If $A : G \times M \rightarrow M$ is a proper action and that $M$ is a metric space, the quotient space $M/G$ is Hausdorff

Question

Def 1: Let $f : X \rightarrow Y$ be a continuous map. $X,Y$ Topological spaces. $f$ is called proper if $f^{-1}(K)$ is compact for every compact $K \subseteq Y$.

Def 2: G : topological group. A continuous action $A: G \times M \rightarrow M$ is called a proper action if the following map is proper: $(A; \text{id}) : G \times M \rightarrow M \times M: (g; m) \mapsto (A(g, m), m) $

Suppose that $A : G \times M \rightarrow M$ is a proper action and that $M$ is a metric space. Prove that the quotient space M/G is Hausdorff

I have been trying for a while now. I've seen some proofs where G was Hausdorff I don't have that here in this case.To prove $M/G$ is Hausdorff I would take distinct points $a, b$ in $M/G$ and tried to find 2 disjoint open sets containing each point separatelly. But this approach has not led me anywhere.

On another hand $U$ is open in $M/G \iff \pi^{-1}(U)$ is open in $M$ where $\pi$ is the canonical projection

How should I proceed?

Note: The followings fact that I have already proven might be useful

-If $Y$ is a metric space and $f: X \rightarrow Y$ is continuous and proper, $f$ is closed

-If G is compact and M is Hausdorff. Every continuous action of G on M is proper

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Edit: as suggested in the comments the following is the proof of the part that interests me: from ( https://mathoverflow.net/questions/55726/properly-discontinuous-action/56490?_gl=1*nkzp9s*_ga*NzY0NDcyMDM3LjE1Njk3NjA3NDU.*_ga_S812YQPLT2*MTY5Nzk5NjE1OC41NDkuMS4xNjk4MDA4NzQyLjAuMC4w#56490 )

Definition. Let G be a topological group acting continuously on a topological space X. The action is called proper if the map ρ:G×X→X×X given by (g,x)↦(x,gx) is proper.

Proposition. If G acts properly on X then X/G is Hausdorff. In particular, each orbit Gx is closed. The stabilizer Gx of each point is compact and the map G/Gx→Gx is a homeomorphism. Moreover, if G is Hausdorff then so is X

Proof. Indeed, the orbit equivalence relation is the image of ρ, hence it is closed. Since the projection X→X/G is open, this implies that X/G is Hausdorff.(...)

However is too succinct to me, so I update the question to:

Could you please explain the details? or link to proofs of these facts? I think they are properties that I don't know

1 Why is the image of $\rho$ closed?

2 I know by a theorem that the projection is open so everything good there, but why does that imply that X/G is Hausdorff?

Answer 1

Your question revolves around the properties of the quotient space ( M/G ) given a proper action of a topological group ( G ) on a metric space ( M ). You're right in seeking to leverage properties that arise from a proper action. Let's tackle the concerns step by step.

1. Closedness of the orbit relation's image:

To understand why the image of ( \rho ) is closed, consider that the map ( \rho: G \times M \rightarrow M \times M ) sends ( (g, m) ) to ( (A(g, m), m) ). Now, since ( M \times M ) is a Hausdorff space and the action is proper, the diagonal ( \Delta = {(m, m) | m \in M} ) is closed. Hence, if ( (x, y) ) and ( (x', y') ) are in distinct orbits, there exist disjoint open neighborhoods around them.

Since the orbits of points under the action of ( G ) are equivalence classes, the orbit relation is an equivalence relation. This implies that the relation is both reflexive and symmetric. Combining this with the fact that the map is proper, it means that the image of the orbit relation is closed.

2. Openness of the projection & Hausdorffness of ( M/G ):

Your intuition is correct regarding the openness of the projection and how it can be leveraged. Consider two distinct points ( x ) and ( y ) in ( M ) that aren't in the same ( G )-orbit. Their image in ( M/G ) will be distinct equivalence classes. Given the closedness of the orbit relation and the Hausdorff property of ( M \times M ), you can find disjoint open sets in ( M \times M ) separating the points from the diagonal. Now, the openness of the projection means these open sets can be "transferred" to the quotient to separate the equivalence classes.

So, given distinct equivalence classes in ( M/G ), you can always find disjoint open sets containing them, verifying the Hausdorff property for the quotient space.

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Your attempt to use properties of proper actions is definitely on the right track. It's worth diving deeper into the interplay between group actions and topology to get a more intuitive understanding. I hope this breakdown clarifies things for you!