Groups with unusual cohomological dimension of direct product

$$\DeclareMathOperator\cd{cd}$$Are there any known examples of non-free groups with a property that $$\cd(G)+1 = \cd(G \times G)$$, or, less restrictive, $$G, H$$ with $$\cd \neq 1, \infty$$ such that $$\cd(H)+1 = \cd(G \times H)$$?

• This question mathoverflow.net/questions/272715/… is relevant Commented Mar 11, 2022 at 13:31
• Although question is relevant, (currently absent) answer to it would not provide an answer to my question as far as I can see and vise versa. Commented Mar 11, 2022 at 14:32
• That questions shows that the "typical" behavior is $cd(G\times G) =2cd(G)$ for $FP_{\infty}$-groups for example and also if you ask $cd(G)=cd(G\times G)$ then $G$ has infinite cohomological dimension. I didn't claim it answers your question. It's for context. Commented Mar 11, 2022 at 15:04
• For the less restrictive you want neither one free right? Commented Mar 11, 2022 at 16:03
• ...I think I need both have cd greater than than one for my possible application (namely, some clarifications/upgrades in Kan-Thurston construction; such groups can provide a source for suspension-like functor on crossed modules with extra structure, in order to get full functor from hoTop to a certain localisation of category of that contraptions), but I guess any example can be pretty illuminating. Commented Mar 11, 2022 at 16:30

1 Answer

Let $$G=(\mathbb{Q},+)$$. Then $${\rm cd}(G)=2$$ and $${\rm cd}(G\times G)=3$$.

• Where does this come from?
– YCor
Commented Mar 12, 2022 at 14:52
• $K(Q,1)$ can be built as the mapping telescope, or homotopy colimit, of the sequence of self maps of $S^1$ of degrees 2,3,4,5... Geometrically this can be realized by gluing together a sequence of cylinders $S^1 \times [0,1]$ by the appropriate maps at the ends. Hence $cd(Q) \leq 2$, and it is not 1 by Swan. $K(Q \oplus Q, 1)$ can be built similarly from cylinders on 2-tori, hence $cd \leq 3$. To see that $cd(Q\oplus Q)$ is not 2, note that the uncountable group $Ext^1_Z(H_2(K(Q \oplus Q,1),Z),Z) = Ext^1_Z(Q\oplus Q, Z)$ injects into $H^3(K(Q\oplus Q,1), Z)$ by universal coefficients. Commented Mar 12, 2022 at 15:00
• According to Kropholler's answer here math.stackexchange.com/questions/2320997/… this works more generally for any solvable group of cohomological dimension 2 Commented Mar 12, 2022 at 15:34