## 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.

For a full explanation, go here:

• Greg Egan, In the chambers with seven thousand symmetries, 2 January 2018.

Layra Idarani has come 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. For the details, go here:

• Layra Idarani, SG-invariant polynomials, 4 January 2018.

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:

• Greg Egan, Kepler orbits with 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.

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

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.

## Quantum Mechanics and the Dodecahedron

31 December, 2017

This is an expanded version of my G+ post, which was a watered-down version of Greg Egan’s G+ post and the comments on that. I’ll start out slow, and pick up speed as I go.

### Quantum mechanics meets the dodecahedron

In quantum mechanics, the position of a particle is not a definite thing: it’s described by a ‘wavefunction’. This says how probable it is to find the particle at any location… but it also contains other information, like how probable it is to find the particle moving at any velocity.

Take a hydrogen atom, and look at the wavefunction of the electron.

Question 1. Can we make the electron’s wavefunction have all the rotational symmetries of a dodecahedron—that wonderful Platonic solid with 12 pentagonal faces?

Yes! In fact it’s too easy: you can make the wavefunction look like whatever you want.

So let’s make the question harder. Like everything else in quantum mechanics, angular momentum can be uncertain. In fact you can never make all 3 components of angular momentum take definite values simultaneously! However, there are lots of wavefunctions where the magnitude of an electron’s angular momentum is completely definite.

This leads naturally to the next question, which was first posed by Gerard Westendorp:

Question 2. Can an electron’s wavefunction have a definite magnitude for its angular momentum while having all the rotational symmetries of a dodecahedron?

Yes! And there are infinitely many ways for this to happen! This is true even if we neglect the radial dependence of the wavefunction—that is, how it depends on the distance from the proton. Henceforth I’ll always do that, which lets us treat the wavefunction as a function on a sphere. And by the way, I’m also ignoring the electron’s spin! So, whenever I say ‘angular momentum’ I mean orbital angular momentum: the part that depends only on the electron’s position and velocity.

Question 2 has a trivial solution that’s too silly to bother with. It’s the spherically symmetric wavefunction! That’s invariant under all rotations. The real challenge is to figure out the simplest nontrivial solution. Egan figured it out, and here’s what it looks like:

The rotation here is just an artistic touch. Really the solution should be just sitting there, or perhaps changing colors while staying the same shape.

In what sense is this the simplest nontrivial solution? Well, the magnitude of the angular momentum is equal to

$\hbar^2 \sqrt{\ell(\ell+1)}$

where the number $\ell$ is quantized: it can only take values 0, 1, 2, 3,… and so on.

The trivial solution to Question 2 has $\ell = 0.$ The first nontrivial solution has $\ell = 6.$ Why 6? That’s where things get interesting. We can get it using the 6 lines connecting opposite faces of the dodecahedron!

I’ll explain later how this works. For now, let’s move straight on to a harder question:

Question 3. What’s the smallest choice of $\ell$ where we can find two linearly independent wavefunctions that both have the same $\ell$ and both have all the rotational symmetries of a dodecahedron?

It turns out to be $\ell = 30.$ And Egan created an image of a wavefunction oscillating between these two possibilities!

But we can go a lot further:

Question 4. For each $\ell,$ how many linearly independent functions on the sphere have that value of $\ell$ and all the rotational symmetries of a dodecahedron?

For $\ell$ ranging from 0 to 29 there are either none or one. There are none for these numbers:

1, 2, 3, 4, 5, 7, 8, 9, 11, 13, 14, 17, 19, 23, 29

and one for these numbers:

0, 6, 10, 12, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28

The pattern continues as follows. For $\ell$ ranging from 30 to 59 there are either one or two. There is one for these numbers:

31, 32, 33, 34, 35, 37, 38, 39, 41, 43, 44, 47, 49, 53, 59

and two for these numbers:

30, 36, 40, 42, 45, 46, 48, 50, 51, 52, 54, 55, 56, 57, 58

The numbers in these two lists are just 30 more than the numbers in the first two lists! And it continues on like this forever: there’s always one more linearly independent solution for $\ell + 30$ than there is for $\ell.$

Question 5. What’s special about these numbers from 0 to 29?

0, 6, 10, 12, 15, 18, 20, 21, 22, 24, 25, 26, 27, 28

You don’t need to know tons of math to figure this out—but I guess it’s a sort of weird pattern-recognition puzzle unless you know which patterns are likely to be important here. So I’ll give away the answer.

Here’s the answer: these are the numbers below 30 that can be written as sums of the numbers 6, 10 and 15.

But the real question is why? Also: what’s so special about the number 30?

The short, cryptic answer is this. The dodecahedron has 6 axes connecting the centers of opposite faces, 10 axes connecting opposite vertices, and 15 axes connecting the centers of opposite edges. The least common multiple of these numbers is 30.

But this requires more explanation!

For this, we need more math. You may want to get off here. But first, let me show you the solutions for $\ell = 6, \ell = 10,$ and $\ell = 15,$ as drawn by Greg Egan. I’ve already showed you $\ell = 6,$ which we could call the quantum dodecahedron:

Here is $\ell = 10,$ which looks like a quantum icosahedron:

And here is $\ell = 15$:

Maybe this deserves to be called a quantum Coxeter complex, since the Coxeter complex for the group of rotations and reflections of the dodecahedron looks like this:

### Functions with icosahedral symmetry

The dodecahedron and icosahedron have the same symmetries, but for some reason people talk about the icosahedron when discussing symmetry groups, so let me do that.

So far we’ve been looking at the rotational symmetries of the icosahedron. These form a group called $\mathrm{A}_5,$ or $\mathrm{I}$ for short, with 60 elements. We’ve been looking for certain functions on the sphere that are invariant under the action of this group. To get them all, we’ll first get ahold of all polynomials on $\mathbb{R}^3$ that are invariant under the action of this group Then we’ll restrict these to the sphere.

