$cd_{\mathbb Q}$ is the cohomological dimension on rational coefficients. Notice that if we replace $\mathbb Q$ with $\mathbb Z$ in $cd_{\mathbb Q}$ the statement is certainly false (think about $\mathbb Z$ and $\mathbb Z / 2 \mathbb Z$).

Suppose that $cd_{\mathbb Q}(G) < \infty$. By a theorem of Fel'dman (see Proposition 2.5 in "Normal subgroups in duality groups and in groups of cohomological dimension 2" of Bieri) if we also assume that $cd_{\mathbb Q}(G/\mathbb Z) < \infty$, we have: $$ cd_{\mathbb Q}(G) = cd_{\mathbb Q}(\mathbb Z) + cd_{\mathbb Q}(G / \mathbb Z) = 1 + cd_{\mathbb Q}(G / \mathbb Z). $$

It follows that $cd_{\mathbb Q}(G/\mathbb Z) < cd_{\mathbb Q}(G)$ if and only if $cd_{\mathbb Q}(G/\mathbb Z) < \infty$.

So the question is: under which condition we can be sure that $G/\mathbb Z$ has finite cohomological dimension? what about $G$ being a poincare group? a CAT(0) group? $\mathbb Z$ being (un)distorted? (those are just random conditions that I put as examples).

Are there known counterexamples (i.e. $cd_{\mathbb Q}(G/ \mathbb Z) = \infty$)?

  • 1
    $\begingroup$ Counterexamples to what? if $H$ has $cd_Q=\infty$ and $G=H\times Z$, then trivially $cd_Q(G/Z)=\infty$. Do you at least mean you assume $cd_Q(G)<\infty$? $\endgroup$
    – YCor
    Apr 15, 2017 at 10:00
  • $\begingroup$ "Let G be a group of finite cohomological dimension.." $\endgroup$
    – fritz
    Apr 15, 2017 at 10:05
  • 11
    $\begingroup$ Ah, assumptions are not supposed to only be in the title, it's normal I didn't see it reading several times. $\endgroup$
    – YCor
    Apr 15, 2017 at 11:08


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