## The 600-Cell (Part 4)

30 November, 2020

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!

## The Kepler Problem (Part 1)

7 January, 2018

Johannes Kepler loved geometry, so of course he was fascinated by Platonic solids. His early work Mysterium Cosmographicum, written in 1596, includes pictures showing how the 5 Platonic solids correspond to the 5 elements:

Five elements? Yes, besides earth, air, water and fire, he includes a fifth element that doesn’t feel the Earth’s gravitational pull: the ‘quintessence’, or ‘aether’, from which heavenly bodies are made.

In the same book he also tried to use the Platonic solids to explain the orbits of the planets:

The six planets are Mercury, Venus, Earth, Mars, Jupiter and Saturn. And the tetrahedron and cube, in case you’re wondering, sit outside the largest sphere shown above. You can see them another picture from Kepler’s book:

These ideas may seem goofy now, but studying the exact radii of the planets’ orbits led him to discover that these orbits aren’t circular: they’re ellipses! By 1619 this led him to what we call Kepler’s laws of planetary motion. And those, in turn, helped Newton verify Hooke’s hunch that the force of gravity goes as the inverse square of the distance between bodies!

In honor of this, the problem of a particle orbiting in an inverse square force law is called the Kepler problem.

So, I’m happy that Greg Egan, Layra Idarani and I have come across a solid mathematical connection between the Platonic solids and the Kepler problem.

But this involves a detour into the 4th dimension!

It’s a remarkable fact that the Kepler problem has not just the expected conserved quantities—energy and the 3 components of angular momentum—but also 3 more: the components of the Runge–Lenz vector. To understand those extra conserved quantities, go here:

• Greg Egan, The ellipse and the atom.

Noether proved that conserved quantities come from symmetries. Energy comes from time translation symmetry. Angular momentum comes from rotation symmetry. Since the group of rotations in 3 dimensions, called SO(3), is itself 3-dimensional, it gives 3 conserved quantities, which are the 3 components of angular momentum.

None of this is really surprising. But if we take the angular momentum together with the Runge–Lenz vector, we get 6 conserved quantities—and these turn out to come from the group of rotations in 4 dimensions, SO(4), which is itself 6-dimensional. The obvious symmetries in this group just rotate a planet’s elliptical orbit, while the unobvious ones can also squash or stretch it, changing the eccentricity of the orbit.

(To be precise, all this is true only for the ‘bound states’ of the Kepler problem: the circular and elliptical orbits, not the parabolic or hyperbolic ones, which work in a somewhat different way. I’ll only be talking about bound states in this post!)

Why should the Kepler problem have symmetries coming from rotations in 4 dimensions? This is a fascinating puzzle—we know a lot about it, but I doubt the last word has been spoken. For an overview, go here:

• John Baez, Mysteries of the gravitational 2-body problem.

This SO(4) symmetry applies not only to the classical mechanics of the inverse square force law, but also the quantum mechanics! Nobody cares much about the quantum mechanics of two particles attracting gravitationally via an inverse square force law—but people care a lot about the quantum mechanics of hydrogen atoms, where the electron and proton attract each other via their electric field, which also obeys an inverse square force law.

So, let’s talk about hydrogen. And to keep things simple, let’s pretend the proton stays fixed while the electron orbits it. This is a pretty good approximation, and experts will know how to do things exactly right. It requires only a slight correction.

It turns out that wavefunctions for bound states of hydrogen can be reinterpreted as functions on the 3-sphere, S3 The sneaky SO(4) symmetry then becomes obvious: it just rotates this sphere! And the Hamiltonian of the hydrogen atom is closely connected to the Laplacian on the 3-sphere. The Laplacian has eigenspaces of dimensions n2 where n = 1,2,3,…, and these correspond to the eigenspaces of the hydrogen atom Hamiltonian. The number n is called the principal quantum number, and the hydrogen atom’s energy is proportional to -1/n2.