To save time, we’ll use the work of Claude Chevalley. He looked at rotation and reflection symmetries of the icosahedron. These form the group $\mathrm{I} \times \mathbb{Z}/2,$ also known as $\mathrm{H}_3,$ but let’s call it $\hat{\mathrm{I}}$ for short. It has 120 elements, but never confuse it with two other groups with 120 elements: the symmetric group on 5 letters, and the binary icosahedral group.

Chevalley found all polynomials on $\mathbb{R}^3$ that are invariant under the action of this bigger group $\hat{\mathrm{I}}.$ These invariant polynomials form an algebra, and Chevalley showed that this algebra is freely generated by 3 homogeneous polynomials:

$P(x,y,z) = x^2 + y^2 + z^2,$ of degree 2.

$Q(x,y,z),$ of degree 6. To get this we take the dot product of $(x,y,z)$ with each of the 6 vectors joining antipodal vertices of the icosahedron, and multiply them together.

$R(x,y,z),$ of degree 10. To get this we take the dot product of $(x,y,z)$ with each of the 10 vectors joining antipodal face centers of the icosahedron, and multiply them together.

So, linear combinations of products of these give all polynomials on $\mathbb{R}^3$ invariant under all rotation and reflection symmetries of the icosahedron.

But we want the polynomials that are invariant under just rotational symmetries of the icosahedron! To get all these, we need an extra generator:

$S(x,y,z),$ of degree 15. To get this we take the dot product of $(x,y,z)$ with each of the 15 vectors joining antipodal edge centers of the icosahedron, and multiply them together.

You can check that this is invariant under rotational symmetries of the icosahedron. But unlike our other polynomials, this one is not invariant under reflection symmetries! Because 15 is an odd number, $S$ switches sign under ‘total inversion’—that is, replacing $(x,y,z)$ with $-(x,y,z).$ This is a product of three reflection symmetries of the icosahedron.

Thanks to Egan’s extensive computations, I’m completely convinced that $P,Q,R$ and $S$ generate the algebra of all $\mathrm{I}$-invariant polynomials on $\mathbb{R}^3.$ I’ll take this as a fact, even though I don’t have a clean, human-readable proof. But someone must have proved it already—do you know where?

Since we now have 4 polynomials on $\mathbb{R}^3,$ they must obey a relation. Egan figured it out:

$S^2 = 500 P^9 Q^2 - 2275 P^6 Q^3 + 3440 P^3 Q^4 - 1728 Q^5 + 200 P^7 Q R$
$- 795 P^4 Q^2 R + 720 P Q^3 R + 4 P^5 R^2 -65 P^2 Q R^2 - R^3$

The exact coefficients depend on some normalization factors used in defining $Q,R$ and $S.$ Luckily the details don’t matter much. All we’ll really need is that this relation expresses $S^2$ in terms of the other generators. And this fact is easy to see without any difficult calculations!

How? Well, we’ve seen $S$ is unchanged by rotations, while it changes sign under total inversion. So, the most any rotation or reflection symmetry of the icosahedron can do to $S$ is change its sign. This means that $S^2$ is invariant under all these symmetries. So, by Chevalley’s result, it must be a polynomial in $P, Q,$ and $R$.

So, we now have a nice description of the $\mathrm{I}$-invariant polynomials on $\mathbb{R}^3,$ in terms of generators and relations. Each of these gives an $\mathrm{I}$-invariant function on the sphere. And Leo Stein, a postdoc at Caltech who has a great blog on math and physics, has kindly created some images of these.

The polynomial $P$ is spherically symmetric so it’s too boring to draw. The polynomial $Q,$ of degree 6, looks like this when restricted to the sphere:

Since it was made by multiplying linear functions, one for each axis connecting opposite vertices of an icosahedron, it shouldn’t be surprising that we see blue blobs centered at these vertices.

The polynomial $R,$ of degree 10, looks like this:

Here the blue blobs are centered on the icosahedron’s 20 faces.

Finally, here’s $S,$ of degree 15:

This time the blue blobs are centered on the icosahedron’s 30 edges.

Now let’s think a bit about functions on the sphere that arise from polynomials on $\mathbb{R}^3.$ Let’s call them algebraic functions on the sphere. They form an algebra, and it’s just the algebra of polynomials on $\mathbb{R}^3$ modulo the relation $P = 1,$ since the sphere is the set $\{P = 1\}.$

It makes no sense to talk about the ‘degree’ of an algebraic function on the sphere, since the relation $P = 1$ equates polynomials of different degree. What makes sense is the number $\ell$ that I was talking about earlier!

The group $\mathrm{SO}(3)$ acts by rotation on the space of algebraic functions on the sphere, and we can break this space up into irreducible representations of $\mathrm{SO}(3).$ It’s a direct sum of irreps, one of each ‘spin’ $\ell = 0, 1, 2, \dots.$

So, we can’t talk about the degree of a function on the sphere, but we can talk about its $\ell$ value. On the other hand, it’s very convenient to work with homogeneous polynomials on $\mathbb{R}^3,$ which have a definite degree—and these restrict to functions on the sphere. How can we relate the degree and the quantity $\ell$?

Here’s one way. The polynomials on $\mathbb{R}^3$ form a graded algebra. That means it’s a direct sum of vector spaces consisting of homogeneous polynomials of fixed degree, and if we multiply two homogeneous polynomials their degrees add. But the algebra of polynomials restricted to the sphere is merely filtered algebra.

What does this mean? Let $F$ be the algebra of all algebraic functions on the sphere, and let $F_\ell \subset F$ consist of those that are restrictions of polynomials of degree $\le \ell.$ Then:

1) $F_\ell \subseteq F_{\ell + 1}$

and

2) $\displaystyle{ F = \bigcup_{\ell = 0}^\infty F_\ell }$

and

3) if we multiply a function in $F_\ell$ by one in $F_m,$ we get one in $F_{\ell + m}.$

That’s what a filtered algebra amounts to.

But starting from a filtered algebra, we can get a graded algebra! It’s called the associated graded algebra.

