Under what conditions the quotient space of a manifold is a manifold?

Solution 1:

As I'm sure you know, the category of smooth (topological?) manifolds is one of those categories where the objects are very nice but the category itself is terrible. I cannot describe the number of times I've heard the algebraic geometers curse the smooth category.

I am not certain this is a total classification, but From Lee's Introduction to Smooth Manifolds, Theorem 9.19:

If $\tilde M$ is a connected smooth manifold and $\Gamma$ is a discrete group acting smoothly, freely, and properly on $\tilde M$, then the quotient $\tilde M/\Gamma$ is a topological manifold and has a unique smooth structure such that $\pi: \tilde M \to \tilde M/\Gamma$ is a smooth covering map.

The manifold portion of this comes from the Quotient Manifold Theorem:

If $G$ is a Lie group acting smoothly, freely, and properly on a smooth manifold $M$, then the quotient space $M/G$ is a topological manifold with a unique smooth structure such that the quotient map $M \to M/G$ is a smooth submersion.

And then applying this to the (zero-dimensional) Lie group of deck transformations.

Edit: The proof of the Hausdorff property is very similar to @useruser43208's response, and uses the properness of the action. Take the orbit set $$ \mathcal O = \{ (g\cdot p,p): g \in G, p \in M \} \subseteq M \times M$$ which is closed under the properness assumption. Any two distinct points $\pi(p)$ and $\pi(q)$ in the image of the quotient map $\pi: M \to M/G$ must have arisen from distinct orbits, so $(p,q) \notin \mathcal O$. Hence we may find a product neighbourhood $U_p\times U_q \subseteq M \times M$ of $(p,q)$ disjoint from $\mathcal O$, hence $\pi(U_p)$ and $\pi(U_q)$ are separating open neighbourhoods (since $\pi$ is open).

Solution 2:

My memory is that Bourbaki has a theorem which says: if $E \subseteq M \times M$ is an equivalence relation such that

• $E$ is a closed submanifold of $M \times M$, and

• The first projection $\pi_1: E \to M$ is a submersion,

then the quotient space $M/E$ inherits from $M$ a smooth manifold structure. Still trying to track down the exact reference.

Edit: Since there has been some commentary below this answer, let me provide a reference to the original Bourbaki. It's section 5.9.5 in Bourbaki's Variétés différentielles et analytiques, (Hermann, 1967). The statement is as I have it above, except that I added that $E$ should be closed in $M \times M$, in order for the quotient $M/E$ to come out Hausdorff.

Note also that Bourbaki formulates it in the form of an if and only if statement.

Solution 3:

There is quite a lot on this problem already known. The study of the Hausdorff (usually metrizable) spaces gotten from taking quotients of manifolds is called decomposition space theory and has a wonderful history.

A special case of this which has gathered a lot of attention is when is the quotient of a manifold $M$ actually homeomorphic to $M$. Bing in the 1950's and 1960's studied cases of this problem extensively and showed a certain quotient of $S^3$ is actually homeomorphic to $S^3$ and used such a decomposition to construct involutions of $S^3$ not conjugate to any smooth ones. Brown proved the generalized Schoenflies theorem using similar techniques. In the 80's Freedman used this theory to prove to 4-dimensional Poincare conjecture.

In your case, if the quotient map has only one nontrivial set $K$ being identified to a point and $K$ is compact, then the quotient space is homeomorphic to a manifold if and only if $K$ is a countably infinite intersection of nested closed balls ($K$ is cellular according to the literature). Such a characterization also works perfectly if there are only finitely many nontrivial sets identified to a point.

If there are infinitely many such sets, things become more complicated. Realizing when a "generalized manifold" is a manifold to my knowledge is not completely understood though there are a lot of partial results.

Daverman gives a nice account of most of this story in his book here including some of the mentioned partial results near the end.