I seem to recall reading this somewhere:

If a Lie group $G$ acts on a topological space $X$ properly, then every compact subset of the orbit space $X/G$ arises as the image of a compact subset of $X$ under the projection map.

Could anyone confirm if this is true/false and why? Cheers.

Edit: by proper I mean that $G\times X\rightarrow X\times X$ given by $(g,x)\mapsto (x,g\cdot x)$ is proper.

  • $\begingroup$ What is your definition of a proper action? (There are several inequivalent notions in the literature.) $\endgroup$ – Moishe Kohan Jun 14 '17 at 13:30
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    $\begingroup$ If you are willing to assume local compactness of $X$ then you do not even need properness of the action and you do not need the assumption that $G$ is a Lie group. $\endgroup$ – Moishe Kohan Jun 14 '17 at 13:48
  • $\begingroup$ Thanks, do you mind elaborating? Cheers. $\endgroup$ – ougoah Jun 14 '17 at 18:08

First note that if $G\times X\to G$ is a topological group action then the quotient map $\pi: X\to X/G$ is open. Now, if $X$ is locally compact (meaning that every point $x$ admits a relatively compact open neighborhood $U_x$ - one can ask for a basis of such neighborhoods but I do not need one), then $\{\pi(U_x): x\in X\}$ is an open cover of $X/G$. By compactness of $X/G$, there are finitely many points $x_1,...,x_n\in X$ such that $$ X/G= \bigcup_{i} \pi(U_{x_i}). $$ Then the compact $$ K= \bigcup_{i} \overline{U_{x_i}}\subset X $$ projects onto $X/G$. qed

What happens without local compactness assumption (but assuming proper discontinuity of the $G$-action, in some form) I do not know. It is a good question though.

P.S. There are several inequivalent notions of proper actions (and proper maps) in the literature. Most people (not educated in France) assume that this means continuity and that preimages of compacts are compact. However, if you like reading Bourbaki - and I do, sometimes - you realize that there is another, and more convincing, definition. See Bourbaki's "General Topology".


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