1
$\begingroup$

Let $M$ be a smooth manifold and $G$ a Lie group, then if $G$ acts smoothly, freely, and properly on $M$ it is a well known result that the quotient $M/G$ is a smooth manifold. In the context of topological spaces in general, I often hear that quotients by free actions are "well behaved", however, when I think of actions on smooth manifolds whose quotients are ugly it's the properness quality which is lacking. For example, $\mathbb{Z}$ acting on $S^1$ by irrational rotation is free but not proper, here the quotient isn't Hausdorff which is what I would call ill behaved. However, if we consider $S^1$ acting on $S^2$ by rotations about the $z$ axis, then the action is not free (the poles are always fixed), but is proper as $S^1$ is compact. The quotient space is simply the closed interval $[-1,1]$ which is a space I would call well behaved, even if it's not necessarily interesting.

Given this, it seems to me that the more crucial property is that the action is proper. So why do we say that quotients by free actions are in general more "well behaved"?

$\endgroup$
7
  • $\begingroup$ Free priper actions give rise to principal fiber bundles. $\endgroup$
    – Moishe Kohan
    Commented Sep 13, 2023 at 0:06
  • $\begingroup$ @MoisheKohan I understand that, my whole point is that it seems that proper is important than free in terms of "well behavedness" $\endgroup$
    – Chris
    Commented Sep 13, 2023 at 0:14
  • $\begingroup$ True, properness is more important. When you say "we say that the opposite is true" you should give a reference. My guess is that you hear this from people who assume properness by default and compare free to nonfree. $\endgroup$
    – Moishe Kohan
    Commented Sep 13, 2023 at 0:17
  • 1
    $\begingroup$ @MoisheKohan introductory lectures on equivariant cohomology, page 30, end of third paragraph “when the action of G on M is free, the quotient M/G is generally well behaved” $\endgroup$
    – Chris
    Commented Sep 13, 2023 at 1:08
  • 1
    $\begingroup$ You should add a more extensive quote (right now, it lacks context) to the post itself, comments are not meant for this. Incidentally, this is a book by Tu, not Bott and Tu. $\endgroup$
    – Moishe Kohan
    Commented Sep 13, 2023 at 1:11

1 Answer 1

Reset to default
3
$\begingroup$

Here is an answer of sorts, based on what you wrote.

When an author claims that "quotients of smooth manifolds by free (but not proper) actions of a Lie group $G$ are well-behaved," this means one of two things:

(a) The author had an (unstated) assumption that the action is actually proper. In this case, as you know, indeed the quotient space $Q$ has a natural smooth manifold structure and the corresponding group action defines a principal $G$-bundle over $Q$.

(b) The author never seriously thought about free non-proper Lie group actions and the stated sentiment can be safely ignored.

By comparison, non-free proper actions are well-behaved comparing to free non-proper actions. Good examples of this phenomenon can be understood by considering isometric actions of finitely generated groups (equipped with discrete topology) on the hyperbolic 3-space:

(a) Such an action can be free but non-proper and essentially nothing is known about a classification of such actions and associated quotient spaces: It is not even clear what the word "classification" can possibly mean in this setting.

(b) In contrast, there are deep classification theorems about non-free proper actions in this setting. In particular, quotient spaces are well-understood, especially if one assumes that the action is orientation-preserving.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .