Let $\Gamma_{g,b}^m$ denote the mapping class group of a genus $g$ surface with $b$ non-permutable parametrised boundary curves and $m$ permutable punctures.

It is clear that in general, presentations for these groups are hard. I only need for the general case a set of generators (I want to show that two morphisms $\varphi,\psi:\Gamma\to G$ are equal and I want to check this on the generators). Partial answers are the following:

  1. If $b,m=0$ (so we have a closed surface), then the group is generated by Dehn twists which can be easily drawn on the surface.
  2. If $g,m=0$, we can declare one boundary curve as “outer” and the group is generated by Dehn twists along the inner boundary curves and the pure braid generators $\alpha_{ij}$.

From Farb–Margalit, I know that there are always finitely many Dehn twists (or half twists) which generate $\Gamma$, but can we say in general where they are?


See the paper

B. Wajnryb, "An elementary approach to the mapping class group of a surface," Geometry & Topology 3 (1999) 405–466.

See also "A finite presentation of the mapping class group of an oriented surface," by Gervais: https://arxiv.org/pdf/math/9811162.pdf.

  • $\begingroup$ Thank you, the Gervais paper is really useful! However, it seems not to cover the case with punctures. Of course, we have a surjection $\Gamma_{g, b+m}\to P\Gamma^m_{g, b}$ onto the pure MCG, so we can also describe generators there, but what about $\Gamma_{g, b}^m$ itself? $\endgroup$ – FKranhold Mar 25 '19 at 14:02
  • $\begingroup$ Well to get the whole group you need all the permutations of the punctures, which you can get by throwing in half twists about curves bounding twice punctured disks. $\endgroup$ – Autumn Kent Mar 25 '19 at 15:51
  • $\begingroup$ Okay, so these are $\binom{m}{2}$ additional generators, right? (maybe not all necessary) $\endgroup$ – FKranhold Mar 25 '19 at 16:06
  • $\begingroup$ You can get away with a transposition (one half twist) and a cycle (which can be done using a mapping class supported on an annulus containing all the punctures). So just two more generators. $\endgroup$ – Autumn Kent Mar 25 '19 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.