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I want to point out that according to an unpublished note of Fabio Simas, if we have an effective isometric action of $SO(3)$ on a positively curved 6-manifold $M^6$ with only isolated fixed points, then the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.
Best Answer
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.