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This is a problem from Lee, Smooth Manifolds (Problem 9.4). It's not an homework problem, I'm just systematically solving every problem of that book, and I got stuck on this one. Usually I try not to ask for help unless I really don't have a clue, but I suspect that this problem is really simple, and it's getting on my nerves.

"Give an example of smooth, proper action of a Lie group on a smooth manifold such that the orbit space is not a topological manifold"

Since the action is assumed to be proper, the orbit space is indeed Hausdorff (and also second countable, of course). Then, the only thing that can go wrong is the orbit space not being locally Euclidean.

Therefore, I was trying to devise smooth action of some group on $\mathbb{R}^2$, whose orbit space is the "cross-shaped" set.

However, I can't figure out any concrete example. In particular, in the examples I've figured out so far, the action turns out to be not proper. A compact group would do the trick.

Perhaps, since there may be some confusion, it's better to provide my (Lee's) definition of proper action. Let $G$ a Lie Group acting on a smooth manifold $M$. The action is proper if the map $G\times M \to M\times M$ given by $(g,p)\to (p,g\cdot p)$ is a proper map. Equivalently, the action is proper iff the following condition is satisfied: "If $\{p_i\}$ is a convergent sequence in $M$ and $\{g_i\}$ is a sequence in $G$ such that $\{g_i \cdot p_i\}$ converges, then a subsequence of $\{g_i\}$ converges".

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Hint: You want an action that isn't free. –  t.b. Oct 2 '11 at 14:51
    
Indeed, that's the point, I'm looking for actions with non trivial stabilizers at some points. Otherwise, by Theorem 9.16 of Lee's "Introduction to Smooth Manifolds", the orbit space would be a smooth manifold. –  Raziel Oct 2 '11 at 14:53
    
For instance $S^1 = SO(2)$ acts on $\mathbb{R}^2$ in an obvious way. By the way: there is a small typo in the addition, that should be $(g,p) \mapsto (p,g\cdotp)$. –  t.b. Oct 2 '11 at 14:54
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I do not include "with boundary" in my definition of topological manifold. –  t.b. Oct 2 '11 at 15:00
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I am pretty sure that Lee wanted to exclude manifolds with boundary from topological-manifoldhood in that exercise. –  Mariano Suárez-Alvarez Oct 2 '11 at 16:54

3 Answers 3

up vote 5 down vote accepted

Here's another example.

Consider $\mathbb{R}P^3$ in the model of a 3-ball with antipodal boundary points identified.

Consider the $G = \mathbb{Z}/2\mathbb{Z}$ action on $\mathbb{R}P^3$ given by the antipodal map sending $(x,y,z)$ to $-(x,y,z)$.

I claim that $\mathbb{R}P^3/G$ is homeomorphic to a cone on $\mathbb{R}P^2$. To see this, use coordinates on $C\mathbb{R}P^2$ given by $([x,y,z],t)$ where we think of $(x,y,z)\in\mathbb{R}^3$ and we're collapsing $\mathbb{R}P^2\times\{0\}$ to a point. Now,map $(x,y,z)$ in $\mathbb{R}P^3$ to $\big([x,y,z], (x^2+y^2+z^2)\big)$ (and map the origin to the cone point). This is clearly continuous away from the origin. It's not too hard to see that it's continuous at the origin as well.

It's also not hard to see that this descends to a bijective map from $\mathbb{R}P^3/G$ to $C\mathbb{R}P^2$, which is therefore a continuous bijection between compact Hausdorff spaces, so is itself a homeomorphism.

Finally, notice that $C\mathbb{R}P^2$ is not a topological manifold (with or without boundary) because of the cone point $p$. A neighborhood $U$ of the cone point $p$ has, by excision, $H_k(U, U-p)\cong H_k(C\mathbb{R}P^2, C\mathbb{R}P^2-p) = H_{k-1}(\mathbb{R}P^2)$, which means $p$ can be neither a manifold point nor a manifold-with-boundary boundary point.

