## Finitely presented subgroups of hyperbolic groups of cohomological dimension 2

Suppose $G$ is a group with group ring $\mathbb{Z}G$. A projective resolution of a $\mathbb{Z}G$–module $M$ is an exact sequence

$\cdots \to P_i\to P_{i-1}\to...\to P_0\to M\to 0$

where each $P_j$ is a projective $\mathbb{Z}G$–module. Regard $\mathbb{Z}$ as a trivial $\mathbb{Z}G$–module. The cohomological dimension $\textup{cd}(G)$ of a group $G$ is the smallest integer $n$ such that $\mathbb{Z}$ admits a projective resolution

$\cdots \to 0 \to 0 \to P_{n} \to...\to P_0\to \mathbb{Z} \to 0$

of $\mathbb{Z}G$–modules. (If there is no such integer, the cohomological dimension is infinity.) For more details, see Cohomology of groups by Kenneth S. Brown.

In this post we shall discuss the proof of Gersten’s result that hyperbolic groups of cohomological dimension $2$ have the property that their finitely presented subgroups are also hyperbolic. This result appears in Gersten’s paper Subgroups of word hyperbolic groups in dimension 2.

Standard norms on modules

Suppose $H$ is a group and $M$ is a finitely generated $\mathbb{Z}H$–module. Let $F$ be a finitely generated free $\mathbb{Z}H$–module, with a basis $\{\alpha_1,...,\alpha_n\}$, such that there is a surjective homomorphism $\nu:F \to M$ of $\mathbb{Z}H$–modules. Then $\{h\alpha_i\mid h\in H,1\leq i\leq n\}$ is a free $\mathbb{Z}$–basis for $F$. Equip $F$ with the $l_1$–norm

$| \sum_{i \in \{1, \ldots, n\}, h \in H} n_i h \alpha_i |_1 = \sum_{i \in \{1, \ldots, n\}, h \in H} | n_i |$.

We give $M$ the norm $|m|=\textup{min} \{|a|_1\mid a\in F, \nu(a)=m\}$. Such a norm $|.|$ on $M$ is, up to linear equivalence, independent on the choices of $F$, $\nu$ and $\{\alpha_1,...,\alpha_n\}$ — that is, if another norm $|.|^{\prime}$ is obtained from another surjective homomorphism from a different finitely generated free $\mathbb{Z}H$–module $F^{\prime}$, then there is a uniform constant $C$ such that $C^{-1}|m|\leq |m|^{\prime}\leq C|m|$ for all $m\in M$. So $|.|$ is called a standard norm.

Lemma 1. Suppose

$0 \to M \stackrel{i}{\to} N \to P\to 0$

is a short exact sequence of $\mathbb{Z}H$–modules. Assume
1) $M$ is finitely generated and equipped with a standard norm $|.|$,
2) $N$ is free and finitely generated and given an $l_1$ norm $|.|_1$ associated to some basis, and
3) $P$ is projective.
Then there is a retraction $\sigma:N\to M$ for $i$ (that is, a map $\sigma:N\to M$ such that $\sigma \circ i$ is the identity on $M$) and a constant $C>0$ so that $|\sigma(x)|\leq C|x|_1$ for all $x\in N$.

Proof. Since $P$ is projective, we know that $N\cong M\oplus P$. We choose a map $\sigma^{\prime}:N\to M$ such that $i\circ \sigma^{\prime}$ is the identity on $M$. Let $I$ be a finitely generated free submodule generated by a subset of the generators of $N$ which contains the image of $M$ under the map $i$. Let $N=Q\oplus I$. Notice that $Q$ is also free, but not necessarily finitely generated. We can modify the retraction $\sigma^{\prime}$ to $\sigma:N\to M$ where $\sigma\mid_{I}=\sigma^{\prime}\mid_{I}$ and $\sigma\mid_{P}=0$. Let $\pi:F\to M$ be a surjective homomorphism of $\mathbb{Z}H$–modules where $F$ is finitely generated, free and based.

