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Wikipedia claims the following:

In mathematics, an action of a group G on a topological space X is cocompact if the quotient space X/G is a compact space or, equivalently, if there is a compact subset K of X such that the image of K under the action of G covers X.

My question is: Isn't this wrong? It is evident that the existence of such a subset K ensures cocompactness, but I am doubting the other direction. How could one possibly choose K? Taking an arbitrary transversal (or its closure) does not work, and I do not see what else could be a candidate.

By the way: Wikipedia points to a specific page in the Handbook of Geometric Topology. This page, however, contains only the definition of a cocompact space, not the claimed equivalence.

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This is a very naive attempt... but why don't taking as $K$ a set of representatives (inside $X$) one for each element of $X/G$. This way you need to make some choice so probably this isn't the best solution. Anyway, it seems to me that the restriction to $K$ of the canonical map from $X\to X/G$ is an homeomorphism... thus $K$ seems to be compact and it clearly covers $X$ under the action of $G$. Does this work? –  Simone Virili Oct 23 '12 at 10:19
    
No, Simone, it does not: $X=R$, $G=Z$ and the circle $R/Z$ is not homeomorphic to any subset of $R$... –  Bugs Bunny Oct 23 '12 at 10:29
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I meant in the subspace topology, off course, Doc –  Bugs Bunny Oct 23 '12 at 10:31
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The concept of cocompact action is usually used in the case of properly discontinuous actions. In this case two definitions of cocompact action are equivalent. –  Misha Oct 23 '12 at 12:45
    
Two definitions are also equivalent for actions on locally compact spaces. –  Misha Oct 23 '12 at 13:01

4 Answers 4

up vote 8 down vote accepted

I agree with @Bugs that some extra assumptions are needed, although I do not have a counter-examples either.

Here is an argument assuming that $X$ is locally compact. For each $y\in Y=X/G$ choose (arbitrarily) a point $y'\in X$ which projects to $y$ (you'd need axiom of choice here). Now, for each $y'$ pick an open neighborhood $U_{y'}$ whose closure in $X$ is compact. Projecting the open sets $U_{y'}$ to $Y$ yields an open covering by the sets $U_y$ (images of $U_{y'}$'s). By compactness, there is a finite set $\{y_1,...,y_n\}\subset Y$ so that the sets $U_{y_i}$ cover $Y$. Then the union $$ \bigcup_{i=1}^n cl(U_{y_i})\subset X $$ is compact and projects onto $Y$.

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Thank you. Does the statement still hold if $X$ is only compactly generated (or compactly generated Hausdorff)? –  Niemi Oct 23 '12 at 16:22

This is a somewhat silly counterexample, but here goes:

Let $G$ be a topological group and let $H$ be a dense subgroup. Then the quotient space $G/H$ has the trivial topology, in particular it is compact.

If $G$ is Polish (separable and completely metrizable) but not locally compact, then every compact subset is nowhere dense. If $H$ is a countable dense subgroup of $G$ then no compact $K \subset G$ can be mapped onto $G/H$ because $G = \bigcup_{h\in H} hK$ would exhibit $G$ as meager, contradicting the Baire category theorem.

For an explicit example, let $G = c_0$ be the additive group of the Banach space of real sequences converging to zero with the supremum norm and let $H$ be the countable subgroup of rational sequences with only finitely many non-zero terms.

This also gives a negative answer to a question raised in the comments: metrizable spaces are compactly generated, so assuming that $X$ is compactly generated is not sufficient for Wikipedia's characterization of cocompactness to hold.

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Having been educated in the French school, my compact spaces are supposed to be Hausdorff - see: fr.wikipedia.org/wiki/Compacité_(mathématiques) so I'm not quite convinced by Gee's example (+1 however for the argument using Baire category) –  Alain Valette Oct 24 '12 at 22:02

Here is a counterexample in which $X/G$ is also Hausdorff.

We give $X:=(\mathbb{N} \times \mathbb{Z}) \cup \{\infty\}$ a topology similar to the Arens-Fort topology. That is, each point $(m,n) \in \mathbb{N} \times \mathbb{Z}$ is isolated and basic neighborhoods of $\infty$ are of the form $$B_{f,k}:=\{\infty\} \cup \{(m,n) : n \geq f(m) \land m \geq k\},$$ where $f:\mathbb{N} \to \mathbb{Z}$ and $k \in \mathbb{N}$.

We let $G:=\mathbb{Z}$ act on $X$ in the natural way: $g(m,n)=(m,n+g)$ and $g(\infty)=\infty$.

Any compact subset of $X$ is finite, so its orbit cannot cover $X$. On the other hand $X/G$ is just a convergent sequence (which is compact and Hausdorff).

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I believe that it is wrong in full generality. I will try to think of a counterexample.

IMHO, you need some additional assumptions. For instance, somewhere at the start of "Representation theory and automorphic functions" Gelfand-Graev-Piatetsky-Shapiro sketch an argument but they assume that $X$ is metrizable with an inner metric, i.e. there exists a point half-distance between two points, and that the actions preserves the metric.

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Seems like the same issue with the difference of a transitive action doesn't ensure that the space is isomorphic to $G/H$, e.g. 2nd countable is enough. –  Marc Palm Oct 23 '12 at 11:58

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