[Math] Properly discontinuous action on manifold

algebraic-topologydifferential-geometrydifferential-topologygeneral-topologymanifolds

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?

Best Answer

Exactly the same group action (but named little differently) is in the book "Differential Geometry"on page 68 written by Marcel Berger and Bernard Gostiaux. enter image description here

Additonaly on the page 70 there is theorem of yours. It is done well with details. enter image description here

Back to your questions.

  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.
  2. Hard question.
  3. Here we come back to the definition. Your group actually acts "properly discontinuously without fixed points (freely)." This additional condition makes the difference. (see the proof in the book I've mentioned)
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