Let $\mathbb{Z}/2$ act on the $m$-sphere $S^m$ freely and properly discontinuously. If the action is not trivial, can we conclude that the action is homotopy equivalent to the antipodal action? That is, the following diagram

enter image description here commutes up to homotopy?

Can we generalize this to $S^1$-actions on $S^{2m+1}$ and $S^3$-actions on $S^{4m+3}$? Is the uniqueness true?

  • 2
    $\begingroup$ Also, note that the properly discontinuous requirement is redundant there. Free action of finite groups on Hausdorff spaces is automatically properly discontinuous. $\endgroup$ – Balarka Sen Dec 10 '15 at 13:10

If $\Bbb Z/2$ acts freely on $S^n$, then acting by the nontrivial element of $\Bbb Z/2$ one gets a map $f:S^n \to S^n$ which has no fixed point.

Thus, $f(x) \neq x$ for all $x \in S^n$. The homotopy $$F(x, t) = \frac{(1-t)x - tf(x)}{|(1-t)x - tf(x)|}$$

between $f$ and the antipodal map $-\text{id}$ is well-defined. Thus the map induced by that action is indeed homotopic to the antipodal map.


No you cannot. For example your action could consist of a rotation by 180 degrees around some hyperplane. That is homotopic to the identity but not the antipodal map.

  • $\begingroup$ That's not a free action. For example if $m = 2$, this action must leave the anitpodal points fixed, hence is not free. $\endgroup$ – Balarka Sen Dec 10 '15 at 11:00
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    $\begingroup$ @BalarkaSen That requirement from the question is added after this answer was posted. $\endgroup$ – user99914 Dec 10 '15 at 11:04

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