TITLE: Fixed component of an $S^1$ action on $S^n$
QUESTION [6 upvotes]: Suppose $S^1$ is acting smoothly on $S^n$ and $M$ is a connected component of the set of fixed points of the action. What can be said about $M$? 
Is it true that $\pi_1(M)=0$? (sorry this first bit of the question is silly since any $S^k$ can appear) Is it true that $M$ has to be homeomorphic to a sphere? If not, what kind of manifolds can one get?

REPLY [2 votes]: It is known that $M$ has the homology of a sphere, see [1] But you probably knew that.
What you also can describe pretty nicely is the cohomology of the orbit space $S^n/G$. To be more precise $S^n/G$ has the cohomology of the join of a sphere and a complex projective space and the dimension of this sphere is exactly the dimension of the fixed point set, see also [1]
[1] Cohomology ofS 1-orbit spaces of cohomology spheres and cohomology complex projective spaces, Mikiya Masudaco 
Mathematische Zeitschrift
1981, Volume 176, Issue 3, pp 405-427