## Notes of Robert Kropholler’s Cambridge lecture 10-01-2017

Hyperbolic groups and their subgroups

A group is of type ${F^n}$ if it admits a classifying space with finite ${n}$-skeleton. For instance, ${F^1}$ iff finitely generated. ${F^2}$ iff finitely presented.

Question (Brady). Do there exist groups of type ${F^n}$ and not ${F^{n+1}}$ with no ${{\mathbb Z}^2}$ subgroups ?

Today, I will focus on looking for examples as subgroups of hyperbolic groups.

1. Results

1.1. ${n=1}$

The answer in case ${n=1}$ follows from this classical thm:

Theorem 1 (Rips) Given a finitely presented group ${Q}$, there exists a hyperbolic group ${G}$ and a surjective map ${G\rightarrow Q}$ with finitely generated kernel.

Furthermore, if ${Q}$ is infinite, ${N}$ is not finitely presented.

1.2. ${n=2}$

Theorem 2 (Brady 1999, Lodha 2017) There exist infinitely many pairwise non isomorphic hyperbolic groups, each containing a subgroup of type ${F^2}$, not ${F^3}$.

No result for ${n\geq 3}$. The following result indicates that it is harder.

Theorem 3 (Gersten) If ${G}$ has cohomological dimension 2, is hyperbolic, any finitely presented subgroup of ${G}$ is hyperbolic.

1.3. Construction

Definition 4 A graph ${\Gamma}$ is sizable if

1. ${\Gamma}$ is bipartite on ${A}$ and ${B}$, ${A=A^+\coprod A^-}$, ${B=B^+\coprod B^-}$.
2. ${\Gamma}$ has no 4-cycles.
3. ${\Gamma(A^s \cup B^t)}$ is connected for all ${s,t=\pm}$.

Such graphs exist, a computer search has produced one with 31 vertices.

Theorem 5 Given sizable graphs ${\Gamma_i}$, ${i=1,2,3}$, ${\Gamma_i\subset A_i *B_i}$, there exists a ${CAT(0)}$ cube complex ${X\subset \prod_{i=1}^3 A_i *B_{i+1}}$ such that ${G=\pi_1(X)}$ is hyperbolic, ${G}$ maps onto ${{\mathbb Z}}$ with ${F^r}$ not ${F^s}$ kernel.

1.4. Observation

One needs hyperbolic groups ${G}$ with high geometric dimension. Here are examples: Januskiewicz-Swiatkowski 2003, 2006. Haglund 2003. Arzhantseva-Bridson-Januskiewicz-Leary-Minasyan-Swiatkowski 2009. All these examples produce systolic groups, they cannot be used, because of following

Theorem 6 (Wise, Zadnik) If ${G}$ is systolic, ${N}$ a finitely presented subgroup of ${G}$, then ${N}$ is systolic (and thus ${F^n}$ for all ${n}$).

Instead, we shall rely on the following family of examples.

Theorem 7 (Osajda 2010) There exist right angled Coxeter groups of arbitrary virtual cohomological dimension.

Such a group ${W_L}$ is hyperbolic iff the graph is 5-large (i.e. triangle and square-free).

5-large graphs are scarce (for instance, no triangulation of an ${n}$-manifold, ${n\geq 5}$, is 5-large). Therefore, I want to relax 5-large.

Definition 8 A pair of ${n}$-partite (i.e. contained in the join of ${n}$ graphs) flag complexes ${A,B}$ are pairwise 5-large if

1. Every 4-cycle in ${A}$ (or ${B}$) is contained in a 2-colored part of ${A}$ (resp. ${B}$).
2. For each pair ${(i,j)}$, either ${A(A_i\cup A_j)}$ or ${B(B_i\cup B_j)}$ has no 4-cycle.

Pairwise 5-large allows to construct ${X\subset \prod_{i=1}^3 A_i *B_{i}}$ with hyperbolic fundamental group. Indeed, the ${n}$-bipartite structure un ${A}$ and ${B}$ allows to get an ${L^1}$-embedding ${\tilde X\rightarrow}$ a product of ${n}$ trees. A flat in ${\tilde X}$, when mapped to one tree factor, falls onto a line, whence a map to ${{\mathbb R}^n}$. We show this is an ${L^2}$-embedding, the image is a plane which cannot be transverse to coordinate hyperplanes, contradiction. Therefore ${\tilde X}$ is hyperbolic.