When do isometric actions exist?

Question

Let $X$ be a metrizable topological space and $G$ be a locally compact group. Given a continuous (left) action of $G$ on $X$, is there a metric on $X$, compatible with the topology, for which the action of $G$ becomes an isometric action? Conversely, given a metric on $X$, is there a nontrivial action of $G$ on $X$ that preserves the metric?

I am looking for the most general necessary and sufficient conditions and any possible obstructions. For the first question, the answer is obviously positive when $G$ is compact: one chooses a metric $d$ on $X$ and simply does an "averaging process along the orbits" by defining $$ \rho(x,y) = \int_{G} d(g^{-1}\cdot x,g^{-1}\cdot y) dg . $$ I suspect that a similar idea would work more generally using a "cut-off" function on $X$ when the action of $G$ on $X$ is proper. Any connections to amenability (of the group or the group action) would also be interesting.

Answer 1

I am sorry That I am getting involved so late. I was away at a meeting for two days. I have an old (1961) paper in the ANNALS OF MATH called "On the Existence of Slices for Actions of Non-Compact Lie Groups" which is quite relevant. In particular, on page 318 you can find the following concerning proper actions of an arbitrary Lie group $G$.

Theorem 4.3.4. Every seperable, metrizeable, proper $G$-space $X$ admits an invariant metric. ...

There are a large number of other theorems there that show that the theory of proper G-spaces for G an arbitrary Lie group is similar to the theory of G-spaces for a compact Lie group. You can find the paper here: http://vmm.math.uci.edu/ExistenceOfSlices.pdf

Answer 2

Regarding the first question: if we take $G=\mathbb{Z}$ or $G=\mathbb{R}$, then there are plenty of group actions that cannot be realised as isometries.

Example 1: Let $G=\mathbb{R}$ and $X=\mathbb{R}$, and consider the flow $\phi_r(x) = e^r x$. Then $G$ is locally compact and $\phi$ defines a continuous $G$-action of $X$, but there is no metric that makes this an isometry; indeed, given any neighbourhood $U\ni 0$ and $0\neq x\in U$, there exists $r_0$ such that $\phi_r(x)\notin U$ for all $r>r_0$. This property holds for any equivalent metric but cannot hold for an isometric action.

If you want $X$ to be compact, you can consider $G=\mathbb{R}$ and $X=[-\pi/2,\pi/2]$, and given $r\in G$, consider the map $\phi_r\colon X\to X$ defined by $\phi_r(x) = \tan^{-1}(r + \tan x)$ for $|x| < \pi/2$ and $\phi_r(x)=x$ for $|x|=\pi/2$. Then $\phi$ is a continuous group action that fixes the two endpoints; one is attracting, the other repelling, and $\phi$ cannot be an isometric action for the same reasons as above.

Example 2: Let $G=\mathbb{Z}$ and $X=\{0,1\}^\mathbb{Z}$ with the product topology; then $G$ acts on $X$ via the shift map. In other words, $\phi_n(x)\_i = x_{i+n}$. Once again, topological considerations using fixed points can be used to show that this action is not isometric for any metric on $X$ that induces the product topology.

So I guess one can sum up those obstructions by saying that in order for $G$ to act isometrically, every fixed point has to be topologically stable. That is, suppose there is $p\in X$ such that $\phi_g(p)=p$ for all $g\in G$, and that there is a neighbourhood $U\ni p$ such that for every $p\neq x\in U$, we have $\phi_g(x)\notin U$ for some $g\in G$. Then $\phi$ is not an isometric action for any equivalent metric on $X$.

Answer 3

This is at best a partial answer but rather too long for a comment (I only adress the last paragraph of the question).

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Indeed, if $X$ is locally compact second countable and the action of $G$ is proper then there exists a $G$-invariant metric compatible with the topology. As you suspect, this can be done by integrating the metric multiplied by a (generalized) Bruhat function (see Bourbaki, Intégration, VII, § 2, No. 4).

More precisely: Let $G$ be a locally compact group and fix a left Haar measure on $G$. The action of $G$ on a locally compact space $X$ (such that $X/G$ is paracompact) is proper if and only if there exists a continuous function $\beta: X \to [0,\infty)$ such that

• For every compact set $K \subset X$ the set $\operatorname{supp}\beta \cap GK$ is compact.
• For all $x \in X$ we have $\int_{G} \beta(g^{-1}x)\\,dg = 1$.

This fact is folklore but it is difficult to locate a simple proof in the literature. Therefore I've given a short one in Appendix E of my thesis, available here. Sometimes these functions are called cut-off functions but I don't like the name.

Using this, the existence of an invariant metric compatible with the topology on a locally compact second countable proper $G$-space $X$ is an easy exercise in integration theory: Pick any metric $d_{X}$ compatible with the topology and replace it by $\frac{d_{X}}{1+d_{X}}$ in case it is unbounded. Then put \[ \delta_{X}(x,y) = \iint_{G \times G} \beta(g^{-1}x)\, \beta(h^{-1} y)\, d_X (g^{-1}x, h^{-1}y)\,dh\,dg \] and verify that $\delta_{X}:X \times X \to [0,1)$ is an invariant metric compatible with the topology.

A similar and detailed argument can be found in one of the first few sections of Koszul's Lectures on groups of transformations (I think it's in the the third section of the first chapter but I can't verify this at the moment).

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Finally, you're asking about the relation to amenability, here I also have at best some comments. Of course, proper actions are known to be topologically amenable (for instance because there is a Bruhat function). In the other direction, I think there is no hope. There are plenty of actions of amenable groups that can't be made into isometric actions (and any continuous action of an amenable group is amenable). Since Vaughn has given some nice examples, I can end this long post now.

Answer 4

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. By a result of Marjanovic (" On topological isometries", Indag. Math. 31(1969), 184–189) this extends to the case when $X$ is locally compact (that is not the way Marjanovic's result is formulated but it is equivalent)

"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 not locally compact then some extensions of this result are known. 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 (update: well now I have looked more seriously and still haven't found anything else in the litterature).

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...

UPDATE (July 4, 2014): I corrected some imprecisions in the text above (about the locally compact case). I logged in to mention that I. Ben Yaacov and I recently worked on this problem again, and proved the following result (there is a preprint on my webpage which supersedes the notes alluded to above):

Assume $X$ is separable metrizable and $G$ is a group of homeomorphisms of $X$. Then there exists a compatible $G$-invariant metric on $X$ if and only if, for any $y \in X$ and any open $V$ containing $y$ there exists an open $W$ containing $y$ and contained in $V$ such that for any $x \in X$ there exists an open $U$ containing $x$ and satisfying $\forall g \in G \ (gU \cap W \ne \emptyset) \Rightarrow gU \subseteq V$.

That is a lot of quantifiers! I think they are really needed; the property above is a uniform version of topological equicontinuity (obtained by switching a universal and an existential quantifier).