If you don’t know all this jargon, don’t worry! All you need to know is this: if we find an eigenfunction of the Laplacian on the 3-sphere, it will give a state where the hydrogen atom has a definite energy. And if this eigenfunction is invariant under some subgroup of SO(4), so will this state of the hydrogen atom!

The biggest finite subgroup of SO(4) is the rotational symmetry group of the 600-cell, a wonderful 4-dimensional shape with 120 vertices and 600 dodecahedral faces. The rotational symmetry group of this shape has a whopping 7,200 elements! And here is a marvelous moving image, made by Greg Egan, of an eigenfunction of the Laplacian on S3 that’s invariant under this 7,200-element group: We’re seeing the wavefunction on a moving slice of the 3-sphere, which is a 2-sphere. This wavefunction is actually real-valued. Blue regions are where this function is positive, yellow regions where it’s negative—or maybe the other way around—and black is where it’s almost zero. When the image fades to black, our moving slice is passing through a 2-sphere where the wavefunction is almost zero.

Layra Idarani came up with a complete classification of all eigenfunctions of the Laplacian on S3 that are invariant under this group… or more generally, eigenfunctions of the Laplacian on a sphere of any dimension that are invariant under the even part of any Coxeter group. Unfortunately he posted it to Google+, so it may be lost now. If you can find it, let me know!

Right now, the best place to look for more information—and pictures—is here:

• Greg Egan, Symmetric waves.

All that is a continuation of a story whose beginning is summarized here:

• John Baez, Quantum mechanics and the dodecahedron.

So, there’s a lot of serious math under the hood. But right now I just want to marvel at the fact that we’ve found a wavefunction for the hydrogen atom that not only has a well-defined energy, but is also invariant under this 7,200-element group. This group includes the usual 60 rotational symmetries of a dodecahedron, but also other much less obvious symmetries.

I don’t have a good picture of what these less obvious symmetries do to the wavefunction of a hydrogen atom. I understand them a bit better classically—where, as I said, they squash or stretch an elliptical orbit, changing its eccentricity while not changing its energy.

We can have fun with this using the old quantum theory—the approach to quantum mechanics that Bohr developed with his colleague Sommerfeld from 1920 to 1925, before Schrödinger introduced wavefunctions.

In the old Bohr–Sommerfeld approach to the hydrogen atom, the quantum states with specified energy, total angular momentum and angular momentum about a fixed axis were drawn as elliptical orbits. In this approach, the symmetries that squash or stretch elliptical orbits are a bit easier to visualize:

This picture by Pieter Kuiper shows some orbits at the 5th energy level, n = 5: namely, those with different eigenvalues of the total angular momentum, ℓ.

While the old quantum theory was superseded by the approach using wavefunctions, it’s possible to make it mathematically rigorous for the hydrogen atom. So, we can draw elliptical orbits that rigorously correspond to a basis of wavefunctions for the hydrogen atom. So, I believe we can draw the orbits corresponding to the basis elements whose linear combination gives the wavefunction shown as a function on the 3-sphere in Greg’s picture above!

We should get a bunch of ellipses forming a complicated picture with dodecahedral symmetry. This would make Kepler happy.

As a first step in this direction, Greg drew the collection of orbits that results when we take a circle and apply all the symmetries of the 600-cell: ### Postscript

To do this really right, one should learn a bit about ‘old quantum theory’. I believe people have been getting it a bit wrong for quite a while—starting with Bohr and Sommerfeld!

If you look at the ℓ = 0 orbit in the picture above, it’s a long skinny ellipse. But I believe it really should be a line segment straight through the proton: that’s what’s an orbit with no angular momentum looks like.

There’s a paper about this:

• Manfred Bucher, Rise and fall of the old quantum theory.

Matt McIrvin had some comments on this:

This paper from 2008 is a kind of thing I really like: an exploration of an old, incomplete theory that takes it further than anyone actually did at the time.

