# “Economic” CW-structure for Eilenberg-MacLane spaces?

The only really "economic" cell structures for $K(\pi,n)$'s that I know is the one with a single cell in each dimension for $K(\mathbb Z/n\mathbb Z,1)$ and the one with a single cell in each even dimension for $K(\mathbb Z,2)$.

Vaguely I remember attending a talk where some lower bounds on numbers of cells in each dimension for Eilenberg-MacLane spaces of cyclic groups were given. Now that I needed this again, I only could find the text "Small CW-models for Eilenberg-MacLane spaces" by Clemens Berger, which contains among other things a cell complex for $K(\mathbb Z/2\mathbb Z,2)$ with 1,0,1,1,2,3,5,8,13,21,... cells (probably the Fibonacci sequence) and a cell complex for $K(\mathbb Z/2\mathbb Z,3)$ with 1,0,0,1,1,2,4,7,13,24,... cells.

What is the current state of the art?

Are there for example any manageable dimensionwise finite cell structures for $K(\mathbb Z,3)$ or $K(\mathbb Z,4)$ or $K(\mathbb Z/n\mathbb Z,2)$ known?

(By manageable I mean... well it is up to you :) )

Is there a geometric construction similar to the real/complex/quaternionic projective spaces known for any other spaces aside of $BO(1)=\mathbb R P^\infty=K(\mathbb Z/2\mathbb Z,1)$, $BU(1)=\mathbb C P^\infty=K(\mathbb Z,2)$ and $B(\textrm{unit quaternions})=\mathbb H P^\infty$? (The latter is of course not any Eilenberg-MacLane space but...)

(Well there are also Grassmanians with their Schubert cells but I mean something as Eilenberg-MacLaneish as possible :) )

Are there any interesting lower bounds on the numbers of cells of each given dimension in a $K(\pi,n)$ known?

And yes of course there is the whole ocean of nonabelian groups with very nice finite-dimensional classifying spaces but I mostly mean $n>1$ and, respectively, abelian groups...

As Jeff Strom and Will Sawin indicate in their answers, homology groups provide the lower bounds, and it is more or less straightforward to arrange for a cell complex with prescribed homology, with minimal possible numbers of cells. Still the question remains (for me) whether such an "absolutely minimal" CW-complex with the correct homotopy type exists.

You can build them using a homology decomposition and read off the number of cells in each dimension from the homology groups.

For any simply-connected space, this will give you a construction by iterated cofiber sequences $M_n \to X(n-1) \to X(n)$ where $M_n$ is a Moore space and $X = \mathrm{colim} X(n)$. If you construct $M_n$ efficiently and give $X(n)$ the inherited CW structure, then this will be the absolute fewest cells possible in each dimension to construct a space with the required homology.

By an efficient construction of $M = M(G,n)$ where $G$ can be generated by $k$ elements but not fewer, and an exact sequence $0\to F_1\to F_0\to G\to 0$ where $F_0$ is free of rank $k$. Then we topologize this to get a cofiber sequence $W_1 \to W_0 \to M$, where $W_0$ is a wedge of $k$ $n$-spheres and $W_1$ is also a wedge of $n$-spheres.

• Yes... but that's not very constructive. – André Henriques Sep 6 '14 at 17:23
• True, but I don't think the question asked for constructive. – Jeff Strom Sep 6 '14 at 17:25
• @JeffStrom well "manageable" is vague enough but still I would like something explicit. But I agree, both your and another answer indicate at least a clear lower bound. – მამუკა ჯიბლაძე Sep 6 '14 at 19:12
• JUst to clarify: The usual construction of a homology decomposition, as in my algebraic topology book for example, shows that for each simply-connected space $X$ with finitely generated homology groups there is a CW complex $Y$ and a weak homotopy equivalence $Y\to X$, where $Y$ has the minimum number of cells in each dimension consistent with the structure of $H_*(X;{\mathbb Z})$, namely, 1 cell for each infinite cyclic summand of the homology and 2 cells for each finite cyclic summand. This applies in particular for $K(\pi,n)$'s with $n>1$ and $\pi$ finitely generated. – Allen Hatcher Sep 6 '14 at 21:37

Some quick observations on the paper you linked, that would not fit into a comment:

(1) It mentions that the generating function of the number of cells for the $CW$ complexes they construct is always a rational function. So this always gives you at worst an exponential upper bound. Also it looks like your Fibonacci guess is correct, since that has generation a rational function.

In fact, it looks like the ones they listed correspond to the sequence of rational functions

$$\frac{1}{1-T}, \frac{1+T}{1+T-T^2}, \frac{1+T+T^2}{1+T+T^2-T^3}, \dots$$

(2) It mentions that the number of cells is equal to the number of pruned level-trees of height $n$. So presumably if one knew what that was, one would be able to compute the rational function and hence an upper bound.

(3) The obvious place to look for a lower bound on the number of cells is the cohomology. In fact, all the ones you mention as economical meet exactly the lower bound coming from cohomology. I think the cohomology of the Eilenberg-Maclane space is fairly well-studied, so you could look there. In the paper, it mentions one computation of the cohomology ring, which with sufficient understanding of the kind of trees they're talking about should tell you a lower bound.

• I see. Another one (for $K(\mathbb Z/2\mathbb Z,3)$) looks like $a_{n+3}=a_n+a_{n+1}+a_{n+2}$. For cyclic groups of prime order cohomology is indeed completely described in the literature. For $\mathbb Z$ - I don't know... – მამუკა ჯიბლაძე Sep 6 '14 at 19:25