Question

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).

Answer 1

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).