Suppose is a group with group ring . A projective resolution of a –module is an exact sequence
where each is a projective –module. Regard as a trivial –module. The cohomological dimension of a group is the smallest integer such that admits a projective resolution
of –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 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 is a group and is a finitely generated –module. Let be a finitely generated free –module, with a basis , such that there is a surjective homomorphism of –modules. Then is a free –basis for . Equip with the –norm
We give the norm . Such a norm on is, up to linear equivalence, independent on the choices of , and — that is, if another norm is obtained from another surjective homomorphism from a different finitely generated free –module , then there is a uniform constant such that for all . So is called a standard norm.
Lemma 1. Suppose
is a short exact sequence of –modules. Assume
1) is finitely generated and equipped with a standard norm ,
2) is free and finitely generated and given an norm associated to some basis, and
3) is projective.
Then there is a retraction for (that is, a map such that is the identity on ) and a constant so that for all .
Proof. Since is projective, we know that . We choose a map such that is the identity on . Let be a finitely generated free submodule generated by a subset of the generators of which contains the image of under the map . Let . Notice that is also free, but not necessarily finitely generated. We can modify the retraction to where and . Let be a surjective homomorphism of –modules where is finitely generated, free and based.
Since is free, there is a map such that . This map can be extended to by setting the images under of the elements of be . It follows this and is supported on .
Now and are both finitely generated and free. So the map is given by a finite matrix with entries in . Therefore there exists a constant such that for all . Since the standard norm on is defined as an infimum over coset representatives, we have that for all . This completes the proof.
Theorem (Eilenberg–Ganea). Let be an arbitrary group and let . Then there exists an -dimensional -complex . If is finitely presented and of finite type, then can be taken to be finite.
If the cohomological dimension of the group is infinite, then the theorem merely asserts the existence of a -complex.
The relation module of a finitely presented group
Let be a finitely presented group. So , where is freely generated by and is the normal closure of the relations . Let the presentation –complex of this group be . The relation module for is a –module defined in the following way. The underlying abelian group is , the abelianization of R , and the –action on it is induced by the conjugation of on . Since acts trivially, the induced action by is well defined. Hence this gives us a –module structure on . This is naturally isomorphic as a module to , where is the -skeleton of the universal cover of the presentation complex of , or in other words, the Cayley graph of .
Proposition 1. If is a word hyperbolic group of cohomological dimension , then is the fundamental group of a finite aspherical –complex such that
1) the relation module is finitely generated and free as a module, and
2) the standard norm on agrees up to linear equivalence with the norm from any basis for the module .
Proof. Since has cohomological dimension , it is torsion free. Torsion free hyperbolic groups are of finite type, so is the fundamental group of a finite asherical complex. In particular, by the Eilenberg-Ganea theorem we know that is the fundamental group of a finite aspherical –complex .
We need the following lemma (page 184 of Cohomology of groups).
Lemma 2. If the cohomological dimension of a –module is , and if is an exact sequence of –modules with each projective, then is projective.
From the resolution,
and the facts that are projective and has cohomological dimension , we get from Lemma 2 that is projective.
The exact sequence
then splits. So we have an isomorphism of –modules . By definition this means that is stably free as a –module. (A –module is stably free if there exist free –modules such that as –modules.)
So we can attach a finite number of -discs trivially at the basepoint of to produce a finite -complex with fundamental group and with the relation module free as a –module. This proves the first part of the result.
The second part of the proposition follows from an application of Lemma .
We are now ready to prove our main result.
Theorem 1. Let be a finitely presented subgroup of the word hyperbolic group , where is of cohomological dimension . Then is word hyperbolic.
Proof. By Proposition , 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 of , the relation module is stably free.
We can choose presentations for and such that the respective presentation complexes and satisfy . Further, by adding trivial letters and corresponding relations to the presentation for (as in a previous proof) we can assume that is free.
Let . An application of the Snake Lemma can be used to show that is projective as a –module. The short exact sequence of –modules arising from the Snake Lemma
satisfies the conditions of Proposition .
It follows that there is a retraction for , such that there is a uniform constant with the property that for all . So if , then .
Let be a circuit in and let be a van Kampen diagram for in of minimal area. The area of is , where is the element of corresponding to . The area of as a circuit in is , where now is in terms of –modules. It is clear that in as a –module is less than or equal to with as a –module. By the discussion in the preceding paragraph, we can conclude that if satisfies a linear isoperimetric inequality, then so does . Since satisfies a linear isoperimetric inequality, it must be word hyperbolic.