Cohomological dimension of $G \times G$

Question

$\DeclareMathOperator\cd{cd}$A question that I have already posted in the Mathematics section, but which seems to be too delicate for that section (see here and here):

Let $\cd(G)$ denote the cohomological dimension of a group $G$, i.e. the minimal length of a projective resolution of Z over $\mathbb{Z} G$. Under which conditions on a group $G$ of finite cohomological dimension is it true that $$\cd(G \times G) > \cd(G) \; ?$$ On the one hand, there are examples of groups with $G \times G \cong G$, though I am not aware of examples of finite cohomological dimension. On the other hand, an affirmative answer follows from the Kuenneth formula for group cohomology as discussed in Chris Gerig's detailed answer to this question. Nevertheless, this line of argument is only applicable if we find a local coefficient system $A$ for which $H_*(G;A)$ is finitely generated and for which there exists $k\in \mathbb{N}$ with $2k>\cd(G)$ and $H^k(G;A)\neq 0$. Under which conditions does such $A$ exist?

Answer 1

One fairly general class of groups $G$ for which $\operatorname{cd}(G\times G)=2\operatorname{cd} (G)$ are the duality groups. A group $G$ is a duality group of dimension $n$ if there is a dualizing $G$-module $D$ and a fundamental class $e\in H_n(G;D)$ such that $$ -\cap e :H^k(G;M)\to H_{n-k}(G;M\otimes D) $$ is an isomorphism for any $k$ and any $G$-module $M$. It follows that $\operatorname{cd}(G)=n$. If the underlying group of $D$ can be taken as $\mathbb{Z}$, then $G$ is a Poincaré duality group. There are duality groups which are not Poincaré duality groups, and therefore can't have a closed aspherical manifold as their classifying space. Finitely generated free groups and knot groups are examples.

A theorem of Bieri and Eckmann states that given an extension $$ 1\to K\to G\to Q\to 1 $$ iwhere $K$ and $Q$ are duality groups of dimensions $m$ and $n$ respectively, then $G$ is a duality group of dimension $m+n$. This implies the claim at the top of my answer. This result and many more useful facts about duality groups can be found in the reference

Bieri, Robert; Eckmann, Beno, Groups with homological duality generalizing Poincaré duality, Invent. Math. 20, 103-124 (1973). ZBL0274.20066.

Answer 2

If $G$ is a non-trivial group with $G\times G\cong G$, then $G$ has infinite cohomological dimension as @YCor suggests. If $H\leq G$, then $cd(H)\leq cd(G)$ because $\mathbb ZG$ is a free $\mathbb ZH$-module and so any $\mathbb ZG$-free resolution of the trivial module is a $\mathbb ZH$-free resolution. If $G$ has an element of finite order, then $G$ has infinite cohomological dimension because finite cyclic groups do. If $G$ has an element of infinite order, then $G\cong G\times G$ implies $\mathbb Z^n$ is a subgroup of $G$ for all $n>0$. Since $cd(\mathbb Z^n)=n$, this means $G$ that $cd(G)\geq n$ for all $n>0$ and hence is infinite.

Answer 3

$\DeclareMathOperator\cd{cd}\newcommand{\FP}{\mathrm{FP}}$Another class of groups for which $\cd(G\times G)=2 \cd(G)$, containing duality groups, is the class of groups of type $\FP_{\infty}$.

If $G$ is of type $\FP_{\infty}$ that is the trivial module has a projective resolution that is finitely generated in each degree, then a theorem of Ken Brown https://www.math.cornell.edu/~kbrown/scan/1975.0050.pdf implies that the functor $H^n(G,-)$ commutes with direct limits for all $n$. Thus if $G$ has cohomological dimension $k$ then there is a finitely generated $G$-module $A$ with $H^k(G,A)\neq 0$ (there is some module for which this cohomology doesn't vanish and any module is a direct limit of its finitely generated submodules). So your Kunneth argument should give $\cd(G\times G)=2 \cd(G)$ if $G$ is of type $\FP_{\infty}$.

This includes the example of duality groups by Corollary 1 of Brown's paper although his argument is essentially Mark Grant's.