# Smoothness of isotropy representation

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?

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 $$GL(T_pM)$$.