To do this, we form

$G_\ell = F_\ell / F_{\ell - 1}$

and let

$\displaystyle{ G = \bigoplus_{\ell = 0}^\infty G_\ell }$

Then $G$ has a product where multiplying a guy in $G_\ell$ and one in $G_m$ gives one in $G_{\ell + m}.$ So, it’s indeed a graded algebra! For the details, see Wikipedia, which manages to make it look harder than it is. The basic idea is that we multiply in $F$ and then ‘ignore terms of lower degree’. That’s what $G_\ell = F_\ell / F_{\ell - 1}$ is all about.

Now I want to use two nice facts. First, $G_\ell$ is the spin-$\ell$ representation of $\mathrm{SO}(3).$ Second, there’s a natural map from any filtered algebra to its associated graded algebra, which is an isomorphism of vector spaces (though not of algebras). So, we get an natural isomorphism of vector spaces

$\displaystyle{ F \cong G = \bigoplus_{\ell = 0}^\infty G_\ell }$

from the algebraic functions on the sphere to the direct sum of all the spin-$\ell$ representations!

Now to the point: because this isomorphism is natural, it commutes with symmetries, so we can also use it to study algebraic functions on the sphere that are invariant under a group of linear transformations of $\mathbb{R}^3.$

Before tackling the group we’re really interested in, let’s try the group of rotation and reflection symmetries of the icosahedron, $\hat{\mathrm{I}}.$ As I mentioned, Chevalley worked out the algebra of polynomials on $\mathbb{R}^3$ that are invariant under this bigger group. It’s a graded commutative algebra, and it’s free on three generators: $P$ of degree 2, $Q$ of degree 6, and $R$ of degree 10.

Starting from here, to get the algebra of $\hat{\mathrm{I}}$-invariant algebraic functions on the sphere, we mod out by the relation $P = 1.$ This gives a filtered algebra which I’ll call $F^{\hat{\mathrm{I}}}.$ (It’s common to use a superscript with the name of a group to indicate that we’re talking about the stuff that’s invariant under some action of that group.) From this we can form the associated graded algebra

$\displaystyle{ G^{\hat{\mathrm{I}}} = \bigoplus_{\ell = 0}^\infty G_\ell^{\hat{\mathrm{I}}} }$

where

$G_\ell^{\hat{\mathrm{I}}} = F_\ell^{\hat{\mathrm{I}}} / F_{\ell - 1}^{\hat{\mathrm{I}}}$

If you’ve understood everything I’ve been trying to explain, you’ll see that $G_\ell^{\hat{\mathrm{I}}}$ is the space of all functions on the sphere that transform in the spin-$\ell$ representation and are invariant under the rotation and reflection symmetries of the icosahedron.

But now for the fun part: what is this space like? By the work of Chevalley, the algebra $F^{\hat{\mathrm{I}}}$ is spanned by products

$P^p Q^q R^r$

but since we have the relation $P = 1,$ and no other relations, it has a basis given by products

$Q^q R^r$

So, the space $F_\ell^{\hat{\mathrm{I}}}$ has a basis of products like this whose degree is $\le \ell,$ meaning

$6 q + 10 r \le \ell$

Thus, the space we’re really interested in:

$G_\ell^{\hat{\mathrm{I}}} = F_\ell^{\hat{\mathrm{I}}} / F_{\ell - 1}^{\hat{\mathrm{I}}}$

has a basis consisting of equivalence classes

$[Q^q R^r]$

where

$6 q + 10 r = \ell$

So, we get:

Theorem 1. The dimension of the space of functions on the sphere that lie in the spin-$\ell$ representation of $\mathrm{SO}(3)$ and are invariant under the rotation and reflection symmetries of the icosahedron equals the number of ways of writing $\ell$ as an unordered sum of 6’s and 10’s.

Let’s see how this goes:

$\ell = 0$: dimension 1, with basis $[1]$

$\ell = 1$: dimension 0

$\ell = 2$: dimension 0

$\ell = 3$: dimension 0

$\ell = 4$: dimension 0

$\ell = 5$: dimension 0

$\ell = 6$: dimension 1, with basis $[Q]$

$\ell = 7$: dimension 0

$\ell = 8$: dimension 0

$\ell = 9$: dimension 0

$\ell = 10$: dimension 1, with basis $[R]$

$\ell = 11$: dimension 0

$\ell = 12$: dimension 1, with basis $[Q^2]$

$\ell = 13$: dimension 0

$\ell = 14$: dimension 0

$\ell = 15$: dimension 0

$\ell = 16$: dimension 1, with basis $[Q R]$

$\ell = 17$: dimension 0

$\ell = 18$: dimension 1, with basis $[Q^3]$

$\ell = 19$: dimension 0

$\ell = 20$: dimension 1, with basis $[R^2]$

$\ell = 21$: dimension 0

$\ell = 22$: dimension 1, with basis $[Q^2 R]$

$\ell = 23$: dimension 0

$\ell = 24$: dimension 1, with basis $[Q^4]$

$\ell = 25$: dimension 0

$\ell = 26$: dimension 1, with basis $[Q R^2]$

$\ell = 27$: dimension 0

$\ell = 28$: dimension 1, with basis $[Q^3 R]$

$\ell = 29$: dimension 0

$\ell = 30$: dimension 2, with basis $[Q^5], [R^3]$

So, the story starts out boring, with long gaps. The odd numbers are completely uninvolved. But it heats up near the end, and reaches a thrilling climax at $\ell = 30.$ At this point we get two linearly independent solutions, because 30 is the least common multiple of the degrees of $Q$ and $R.$

It’s easy to see that from here on the story ‘repeats’ with period 30, with the dimension growing by 1 each time:

$\mathrm{dim}(G_{\ell+30}^{\hat{\mathrm{I}}}) = \mathrm{dim}(G_{\ell}^{\hat{\mathrm{I}}}) + 1$

Now, finally, we are to tackle Question 4 from the first part of this post: for each $\ell,$ how many linearly independent functions on the sphere have that value of $\ell$ and all the rotational symmetries of a dodecahedron?

