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Let $F$ be a finite group.

Is there a model for $BF$ as a simplicial set such that the number of nondegenerate $n$-simplices grows at most polynomially?

For example the Bar construction has the property that there are exactly $(|F|-1)^n$ nondegenerate $n$-simplices. This answers the question affirmatively for $\mathbb{Z}/2$, but for other groups it still grows exponentially.

A lower bound for the number of such simplices is of course given by the rank of the group homology and in all examples that I know this only grows polynomially.

Of course it would be nice to have a functorial model, but that might be a follow up.

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    $\begingroup$ The ranks of the group homology of a finite group always grow polynomially. This follows from a much stronger fact: the cohomology algebra of a finite group with coefficients in any Noetherian ring $R$ is a finitely generated $R$-algebra. This was basically proved by Venkov (though he didn't state the full result). For a complete proof and references/history, see Evens, Leonard The cohomology ring of a finite group. Trans. Amer. Math. Soc. 101 (1961), 224–239. $\endgroup$
    – Andy Putman
    Jul 22, 2021 at 15:48
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    $\begingroup$ (it seems reasonable to hope that contemplating the proof of this might give the result you want, but I'm not really sure) $\endgroup$
    – Andy Putman
    Jul 22, 2021 at 15:49
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    $\begingroup$ By the way for $\mathbb{Z}/3$ is an example for a group, where more nondegenerate simplices are needed than cells (in the analogous question for CW-complexes). There one can construct a CW-complex with one cell in each dimension, e.g. just the rank of the homology with coefficients in $\mathbb{F}_3$ suffices. However, for any such simplicial set, the differential on $C^*(B\mathbb{Z}/3)$ cannot be zero, since it has a DGA structure and hence we can compute Massey-products on it. And these do not vanish (which they would if we could achieve that the differential was zero). $\endgroup$
    – HenrikRüping
    Jul 22, 2021 at 16:36
  • $\begingroup$ I suppose the case where $F$ is finite abelian is clear. Is the case where $F$ is finite nilpotent clear? Is the class of $F$ such that $BF$ has a polynomial model closed under extensions? Is it easier to address subexponentiality than polynomiality? $\endgroup$
    – Tim Campion ♦
    Jul 22, 2021 at 18:02
  • $\begingroup$ At least to me the case of $\mathbb{Z}/3$ is not clear at all. What is the polynomially growing model or why isnt there one? $\endgroup$
    – HenrikRüping
    Jul 22, 2021 at 18:28

3 Answers 3

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Going out on a limb (I may well be messing up badly), I think the answer is yes at least for the CW structure version of the question.

Proof:

Choose a finite presentation of $F$ with generating set $G$ and relation set $R$, and consider the induced $\pi_1$-isomorphism $X = (S^1)^{\vee G} \cup_{(S^1)^{\vee R}} \ast \to BF$. Note that $X$ has finitely many cells. I believe that the rank over $\mathbb Z[F]$ of $H_\ast(\tilde X ; \underline{\mathbb Z[F]})$ grows polynomially in $\ast$ (if it doesn't, then we should have a homology obstruction, giving a negative answer to the question). Then we should be able to mimic the construction showing that the homology bound on number of cells is realized for a simply-connected space. That is, we attach free $F$-equivariant cells (i.e. cells of the form $\vee^{F} D^n_+$, with $n \geq 2$) to $\tilde X$ one-by-one, building up a space $\tilde X'$ which has the same $\underline{\mathbb Z[F]}$-homology as $\tilde{BF}$ and conclude by homology Whitehead (with coefficients) that $X' \to BF$ is a weak homotopy equivalence, where $X' = \tilde X'_{hF}$ has one cell for each cell of $X$ plus a cell for each $F$-equivariant cell we attached.

(I'm not 100% sure though -- in the simply-connected case, we use the Hurewicz theorem to be sure that we can always map to a homology class with a sphere... perhaps this breaks down if the relevant $F$-equivariant Hurewicz theorem fails?)

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There is an algebraic result that is relevant, due to Benson and Carlson and stated as Corollary 4.5 in `Complexity and Multiple Complexes' Math. Z. vol 195 (1987) 221--238. Given a finite group $G$, let $n$ be the maximum of the $p$-ranks of $G$ over all primes. Then there is a free resolution of $\mathbb{Z}$ over $\mathbb{Z}G$ that is the tensor product of $n$ non-negative periodic complexes. The sizes of the modules in this resolution grow as a polynomial of degree $n-1$, and one can think of the resolution as being built from copies of a finite chain complex of free $\mathbb{Z}G$-modules: the tensor product of the period pieces for the $n$ periodic complexes. This is the same sort of thing as you would get if the group acted freely trivially on homology on a product of $n$ spheres (possibly of different dimensions).

There are two issues with promoting this to the result that you want: firstly realizing the periodic pieces as the chain complexes of simply connected $G$-CW-complexes, for which Tim Campion's answer is relevant; secondly somehow realizing the whole complex as a simplicial set in such a way that you don't need any extra low-dimensional non-degenerate simplices in the higher-dimensional copies of the periodic pieces. For the second of these, the cyclic group $C_n$ (where the periodic piece should be the chain complex for the circle with the group acting freely by rotation) is an important test case.

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This question may be somewhat relevant: Small simplicial complexes with torsion in their homology. David Speyer's answer there shows that one can build a simplicial complex $X$ with $H_1(X)=\mathbb{Z}/p$ where the number of simplices of $X$ is $O(\log(p))$. It seems unlikely that one can do much better than that in the world of simplicial sets. With CW complexes, you only need a single $0$-cell, a single $1$-cell and a single $2$-cell. This gives an initial picture of how the simplicial set version of the question might deviate from the CW complex version.

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