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Let $G$ be a topological group, then it has a classifying space $BG$.

When is $BG$ a topological group?

My motivation for asking this question is that I was thinking about the $B$-analogue of my previous question and realised it's not clear whether $B^kG$ is defined for $k > 1$.

One case where I know it is true is when $G$ is an Eilenberg-Maclane space, i.e. $G = K(H, n)$. First of all, they are topological groups (see here) and by considering the long exact sequence in homotopy associated to the fibre bundle $G \to EG \to BG$, we see that $BG \cong K(H, n+1)$ and is therefore a topological group. It is also true when $G$ is a discrete group for similar reasons.

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Michael Albanese
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1 Answers1

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Up to homotopy, the answer is if and only if $G$ has the additional structure of an $E_2$ space. (This is exactly the structure that a double loop space has; we need this since a topological group has a classifying space, so if $BG$ has a classifying space then $G$ is a double loop space.)

This is a certain higher categorical version of abelianness. If $G$ is discrete, it must be abelian (by the Eckmann-Hilton argument, which is really about $E_2$ structures), but there are interesting examples beyond abelian or topological abelian groups, such as (again, up to homotopy) the stable unitary group $U$.

The easiest case past the discrete case is the case that $G$ is a groupoid; then $E_2$ means $G$ is a braided monoidal groupoid where every object is invertible. These have a cohomological classification.

Qiaochu Yuan
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  • Do you have a reference for what an $E_2$ space is? Also, would I be right in guessing that if $G$ is an $E_k$ space, $B^kG$ is defined? – Michael Albanese Apr 05 '16 at 22:46
  • @Michael: there are several ways to say what it is on a point-set level, although it is really a homotopical thing; morally it means that $G$ has two monoid structures (up to coherent homotopy) which commute (up to coherent homotopy). For a double loop space the two monoid structures are loop composition in the two loop directions. Traditionally you can use gadgets called $E_2$ operads; May's Geometry of Iterated Loop Spaces is the canonical reference (May introduced operads in this paper for this purpose). Nowadays other ways of describing this structure are also known. And yes. – Qiaochu Yuan Apr 06 '16 at 05:44
  • So if $G$ is a topological group satisfying $G \cong B^kG$ for some $k \geq 1$, then $G$ must be an $E_{\infty}$ space? If I'm not mistaken, by using the long exact sequence of homotopy applied to the fibration $G \to EG \to BG$, one can show that any topological group $G$ satisfying $G \cong B^kG$ must be weakly contractible. – Michael Albanese Apr 06 '16 at 15:18
  • @Michael: yes and yes. The problem is that $G$ cannot have a lowest nonzero homotopy group, since taking $B$ shifts up the homotopy groups by one. – Qiaochu Yuan Apr 06 '16 at 15:41
  • @Michael: incidentally, another name for $E_{\infty}$ space, at least up to a mild additional condition of being "grouplike," is "infinite loop space." You'll get a lot more search results this way. – Qiaochu Yuan Apr 06 '16 at 16:54
  • I saw this on nLab, but I didn't understand what it means to be grouplike in this situation. – Michael Albanese Apr 06 '16 at 16:56
  • @Michael: it just means that $\pi_0$ is a group. – Qiaochu Yuan Apr 06 '16 at 18:23
  • $\pi_0$ is a group with the multiplication induced by the $E_\infty$ multiplication (which always gives a commutative monoid). –  Apr 07 '16 at 09:39