We just need to repeat our analysis with $\mathrm{I},$ the group of rotational symmetries of the dodecahedron, replacing the bigger group $\hat{\mathrm{I}}.$

We start with algebra of polynomials on $\mathbb{R}^3$ that are invariant under $\mathrm{I}$. As we’ve seen, this is a graded commutative algebra with four generators: $P,Q,R$ as before, but also $S$ of degree 15. To make up for this extra generator there’s an extra relation, which expresses $S^2$ in terms of the other generators.

Starting from here, to get the algebra of $\mathrm{I}$-invariant algebraic functions on the sphere, we mod out by the relation $P = 1.$ This gives a filtered algebra I’ll call $F^{\mathrm{I}}.$ Then we form the associated graded algebra

$\displaystyle{ G^{\mathrm{I}} = \bigoplus_{\ell = 0}^\infty G_\ell^{\mathrm{I}} }$

where

$G_\ell^{\mathrm{I}} = F_\ell^{\mathrm{I}} / F_{\ell - 1}^{\mathrm{I}}$

What we really want to know is the dimension of $G_\ell^{\mathrm{I}},$ since this is the space of functions on the sphere that transform in the spin-$\ell$ representation and are invariant under the rotational symmetries of the icosahedron.

So, what’s this space like? The algebra $F^{\mathrm{I}}$ is spanned by products

$P^p Q^q R^r S^t$

but since we have the relation $P = 1,$ and a relation expressing $S^2$ in terms of other generators, it has a basis given by products

$Q^q R^r S^s$ where $s = 0, 1$

So, the space $F_\ell^{\mathrm{I}}$ has a basis of products like this whose degree is $\le \ell,$ meaning

$6 q + 10 r + 15 s \le \ell$ and $s = 0, 1$

Thus, the space we’re really interested in:

$G_\ell^{\mathrm{I}} = F_\ell^{\mathrm{I}} / F_{\ell - 1}^{\mathrm{I}}$

has a basis consisting of equivalence classes

$[Q^q R^r S^s]$

where

$6 q + 10 r + 15 s = \ell$ and $s = 0, 1$

So, we get:

Theorem 2. The dimension of the space of functions on the sphere that lie in the spin-$\ell$ representation of $\mathrm{SO}(3)$ and are invariant under the rotational symmetries of the icosahedron equals the number of ways of writing $\ell$ as an unordered sum of 6’s, 10’s and at most one 15.

Let’s work out these dimensions explicitly, and see how the extra generator $S$ changes the story! Since it has degree 15, it contributes some solutions for odd values of $\ell.$ But when we reach the magic number 30, this extra generator loses its power: $S^2$ has degree 30, but it’s a linear combination of other things.

$\ell = 0$: dimension 1, with basis $[1]$

$\ell = 1$: dimension 0

$\ell = 2$: dimension 0

$\ell = 3$: dimension 0

$\ell = 4$: dimension 0

$\ell = 5$: dimension 0

$\ell = 6$: dimension 1, with basis $[Q]$

$\ell = 7$: dimension 0

$\ell = 8$: dimension 0

$\ell = 9$: dimension 0

$\ell = 10$: dimension 1, with basis $[R]$

$\ell = 11$: dimension 0

$\ell = 12$: dimension 1, with basis $[Q^2]$

$\ell = 13$: dimension 0

$\ell = 14$: dimension 0

$\ell = 15$: dimension 1, with basis $[S]$

$\ell = 16$: dimension 1, with basis $[Q R]$

$\ell = 17$: dimension 0

$\ell = 18$: dimension 1, with basis $[Q^3]$

$\ell = 19$: dimension 0

$\ell = 20$: dimension 1, with basis $[R^2]$

$\ell = 21$: dimension 1, with basis $[Q S]$

$\ell = 22$: dimension 1, with basis $[Q^2 R]$

$\ell = 23$: dimension 0

$\ell = 24$: dimension 1, with basis $[Q^4]$

$\ell = 25$: dimension 1, with basis $[R S]$

$\ell = 26$: dimension 1, with basis $[Q R^2]$

$\ell = 27$: dimension 1, with basis $[Q^2 S]$

$\ell = 28$: dimension 1, with basis $[Q^3 R]$

$\ell = 29$: dimension 0

$\ell = 30$: dimension 2, with basis $[Q^5], [R^3]$

From here on the story ‘repeats’ with period 30, with the dimension growing by 1 each time:

$\mathrm{dim}(G_{\ell+30}^{\mathrm{I}}) = \mathrm{dim}(G_{\ell}^{\mathrm{I}}) + 1$

So, we’ve more or less proved everything that I claimed in the first part. So we’re done!

### Postscript

But I can’t resist saying a bit more.

First, there’s a very different and somewhat easier way to compute the dimensions in Theorems 1 and 2. It uses the theory of characters, and Egan explained it in a comment on the blog post on which this is based.

Second, if you look in these comments, you’ll also see a lot of material about harmonic polynomials on $\mathbb{R}^3$—that is, those obeying the Laplace equation. These polynomials are very nice when you’re trying to decompose the space of functions on the sphere into irreps of $\mathrm{SO}(3).$ The reason is that the harmonic homogeneous polynomials of degree $\ell,$ when restricted to the sphere, give you exactly the spin-$\ell$ representation!

If you take all homogeneous polynomials of degree $\ell$ and restrict them to the sphere you get a lot of ‘redundant junk’. You get the spin-$\ell$ rep, plus the spin-$(\ell-2)$ rep, plus the spin-$(\ell-4)$ rep, and so on. The reason is the polynomial

$P = x^2 + y^2 + z^2$

and its powers: if you have a polynomial living in the spin-$\ell$ rep and you multiply it by $P,$ you get another one living in the spin-$\ell$ rep, but you’ve boosted the degree by 2.

