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I want to point out that according to an unpublished note of Fabio Simas, if we have an effective isometric action of $SO(3)$ on a positively curved 6-manifold $M^6$ with only isolated fixed points, then the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

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    $\begingroup$ Tangentially related, Fabio Simas's thesis classifies $5$-manifolds admitting metrics of non-negative sectional curvature invariant under a non-trivial $SU(2)$ or $SO(3)$ action. See here: arxiv.org/pdf/1212.5022.pdf $\endgroup$ – Jason DeVito Apr 15 '17 at 0:39
  • $\begingroup$ Thank you. I've seen Fabio's thesis. It seems he classifies SU(2) or SO(3) actions on 5-manifolds topologically first, and then imposes the curvature condition. I don't know if the same scheme carries over to 6-manifolds. $\endgroup$ – Yuhang Liu Apr 15 '17 at 13:55

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.

  • $\begingroup$ Could you clarify your example, are you acting on $S^{3}$ through the standard embedding $SO(3) \rightarrow SO(4)$? In this case the action on $S^{3}$ has two fixed points, hence the action on $M$ fixes $2$ copies of $\#^{k} (S^{1} \times S^{2})$, so this action does not have discrete fixed points. $\endgroup$ – Nick L Apr 14 '17 at 16:59
  • $\begingroup$ I guess you make a point here. My example works for $SU(2)$ instead of $SO(3)$. If you really want $SO(3)$, it is maybe better to replace $S^3$ by $\mathbb RP^3 \cong SO(3)$ which has the same rational cohomology as $S^3$, on which $SO(3)$ acts by left translation. $\endgroup$ – Jens Reinhold Apr 14 '17 at 17:29
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    $\begingroup$ I'm confused since there are examples with $b_{3}(M) = 0$. The standard action of $SO(3)$ on $S^{2}$ is effective and fixed point free. Hence there is an effective action of $SO(3)$ on $S^{2} \times S^{4}$ without fixed points, satisfying $b_{3}(M) = 0$ and indeed $0 = b_{1} \neq b_{2} = 1$. $\endgroup$ – Nick L Apr 14 '17 at 18:04
  • $\begingroup$ Okay, I see. I actually misread the question, instead of "discrete fixed point set" I read "discrete stabilizers", i.e. I assumed all orbits would be $H\backslash {SO(3)}$ with $H$ a finite group. In this case I think my proof works. I do not know what happens when we only know that the fixed points form a discrete set. $\endgroup$ – Jens Reinhold Apr 14 '17 at 18:07
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    $\begingroup$ Thanks. Discrete stabilizer would also be an interesting case, I think some people call it an "almost-free" action, in which case the quotient space would be an orbifold (which is good). I'm wondering if in this case one can say anything more if assuming positive sectional curvature. (Yeah the whole point is about "positive curvature".. more about geometry) $\endgroup$ – Yuhang Liu Apr 14 '17 at 18:24

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.

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    $\begingroup$ Does S^2*S^2*S^2 admit positive sectional curvature? That's basically the Hopf conjecture.. Positivity of curvature puts strong restriction on topology. $\endgroup$ – Yuhang Liu Apr 14 '17 at 18:09
  • $\begingroup$ Ahh this is more subtle than I thought. I do not know the answer to your question. $\endgroup$ – Nick L Apr 14 '17 at 18:14
  • $\begingroup$ I would add that this action preseves a metric with positive definite Ricci curvature (Since $S^{2} \times S^{2} \times S^{2}$ is a Fano 3-fold), which is still weaker than what you ask. $\endgroup$ – Nick L Apr 14 '17 at 19:59
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    $\begingroup$ Yes. Taking product preserves positive Ricci curvature but not positive sectional curvature. $S^3\times S^3$ would also be an example. It seems somehow Fang could rule out those product manifolds(assuming SU(2) or SO(3) symmetry), leaving $S^2\times S^4$ as a possible exception. However I don't quite understand his arguments.. $\endgroup$ – Yuhang Liu Apr 15 '17 at 14:02

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.


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