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Let $M$ be a compact connected manifold with an $S^1$-action. Suppose that $S^1$ has a fixed point in $M$. Is it true that $\pi_1(M)=\pi_1(M/S^1)$?

I is there some reference or a short proof of this fact?

PS. I am sorry for amending the question. In reality I only want to know that the kernel of the map $\pi_1(M)\to \pi_1(M/S^1)$ is finite. Is at least this fact true?

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    $\begingroup$ Unless I've misunderstood the question, the real projective plane with $S^1$ acting on it by rotation around a point is a counterexample. $\endgroup$ – Anton Malyshev Apr 7 '14 at 23:59
  • $\begingroup$ Thank you, indeed, this is a counterexample. I amended the question, because I was asking for more than I really need. $\endgroup$ – aglearner Apr 8 '14 at 0:09
  • $\begingroup$ Can we have the exact sequence $\pi_1(S^1)\to\pi_1(M)\to \pi_1(M/S^1)\to1$? $\endgroup$ – DLIN Dec 25 '19 at 12:45
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The answer to the amended question still seems to be NO For a source of counterexamples, check out Frank Raymond's 1968 paper on circle actions on 3-manifolds.

(it seems that the fundamental group of the quotient is always free, but the manifolds admitting such actions admit much more complicated than free-by-finite fundamental groups).

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  • $\begingroup$ Thank you Igor! I will check the paper once I wake up :) $\endgroup$ – aglearner Apr 8 '14 at 0:55
  • $\begingroup$ Igor, I will accept your answer. I think indeed, that explicit counterexamples can be constructed for example as follows. Take the linear S^1 action on S^3 that fixes $S^1$. Then the quotient is $D^2$. Now take as many $S^1$ fibers in $S^3$ as you want and make a Dehn surgery at these fibres. Then the quotient will become a disk with orbi-points, and I guess the obrifold fundamental group of the disk will be related to the fundamental group of the surgered $3$-manifold. Would you agree with this? $\endgroup$ – aglearner Apr 8 '14 at 8:28
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In general, the answer to the original question is no.

In fact, $S^1$ acts on $\mathbf{RP}^2$ in such a way that that the action has a fixed point. The quotient space is homeomorphic to a closed interval, hence $\pi_1(\mathbf{RP}^2/S^1)=\{1\}$ whereas $\pi_1(\mathbf{RP}^2) = \mathbf{Z} / 2 \mathbf{Z}$.

The description of the action is the following. We see $\mathbf{RP}^2$ as the quotient of $S^2$ by the antipodal map $x \mapsto -x$. Then the action of $S^1$ on $S^2$ given by rotations around the vertical axis descends to $\mathbf{RP}^2$. All the points have trivial stabilizer, except the point corresponding to the class of the two poles (which is fixed) and the points corresponding to classes on the equatorial circle (whose stabilizers have order $2$).

Hovewer, the answer is yes when $M$ is simply connected. This is a consequence of a more general result of Armstrong concerning actions of compact Lie groups on simply connected spaces, see the paper

M-A. Armstrong, Calculating the fundamental group of an orbit space, Proceedings of the American Mathematical Society 84 (1982), 267-271, in particular Example 4.

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  • $\begingroup$ Francesco, thank you for this example. I amended the question to ask if at least the kernel of the map is finite. This is what I really want... $\endgroup$ – aglearner Apr 8 '14 at 0:13
  • $\begingroup$ Here is a follow up question. Chapter 11 of "Topology and Groupoids" deals with orbit spaces and actions of groups on groupoids, obtaining a theorem that for a discrete, group $G$ acting discontinuously on a Hausdorff space $X$ with some good local properties, the fundamental groupoid of the orbit space $X/G$ is isomorphic to the orbit groupoid of the fundamental groupoid $\pi_1 X$ by the action of $G$. This result includes work of A. Armstrong. Question: Are there possible extensions to actions of topological groups, such as $S^1$? $\endgroup$ – Ronnie Brown Apr 8 '14 at 17:20

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