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In Introduction to Riemannian Manifolds John Lee states:

Corollary 3.18. If a Lie group G acts smoothly and transitively on a smooth manifold M with compact isotropy groups, then there exists a G-invariant Riemannian metric on M.

From what I understand the proof of this corollary follows like this:

There exist a point $p\in M$ such that $G_p$ is compact, therefore because the isotropy representation $I_p:G_p \rightarrow GL(T_pM)$ is smooth then $I_p(G_p)$ is compact and because of theorem 3.17 there exists a G-invariant Riemannian metric on M.

Is this proof correct? And if so why is the isotropy representation smooth or even the isotropy group a submanifold with differential structure?

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The proof is correct. By construction (as a stabilizer of a point), the isotropy group $G_p$ is closed in $G$ and hence a Lie subgroup by the closed subgroup theorem. Hence you can restrict the action of $G$ to $G_p$, which defines a smooth map $G_p\times M\to M$. Smoothness of the tangent map of this easily implies that via a "partial derivative" you get a smooth map $G_p\times TM\to TM$. Restricting this to $G_p\times T_pM$, the values lie in $T_pM$ and what you get is exactly $(g,X)\mapsto I_p(g)(X)$, so the latter map is smooth by construction. Applying this to a basis of $T_pM$ and expanding the result in this basis, you see that the matrix describing $I_p(g)$ depends smoothly on $G$ and since these matrices are all invertible, $I_p$ is smooth as a map to $GL(T_pM)$.

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