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Is it true that every faithful and locally smooth action $S^1 \curvearrowright T^n$ is free?

I know such an action must induce an injection $\rho:\pi_1(S^1)\to\pi_1(T^n)$. Another related question is: Is the image of $\rho$ saturated?

Thanks in advance!

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  • $\begingroup$ I assume that your action is smooth? If so, it seems reasonable to guess that a smooth, faithful, locally free action has a "global section" -- a torus $T^{n-1}$ which meets all orbits, exactly once, transversely. If there is a section, then the action should be conjugate to the action which rotates the first coordinate and fixes the others. $\endgroup$
    – Sam Nead
    Jan 7, 2023 at 8:34
  • $\begingroup$ Yes I assume the action is smooth, but I am not sure whether there exists a transversal $T^{n-1}$. In fact, I am not sure $T^n/S^1$ is smooth. $\endgroup$ Jan 7, 2023 at 9:17
  • $\begingroup$ This is true when $n\le 3$, I do not know about dimension 4 and higher. $\endgroup$ Jan 7, 2023 at 14:42
  • $\begingroup$ @MoisheKohan Could you please briefly explain the argument for dim 3? $\endgroup$ Jan 7, 2023 at 23:39
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    $\begingroup$ In a little while: It comes from the essential uniqueness of Seifert fibrations on 3-manifolds. $\endgroup$ Jan 7, 2023 at 23:44

1 Answer 1

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This follows from Theorem 9.3 (page 216) of the book "Compact tranformation groups" by Bredon (note that it is freely available online).

More is true: Any effective group action of a torus on a torus is free. The proof starts along the lines you mentioned via fundamental group considerations.

The result is originally due to Conner and Montgomery:

Conner, P. E., and Montgomery, DC. Transformation groups on a K(n, l), I. Michigan Math. J. 6 (1959), 405-412.

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  • $\begingroup$ (I guess "effective" is an old synonym of "faithful".) $\endgroup$
    – YCor
    Oct 13, 2023 at 11:02

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