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What is the CW-complex of Eilenberg-MacLane space $K(\mathbb{Z}_2,2)$?

What is the CW-complex of Eilenberg-MacLane space $K(\mathbb{Z}_n,d)$?

What is the CW-complex of Eilenberg-MacLane space $K(\mathbb{Z}_n\times \mathbb{Z}_m,d)$?

For example, I like to know the number of cells in each dimensions, and the chain complex formed by those cells, so that one can compute cohomology. (I like to know the chain complex, in addition to the cohomology).

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    $\begingroup$ There is a standard construction for these spaces. What exactly is not satisfactory about that construction to you? Maybe you should include more details as to why exactly you are interested in these spaces. $\endgroup$ – Johannes Hahn Feb 1 '18 at 14:05
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    $\begingroup$ mathoverflow.net/questions/180236/… $\endgroup$ – Thomas Rot Feb 1 '18 at 14:09
  • $\begingroup$ I like to know the results of those standard constructions. How many cells in each dimension? How the cells are related? What is the resulting chain complex? $\endgroup$ – Xiao-Gang Wen Feb 1 '18 at 14:10
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    $\begingroup$ mathoverflow.net/questions/180236/… is very nice, but not explicit enough. $\endgroup$ – Xiao-Gang Wen Feb 1 '18 at 14:20
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    $\begingroup$ There is no such thing as "the CW-complex". Take a CW-complex representing $K(\mathbb{Z}_n, d)$, cross it with $[0,1]$ -- boom, you've got a new one. $\endgroup$ – Najib Idrissi Feb 2 '18 at 16:23
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There is a completely explicit simplicial set realizing to $K(A,n)$ for each $n$, coming from the Dold-Kan correspondence.

It is defined as $$\bar{A}[S^n]=\mathrm{ker}(A[S^n]\to A[*])$$ (the kernel is taken levelwise, and $A[X]$ for each simplicial set $X$ is obtained by taking the free $A$-module levelwise). Here I let $S^n=\Delta^n/\partial \Delta^n$.

I don't think you can get more explicit than that.

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This is mostly an addendum to Denis' answer.

The "standard model" of the Eilenberg-Mac Lane space $K(A,n)$ that Denis mentioned has the following number of $k$-dimensional cells: $$ \sum_{m=0}^k \binom{k}{m} (-1)^{k-m} |A|^{\binom{m}{n}} $$

When $n=1$, this is $$ \sum_{m=0}^k \binom{k}{m} (-1)^{k-m} |A|^m = (|A|-1)^k $$ by the binomial theorem, but I don't know a closed-form formula for general $n$. For a chosen $A$ and $n$, there is usually a model with fewer cells, but these have to be constructed "by hand" and there are not many systematic answers.

If you want more of this type of method, you can look at Eilenberg-Mac Lane's papers "On the groups $H(\Pi,n)$" (volume I, volume II, volume III), which appeared in the Annals of Mathematics back in the early 1950s. These cochain-level methods for computing are very difficult and have been heavily supplemented by less explicit machinery, such as Serre's methods that other commenters have mentioned.

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This article of C. Berger gives a general principle to compute the number of $n+k$-cells of $K(\pi,n)$ for an abelian group $\pi$ in terms of "generalised Fibonacci numbers" (see section 4.10 in loc. cit.). The CW-complex structure comes from the language of $n$-categories, which is used to describe iterated loop spaces (the purpose of the article is mainly to make sense of this last sentence).

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The model for $K(Z,2)$ is the infinite-dimensional complex projective space $\mathbb CP^\infty$ with a single cell in each even dimension. To obtain $K(Z_2,2)$ from that you would have first to eliminate twice the 2-dimensional generator by attaching a cell, etc. In very low dimensions you can still get specific information about the number of cells, but to go further you would need detailed information about homotopy groups of spheres which quickly gets hairy, so "not explicit enough" will have to be good enough already in this case.

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It is not too hard to see that for general groups $K(G\times H,d)\cong K(G,d)\times K(H,d)$. So three follows from the first two with the product CW structure.

Edit:

Computing the cohomology of the $K(G,n)$'s on the nose is difficult I think.

I think I read somewhere that the full computation of the cohomology was done by Serre (edit Cartan apparently).

To get a grasp on the cohomology you look at the pathspace fibration:

$\Omega(K(g,n))\rightarrow PK(g,n)\rightarrow K(g,n)$

and the associated Leray-Serre spectral sequence. As the loopspace of $K(G,n)$ is a model of $K(G,n-1)$ and the pathspace is contractible one gets an inductive description of the cohomology.

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    $\begingroup$ I don't think Serre computed the integral cohomology of a general $K(G,n)$. It is possible this was done by Cartan in one of his papers, but I am skeptical a general answer is even known. $\endgroup$ – Denis Nardin Feb 1 '18 at 16:29
  • $\begingroup$ mathoverflow.net/a/24759/12156 , but indeed it seems I misremembered $\endgroup$ – Thomas Rot Feb 1 '18 at 16:54
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    $\begingroup$ Evidently I must pay penance for having doubted Cartan's skill and determination $\endgroup$ – Denis Nardin Feb 1 '18 at 17:28

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