# Finite group acting on a smooth manifold $M$ by diffeomorphisms. Fixed Point set is Smooth Manifold [duplicate]

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?

• 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 Apr 16 at 15:35
• 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 $M$. Apr 16 at 15:40
• $gm=m$ for all $g \in G$.Also I have not studied anything about metric on manifolds Apr 16 at 15:40
• Also pre-image of $\{0\}$ is a closed set. I don't think it helps me. Apr 16 at 15:47
• 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. Apr 17 at 14:16