# Orbits of properly discontinuous actions

Definition Let $G$ be a group and $X$ a topological space. Let $G\curvearrowright X$ by homeomorphisms. We call the action properly discontinuous if for all $x\in X$ there exists an open neighborhood $U\ni x$ such that for all $g\in G\smallsetminus\{1_G\}$ we have $gU\cap U=\varnothing$, $gU$ being the image of $U$ by the action of $G$, i.e. $gU=\{g\cdot x:x\in U\}$, having denoted the action by $\cdot$.

Remark If $G\curvearrowright X$ is a properly discontinuous action, then its orbits are discrete and closed.

That is what I have on my Geometry 3 notes. I tried to prove the remark, whose proof was left as an exercise, and got stuck on the closed part. Let me show you what I have managed.

Claim 1 The orbits are discrete, i.e. the subspace topology on the orbits is the discrete one. Which means all subsets of the orbits are open in the subspace topology the orbit inherits from $X$.

Proof Let $x\in X$. There exists, by definition, a neighborhood $U\ni x$ such that if $g\neq1_G$, $gU\cap U=\varnothing$. This means $U\cap(G\cdot x)=\{x\}$, so $\{x\}$ is open in the orbit. The same line of thought can be conducted for any element of the orbit. Therefore, the orbit is discrete. $\hspace{1cm}\square$

Claim 2 If $X$ is not Hausdorff, I cannot prove the orbits are closed, since in general they are not.

Proof If $X$ is not Hausdorff, there exists at least a couple of topologically indistinguishable points, i.e. there exist $x,y\in X$ such that if $U\ni x$ is a neighborhood of $x$, then $y\in U$, and viceversa. There can be no $g\in G$ such that $gx=y$, because we have the neighborhood in the definition which separates $x$ from the rest of the orbit, and if $y$ were in the same orbit as $x$ we would have a contradiction. Let me say this better. Suppose $y=gx$ for some $g\in G$. We know $g\neq1_G$ as $x\neq y$. But then there exists $U\ni x$ a neighborhood such that $gU\cap U=\varnothing$ for all $g\in G\smallsetminus\{1_G\}$, in particular for the element sending $x$ to $y$, let it be $\overline{g}$. But $y\in\overline{g}U$, and $x\in U$, so that would separate $x,y$, against the hypothesis of topological indistinguishability, since $x\mapsto\overline{g}x$ is a homeomorphism and thus $\overline{g}U$ is still open. But if there exist such two points, then $y$ is an accumulation point of $G\cdot x$ and viceversa, so $G\cdot x$ and $G\cdot y$ are not closed. $\hspace{2cm}\square$

Claim 3 If $x\in X$, and $y$ is an accumulation point of $G\cdot x$, that is for all neighborhoods $U\ni y$ there exists $g:gx\in U\smallsetminus\{y\}$, then $y\not\in G\cdot x$.

Proof Suppose $y=\overline{g}x$ for $\overline{g}\in G$. By definition of p.d. actions, we can find a neighborhood $U\ni y$ such that $gU\cap U=\varnothing$ for all $g\in G\smallsetminus\{1_G\}$. This $U$ will contain at least a point $\tilde x$ of $G\cdot x$ different from $y$. $G$ is transitive over $G\cdot x$, so we can find $\tilde g:\tilde gy=\tilde x$. But then $\tilde gU\cap U\ni\tilde x$, against the definition of p.d. action. $\hspace{6cm}\square$

Claim 4 If $|G|<\infty$, then the orbits are closed.

Proof If $|G|<\infty$, then $|G\cdot x|\leq|G|<\infty$. Since by now we have the hypothesis that $X$ is Hausdorff, as we have show above that the orbits are, in general, not closed in a non-Hausdorff space, we conclude that for all $g\in G,y\not\in G\cdot x$ there exist neighborhoods $U_g\ni gx,V_g\ni y$ such that $U_g\cap V_g=\varnothing$. Let $V=\bigcap_{g\in G}V_g$. This is a finite intersection of neighborhoods, so is still a neighborhood of $y$, and it is disjoint from the orbit since we have excluded every $gx$ from it. Btw, $gx$ is $g\cdot x$, so I'm implying the $\cdot$. So $y$ is in the interior of the orbit's complement, which is therefore open. $\hspace{11cm}\square$

And here is where I got stuck: with infinite groups.