Layra Idarani pointed out that this is part of a nice general theory. But I found all this stuff slightly distracting when I was trying to prove Theorems 1 and 2 assuming that we had explicit presentations of the algebras of $\hat{\mathrm{I}}$– and $\mathrm{I}$-invariant polynomials on $\mathbb{R}^3.$ So, instead of introducing facts about harmonic polynomials, I decided to use the ‘associated graded algebra’ trick. This is a more algebraic way to ‘eliminate the redundant junk’ in the algebra of polynomials and chop the space of functions on the sphere into irreps of $\mathrm{SO}(3).$

Also, Egan and Idarani went ahead and considered what happens when we replace the icosahedron by another Platonic solid. It’s enough to consider the cube and tetrahedron. These cases are actually subtler than the icosahedron! For example, when we take the dot product of $(x,y,z)$ with each of the 10 vectors joining antipodal face centers of the cube, and multiply them together, we get a polynomial that’s not invariant under rotations of the cube! Up to a constant it’s just $x y z,$ and this changes sign under some rotations.

People call this sort of quantity, which gets multiplied by a number under transformations instead of staying the same, a semi-invariant. The reason we run into semi-invariants for the cube and tetrahedron is that their rotational symmetry groups, $\mathrm{S}_4$ and $\mathrm{A}_4,$ have nontrivial abelianizations, namely $\mathbb{Z}/2$ and $\mathbb{Z}/3.$ The abelianization of $\mathrm{I} \cong \mathrm{A}_5$ is trivial.

Egan summarized the story as follows:

Just to sum things up for the cube and the tetrahedron, since the good stuff has ended up scattered over many comments:

For the cube, we define:

A of degree 4 from the cube’s vertex-axes, a full invariant
B of degree 6 from the cube’s edge-centre-axes, a semi-invariant
C of degree 3 from the cube’s face-centre-axes, a semi-invariant

We have full invariants:

A of degree 4
C2 of degree 6
BC of degree 9

B2 can be expressed in terms of A, C and P, so we never use it, and we use BC at most once.

So the number of copies of the trivial rep of the rotational symmetry group of the cube in spin ℓ is the number of ways to write ℓ as an unordered sum of 4, 6 and at most one 9.

For the tetrahedron, we embed its vertices as four vertices of the cube. We then define:

V of degree 4 from the tet’s vertices, a full invariant
E of degree 3 from the tet’s edge-centre axes, a full invariant

And the B we defined for the embedding cube serves as a full invariant of the tet, of degree 6.

B2 can be expressed in terms of V, E and P, so we use B at most once.

So the number of copies of the trivial rep of the rotational symmetry group of the tetrahedron in spin ℓ is the number of ways to write ℓ as a sum of 3, 4 and at most one 6.

All of this stuff reminds me of a baby version of the theory of modular forms. For example, the algebra of modular forms is graded by ‘weight’, and it’s the free commutative algebra on a guy of weight 4 and a guy of weight 6. So, the dimension of the space of modular forms of weight $k$ is the number of ways of writing $k$ as an unordered sum of 4’s and 6’s. Since the least common multiple of 4 and 6 is 12, we get a pattern that ‘repeats’, in a certain sense, mod 12. Here I’m talking about the simplest sort of modular forms, based on the group $\mathrm{SL}_2(\mathbb{Z}).$ But there are lots of variants, and I have the feeling that this post is secretly about some sort of variant based on finite subgroups of $\mathrm{SL}(2,\mathbb{C})$ instead of infinite discrete subgroups.

There’s a lot more to say about all this, but I have to stop or I’ll never stop. Please ask questions and if you want me to say more!

## 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.

## The 600-Cell (Part 2)

24 December, 2017

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.

## The 600-Cell (Part 1)

16 December, 2017

I can’t stop thinking about the 600-cell:

It’s a ‘Platonic solid in 4 dimensions’ with 600 tetrahedral faces and 120 vertices. One reason I like it is that you can think of these vertices as forming a group: a double cover of the rotational symmetry group of the icosahedron. Another reason is that it’s a halfway house between the icosahedron and the $\mathrm{E}_8$ lattice. I explained all this in my last post here:

I wrote that post as a spinoff of an article I was writing for the Newsletter of the London Mathematical Society, which had a deadline attached to it. Now I should be writing something else, for another deadline. But somehow deadlines strongly demotivate me—they make me want to do anything else. So I’ve been continuing to think about the 600-cell. I posed some puzzles about it in the comments to my last post, and they led me to some interesting thoughts, which I feel like explaining. But they’re not quite solidified, so right now I just want to give a fairly concrete picture of the 600-cell, or at least its vertices.

This will be a much less demanding post than the last one—and correspondingly less rewarding. Remember the basic idea:

Points in the 3-sphere can be seen as quaternions of norm 1, and these form a group $\mathrm{SU}(2)$ that double covers $\mathrm{SO}(3).$ The vertices of the 600-cell are the points of a subgroup $\Gamma \subset \mathrm{SU}(2)$ that double covers the rotational symmetry group of the icosahedron. This group $\Gamma$ is the famous binary icosahedral group.

Thus, we can name the vertices of the 600-cell by rotations of the icosahedron—as long as we remember to distinguish between a rotation by $\theta$ and a rotation by $\theta + 2\pi.$ Let’s do it!

• 0° (1 of these). We can take the identity rotation as our chosen ‘favorite’ vertex of the 600-cell.

• 72° (12 of these). The nearest neighbors of our chosen vertex correspond to the rotations by the smallest angles that are symmetries of the icosahedron; these correspond to taking any of its 12 vertices and giving it a 1/5 turn clockwise.

• 120° (20 of these). The next nearest neighbors correspond to taking one of the 20 faces of the icosahedron and giving it a 1/3 turn clockwise.

• 144° (12 of these). These correspond to taking one of the vertices of the icosahedron and giving it a 2/5 turn clockwise.

• 180° (30 of these). These correspond to taking one of the edges and giving it a 1/2 turn clockwise. (Note that since we’re working in the double cover $\mathrm{SU(2)}$ rather than $\mathrm{SO}(3),$ giving one edge a half turn clockwise counts as different than giving the opposite edge a half turn clockwise.)

