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As a motivation for the question, note that $\mathbb{C}P^\infty$ is at the same time a $BU(1)$ and a $K(\mathbb{Z},2)$; therefore, $H^2(X,\mathbb{Z})$ classifies complex line bundles on a space $X$. By the same line of arguments, $BO(1) = \mathbb{R}P^\infty = K(\mathbb{Z}/2\mathbb{Z},1)$, hence real line bundles are classified by $H^1(X,\mathbb{Z}/2\mathbb{Z})$. Further examples, with a somehow trivial flavour, can be obtained as follows: for any discrete group $G$ we have $BG = K(G,1)$. From the opposite point of view, we could say that for any group $\pi$, $K(\pi,1) = B\pi_{disc.}$. This gives rise to a lot of examples and I wonder:

Can we classify the triples $(G,\pi,n)$ of sufficiently nice topological groups1) $G$, arbitrary groups $\pi$ and positive integers $n$ such that $BG = K(\pi,n)$?

1) Please replace sufficiently nice topological groups by whatever setting you can deal with such that the question is interesting. It would be nice to cover the classical compact Lie groups. It will depend on the setting which groups we want to consider isomorphic.

As Tyrone pointed out in the comments, the question is essentially equivalent to the question which topological groups are Eilenberg-MacLane spaces.

Here are some observations, resulting in the homotopical classification:

  • If $G$ is commutative and discrete or an $E_n$-space, so that the $n$-th iteration $B^nG = B(B^{n-1}G)$ exists, then we have $K(G,n) = B^nG$ as shown here.
  • As discussed in the comments to this question, if $BG$ and $BH$ are homotopy equivalent, then $G$ and $H$ are homotopy equivalent as H-spaces.

Combining these two observations, we conclude that if $BG = K(\pi,n)$, then, understanding $\pi$ as a discrete group, $G$ is H-space homotopy equivalent to $B^{n-1}\pi$. Thus, up to homotopy, we get the complete list as $(K(\pi,n-1),\pi,n)$, but I was hoping to get something finer.

Ben
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  • I don'y understand your question. Your classification is just the triples triples $(K(\pi,n-1),\pi,n)$ with $\pi$ a discrete group and abelian if $n\geq 2$. – Tyrone Jul 06 '17 at 13:25
  • Dear @MikeMiller, you're right, this was stupid. Thank you. I'll remove this sentence. – Ben Jul 06 '17 at 14:01
  • No worries, I deleted my comment. –  Jul 06 '17 at 14:03
  • Dear @Tyrone, this gives a list of all possibilities up to homotopy, which is a reasonable interpretation. But I would like to read the question from the opposite point of view: given a topological group $G$, how do I know if it's a $K(\pi,n-1)$? – Ben Jul 06 '17 at 14:07
  • If this in infeasible, I would be happy to accept an answer restricting to classical Lie groups. – Ben Jul 06 '17 at 14:08
  • E-M spaces are only defined up to homotopy. If $BG\simeq K(G,n)$ then $G\simeq K(G,n-1)$, so $G$ is an E-M space. – Tyrone Jul 06 '17 at 14:12
  • Yes, if $BG$ is Eilenberg-MacLane, then so is $G$, but not each topological group is an Eilenberg-MacLane space, is it? So, if you want to put it like this, it can be rephrased as asking which topological groups are Eilenberg-MacLane. – Ben Jul 06 '17 at 14:16
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    @Ben For $n\geq 2$ and $\pi$ non-trivial $K(\pi, n)$ is never finite-dimensional. (for the $n=1$ case) Assume $G$ is connected (any discrete $G$ obviously works). As the universal cover of your group will always be contractible, the only groups you have are ones which are covered by groups diffeomorphic to $\Bbb R^n$ (as those are the only contractible Lie groups). – PVAL-inactive Jul 06 '17 at 14:30
  • Thank you, @PVAL-inactive, this indeed reduced the list of candidates a lot. – Ben Jul 06 '17 at 14:34
  • @PVAL-inactive: Do you have a reference for the fact that $K(\pi, n)$ is not finite-dimensional when $n \geq 2$ and $\pi$ is non-trivial? – Michael Albanese Feb 22 '18 at 20:25
  • @MichaelAlbanese I've seen a variety of proofs for this when $\pi$ is finitely generated (you can just use the fundamental theorem of finitely generated groups and compute the cohomology rings of $K(\Bbb Z,n)$ and $K(\Bbb Z/p,n)$. There is an argument given here https://math.stackexchange.com/questions/84422/finite-dimensional-eilenberg-maclane-spaces . If $G$ is is not finitely generated I don't know if its true (I may have had a reference when I wrote that comment but do not know of one now). – PVAL-inactive Feb 26 '18 at 05:23
  • @MichaelAlbanese I think you can also (in the the finitely generated case) appeal to the nontriviality of certain cohomology operations and the correspondence between cohomology operations and elements of the cohomology of $K(\pi,n)$, but I didn't dig through the proof for each case. – PVAL-inactive Feb 26 '18 at 05:25
  • In re:As discussed in the comments to this question, if and are homotopy equivalent, then . And much more: as A_\infty spaces and H are homotopy equivalent as H-spaces. – Jim Stasheff Oct 13 '20 at 20:53

