SO(3) action on (simply connected) 6 manifold with discrete fixed point If a 6-dimensional orientable smooth manifold $M$ admits a smooth effective $SO(3)$ action with discrete fixed point set, what can we say about the topology of $M$? What if we assume that in addition M is a Riemannian manifold with nonnegative/positive sectional curvature, and the group action is isometric?
The question is motivated by Fuquan Fang's paper: Positively curved 6-manifolds with simple symmetry groups, in which he tried to classify all such 6-manifolds. But the author overlooked the possibility of finite isotropy groups, and his proof turned out to have a gap. Since finite isotropy groups occur for the $SO(3)$ action on $SU(3)/\mathbb{T}^2$, I am trying to look at this one particular case that he missed. 
I am also aware that some people had worked on similar things on 4-manifolds with $\mathbb{S}^{1}$ symmetry, e.g. https://arxiv.org/pdf/1703.05464.pdf. Through studying local isotropy representation and applying some signature formula, the author was able to classify the fixed point data of circle actions on 4-manifolds with discrete fixed point when there are few fixed points. 
I am wondering if one can do analogous things on 6-manifolds with $SO(3)$ symmetry and discrete fixed point set. The isotropy representation near an isolated fixed point on $M^6$ must be $\mathbb{R}^{3}\oplus \mathbb{R}^{3}$, on which SO(3) acts diagonally. Unfortunately, this is the only piece of related work I know so far.
I want to point out that according to an unpublished note of Fabio Simas, if we have an effective isometric action of $SO(3)$ on a positively curved 6-manifold $M^6$ with only isolated fixed points, then the number of fixed points is at most 3. This follows from a "q-extent" argument in comparison geometry. I'm really most interested in the case where M carries a metric with positive sectional curvature. In this case, if we assume M admits an $SO(3)$ action with isolated fixed points, then we know the orbit space $M/SO(3)$ is a 3-dim Alexandrov space. I want to find a way of reconstructing the manifold M from the structure of the quotient $M/SO(3)$ and information about stabilizer groups,and information of "gluing maps" which identify the boundaries of different "orbit types" in M. But so far I'm still trying to work it out. 
I have 2 examples for such actions. One is the linear $SO(3)$-action on $\mathbb{S}^6$, induced from the 7-dimensional representation $\mathbb{R}^3\oplus \mathbb{R}^3\oplus \mathbb{R}$, where $SO(3)$ acts diagonally and trivially on the last factor $\mathbb{R}$. This action has 2 isolated fixed points. Another is linear $SO(3)$-action on $\mathbb{CP}^3$, induced from the 4-dimensional complex representation $\mathbb{C}^3\oplus \mathbb{C}$, where $SO(3)$ acts trivially on $\mathbb{C}$. This action has 1 isolated fixed point. I'm wondering if there exists an example with 3 isolated fixed points, but yet still positively curved.
 A: There is a vast amount of examples for smooth actions: $SO(3)$ acts on itself effectively and without fixed points by it's group law. Hence it acts effectively and without fixed points on $M= SO(3) \times N$ where $N$ is any smooth $3$-manifold. So maybe you should specify that the fixed point set is non-empty and discrete for more finiteness. (I would add this as a comment but I am not yet allowed). 
Also, I don't see why the example $M = S^2 \times S^{2} \times S^{2}$ with the diagonal action of $SO(3)$ does not appear in the statement of theorem $C$ of Fang's paper. This action also has no (hence discrete) fixed points and seems to preserve a metric with positive curvature.
A: So I am answering my own question. I have been thinking about this question and other related questions for months, and now I have some results.
For this question, if we have a positively curved 6-manifold $M$ with $SO(3)$-action with discrete fixed point set, then the orbit space $M/SO(3)$ is homeomorphic to a 3-ball. For a priori it is a simply connected compact 3-manifold with boundary, thus it is homeomorphic to 3-ball minus finitely many open disks. But it is also an Alexandrov space with positive curvature: now by the soul theorem, it has at most one boundary component; so it is either a 3-sphere or a 3-ball. Now by the slice representation of $SO(3)$ around the discrete fixed point, the orbit space has boundary and the boundary orbit types mainly consist of $SO(3)/SO(2)$.
