6-manifolds admitting SO(3) action with 2 orbit types

Question

Let $M^6$ be a 6-dimensional smooth manifold, on which the group $G=SO(3)$ acts smoothly with 2 orbit types $SO(3)/SO(2)$ and $SO(3)$, such that the orbit space $X=M/SO(3)$ is a 3-ball $B^3$, whose boundary 2-sphere corresponds to the singular isotropy group (stabilizer) $K=SO(2)$ and interior corresponds to the principal isotropy group $H=id$. The problem is to classify all such 6 manifolds admitting such actions.

In Bredon's book "Introduction to Compact Tranformation Groups", he has a way to classify group actions with 2 orbit types, which was initially due to Janich. According to Corollary 6.2 on page 257, Chapter V.6 of the book, the class of such 6-manifolds is in bijection with the set $[S^2,N(H)/(N(H)\cap N(K))]/\pi_0(N(H)/H)=\pi_2(N(H)/(N(H)\cap N(K)))/\pi_0(N(H)/H)=\pi_2(\mathbb{RP^2})=\mathbb{Z}.$

Here $K=SO(2),\ H=id$ are the isotropy groups (stabilizers) given in the first paragraph, and $N(H)=id,\ N(K)=O(2)$ are their normalizers in $G=SO(3)$. Thus such actions are parametrized by integers.

But Bredon's construction in the book is rather involved. I'm aiming at simple explicit description of this family of $SO(3)$-actions. Moreover, I'm wondering which of those manifolds are simply connected. I'm mainly interested in simply connected examples.

Two more remarks:

Remark 1. If we ask the orbit space to be a 2-disk instead of 3-ball, then in Bredon's book he gave explicit classification of such $G$-spaces. More precisely, all $O(n)$-spaces with isotropy groups $O(n-1)$ or $O(n-2)$ and orbit space $D^2$ are given by $\Sigma_k^{2n-1}=S^{n-1}\times D^n \cup_{\varphi^k} S^{n-1}\times D^n$, where $\varphi$ is an $O(n)$-equivariant gluing map. For details see Chapter I.7 and V.6 of Bredon's book "Introduction to Compact Tranformation Groups". Those spaces have alternative description: the Brieskorn varieties $B^{2n-1}_k=\{z_0^k+z_1^2+\cdots z_n^2=0\}\cap S^{2n+1}$, see Chapter V.9 of the book. My goal is to find such explicit description in the case when the orbit space is a 3-ball and the group $G=SO(3)$.

Remark 2. I have one example of such actions. Consider the 4-dimensional complex representation $\mathbb{C}^2\oplus \mathbb{C}^2$ of $SU(2)$, take its projectivization, then we get a linear $SU(2)$-action on $\mathbb{CP}^3$, which is ineffective (not faithful) and descends to an effective $SO(3)$-action. This action satisfies the requirement. And my intuition tells me that this action corresponds to the integer parameter 0. But I don't know how to show it.

I'd like to point out that this question is related to another question I asked: SO(3) action on (simply connected) 6 manifold with discrete fixed point. They are both about $SO(3)$-actions on 6-manifolds, but under different conditions.

Update: I suspect they are $S^2$ bundles over $S^4$, as I construct such actions on $\mathbb{CP}^3$ and $S^2\times S^4$.

Answer 1

Disclaimer: I have not checked Bredon's book. I am trying to describe $M$ as explicitly as I can, but maybe I am simply reproducing Bredon's description.

Let $N$ denote the union of singular orbits. From your description, $N$ is an $S^2$-bundle over $S^2$. Let $\nu\to N$ denote the normal bundle in $M$. The group $SO(3)$ acts simply transitively on the fibres of the unit normal bundle, which therefore forms an $SO(3)$ principal bundle $P\to S^2$.

On the other hand, the regular orbits form an $SO(3)$-principal bundle over the open ball, which is necessarily trivial. Because both are glued together, we conclude that $P$ is trivial, too. We thus have to glue two trivial $SO(3)$-principal bundles over $S^2$. Two different gluing maps differ by a map from $S^2$ to $SO(3)$. Because $\pi_2(SO(3))=\pi_2(\mathbb RP^3)=0$, there is only one choice.

The only way to produce different manifolds comes from the fact that $SO(3)$ contains an $\mathbb RP^2$ worth of one-parameter subgroups, which are all isomorphic to $SO(2)$. The reason is that each unit vector $X$ in the Lie algebra $\mathfrak{so}(3)$ generates one of these groups, and $\pm X$ generate the same subgroup. Hence let $f\colon S^2\to\{\,H\subset SO(3)\mid H$ is a subgroup, $H\cong SO(2)\,\}\cong\mathbb RP^2$ be a map, then we construct the unit disc bundle in the normal bundle over $N$ by dividing the group $f(p)$ out of $P_p=SO(3)$ for each $p\in S^2$. Because $\pi_2(\mathbb RP^2)\cong\mathbb Z$, we get $\mathbb Z$-many homotopy classes of such quotients. We should still show that different homotopy classes lead to different manifolds.

The following construction seems to be equivalent and maybe easier to digest. It should give at least half of the manifolds. There are only two $S^2$-bundles over $S^2$ because $\pi_2(BSO(3))\cong\mathbb Z/2$. So I assume that $N=S^2\times S^2\to S^2$ is the trivial bundle. The normal bundle $\nu\to N$ in $M$ can be regarded as a complex line bundle over $N$. It is determined by $c_1(\nu)\in H^2(N)\cong\mathbb Z^2$. Over each fibre $S^2$, it has degree $2$. On the other hand, its degree $d$ over the base $S^2$ has still to be chosen. Again, this would give rise to $\mathbb Z$-many manifolds. Here, any two of them differ as $SO(3)$-manifolds because $d$ can be read off from the way that $SO(3)$ acts.

Let $D\nu\to N$ denote its unit disk bundle. We should check that $\partial D\nu$ is a trivial $SO(3)$-principal bundle, so that we can glue it to $D^3\times SO(3)$. It obviously is a bundle with fibre $\mathbb R P^3$, and therefore classified by a map $S^2\to BSO(4)$. But $\pi_2(BSO(4))=\mathbb Z/2$, so for even $d$ we obtain a trivial bundle.

Finally, all of these manifolds are simply connected by Seifert-van Kampen. The normal disc bundle $D\nu$ has an $S^2$-bundle over $S^2$ as a deformation retract, which is simply connected. The fundamental group of $D^3\times SO(3)$ comes from $S^2\times SO(3)$, so it vanishes under gluing.