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.
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$\begingroup$ Perhaps you can perform repeated subdivisions on the cell structure of $X$ to arrive at the cellular case? I'm not claiming this is always possible, but it may be. $\endgroup$– Mark GrantCommented Oct 15, 2012 at 14:10
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3$\begingroup$ Here is a suggestion for proving that $X/G$ is homotopy equivalent to a CW-complex in a special case when the complex $X$ is countable and locally finite. Any metrizable ANR is homotopy equivalent to a CW-complex. If the complex $X$ is countable and locally finite, then it is a metrizable separable ANR, and I suspect metrizability and separability are inherited by $X/G$. Now a metrizable separable space that is locally an ANR is globally an ANR, so $X/G$ would then be a metrizable ANR. $\endgroup$– Igor BelegradekCommented Oct 15, 2012 at 14:28
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1$\begingroup$ In seeing whether $X/G$ is homeomorphic to a CW-complex, even the case when $X$ is a smooth manifold is unclear. Indeed, if the $G$-action is nonsmoothable, then $X/G$ would only be a topological manifold, and in general it is unclear to me whether $X/G$ is homeomorphic to a CW-complex. (I think the existence of a CW structure on a topological manifold is unknown in dimension 4 and also for noncompact manifolds in higher dimensions, at least the proof in Kirby-Siebenmann's book is for compact case only). $\endgroup$– Igor BelegradekCommented Oct 15, 2012 at 14:35
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$\begingroup$ I would see if their is some sort of Borel construction. At the very least, you can take the singular chains on $X$, make the group action free with a simplicial Borel construction, then take the geometric realization of that. After this process, you will be in possession of a CW complex with the right weak homotopy type. I suspect that if you began with a CW complex, you will have a homotopy equivalence. $\endgroup$– Spice the BirdCommented Oct 15, 2012 at 15:27
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1$\begingroup$ @Igor: I think you should write your comments as an answer. Note that if $X$ is metrizable, so is $X/G$ (by averaging the distance function on $X$ under the group action: Sum of distance functions is again a distance function). Local finiteness passes to the quotient, separability too. $\endgroup$– MishaCommented Oct 15, 2012 at 18:07
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4 Answers
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Lemma 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.
Proof Any metrizable ANR is homotopy equivalent to a CW-complex (I am not sure who proved it first but see Theorem 3.6.1 here. 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.
Remark 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.
answered Oct 15, 2012 at 21:09
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$\begingroup$ I wonder when $G$ is finite and the CW-complex $X$ is of dimension $n$, can we choose the CW-complex homotopic to the orbit space $X/G$ to be $n$-dimensional too? $\endgroup$– Li YuCommented Oct 16, 2012 at 2:48
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$\begingroup$ @Igor: I do not think you need compactness for this. The point is that you can exhaust an open topological manifold $X$ by an increasing sequence of codimension $0$ compact submanifolds with boundary $X_i$. Let $B_i=\partial X_i$. Then, inductively, the handle structure extends from the collar $B_i\times I\subset X_{i+1}$ to the rest of $X_{i+1}$ (all the existence results are relative). Incidentally, the handle decomposition of Kirby-Siebenmann works in dimensions $\ge 6$; it is extended to dimension $5$ by Frank Quinn in "Ends of Maps-III". $\endgroup$– MishaCommented Oct 16, 2012 at 5:00
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If $G$ (finite or more generally discrete) acts cellularly on $X$, i.e.
- if $\sigma$ is an open cell of $X$ then $g\sigma$ is again an open cell in $X$ for all $g \in G$
- 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 $x \in \sigma$)
then $X/G$ is a CW-complex. This follows from Prop. 1.15 and Ex. 1.17(2) of tom Dieck: Transformation Groups
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2$\begingroup$ What I want to know is exactly the case when G does not acts cellularly! $\endgroup$– Li YuCommented Oct 15, 2012 at 9:15
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2$\begingroup$ Yes, but when he answered this, you had not made that clear $\endgroup$ Commented Oct 15, 2012 at 13:38
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1$\begingroup$ I am very sorry. I thought the cellularly action case is easy. So I did not mention it at the beginning. But after your answer, I realize I should emphasize where the difficulty of the question lies. $\endgroup$– Li YuCommented Oct 15, 2012 at 13:56
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1$\begingroup$ @David White: your comment seems unfair. If you look in the edits, the first version of the question did not assume the action was cellular; it said "Suppose a finite group G acts freely and continuously on a CW-complex X. Then can we conclude that the orbit space of this action is still a CW-complex? (or homotopy equivalent to a CW-complex?)". In my view this is clearly stated, and it was Ralph who misread the question; why blame the OP? $\endgroup$ Commented Oct 15, 2012 at 16:18
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2$\begingroup$ @Igor: 1) Right, the question in its original version makes no assumption on cellularity. Hence there are two case: a) cellular action b) non-cellular action. My answer obviously treats case a). So how do you conclude I misread the question ? 2) If one is only interested in case b) then it's good style to point it out, as mentioned by David. - Anyway, now it's clear and I'm curious about the answer of this interesting problem. $\endgroup$– RalphCommented Oct 15, 2012 at 17:45
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The 3-sphere gives an example of an action with fixed points. If one takes the solid Alexander horned sphere, 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 answer to this question on the Alexander horned sphere. 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.
answered Oct 15, 2012 at 21:02
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$\begingroup$ This action is not free, as requested, the subspace of fixed points is Alexander's horned sphere (the common boundary). $\endgroup$ Commented Oct 15, 2012 at 21:10
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$\begingroup$ Ian, does not your involution have fixed points? $\endgroup$ Commented Oct 15, 2012 at 21:11
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$\begingroup$ I also missed the assumption that the action was supposed to be free. However, the question is still a good one without that assumption and Ian's answer is instructive so I hope he leaves it up. $\endgroup$ Commented Oct 15, 2012 at 21:24
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$\begingroup$ I missed the freeness assumption, but as Neil suggests I'll leave it up. $\endgroup$– Ian AgolCommented Oct 15, 2012 at 21:46
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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 $f\colon\mathbb{R}^n\to\mathbb{R}$.
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).
answered Oct 15, 2012 at 21:01
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$\begingroup$ Neil: Even more, if $G$ acts topologically freely and properly discontinuously on a topological manifold $X$ of dimension $\ne 4$ then the quotient $X/G$ is homeomorphic to a CW-complex. The same applies if $X$ is merely a simplicial complex of dimension $\le 3$ and $G$ acts topologically. $\endgroup$– MishaCommented Oct 15, 2012 at 22:04
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$\begingroup$ Misha: is your $X$ compact? If not, would you give a reference for the first claim, say in higher dimensions? $\endgroup$ Commented Oct 15, 2012 at 22:42