quotient by a group that acts almost freely

Question

How can I show that if a compact lie group G acts almost freely and smoothly on a manifold M, then M/G is Hausdorff? (an action is almost free if $G_x$ is finite for all x $\in$ M)

Answer 1

I think I can prove...

Suppose a compact Lie group $G$ acts smoothly on a manifold $M$. Them $M/G$ is Hausdorff.

(So, no assumption of almost freeness).

The key ingredient is the Slice Theorem:

Suppose a compact Lie group $G$ acts smoothly on smooth manifold $M$. Then for every $x\in M$, there is a "nice" slice $S$ at $x$.

By definition, a slice $S$ is a subset of $M$ for which the following conditions hold:

1. $G_x \cdot S \subseteq S$

2. $G\cdot S$ is open in $M$.

3. $S$ is closed in $G\cdot S$ and $gS\cap S\neq \emptyset$ iff $g\in G_x$.

In this context, "nice" means one can find an embedding of the normal bundle of $G\cdot x$ into $M$ for which the fibers are slices.

Since the projection $\pi:M\rightarrow M/G$ is open (since we're quotienting by a group action), and since $G\cdot S$ is open in $M$, this gives a way of thinking about open subsets of $M/G$: every $\pi(G\cdot S) = \pi(S)$ is open in $M/G$.

Now, suppose for a moment, that given $x$ and $y$ in $M$ which are not in the same orbit (i.e., $\pi(x)\neq \pi(y)$), we can find slices $S_x$ and $S_y$ for which $G\cdot S_x \cap G\cdot S_y = \emptyset$. Then it easily follows that $\pi(S_x)\cap \pi(S_y) = \emptyset.$ Thus, we would know $M/G$ is Hausdorff.

So, given $x$ and $y$ which are not in the same orbit, how do we find slices whose orbits are disjoint? Well, let's pick a background Riemannian metric. By average the metric over the group action, we may assume without loss of generality that the metric is $G$-invariant.

Now, the orbits $G\cdot x$ and $G\cdot y$ are compact subsets of $M$, because they are the image of the compact set $G\times \{x\}$ under the group action map. They are disjoint by assumption, and hence $d(G\cdot x, G\cdot y) := r > 0$.

Now, given $S_x$, let $T_x\subseteq S_x$ be the set of all elements in $S_x$ a distance at most $r/2$ from $x$. I claim that $T_x$ is a slice. Believing this for a moment, the triangle inequality proves that $G\cdot T_x \cap G\cdot T_y = \emptyset.$ For if $p \in G\cdot T_x$, then $d(p,G\cdot x) < r/2$. Likewise, $d(p,G\cdot y) < r/2$. But then, \begin{align*} r &= d(G\cdot x, G\cdot y)\\ &\leq d(G\cdot x, p) + d(p,G\cdot y)\\ &< r/2 + r/2,\end{align*} contradicting the fact that $r > 0$.

Now, let's see why $T=T_x$ is a slice.

For 1, if $g\in G_x$ and $p\in T$, then $d(gp,x) = d(gp,gx) = d(p,x) < r/2$ since the action is isometric. This proves $G\cdot T \subseteq T$.

For 2, $G\cdot T$ is open because it's clearly the image of an open subset of the normal bundle, which is embedded into $M$.

For 3, We first show $gT\cap T \neq emptyset $ iff $g\in G_x$. If $g\in G_x$, then $gx = x \in gT\cap T$. Conversely, if $gT\cap T\neq \emptyset$, then $gS\cap S\neq \emptyset $ since $gT\subseteq gS$ and $T\subseteq S$. Thus, since $S$ is a slice, $g\in G_x$.

We now show $T$ is closed in $G\cdot T$. If $y_n$ is a sequence converging to $y\in G\cdot T\subseteq G\cdot S\subseteq M$ with $y_n\in T$ for all $n$, then $y_n$ is a sequence in $S$. Since $S$ is closed in $G\cdot S$, this implies $y\in S$. Now, $y\in G\cdot T$ is of the form $y= gt$ for some $g\in G$ and $t\in T$. Since $t\in S$ and $gt = y \in S$, $S\cap gS \neq \emptyset$, so $g\in G_x$ since $S$ is a slice. But we showed in 1 that $G\cdot T\subseteq T$, so this shows $y = gt\in T$.