Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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)

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.