most recent 30 from http://mathoverflow.net 2013-05-22T17:56:31Z http://mathoverflow.net/feeds/question/81896 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81896/free-mathbbz-2-actions-match-at-some-point Joseph O'Rourke 2011-11-25T15:23:44Z 2011-11-25T17:16:20Z

<p>I have in front of me a proof of this lemma:</p> <blockquote> <p>If $f$ and $g$ are free $\mathbb{Z}_2$-actions on $S^1$, then $f(x)=g(x)$ for some $x \in S^1$.</p> </blockquote> <p>A $\mathbb{Z}_2$-action on the unit circle $S^1$ is a homeomorphism $f \;:\; S^1 \rightarrow S^1$ such that $f(f(x))=x$ for all $x \in S^1$; and $f$ is free if $f(x) \neq x$ for all $x \in S^1$.</p> <p>The proof (in a paper I'm refereeing) is clear but somewhat laborious. It would be nice to either have a succinct proof, or a reference, rather than a detailed proof from first principles. Has anyone seen this before? Thanks!</p>

http://mathoverflow.net/questions/81896/free-mathbbz-2-actions-match-at-some-point/81899#81899 Vitali Kapovitch 2011-11-25T15:39:48Z 2011-11-25T17:16:20Z

<p>we can clearly assume that $f(z)=-z$ in the standard metric on $S^1\subset \mathbb C$ (as we can assume that $f$ is isometric with respect to some Riemannnian metric on $S^1$). Then $g(z)=z\cdot e^{i\alpha(z)}$ with $0&lt;\alpha(z)&lt;2\pi$. If $\alpha(z_0)&lt;\pi$ then $\alpha(g(z_0))>\pi$ and there is a point $z_1$ with $\alpha(z_1)=\pi$ by the intermediate value theorem.</p>