# When do isometric actions exist?

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.

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I am having trouble making sense of your second question. Could you just take $G = U(1)$ and $X$ any of a large collection of spaces (e.g., finite metrized set, genus two surface, etc.) as a counterexample? – S. Carnahan Feb 26 '11 at 15:43
Yes, Scott. There are plenty of counterexamples to both questions, and it will be very instructive to find them, indeed. However, I need some necessary and/or sufficient conditions on the given data when the answer is positive. – Kamran Reihani Feb 26 '11 at 17:58

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

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Thanks, Dick! But do we have to be in the smooth category to have this fact? As is stated above, using "cut-off" (or generalized Bruhat) functions, one can easily show that the $G$-space $X$ admits an invariant metric inducing the same topology on $X$, where $G$ is a locally compact group, which acts properly on the locally compact metrizable space $X$. – Kamran Reihani Feb 28 '11 at 6:52
No, I'm pretty sure that you can use the fact that locally compact groups are suitable limits of Lie groups to get the result for the general case. My interest in that paper was primarily in proving the important slice theorem, and for that one does need smoothness. BTW, that paper was probably the first place where the concept of a proper group action appears in a published paper. I did NOT invent the term. In fact I worked primarily with a more general concept I called a Cartan G-space. It was Armand Borel who told me he had been investigating proper actions and suggested I use that term. – Dick Palais Feb 28 '11 at 17:54

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 $X$.

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Very nice counterexamples. Your last paragraph on the necessary condition is what I like the most, though. – Kamran Reihani Feb 26 '11 at 18:00

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).

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Thanks for the very interesting references and for your French notes. It seems that we have generalized some results of Marjanovic (which was published in 1969 not 1960) to the context of spectral triples in noncommutative geometry with Bellissard and Marcolli (Theorem 1 in the article "Dynamical Systems on Spectral Metric Spaces", arXiv:1008.4617), which is equivalent to almost periodicity of the action. Please say hi to Johannes Kellendonk and Jean Savinien:-) – Kamran Reihani Feb 27 '11 at 19:44
Here is the theorem: Theorem 1 (Arzel`a-Ascoli theorem). Let $X = (A, H, D)$ be a compact spectral metric space. Let $G\subset{\rm Aut}(A)$ be a quasi-isometric subgroup. Then $G$ is equicontinuous if and only if it has a compact closure. – Kamran Reihani Feb 27 '11 at 19:51
Ah, I'll have to ask my colleagues about what spectral metric spaces are... Thanks for catching the mistake in the reference to Marjanovic's article, I edited my answer accordingly. – Julien Melleray Mar 4 '11 at 9:09

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

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).

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.

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Many thanks for the formula! I was just playing around with the sums of the Bruhat functions instead of their products. I was quite sure this was folklore, and I agree that it is hard to locate it in the literature. The term "cut-off" function, however, was used by Jean-Louis Tu in his paper "La conjecture de Novikov pour les feuilletages hyperboliques", published in 1999 in K-theory. He proved the existence of such a function in his Proposition 6.11 for any locally compact proper groupoid with Haar system. I couldn't find any earlier reference, though. He might be the first one publishing it. – Kamran Reihani Feb 27 '11 at 20:18
Thanks for the reference, good to know that! The formula as well as the construction of a Bruhat function is implicitly contained in Koszul's lectures, I think (at least it was a major source of inspiration for me). The main idea definitely goes back to Bruhat. I really prefer to think of Bruhat functions as sort of transverse partitions of unity. The extension to groupoids is (as usual) not so difficult, as soon as you have the case of transformation groups right (I found it myself when I was thinking about amenable groupoids some years ago). – Theo Buehler Feb 27 '11 at 21:02
I also learned from the book "Continuous bounded cohomology of locally compact groups" by Nicolas Monod (P. 44, Lemma 4.5.4) that the existence of the generalized Bruhat functions was explicitly stated in Bourbaki, "INTEGRATION", published in 1963. It is the Proposition 8 on p. 51 (French version). (I think Koszul lectures were published in 1965.) – Kamran Reihani Feb 28 '11 at 4:58
Monod's thesis is where I learned about it from (Monod and I are students of Marc Burger and Monod was one of the co-examiners of my thesis) :) – Theo Buehler Feb 28 '11 at 8:41
But we have $\delta_X(x,x)>0$. Am I missing something? – Kamran Reihani Mar 4 '11 at 7:35