Problem If $G$ is infinite, how do I show that there cannot be a point $y\in X$ such that if $U\ni y$ is open, $U\cap(G\cdot x)\neq\varnothing,\{y\}$?

For the moment, I have shown that if there exists such a $y$, besides being outside the orbit by claim 3, there are infinitely many such points. More precisely, I have the following two further claims.

Claim 5 If $y$ is a point as described above, then $gy$ is another such point, for all $g\in G$.

Proof If $U$ is a neighborhood of $gy$, then $g^{-1}U$ is a neighborhood of $y$. $y$ is as above, so $g^{-1}U$ contains a point of the orbit $\overline{g}x$. But then $g\overline{g}\in gg^{-1}U=U$, proving $U$ contains a point of the orbit. Thus, $gy$ is as above, just like $y$. $\hspace{8cm}\square$

Claim 6 If an action is properly discontinuous, then its orbits are in bijection with the group $G$.

Proof Suppose it is not so. This means there exist $g_1,g_2\in G,x\in X$ such that $g_1x=g_2x$. But then we have $g_2^{-1}g_1x=x$. So without loss of generality, we can assume there exists $g\in G:gx=x$. But this cannot be, since there is a neighborhood of $x$ mapped by $g$ into a disjoint neighborhood of $gx$. $\hspace{8cm}\square$

That is all. I did not manage to show there cannot be a $y$ as above. I tried reaching a contradiction by supposing there was, but ended up with an orbit of such $y$s and an infinite number of subsets of $G\cdot x$ accumulating each around a different point of $G\cdot y$. I am tempted to say well, but then there are more points in $G\cdot x$ than in $G\cdot y$, but that cannot be since they are both in bijection with $G$. But then, $G$ is infinite, and this arguments works terribly with infinity. For example, a countable union of countable subsets would be more than countable with that argument, but in fact it is countable. So how do I proceed?

• Pardon the terrible placement of qed symbols but I have neither the \hfil command nor the proof environment and it is unpredictable how these get placed. – MickG Mar 29 '15 at 14:44
• I tried taking $y\not\in G\cdot x$ and proving it is in the interior of $(G\cdot x)^C$, but I did not manage to reach a contradiction of any kind. – MickG Apr 1 '15 at 16:58
• The above assumption proves there is a sequence (or net) converging to $y$ with values in the orbit. If I could conclude that such a net has to converge to an element of the orbit the contradiction would be here. BUt can I now? – MickG Apr 1 '15 at 16:59
• Another route is to say well, suppose we have that net, then for every neighborhood of $y$ you find an infinite number of terms of the net in there, but the net's elements are separated by neighborhoods all homeomorphic to one another, so there can't be infinitely many all in a neighborhood as "small" as we may want. This is modeled on the action $\mathbb{Z}\curvearrowright\mathbb{R}$ by translations. My objection to this is: how can we prove it is impossible that these homeomorphic neighborhood "reduce their sizes" more and more so as to fit into the "infinitesimal" neighborhoods? – MickG Apr 1 '15 at 17:20
• It seems an action as in the question is usually termed as wandering rather than properly discontinuous. Who knows why my teacher called it p.d.… – MickG May 11 '15 at 16:55

You need your points to be closed. Otherwise assume there is a nonclosed point $x$. Then the trivial group acting on $X$ in the only possible way gives a properly discontinuous action. The orbit of each point is the same point, and if a point is not closed, then its orbit won't be closed.

A less trivial example: Take $X:=X_G\cup Y_G$, where $$X_G:=\{ x_g\ :\ g\in G\}\quad\text{and}\quad Y_G:=\{ y_g\ :\ g\in G\},$$ and take the topology generated by the sets $\{x_g,y_g\}$ for $g\in G$. Take the action $g\cdot x_h:=x_{gh}$ and $g\cdot y_h:= y_{gh}$. Then this is a properly discontinuous action, the orbits are discrete, but are not closed.

Now, if you assume that the points are closed (for example if $X$ is Hausdorff) and have already shown that the orbits are discrete, then it is not so difficult to show that they are also closed: Take an orbit $Gx$. We will show that there are no limit points of $Gx$ outside of $Gx$.

