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Let $G$ be a topological abelian group. Its classifying space $BG$ is (at least sometimes) also a topological group. For example if $G$ is a finite abelian group, higher Eilenberg-MacLane spaces are $K(G,n)=B^nG=B(B^{n-1}G)$, implying that $B^{n-1}G$ must be a group.

  1. When is it true that $BG$ has a group structure?

  2. How can we describe the multiplication in $BG$ in terms of the multiplication in $G$?

I would like to understand this directly from the realization of $BG$ as the quotient $EG/G$, where $EG$ is a contractible space with a free a action of $G$. Or at least I would like to see the explicit connection of that model with a one more suitable for providing a product structure.

Andrea Antinucci
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  • Related: https://math.stackexchange.com/a/1729555/19006 – diracdeltafunk May 23 '23 at 22:01
  • Your edit doesn't really address the close reason. It would be helpful if, for example, you told us where you heard this - in a research seminar, or you think you read it in a book which you can't find anymore? Maybe you could also say something about why you care about this. – user1729 May 31 '23 at 07:47
  • Thank you for your comment, I added some clarification on this point and I made the question more precise – Andrea Antinucci May 31 '23 at 08:02

2 Answers2

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The space $BG$ has a model as a labeled configuration space. The points of $BG$ consist of configurations of points in $\mathbb{R}$ labeled with elements of $G$ topologized such that if two points collide we multiply their values according to the order of the collision. Further, elements labeled by the identity are allowed to disappear or reappear, and points in a configuration may disappear to infinity. It is not difficult to argue this is homeomorphic to the nerve of the one object category associated to G.

If $G$ is abelian, then the overlay map $BG \times BG \rightarrow BG$ is well defined since if points in our two configurations coincide, we may add them in whatever order we see fit and get a single output. This operation makes $BG$ into an abelian monoid.

Connor Malin
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  • This is a definition of $BG$ that I didn't know. I only know the definition as $BG=EG/G$. How one can see that the two things coincide? – Andrea Antinucci May 24 '23 at 09:06
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    Imagine $\mathbb{R}$ as $(0,1)$, you can get the coordinates of a point in an $n$-simplex out of a configuration of $n$ points by taking the consecutive distances between points (including the end points). The labels of the points on the simplex are given by the labels of the configuration. This defines a map to the nerve construction which is well defined based on the fact collisions multiply points and identities can appear, corresponding to the face and degeneracy maps in the nerve. This is a homemorphism. – Connor Malin May 24 '23 at 13:18
  • Does this work for non-discrete $G$? – diracdeltafunk May 24 '23 at 22:27
  • @diracdeltafunk Yes, it works; I think due to Segal. – Connor Malin May 25 '23 at 16:19
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Basically this depends on how you set up the classifying space. A good choice is to define $BG$ to be the geometric realization of the nerve of $G$. Then $B$ is a monoidal functor from topological groups to topological spaces, and so it sends group objects (topological abelian groups) to group objects (topological groups).

diracdeltafunk
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