Showing that the set of fixed points is a smooth manifold

Question

Let $M$ be a smooth manifold and Let $G$ be a finite group acting on $M$ by diffeomorphisms. Show that the set of fixed point $F=\{m \in M: g.m=m \}$ is a smooth manifold.

I am unable to deal with the set of fixed points. It doesn't seem to be an open set. It would have been trivial if it was open. Any hints?

Answer 1

On the contrary, by its definition via an equality, $F$ will always by closed. You will need to show the properties of a smooth manifold step by step. For "locally Euclidean", you may want to use the implicit function theorem.

After that, the transition maps of the atlas may not "fit" the subsets a priori, but we haven't used finiteness of $G$ yet. Note how the action of $G$ on $M$ induces an action on transition maps near points $\in F$. By averaging over the finitely many transition maps in the $G$-orbit of an arbitrary transition map, you can obtain a $G$-invariant transition map. Being $G$-invariant, it respects $F$ and thus is a valid transition map for $F$.

Answer 2

You may assume $M$ is a Riemannian manifold with some metric $g_0$. First, observe that $M$ admits a new Riemannian metric $g$ that is invariant under the group action: for example define $$ g=\sum_{\tau\in G} \tau^*g_0; $$ this can be defined since $G$ is a finite group.

In general, for any finitely many elements $\tau_1, \tau_2, ..., \tau_n$ assume $p$ is a fixed point of them, i.e. $\tau_j p=p$ for $j=1, ..., n$. Note that since $g$ is an invariant metric, the differential $d\tau_j$ is an orthogonal linear transformation on the tangent space $T_p$.

If there is no tangent vector $X$ at $p$ that is a common eigenvector of all $d\tau_j$ with eigenvalue $1$, then $p$ is an isolated fixed point: in fact, $\tau_j$ maps any geodesic from $p$ to a geodesic from $p$, so if a tangent vector $X$ is not an eigenvector with eigenvalue $1$ for some $d\tau_j$, then $\tau_j$ moves the geodesic $\exp_p(tX)$ (here $\exp_p$ is the exponential map from $p$) to a different geodesic, so there is no fixed point on $\exp_p(tX)$ for $0<t<\delta$.

Now assume the common eigenvectors of all $d\tau_j$ at $p$ with eigenvector $1$ form a linear subspace $V$ in $T_p$. Then as above we see the manifold $\exp_p V$ are all fixed points; on the other hand, any other point $q$ near $p$ can be written as $\exp_ptX$ so that $X\notin V$ and $t$ small, as above we see this point is moved to some other point by some $\tau_j$. So near $p$ the fixed point set is exactly $\exp_tV$.