# “Economic” Eilenberg-MacLane topological abelian groups

This might be regarded as a sequel to my previous "Economic" CW-structure for Eilenberg-MacLane spaces? However the content seems to be quite different.

I believe it is easy to prove that for any topological abelian group $A$ there is a model of $BA$ (that is, a space $X$ with $\Omega X$ homotopy equivalent to $A$) which is a topological abelian group; in particular, for any (discrete) abelian group $\pi$ and any $n\ge0$ there is a model for $K(\pi,n)$ which is a topological abelian group.

This latter can be actually quite easily seen using the Dold-Kan correspondence: take the chain complex with $\pi$ in degree $n$ and zero everywhere else; take the corresponding simplicial abelian group; take its geometric realization. Since the latter preserves products, it will be a topological abelian group.

However I somehow feel that this is not the most economic way to do it; in any case I do not quite "see" the result. In fact the only case when I know a satisfactory description (for me) is $K(\mathbb Z/2\mathbb Z,1)$ built from the infinite-dimensional sphere. It follows e. g. from the wonderful paper "Using the generic interval" by Gavin Wraith that $S^\infty$ has a topological $\mathbb F_2$-vector space structure. Since it is contractible, its quotient by any two element subgroup is a $K(\mathbb Z/2\mathbb Z,1)$ which is a topological elementary abelian 2-group.

I don't know if $S^\infty$ has any other topological abelian group structure which would turn the infinite lens spaces into subgroup quotients so as to obtain a $K(\mathbb Z/p\mathbb Z,1)$ topological abelian group. Is something like this known?

Infinite symmetric power of $S^n$ is a $K(\mathbb Z,n)$ (related discussion here on MO is in Why the Dold-Thom theorem?); it is a commutative topological monoid. Can it be made into a group? Is its group of quotients again a $K(\mathbb Z,n)$?

Also here on MO there is the question about H-space structure on infinite projective spaces where something similar is done but honestly speaking I could not understand from answers there whether there is a topological abelian group structure on, say, $\mathbb C\mathrm P^\infty$ or not. There, they also mention the page by John Baez about Classifying Spaces Made Easy but again, I could not quite find answers to my questions there.

Other than that, I've seen Stephan Stolz using the projective unitary group of an infinite-dimensional separable Hilbert space as a model for $K(\mathbb Z,2)$ (in "A conjecture concerning positive Ricci curvature and the Witten genus", Math. Ann. 304(1) 1996, page 795) but that is very-very nonabelian.

Not counting the circle as $K(\mathbb Z,1)$, and lots of nice $K(\pi,1)$s for (mostly) nonabelian $\pi$, these are basically the only ones that I know.

Is there any systematic construction of $K(\pi,n)$ topological abelian groups which would be minimal in the sense that they do not have contractible [closed] subgroups or contractible [continuous] quotients? Are there any restrictions on the resulting groups? For example, what kind of torsion they should have if $\pi$ is cyclic? Obviously the case when $\pi$ is finitely generated reduces to the investigation of cyclic $\pi$'s. What about, say, $\mathbb Q$? Are there any nice topological abelian $K(\mathbb Q,n)$ groups known?

For $\pi$ a (discrete) abelian group and $n \geq 1$, one can build an interesting model of a $K(\pi,n)$ using configuration spaces. Namely, for each $k \geq 0$ let $X_k$ be the space of configurations of $k$ points in $D^n$ (the closed $n$-ball), where the points are required to lie in the open ball ${\overset{\circ}{D^n}}$, and each point carries a label in $\pi$. Now construct a space $X$ as a suitable quotient of $$\coprod_{k \geq 0} X_k$$ where we allow points to collide (in which case we add their labels) or go to the boundary $S^{k-1} = \partial D^k$, in which case they vanish. It is possible to write down a formal definition of this process, but I will not make the effort to describe it here. The collision of points does not cause any problems since $\pi$ is abelian. (If instead of $\pi$ you wanted to plug in any space $Y$, this construction would only make sense if $Y$ is an $E_n$ algebra.) One can now check that the homotopy type of $X$ will be $K(\pi,n)$.

Now the group structure on $X$ is (up to homotopy) given by stacking two configurations next to each other, in a similar spirit to how the addition in homotopy groups is defined. But since $\pi$ is abelian and we do not have to worry about any issues that usually occur, I think (but I might be wring) that we could also define the addition by just putting two configurations on top of each other, possibly adding labels if points collide. That should give a honest model of $K(\pi,n)$ as a topological group.

The whole thing I described can be done in a far more general fashion, replacing the abelian group $\pi$ by an $E_n$-algebra and the disk $D^n$ by some $n$-dimensional (framed) manifold, it is then referred to by the name 'factorization homology'. There has been done quite some research on these kinds of things in the last decade.

