The 600-Cell (Part 4)

I get really bored some evenings these days, after I run out of energy to work on my own projects, and before I lie in bed and read Whittaker’s mammoth tome, A History of the Theories of Aether & Electricity. So I’ve taken up browsing the arXiv. It can be quite entertaining! Here’s something I found last night:

• Tomme Denney, Da’Shay Hooker, De’Janeke Johnson, Tianna Robinson, Majid Butler and Sandernisha Claiborne, The geometry of H4 polytopes.

It mentions some cool facts that call for a new installment of this series of mine:

• The 60-cell: Part 1, Part 2, Part 3.

Remember that the 24-cell is a four-dimensional regular polytope with 24 vertices and 24 octahedral faces:

The 600-cell is a four-dimensional regular polytope with 120 vertices and 600 tetrahedral faces:

Since 120/24 = 5, you might hope that there’s a way to partition the 600-cell’s vertices into the vertices of five 24-cells. And indeed there is!

So we get a compound of five 24-cells. It’s a kind of four-dimensional analogue of this picture by Greg Egan, showing a compound of five tetrahedra:

How many ways are there to inscribe a compound of 24-cells in the 600-cell? That is: how many ways are there to partition the 600-cell’s vertices into the vertices of five 24-cells?

This question has an interesting history, which I explained in Part 2. A fellow named P. H. Schoute claimed in 1905 that the answer is 10. In 1933 the famous geometer Coxeter publicly doubted this claim, writing that surely there should be just 5. Later he changed his mind and agreed that Schoute was correct… but still gave no proof. In 2017 David Roberson verified it using computer calculations. But the paper I’m talking finally offers a human-readable proof.

But they show something even better! First: there are exactly 25 ways to inscribe a 24-cell into a 600-cell—that is, ways to find a subset of the 600-cell’s vertices that form the vertices of a 24-cell.

But now for the cool part: we can list these 25 in a 5 × 5 square, so that each row and each column give a different way to inscribe a compound of 24-cells in the 600-cell. So we get a total of 10.

I hope you understood that. If not, maybe the paper’s summary will be clearer:

The 25 24-cells can be placed in a 5×5 array, so that each row and each column of the array partition the 120 vertices of the 600-cell into five disjoint 24-cells. The rows and columns of the array are the only ten such partitions of the 600-cell.

This too was claimed without proof by P. H. Schoute in 1905. A proof is in the paper by Denney, Hooker, Johnson, Robinson, Butler and Claiborne!

There’s a lot more cool stuff in this paper, as hinted at in the abstract:

Abstract. We describe the geometry of an arrangement of 24-cells inscribed in the 600-cell. In §7 we apply our results to the even unimodular lattice E8 and show how the 600-cell transforms E8/2E8, an 8-space over the field F2, into a 4-space over F4 whose points, lines and planes are labeled by the geometric objects of the 600-cell.

Yes, if you take the E8 lattice and mod out by the vectors that are two times vectors in that lattice, you get an 8-dimensional vector space over the field with 2 elements. But you can think of it as a 4-dimensional vector space over 4 elements. How many 1-dimensional subspaces does this vector space have? You can count them, and the answer is

$\frac{4^4 - 1}{4 - 1} = 85$

The paper shows how these correspond to the 60 pairs of opposite vertices in the 600-cell together with the 25 24-cells inscribed in the 600-cell! Wow!

11 Responses to The 600-Cell (Part 4)

1. Jamie Vicary says:

Wow, so that means you get something a bit like a Latin square. In this case it’s more general than a Latin square, since the grid entries hold subsets, rather than elements of the set. I bet there’s more to say about what’s going on here!

2. John Baez says:

Yeah, something sort of like a Latin square!

The trick is that the vertices of the 600-cell form a group: the ‘binary icosahedral group’, called $2I.$ There’s a way to inscribe a 24-cell in it so that its vertices form a subgroup called the ‘binary tetrahedral group’, $2T.$ Then we can choose an element $g \in 2I$ of order 5 such that the 25 different 24-cells are of the form

$g^i 2T g^j$

where $i, j = 0,1,2,3,4.$ So that gives us a 5 × 5 array of them!

• John Baez says:

Hmm, this seems to be studied in somewhat different language here:

• Steve Fisk, Coloring the 600-cell.

Abstract. The 600 cell S has exactly 10 5-colorings. From these colorings we can construct the space of colorings B(S). This complex has 1344 colorings, and is isomorphic to the space of 5 by 5 Latin squares. These simplices split into 4 copies of a quotient of S by an involution, and two copies of a space made up of even Latin squares.

His space B(S) has 25 vertices which seem to be the 25 24-cells inscribed in a 600-cell.

3. Hendrik Boom says:

I once wondered about using the 600-cell as a game map for a space travel game. The tetrahedra could be divided into a coordinate system based on the tetrahedron-octahedron tesselation. But I never got around to actually doing it.

I always thought that space-exploration games with bounded flat two-dimensional universe were kind of stupid.

— hendrik

• Reminds me of a project to use the vertices and edges of various uniform polychora as Go boards.

• John Baez says:

Someday people will have a lot of fun with such games.

4. Victor V Albert says:

In our paper “Robust encoding of a qubit in a molecule” on quantum error correction in group spaces, and, thanks to this blog and Greg Egan’s posts, mentioned codes constructed from tetrahedra inscribed in an icosahedron. Seems like a natural 4D extension can hold…

5. Billb says:

So, reading Whittaker late at night can be very satisfying. It’s a beautiful flight path from past to future, never ends.

• John Baez says:

The early history of research in electromagnetism and optics is wonderful. It’s amazing, for example, how much people learned about electric current back when the only way of measuring its strength was feeling how strong the electrical shock you got from it.

6. John
I might re-read Whittaker’s tome to see what dimension it transports me to!!
Cheers
Peter

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