## The 600-Cell (Part 2)

This is a compound of five tetrahedra:

It looks like a scary, almost random way of slapping together 5 regular tetrahedra until you realize what’s going on. A regular dodecahedron has 20 vertices, while a regular tetrahedron has 4. Since 20 = 4 × 5, you can try to partition the dodecahedron’s vertices into the vertices of five tetrahedra. And it works!

The result is the compound of five tetrahedra. It comes in in two mirror-image forms.

I want to tell you about a 4-dimensional version of the same thing. Amazingly, the 4-dimensional version arises from studying the symmetries of the 3-dimensional thing you see above! The symmetries of the tetrahedron gives a 4-dimensional regular polytope called the ’24-cell’, while the symmetries of the dodecahedron give one called the ‘600-cell’. And there’s a way of partitioning the vertices of the 600-cell into 5 sets, each being the vertices of a 24-cell! So, we get a ‘compound of five 24-cells’.

To see how this works, we need to think about symmetries.

Any rotational symmetry of the dodecahedron acts to permute the tetrahedra in our compound of five tetrahedra. We can only get even permutations this way, but we can get any even permutation. Furthermore, knowing this permutation, we can tell what rotation we did. So, by the marvel of mathematical reasoning, the rotational symmetry group of the dodecahedron must be the alternating group $\mathrm{A}_5.$

On the other hand, the rotational symmetry group of the tetrahedron is $\mathrm{A}_4,$ since any rotation gives an even permutation of the 4 vertices of the tetrahedron.

If we pick any tetrahedron in our compound of five tetrahedra, its rotational symmetries give rotational symmetries of the dodecahedron. So, these symmetries form a subgroup of $\mathrm{A}_5$ that is isomorphic to $\mathrm{A}_4.$ There are exactly 5 such subgroups—one for each tetrahedron.

So, to the eyes of a group theorist, the tetrahedra in our compound of five tetrahedra are just the $\mathrm{A}_4$ subgroups of $\mathrm{A}_5.$ $\mathrm{A}_5$ acts on itself by conjugation, and this action permutes these 5 subgroups. Indeed, it acts to give all the even permutations—after all, it’s $\mathrm{A}_5.$

So, the compound of five tetrahedra has dissolved into group theory, with each figure becoming a group, and the big group acting to permute its own subgroups!

All this is just the start of a longer story about compounds of Platonic solids:

Dodecahedron with 5 tetrahedra, Visual Insight, 15 May 2015.

But only recently did I notice how this story generalizes to four dimensions. Just as we can inscribe a compound of five tetrahedra in the dodecahedron, we can inscribe a compound of five 24-cells in a 600-cell!

Here’s how it goes.

The rotational symmetry group of a tetrahedron is contained in the group of rotations in 3d space:

$\mathrm{A}_4 \subset \mathrm{SO}(3)$

so it has a double cover, the binary tetrahedral group

$2\mathrm{T} \subset \mathrm{SU}(2)$

and since we can see $\mathrm{SU}(2)$ as the unit quaternions, the elements of the binary tetrahedral group are the vertices of a 4d polytope! This polytope obviously has 24 vertices, twice the number of elements in $\mathrm{A}_4$—but less obviously, it also has 24 octahedral faces, so it’s called the 24-cell:

Similarly, the rotational symmetry group of a dodecahedron is contained in the group of rotations in 3d space:

$\mathrm{A}_5 \subset \mathrm{SO}(3)$

so it has a double cover, usually called the binary icosahedral group

$2\mathrm{I} \subset \mathrm{SU}(2)$

and since we can see $\mathrm{SU}(2)$ as the unit quaternions, the elements of the binary icosahedral group are the vertices of a 4d polytope! This polytope obviously has 120 vertices, twice the number of elements in $\mathrm{A}_5$—but less obviously, it has 600 tetrahedral faces, so it’s called the 600-cell:

Each way of making $\mathrm{A}_4$ into a subgroup of $\mathrm{A}_5$ gives a way of making the binary tetrahedral group $2\mathrm{T}$ into a subgroup of the binary dodecahedral group $2\mathrm{I}$… and thus a way of inscribing the 24-cell in the 600-cell!

Next, since 120 = 24 × 5, you can try to partition the 600-cell’s vertices into the vertices of five 24-cells. And it works!

And it’s easy: just take a $2\mathrm{T}$ subgroup of $2\mathrm{I}$, and consider the cosets of this subgroup. Each coset gives the vertices of a 24-cell inscribed in the 600-cell, and there are 5 of these cosets, all disjoint.

So, we get a compound of five 24-cells, whose vertices are those of the 600-cell.

This leads to another question: how many ways can we fit a compound of five 24-cells into a 600-cell?

The answer is 10. It’s easy to get ahold of 10, so the hard part is proving there are no more. Coxeter claims it’s true in the footnote in Section 14.3 of his Regular Polytopes, in which he apologizes for criticizing someone else who earlier claimed it was true:

Thus Schoute (6, p. 231) was right when he said the 120 vertices of {3,5,3} belong to five {3,4,3}’s in ten different ways. The disparaging remark in the second footnote to Coxeter 4, p. 337, should be deleted.

I believe neither of these references has a proof! David Roberson has verified it using Sage, as explained in his comment on a previous post. But it would still be nice to find a human-readable proof.

To see why there are at least 10 ways to stick a compound of five 24-cells in the 600-cell, go here:

• John Baez, How many ways can you inscribe five 24-cells in a 600-cell, hitting all its vertices?, MathOverflow, 15 December 2017.

Puzzle. Are five of these ways mirror-image versions of the other five?

### Image credits

You can click on any image to see its source. The first image of the compound of five tetrahedra was made using Robert Webb’s Stella software and placed on Wikicommons. The rotating compound of five tetrahedra in a dodecahedron was made by Greg Egan and donated to my blog Visual Insight. The rotating 24-cell and 600-cell were made by Jason Hise and put into the public domain on Wikicommons.

### 7 Responses to The 600-Cell (Part 2)

• as well as here with this animation:

• John Baez says:

Pictures sometimes don’t show up when people other than me try to post them as comments here. I can add the animation if you tell me the link.

• jgmoxness says:

a normal link to just the animated gif here

and another try at an embedded img link:

• John Baez says:

Thanks! Now everyone can see what works and what doesn’t.

1. John Baez says:

It turns out David Richter built a model of the compound of five 24-cells. He has a nice discussion of it, but the final product is here:

2. First, look at the orthoplexes sitting inside the 24-cell! We’ve got 8-element subgroup of a 24-element group:

$\mathrm{Q} \subset 2\mathrm{T}$

so it has three right cosets, each forming the vertices of an orthoplex inscribed in the 24-cell. So, we get compound of three orthoplexes: a way of partitioning the vertices of the 24-cell into those of three orthoplexes.

Second, look at the orthoplexes sitting inside the 600-cell! We’ve got 8-element subgroup of a 120-element group:

$\mathrm{Q} \subset 2\mathrm{I}$

so it has 15 right cosets, each forming the vertices of an orthoplex inscribed in the 600-cell. So, we get a compound of 15 orthoplexes: a way of partitioning the vertices of the 600-cell into those of 15 orthoplexes.

And third, these fit nicely with what we saw last time: the 24-cells sitting inside the 600-cell! We saw a 24-element subgroup of a 120-element group

$2\mathrm{T} \subset 2\mathrm{I}$

so it has 5 right cosets, each forming the vertices of a 24-cell inscribed in the 600-cell. That gave us the compound of five 24-cells: a way of partitioning the vertices of the 600-cell into those of five 24-cells.

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