• Is not this the same as $A[D^n]/A[S^{n-1}]$, where $A[X]$ is $A\otimes\operatorname{SP}^\infty(X)$, where $\operatorname{SP}^\infty$ is the infinite symmetric power? If so, I believe it is not the same as factorization structure. – მამუკა ჯიბლაძე Mar 3 '17 at 6:03
• In other words, $A[X]$ is determined by the following universal property: it is a topological abelian group with a continuous map $i:X\to A[X]$, and for any other continuous map $i':X\to A'$ to a topological abelian group, there is a unique continuous homomorphism $\varphi:A[X]\to A'$ with $i'=\varphi\circ i$. – მამუკა ჯიბლაძე Mar 3 '17 at 6:37

For a discrete abelian group $\pi$ the following simplicial abelian group models for $K(\pi,1)$ are isomorphic, I believe:

Simplicial abelian group whose normalized chain complex is $0\leftarrow \pi\leftarrow 0\leftarrow 0\leftarrow \dots$.

Nerve of category having one object and having morphisms $\pi$ (with obvious addition law since $\pi$ is abelian).

Universal example of simplicial abelian group $A$ equipped with a based map of simplicial sets from $S^1=\Delta^1/\partial$ to the simplicial set $Hom(\pi,A)$.

When realized as a topological abelian group, this becomes the universal example of a topological abelian group $A$ equipped with a continuous based map from the topological space $S^1$ to the space of homomorphisms from $\pi$ to $A$.

This latter space can also be described using configurations as in the previous answer.

In each of these constructions of $B\pi$, the universal cover $E\pi$ has its own corresponding description.

I believe that when $\pi$ has order two then $E\pi$ is homeomorphic to $S^\infty$, but I don't know if this abelian group structure on $S^\infty$ is the one referred to in the question.

$E\pi$ as a space, or even as a simplicial set, does not depend on the group structure in $\pi$; if you have a set $\pi$ then you can make $E\pi$ using the nerve of the category that has one object for each element of $\pi$ and one morphism between any two objects. If $\pi$ is nonempty then this is contractible, and if $\pi$ is a group then it gets an obvious free group action. It seems likely that if it is homeomorphic to $S^\infty$ when $\pi$ has two elements then the same is true when it has more than two (but finitely many) elements.

These don't seem to have the kind of strong minimality property that you are asking about: when $\pi$ has order two then there are plenty of contractible closed subgroups of $B\pi$ (for which the quotient is isomorphic to $B\pi$). But maybe that property is too much to wish for.

• Some questions, please? I don't understand two things: (1) You name three simplicial abelian groups; I don't see how the third one depends on $\pi$. (2) At least when $\pi$ is very large, presumably $E\pi$ cannot be homeomorphic to $S^\infty$ by cardinality reasons. Are there some similar "large spheres", something like $S^{\text{some cardinal}}$ which would work? – მამუკა ჯიბლაძე Mar 3 '17 at 18:02
• Also a separate question concerning the last paragraph (although this probably should go into the main question). Consider the category with objects all $K(\pi,n)$ topological abelian groups and morphisms continuous homomorphisms which induce isomorphism of homotopy groups; I believe this means having contractible kernels and cokernels. Do you expect that this category does possess neither initial nor terminal object? Could it have weak terminals or initials (existence without uniqueness)? – მამუკა ჯიბლაძე Mar 3 '17 at 18:02
• Oh sorry one more thing. Say $\pi$ is not discrete; one can still form $E\pi$ as the geometric realization of the simplicial space $$\pi\begin{smallmatrix}\leftarrow\\\rightarrow\\\leftarrow\end{smallmatrix} \pi^2\begin{smallmatrix}\leftarrow\\\rightarrow\\\leftarrow\\\rightarrow\\\leftarrow\end{smallmatrix}\pi^3\begin{smallmatrix}\leftarrow\\\vdots\\\leftarrow\end{smallmatrix}\cdots;$$does this have chance to be something like infinite-dimensional sphere of sorts?? This would say something about $K(\pi,n)$s with $n>1$... – მამუკა ჯიბლაძე Mar 3 '17 at 18:10
• What you say about contractible kernels and cokernels is not correct. You can make a chain map from a contractible chain complex to another contractible complex such that the kernel and the cokernel have nontrivial homology. (The image will have to also have nontrivial homology.) – Tom Goodwillie Mar 3 '17 at 19:23
• Yes, the $E\pi$ that I was thinking of has an abelian group structure if $\pi$ is an abelian group. Think of the exact sequence of chain complexes $0\to (\pi\leftarrow 0) \to (\pi\leftarrow\pi)\to (0\leftarrow\pi)\to 0$. – Tom Goodwillie Mar 3 '17 at 19:25