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.



How many cells needed to build the classifying space $BG$?
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](https://mathscinet.ams.org/mathscinet-getitem?mr=1465330). 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](https://doi.org/10.1215/00127094-2018-0023) to compute this. However, the [rank and the Heegaard genus can differ](https://doi.org/10.1090/S0894-0347-2013-00767-5). 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](https://doi.org/10.2140/gt.2005.9.375). 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.