Since $I$ is free, there is a map $\rho:I\to F$ such that $\sigma\mid_I=\pi\circ \rho$. This map $\rho$ can be extended to $I\oplus Q$ by setting the images under $\rho$ of the elements of $Q$ be $0$. It follows this $\sigma=\pi\circ \rho$ and $\rho$ is supported on $I\oplus Q$.

Now $I$ and $F$ are both finitely generated and free. So the map $\rho$ is given by a finite matrix with entries in $\mathbb{Z}H$. Therefore there exists a constant $C>0$ such that $|\rho(u)|_1\leq C|u|_1$ for all $u\in N$. Since the standard norm on $M$ is defined as an infimum over coset representatives, we have that $|\sigma(u)|=|\pi\circ \rho(u)|\leq |\rho(u)|_1\leq C|u|_1$ for all $u\in N$. This completes the proof.

Theorem (Eilenberg–Ganea). Let $\Gamma$ be an arbitrary group and let $n=\textup{max}\{\textup{cd}(\Gamma),3\}$. Then there exists an $n$-dimensional $K(\Gamma,1)$-complex $Y$. If $\Gamma$ is finitely presented and of finite type, then $Y$ can be taken to be finite.

If the cohomological dimension of the group is infinite, then the theorem merely asserts the existence of a $K(\Gamma,1)$-complex.

The relation module of a finitely presented group

Let $G= \langle a_1,...,a_n\mid r_1,...,r_m\rangle$ be a finitely presented group. So $G=F/R$, where $F$ is freely generated by $a_1,...,a_n$ and $R$ is the normal closure of the relations $r_1,...,r_m$. Let the presentation $2$–complex of this group be $Y$. The relation module for $G$ is a $\mathbb{Z}G$–module defined in the following way. The underlying abelian group is $R_{ab}$, the abelianization of R , and the $G$–action on it is induced by the conjugation of $F$ on $R_{ab}$. Since $R$ acts trivially, the induced action by $G$ is well defined. Hence this gives us a $\mathbb{Z}G$–module structure on $R_{ab}$. This is naturally isomorphic as a $\mathbb{Z}G$ module to $H_1(\tilde{Y}^{(1)})$, where $\tilde{Y}^{(1)}$ is the $1$-skeleton of the universal cover of the presentation complex $Y$ of $G$, or in other words, the Cayley graph of $G$.

Proposition 1. If $G$ is a word hyperbolic group of cohomological dimension $2$, then $G$ is the fundamental group of a finite aspherical $3$–complex $Y$ such that
1) the relation module $H_1(\tilde{Y}^{(1)})$ is finitely generated and free as a $\mathbb{Z}G$ module, and
2) the standard norm on $H_1(\tilde{Y}^{(1)})$ agrees up to linear equivalence with the $l_1$ norm from any basis for the $\mathbb{Z}G$ module $H_1(\tilde{Y}^{(1)})$.

Proof. Since $G$ has cohomological dimension $2$, it is torsion free. Torsion free hyperbolic groups are of finite type, so $G$ is the fundamental group of a finite asherical complex. In particular, by the Eilenberg-Ganea theorem we know that $G$ is the fundamental group of a finite aspherical $3$–complex $Y_1$.

We need the following lemma (page 184 of Cohomology of groups).

Lemma 2. If the cohomological dimension of a $\mathbb{Z}G$–module $M$ is $n$, and if $0\to K\to P_{n-1}\to...\to P_0\to M\to 0$ is an exact sequence of $\mathbb{Z}G$–modules with each $P_i$ projective, then $K$ is projective.

From the resolution,

$0\to H_1(\tilde{Y}^{(1)})\to C_1(\tilde{Y_1})\to C_0(\tilde{Y_1})\to \mathbb{Z}\to 0$

and the facts that $C_i(\tilde{Y}^{(1)})$ are projective and $G$ has cohomological dimension $2$, we get from Lemma 2 that $H_1(\tilde{Y_1}^{(1)})$ is projective.

