16
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.