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.

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? 

