# Manifolds and actions: a counterexample

Let $X$ be an $m$-dimensional simply connected manifold that is also a G-space. If the action is properly discontinuous, is $X/G$ necessarily a manifold?

Do you have a hint for this one?

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If you don't suppose that the action is free then $X/G$ may not be a manifold. For example, $\mathbb{R}/\{\pm 1\}=\mathbb{R}_{\geq0}$ (action by multiplication). To get something that is not even a manifold with boundary, take e.g. $Y=\mathbb{R^3}/\{\pm 1\}$ (to prove it, notice e.g. that the relative homology $H_2(Y,Y-\{0\})\cong H_1(\mathbb{RP}^2)$ is non-trivial, or use the local fundamental group of $Y-{0}$).

(if the action is free and properly discontinuous then $X/G$ is certainly a manifold)

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Here is a precise result:

Theorem
Let a Lie group $G$ act smoothly, freely and properly on the Hausdorff manifold $X$. Then the quotient topological space $X/G$ is Hausdorff and admits of a differential structure such that the quotient map $X\to X/G$ is a smooth submersion.

Remarks
1) Simple connectedness of $X$ is irrelevant.

2) This is quite a difficult theorem very well treated in Chapter 9 of John M. Lee's book.

3) If $G$ is compact, properness of a smooth free action is automatic.

4) The smooth map $\pi: X\to X/G$ is a categorical quotient in the sense that, for any differential manifold $Y$, composition with $\pi$ yields a bijection $$\mathcal C^\infty (X/G,Y)\stackrel {\cong}{\to} \mathcal C^\infty_G (X,Y)$$
where $\mathcal C^\infty_G (X,Y)$ denotes $G$-equivariant smooth maps.

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