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When I am reading one paper, I have met the following statement:

It is impossible to define a $Z_{2}\times Z_{2}$ action on a connected closed curve on a compact Riemann surface.

The claim is equivalent to say that the existence of two fixed point free involutions on a circle is not true.

The author just took this statement as an obvious fact and have used it to prove a lemmer about the antiholomorphic involutions on the compact Riemann surface. It seems the claim is pretty simple, but I cannot find an elegant way to prove this is indeed true.

I have tried to use the the following fact to prove it:

the one-dimensional manifold can be considered as a circle under the homeomorphism, and it is diffeomorphic to the one dimensional real projective plane, which inherits the automorphism group of $PGL(2, \mathbb{R})$ from $\mathbb{R}^{2}.$ And in $PGL(2, \mathbb{R}),$ the element corresponding to the antipodal map is unique, so the above statement is true.

Is this a correct proof?

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There must be an additional hypothesis here (for a $\mathbb{Z}_2\times\mathbb{Z}_2$ action on $S^1$, because I can define an action which is trivial on the first summand and acts by antipodal-map ($z \rightarrow -z$) on the 2nd summand –  Chris Gerig Feb 18 '11 at 5:30
Notice that there are many, many fixed-point free involutions on a circle... –  Mariano Suárez-Alvarez Feb 18 '11 at 5:46

2 Answers 2

up vote 5 down vote accepted

Assuming you want fixed-point free actions...

Since $\pi_1(S^1)$ is cyclic, all connected coverings of $S^1$ have cyclic group of covering transformations. Now, if $\mathbb Z_2^2$ acted without fixed points on $S^1$, the corresponding quotient map would be a covering $S^1\to S^1$ of group $\mathbb Z_2^2$, which is impossible.

Later: a bit less technological: suppose $G=\mathbb Z_2^2$ acts freely on $S^1$ and pick a point $x$. The orbit of $x$ cuts $S^1$ in $4$ segments, and one of them, call it $I$, has $x$ and $\sigma(x)$ as endpoints for some $\sigma\in G\setminus\{1\}$. But the map $\sigma$ maps $I$ to itself, so by continuity, $\sigma$ fixes a point in $I$, which it didn't.

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If you mean the action is a free action:

If a group $G$ acts freely on an odd-dimensional sphere $S^{2k-1}$ then $G$ has periodic cohomology of period $2k$. And if $G$ is abelian but not cyclic, then $G$ does $\textit{not}$ have periodic cohomology. Thus your group does not act freely on the circle.

[[edit]] To see the latter claim: a calculation using the Kunneth formula shows that $H^n(\mathbb{Z}_2\times\mathbb{Z}_2,\mathbb{Z}_2)$ has $\mathbb{Z}_2$-dimension $n+1$ for $n\ge 0$ and hence is not periodic.

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