Let $G$ be a group acting on some manifold $M$. For s subset $A\subseteq M$, let $G[A]:=\{g\in G\mid A\cap g.A\neq\emptyset\}$. We assume that $G$ acts freely and properly discontinuously on $M$, that is:
- Proper discontinuous: For every $x\in M$, there is an open neighborhood $U$ of $x$ for which $G[U]$ is finite;
- Free: For very $g\in G$, if there is $x\in M$ for which $g.x=x$, then necessarily $g=\mathrm{id}_G$.
In order to prove that $M/G$ also has the structure of a manifold, we use the following:
Lemma.
- If $K\subseteq M$ is compact, then $G[K]$ is finite.
- Every $x\in M$ has an open neighborhood $U$ for which $G[U]=\mathrm{id}_G$.
I am trying to prove this result.
For 1., I assume that there are distinct $g_k\in G[K]$ $(k\geq 0)$. Then we find $x_k\in K$ for which also $g_k.x_k\in K$. Since $G$ acts freely on $M$, also the $x_k$ are distinct. Both sequences $(x_k)_{k\geq 0}$ and $(g_k.x_k)_{k\geq 0}$ have a limit point because $K$ is compact. Am I on the right track? How can I continue from here?
For 2., I feel like we should start with the open neighborbood $U$ of $x$ given by proper discontinuity, and in case $G[U]$ contains other elements than the identity, we remove them one by one until only the identity is left. However, I don't see how I can put this into practice...
