Suppose that $G$ is a Lie group acting smoothly on a manifold $M,$ does the Borel $M \times_G EG$ construction have the homotopy type of a CW-complex? If not, under what conditions would this be true? (I mostly care about the case of discrete stabilizers). More generally, if $G$ is any topological group acting on a CW-complex $X$, under what conditions does $X \times_G EG$ have the homotopy type of a CW-complex?
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2$\begingroup$ Just to clarify: do you really mean CW-complex and not finite CW-complex? $\endgroup$– John PardonCommented Mar 26, 2015 at 10:26
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$\begingroup$ Presumably this depends on a choice of $EG$. What $EG$ are you choosing? $\endgroup$– Qiaochu YuanCommented Mar 26, 2015 at 16:45
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$\begingroup$ @JohnPardon: Yes, I really mean a CW-complex. I just want it to satisfy Whitehead's theorem. $\endgroup$– David CarchediCommented Mar 26, 2015 at 20:40
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$\begingroup$ @QiaochuYuan: See my comment on Tyler's answer below. $\endgroup$– David CarchediCommented Mar 26, 2015 at 20:41
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Yes, this is true. It suffices for $X$ to have the homotopy type of a CW-complex (this is true of smooth manifolds; see e.g. here or here).
I'm going to assume that you're using a definition of $EG$ that includes something like: $EG$ is a $G$-CW-complex, so that it is built by iteratively taking pushouts of diagrams of the form $$ D^n \times G \leftarrow S^{n-1} \times G \rightarrow Y. $$ As a result, the space $EG \times_G M$ is formed by an iterated sequence of pushouts $$ D^n \times M \leftarrow S^{n-1} \times M \rightarrow Z. $$ (This comes with the standard warnings about probably having to use compactly generated spaces so that products, quotients, and the direct limit topology interact well.)
Each of these pushouts is the mapping cone of the map $S^{n-1} \times M \to Z$. This mapping cone would not necessarily be a CW-complex even if $M$ and $Y$ were (the map would have to be cellular for that), but if $M$ and $Z$ both have the homotopy type of CW-complexes, the cone does have the homotopy type of a CW-complex (it's homotopy equivalent to a cellular map, and that equivalence carries across to an equivalence on mapping cones). By inducting on the cell structure of $EG$, we can assume that $Z$ has the homotopy type of a CW-complex and find that this next pushout map $Z \to Z'$ is homotopy equivalent to a cell inclusion of CW-complexes. Taking colimits we get the desired result.
answered Mar 26, 2015 at 17:21
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$\begingroup$ Thanks! The model for $EG$ I am using is the fat geometric realization of the (topologically enriched) nerve of the topological action groupoid $G \ltimes G$. Please correct me if I'm wrong, but I imagine that the same argument goes through by using the skeleta filtration of its fat geometric realization. $\endgroup$ Commented Mar 26, 2015 at 20:40
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$\begingroup$ P.S. I think all topological manifolds also have the homotopy type of a CW-complex as well (when it's 2nd countable). $\endgroup$ Commented Mar 26, 2015 at 20:47
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$\begingroup$ Actually, using your mapping cone idea, it's probably easiest to work directly with the fat geometric realization of the nerve of the action groupoid $G \ltimes X$. Each of its skeleta, by your above observation, will have the homotopy type of a CW-complex. $\endgroup$ Commented Mar 26, 2015 at 21:03
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1$\begingroup$ @DavidCarchedi Ah, great, that should work perfectly as well. (Unfortunately above I used the word "mapping cone" incorrectly; it merely has the homotopy type of a cellular inclusion.) $\endgroup$ Commented Mar 27, 2015 at 0:44
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$\begingroup$ Thanks, I was just about to send you an email asking about that, I assume you meant double mapping cylinder? I think I'll just send the email :). $\endgroup$ Commented Mar 27, 2015 at 1:16