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Since the diffeomorphism group is not locally compact; is it true that there is no proper action of an infinite-dimensional diffeomorphism group on a finite-dimensional smooth manifold?

Edit: The group is the $C^\infty$ self-diffeomorphisms of a finite-dimensional smooth open manifold. Let's say that the group is endowed with the Whitney $C^\infty$-topology.

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  • $\begingroup$ Properness will depend on the topology of the diffeomorphism group. And the whole question depends on the definition of diffeomorphism you choose (groups of $C^k$ diffeomorphisms of the circle, for instance, are very different for $k=1$, $k=2$, $k=\infty$, $k=\omega$...) $\endgroup$
    – YCor
    Commented Feb 20, 2016 at 19:25

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Let $G$ be a Hausdorff topological group acting continuously properly on a locally compact space $X$. Then $G$ is locally compact.

Indeed, fix $x\in X$ and let $Y$ be a compact neighborhood of $x$. By definition of properness, there is a compact subset $K$ of $G$ such that $g\notin K$ implies $gY\cap Y=\emptyset$. By continuity of the action, there is a neighborhood $V$ of 1 in $G$ such that $g\in V$ implies $gx\in Y$. In particular, $V\subset K$, so $K$ is a compact neighborhood of 1 in $G$. It follows that $G$ is locally compact.

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  • $\begingroup$ yes but this does not answer my question, I'm not asking if the group is locally compact. The diffeomorphism group is not locally compact. $\endgroup$
    – s k
    Commented Feb 20, 2016 at 20:28
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    $\begingroup$ Yes, it does answer your question. By contraposition: if the group is not locally compact, then it can't act properly on any manifold. $\endgroup$
    – YCor
    Commented Feb 20, 2016 at 20:45

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