Is M/SO(3) a manifold when the orbits are spheres?

Question

$M$ is a smooth 4-manifold equipped with an $SO(3)$-Lie group action, such that each orbit is diffeomorphic to the sphere $\mathbb{S}^2$. Prove or disprove that $M/SO(3)$ is a smooth manifold.

Due to $SO(3)$, the action has a large fixed point set. So the usual Quotient Manifold Theorem doesn't apply. (I'm trying to prove the Penrose Diagram of a spherically symmetric space-time is a manifold. Any reference suggestions?)

Answer 1

Yes, it will be a manifold.

The most general result I know along these lines is given by the following:

Theorem: Suppose $G$ is a compact Lie group acting smoothly on a manifold such that all isotropy groups are conjugate. Then there is a unique smooth structure on the orbit space $M/G$ for which the canonical projection $M\rightarrow M/G$ is a submersion.

Notice that in the case of a free action, all the isotropy groups are trivial, and thus conjugate. So, this result has the more well-known free case as a corollary. I should also add that one can probably replace "$G$ is compact" with "$G$ acts properly." I never think about non-compact Lie group actions, so I'm not certain of this, though.

The idea of the proof is as follows:

1. By the Slice Theorem, small enough neighborhoods around the orbit have the form $(G\times B)/G_p$ where $B\subseteq T_p M$ is a ball of dimension $\dim M - \dim (G\cdot p)$, and where the isotropy group $G_p$ acts linearly on $B$.

2. In particular, in $M/G$, one has a neighborhood basis of the form $B/G_p$ around the image of $p$.

3. If the isotropy group $G_p$ is minimal, in the sense that it is conjugate to a subgroup of all other isotropy groups, then the $G_p$ action on $B$ is trivial. (The idea of the proof: If $g\in G_p$ moves some non-zero point $q\in B$, then the isotropy group $G_q$ is too small to contain $G_p$, contradicting minimality.)

4. If all isotropy groups are conjugate, then they are all minimial in the above sense.

Armed with these one easily obtains a topological manifold structure on $M/G$, and with a bit more work, one can see it is smooth.

In your specific case, we still need to check that all isotropy groups are conjugate. This is where the assumption that all orbits are $S^2$ enters the picture.

Proposition: Suppose $SO(3)$ acts on $S^2$ smoothly and transitively. Then the isotropy group at any point is conjugate to the usual $SO(2)\subseteq SO(3)$.

Proof: By counting dimensions, the isotropy groups must all be one dimensional and compact, and hence, a union of circles. If the isotropy group is disconnected, it is easy to see that $G/G_p$ has non-trivial fundamental group, so is not $S^2$. Thus, the isotropy groups are all circles. Thus, each isotropy group is a maximal torus of $SO(3)$, so they are all conjugate. $\square$

Lastly, I wanted to know that for other group actions, merely having all orbits diffeomorphic is not sufficient. For example, viewing the Klein bottle as a connect sum of two copies of $\mathbb{R}P^2$, one sees there is a circle action on $K$ where all orbits are diffeomorphic to $S^1$. Yet the quotient is an interval $[0,1]$, so is not a smooth manifold.