# Properly discontinuous action on manifold

I am actually not familiar with topology, but since we had a short outlook on these things in our differential geometry lecture today, I would appreciate some general remarks:

Let $M$ be a smooth manifold and $\Gamma$ a group of diffeomorphisms on $M$, then

$\Gamma$ is called properly discontinuous, if 1.) and 2.) hold.

$1.) \forall p \in M \exists p\in U$ (open nbh.)$\forall \phi \in \Gamma \backslash\{id\}: \phi(U) \cap U = \emptyset.$

This seems to tell me that there is basically no fixed point under such a non-trivial diffeomorphism (even more, we get that we can separate the image by an open nbh.)

$2.) \forall p,q \in M,$where $q$ is not in the orbit of $p$ there are open nbhs $U(p),U(q)$ such that $\phi(U(p)) \cap U(q) = \emptyset$ for all $\phi \in \Gamma.$

First question: I feel that the second property is often skipped in textbooks, but don't know why?! To me, it seems to be similar to a Hausdorff property in $M/\Gamma.$

Now, we showed that such a group $\Gamma$ implies that $M/\Gamma$ is again a manifold and the identity $id: M \rightarrow M/\Gamma$ is a covering map. In this sense $\Gamma$ can be regarded as the group of deck transformations. If $M$ is simply connected, then this group can be also considered as the fundamental group of $M / \Gamma.$

Second question: If I understand this correctly, then the converse also holds. $M/\Gamma$ is a manifold, only if $\Gamma$ acts properly discontinuous on $M$?

Third question: In this thread Lee Mosher argues that properly discontinuous is not sufficient to conclude that $M \rightarrow M/\Gamma$ is a covering map (see his comments). Is this also true for my definition? Actually I don't understand why free action is not a corollary of my first part of the definition, cause it just means that the only group element that is allowed to have fixed points is the identity?

• Did you check out this math.OF thread? I believe property 1 corresponds to Type D there. Jun 10 '15 at 18:13
• See my answer to this question. Jun 10 '15 at 19:08
• @JackLee thanks that clarifies it, I bet you also know the answer to question 2.) Jun 10 '15 at 19:19
• @JackLee could you please give me also a hint regarding the second question? Jun 11 '15 at 7:22
• The answer to your second question, as you posed it, is no. For example, the action of $\mathbb R$ on $\mathbb R^2$ by vertical translations has a quotient diffeomorphic to $\mathbb R$, but it's not properly discontinuous. But if you assume it's a covering space action (meaning that it satisfies your condition 1), then the quotient is a manifold iff the action satisfies condition 2. Jun 11 '15 at 14:36

1. You are perfectly right. In the proof of this theorem condition 2) starts to work, when we want to show that $M/\Gamma$ is Hausdorff.