The Binary Octahedral Group (Part 2)

Part 1 introduced the ‘binary octahedral group’. This time I just want to show you some more pictures related to this group. I’ll give just enough explanation to hint at what’s going on. For more details, check out this webpage:

• Greg Egan, Symmetries and the 24-cell.

Okay, here goes!

You can inscribe two regular tetrahedra in a cube:

Each tetrahedron has 4! = 24 symmetries permuting its 4 vertices.

The cube thus has 48 symmetries, twice as many. Half map each tetrahedron to itself, and half switch the two tetrahedra.

If we consider only rotational symmetries, not reflections, we have to divide by 2. The tetrahedron has 12 rotational symmetries. The cube has 24.

But the rotation group SO(3) has a double cover SU(2). So the rotational symmetry groups of tetrahedron and cube have double covers too, with twice as many elements: 24 and 48, respectively.

But these 24-element and 48-element groups are different from the ones mentioned before! They’re called the binary tetrahedral group and binary octahedral group—since we could have used the symmetries of an octahedron instead of a cube.

Now let’s think about these groups using quaternions. We can think of SU(2) as consisting of the ‘unit quaternions’—that is, quaternions of length 1. That will connect what we’re doing to 4-dimensional geometry!

The binary tetrahedral group

Viewed this way, the binary tetrahedral group consists of 24 unit quaternions. 8 of them are very simple:

These form a group called the quaternion group, and they’re the vertices of a shape that’s the 4d analogue of a regular octahedron. It’s called the 4-dimensional cross-polytope and it looks like this:

The remaining 16 elements of the binary tetrahedral group are these:

They form the vertices of a 4-dimensional hypercube:

Putting the vertices of the hypercube and the cross-polytope together, we get all 8 + 16 = 24 elements of the binary tetrahedral group. These are the vertices of a 4-dimensional shape called the 24-cell:

This shape is called the 24-cell not because it has 24 vertices, but because it also has 24 faces, which happen to be regular octahedra. You can see one if you slice the 24-cell like this:

The slices here have real part 1, ½, 0, -½, and -1 respectively. Note that the slices with real part ±½ contain the vertices of a hypercube, while the rest contain the vertices of a cross-polytope.

And here’s another great way to think way about binary tetrahedral group. We’ve seen that if you take every other vertex of a cube you get the vertices of a regular tetrahedron. Similarly, if you take every other vertex of a 4d hypercube you get a 4d cross-polytope. So, you can take the vertices of a 4d hypercube and partition them into the vertices of two cross-polytopes.

As a result, the 24 elements of the binary tetrahedral group can be partitioned into three cross-polytopes! Greg Egan shows how it looks:

The binary octahedral group

Now that we understand the binary tetrahedral group pretty well, we’re ready for our actual goal: understanding the binary octahedral group! We know this forms a group of 48 unit quaternions, and we know it acts as symmetries of the cube—with elements coming in pairs that act on the cube in the same way, because it’s a double cover of the rotational symmetry group of the cube.

So, we can partition its 48 elements into two kinds: those that preserve each tetrahedron in this picture, and those that switch these two tetahedra:

The first 24 form a copy of the binary tetrahedral group and thus a 24-cell, as we have discussed. The second form another 24-cell! And these two separate 24-cells are ‘dual’ to each other: the vertices of one hover right above the faces of the other.

Greg has nicely animated the 48 elements of the binary octahedral group here:

He’s colored them according to the rotations of the cube they represent:

• black: identity • red: ±120° rotation around a V axis • yellow: 180° rotation around an F axis • blue: ±90° rotation around an F axis • cyan: 180° rotation around an E axis

Here ‘V, F, and E axes’ join opposite vertices, faces, and edges of the cube.

Finally, note that because

• we can partition the 48 vertices of the binary octahedral group into two 24-cells

and

• we can partition the 24 vertices of the 24-cell into three cross-polytopes

it follows that we can partition the 48 vertices of the binary octahedral group into six cross-polytopes.

I don’t know the deep meaning of this fact. I know that the vertices of the 24-cell correspond to the 24 roots of the Lie algebra I know that the the famous ‘triality’ symmetry of permutes the three cross-polytopes in the 24-cell, which are in some rather sneaky way related to the three 8-dimensional irreducible representations of I also know that if we take the two 24-cells in the binary octahedral group, and expand one by a factor of so the vertices of other lie exactly at the center of its faces, we get the 48 roots of the Lie algebra But I don’t know how to extend this story to get a nice story about the six cross-polytopes in the binary octahedral group.

All I know is that if you pick a quaternion group sitting in the binary octahedral group, it will have 6 cosets, and these will be six cross-polytopes.