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? 



Fixed component of an $S^1$ action on $S^n$
The fixed point set need not be simply connected in general. If $M$ is any smooth homology $(n-2)$-sphere that bounds a smooth contractible $(n-1)$-manifold $W$ (such exist in abundance), then $S^1$ acts smoothly on the $(n+1)$-disk $W\times D^2$, by rotations in the $D^2$ factor, with fixed point set $W$ and therefore on the boundary $n$-sphere with fixed point set $M$. 


To see that $W\times D^{2}$ is a disk we invoke the $h$-cobordism theorem. Note that it is a contractible manifold with simply connected boundary by the Seifert-van Kampen theorem. By duality one can see that the boundary has the homology, hence the homotopy type of a sphere. Removing the interior of a small ball from the manifold provides an $h$-cobordism between the boundary and the boundary of the small ball. The $h$-cobordism theorem shows that the cobordism is a product (in higher dimensions of course).


The smallest dimension in which any of this happens is $n=5$. In that case the simplest examples of such 4-manifolds $W$ are the Mazur 4-manifolds that consist of a 4-ball with a 1-handle and then a 2-handle attached in such a way that the attaching map kills $\pi\_{1}$ but is not isotopic to the attaching map of a handle that just goes once over the 1-handle. On the other hand, crossing with $[0,1]$ gives enough room to isotope the 2-handle so that it goes once across the 1-handle, showing that in fact $W\times [0,1]$ is a 5-ball.


