The Binary Octahedral Group

Azimuth 2019-08-28

The complex numbers together with infinity form a sphere called the Riemann sphere. The 6 simplest numbers on this sphere lie at points we could call the north pole, the south pole, the east pole, the west pole, the front pole and the back pole. They’re the corners of an octahedron!

On the Earth, I’d say the “front pole” is where the prime meridian meets the equator at 0°N 0°E. It’s called Null Island, but there’s no island there—just a buoy. Here it is:

Where’s the back pole, the east pole and the west pole? I’ll leave two of these as puzzles, but I discovered that in Singapore I’m fairly close to the east pole:

If you think of the octahedron’s corners as the quaternions $\pm i, \pm j, \pm k,$ you can look for unit quaternions $q$ such that whenever $x$ is one of these corners, so is $qxq^{-1}$ . There are 48 of these! They form a group called the binary octahedral group.

By how we set it up, the binary octahedral group acts as rotational symmetries of the octahedron: any transformation sending $x$ to $qxq^{-1}$ is a rotation. But this group is a double cover of the octahedron’s rotational symmetry group! That is, pairs of elements of the binary octahedral group describe the same rotation of the octahedron.

If we go back and think of the Earth’s 6 poles as points $0, \pm 1,\pm i, \infty$ on the Riemann sphere instead of $\pm i, \pm j, \pm k$ , we can think of the binary octahedral group as a subgroup of $\mathrm{SL}(2,\mathbb{C}$ ), since this acts as conformal transformations of the Riemann sphere!

If we do this, the binary octahedral group is actually a subgroup of $\mathrm{SU}(2)$ , the double cover of the rotation group—which is isomorphic to the group of unit quaternions. So it all hangs together.

It’s fun to actualy see the unit quaternions in the binary octahedral group. First we have 8 that form the corners of a cross-polytope (the 4d analogue of an octahedron):

$\pm 1, \pm i , \pm j , \pm k$

These form a group on their own, called the quaternion group. Then we have 16 that form the corners of a hypercube (the 4d analogue of a cube, also called a tesseract or 4-cube):

$\displaystyle{ \frac{\pm 1 \pm i \pm j \pm k}{2} }$

These don’t form a group, but if we take them together with the 8 previous ones we get a 24-element subgroup of the unit quaternions called the binary tetrahedral group. They’re also the vertices of a 24-cell, which is yet another highly symmetrical shape in 4 dimensions (a 4-dimensional regular polytope that doesn’t have a 3d analogue).

That accounts for half the quaternions in the binary octahedral group! Here are the other 24:

$\displaystyle{ \frac{\pm 1 \pm i}{\sqrt{2}}, \frac{\pm 1 \pm j}{\sqrt{2}}, \frac{\pm 1 \pm k}{\sqrt{2}}, }$

$\displaystyle{ \frac{\pm i \pm j}{\sqrt{2}}, \frac{\pm j \pm k}{\sqrt{2}}, \frac{\pm k \pm i}{\sqrt{2}} }$

These form the vertices of another 24-cell!

The first 24 quaternions, those in the binary tetrahedral group, give rotations that preserve each one of the two tetrahedra that you can fit around an octahedron like this:

while the second 24 switch these tetrahedra.

The 6 elements

$\pm i , \pm j , \pm k$

describe 180° rotations around the octahedron’s 3 axes, the 16 elements

$\displaystyle{ \frac{\pm 1 \pm i \pm j \pm k}{2} }$

describe 120° clockwise rotations of the octahedron’s 8 triangles, the 12 elements

$\displaystyle{ \frac{\pm 1 \pm i}{\sqrt{2}}, \frac{\pm 1 \pm j}{\sqrt{2}}, \frac{\pm 1 \pm k}{\sqrt{2}} }$

describe 90° clockwise rotations holding fixed one of the octahedron’s 6 vertices, and the 12 elements

$\displaystyle{ \frac{\pm i \pm j}{\sqrt{2}}, \frac{\pm j \pm k}{\sqrt{2}}, \frac{\pm k \pm i}{\sqrt{2}} }$

describe 180° clockwise rotations of the octahedron’s 6 opposite pairs of edges.

Finally, the two elements

$\pm 1$

do nothing!

So, we can have a lot of fun with the idea that a sphere has 6 poles.