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The braid group on 3 strands is a central extension of the modular group. By definition, \[ B_3 = \langle \sigma_1, \sigma_2: \sigma_1\sigma_2\sigma_1=\sigma_2\sigma_1\sigma_2 \rangle \] This group has a central element (commuting with both $\sigma1$ and $ \sigma_2$): \[ \sigma_1\sigma_2\sigma_1\sigma_2\sigma_1\sigma_2\] The coset get mapped to elements of PSL(2,Z) (which can act on the hyperbolic plane). \[ [\sigma_1] = \left[ \begin{array}{cc} 1 & 1 \\\\ 0 & 1\end{array}\right] \text{ and } [\sigma_2] = \left[ \begin{array}{cc} 1 & 0 \\\\ -1 & 1\end{array}\right] \] I wonder, in terms of the hyperbolic plane, what is being braided here (modulo the garside elements).

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If something were being braided, I'd think the relevant group would have a map to $B_3$, rather than from $B_3$. –  S. Carnahan Nov 1 '11 at 18:28
    
Related: mathoverflow.net/questions/20281/… –  Qiaochu Yuan Nov 1 '11 at 21:03
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Without thinking about this too carefully: I think what's getting braided are three of the Weierstrass points of an elliptic curve. More precisely: consider the space of distinct 3-tuples of points p,q,r on A^1. On the one hand, you can braid these points around. On the other hand, every path in this space (i.e. every braid) gives a family of elliptic curves

y^2 = (x-p)(x-q)(x-r)

and you can ask what the braid does to the homology of the elliptic curve; that's an element of SL_2(Z).

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Don't you need some other identification between the homology of two elliptic curves in a family to get an element of $\text{SL}_2(\mathbb{Z})$? Otherwise you just get a homomorphism between two groups which are abstractly isomorphic to $\mathbb{Z}^2$. –  Qiaochu Yuan Nov 1 '11 at 21:06
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@Qiaochu: That's what you get from the Gauss-Manin connection. –  Dan Petersen Nov 1 '11 at 21:24
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