Let $$ Br_3 := \langle \tau_1,\tau_2\ :\ \tau_1 \tau_2 \tau_1 = \tau_2 \tau_1 \tau_2 \rangle $$ be the braid group on three strands, and consider the surjection $$\phi : Br_3 \twoheadrightarrow SL_2(\mathbb Z), \qquad \tau_1 \mapsto \begin{pmatrix} 1&0\\ 1&1\end{pmatrix}, \quad \tau_2 \mapsto \begin{pmatrix} 1&-1\\ 0&1\end{pmatrix}. $$ According to Wikipedia, this is the universal central extension. Moreover, it arises as the fiber product of the inclusion $SL_2(\mathbb Z) \hookrightarrow SL_2(\mathbb R)$ and the universal cover $\widetilde{SL_2(\mathbb R)} \twoheadrightarrow SL_2(\mathbb R)$.

Is there a satisfying reason why there should be a map $\phi$?

Perhaps something Galois-theoretic, based on $Br_3$ being the fundamental group of the configuration space of $3$ points in the plane?

Feel free to retag. If there are many satisfying reasons, I'll make it community wiki!

ADDED: This is, if not the same question, at least really close to Details for the action of the braid group B_3 on modular forms , and Dylan Thurston's answer is pretty satisfying (even if he isn't satisfied).

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    $\begingroup$ Modulo its center, this braid group is the stabilizer of one puncture in the mapping class group of the 4 times punctured sphere. The latter is is the quotient of the mapping class group of the torus by $\pm 1$ (Divide the torus by the involution). This gives you a natural homonorphism of the braid group to $PSL(2,Z)$. $\endgroup$
    – Misha
    Dec 22, 2014 at 4:37
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    $\begingroup$ There's an elegant discussion of this on pages 83-85 of Milnor's book on algebraic k-theory. $\endgroup$ Dec 22, 2014 at 5:23
  • $\begingroup$ @AndyPutman except that Milnor uses the fact that $Br_3$ is the fundamental group of the trefoil knot complement rather than braid group. $\endgroup$ Dec 22, 2014 at 6:47
  • $\begingroup$ ...and in fact there are generalized braid groups corresponding to Coxeter groups (see e. g. this question) $\endgroup$ Dec 22, 2014 at 7:38

1 Answer 1


This answer is essentially equivalent to Misha's and Dylan's, but phrased in terms of configuration spaces rather than mapping class groups. As you note, $Br_3$ is the fundamental group of the configuration space of 3 points in the (complex) plane, a 3-complex dimensional space. One may take a subspace whose center of mass is at the origin, splitting off $\mathbb{C}$. The conformally equivalent configurations are then related by the action of $\mathbb{C}^\times$. If we only quotient by $\mathbb{R}_+$ (say by normalizing the moment of inertia to be $1$), then we get a 3-dimensional space whose fundamental group is still $B_3$. However, if we quotient by $S^1$, then we get a 2-dimensional orbifold which is the conformal classes of the 4-punctured sphere with a single marked point at $\infty$ (or thrice-punctured $\mathbb{C}$).

The 2-fold branched cover of a 4-punctured $\mathbb{CP}^1$ ramified over each puncture is a torus with a marked point, and covering translation given by the elliptic involution. So $SL_2(\mathbb{Z})$ acts on this space as the usual quotient of the affine action on $\mathbb{R}^2$ descending to $\mathbb{R}^2/\mathbb{Z}^2$. Note however that the orbifold fundamental group of the conformal classes of 3-punctured $\mathbb{C}$ is actually $PSL_2(\mathbb{Z})$, rather than $SL_2(\mathbb{Z})$, since we've quotiented by the center corresponding to the elliptic involution. However, the universal central extention of $PSL_2(\mathbb{Z})$ by $\mathbb{Z}$ certainly maps to the universal $\mathbb{Z}/2\mathbb{Z}$ central extension of $PSL_2(\mathbb{Z})$. But this point makes the geometric correspondence a bit less natural, since the hyperelliptic involution is the lift of the element of $\pi_1$ of the configuration space which rotates a triple of points by $2\pi$.

Maybe another remark, that $PSL_2(\mathbb{Z}) \cong \mathbb{Z}/2\ast \mathbb{Z}/3$, so $H^2(PSL_2(\mathbb{Z}),\mathbb{Z})= \mathbb{Z}/6$. So I suppose there are two universal central extensions, but I guess these are differing by the outer automorphism of $\mathbb{Z}/3$ applied to $\mathbb{Z}/2\ast \mathbb{Z}/3$. If we think of $PSL_2(\mathbb{Z})$ again as the fundamental group of the modular orbifold, then its universal $\mathbb{Z}$-central extension is the fundamental group of the unit tangent bundle to the orbifold. I think the other central extension is also obtained by the action on the lower half-plane, and taking the unit tangent bundle to this orbifold.

Here's another geometric description going in the other direction. Recall that a conformal class of torus is obtained from a point $z\in\mathbb{H}^2$ by taking the quotient $\mathbb{C}/\langle 1,z\rangle$. The unit tangent bundle then gives a unit tangent vector at $z$. Taking the quotient by the elliptic involution, this also gives a unit tangent vector at $\infty$ on the 4-punctured $\mathbb{CP}^1$. There is a unique conformal map to $\mathbb{C}$ which has this vector pointing in a fixed direction, and having center of mass at the origin and normalized moment of inertia. However, again this only gives a natural map to $PSL_2(\mathbb{Z})$, not $SL_2(\mathbb{Z})$.


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