Recall that an Eilenberg-Maclane space $K(G, n)$ is characterized by $\pi_i(K(G,n)) = G$ if $i=n$ and is trivial otherwise. (Of course $G$ should be abelian if $n>1$.)

Let $G$ be a finite abelian group.

Below I describe cell complexes $X_1$ and $X_2$ with $\pi_2(X_i) = G$ and $\pi_0(X_i)$ and $\pi_1(X_i)$ both trivial. By standard results it is possible to add 4-cells to $X_i$ to kill off $\pi_3$, then add 5-cells to kill off $\pi_4$, and so on.

My questions:

(1.i) Does there exist in the literature an explicit description of the 4- and 5-cells one would need to add to $X_i$ in order to turn it into a $K(G,2)$? (I'm only interested in dimensions 4 and 5, not higher.)

(2) More generally, are there explicit descriptions of $K(G,2)$ in the literature? (I'm already aware of making $K(G, 1)$ into a group and then applying the bar construction.)

Definition of $X_1$: A single 0-cell. A 2-cell $c_g$ for each element $g\in G$. A 3-cell $d_{g,h}$ for each $(g,h)\in G\times G$, with $\partial d_{g,h} = c_g + c_h - c_{gh}$.

(This starts out similarly to a standard construction of $K(G, 1)$, but the higher dimensional cells will necessarily be more complicated. Obvious candidates for the boundaries of 4-cells would include $d_{g,h} - d_{fg,h} + d_{f,gh} - d_{f,g}$ for all $(f,g,h)\in G\times G\times G$, and also Hopf maps to the 2-cells $c_g$ for each $g$, and also $d_{g,h} - d_{h,g} + x$, where $x$ is a map to $c_h\cup c_h$ which exhibits the commutativity of $\pi_2(c_g\cup c_h)$.)

Definition of $X_2$: Let $G = \mathbb Z/k_1 \times\cdots\times \mathbb Z/k_m$, a prodict of cyclic groups. $X_2$ has $m$ 2-cells $e_1,\ldots, e_m$ and $m$ 3-cells $f_1,\ldots, f_m$, with $\partial f_i = k_i\cdot e_i$.


To expand on Tim's answer:

The simplicial group whose realization is $K(G,2)$ looks as follows:

$$ * \begin{matrix} \stackrel{\displaystyle\leftarrow} \leftarrow \end{matrix} * \begin{matrix} \stackrel{\displaystyle\leftarrow} {\stackrel{\displaystyle\leftarrow} \leftarrow} \end{matrix} G \begin{matrix} \stackrel{\displaystyle\leftarrow} {\stackrel{\displaystyle\leftarrow} {\stackrel{\displaystyle\leftarrow} \leftarrow}} \end{matrix} G^3 \begin{matrix} \stackrel{\displaystyle\leftarrow} {\stackrel{\displaystyle\leftarrow} {\stackrel{\displaystyle\leftarrow} {\stackrel{\displaystyle\leftarrow} \leftarrow}}} \end{matrix} G^6 \begin{matrix} \stackrel{\displaystyle\leftarrow} {\stackrel{\displaystyle\leftarrow} {\stackrel{\displaystyle\leftarrow} {\stackrel{\displaystyle\leftarrow} {\stackrel{\displaystyle\leftarrow} \leftarrow}}}} \end{matrix} G^{10} \begin{matrix} \stackrel{\displaystyle\leftarrow} {\stackrel{\displaystyle\leftarrow} {\stackrel{\displaystyle\leftarrow} {\stackrel{\displaystyle\leftarrow} {\stackrel{\displaystyle\leftarrow} {\stackrel{\displaystyle\leftarrow} \leftarrow}}}}} \end{matrix} G^{15}\ldots $$

and has the group of closed $G$-valued 2-cocycles on $\Delta^n$ in simplicial degree $n$.

The first place in which this differs from Kevin's proposed model is in dimension 3, where there are $G\times G\times G$ cells, and the boundary of a cell $d_{g,h,k}$ is given by $c_g - c_h + c_k - c_{g-h+k}$.

