Let $\Gamma_{g,n}$ denote the mapping class group of an oriented surface of genus $g$ and with $n$ marked points. We assume that elements of $\Gamma_{g,n}$ are not allowed to permute the marked points. I am interested in the case $g=0$.

In Farb & Margalit, on page 114, it is claimed that $\Gamma_{g,n}$ can be generated by $2g+n$ Dehn twists along the curves drawn on Figure 4.10. I was wondering if this statement is also true in the case $g=0$.

In math/9912248, Wajnryb exhibits a family of generators for $\Gamma_{0,n}$ in Lemma 23. The curves $\alpha_{i,j}$ illustrated on Figure 12 are a family of $n(n-1)/2$ generators.

My questions are the following:

  1. Is the minimal number of Dehn twists generators of $\Gamma_{0,n}$ known ?
  2. What would be the $n$ generators of $\Gamma_{0,n}$ if the construction of Farb-Margalit cited above applies in the case $g=0$ ?
  • $\begingroup$ The question in the title is distinct from the question in main text (minimal cardinal of generating subset vs minimal cardinal of generating subset consisting of Dehn twists). $\endgroup$ – YCor Oct 16 '20 at 21:42
  • $\begingroup$ Yeah sorry, you are right, I am interesting only in Dehn twist generators $\endgroup$ – Arnaud Maret Oct 17 '20 at 10:49

The minimum number of Dehn twist generators (and in fact the minimum number of generators of any kind) for $\Gamma_{0,n}$ is ${n-1 \choose 2} - 1$. Here's why.

A presentation for $\Gamma_{0,n}$ is known, and can be found in Lemma 4.1 of this paper by Rebecca R. Winarski and myself. In the paper, $\operatorname{PMod}(\Sigma_0,\mathcal B(n))$ is the group $\Gamma_{0,n}$.

Number the $n$ marked points. The generators $A_{i,j}$ are Dehn twists about curves that surround only the $i$th and $j$th marked points (see Figure 3 from the paper). These are essentially the same curves in the paper by Wajnryb that is linked in the question.

The generating set used in our paper is the set $\{A_{i,j} \mid 1 \leq i < j \leq n-1\}$, and one of the relations (relation (5)) is $$(A_{1,2}A_{1,3} \cdots A_{1,n-1})\cdots(A_{n-3,n-2}A_{n-3,n-1})(A_{n-2,n-1}) = 1. $$ In this relation, each $A_{i,j}$ appears exactly once, so you can use a Tietze transformation to eliminate one of the generators. We are now left with a generating set consisting of ${{n-1}\choose{2}} - 1$ Dehn twists.

The other 4 relations are all commutation relations (that is, of the form $[W,X] = 1$), so we can conclude that the abelianization of $\Gamma_{0,n}$ is a free abelian group of rank ${n-1 \choose 2} - 1$. Therefore $\Gamma_{0,n}$ cannot be generated by less than ${n-1 \choose 2} - 1$ elements.

I guess this also answers your question 2, in the sense that the result stated in Farb & Margalit does not hold for genus 0.


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.