It has to do with the Bohr-Sommerfeld “old quantum theory”, in which electrons followed definite orbits in the atom, but these were quantized–not all orbits were permitted. Bohr managed to derive the hydrogen spectrum by assuming circular orbits, then Sommerfeld did much more by extending the theory to elliptical orbits with various shapes and orientations. But there were some problems that proved maddeningly intractable with this analysis, and it eventually led to the abandonment of the “orbit paradigm” in favor of Heisenberg’s matrix mechanics and Schrödinger’s wave mechanics, what we know as modern quantum theory.

The paper argues that the old quantum theory was abandoned prematurely. Many of the problems Bohr and Sommerfeld had came not from the orbit paradigm per se, but from a much simpler bug in the theory: namely, their rejection of orbits in which the electron moves entirely radially and goes right through the nucleus! Sommerfeld called these orbits “unphysical”, but they actually correspond to the s orbital states in the full quantum theory, with zero angular momentum. And, of course, in the full theory the electron in these states does have some probability of being inside the nucleus.

So Sommerfeld’s orbital angular momenta were always off by one unit. The hydrogen spectrum came out right anyway because of the happy accident of the energy degeneracy of certain orbits in the Coulomb potential.

I guess the states they really should have been rejecting as “unphysical” were Bohr’s circular orbits: no radial motion would correspond to a certain zero radial momentum in the full theory, and we can’t have that for a confined electron because of the uncertainty principle.

## The 600-Cell (Part 3)

28 December, 2017

There are still a few more things I want to say about the 600-cell. Last time I described the ‘compound of five 24-cells’. David Richter built a model of this, projected from 4 dimensions down to 3:

It’s nearly impossible to tell from this picture, but it’s five 24-cells inscribed in the 600-cell, with each vertex of the 600-cell being the vertex of just one of these five 24-cells. The trick for constructing it is to notice that the vertices of the 600-cell form a group sitting in the sphere of unit quaternions, and to find a 24-cell whose vertices form a subgroup.

The left cosets of a subgroup $H \subset G$ are the sets $gH = \{gh : \; h \in H\}$

They look like copies of $H$ ‘translated’, or in our case ‘rotated’, inside $G.$ Every point of $G$ lies in exactly one coset.

In our example there are five cosets. Each is the set of vertices of a 24-cell inscribed in the 600-cell. Every vertex of the 600-cell lies in exactly one of these cosets. This gives our ‘compound of five 24-cells’.

It turns out this trick is part of a family of three tricks, each of which gives a nice compound of 4d regular polytopes. While I’ve been avoiding coordinates, I think they’ll help get the idea across now. Here’s a nice description of the 120 vertices of the 600-cell. We take these points: $\displaystyle{ (\pm \textstyle{\frac{1}{2}}, \pm \textstyle{\frac{1}{2}},\pm \textstyle{\frac{1}{2}},\pm \textstyle{\frac{1}{2}}) }$ $\displaystyle{ (\pm 1, 0, 0, 0) }$ $\displaystyle{ \textstyle{\frac{1}{2}} (\pm \Phi, \pm 1 , \pm 1/\Phi, 0 )}$

and all those obtained by even permutations of the coordinates. So, we get $2^4 = 16$

points of the first kind, $2 \times 4 = 8$

points of the second kind, and $2^3 \times 4! / 2 = 96$

points of the third kind, for a total of $16 + 8 + 96 = 120$

points.

The 16 points of the first kind are the vertices of a 4-dimensional hypercube, the 4d analogue of a cube:

The 8 points of the second kind are the vertices of a 4-dimensional orthoplex, the 4d analogue of an octahedron:

The hypercube and orthoplex are dual to each other. Taking both their vertices together we get the 16 + 8 = 24 vertices of the 24-cell, which is self-dual:

The hypercube, orthoplex and 24-cell are regular polytopes, as is the 600-cell.

Now let’s think of any point in 4-dimensional space as a quaternion: $(a,b,c,d) = a + b i + c j + d k$

If we do this, we can check that the 120 vertices of the 600-cell form a group under quaternion multiplication. As mentioned in Part 1, this group is called the binary icosahedral group or $2\mathrm{I},$ because it’s a double cover of the rotational symmetry group of an icosahedron (or dodecahedron).

