Uniqueness of smooth structure on the Isometry group from Steenrod-Myers theorem

Question

Let $(M,g)$ be a connected Riemannian manifold. The theorem of Steenrod and Myers states that the group $G$ of isometries $M \to M$ is a Lie group with the compact-open topology such that evaluation map $$G \times M \to M, \qquad (f,p) \mapsto f(p)$$ is smooth.

Question: Is the smooth structure on $G$ uniqueley specified by this property, similar to how the compact-open topology is the coarsest topology such that the evaluation map is continuous? Or can there be multiple such smooth structures which make this map smooth?

Answer 1

After re-reading, I'm fairly confident that I've understood the question, and it means the following:

Give $G:=\text{Aut}(M)$ the compact open topology inside of $\text{Cont.}(M,M)$. Then, $G$ admits a Lie group structure such that $G\times M\to M$ is smooth. Is this Lie group structure 'unique' in any way?

The answer is, well, yes. But, not for anything specific to this problem.

Indeed:

Theorem: Let $G$ and $H$ be Lie groups and $f:G\to H$ a continuous group map. Then, $f:G\to H$ is smooth.

Proof(Sketch): Consider the graph $\Gamma_f\subseteq G\times H$. This is a closed subgroup of $G\times H$ and thus, from the basic theory of Lie groups, a closed (embedded) submanifold. The map $f$ is the composition of the smooth map $G\times H\to H$ (the projection) with the inverse of the projection map $\Gamma_f\to H$. So, it suffices to show that $\Gamma_f\to H$ is smooth (and thus its inverse is smooth). But, this follows since it's a continuous group isomorphism and so is generically smooth by Sard's theorem and so, by the homogenenity of group maps, smooth. $\blacksquare$

From this we deduce the following:

Theorem: Let $G$ be a topological group. Then, any Lie group structure on $G$ is unique (up to isomorphism).

Proof: Apply the previous theorem to $\text{id}_G$. $\blacksquare$

This then handily answers your question in the nicest way possible--yes, it's unique, for no fancy reason. :)