Now I need a few extra assumptions. I need to assume that there are no exceptional orbits, or in other case, there are no finite non-trivial stabilizer groups. Under this assumption, I can show that the boundary 2-sphere of the orbit space has only 2 fixed points, and no other orbit types. The argument is as follows:
Take $S^1$-fixed point $M^{S^1}$ of this action, where $S^1$ is any maximal torus of $SO(3)$. The $S^1$-fixed point component $M^{S^1}_0$ above the boundary of the orbit space is a branched double cover of $S^2$, since in each $SO(3)/SO(2)$-orbit, the $SO(2)$-fixed point set is a set of 2 elements. But $M^{S^1}_0$ is also a 2-sphere since it is a totally geodesic submanifold of $M$, thus it also has positive curvature. By Riemann-Hurwicz formula, a branched double cover of $S^2$ over $S^2$ has 2 branched points. Thus we have 2 points on the boundary of the orbit space which are "more singular" than $SO(3)/SO(2)$. These could a priori be fixed points or $SO(3)/O(2)$. But from the representation of $O(2)$, the existence of $SO(3)/O(2)$ orbit will force the existence of finite non-trivial stabilizer group, violating my assumption. Thus the 2 branched points must be 2 fixed points.
Now our picture of the orbit space is clearer. It is a 3-ball, whose principal isotropy group in the interior is trivial, and the boundary orbit types are $SO(3)/SO(2)$ and fixed points. There are 2 fixed points. We have two cases: Case I: interior orbits are all $SO(3)$; Case II: there is one singular orbit $SO(3)/SO(2)$ in the interior of the orbit space. There can't be more singular orbits in the interior, because by the q-extent argument, the total number of interior singular orbits and boundary fixed points is at most 3.
I can solve Case I. In this case, $M^6$ is a suspension of a 5-dim $SO(3)$-space $N^5$ whose orbit space corresponds to the equator disk of $M/SO(3)$. In other words, $N^5/SO(3)$ is a 2-disk with two orbit types $SO(3)$ and $SO(3)/SO(2)$, and all the singular orbits lie on the boundary. According to the "second classification theorem" in Bredon's book , Ch.5, Theorem 6.1 and Corollary 6.2, there are 2 such $SO(3)$-spaces, and they are $S^5$ and $S^2\times S^3$. $M^6$ is suspension of one of these two, but only suspension of $S^5$ is a manifold. Thus $M^6$ is homeomorphic to $S^6$. Case I is done.
For Case II and more general cases where one allows exceptional orbits, I do not know what to do so far. I can not get a homeomorphism classfication in these cases, but still I can say something about the (co)homology groups, using tools like Mayer-Vietoris sequence. 
A: EDIT: As explained in my comment below, this answer does not really address the question, but rather the changed question where we have the (stronger) assumption that all stabilizers are discrete.
Suppose $M$ is closed and orientable. We can then consider the Borel fiber sequence
$$M \to M \times_{SO(3)} ESO(3) \to BSO(3);$$
in case that the given action $SO(3) \curvearrowright M$ has discrete stabilizers, the map $M \times_{SO(3)} ESO(3) \to M/SO(3) = X$ is a rational equivalence. Now $X$ is a finite complex, and $H^{\ast}(BSO(3);\mathbb Q) = \mathbb Q[p_1]$, with $p_1$ the first Pontryagin class in degree $4$. Ananalyzing the Serre spectral sequence of the above fibration then yields 
$b_1(M) = b_4(M)$, $b_2(M) = b_5(M)$ and $b_3(M) = b_0(M)+b_6(M) = 2$, where $b_i(M) = \text{dim}_{\mathbb Q}H^i(M)$ stands for the $i^{\text{th}}$ Betti number of $M$. Using Poincaré duality, we can finally deduce $b_1 = b_2 = b_4 = b_5$. All possible values indeed arise, simply take
$$M = SO(3) \times \#^k(S^1 \times S^2),$$
and let $SO(3)$ act on the first factor by left translation, and trivially on the second factor.
