On Properly Discontinuous Groups

Question

I need to prove the following:

Every group of diffeomorphims that act properly discontinuous in a compact smooth manifold is finite.

I've been looking for it in some many references but couldn't find. Here the diffeomorphims are $g:M\rightarrow M$. Some remarks:

Definition 1: $G$, a group of diffeomorphims, act freely in $M$ if for all $g\in G\ -\{e\}$ , $g$ has not fixed points, where $e$ is the identity.

Definition 2: $G$ is properly discontinuous in $M$ if:

(i) $G$ act freely in $M$.

(ii) for all $x, y\in M$, such that $Gx\neq Gy$, there are open subsets of $M$, $x\in U$, $y\in V$ such that $U \cap g(V)= \emptyset$, for all $g\in G$.

(iii) for each $x\in M$, there is an open subset of $M$, $x\in V$, such that $g(V)\cap V=\emptyset$ for all $g\in G-\{ e \}$.

Can anyone help me? Thanks.

Answer 1

Suppose that $G$ acts properly and discontinuously on the compact manifold $M$ and $G$ is infinite. Let $g_n,n\in\mathbb{N}$ distinct elements of $G$ and $x\in M$. Since $M$ is compact,there exists a subsequence $g_{n_k}(x)$ of $g_n(x)$ which converges towards $y$.

We deduce that for every open subsets $x\in U, y\in V$, there exists $n_k$ with $g_{n_k}(x)\in V$, this implies that $g_{n_k}(U)\cap V$ is not empty. The condition $(ii)$ implies that $Gx=Gy$ and $y=gx$. We deduce that $g^{-1}g_{n_k}(x)$ converges towards $x$ and for every open subset $V$ containing $x$, there exists $N$ such that for $n_k>N$, $g^{-1}g_{n_k}(x)\in V$, we can find $n_k>N$ with $g^{-1}\neq g_{n_k}$ since the $g_{n_k}$ are distinct, this is in contradiction with $(iii)$.

Answer 2

This is the same basic idea as Tsemo Aristide's answer, but perhaps a bit more geometric.

Let's rephrase the conditions more geometrically.

(i) The map $G\to Gx$ defined by $g\mapsto gx$ is an injection for all $x\in M$.

(ii) This condition is actually stronger, but all we need from it is that $Gx$ is closed.

(iii) Again, this condition is actually stronger, but again, all we need is that $Gx$ is discrete.

Then condition (i) says $Gx$ is infinite. Since $M$ is compact, it has a limit point, which by (ii) is in $Gx$, contradicting (iii).