• 216° (12 of these). These correspond to taking one of the vertices of the icosahedron and giving it a 3/5 turn clockwise. (Again, this counts as different than rotating the opposite vertex by a 2/5 turn clockwise.)

• 240° (20 of these). These correspond to taking one of the faces of the icosahedron and giving it a 2/3 turn clockwise. (Again, this counts as different than rotating the opposite vertex by a 1/3 turn clockwise.)

• 288° (12 of these). These correspond to taking any of the vertices and giving it a 4/5 turn clockwise.

• 360° (1 of these). This corresponds to a full turn in any direction.

Let’s check:

$1 + 12 + 20 + 12 + 30 + 12 + 20 + 12 + 1 = 120$

Good! We need a total of 120 vertices.

This calculation also shows that if we move a hyperplane through the 3-sphere, which hits our favorite vertex the moment it touches the 3-sphere, it will give the following slices of the 600-cell:

• Slice 1: a point (our favorite vertex),

• Slice 2: an icosahedron (whose vertices are the 12 nearest neighbors of our favorite vertex),

• Slice 3: a dodecahedron (whose vertices are the 20 next-nearest neighbors),

• Slice 4: an icosahedron (the 12 third-nearest neighbors),

• Slice 5: an icosidodecahedron (the 30 fourth-nearest neighbors),

• Slice 6: an icosahedron (the 12 fifth-nearest neighbors),

• Slice 7: a dodecahedron (the 20 sixth-nearest neighbors),

• Slice 8: an icosahedron (the 12 seventh-nearest neighbors),

• Slice 9: a point (the vertex opposite our favorite).

Here’s a picture drawn by J. Gregory Moxness, illustrating this:

Note that there are 9 slices. Each corresponds to a different conjugacy class in the group $\Gamma.$ These in turn correspond to the dots in the extended Dynkin diagram of $\mathrm{E}_8,$ which has the usual 8 dots and one more.

The usual $\mathrm{E}_8$ Dynkin diagram has ‘legs’ of lengths $5, 2$ and $3:$

The three legs correspond to conjugacy classes in $\Gamma$ that map to rotational symmetries of an icosahedron that preserve a vertex (5 conjugacy classes), an edge (2 conjugacy classes), and a (3 conjugacy classes)… not counting the element $-1 \in \Gamma.$ That last element gives the extra dot in the extended Dynkin diagram.

### Image credits

You can click on an image to see its source. The shiny ball-and-strut picture of the 120-cell was made by Tom Ruen using Robert Webb’s Stella software and placed on Wikicommons. The picture of slices of the 120-cell was drawn by J. Gregory Moxness and placed on Wikicommons under a Creative Commons Attribution-Share Alike 4.0 International license.

## From the Icosahedron to E8

10 December, 2017

Here’s a draft of a little thing I’m writing for the Newsletter of the London Mathematical Society. The regular icosahedron is connected to many ‘exceptional objects’ in mathematics, and here I describe two ways of using it to construct $\mathrm{E}_8.$ One uses a subring of the quaternions called the ‘icosians’, while the other uses Patrick du Val’s work on the resolution of Kleinian singularities. I leave it as a challenge to find the connection between these two constructions!

You can see a PDF here:

Here’s the story:

### From the Icosahedron to E8

In mathematics, every sufficiently beautiful object is connected to all others. Many exciting adventures, of various levels of difficulty, can be had by following these connections. Take, for example, the icosahedron—that is, the regular icosahedron, one of the five Platonic solids. Starting from this it is just a hop, skip and a jump to the $\mathrm{E}_8$ lattice, a wonderful pattern of points in 8 dimensions! As we explore this connection we shall see that it also ties together many other remarkable entities: the golden ratio, the quaternions, the quintic equation, a highly symmetrical 4-dimensional shape called the 600-cell, and a manifold called the Poincaré homology 3-sphere.

Indeed, the main problem with these adventures is knowing where to stop! The story we shall tell is just a snippet of a longer one involving the McKay correspondence and quiver representations. It would be easy to bring in the octonions, exceptional Lie groups, and more. But it can be enjoyed without these esoteric digressions, so let us introduce the protagonists without further ado.

The icosahedron has a long history. According to a comment in Euclid’s Elements it was discovered by Plato’s friend Theaetetus, a geometer who lived from roughly 415 to 369 BC. Since Theaetetus is believed to have classified the Platonic solids, he may have found the icosahedron as part of this project. If so, it is one of the earliest mathematical objects discovered as part of a classification theorem. It’s hard to be sure. In any event, it was known to Plato: in his Timaeus, he argued that water comes in atoms of this shape.

The icosahedron has 20 triangular faces, 30 edges, and 12 vertices. We can take the vertices to be the four points

$\displaystyle{ (0 , \pm 1 , \pm \Phi) }$

and all those obtained from these by cyclic permutations of the coordinates, where

$\displaystyle{ \Phi = \frac{\sqrt{5} + 1}{2} }$

is the golden ratio. Thus, we can group the vertices into three orthogonal golden rectangles: rectangles whose proportions are $\Phi$ to 1.

In fact, there are five ways to do this. The rotational symmetries of the icosahedron permute these five ways, and any nontrivial rotation gives a nontrivial permutation. The rotational symmetry group of the icosahedron is thus a subgroup of $\mathrm{S}_5.$ Moreover, this subgroup has 60 elements. After all, any rotation is determined by what it does to a chosen face of the icosahedron: it can map this face to any of the 20 faces, and it can do so in 3 ways. The rotational symmetry group of the icosahedron is therefore a 60-element subgroup of $\mathrm{S}_5.$ Group theory therefore tells us that it must be the alternating group $\mathrm{A}_5.$

The $\mathrm{E}_8$ lattice is harder to visualize than the icosahedron, but still easy to characterize. Take a bunch of equal-sized spheres in 8 dimensions. Get as many of these spheres to touch a single sphere as you possibly can. Then, get as many to touch those spheres as you possibly can, and so on. Unlike in 3 dimensions, where there is ‘wiggle room’, you have no choice about how to proceed, except for an overall rotation and translation. The balls will inevitably be centered at points of the $\mathrm{E}_8$ lattice!

