Timeline for What is π_1(BG) for an arbitrary topological group $G$?
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| Jul 11, 2018 at 3:17 | comment | added | Dan Ramras | @JeremyBrazas I never found a proof of Andre's general statement, so I gave a short proof in the appendix to my preprint. (I prove the more general statement for arbitrary topological monoids with $\pi_0 G$ is replaced by its Grothendieck group.) In addition to using the thick realization, I needed to use the compactly generated topology on $G^n$ when forming the nerve $NG$; I don't know if this matters. I haven't thought about whether my argument says anything about the topologized fundamental group. | |
| Jul 11, 2018 at 3:00 | comment | added | Dan Ramras | @jdc: One advantage of the thin realization is that it commutes with products (Milnor's theorem) so long as you work in the category of compactly generated spaces. Another is that level-wise constructions, like level-wise loop spaces, give the right homotopy type (this is in May's Geometry of Iterated Loop Spaces). I would agree with Segal (see the Appendix to Categories and Cohomology Theories) that the thick realization gives the correct homotopy type in general, and the thin version is a useful tool (e.g. for the previous reasons) when it's homotopy equivalent to the thick version. | |
| Mar 5, 2016 at 17:43 | comment | added | jdc | I have a very retroactive, naive question about this answer. It seems like the fat realization is generally better behaved with the sole exception of being much bigger. Why is the thin realization insisted on so frequently? | |
| Nov 25, 2011 at 19:53 | comment | added | John Klein | To get a better homotopy type, one can alternatively replace $G$ by the realization $G' :=|S.G|$, which is a topological group in the compactly generated topology ($S.$ = total singular complex, $| \quad |$ = realization). Then the map of topological groups $G' \to G$ is a weak homotopy equivalence. Then the fat and thin realizations of $N.G'$ coincide up to homotopy. | |
| Nov 7, 2010 at 12:58 | comment | added | André Henriques | @Chris: I believe that you are correct. But viewing things up to weak homotopy equivalence would tend to forget the topology on $\pi_0(G)$, and I think that Jeremy cares about the latter. | |
| Oct 12, 2010 at 17:04 | comment | added | Chris Schommer-Pries | @Andre: As you know, the usual geometric realization is given as a certain colimit (coend). I think it is possible to show that the fat geometric realization the same thing as (i.e. weakly equivalent to) taking the homotopy colimit of this same diagram. Right? If so, doesn't your claim follow from the weak homotopy invariance of the homotopy colimit? You simply replace G be a CW approximation. | |
| Oct 11, 2010 at 13:49 | vote | accept | Jeremy Brazas | ||
| Oct 11, 2010 at 9:41 | comment | added | André Henriques | I don't think that this "known". But I think that it's not too difficult prove it using transversality w.r.t the subset of centers of simplices. | |
| Oct 10, 2010 at 15:03 | comment | added | Jeremy Brazas | Thanks very much Andre! The business about the geometric realization makes sense. I am not that familiar with fat geometric realizations but I believe the difference comes from restricting the identifications to only strictly increasing maps in $\Delta$ (correct me if I am wrong). Is there somewhere I can find a proof of the assertion about the group isomorphism? | |
| Oct 10, 2010 at 9:34 | comment | added | BS. | Could you explain the subtelty between fat and ordinary geometric realization ? | |
| Oct 9, 2010 at 22:25 | history | edited | André Henriques | CC BY-SA 2.5 |
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| Oct 9, 2010 at 22:15 | history | edited | André Henriques | CC BY-SA 2.5 |
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| Oct 9, 2010 at 22:05 | history | answered | André Henriques | CC BY-SA 2.5 |