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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 $\partial f_i = k_i\cdot e_i$.

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4 Answers

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 $c_g - c_h + c_k - c_{g-h+k}$.

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Yes, but you could also pick a presentation of $G$ and run that through the Dold-Kan machine, which might give a smaller model. –  Tim Porter Apr 13 '11 at 17:24
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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. –  Daniel Moskovich Apr 13 '11 at 20:29
    
@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. –  André Henriques Apr 13 '11 at 20:58
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+1 for $n$-tuply stacked arrows. –  Theo Johnson-Freyd Apr 14 '11 at 1:05
    
@André Good point. I think it is true that it could be used to give a smaller model. –  Tim Porter Apr 14 '11 at 6:05
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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.)

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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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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 $0$.

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