You will likely find Section III. CONSTRUCTING E8 FROM THE H4 600-CELL interesting, since it has references and discussion related to this post (and cites your paper, along with Dechant’s on the same topic):

[4] P.-P. Dechant, Proceedings of the Royal Society of London Series A 472, 20150504 (2016), 1602.05985.

[5] J. C. Baez, ArXiv e-prints (2017), 1712.06436.

I still don’t know the answers to puzzles 1 and 2. I knew the answers to puzzles 3, 5 and 6, though I only knew the answer to puzzle 6 due to a footnote in Coxeter’s *Regular Polytopes* which doesn’t contain any proof, just a statement of the answer, so it’s reassuring to have some confirmation. I didn’t know the answer to puzzle 4; now I do! So there are 280 ways to partition the 120 vertices of the 600-cell into the vertices of 15 orthoplexes! Great.

I don’t mind someone using a computer for these. *Getting the answer at all* is hard enough. It’s often easier to come up with a human-readable proof after you already know the answer.

Merry Christmas! It’ll take me a while to really digest what you wrote, since I’ve never heard of ‘association schemes’, but I wanted to reply right away.

]]>This makes things a bit easier (for me) to construct the classes of this scheme in Sage. The graph of the 600-cell is built into Sage, and then I can either find the orbitals of its automorphism group, or I can compute the projection onto its second largest eigenspace. This eigenspace has dimension four, and the projection onto it is (up to a scalar) the Gram matrix of the 120 vectors above. So you can use the entries of this matrix to construct the scheme.

You can use this scheme to help with puzzles 3-6, though it seems you mostly already know the answers. I will write some things anyways in case it is of interest.

For puzzles 3 and 4, you can consider the union of all of the classes of the scheme except those corresponding to or . Call this graph . Then an orthoplex will be an independent set of size 8 in . Conversely, it is not hard to see that any independent set of size 8 in this graph must correspond to an orthoplex, since any two vectors in an independent set must be either orthogonal or antipodal. You can ask Sage and it will tell you that there are exactly 75 independent sets of size 8 in this graph, so there are 75 orthoplexes in the 600-cell, but you probably already knew this. Moreover, any partition of the 600-cell into orthoplexes corresponds to a coloring of this graph with 15 colors and vice versa. Sage tells me that there are 280 such colorings. Not exactly a proof though.

For puzzles 5 and 6, you can consider the union of all the classes of the scheme except those with . Call this graph . A 24-cell corresponds to an independent set of size 24 in . It is not as easy to see that any independent set of size 24 necessarily corresponds to a 24-cell, but we will get some help with this later. A partition of the 600-cell into five 24-cells corresponds to a 5-coloring of . This is an optimal coloring of and Sage tells me that there are 10 of these (actually, the graph of the 600-cell also has exactly 10 5-colorings). So there are at most 10 partitions of the 600-cell into 24-cells, but you already know that there are at least 10 by the coset construction, and so every 5-coloring of does actually come from one of these partitions. In this case, we can use this to show that the independent sets of size 24 in do in fact correspond to 24-cells. This is because (and in fact all the graphs in the scheme) is a “normal” Cayley graph. This means that its connection set is closed under conjugation by any element of the group, i.e., it is a union of conjugacy classes. Actually every single class in the scheme corresponds exactly to a single conjugacy class in this way. I put “normal” in quotes because some people use “normal Cayley graph” to mean something else. Anyways, is a normal Cayley graph and it has clique number 5 and independence number 24, and 5*24 = 120 which is the number of vertices in . This means that if and are maximum cliques and maximum independent sets respectively, then the sets for form a partition of into maximum independent sets, i.e., they give a 5-coloring of . If there were any maximum independent set in that did not correspond to one of the cosets from your construction, then we could use the above construction to give a 5-coloring of that would necessarily be different (no color class forms a subgroup) from the 10 you constructed. Therefore, every maximum independent set of must correspond to a 24-cell. Sage tells me that there are 25 maximum independent sets of , and so there are 25 24-cells contained in the 600-cell. Interestingly, the graph of the 600-cell (which is a subgraph of ) also has exactly 25 independent sets of size 24 (which is maximum) and so these must be the same.

You don’t really need to use association schemes for any of this, you could have just constructed the graphs and directly. But it is interesting that there is an association scheme in the background, and it could maybe be useful for turning some of these computations into human proofs, since there is a lot of stuff known about independent sets in association schemes.

By the way, for finding and counting colorings I used a Gap package called Digraphs because Sage’s built-in coloring functions are not nearly fast enough.

Merry Christmas!

P.S. John, I think we briefly shared an office at CQT in Singapore. Unfortunately, I don’t think we spoke to each other (usually I was not in the office). Sorry for being antisocial.

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