was more curiosity than serious research on my part.

I did obtain the following: assume $G$ is abelian (this is the only case I thought about, as I was mostly interested in the case when $G$ is generated by $1$ element), that $X$ is Polish and that the action of $G$ on $X$ is topologically transitive. Then there is a compatible $G$-invariant metric if, and only if, $G$ is a topologically equicontinuous group of homeomorphisms of $X$. Under the same assumptions, there is a complete invariant metric iff any invariant metric is complete.

In case you can read French there are some notes on this on my webpage - I wrote them for myelf so probably you should not take anything that's written on faith...