We can also characterize the $\mathrm{E}_8$ lattice as the one giving the densest packing of spheres among all lattices in 8 dimensions. This packing was long suspected to be optimal even among those that do not arise from lattices—but this fact was proved only in 2016, by the young mathematician Maryna Viazovska [V].

We can also describe the $\mathrm{E}_8$ lattice more explicitly. In suitable coordinates, it consists of vectors for which:

1) the components are either all integers or all integers plus $\textstyle{\frac{1}{2}},$ and

2) the components sum to an even number.

This lattice consists of all integral linear combinations of the 8 rows of this matrix:

$\left( \begin{array}{rrrrrrrr} 1&-1&0&0&0&0&0&0 \\ 0&1&-1&0&0&0&0&0 \\ 0&0&1&-1&0&0&0&0 \\ 0&0&0&1&-1&0&0&0 \\ 0&0&0&0&1&-1&0&0 \\ 0&0&0&0&0&1&-1&0 \\ 0&0&0&0&0&1&1&0 \\ -\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2} \end{array} \right)$

The inner product of any row vector with itself is 2, while the inner product of distinct row vectors is either 0 or -1. Thus, any two of these vectors lie at an angle of either 90° or 120°. If we draw a dot for each vector, and connect two dots by an edge when the angle between their vectors is 120° we get this pattern:

This is called the $\mathrm{E}_8$ Dynkin diagram. In the first part of our story we shall find the $\mathrm{E}_8$ lattice hiding in the icosahedron; in the second part, we shall find this diagram. The two parts of this story must be related—but the relation remains mysterious, at least to me.

### The Icosians

The quickest route from the icosahedron to $\mathrm{E}_8$ goes through the fourth dimension. The symmetries of the icosahedron can be described using certain quaternions; the integer linear combinations of these form a subring of the quaternions called the ‘icosians’, but the icosians can be reinterpreted as a lattice in 8 dimensions, and this is the $\mathrm{E}_8$ lattice [CS]. Let us see how this works.

The quaternions, discovered by Hamilton, are a 4-dimensional algebra

$\displaystyle{ \mathbb{H} = \{a + bi + cj + dk \colon \; a,b,c,d\in \mathbb{R}\} }$

with multiplication given as follows:

$\displaystyle{i^2 = j^2 = k^2 = -1, }$
$\displaystyle{i j = k = - j i \textrm{ and cyclic permutations} }$

It is a normed division algebra, meaning that the norm

$\displaystyle{ |a + bi + cj + dk| = \sqrt{a^2 + b^2 + c^2 + d^2} }$

obeys

$|q q'| = |q| |q'|$

for all $q,q' \in \mathbb{H}.$ The unit sphere in $\mathbb{H}$ is thus a group, often called $\mathrm{SU}(2)$ because its elements can be identified with $2 \times 2$ unitary matrices with determinant 1. This group acts as rotations of 3-dimensional Euclidean space, since we can see any point in $\mathbb{R}^3$ as a purely imaginary quaternion $x = bi + cj + dk,$ and the quaternion $qxq^{-1}$ is then purely imaginary for any $q \in \mathrm{SO}(3).$ Indeed, this action gives a double cover

$\displaystyle{ \alpha \colon \mathrm{SU}(2) \to \mathrm{SO}(3) }$

where $\mathrm{SO}(3)$ is the group of rotations of $\mathbb{R}^3.$

We can thus take any Platonic solid, look at its group of rotational symmetries, get a subgroup of $\mathrm{SO}(3),$ and take its double cover in $\mathrm{SU}(2).$ If we do this starting with the icosahedron, we see that the 60-element group $\mathrm{A}_5 \subset \mathrm{SO}(3)$ is covered by a 120-element group $\Gamma \subset \mathrm{SU}(2),$ called the binary icosahedral group.

The elements of $\Gamma$ are quaternions of norm one, and it turns out that they are the vertices of a 4-dimensional regular polytope: a 4-dimensional cousin of the Platonic solids. It deserves to be called the “hypericosahedron”, but it is usually called the 600-cell, since it has 600 tetrahedral faces. Here is the 600-cell projected down to 3 dimensions, drawn using Robert Webb’s Stella software:

Explicitly, if we identify $\mathbb{H}$ with $\mathbb{R}^4,$ the elements of $\Gamma$ are the 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 those obtained from these by even permutations of the coordinates. Since these points are closed under multiplication, if we take integral linear combinations of them we get a subring of the quaternions:

$\displaystyle{ \mathbb{I} = \{ \sum_{q \in \Gamma} a_q q : \; a_q \in \mathbb{Z} \} \subset \mathbb{H} .}$

Conway and Sloane [CS] call this the ring of icosians. The icosians are not a lattice in the quaternions: they are dense. However, any icosian is of the form $a + bi + cj + dk$ where $a,b,c,$ and $d$ live in the golden field

$\displaystyle{ \mathbb{Q}(\sqrt{5}) = \{ x + \sqrt{5} y : \; x,y \in \mathbb{Q}\} }$

Thus we can think of an icosian as an 8-tuple of rational numbers. Such 8-tuples form a lattice in 8 dimensions.

In fact we can put a norm on the icosians as follows. For $q \in \mathbb{I}$ the usual quaternionic norm has

$\displaystyle{ |q|^2 = x + \sqrt{5} y }$

for some rational numbers $x$ and $y,$ but we can define a new norm on $\mathbb{I}$ by setting

$\displaystyle{ \|q\|^2 = x + y }$

With respect to this new norm, the icosians form a lattice that fits isometrically in 8-dimensional Euclidean space. And this is none other than $\mathrm{E}_8!$

### Klein’s Icosahedral Function

Not only is the $\mathrm{E}_8$ lattice hiding in the icosahedron; so is the $\mathrm{E}_8$ Dynkin diagram. The space of all regular icosahedra of arbitrary size centered at the origin has a singularity, which corresponds to a degenerate special case: the icosahedron of zero size. If we resolve this singularity in a minimal way we get eight Riemann spheres, intersecting in a pattern described by the $\mathrm{E}_8$ Dynkin diagram!