Assume by contradiction that $y\notin Gx$ is a limit point. Then there is an open neighborhood $U$ of $y$ such that $U\cap (G\setminus\{1\})U=\emptyset$. Then there is a point $gx\in U$, since $y$ is a limit point of $Gx$. Now $U\setminus \{gx\}$ is also an open neighborhood of $y$ (the points are closed), hence there is an $hx\in U\setminus \{gx\}$. But then $(hg^{-1})gx=hx\in U$ and $(hg^{-1})gx\in (G\setminus\{1\})U$, since $hg^{-1}\in (G\setminus\{e\})$ and $gx\in U$. This contradicts $U\cap (G\setminus\{1\})U=\emptyset$ and proves that $Gx$ is closed.

• Do you have an argument in mind for "closed points and discrete orbits" $\Rightarrow$ "closed orbits"? – Kyle May 11 '15 at 17:46
• No, that is not true. You have to use the property of being properly discontinuous. – san May 11 '15 at 17:49
• Right. I meant in this context. – Kyle May 11 '15 at 17:51
• Got it. I guess by $GU$ you mean the union of the images of $U$ under nontrivial elements of $G$. I think that would be better denoted by $(G\smallsetminus\{1\})U$, 1 being the identity of $G$. And a note to self: since $hx\in U\smallsetminus\{gx\}$, $h$ cannot be $g$, so $hg^{-1}\neq1$. – MickG May 11 '15 at 18:23
• Ok. I corrected the typo. – san May 11 '15 at 18:31

Our setup consists of a space $X$, a group $G$ equipped with the discrete topology, and a continuous proper map $\rho: G\times X \to X \times X$ defined by $(g,x) \mapsto (g\cdot x, x)$.

Claim. If $X$ (hence $X \times X$) is a locally compact and Hausdorff, the orbit $G_x$ is closed.

Proof. The inclusion $\iota_x: X \to X \times X$ defined by $y \mapsto (y,x)$ is clearly continuous. Since $G_x$ is simply $\iota_x^{-1}(G_x \times \{x\})$, it suffices to show that $G_x \times \{x\}$ is closed. We translate our problem yet again by noting that $G_x \times \{x\}$ is the image of $G \times \{x\}$ under $\rho$. Since $\rho$ is a continuous proper map and $X \times X$ is locally compact, the map $\rho$ is closed. Since $G \times \{x\}$ is closed in $G \times X$, its image under $\rho$, i.e. $G_x \times \{x\}$, is closed. As argued above, this implies that $G_x$ is closed. $\square$

Another perspective may be more enlightening: The quotient map $\pi: X \to X/G$ is continuous, hence the orbit $G_x=\pi^{-1}([x])$ is closed if $\{[x]\}$ is closed in $X /G$. When is $\{[x]\}$ closed in $X/G$? You'll probably want to shoot for $X/G$ Hausdorff. Such is the case when $X$ is locally compact Hausdorff and the action of $G$ is properly discontinuous.

I'll update if I happen upon a proof or counterexample for $X$ Hausdorff but not locally compact.

• I don't really see how continuous + proper + with locally compact codomain gives me closed. Plus, I think you may want to read my definition of p.d. action since it is different from "the map \rho is proper. What I have as p.d. seems to be sometimes called "wandering". In any case, how does my definition of p.d. action imply the map above is proper? – MickG May 11 '15 at 18:11
• Ah, I didn't realize you had a different definition of a "properly discontinuous" group action. I believe the standard definition of p.d. is what I wrote in the first line. See math.stackexchange.com/questions/501510/… for an explanation of why $\rho$ is closed. On the other hand, your definition does not imply that $\rho$ is closed. – Kyle May 11 '15 at 18:21
• @MickG: You may find the following discussion interesting: mathoverflow.net/questions/55726/properly-discontinuous-action . Note that "Type D" is similar to your setup. – Kyle May 11 '15 at 18:25
• Indeed it seems my definition of p.d. is something that in Hausdorff spaces is equivalent to a wandering action as Wikipedia defines that here. So "standardly defined" proper discontinuity is either stronger than my definition or unrelated to it. Which of these two? – MickG May 11 '15 at 18:31
• Indeed I saw that discussion, but the long answer is a bit too long for me and anyway I don't get what the poster meant by "replace finite by compact in Type A and Type B". The short answer establishes equivalence of Type tweaked D with C and (IIRC) E, so is not too much of use. I may give it a better look, but for now I think I'd better concentrate on the hardest exam of three years of Math degree :). – MickG May 11 '15 at 18:34