Does there exist smooth circle action on manifolds M^{4n} with exactly three fixed points such that n\neq 1 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.
 A: 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? 
A: 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 $\mathbb{H}\mathbb{P}^2$.
