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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

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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/… –  Daniel Moskovich May 19 '11 at 12:02
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By the way, this fact was known long before Ishida - it is in Thurston's notes from 1976! –  Dan Margalit Mar 11 '13 at 14:54
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up vote 6 down vote accepted

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.

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You also need to argue that $M_{1,1} \to Mod(S)$ is injective. –  Oscar Randal-Williams Feb 23 '11 at 15:51
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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 $\Sigma$. –  aaron 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. –  Andy Putman 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 $\pi_1(S)$. –  Andy Putman 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 $\text{Mod}(T^2)\cong SL(2,\mathbb{Z})$) ? –  Steve D Feb 23 '11 at 16:12
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