Proper, smooth action with singular orbit space

Question

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".

Answer 1

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.

Answer 2

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 $$t\cdot(x,y)=(e^tx,e^{-t}y).$$

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.

Answer 3

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 $S^1$.