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The antipodal involution of $\mathbb{S}^2$ clearly has no fixed points. However I cannot think up an example of homeomorphism of order 4 which has no fixed points. Could you help me?

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    $\begingroup$ Try combining multiple symmetries of $S^2$. You should be able to find a symmetry with small order that you can combine with the antipodal involution. $\endgroup$
    – Kyle
    Jul 20, 2016 at 18:49

3 Answers 3

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Pick any non-identity involution $i$ of $\Bbb{RP}^2$. It has two lifts to a map $S^2 \to S^2$; pick the one whose square is the antipodal map. Because the antipodal map has no fixed points and is order 2, your lift must have no fixed points and be order 4.

Note that necessarily the involution $i$ downstairs must have fixed points; there is no surface with fundamental group of order 4.

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Consider the linear transformation of $\Bbb{R}^3$ corresponding to $$ A=\left(\begin{array}{ccc}0&-1&0\\1&0&0\\0&0&-1\end{array}\right). $$ The transformation is orthogonal, hence preserves lengths, hence maps $S^2$ to itself. It is clearly of order four. The eigenvalues of $A$ are $i,-i,-1$, so it has no fixed points.


This is in the set of homeomorphisms described by Antonio: a 90 degree rotation followed by a reflection.

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You can build a self-diffeomorphism $T$ of $S^2$ with no fixed point and order $2n$ as follows.

Think $S^2$ has the unit sphere in $\mathbb{R}^3$. Denote by $R$ the rotation of angle $\theta=\pi/n$ about the $z$-axis and by $S$ the reflection about the $xy$-plane. Notice that $T=S \circ R$ has no fixed point (since it interchanges the north and the south hemispheres of $S^2$ and acts as a rotation on the equatorial $S^1 \subset S^2$) and that $$ T^{2n}= (S \circ R)^{2n} = (S^2)^n \circ R^{2n}= id .$$ In fact it is easy to check that $T$ has exactly order $2n$.

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  • $\begingroup$ Yes this is a coincidence, due to the fact that for $n=2$ both $R$ and $S$ have order 2. For $n>2$ everything goes smoothly. $\endgroup$ Jul 27, 2016 at 10:24
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    $\begingroup$ I did an edit. If you take $\theta = \pi/n$ instead of $\theta = 2 \pi /n$ it should be ok also for $n=2$. $\endgroup$ Jul 27, 2016 at 12:01

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