  • $\begingroup$ Yes, but you could also pick a presentation of $G$ and run that through the Dold-Kan machine, which might give a smaller model. $\endgroup$ – Tim Porter Apr 13 '11 at 17:24
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    $\begingroup$ To add a citation: I believe this description is due to Adrien Douady, and is in Expos\'e 9 of the 1958/59 S\'eminaire Henri Cartan vol. 1. It's my favourite (therefore the only one I ever remember) because it makes it most obvious that $K(G,n)$ indeed represent cohomology. $\endgroup$ – Daniel Moskovich Apr 13 '11 at 20:29
  • $\begingroup$ @Tim Porter: picking a presentation of $G$ would produce a bigger model. However, that bigger model would be a free $\mathbb Z$-module in each simplicial degree. $\endgroup$ – André Henriques Apr 13 '11 at 20:58
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    $\begingroup$ +1 for $n$-tuply stacked arrows. $\endgroup$ – Theo Johnson-Freyd Apr 14 '11 at 1:05
  • $\begingroup$ @André Good point. I think it is true that it could be used to give a smaller model. $\endgroup$ – Tim Porter Apr 14 '11 at 6:05

One possible way is to take a chain complex model of a K(G,2), pass via Dold-Kan to the corresponding simplicial Abelian group and then take the classifying space of that. (Each part is explicit, but is likely to generate some extra redundant cells.)


As an explicit construction (not obviously as a CW complex one, but I think it's one, though), I would suggest the Dold-Thom construction. Unfortunately I can't find a link to the precise construction I have in mind (can't reach wikipedia right now!?).

Simply put, $K(G,2)$ is obtained as a component of the space $G[S^2]$ of (finite) formal combination of points on $S^2$ with $G$ coefficients, with a topology such that e.g. $ax+by\to (a +b)z$ when $x,y\to z$ in $S^2$ and $a,b$ in $G$.

Then as $K(G,2)$ you can take the component where the sum of all coefficients is $0$ ("neutral configurations of particles"). This is an abelian topological group.

I think this is due to Dusa McDuff, Configuration spaces of positive and negative particles, Topology 14 (1975), 91-107, elaborating on a slighltly different construction by Dold and Thom. The original construction was an infinite symmetric product with of $S^2$, meaning the topological monoid of combinations with nonnegative coefficients, and a base point $*$ identified to $0$.


This is a comment relating the other answers more than anything else.

Following are three isomorphic simplicial abelian groups which are Eilenberg-Maclane spaces $K(G,n)$.

  • The result of applying the Dold-Kan correspondence to the chain complex which has a copy of $G$ in degree $n$ and zero in all other degrees. This is an instance of the answer given by Tim Porter.

  • The result of taking the levelwise tensor product of $G$ with the free simplicial abelian group on the pointed simplicial set $\Delta^n/\partial\Delta^n$. This is an instance of the simplicial variant of the answer by BS.

  • The simplicial abelian group $\tilde{H}^{n-1}(sk^{n-1}\Delta^\bullet,G)$, where $\tilde{H}^\ast$ is reduced cohomology, $sk^k$ denotes the k-th skeleton, and $\Delta^\bullet$ stands for the standard functor from the simplicial category $\Delta$ into topological spaces.

You can prove that these are all isomorphic by showing that the normalized chain complex associated to each of them is the chain complex concentrated in degree $n$ with $G$ at that level. This is obvious for the first one, and easy for the second one. It is also not terribly difficult for the last one: it starts with observing that $sk^k \Delta^n$ is homotopy equivalent to a wedge of $\binom{n}{k+1}$ spheres of dimension $k$ (the number $\binom{n}{k+1}$ of spheres is also the number of monomorphisms of $\Delta^k$ into $\Delta^{n-1}$).

This also recovers the number of copies of $G$ appearing in Andre's answer: $\binom{n}{2}$ copies of $G$ for the $n$-simplices. This binomial coefficient is also easily seen to be the number of non-degenerate $n$-simplices in $\Delta^2/\partial\Delta^2$ for $n>0$, so the same answer could be easily obtained from the second description above.


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