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Let $X$ be a compact smooth manifold. Then we can endow the set of diffeomorphisms of $X$ with $C^1$ topology. Are there any examples when $\mathrm{Diff}(X)$ is compact (as a topological space)?

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    $\begingroup$ No, unless $dim(X)=0$. $\endgroup$ – Moishe Kohan Jan 3 '18 at 16:24
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    $\begingroup$ Consider the case $X = \mathbb{R}^n$, then look at diffeomorphisms that are trivial off a small neighborhood of a point. $\endgroup$ – anomaly Jan 3 '18 at 16:34
  • $\begingroup$ @anomaly: yes, indeed $\mathrm{Diff}(\mathbb{R}^n)$ embeds into $\mathrm{Diff}(X)$ for a smooth manifold $X$ of dimension $n$. Apparently, I was a little confused with my question $\endgroup$ – user285001 Jan 3 '18 at 16:39
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It is very far from being compact, its Lie algebra is the set of vector fields which is infinite dimensional if $dim>0$ so it is an infinite dimensional Lie group.

https://en.wikipedia.org/wiki/Diffeomorphism#Diffeomorphism_group

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