I am currently re-reading a course on basic algebraic topology, and I am focussing on the parts that I feel I had very little understanding of. There is one exercise in the chapter devoted to groups acting on topological spaces (preceding the chapter on covering spaces) that I have spent several frustrating hours on, but I just can't crack.

I will state the question below, but PLEASE, DO NOT POST A SOLUTION OR A HINT. I am only interested in knowing wether the statement in question is true. I have looked on the internet for a reference about this fact, but didn't get anywhere. If you know a book that gives proves it, or a document online where this is discussed, I would like to get the reference, in case I continue to fail at giving a proof of this the next week.

I recall some terminology first: let $X,Y$ be two topological spaces, and $f:X\rightarrow Y$ a map that isn't supposed to be continuous. The author defines such a map to be $\mathrm{PROPER}$ whenever the following two properties are satisfied : $f$ is a closed map, and $\forall y\in Y, ~f^{-1}(\lbrace y\rbrace)$ is a compact subspace of $X$. It then follows that for all compact subsets $K$ of $Y,~f^{-1}(K)$ is a compact subset of $X$. Also, if $X$ is Hausdorff, and $Y$ is a locally compact Hausdorff, then properness is equivalent to this property.

Let $X$ be a topological space, and $G$ a topological group. Suppose there is a continuous left group action $\rho: G\times X\rightarrow X,~(g,x)\mapsto g\cdot x$. Let $\theta:G\times X\rightarrow X\times X, ~(g,x)\mapsto (x,g\cdot x)$. The author defines the group action to be $\mathrm{proper}$ if $\theta$ is a proper map.

Here is the question: "Show the action of a compact Hausdorff group $G$ on a Hausdorff space $X$ is always proper".

As I said, I have struggled with this for days (since friday). IS THAT STATEMENT TRUE? There are no further hypothesis, $X$ is not supposed to be locally compact, and the action is completely arbitrary (continuous of course, but not supposed free, or other things).

Thank you for your time!

  • $\begingroup$ I think it is a very unusual, and hence confusing, idea not to include "continuous" in the definition of "proper". Anyway, Theo certainly makes this continuity assumption in the fine text (mentioned in his answer) that he wrote for MathOverflow. $\endgroup$ – Georges Elencwajg May 23 '11 at 10:19

Since that's all you want to hear in terms of mathematics: Yes, it's true.

You can find the argument in Bourbaki's Topologie générale, Chapitre III. More precisely, it's the statement of Proposition 2 a) of §4 in Ch. III on page TG III.28 of my edition.

Some time ago, I wrote a résumé on proper actions on MO, which you can find here.

| cite | improve this answer | |
  • $\begingroup$ Do you recall the proof to be straight forward? Am I missing something obvious? $\endgroup$ – Olivier Bégassat May 23 '11 at 8:22
  • $\begingroup$ @Olivier: I think it's rather straightforward and no trickery is required. I'm reluctant to say anything more, since you don't want any hints... However, you can try to show first that the projection map $G \times X \to X$ and the action map $\rho$ are proper (since $G$ is compact). Their "combination" $\theta$ is then quite easily seen to be proper, as $X$ is Hausdorff. $\endgroup$ – t.b. May 23 '11 at 15:08
  • $\begingroup$ Thanks Theo, I was able to look it up, but I haven't gone through all the details yet. Somehow I can't get a good understanding of the notion of proper maps... even when continuity is added, I don't understand what makes this notion important. There are some reasons I get, like the fact that you can show that certain quotients are Hausdorff, and that you can compose proper maps and stay proper, but otherwise I am struggling to understand the importance of this notion. Do you know of good reasons why this is an important notion, or is it a simple technical requirement that makes life easier? $\endgroup$ – Olivier Bégassat May 25 '11 at 22:48
  • 1
    $\begingroup$ Since I don't know much about your background I find it a bit hard to give a convincing answer. Let me restrict to actions. For one thing, Hausdorff quotients correspond to well-behaved orbits, in some sense. If you look at discrete groups, proper actions (=properly discontinuous actions) are "spread out". On the hyperbolic plane for example, the groups acting properly by isometries are precisely the closed subgroups of the isometry group. I'd say in a nutshell: proper actions are "the most geometric ones". Proper actions allow you to have invariant metrics, forms, measures etc. $\endgroup$ – t.b. May 25 '11 at 22:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.