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When is the group of homeomorphisms of a compact space locally compact?

I am interested in finding out when the group of homeomorphisms of a compact topological space $X$ (with appropriate topology e.g. 'weak' or compact-open) is a locally compact space.

What extra conditions might we be able to put on $X$ to ensure that it is so?... What if $X$ is, say, a metric space and we ask when the isometry group is locally compact?

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    $\begingroup$ What is $Aut(X)$? If you mean homeomorphisms, then $Aut(X)$ is almost never locally compact. If you mean isometries, then $Aut(X)$ is compact with the compact-open topology by a straightforward application of the Arzela-Ascoli theorem. $\endgroup$
    – Johannes Ebert
    Commented Dec 1, 2010 at 15:01
  • $\begingroup$ Have edited to clear up that ambiguity. I guess I just meant 'structure preserving maps' but sorry if that was vague. $\endgroup$
    – Spencer
    Commented Dec 1, 2010 at 19:16
  • $\begingroup$ What is the context of this question? Do you have already nice examples of compact spaces whose groups of homeomorphisms are locally compact? It is easy to show that any (non 0-dimensional) manifold will have a non locally compact group of homeomorphisms, so examples of such compact spaces will probably not be as nice as manifolds. $\endgroup$
    – Guillaume Brunerie
    Commented Dec 1, 2010 at 19:56
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    $\begingroup$ Now that the OP has clarified this question, this is completely off-topic, but Julien Melleray has proved the converse of Johannes Ebert's comment in his nicely titled "Compact metrizable groups are isometry groups of compact metric spaces" (Proc. AMS 136 (2008), also available on his webpage). $\endgroup$
    – Maxime Bourrigan
    Commented Dec 1, 2010 at 21:30

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I do not know what you mean by automorphism group, I guess you mean homeomorphisms. In that case the answer is no:

For instance, the homeomorphisms of the circle are in one-to-one correspondence with continuous strictly monotone functions $[0,1] \to \mathbb{R}$ such that $f(0) \in [0,1)$ and $f(1) = f(0)\pm 1$. Compact-open topology just means uniform convergence, and this obviously is not a locally compact space.

As for local compactness of the isometry group, it follows from the Arzelà-Ascoli theorem that that the isometry group of a proper metric space (i.e., closed balls are compact) is locally compact.

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For a (connected) smooth Riemannian manifold $M$, it has been shown by Myers and Steenrod that that the group of isometries is a Lie group, hence is locally compact. On the other hand the group of homeomorphisms of a smooth manifold $M$ is never locally compact. When the dimension is at least $2$, this group acts $k$-transitively for any $k$ on $M$ and from here I think it should be easy to show that the groups is not locally compact.

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