Question

3. Let M be a manifold and let G C Homeo(M) be a group acting on M. Suppose that this group action is properly discontinuous and free prove that the quotient space M/G is a manifold. For this problem properly discontinuous means that if K c M is compact then the set {ge G | g(K) n/Kメ0) is finite) and free means the only element of g that fixes any point of M is the identity.

Answer 1

We given that
G C Homeo(M)
is some group of homeomorphism, whose action on the manifold is properly discontinuous and free. We have to prove that the quotient space
M/G
is again a manifold. Suppose, $q:M\to M/G$ is the quotient map.

First let us observe the following. Suppose for $x\in M$ we choose some locally euclidean neighborhood $x\in U\subset M$. Now say for some non-identity element
gEG
, we have that $U\cap g(U) \ne \emptyset$ . Since the action is free, we have that $x \ne gx$. Since is Hausdorff, shrinking the neighborhood if necessary, we can assume that $gx\not\in U$. Now since is locally euclidean, it is in particular locally compact. So we can find another euclidean neighborhood $x\in V\subset U$ such that the closure $\bar{V}$ is compact and $\bar{V}\subset U$ . Then we see that, $gx\in g(U) \setminus \bar{V}$ , which is open in , as $\bar{V}$ is closed. Then we can find some euclidean neighborhood $gx \in W_1 \subset g(U)\setminus \bar{V}$ . Consider the open set $g^{-1}(W_1)$. Clearly, $x\in V\cap g^{-1}(W_1)$ and we can find some euclidean neighborhood $x\in U_1\subset V\cap g^{-1}(W_1)$ . Then by construction we have, $g(x) \in g(U_1)\subset W_1 \subset g(U) \setminus \bar{V}$ and as $U_1\subset V\subset \bar{V}$ , we get $g(U_1) \cap U_1=\emptyset$ .

Now we prove that
M/G
is in fact locally euclidean and Hausdorff. Say, $y\in M/G$. Then is an equivalence class of some $x\in M$. Get some locally euclidean neighborhood $x\in U\subset M$ such that $\bar{U}$ is compact. Since the action is properly discontinuous, we have that there are finitely many elements, say, $\{g_1,\ldots, g_r\} \subset G$ such that $\bar{U}\cap g_i (\bar{U}) \ne \emptyset$ . Now, if $U\cap g_i (U) \ne \emptyset$ , by the preceding observation we can get some euclidean neighborhood $x\in U_i\subset U$ such that $U_i \cap g(U_i) =\emptyset$ . If  $U\cap g_i (U) = \emptyset$, then we set . Consider $\cap_{i=1}^r U_i$ . Since this is a finite intersection, it is open. Get some euclidean neighborhood $x\in V\subset \cap_{i=1}^r U_i$ . Then by construction we have, $V\cap g_i(V) =\emptyset$ for every $i=1,\ldots, r$. Then we see that are disjoint from each other for every
gEG
. Now, set $W=q(V)$. Then $q^{-1}(W) = \sqcup g(V)$ is an open set. So, is an open neighborhood of . Also, $q|_V:V\to W$ clearly a homeomorphism and hence is in fact an euclidean neighborhood of . Thus,
M/G
is locally euclidean.

Now to show that
M/G
is Hausdorff. Suppose, $y_1 \ne y_2\in M/G$ . Say, for $x_1,x_2\in M$. Clearly, $x_2\ne gx_1$ for any
gEG
. As before we can get neighborhoods, $y\in W_1 = \sqcup g(V_1), y_2\in W_2=\sqcup g(V_2)$ , where $x_1\in V_1, x_2\in V_2$ . Now since is Hausdorff, we can get $x_1\in B_1 \subset V_1, x_2\in B_2 \subset V_2$ such that $B_1\cap B_2 =\emptyset$ . Then the images $q(B_1)\cap q(B_2) =\emptyset$ and they are open, since $q^{-1}(q(B_i)) = \sqcup g(B_i)$ . Thus,
M/G
is Hausdorff.

Hence we have proved, is a manifold.

M/G

Hope this helps. Comment if you need further clarifications. Cheers!