Let $M$ be a smooth manifold and let $G$ be a finite group acting on $M$ by diffeomorphisms. Show that the set of fixed points $M^G := \{m \in M \mid g . m = m, \forall g \in G\}$ is a smooth manifold

What I tried

Since $M$ is a smooth manifold, then there exist a maximal smooth atlas $\mathfrak{A}=\{(U_\alpha, V_\alpha, \phi_\alpha)\}_{\alpha \in I}$. Now I want to prove that the fixed point set $M^G$ defined above is a smooth manifold and I want to establish a chart for the same. I am unable to understand how will I use the fact that the group $G$ is finite here. Also I think it is diffcult for me to get some charts for $M^G$. Can anyone help me this way?

  • $\begingroup$ Also I do not want to use any riemanian metric , as I have not covered that. Hence other posts on this site won't help me much $\endgroup$ Apr 16 at 15:35
  • $\begingroup$ Is $g$ here a fixed $g \in G$, or do you want $g.m=m$ for all $g \in G$. In any case, just as a guess, could you use something like the fact that the preimage of $\{0\}$ is a smooth manifold under the map $f \colon M \to \mathbb{R}$ given by $f(m) = d(g.m,m)$ which should be a smooth map for some smooth metric $d$ on $M$. $\endgroup$
    – Dan Rust
    Apr 16 at 15:40
  • $\begingroup$ $gm=m $ for all $g \in G$.Also I have not studied anything about metric on manifolds $\endgroup$ Apr 16 at 15:40
  • $\begingroup$ Also pre-image of $\{0\}$ is a closed set. I don't think it helps me. $\endgroup$ Apr 16 at 15:47
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    $\begingroup$ Does this answer your question? Showing that the set of fixed points is a smooth manifold. Also, the answer depends on your definition of a manifold. $\endgroup$ Apr 17 at 14:16