I have a question in mind for some time. That is, does there exist a smooth circle action on a closed manifold M^{4n} $M^{4n}$ (n\geq 2$n\geq 2$) with exactly three fixed points? RemmarksRemarks:(1)For For n=1, the examples are obvious (standard linear circle action on CP^2$CP^2$). (2)If 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.