Let $G$ be a finite abelian group (of odd order if it's significant) acting on sphere $S^2\subset\mathbb{R}^3$. So my question: is it true that $G$ has a fixed point?
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1$\begingroup$ This is a small homework exercise in an Algebraic Topology class -- if a finite group $G$ acts freely on $S^{2n}$ then $G=\mathbb{Z}_2$ (using degree theory). $\endgroup$– Chris GerigCommented Apr 14, 2014 at 19:08
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2$\begingroup$ @ChrisGerig: A group action may be nonfree, yet also have no fixed point (there is a difference between having a fixed point and having a point with nontrivial stabilizer). $\endgroup$– Jason StarrCommented Apr 14, 2014 at 19:10
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1$\begingroup$ Ah good catch. $\mathbb{Z}_2\times\mathbb{Z}_2$ acts on $S^2$ by rotation through two orthogonal axes (one per summand) with no fixed point, and the "4 poles" each have $\mathbb{Z}_2$-stabilizer. $\endgroup$– Chris GerigCommented Apr 14, 2014 at 19:47
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$\begingroup$ Subsequently, unless I'm thinking too quickly again, $\mathbb{Z}_3\times\mathbb{Z}_3$ acts similarly and provides the smallest odd-order counterexample to the question. $\endgroup$– Chris GerigCommented Apr 14, 2014 at 20:05
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$\begingroup$ @ChrisGerig you're thinking too quick again: $\mathbb{Z}_2\times\mathbb{Z}_2$ works because the rotations (by $\pi$, of course) on the two axes commute with each other; when you go to $\mathbb{Z}_3\times\mathbb{Z}_3$ this no longer holds for the rotations by $\frac{2\pi}{3}$, and so e.g. products like $abab$ don't correspond to $a^2b^2$. (In fact, I believe the group you provide is actually the octahedral group $O$, of even order.) $\endgroup$– Steven StadnickiCommented Apr 14, 2014 at 21:33
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See
Fixed points of abelian actions on S2 JOHN FRANKSa1, MICHAEL HANDELa2 and KAMLESH PARWANIa3
(Ergodic Theory and Dynamical Systems, 2007)
answered Apr 14, 2014 at 18:44