# Subgroup of mapping class group generated by two Dehn twists

Let $S$ be a surface (possibly with boundary, and punctures), and let $\alpha,\beta$ be two simple closed curves on $S$ which intersect once. If $a,b$ denote the isotopy classes of $\alpha,\beta$, respectively, then why is the subgroup of $\text{Mod}(S)$ generated by $T_a,T_b$ isomorphic to the braid group $B_3$? [Here, $T_a$ is the Dehn twist around $\alpha$, and $T_b$ the Dehn twist around $\beta$.] I understand why the relation $T_aT_bT_a=T_bT_aT_b$ holds, but why is this the only relation?

If it makes any difference, I am reading the "Primer on Mapping Class Groups" by Farb and Margalit (available here); they claim this is true, but give no proof. The relevant section in that PDF is 3.5, specifically pages 91--94 (in the PDF).

Thanks for any help, Steve

• More generally, if the curves intersect zero times, they generate an abelian group. If once, they generate B<sub>3</a>, by the Birman-Hilden argument which Andy Putman outlined. And if $\geq 2$ times, there are no relations between them by Ishida: projecteuclid.org/DPubS/Repository/1.0/… May 19 '11 at 12:02
• By the way, this fact was known long before Ishida - it is in Thurston's notes from 1976! Mar 11 '13 at 14:54

A regular neighborhood of these two curves is a 1-holed torus, so you're question is equivalent to asking why the mapping class group $M_{1,1}$ of a 1-holed torus is isomorphic to the 3-strand braid group $B_3$. The key observation is that you can construct a homomorphism $f : M_{1,1} \rightarrow B_3$ as follows. Let $i : \Sigma_{1,1} \rightarrow \Sigma_{1,1}$ be the hyperelliptic involution. This is not an element of $M_{1,1}$ since it rotates the boundary component by $\pi$, but it lies in a degree 2 extension $M_{1,1}'$ of $M_{1,1}$. In fact, $i \in M_{1,1}'$ lies in the center of $M_{1,1}'$. It follows that $M_{1,1} \cong M_{1,1}' / \langle i \rangle$ acts (modulo homotopy) on the quotient $\Sigma_{1,1}/i$, which is a disc with $3$ punctures. Since $B_3$ is the mapping class group of a $3$-punctured disc, this give you a homomorphism $f : M_{1,1} \rightarrow B_3$. The dehn twists $T_{\alpha}$ and $T_{\beta}$ go to the standard generators of $B_3$, so the proof is completed by observing that they satisfy the braid relation.
• You also need to argue that $M_{1,1} \to Mod(S)$ is injective. Feb 23 '11 at 15:51
• I suppose you also need to know that if $S$ is a subsurface (namely, the neighborhood of the curves) of $\Sigma$ then the mapping class group of $S$ injects into the mapping class group of $\Sigma$. Feb 23 '11 at 15:51
• True! Otherwise the argument would work for the closed torus. This, however, is an general fact about the mapping class group. Let $S' \hookrightarrow S$ be a connected subsurface. There is an induced map $Mod(S') \rightarrow Mos(S)$ which is injective unless some component of $S \setminus S'$ is homeomorphic to a disc or an annulus both of whose boundary components lie in $S'$. I'm sure this is proven somewhere in the Primer, but in case it is not it is contained in the paper "Geometric subgroups of mapping class groups" by Rolfsen and Paris. Feb 23 '11 at 16:00
• Actually, if you believe the Dehn-Nielsen-Baer theorem (a special case of which says that the map $Mod(S) \rightarrow Out(\pi_1(S))$ is injective), then you could also deduce this from the fact by examining how $M_{1,1}$ acts on $\pi_1(S)$. Feb 23 '11 at 16:07
• Actually, the determination of $\text{Mod}(S_{1,1}$ is done in the Primer, at the end of Chapter 3. The result Andy mentions in his comment is also there (called the inclusion homomorphism), where a complete description of the kernel in general is given. I guess I just didn't see how to put it together! But does this work for the closed torus? Shouldn't the kernel of the map you desribed be $(T_aT_b)^6$, that is, $\langle T_a, T_b\rangle$ is not isomorphic to $B_3$ (but in fact all of $\text{Mod}(T^2)\cong SL(2,\mathbb{Z})$) ? Feb 23 '11 at 16:12