1 Answers1

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It's time to get this off the unanswered list; it was essentially answered in the comments. As user PVAL-inactive noted, a $K(\pi,n)$ is infinite-dimensional as soon as $n\geq 2$. Therefore, restricting to classical groups (hence, finite-dimensional manifolds) we may consider the question which of classical Lie groups are $K(\pi,1)$s or $K(\pi,0)$s.

Suppose $G$ is a $K(\pi,0)$. Then $G\simeq \pi_{\text{disc.}}\cong \pi_0(G)$ and almost all the classical groups with normalised determinant are connected, in fact, all but $\mathrm{SO}(p,q)$ with $p,q\geq 1$, which has two components. Since $\pi$ can be assumed to be non-trivial, we conclude that for those groups, $\pi$ has to be the group of order two. For the real classical groups, not normalising the determinant results in a product of $\pi_0$ with $\mathbb{Z}/2\mathbb{Z}$. Therefore, we get:

  • The groups with two contractible components, $\mathrm{Spin}(1) = \mathrm{O}(1)\simeq\mathrm{GL}(1,\mathbb{R})$, which give a $K(\mathbb{Z}/2\mathbb{Z},0)$
  • and $\mathrm{O}(1,1)$, which is a $K(\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z})$.

In particular, $B\mathrm{O}(1)$ is a $K(\mathbb{Z}/2\mathbb{Z},1)$, as noted in the question text, resulting in the fact that $H^1(X,\mathbb{Z}/2\mathbb{Z})$ classifies real line bundles, and (for what it’s worth) we see that $B\mathrm{O}(1,1)$ is a $K(\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z},1)$ and so $H^1(X,\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z})$ classifies $\mathrm{O}(1,1)$-principal bundles.

Now suppose $G$ is a $K(\pi,1)$. Then $G$ is connected and $\pi_1(G) = \pi$. Looking at the list reveals that the only possibility is $\pi = \mathbb{Z}$ and in fact we find $\mathrm{SL}(2,\mathbb{R})$, $\mathrm{SO}(2)$, $\mathrm{SO}(2,\mathbb{C})$, $\mathrm{Sp}(2,\mathbb{R})$, $\mathrm{SU}(1,1)$, $\mathrm{SO}^{*}(2)$ and $\mathrm{SU}(1)$, all homotopy equivalent to $\mathrm{U}(1) = S^1$, hence a $K(\mathbb{Z},1)$, in accordance with $B\mathrm{U}(1)$ being a $K(\mathbb{Z},2)$ as noted in the question text.

Ben
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