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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 $g=0$ ?
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  • $\begingroup$ 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
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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.

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