The Torelli group is defined as the kernel of the symplectic representation

It is known to be finitely generated, but the question of finite presentation is still open. Given that the mapping class group is well known to be finitely presented, a natural question to ask is then, under what conditions, if any, does every finitely generated subgroup of a given group permit a finite presentation? A group satisfying this property is said to be coherent. Of course if the mapping class group could be shown to be coherent then the question of finite presentation of the Torelli group is trivially solved. All I have done is said; if every finitely generated subgroup of the mapping class group is finitely presented, then the Torelli group is finitely presented, which isn’t very helpful. But the question of coherence could provide a roundabout way of solving this problem. More importantly however it serves as motivation for the question; under what conditions can a group be said to be coherent.

Let’s show that some groups are coherent and that some groups aren’t, to make sure that there is actually a question here:

*Example: Some non-coherent groups*

Let be the homomorphism sending the generators of one copy of , and of the other copy all to 1. Then is finitely generated, but not finitely presented, with presentation where . Details of this can be found here. This not only serves as an example of an incoherent group, but also provides a tool for showing other groups are not coherent, simply by embedding copies of in them. For instance, we can embed in as it is generated by

Using this we can embed in with generators , , and . And similarly we can embed in with . So we have that these linear groups are also not coherent.

*Example: Surface groups are coherent*

Surface groups are 1-relator groups. Moreover, any subgroup of a surface group is another surface group, and thus is finitely presented. This is a nice easy case as surfaces have a powerful underlying topological restriction, namely the classification of surfaces.

In fact it is also known that the fundamental groups of 3-manifolds are coherent, and similarly it is the underlying topological properties that allow this to be shown, specifically the Scott core theorem.

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Hey Amin, nice blog!

A question:

What if we let X be the wedge of two circles, then take Y to be the direct product of two copies of X. If we thicken Y slightly we obtain a 3-manifold. However the fundamental group of Y is unaffected by this thickening, remaining F2 x F2, which as you have already said is incoherant?

Don’t know if you are still reading this, but the mapping class group contains copies of $F_2 \times F_2$ once the genus is at least $2$, so it is not coherent.

Hey – thanks for the reply! I know how to get an sitting in there (take Dehn twists about curves with and use ping pong lemma). I guess then with genus 2 or higher we have enough room to do this twice disjointly?

Yes, that is correct. By the way, you don’t need the ping pong lemma in this case (i.e. you don’t need to take powers); see

MR1435728 (97m:57015) Reviewed

Ishida, Atsushi

The structure of subgroup of mapping class groups generated by two Dehn twists.

Proc. Japan Acad. Ser. A Math. Sci. 72 (1996), no. 10, 240–241.