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Let $G$ be a finitely presented group of cohomological dimension $n$.

Apart from the unresolved ambiguity pertaining to the Eilenberg--Ganea conjecture, it is known that we can find an $n$-dimensional model of the classifying space $BG = K(G,1)$.

It is also not hard to see that $BG$ needs to have cells in every dimension $j \leq n$: otherwise $EG^{(j-1)}$ would be a simply-connected acyclic space, hence already equal to $EG$, which is impossible by the assumption on $\text{cd}(G)$.

My question now is: what is the smallest number of $j$-cells that $BG$ can have?

For $G = \mathbf{Z}^n$ the count is $\binom{n}{j}$, and in general the number of $1$- and $2$-cells are the minimal number of generators and relations of a presentation of $G$, so can be $2$ (except for $G = \mathbf Z$ where $n = 1$) and $1$, respectively.

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    $\begingroup$ $G=\mathbb{Z}^n$ is exceptional in that the minimal $BG$ model is a compact orientable manifold, and therefore satisfies Poincare duality, Morse inequalities, etc., and in which Lusternick-Schnirelmann category is well studied. But in most cases, especially if $G$ is a Bieri-Eckmann duality group with dualizing $\mathbb{Z}G$-module $\mathbb{D}$ concentrated at a homological dimension $\nu>0$, then i don't think anybody knows how to count cells of any $BG$ model, nevermind the minimal $BG$ models which are no longer manifolds but necessarily singular topological objects. $\endgroup$
    – JHM
    Jan 19, 2021 at 14:29
  • $\begingroup$ Evenmore explicit geometric $BG$ models from geometric topology (arithmetic, knots, surface groups, etc) are typically neither minimal nor simplicial. An exception is the well-rounded retract model $W=W_n$ of $PGL(\mathbb{Z}^n)\backslash PGL(\mathbb{R}^n)/SO(n)$, consisting of all flat $n$-dimensional tori whose $1$-systoles generate $\mathbb{Q}^n$ in rational homology. Here the cells are naturally parameterized by certain collections of simple closed curves on the torus. But the $G$-orbits of these collections is equivalent to the explicit $\mathbb{Z}G$-module structure of $\mathbb{D}$. $\endgroup$
    – JHM
    Jan 19, 2021 at 15:02
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    $\begingroup$ I not aware of a useful exact formula in the generality of the question, but for any skew field $K$ and homomorphism $\phi: G \to K^\times$ there is a lower bound of $\dim_K H_j(BG;K^\phi)$. $\endgroup$
    – user171227
    Jan 20, 2021 at 0:02

1 Answer 1

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A group $G$ is of type $\mathcal{F}_n$ if it has a $K(G,1)$ with finite $n$-skeleton. Let $F_2$ be the free group on $2$ generators. Consider the kernel of the map $F_2\times F_2 \times \cdots \times F_2 \to \mathbb{Z}$, sending each generator to $1$. Then the kernel is of type $\mathcal{F}_{n-1}$ but not of type $\mathcal{F}_n$. This was proved by Stallings for $n=2$ and Bieri for $n>2$, see the discussion and links here. Taking $n\geq 3$, one gets finitely presented groups of finite cohomological dimension which do not have finite $n$-skeleton.

Nevertheless, one can ask your question for groups of cohomological dimension $n$ and type $\mathcal{F}_n$. For surface groups, and more generally 1-relator groups without torsion, this is known. A 1-relator group $G$ without torsion has presentation complex a $K(G,1)$, and cannot have a complex with fewer cells unless it is free.

For 3-manifold groups, this is a well-studied question when the $K(G,1)$ is a manifold and the cell structure is a handle structure. In this case, the number of cells is determined by the Heegaard genus of the manifold. There are known algorithms to compute this. However, the rank and the Heegaard genus can differ. So there can be a $K(G,1)$ with fewer $1$-cells than the Heegaard genus. Most likely the rank of $3$-manifolds (so the minimal number of $1$-cells in a $K(G,1)$) is computable for $3$-manifolds, but it has only been proved in special cases. I'm not sure though whether the minimality of the number of 2-cells has been addressed. The minimal number of 3-cells is $1$. An interesting possibility is that for a CW complex with a minimal number of $2$-cells, the number of $3$-cells or 1-cells might not be minimal.

In general, I'm not sure how much is known about this question, I'm just addressing the few special cases that I am familiar with to show how complex the answer can be.

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    $\begingroup$ Thanks for this nice answer from which I learnt a lot! I am mostly interested in the case of bigger $n$, though. Another (related) more specialized question one could ask there is the following: if $M^n$ is a closed aspherical manifold, can it happen that the minimal number of $j$-cells in a CW-model of $M$ is smaller than $\binom{n}{j}$? $\endgroup$ Jan 20, 2021 at 7:30
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    $\begingroup$ @JensReinhold The Heisenberg manifold (compact three dimensional nilmanifold) does not satisfy your binomial coefficient bound. Compare Pete L. Clark's answer math.stackexchange.com/questions/434384/the-heisenberg-manifold. More generally, you should compare your binomial coefficient bound with the Toral Rank Conjecture. Manifolds with $M^n$ with toral rank $<n$ are likely going to be counterexamples to your $nCj$ lower bound. $\endgroup$
    – JHM
    Jan 20, 2021 at 14:33
  • $\begingroup$ @JHM right, and cross with tori or with itself to get examples in all dimensions. $\endgroup$
    – Ian Agol
    Jan 20, 2021 at 14:50

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