Let $G$ be a discrete countable group and let $X$ be $G$-Hausdorff-space. We call $X$ proper, iff the map $$ G \times X \rightarrow X \times X , \qquad (g,x) \mapsto (x, gx) $$ is proper and we call $X$ cocompact, iff the quotient-space $G\backslash X$ is compact. So if $X$ is both cocompact and proper, is $X$ then locally compact? Thank you.
Is a cocompact proper $G$-space locally compact`?
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general-topology
algebraic-topology
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0You mean that $G$ is a discrete countable group, and not $X$ right? And for cocompact you are looking at $X/G$ and not $G/X$ right? – 2017-02-19
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0Well, I meant $G$ to be discrete and countable, yes. I consider a left $G$-action on $X$, so the elements in $G\backslash X$ are supposed to be left orbits and therefore I wrote $G\backslash X$. Now, some authors write $G/X$ but in that sense, if $H < G$ is a subgroup then $G/H$ denotes the orbits of the Right-$H$ action on $G$ by multiplication. – 2017-02-19
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0The standard notion of a proper action without local compactness assumption is not particularly useful. A better notion of properness in this case is the one defined by Richard Palais, see my answer here: http://math.stackexchange.com/questions/560371/orbit-space-of-a-free-proper-g-action-principal-bundle/1754706#1754706. If you replace "proper" with "Palais-proper" and assume complete regularity, then the claim should follow from the Palais' theorem on existence of slices. Take a look. – 2017-02-19