Manifold of fixed points

Question

Let $M$ be a smooth manifold and let $G$ be a Lie group smoothly acting on $M$.

Then, under suitable assumptions (if $G$ acts freely and properly on $M$) we have a new smooth manifold $M/G$ corresponding to the orbits of the action.

I would like to know if there is a theorem that states (under suitable assumptions) that the set of fixed points $M^G$ can be equipped with a smooth manifold structure.

I suppose there is such a theorem, because $M^G$ is also the zero set of the infinitesimal generator of the action, which is a smooth vector field, so we have "smooth equations" describing it.

I will greatly appreciate any reference about this topic.

Thanks for your help!

Answer 1

I'll write the action as a morphism of groups $$ \begin{array}{ccc} G & \longrightarrow & Diff(M)\\ g & \longmapsto & \psi_g \end{array}. $$ If $x\in M^G$ we have that $d_x\psi_g\in End(T_xM)$, so we get a linear action of $G$ on $T_xM$.

If $\rho$ is a $G$-invariant metric (it exists if you take $G$ compact, for example) on $M$, then the exponential map $$\exp^\rho_x:T_xM\rightarrow M$$ is equivariant with respect to the two actions above: $\exp^\rho_x(d_x\psi_g(v))=\psi_g(\exp_x^\rho(v))$.

Recall that $\exp_x^\rho$ gives a diffeomorphism between a neighbourhood $U_0$ of $0\in T_xM$ and a neighbourhood $U_x$ of $x\in M$. Consider $U_0^G$ and $U_x^G$ the corresponding sets of fixed points with respect to the actions above. Since $\exp^\rho_x$ is equivariant it also defines a diffeomorphism between $U_0^G$ and $U_x^G$. Finally, since the action of $G$ on $T_xM$ is linear, its fixed points form a linear subspace, and then $U_0^G$ is an open subset of a vector space.

Summing it all up, $(\exp_x^\rho)^{-1}:U_x^G\rightarrow U_0^G$ is a chart about $x\in M^G$.