We can also check that the 24 vertices of the 24-cell form a group under quaternion multiplication. As mentioned in Part 1, this is called the binary tetrahedral group or $2\mathrm{T},$ because it’s a double cover of the rotational symmetry group of a tetrahedron.

All this is old news. But it’s even easier to check that the 8 vertices of the orthoplex form a group under quaternion multiplication: they’re just $\pm 1, \pm i, \pm i, \pm k$

This group is often called the quaternion group or $\mathrm{Q}.$ It too is a double cover of a group of rotations! The 180° rotations about the $x, y$ and $z$ axes square to 1 and commute with each other; up in the double cover of the rotation group (the unit quaternions, or $\mathrm{SU}(2)$) they give elements that square to -1 and anticommute with each other.

Furthermore, the 180° rotations about the $x, y$ and $z$ axes are symmetries of a regular tetrahedron! This is easiest to visualize if you inscribe the tetrahedron in a cube thus:

So, up in the double cover of the 3d rotation group we get a chain of subgroups $\mathrm{Q} \subset 2\mathrm{T} \subset 2\mathrm{I}$

which explains why we’re seeing an orthoplex inscribed in a 24-cell inscribed in a 600-cell! This explanation is more satisfying to me than the one involving coordinates.

Alas, I don’t see how to understand the hypercube inscribed in the 24-cell in quite this way, since the hypercube is not a subgroup of the unit quaternions. It certainly wasn’t in the coordinates I gave before—but worse, there’s no way to rotate the hypercube so that it becomes a subgroup. There must be something interesting to say here, but I don’t know it. So, I’ll forget the hypercube for now.

Instead, I’ll use group theory to do something nice with the orthoplex.

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.

There are some nontrivial counting problems associated with each of these three compounds. David Roberson has already solved most of these.

1) How many ways are there of inscribing an orthoplex in a 24-cell?

2) How many ways are there of inscribing a compound of three orthoplexes in a 24-cell?

3) How many ways are there of inscribing an orthoplex in a 600-cell? David used a computer to show there are 75. Is there a nice human-understandable argument?

4) How many ways are there of inscribing a compound of 15 orthoplexes in a 600-cell? David used a computer to show there are 280. Is there a nice human-understandable argument?

5) How many ways are there of inscribing a 24-cell in a 600-cell? David used a computer to show there are 25. Is there a nice human-understandable argument?

4) How many ways are there of inscribing a compound of five 24-cells in a 600-cell? David used a computer to show there are 10. Is there a nice human-understandable argument? (It’s pretty easy to prove that 10 is a lower bound.)

For those who prefer visual delights to math puzzles, here is a model of the compound of 15 orthoplexes, cleverly projected from 4 dimensions down to 3, made by David Richter and some friends:

It took four people 6 hours to make this! Click on the image to learn more about this amazing shape, and explore David Richter’s pages to see more compounds.

So far my tale has not encompassed the 120-cell, which is the dual of the 600-cell. This has 600 vertices and 120 dodecahedral faces:

Unfortunately, like the hypercube, the vertices of the 120-cell cannot be made into a subgroup of the unit quaternions. I’ll need some other idea to think about them in a way that I enjoy. But the 120-cell is amazing because every regular polytope in 4 dimensions can be inscribed in the 120-cell.

For example, we can inscribe the orthoplex in the 120-cell. Since the orthoplex has 8 vertices while the 120-cell has 600, and $600/8 = 75$

we might hope for a compound of 75 orthoplexes whose vertices, taken together, are those of the 120-cell. And indeed it exists… and David Richter and his friends have built a model!

### Image credits

You can click on any image to see its source. The photographs of models of the compound of five 24-cells and the compound of 15 orthoplexes are due to David Richter and friends. The shiny ball-and-strut pictures of the tetrahedron in the cube and the 120-cells were made by Tom Ruen using Robert Webb’s Stella software and placed on Wikicommons. The 2d projections of the hypercube, orthoplex and 24-cell were made by Tom Ruen and placed into the public domain on Wikicommons.