Regular Irregular II

The double covering of the Klein quartic is described using the binary octahedral group, a subgroup of the quaternions. As Tony Smith points out, this group is given by the 48 elements

$\pm 1, \pm i, \pm j, \pm k$
$\frac{1}{2} (\pm 1 \pm i \pm j \pm k)$
$\frac{1}{2} (\pm 1 \pm i)$, $\frac{1}{2} (\pm 1 \pm j)$, $\frac{1}{2} (\pm 1 \pm k)$
$\frac{1}{2} (\pm i \pm j)$, $\frac{1}{2} (\pm j \pm k)$, $\frac{1}{2} (\pm k \pm i)$

of a 24 element binary tetrahedral group along with its dual. This is nice, because we were hoping that the genus 3 surface would have something to do with quaternions, since the real dimension of the moduli is 12, the same dimension as $\mathbb{HP}^3$. The binary tetrahedral group can be described by two generators $a$ and $b$ satisfying $a^3 = b^3 = (ab)^2$. There is a quaternion $q$ that satisfies $aq = qb$, namely $q = i \textrm{exp}(\frac{\pi j}{4})$. The binary octahedral group is described by generators $A, B$ and $q$ and relations $A^4 = B^3 = q^2 = ABq$.

This is all secretly about $SL(2,7)$, of order 336. Now it so happens that $PSL(2,7)$, of order 168 (recall the $24 \times 7$ from the last post), is about products for octonions. $PSL(2,7)$ is the group of linear fractional transformations of a heptagon, or the Fano plane. Remember that there are 7 imaginary octonions, just like there were 3 for the quaternions. That means $2^7 = 128$ possible sign combinations. The 480 octonion products come from
$480 = \frac{128 \times 7!}{8 \times 168}$