Is it true that the orbit space of a free finite group action on a CW-complex is also a CW-complex? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T10:18:35Z http://mathoverflow.net/feeds/question/109687 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109687/is-it-true-that-the-orbit-space-of-a-free-finite-group-action-on-a-cw-complex-is Is it true that the orbit space of a free finite group action on a CW-complex is also a CW-complex? Li Yu 2012-10-15T06:20:07Z 2012-10-15T21:45:19Z <p>Suppose a finite group G acts freely and continuously on an n-dimensional CW-complex X. Then can we conclude that the orbit space of this action is still an n-dimensional CW-complex? (or homotopy equivalent to an n-dimensional CW-complex?) In particular, we do not assume G acts cellularly on X.</p> http://mathoverflow.net/questions/109687/is-it-true-that-the-orbit-space-of-a-free-finite-group-action-on-a-cw-complex-is/109689#109689 Answer by Ralph for Is it true that the orbit space of a free finite group action on a CW-complex is also a CW-complex? Ralph 2012-10-15T07:26:18Z 2012-10-15T07:26:18Z <p>If $G$ (finite or more generally discrete) acts cellularly on $X$, i.e. </p> <ul> <li>if $\sigma$ is an open cell of $X$ then $g\sigma$ is again an open cell in $X$ for all $g \in G$</li> <li>if $g \in G$ fixes an open cell $\sigma$ (i.e. $g\sigma=\sigma$), then it fixes $\sigma$ pointwise (i.e. $gx=x$ for all $x \in \sigma$)</li> </ul> <p>then $X/G$ is a CW-complex. This follows from Prop. 1.15 and Ex. 1.17(2) of <a href="http://books.google.de/books?id=azcQhi6XeioC&amp;pg=PA103&amp;lpg=PA103&amp;dq=G-CW-complex+tomdieck+transformation+groups&amp;source=bl&amp;ots=sdu_oD5B32&amp;sig=eqAYeGeXEaGmjJeCeAoK3P_qwlI&amp;hl=de&amp;sa=X&amp;ei=ArV7UO2MFY7GtAaF14CQCw&amp;ved=0CFcQ6AEwBw#v=onepage&amp;q=G-CW-complex%2520tomdieck%2520transformation%2520groups&amp;f=false" rel="nofollow">tom Dieck: Transformation Groups</a></p> http://mathoverflow.net/questions/109687/is-it-true-that-the-orbit-space-of-a-free-finite-group-action-on-a-cw-complex-is/109760#109760 Answer by Neil Strickland for Is it true that the orbit space of a free finite group action on a CW-complex is also a CW-complex? Neil Strickland 2012-10-15T21:01:03Z 2012-10-15T21:01:03Z <p>This is not really an answer, but a comment about an interesting special case. Suppose that $G$ acts smoothly on $S^2$. By averaging we can choose a $G$-invariant Riemannian metric. This gives $S^2$ a conformal structure, making it a Riemann surface. Any Riemann surface homeomorphic to $S^2$ is conformally equivalent to the standard Riemann sphere. Thus, we can reduce to the case where $G$ acts on $\mathbb{C}\cup\{\infty\}$ by conformal and anticonformal maps, which must have the form $z\mapsto (az+b)/(cz+d)$ or $z\mapsto (a\overline{z}+b)/(c\overline{z}+d)$. I think it even works out here that the quotient $(\mathbb{C}\cup\{\infty\})/G$ is always either a sphere or a disc. Thus, one cannot get any local pathology in this context. This contrasts with other settings where smooth functions can generate topological pathology: for example, any closed subset of $\mathbb{R}^n$, however fractal, can be expressed as the zero set of a smooth function $f\colon\mathbb{R}^n\to\mathbb{R}$.</p> <p>Along somewhat similar lines, I think one can show that when $X$ is a one-dimensional CW complex with continuous action of a finite group $G$, then $X/G$ is again a one-dimensional CW complex (up to homeomorphism, not just homotopy equivalence).</p> http://mathoverflow.net/questions/109687/is-it-true-that-the-orbit-space-of-a-free-finite-group-action-on-a-cw-complex-is/109761#109761 Answer by Ian Agol for Is it true that the orbit space of a free finite group action on a CW-complex is also a CW-complex? Ian Agol 2012-10-15T21:02:11Z 2012-10-15T21:45:19Z <p>The 3-sphere gives an example of an action with fixed points. If one takes the <a href="http://en.wikipedia.org/wiki/Alexander_horned_sphere" rel="nofollow">solid Alexander horned sphere</a>, then Bing proved that its double is homeomorphic to the 3-sphere. So the quotient of the involution acting on $S^3$ is the solid Alexander horned sphere. However, the solid horned sphere is not homeomorphic to a CW complex. This follows from the <a href="http://mathoverflow.net/questions/98750/can-the-alexander-horned-sphere-arise-as-a-cell-boundary-in-a-finite-cw-sphere/98762#98762" rel="nofollow">answer to this question on the Alexander horned sphere</a>. If the solid Alexander horned sphere were a CW complex, then one could attach the exterior 3-ball to get a CW structure on $S^3$ with the Alexander horned sphere being the boundary of the closure of a 3-cell, which is a contradiction to the other question. </p> http://mathoverflow.net/questions/109687/is-it-true-that-the-orbit-space-of-a-free-finite-group-action-on-a-cw-complex-is/109763#109763 Answer by Igor Belegradek for Is it true that the orbit space of a free finite group action on a CW-complex is also a CW-complex? Igor Belegradek 2012-10-15T21:09:03Z 2012-10-15T21:09:03Z <p><b>Lemma</b> If $X$ is a countable locally finite CW-complex and $G$ acts freely and properly discontinuously on $X$, then $X/G$ is homotopy equivalent to a CW-complex.</p> <p><b>Proof</b> Any metrizable ANR is homotopy equivalent to a CW-complex (I am not sure who proved it first but see Theorem 3.6.1 <a href="http://image.diku.dk/aasa/oldpage/aasa.pdf" rel="nofollow">here</a>. Since $X$ is countable and locally finite, it is a metrizable separable ANR. As Misha remarks in comments averaging the metric over the group action implies that $X/G$ is metrizable. Also a countable dense subset of $X$ projects to a countable dense subset of $X/G$. Finally, if a metrizable separable space is locally ANR, it is an ANR (see Borsuk's "Theorey of Retracts", Corollary 10.4, Chapter IV). It follows that $X/G$ is a metrizable ANR as desired.</p> <p><b> Remark</b> In seeing whether $X/G$ is homeomorphic to a CW-complex, even the case when $X$ is a PL manifold is unclear. The difficulty is that it seems unknown which topological manifolds are homeomorphic to CW-complexes (Kirby-Siebenmann prove this for compact manifolds of dimension $\ge 6$ (or maybe $\ge 5$?, but certainly not $4$). So there might exist manifolds not homeomorphic to CW-complexes but whose finite covers are PL. </p>