A torsionfree group with infinite cohomological dimension and no infinitely generated free abelian subgroup - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T07:55:43Z http://mathoverflow.net/feeds/question/60556 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60556/a-torsionfree-group-with-infinite-cohomological-dimension-and-no-infinitely-gener A torsionfree group with infinite cohomological dimension and no infinitely generated free abelian subgroup tmf 2011-04-04T14:03:29Z 2011-04-04T16:09:19Z <p>Recently I've been reading about cohomological finiteness conditions for groups, my main source being Brown's book "Cohomology of Groups". </p> <p>One of the first things one learns is that a group with finite cohomological dimension necessarily is torsionfree. It is easy to come up with an example of a torsionfree group with infinite cohomological dimension: non-finitely generated free abelian group immediately springs to mind. A little bit of googling revealed an example of an infinite-dimensional torsionfree $FP_{\infty}$ group.$^1$ (A group is said to be of type $FP_{\infty}$ if there exists a projective $ZG$-resolution {$P_i$} of $Z$ such that each $P_i$ is finitely generated.) This example is Thompson's group $F$. However, it is well-known that $F$ contains a non-finitely generated free abelian group. </p> <p>So, the question is: what is an example of a torsionfree group with infinite cohomological dimension and no inifinitely generated free abelian subgroup (that is, if there exists one)? </p> <p>$^1$ In case anyone is interested: K. S. Brown, R. Geoghegan, Invent. Math. 77, 367--381.</p> http://mathoverflow.net/questions/60556/a-torsionfree-group-with-infinite-cohomological-dimension-and-no-infinitely-gener/60565#60565 Answer by Ian Agol for A torsionfree group with infinite cohomological dimension and no infinitely generated free abelian subgroup Ian Agol 2011-04-04T16:09:19Z 2011-04-04T16:09:19Z <p>You can take a union of torsion-free word-hyperbolic groups with cohomological dimension approaching infinity to get such an example. Word-hyperbolic groups are rank one, so one can show that any abelian subgroup must intersect each subgroup in a cyclic group, and therefore the subgroup cannot be an infinitely generated free abelian subgroup. </p> <p>Here's an explicit sequence. Take a sequence of cocompact torsion-free lattices in $\Gamma_n&lt; Isom(\mathbb{H}^n)$, such that $\Gamma_n &lt; \Gamma_{n+1}$ from the natural embedding $Isom(\mathbb{H}^n) &lt; Isom(\mathbb{H}^{n+1})$. The limit $\Gamma_{\infty}=\underset{n\to\infty}{\lim} \Gamma_n$ is a group of cohomological dimension $\infty$ since $\Gamma_n$ is of cohomological dimension $n$. </p> <p>For example, consider the quadratic form over $\mathbb{Q}(\sqrt{5})$ which is $q_n(x_0,x_1,\ldots,x_n)=((1-\sqrt{5})/2 )x_0^2+x_1^2+\cdots+x_n^2$. This gives a quadratic form over $\mathbb{R}$ of signature $(n,1)$. Consider the group $O(q_n,\mathbb{Z}[\phi])&lt; GL(n+1,\mathbb{Q}(\sqrt{5}))$, where $\phi=(1+\sqrt{5})/2$, the group of integral matrices preserving this quadratic form. Then there exists a prime ideal $\mathcal{P}&lt;\mathbb{Z}[\phi]$ such that $\Lambda_n=Ker\{ GL(n+1,\mathbb{Z}[\phi])\to GL(n+1,\mathbb{Z}[\phi]/\mathcal{P})\}$ is torsion-free for all $n$. Let $\Gamma_n=O(q_n,\mathbb{Z}[\phi])\cap \Lambda_n$. Then $\Gamma_n$ is a torsion-free hyperbolic group acting cocompactly on $\mathbb{H}^n$ (by considering the Lorentzian model of hyperbolic space), and clearly $\Gamma_n&lt;\Gamma_{n+1}$. </p>