I have a question in mind for some time. That is, does there exist a smooth circle action on a closed manifold $M^{4n}$ ($n\geq 2$) with exactly three fixed points? Remarks:(1) For n=1, the examples are obvious (standard linear circle action on $CP^2$). (2) If a manifold admits a circle action with 3 fixed points, then the signature of this manifold is 1, so the dimension of this manifold is necessarily divisible by 4.
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1$\begingroup$ Is this known for $Z_p$ actions with prime $p$? I can show using rather elementary tools that a $Z_p$ action can not have exactly one fixed point (I'm sure this must be very well-known) but I don't know about 3. Another obvious observation is that an $S^1$-action like this can not be semi-free for $n\ge 2$. $\endgroup$– Vitali KapovitchCommented Nov 7, 2011 at 19:23
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$\begingroup$ Thanks very much for your remark. In fact, any circle or Z_P action on a manifold cannot have exactly one fixed point, which is quite well-known. If they have two fixed points, then the representations on the tangent spaces are isomorphic and this case have examples in each dimension (rotation on S^{2n}). So the next case should be with three fixed points. As far as I know, there are known examples in dimensions other than 4. It can also be shown that if a manifold admits a semi-free circle action with isolated fixed points, then the number of this action must be even (Bott residue formula). $\endgroup$– PingCommented Nov 7, 2011 at 23:57
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$\begingroup$ Sorry "there are known examples in dimensions other than 4" should be "there are no known examples in dimensions other than 4" $\endgroup$– PingCommented Nov 7, 2011 at 23:58
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$\begingroup$ In fact, I showed in mathjournals.org/mrl/2011-018-003/2011-018-003-005.pdf that if a manifold admits a semi-free circle action with isolated fixed points, then the manifolds bound, i.e., the Pontrjagin numbers and Stiefiel-Whitney numbers of the manifold vanish (Theorem 1.6). $\endgroup$– PingCommented Nov 8, 2011 at 0:04
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$\begingroup$ Thanks, this is a nice theorem. I had something much easier in mind - if the action of $S^1$ is semi-free and the number of fixed points is odd then you can get rid of them in pairs by cutting out little disks around them and gluing together the boundary $S^{4n−1}$'s preserving the $S^1$ action (this is possible since the action is semi-free). In the end you'll end up with a semifree $S^1$ action on some manifold with a single isolated fixed point. the manifold you get might be non-orientable but that doesn't matter when you look at the semi-free $Z_2$ action. $\endgroup$– Vitali KapovitchCommented Nov 8, 2011 at 4:40
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2 Answers
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I think the answer to your question is yes, but perhaps you meant to ask a slightly different question than the precise one you pose.
For example, take the model of $\mathbb RP^{2n}$ where you view it as the unit ball in $\mathbb C^n$ modulo the antipodal map on the boundary.
So the action of $S^1$ on $\mathbb C^n$ factors to an action on $\mathbb RP^{2n}$, and it has precisely one fixed point.
Take the disjoint union of three of these $S^1$ spaces. By design, this has three fixed points.
Perhaps you wanted the manifold to be connected? So do an equivariant surgery, pairwise along a free orbit, drilling out $D^{2n-1} \times \{0,1\} \times S^1$ and gluing in an $S^{2n-2} \times [0,1] \times S^1$. Do this twice, and now you have a connected $S^1$-manifold with precisely three fixed points.
So my first guess is that you would prefer the manifold to be orientable?
answered Nov 8, 2011 at 7:52
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$\begingroup$ Thank you Ryan for your detailed construction. Your last sentence is the key: I need the manifold to be closed and orientable! Regarding your construction, I need sometime to absorb:-) $\endgroup$– PingCommented Nov 8, 2011 at 9:03
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There is an $S^1$-action on the Quaternionic projective plane $\mathbb{H}\mathbb{P}^2$ with exactly 3 fixed points. They are not hard to construct (done in a similar way to the standard $S^1$-actions on $\mathbb{C}\mathbb{P}^2$) the details of such an action are contained here https://arxiv.org/pdf/1401.4731.pdf. There Kustarev proves that any $S^1$-action on a oriented smooth $8$-manifold with three fixed points must have the same weights as some linear action on $\mathbb{H}\mathbb{P}^2$ and under mild hypothesis must be diffeomorphic to $\mathbb{H}\mathbb{P}^2$.
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$\begingroup$ The octonionic projective plane $\mathbb{OP}^2\cong F_4/\mathrm{Spin}(9)$ should also work. $\endgroup$ Commented Jul 5, 2017 at 18:38
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$\begingroup$ @SebastianGoette where can I get more information about the $S^1$ action on the octonionic projective plane $OP^2$? Thanks. $\endgroup$– hao dongCommented Nov 13, 2023 at 2:16