Let $G$ be group and let $X$ be a topological space on which $G$ acts continuously. Now let us consider the following two properties relative to the group action:

(a) For every compact subset $K\subseteq X$ we have that $|\lbrace g\in G:gK\cap K\neq\emptyset\rbrace|<\infty$

(b) For all sequence $\{g_n\}_{n\geq 1}$ of pairwise distinct elements of $G$ and every $x\in X$ the sequence $\{g_n x\}_{n\geq 1}$ has no limit point in $X$.

It is easy to see that $(a)\Rightarrow (b)$. What about the converse?

Under the following assumptions one may show that $(b)\Rightarrow (a)$:

(*) Assume that $X$ is a locally compact metric space where the distance function is denoted by $d$. Assume that there is an absolute constant $C>0$ such that for all $x\in X$ there exists a neighborhood $U_x$ of $x$ such that for all $g\in G$ and all $u,v\in U_x$ one has that $d(gu,gv) < C\cdot d(u,v)$.

For example if $G$ acts through isometries on $X$ on a locally compact space then $(a)$ is equivalent to $(b)$.

  1. Is it possible to weaken Assumption $(*)$ ?
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    $\begingroup$ Hugo -- it is not very clear (to me) what these questions have to do with differential geometry. $\endgroup$ – algori Dec 22 '10 at 5:41
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    $\begingroup$ I am confused by your second sentence. If $Aut(X)$ has the discrete topology and $\rho$ is continuous, doesn't this mean that some n.h. of $1$ in $G$ has to mapped to the identity in $Aut(X)$, i.e. that some n.h. of $1$ in $G$ acts trivially on all of $X$. Is this part of your assumptions, or am I misunderstanding what you wrote? $\endgroup$ – Emerton Dec 22 '10 at 5:43
  • $\begingroup$ Hi algori, well these two questions arose naturally to me in the context of Hermitian symmetric domains. I would like to see how this symmetric condition can fail even though the homogeneous condition holds. $\endgroup$ – Hugo Chapdelaine Dec 22 '10 at 13:35
  • $\begingroup$ Yes @Matthew you were right these 2 sentences were confusing I deleted them. The point is if $X$ il locally compact then the group $Aut(X)$ can be turned into a topological group in a natural way using the compact-open topology. $\endgroup$ – Hugo Chapdelaine Dec 22 '10 at 17:21

Let's assume that $X$ is locally compact and instead of assuming that $G$ is a discrete group, assume more generally that it is locally compact and the action of $G$ on $X$ is continuous (i.e., the map $(g,x) \mapsto gx$ is continuous). Then the natural generalization of (a) is that if $K$ is compact, then the set $((K,K)) = \{ g\in G \mid gK \cap K \ne \emptyset\}$ is relatively compact. If this condition holds then $X$ is called a proper $G$-space, and it is known that this is equivalent to the condition that the map $(g,x) \mapsto (gx,x)$ of $G\times X \to X \times X$ is a proper map (in the sense that the inverse image of a compact set is compact), which is probably what you should assume instead of (b). The theory of proper group actions is a very old and rich one and I think if you do some googling and looking in Wikipedia you will find a lot along the lines that your questions suggest that you are interested in. (In particular, you might want to look at an old Annals paper of mine called "On the Existence of Slices for Actions of Non-Compact Lie Groups".)

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  • $\begingroup$ Thanks @Dick. So I agree that the equivalence between $X$ being a proper $G$-space iff the map $G\times X\rightarrow X\times X$ is proper not only answers Q1 but generalizes it. So where can I find a proof of this equivalence? $\endgroup$ – Hugo Chapdelaine Dec 23 '10 at 2:19
  • $\begingroup$ @Hugo Chapdelaine. It's in my Annals article, but that may be hard to get. It is also in an article by Abels, Manousssos, and Noskov called "Proper actions and proper invariant metrics" that is available as pdf online (see Proposition 2.3, page 4). $\endgroup$ – Dick Palais Dec 23 '10 at 3:05

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