The exact sequence

$0\to C_3(\tilde{Y_1})\to C_2(\tilde{Y_1})\to H_1(\tilde{Y_1}^{(1)})\to 0$

then splits. So we have an isomorphism of $\mathbb{Z}H$–modules $C_2(\tilde{Y_1})\cong H_1(\tilde{Y_1}^{(1)})\oplus C_3(\tilde{Y_1})$. By definition this means that $H_1(\tilde{Y_1}^{(1)})$ is stably free as a $\mathbb{Z}G$–module. (A $\mathbb{Z}G$–module $M$ is stably free if there exist free $\mathbb{Z}G$–modules $N,P$ such that $M\oplus N\cong P$ as $\mathbb{Z}G$–modules.)

So we can attach a finite number of $2$-discs trivially at the basepoint of $Y_1$ to produce a finite $3$-complex $Y$ with fundamental group $G$ and with the relation module $H_1(\tilde{Y})$ free as a $\mathbb{Z}G$–module. This proves the first part of the result.

The second part of the proposition follows from an application of Lemma $1$.

We are now ready to prove our main result.

Theorem 1. Let $H$ be a finitely presented subgroup of the word hyperbolic group $G$, where $G$ is of cohomological dimension $2$. Then $H$ is word hyperbolic.

Proof. By Proposition $2$, $G$ has a presentation for which the relation module is free and finitely generated.

Using this fact and Schanuel’s Lemma one can establish that for any finite presentation $\langle F \mid R \rangle$ of $G$, the relation module $R_{ab}$ is stably free.

We can choose presentations for $H$ and $G$ such that the respective presentation complexes $X$ and $Y$ satisfy $X\subseteq Y$. Further, by adding trivial letters and corresponding relations to the presentation for $G$ (as in a previous proof) we can assume that $H_1(\tilde{Y}^{(1)})$ is free.

Let $Q=\tilde{Y}^{(1)}/\tilde{X}^{(1)}$. An application of the Snake Lemma can be used to show that $H_1(Q)$ is projective as a $\mathbb{Z}H$–module. The short exact sequence of $\mathbb{Z}H$–modules arising from the Snake Lemma

$0\to H_1(\tilde{X}^{(1)})\to H_1(\tilde{Y}^{(1)})\to H_1(Q)\to 0$

satisfies the conditions of Proposition $1$.

It follows that there is a retraction $p:H_1(\tilde{Y}^{(1)})\to H_1(\tilde{X}^{(1)})$ for $i: H_1(\tilde{X}^{(1)})\to H_1(\tilde{Y}^{(1)})$, such that there is a uniform constant $C>0$ with the property that $|p(x)|\leq C|x|_1$ for all $x\in H_1(\tilde{Y}^{(1)})$. So if $y\in H_1(\tilde{X}^{(1)})$, then $|y|=|p\circ i(y)|\leq C|i(y)|$.

Let $w$ be a circuit in $\tilde{X}^{(1)}$ and let $S$ be a van Kampen diagram for $w$ in $\tilde{X}$ of minimal area. The area of $S$ is $|y|$, where $y$ is the element of $H_1(\tilde{X}^{(1)})$ corresponding to $w$. The area of $w$ as a circuit in $\tilde{Y}^{(1)}$ is $|i(y)|$, where now $|.|$ is in terms of $\mathbb{Z}G$–modules. It is clear that $|i(y)|$ in $H_1(\tilde{Y}^{(1)})$ as a $\mathbb{Z}G$–module is less than or equal to $|i(y)|$ with $H_1(\tilde{Y}^{(1)})$ as a $\mathbb{Z}H$–module. By the discussion in the preceding paragraph, we can conclude that if $G=\pi_1(Y)$ satisfies a linear isoperimetric inequality, then so does $H=\pi_1(X)$. Since $H$ satisfies a linear isoperimetric inequality, it must be word hyperbolic.