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?

  • 2
    $\begingroup$ There is an obvious circle action on $S^3$ with fixed point set a circle (suspend twice the natural action of $S^1$ on itself). This answers your first question. $\endgroup$ Aug 28, 2013 at 20:03
  • $\begingroup$ Dan thanks, of course you are right I knew that any $S^k$ can be realized, but I missed $S^1$ :) .... $\endgroup$
    – aglearner
    Aug 28, 2013 at 20:12

2 Answers 2


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.

  • $\begingroup$ Allan thank you. I have a problem to understand your answer fully. It seems to me that there is a misprint in line 3: "$(n+1)$-disk $W\times D^2$"? $\endgroup$
    – aglearner
    Aug 28, 2013 at 20:20
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    $\begingroup$ Ah, the misprint started with saying $M$ was $(n-1)$-dimensional, instead of $(n-2)$-dimensional, and $W$ should be $(n-1)$ dimensional. I think I got it fixed, now. $\endgroup$ Aug 28, 2013 at 21:12
  • $\begingroup$ Allan, thanks again! So you say that $W\times D^2$ is a $(n+1)$-disk. As far as I understand you deduce this from the fact that $W$ is contractible. Could you tell me please what theorem one has to use here? (or maybe I miss again something simple) $\endgroup$
    – aglearner
    Aug 28, 2013 at 21:31
  • $\begingroup$ I'll edit the answer to provide a bit more detail on this point. It uses the h-cobordism theorem. $\endgroup$ Aug 29, 2013 at 1:02
  • $\begingroup$ this is very nice! Thank you very much! $\endgroup$
    – aglearner
    Aug 29, 2013 at 10:21

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


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