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I forgot to argue why this action is proper. A compact subset of $\mathbb{R}P^3\times\mathbb{R}P^3$ is closed. Since the action is continuous, the inverse image is a closed subset of $\mathbb{Z}/2\mathbb{Z}\times\mathbb{R}P^3$. But this space is compact, so every closed subset of it is compact. –  Jason DeVito Oct 4 '11 at 0:50
    
That's a nice example. It seems to me that the kind of singularities you can create by a proper action are quite generically of this type, at least in all the examples I could come up with. I could prove this to some extent (under some additional assumptions), but the arguments got too involved to post here. By the way, every action of a compact group is proper. –  t.b. Oct 4 '11 at 1:06
    
@t.b. I only ever work with compact stuff, so I always forget how weak you can make things and still get a true statement. Thanks for the heads up. –  Jason DeVito Oct 4 '11 at 1:08

N.B. This example does not work: it is not really proper, as observed by Theo and Zev in the comments!

$\newcommand\RR{\mathbb R}$Consider the action of $\RR$ on $\RR^2\setminus\{0\}$ such that $$t\cdot(x,y)=(e^tx,e^{-t}y).$$

A saddle point

More generally, I think that if $X$ is a smooth vector field on $\RR^2$ and we set $Z=\{p\in\RR^2:X(p)=0\}$, then the action of $\RR$ on $\RR^2\setminus Z$ by the flow of $X$ is proper if there are no closed orbits. Each saddle point in the field will give problems in the quotient.

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I must be missing something. The quotient space doesn't look Hausdorff to me, so the action doesn't seem to be proper. Your conditions seem to guarantee that the action is free. –  t.b. Oct 2 '11 at 17:58
    
@Mariano, am I correct that the orbit space of this action not a topological manifold because it's not Hausdorff, e.g. $\overline{(0,1)}$ and $\overline{(1,0)}$ can't be separated? –  Zev Chonoles Oct 2 '11 at 18:03
    
@t.b.: I was wondering that too, for example, taking the (compact) annulus $$K=\{(x,y)\in\mathbb{R}^2\mid 1\leq x^2+y^2\leq 2\}$$ then I think $$\{t\in\mathbb{R}\mid tK\cap K\neq\varnothing\}=\mathbb{R} \text{ (which is not compact)},$$ right? But there are so many different notions of proper, I can't keep straight which is equivalent to which. –  Zev Chonoles Oct 2 '11 at 18:05
    
@Zev: Yes, I agree with that. I tried to clarify the dependence between the definitions a bit in this thread on MO, but it is a bit dense. I learned the material from J.L. Koszul's Lectures on groups of transformations which was recently re-typed (I loved the original typoscript, but haven't really looked at the new version). Palais's original work is very tough for me to read. –  t.b. Oct 2 '11 at 18:10
    
Ah. Of course, you are all right. I did not check correctly for properness! –  Mariano Suárez-Alvarez Oct 2 '11 at 18:13

For a really simple example you can just take $M = S^1\times 0 \cup S^1 \times 1$ the disjoint union of two copies of $S^1$ and let the action of $G=S^1$ be given by

$$g\cdot (p,\epsilon) = \begin{cases} (gp,0) & (\epsilon = 0) \\ (p,1) & (\epsilon = 1) \end{cases} $$

Then the quotient is the disjoint union of a point and $S^1$.

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This probably doesn't count, as the connected components are manifolds :) –  t.b. Oct 2 '11 at 21:15
    
=) Yeah, it's a bit of a cheat. –  Sam Oct 2 '11 at 21:49
    
This indeed does not count. The orbit space is a non-connected manifold, but still a manifold. –  Raziel Oct 3 '11 at 7:32
    
The orbit space is certainly not a manifold by Lee's definition (he doesn't even mention manifolds which do not have the same dimension everywhere, but instead only talks about $n$-manifolds for given $n$.). In any case, it definitely is a counterexample to the quotient manifold theorem (dropping the assumption that $G$ acts freely). –  Sam Oct 3 '11 at 8:48
    
Btw: I don't consider this counterexample to be a good one by any means. It misses the point of your question, of course. Nevertheless, I feel that it is an acceptable counterexample for the exercise and shows at least one thing that might go wrong in an attempt of a generalization of the theorem. –  Sam Oct 3 '11 at 9:30

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