# Small simplicial set models for BG

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.

• 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. Jul 22 at 15:48
• (it seems reasonable to hope that contemplating the proof of this might give the result you want, but I'm not really sure) Jul 22 at 15:49
• 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). Jul 22 at 16:36
• 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? Jul 22 at 18:02
• 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? Jul 22 at 18:28

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?)

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.

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.