Suppose $M$ is a smooth manifold. Suppose that $G$ is a finite group. Consider the left action of $G$ on $M$. Suppose that this action is free. Prove that the quotient $\frac{M}{G}$ is a smooth manifold.
I understand that this can be done by Quotient Manifold Theorem(Lee). But I want a clear solution. What I understood from my attempt is that $\frac{M}{G}$ is the collection of orbits the action, which partitions $M$. We use that partition to get a topology.
But I am not able to prove that $\frac{M}{G}$ is a smooth manifold. Any help would be appreciated.
So $M/G$ has the quotient topology and $p: M\rightarrow M/G$ is surjective continuous, with $M$ compact and $M/G$ Hausdorff, so that $M/G$ is Hausdorff compact.
Moreover, if $U \subset M$ is open, then $p^{-1}(p(U))=\bigcup_{g \in G}{gU}$ is open so that $p$ is an open continuous map.
Let’s prove that $p$ is a covering map.
Let $x \in M$. For each $g \in G$ that isn’t the identity, the image at $(x,x)$ of $(g,id): M \times M \rightarrow M \times M$ isn’t on the diagonal so that there exists an open subset $(x,x) \in U \times V$ of $M \times M$ so that $(g,id)(U \times V)$ doesn’t meet the diagonal either.
Write $U_{g,x}=U \cap V$: it’s an open subset containing $x$ and such that if $y,z \in U_{g,x}$, $gy \neq z$. Thus, the intersection $U_x$ of the $U_{g,x}$ is an open subset of $M$ containing $x$ such that any two distinct points of $U_x$ have disjoint orbits and thus the $gU_x$, $g \in G$ are pairwise disjoint, and $p: \coprod_g{gU_x}=p^{-1}(p(U_x)) \rightarrow p(U_x)$ is continuous, open, and its restriction to any $gU_x$ is injective and surjective, so $p$ is a homeomorphism from any $gU_x$ to $p(U_x)$ which is open in $M/G$, hence the conclusion.
Now, we want to say that $p$ endows $M/G$ with the structure of a smooth manifold. We reduce each $U_x$ around $x$ so that $\psi_x:U_x \rightarrow B$ is a diffeomorphism, where $B$ is an open ball. For each $x$, let $\phi_x: p(U_x) \rightarrow U_x \rightarrow B$ be a homeomorphism (let $i_x:p(U_x) \rightarrow U_x$ be the inverse of $p$), and let’s show that the set of $\phi_x$ is an atlas for $M/G$.
So let $x,y$ be such that $p(U_x)$ and $p(U_y)$ meet, and we want to show that $\phi_y \circ \phi_x^{-1}:\phi_x(p(U_x) \cap p(U_y)) \subset B \rightarrow \phi_y(p(U_y)\cap p(U_x))$ is smooth. Clearly, it is enough to show that the map $f: U_x \cap p^{-1}(p(U_y)) \rightarrow U_y \cap p^{-1}(p(U_x))$ given by applying $p$ and then $i_y$ is smooth (they’re open subsets of $M$).
This map is continuous and has the following property: for every $z$ in its domain $D$, $f(z)=g\cdot z$ for a certain (unique) $g \in G$. Now, let $z \in D, g \in G$ such that $f(z)=g \cdot z$. Let’s show that there is an open subset $z \in V \subset D$ such that for every $t \in D$, $f(t)=g\cdot t$: it will follow that $f$ is smooth on $V$ – and thus that $f$ is smooth.
Assume the opposite statement, for the sake of contradiction. Siince $G$ is finite, there is a sequence $t_n \in D$ converging to $z$ and a sequence $g_n$ of elements of $G \backslash \{g\}$ such that $f(t_n)=g_n \cdot t_n$. Up to taking a subsequence (since $G$ is finite), we can assume that $g_n$ is constant equal to $h\neq g$, and $f(t_n)=h \cdot t_n$. Taking limits, it follows that $f(z)=h \cdot z$, a contradiction since the action is free.