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If $H$ is an infinite index subgroup of the braid group $\mathcal{B}_n$, is there a way to find a presentation for $H$ ?

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  • $\begingroup$ Thanks Mathew. Maybe the braid group is not an easier route after all. To start with, $H$ is actually a subgroup of $Mod(S)$ generated by three or more Dehn twists, where $S$ is a compact, orientable surface with nonempty boundary. $H$ also happens to be a subgroup of the braid group $\mathcal{B}_n$. The braid and disjointness relations are obvious for the generating Dehn twists of $H$. It is not clear, however, whether these are all the defining relations. $\endgroup$
    – Jim B
    Apr 30, 2010 at 1:23
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    $\begingroup$ Do you know them as words in standard generators? Perhaps you could experiment a bit with Magma or Gap using a known presentation. $\endgroup$ Apr 30, 2010 at 1:32
  • $\begingroup$ Thanks Mathew! GAP is not very useful to me. Based on your suggestion, I've played with Magma and got some interesting results. I still couldn't find the presentation I want though. $\endgroup$
    – Jim B
    May 4, 2010 at 3:43
  • $\begingroup$ Your question isn't well-posed. The group multiplication table is a presentation of any group. Are you ruling out things like that, and if so, how? $\endgroup$ Jul 5, 2012 at 6:25

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Unless you have a specific type of subgroup, like one that acts cocompactly on something, you're in deep trouble. Braid groups are incoherent (see Artin groups, 3-manifolds and coherence by Cameron Gordon). That is, there are finitely generated subgroups that are not finitely presented.

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  • $\begingroup$ there are also infinitely generated subgroups. $\endgroup$
    – Ian Agol
    Jul 5, 2012 at 14:33
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I'll address the question in your comments, can one determine a presentation for a subgroup of the mapping class group generated by Dehn twists? Given a bunch of simple closed curves on a surface, there is a minimal essential subsurface containing the curves. Then the Dehn twists will generate a subgroup of the mapping class group of this subsurface, which will have infinite index in the mapping class group of the full surface if it is a proper subsurface.

If it is finite-index in this mapping class group of the subsurface, then there is a procedure which will terminate with a presentation. The problem is that the procedure may never stop if the subgroup is infinite index in this subgroup. The idea is to enumerate finite-index subgroups and their generators via Reidemeister-Schreier, and then see if your elments can generate the generators for these subgroups. If you can, then you know your group is finite-index, and you can test all intermediate subgroups to see if your elements lie in them by looking at their image in a finite group quotient.

If the subsurface is planar, then the subgroup generated by Dehn twists about the curves will be a subgroup of a central extension of the braid group. If your curves are labelled $c_1, c_2, \ldots, c_k$, such that $c_i\cap c_j=\emptyset$ if $|i-j|>1$, and $|c_i\cap c_{i+1}|=2$, and the subsurface is planar, then the group will generate the mapping class of the planar subsurface, since these are the form of the standard generators for the braid group (the central extension comes from Dehn twists around boundary components).

For subgroups generated by 3 Dehn twists, see this paper.

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