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 H_{4} 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 E_{8}and show how the 600-cell transforms E_{8}/2E_{8}, an 8-space over the field F_{2}, into a 4-space over F_{4}whose points, lines and planes are labeled by the geometric objects of the 600-cell.

Yes, if you take the E_{8} 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

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!