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Let $M$ be a differentiable manifold. The diffeomorphism group of $M$ is the group of all $C^{\infty}$ diffeomorphisms of $M$ to itself, denoted by $\text{Diff}(M)$. This space of diffeomorphism $\text{Diff}(M)$ is considered as a Lie Group equipped with composition of functions ($\circ$) as a group operation. According to the definition of Lie Group, this implies that $\text{Diff}(M)$ is a manifold. I was wondering what is the intuition or proof that shows that $\text{Diff}(M)$ is indeed a manifold?

User3113446
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Swakshar Deb
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    I just wanted to say that I believe $Diff(M)$ is infinite dimensional, so we are talking about and infinite dimensional Lie Group. So, it isn't a typical finite dimensional manifold. – Jeff Feb 18 '24 at 18:39
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    Yes, indeed $Diff(M)$ is an infinite-dimensional Lie Group. – Swakshar Deb Feb 18 '24 at 19:52
  • Does this answer your question? Is the group of diffeomorphisms a Lie Group? – Marco Feb 18 '24 at 20:46
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    The discussion in the answers to the linked question is not quite right. I think, there is an implicit assumption there that the manifold is compact. For noncompact manifolds I am not even sure what is the right topology to make it a topological group. – Moishe Kohan Feb 18 '24 at 20:55
  • According to Wikipedia, this group has two standard topologies (weak and strong). If the manifold $M$ is not compact, then neither topology yields a locally contractible space. In particular, $Diff(M)$ is not an infinite dimensional manifold. – Moishe Kohan Feb 19 '24 at 19:43

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