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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...

(added after two answers below)

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.

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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.

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    $\begingroup$ Yes... but that's not very constructive. $\endgroup$ – André Henriques Sep 6 '14 at 17:23
  • $\begingroup$ True, but I don't think the question asked for constructive. $\endgroup$ – Jeff Strom Sep 6 '14 at 17:25
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    $\begingroup$ @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. $\endgroup$ – მამუკა ჯიბლაძე Sep 6 '14 at 19:12
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    $\begingroup$ 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. $\endgroup$ – Allen Hatcher Sep 6 '14 at 21:37
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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.

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  • $\begingroup$ 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... $\endgroup$ – მამუკა ჯიბლაძე Sep 6 '14 at 19:25

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