7

Let $G$ be a topological group and $BG$ its classifying space. From the LES of the universal bundle, we get $\pi_i(BG)\cong\pi_{i-1}(G)$, so given the classifying space, we know all homotopy groups of $G$. If $G$ is discrete, then $\pi_0(G)\cong G$, which shows that we can reconstruct the group $G$ from its classifying space.

Surely this does not work for topological groups, but what is a small concrete example, where $BG$ and $BH$ coincide, but $G$ and $H$ do not?

I'm intentionally unprecise what I mean by "coincide", since I'm curious for different answers.

Reinhard
  • 71
  • 1
    Note that $G \xrightarrow{\sim} \Omega BG$, so if $BG$ and $BH$ "coincide" (for pretty much any reasonable definition of "coincide") then $G$ and $H$ will at least be homotopy equivalent (and not simply have the same homotopy groups). – Najib Idrissi Jun 10 '15 at 08:44
  • Thanks, I should have thought more by myself before asking. So if $BG$ and $BH$ are homotopy equivalent, then $G$ and $H$ are equivalent as H-groups. As a moral, we can reconstruct the homotopy type of the space and the group multiplication up to homotopy. Did I say anything stupid? – Reinhard Jun 10 '15 at 08:47
  • I don't think so (that you said anything stupid), I believe that yes, you can completely reconstruct $G$ up to homotopy from $BG$ (I'm not really an expert on this though). – Najib Idrissi Jun 10 '15 at 08:49
  • What about the case of topological monoids which are not group like, i.e. the map $M\rightarrow \Omega BM$ is not an equivalence. We'll probably loose more information going from $M$ to $BM$. Do you have any thoughts on that? – Reinhard Jun 10 '15 at 08:51
  • I don't really know, to be honest. I would expect the information to be lost when you take the loop space, not the classifying space (so $BM$ would still contain enough information). But it's probably best to wait for someone who does know for sure. – Najib Idrissi Jun 10 '15 at 08:54
  • Thanks for your help anyway! – Reinhard Jun 10 '15 at 09:09
  • The transition $M \mapsto BM$ is a bit mysterious (to me at least) when $M$ is not a group. There is the old theorem of MacDuff and Segal (I believe) that any connected homotopy type can be expressed as $BM$ of a discrete monoid $M$. So, I'm not sure if information is lost so much as the theory of monoids is deep and not well-understood. – Justin Young Jun 11 '15 at 11:13

0 Answers0