11
$\begingroup$

It is well-known that if $G$ is a discrete group, then $BG=K(G,1)$. I'm interested in comparing classifying spaces of topological groups with the classifying spaces of the same groups but equipped with the discrete topology. They will be quite different in general, that much is clear.

For instance, $S^1$ can be thought of as a topological group (the usual way) and as a discrete group, denoted by, say, $S^1_d$. On one hand we have that $BS^1=\mathbb{C}P^∞(=K(\mathbb{Z},2))$, and on the other we should have $K(S^1_d,1)$, which I guess we could construct by hand (as in Hatcher's book, for example), but other than that I can't really say anything about it. Perhaps they usually are quite messy.

What is known about this space? Or, more generally, about $K(G,1)$ where $G$ is infinite and discrete (and not $\mathbb{Z}$)? Are there any references about this?

Any thoughts on the relation between the classifying spaces of top. groups vs E-M spaces of the same (discrete) groups (if any) would be greatly appreciated.

Thanks!

On another matter: this is a cross-post of https://math.stackexchange.com/questions/480437/what-is-k-s1-1, which I asked quite some time ago, and didn't get any answers there. Should I delete that question?

|cite|improve this question
$\endgroup$
  • $\begingroup$ Why are you interested about $S^1$ with the discrete topology? What kind of property do you want to study for $K(G,1)$? For example are you interested in its homology? $\endgroup$ – Denis Nardin Apr 12 '14 at 18:08
  • $\begingroup$ Originally I started thinking about the fact that $BS^1$ is not $K(S^1,1)$, and then I got curious about what the E-M spaces of $G=S^1$ should look like, if there are any specific models for it. I guess that the answer to your first question would be "just curious". Any information (homology, cohomology, homotopy type) would be very welcome. Since I wasn't able to find any references, I thought of asking in it here (well, first at MSE and now here). $\endgroup$ – Aldo Guzmán Sáenz Apr 12 '14 at 18:17
  • 1
    $\begingroup$ I think "infinite and discrete and not $\mathbb{Z}$" is too broad for the question you actually want to ask. There are lots of reasonable $K(G, 1)$ in this category, e.g. aspherical manifolds. Maybe uncountable $G$? $\endgroup$ – Qiaochu Yuan Apr 12 '14 at 18:18
  • $\begingroup$ There is an explicit simplicial model for $K(G,1)$ for any discrete $G$ (this is what the construction 'á la Hatcher' implicitly does) and using that you can prove that the (co)homology of $K(G,1)$ is the same thing as the group (co)homology for $G$ and you can probably get informations for $S^1$ using the short exact sequence $1\to\mathbb{Z}/n\to S^1\to S^1\to 1$. That said in general I fear that these groups will be intractable for computations. $\endgroup$ – Denis Nardin Apr 12 '14 at 18:23
  • 7
    $\begingroup$ These are more natural things to consider than the comments seem to indicate. For a reasonable topological group $G$, the space $BG$ classifies principal $G$-bundles, while the space $BG^{\delta}$ classifies flat principal $G$-bundles (here $G^{\delta}$ is $G$ considered as a discrete group). So the cohomology of $BG^{\delta}$ consists of characteristic classes for flat bundles; obviously these of great interest. $\endgroup$ – Andy Putman Apr 12 '14 at 20:24
11
$\begingroup$

Let me just point out that if you're interested in, say, homology, then discrete $S^{1}$ is not as complicated as it might seem. The resulting invariants will be huge, of course, but one should be able to compute them explicitly.

The point is that $S^{1} = \mathbb{R} / \mathbb{Z}$ and $\mathbb{R} \simeq \bigoplus \mathbb{Q}$ (as an abelian group), thus $S^{1} \simeq (\mathbb{Q} / \mathbb{Z}) \bigoplus (\oplus \mathbb{Q})$. Any direct sum of groups can be written as a filtered colimit of its finite subsums and finite direct sums coincide with finite products.

Taking classifying spaces commutes with both filtered colimits and finite products and so the classifying space of discrete $S^{1}$ can be described in terms of classifying spaces of $\mathbb{Q} / \mathbb{Z}$ and $\mathbb{Q}$. You can use this to compute homology (it commutes with filtered colimits) by using Künneth formula to deal with products.

I imagine classifying spaces of $\mathbb{Q} / \mathbb{Z}$ and $\mathbb{Q}$ are not difficult to describe, as $\mathbb{Q} / \mathbb{Z}$ falls apart into a direct sum of $p$-torsion parts (which I imagine are limits of finite cyclic $p$-groups?) and $\mathbb{Q} = colim (\mathbb{Z} \rightarrow \mathbb{Z} \rightarrow \ldots)$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I see. I'll try to find out more about the classifying spaces of $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$, thanks. $\endgroup$ – Aldo Guzmán Sáenz Apr 13 '14 at 2:57
9
$\begingroup$

One place to start might be:

Milnor, J. On the homology of Lie groups made discrete. Comment. Math. Helv. 58 (1983), no. 1, 72–85. http://www.ams.org/mathscinet-getitem?mr=699007

|cite|improve this answer
$\endgroup$
  • 6
    $\begingroup$ The conjecture in this paper has recently been proved in most cases by F. Morel; see Corollary 2 of mathematik.uni-muenchen.de/~morel/FriedlanderMilnorNew.pdf $\endgroup$ – Andy Putman Apr 12 '14 at 20:20
  • 1
    $\begingroup$ Thank you for your links. I have started reading the paper by Milnor, and in the appendix there is a result (Lemma 8) that says that for $G$ compact, then the homomorphism $H_iBG^{\delta} \rightarrow H_iBG$ is zero for $i>0$ (with real or rational coefficients). Does this mean that one has to compute $H_i(BG^{\delta};\mathbb{Q})$ by other means? (I apologize if this question is addresed somewhere else in the paper, I haven't read it thoroughly. I will keep reading it, of course.) $\endgroup$ – Aldo Guzmán Sáenz Apr 13 '14 at 2:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.