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Let $M$ be a compact $n$-dimensional manifold let $H(M)$ denote the homeomorphism group of $M$.

If $n=2$ then $H(M)$ enjoys nice properties such as being an ANR, is locally contractible, separable. The dimension restriction for $n=2$ is superfluous for local contractibility.

However, is $H(M)$ locally compact and in general finite dimensional? Is it a Lie group in general?

In the non-compact case, consider $\mathbb{R}^2$. Then, $H(\mathbb{R}^2)$ is homotopy equivalent to $O(2)$; albeit this doesn't say much.

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    $\begingroup$ Do you mean the isometry group of a manifold with metric, instead of homeomorphism group? There are many many more homeos of $\mathbb R^2$ than $O(2)$. $\endgroup$
    – Kevin Casto
    Commented Jul 22, 2023 at 19:13
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    $\begingroup$ If the manifold is infinite, the homeomorphism group is not locally compact, since every neighborhood of the identity contains a closed, noncompact subgroup. $\endgroup$
    – YCor
    Commented Jul 22, 2023 at 20:01
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    $\begingroup$ @SomePerson no: I mean when the number of points is infinite. That is, either there is a positive-dimensional component, or there are infinitely many components $\endgroup$
    – YCor
    Commented Jul 22, 2023 at 20:27
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    $\begingroup$ Local contractibility (when equipped with the compact open topology) holds in all dimensions: This is due to Chernavsky. $\endgroup$
    – Moishe Kohan
    Commented Jul 22, 2023 at 21:25
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    $\begingroup$ Deciding if $Homeo(M)$ is an ANR is a well-known open problem (except as you say if $\dim(M)\le 2$). Separability of $Homeo(M)$ is immediate because it is a subspace of the space of continuous self-maps of $M$, which is is metrizable and second countable, and these two properties are inherited by subspaces. $\endgroup$
    – Igor Belegradek
    Commented Jul 23, 2023 at 2:44

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