The spherical space form conjecture (now theorem) asserts that every finite group acting on the 3-sphere is conjugate to a subgroup of $\mathrm{SO}(4)$. Is the action assumed to be smooth? Can anything similar be said for topological actions? Also, I would appreciate a reference about the spherical space form conjecture.
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$\begingroup$ I think it's been known for a while (long before Hamilton–Perelman and even Thurston) that in dimension 3 the smooth and topological Poincaré conjectures are equivalent. Unfortunately I don't even seen this mentioned in Wikipedia's Poincaré conjecture page. Hopefully somebody will provide a reference and more precise statement. $\endgroup$– YCorCommented Sep 17, 2020 at 8:50
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4$\begingroup$ The SFS conjecture is about free finite group actions. If you drop the freeness assumption then in the smooth category every finite group action is still conjugate to an orthogonal action, but in the topological category this is no longer true (Bing's examples).It is a rather old theorem (Moise+Munkres) that in dimension 3 TOP=DIFF: every topological manifold admits a smooth structure and this structure is unique up to diffeomorphism. See the references I gave here. $\endgroup$– Moishe KohanCommented Sep 17, 2020 at 9:46
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$\begingroup$ @MoisheKohan Thank you. Does that mean that the spherical space form conjecture holds for topological, non-smooth actions over S^3 as well? $\endgroup$– George KCommented Sep 17, 2020 at 11:04
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2$\begingroup$ No. To rephrase Moishe: If your action is free, then it is topologically conjugate to a linear action (smoothly conjugate if the action was smooth to begin with). Non-free topological actions need not be smoothable. Eg, the double of the (non-manifold) Alexander horned ball is homeomorphic to $S^3$, and hence $S^3$ carries an involution with non-locally-flat fixed set, so this action is not even topologically conjugate to a smooth action. $\endgroup$– mmeCommented Sep 17, 2020 at 11:11
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1$\begingroup$ @IgorBelegradek Yes, one can prove that topological 3d orbifolds are smoothable. I am not sure if this is in the literature though. But it should be a good separate question. $\endgroup$– Moishe KohanCommented Sep 17, 2020 at 14:57
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