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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?

Shaun
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permutation_matrix
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  • 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 – permutation_matrix Apr 16 '21 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 $M$. – Dan Rust Apr 16 '21 at 15:40
  • $gm=m $ for all $g \in G$.Also I have not studied anything about metric on manifolds – permutation_matrix Apr 16 '21 at 15:40
  • Also pre-image of ${0}$ is a closed set. I don't think it helps me. – permutation_matrix Apr 16 '21 at 15:47
  • Have you tried to prove the case where $G$ is a finite cyclic group? Maybe that is a good starting point – Aitor Iribar Lopez Apr 16 '21 at 15:48
  • Alright, but how does $G$ is finite helps? Basically I am asking in general. How does $G$ is finite impact the group action $G \times M \to M$ – permutation_matrix Apr 16 '21 at 15:50
  • I mean you would expect the preimage of ${0}$ to be closed, yes. If you have a convergent sequence of fixed points, their limit will also be fixed, by continuity. The preimage of ${0}$ is precisely the set of fixed points (for a single $g$, which I initially assumed. You need to modify to get it to work for the whole of $G$). – Dan Rust Apr 16 '21 at 15:51
  • Large group actions can be quite badly-behaved. See: https://math.stackexchange.com/a/1739768/29059 . The main nice thing about finite actions is that you know the orbit of any point is finite and hence is a discrete subset of $M$, so such actions are also proper. Moreover, the quotient of the action will itself be a manifold by the slice theorem. – Dan Rust Apr 16 '21 at 16:00
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    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. – Moishe Kohan Apr 17 '21 at 14:16

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