# Minimal number of (Dehn twists) generators of the mapping class group of a marked sphere

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 $$g=0$$ ?
• 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). – YCor Oct 16 '20 at 21:42
• Yeah sorry, you are right, I am interesting only in Dehn twist generators – 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.