First let's recall some definitions. Let $G$ be a perfect group, so that

$$H^2(G, A) \cong \text{Hom}(H_2(G), A)$$

for all abelian groups $A$ by universal coefficients. This means that when $A = H_2(G, \mathbb{Z})$ there is a distinguished class in $H^2(G, A)$ corresponding to the identity $H_2(G) \to H_2(G)$. The universal central extension of $G$ is the central extension classified by this map; it fits into a short exact sequence

$$1 \to H_2(G) \to \widetilde{G} \to G \to 1.$$

Now, you can find it claimed in many places that the braid group

$$B_3 \cong \langle a, b \mid a^2 = b^3 \rangle$$

is the universal central extension of the modular group

$$\Gamma \cong PSL_2(\mathbb{Z}) \cong \langle a, b \mid a^2 = b^3 = e \rangle.$$

But there's something fishy about this claim: $\Gamma$ isn't a perfect group! In fact, since $\Gamma \cong \mathbb{Z}_2 \ast \mathbb{Z}_3$, it's clear that $H_1(\Gamma) \cong \mathbb{Z}_6$ (and that $H_2(\Gamma) \cong 0$). So:

What is meant by the claim that $B_3$ is the universal central extension of $\Gamma$?

We have that

$$H^2(\Gamma, \mathbb{Z}) \cong \text{Ext}^1(\mathbb{Z}_6, \mathbb{Z}) \cong \mathbb{Z}_6$$

so presumably $B_3$ is the central extension classified by a generator of this group. But I don't understand in what sense this central extension is universal.

  • $\begingroup$ It has something to do with the fact that it sits (as a lattice) inside the universal cover of $PSL_2(\mathbb{R})$ (as topological group), and we have the commutative diagram of short exact sequences of groups (with the same factor of $\mathbb{Z}$). Googling more, there is this book Moonshine beyond the Monster which argues that $B_3$ is "universal" in some sense, and Baez tries to explain it in his blog post here: math.ucr.edu/home/baez/week233.html $\endgroup$ – Chris Gerig Nov 21 '15 at 19:24
  • $\begingroup$ @Chris: I don't buy this. For example, $\frac{1}{n} \mathbb{Z}/\mathbb{Z}$ sits as a lattice inside $S^1$. The corresponding subgroup of the universal cover $\mathbb{R}$ is $\frac{1}{n} \mathbb{Z}$, which is in no sense (that I'm familiar with) the universal central extension of $\frac{1}{n} \mathbb{Z}/\mathbb{Z}$. $\endgroup$ – Qiaochu Yuan Nov 21 '15 at 19:26
  • 5
    $\begingroup$ Of the possible central extensions by $\mathbb{Z}$ of $\mathbb{Z}/2\mathbb{Z}\ast \mathbb{Z}/3\mathbb{Z}$, this is the only one (up to isomorphism) which is torsion-free. $\endgroup$ – Ian Agol Nov 21 '15 at 23:48

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