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.

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 wellknown) but I don't know about 3. Another obvious observation is that an $S^1$action like this can not be semifree for $n\ge 2$. $\endgroup$ – Vitali Kapovitch Nov 7 '11 at 19:23

$\begingroup$ 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 wellknown. 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 semifree circle action with isolated fixed points, then the number of this action must be even (Bott residue formula). $\endgroup$ – Ping Nov 7 '11 at 23:57

$\begingroup$ Sorry "there are known examples in dimensions other than 4" should be "there are no known examples in dimensions other than 4" $\endgroup$ – Ping Nov 7 '11 at 23:58

$\begingroup$ In fact, I showed in mathjournals.org/mrl/2011018003/2011018003005.pdf that if a manifold admits a semifree circle action with isolated fixed points, then the manifolds bound, i.e., the Pontrjagin numbers and StiefielWhitney numbers of the manifold vanish (Theorem 1.6). $\endgroup$ – Ping Nov 8 '11 at 0:04

$\begingroup$ Thanks, this is a nice theorem. I had something much easier in mind  if the action of $S^1$ is semifree 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 semifree). 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 nonorientable but that doesn't matter when you look at the semifree $Z_2$ action. $\endgroup$ – Vitali Kapovitch Nov 8 '11 at 4:40
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 $D^{2n1} \times \{0,1\} \times S^1$ and gluing in an $S^{2n2} \times [0,1] \times S^1$. 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?

$\begingroup$ 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:) $\endgroup$ – Ping Nov 8 '11 at 9:03
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 $\mathbb{H}\mathbb{P}^2$.

$\begingroup$ The octonionic projective plane $\mathbb{OP}^2\cong F_4/\mathrm{Spin}(9)$ should also work. $\endgroup$ – Sebastian Goette Jul 5 '17 at 18:38