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

$480 = \frac{128 \times 7!}{8 \times 168}$

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## kneemo said,

February 6, 2007 @ 6:39 am

24 of the 48 elements of the binary octahedral group you described are units of the E8 lattice. If you work with octonions, there are 24 other E8 units that complete the units you wrote down. Together, they are: ±1, ±i, ±j, ±k, ±e, ±ie, ±je, ±ke, (±1±ie±je±ke)/2, (±e±i±j±k)/2 (see Tony’s E8 page).

If you want the full 240 units of the E8 lattice just include the cyclical permutations of (±1±ie±je±ke)/2, (±e±i±j±k)/2, yielding 7×16 + 7×16 = 224 units. Adding the remaining ±1, ±i, ±j, ±k, ±e, ±ie, ±je, ±ke units gives 240 units.

Perhaps one can describe a sector of the heterotic string where the periodic bosons are restricted to a binary octahedral lattice.

## kneemo said,

February 6, 2007 @ 6:26 pm

Oh ok, the first 24 quaternionic vectors you listed are vertices of a 24-cell centered at the origin of 4-space. The other 24 are vertices of the dual 24-cell.

Collectively, the 48 vertices of the 24-cell and its dual give the root vectors for F4.

## Kea said,

February 6, 2007 @ 8:06 pm

Hi kneemo! Yeah, but I’m glad you mentioned the E8 lattices and gave a link to Tony’s page on them.

## CarlBrannen said,

February 7, 2007 @ 11:20 pm

The numbers you guys are getting from this are very interesting. In particular, 7/9 and 2/9 are Koide type numbers. And I should point out that 9×9 – 7×7 = 2^5.

## Doug said,

February 7, 2007 @ 11:53 pm

Check out the John Baez TWF week 215 blog on Klein’s quartic curve [Greg Egan] with both static and dynamic depictions. I think Tony Smith refers to this blog.

What is fascinating from my perspective is the static genus-3-torus is somewhat reminiscent of the chordate semicircular canals.

When, perhaps if, TOE or GUT is achieved, biophysics will unify with HEP?

This may be a bimatrix with each matrix hacing 24 elements?

## Kea said,

February 8, 2007 @ 2:58 am

Hi guys

Yes, we can be hopeful, but it isn’t all worked out yet!