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The Eilenberg-Maclane space $K(\mathbb{Z}/2\mathbb{Z}, 1)$ has a particularly simple cell structure: it has exactly one cell of each dimension. This means that its "Euler characteristic" should be equal to $$1 - 1 + 1 - 1 \pm ...,$$ or Grandi's series. Now, we "know" (for example by analytic continuation) that this sum is morally equal to $\frac{1}{2}$. One way to see this is to think of $K(\mathbb{Z}/2\mathbb{Z}, 1)$ as infinite projective space, e.g. the quotient of the infinite sphere $S^{\infty}$ by antipodes. Since $S^{\infty}$ is contractible, the "orbifold Euler characteristic" of the quotient by the action of a group of order two should be $\frac{1}{2}$.

More generally, following John Baez $K(G, 1)$ for a finite group $G$ should be "the same" (I'm really unclear about what notion of sameness is being used here) as $G$ thought of as a one-object category, which has groupoid cardinality $\frac{1}{|G|}$. In particular, $K(\mathbb{Z}/n\mathbb{Z}, 1)$ should have groupoid cardinality $\frac{1}{n}$. I suspect that $K(\mathbb{Z}/n\mathbb{Z}, 1)$ has $1, n-1, (n-1)^2, ...$ cells of each dimension, hence orbifold Euler characteristic

$$\frac{1}{n} = 1 - (n-1) + (n-1)^2 \mp ....$$ Unfortunately, I don't actually know how to construct Eilenberg-Maclane spaces...

Question 1a: How do I construct $K(\mathbb{Z}/n\mathbb{Z}, 1)$, and does it have the cell structure I think it has? (I've been told that one can write down the cell structure of $K(G, 1)$ for a finitely presented group $G$ explicitly, but I would really appreciate a reference for this construction.)

Question 2: $K(\mathbb{Z}/2\mathbb{Z}, 1)$ turns out to be "the same" as the set of all finite subsets of $(0, 1)$, suitably interpreted; the finite subsets of size $n$ form the cell of dimension $n$. Jim Propp and other people who think about combinatorial Euler characteristic would write this as $\chi(2^{(0, 1)}) = 2^{\chi(0, 1)}$. Is it true more generally that $K(\mathbb{Z}/n\mathbb{Z}, 1)$ is "the same" as the set of all functions $(0, 1) \to [n]$, suitably interpreted?

Question 3: What notion of "sameness" makes the above things I said actually true?

Question 4: Let $G$ be a finite group and let $K(G, 1)$ be constructed using the standard construction I asked about in Question 1. If $c_n$ denotes the number of cells of dimension $n$, let $f_G(z) = \sum_{n \ge 0} c_n z^n$. Can $f_G$ always be analytically continued to $z = -1$ so that $f_G(-1) = \frac{1}{|G|}$?

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It's the infinite lens space. I believe it's in Hatcher's book. The cell structure you build inductively just like with real projective space. Q2, see this recent thread:… Q3: see the linked thread. – Ryan Budney Apr 2 '10 at 18:35
For the lens space, see Hatcher, Example 2.43. – Dylan Thurston Apr 2 '10 at 18:40
Hum, you're right; I should've checked Hatcher more thoroughly before asking this question. – Qiaochu Yuan Apr 3 '10 at 1:49
up vote 13 down vote accepted

There are multiple possible cell structures on K(Z/n,1).

One is generic. For any finite group G there is a model for BG that has (|G|-1)k new simplices in each nonzero degree k. This is the standard simplicial bar construction of K(G,1). This gives you that BG has Euler characteristic 1/|G|, if you like.

One is more specific. There is another cell structure on K(Z/n,1), viewing it as a union of generalized lens spaces, that has exactly one cell in each degree. This is a topological avatar of the "simple" resolution of Z by free Z[Z/n]-modules. Obviously this doesn't give you the Euler characteristic argument you're seeking - one needs to keep track of more intricate information about the cell attachments in order to extract something.

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The answer to something like question 4 is "yes". It not only has an analytic continuation to $-1$, it has a unique such. In fact, it's a rational function. Something similar is true not only for finite groups but for "most" finite categories. See this paper, especially Example 2.4.

(I say "something like question 4" because I'm taking $c_n$ to be the number of nondegenerate $n$-simplices in the nerve of $G$, which I think isn't quite what you have in mind.)

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