This remarkable story starts around 1884 with Felix Klein’s Lectures on the Icosahedron [Kl]. In this work he inscribed an icosahedron in the Riemann sphere, $\mathbb{C}\mathrm{P}^1.$ He thus got the icosahedron’s symmetry group, $\mathrm{A}_5,$ to act as conformal transformations of $\mathbb{C}\mathrm{P}^1$—indeed, rotations. He then found a rational function of one complex variable that is invariant under all these transformations. This function equals $0$ at the centers of the icosahedron’s faces, 1 at the midpoints of its edges, and $\infty$ at its vertices.

Here is Klein’s icosahedral function as drawn by Abdelaziz Nait Merzouk. The color shows its phase, while the contour lines show its magnitude:

We can think of Klein’s icosahedral function as a branched cover of the Riemann sphere by itself with 60 sheets:

$\displaystyle{ \mathcal{I} \colon \mathbb{C}\mathrm{P}^1 \to \mathbb{C}\mathrm{P}^1 .}$

Indeed, $\mathrm{A}_5$ acts on $\mathbb{C}\mathrm{P}^1,$ and the quotient space $\mathbb{C}\mathrm{P}^1/\mathrm{A}_5$ is isomorphic to $\mathbb{C}\mathrm{P}^1$ again. The function $\mathcal{I}$ gives an explicit formula for the quotient map $\mathbb{C}\mathrm{P}^1 \to \mathbb{C}\mathrm{P}^1/\mathrm{A}_5 \cong \mathbb{C}\mathrm{P}^1.$

Klein managed to reduce solving the quintic to the problem of solving the equation $\mathcal{I}(z) = w$ for $z.$ A modern exposition of this result is Shurman’s Geometry of the Quintic [Sh]. For a more high-powered approach, see the paper by Nash [N]. Unfortunately, neither of these treatments avoids complicated calculations. But our interest in Klein’s icosahedral function here does not come from its connection to the quintic: instead, we want to see its connection to $\mathrm{E}_8.$

For this we should actually construct Klein’s icosahedral function. To do this, recall that the Riemann sphere $\mathbb{C}\mathrm{P}^1$ is the space of 1-dimensional linear subspaces of $\mathbb{C}^2.$ Let us work directly with $\mathbb{C}^2.$ While $\mathrm{SO}(3)$ acts on $\mathbb{C}\mathrm{P}^1,$ this comes from an action of this group’s double cover $\mathrm{SU}(2)$ on $\mathbb{C}^2.$ As we have seen, the rotational symmetry group of the icosahedron, $\mathrm{A}_5 \subset \mathrm{SO}(3),$ is double covered by the binary icosahedral group $\Gamma \subset \mathrm{SU}(2).$ To build an $\mathrm{A}_5$-invariant rational function on $\mathbb{C}\mathrm{P}^1,$ we should thus look for $\Gamma$-invariant homogeneous polynomials on $\mathbb{C}^2.$

It is easy to construct three such polynomials:

$V,$ of degree 12, vanishing on the 1d subspaces corresponding to icosahedron vertices.

$E,$ of degree 30, vanishing on the 1d subspaces corresponding to icosahedron edge midpoints.

$F,$ of degree 20, vanishing on the 1d subspaces corresponding to icosahedron face centers.

Remember, we have embedded the icosahedron in $\mathbb{C}\mathrm{P}^1,$ and each point in $\mathbb{C}\mathrm{P}^1$ is a 1-dimensional subspace of $\mathbb{C}^2,$ so each icosahedron vertex determines such a subspace, and there is a linear function on $\mathbb{C}^2,$ unique up to a constant factor, that vanishes on this subspace. The icosahedron has 12 vertices, so we get 12 linear functions this way. Multiplying them gives $V,$ a homogeneous polynomial of degree 12 on $\mathbb{C}^2$ that vanishes on all the subspaces corresponding to icosahedron vertices! The same trick gives $E,$ which has degree 30 because the icosahedron has 30 edges, and $F,$ which has degree 20 because the icosahedron has 20 faces.

A bit of work is required to check that $V,E$ and $F$ are invariant under $\Gamma,$ instead of changing by constant factors under group transformations. Indeed, if we had copied this construction using a tetrahedron or octahedron, this would not be the case. For details, see Shurman’s book [Sh], which is free online, or van Hoboken’s nice thesis [VH].

Since both $F^3$ and $V^5$ have degree 60, $F^3/V^5$ is homogeneous of degree zero, so it defines a rational function $\mathcal{I} \colon \mathbb{C}\mathrm{P}^1 \to \mathbb{C}\mathrm{P}^1.$ This function is invariant under $\mathrm{A}_5$ because $F$ and $V$ are invariant under $\Gamma.$ Since $F$ vanishes at face centers of the icosahedron while $V$ vanishes at vertices, $\mathcal{I} = F^3/V^5$ equals $0$ at face centers and $\infty$ at vertices. Finally, thanks to its invariance property, $\mathcal{I}$ takes the same value at every edge center, so we can normalize $V$ or $F$ to make this value 1.

Thus, $\mathcal{I}$ has precisely the properties required of Klein’s icosahedral function! And indeed, these properties uniquely characterize that function, so that function is $\mathcal{I}.$

### The Appearance of E8

Now comes the really interesting part. Three polynomials on a 2-dimensional space must obey a relation, and $V,E,$ and $F$ obey a very pretty one, at least after we normalize them correctly:

$\displaystyle{ V^5 + E^2 + F^3 = 0. }$

We could guess this relation simply by noting that each term must have the same degree. Every $\Gamma$-invariant polynomial on $\mathbb{C}^2$ is a polynomial in $V, E$ and $F,$ and indeed

$\displaystyle{ \mathbb{C}^2 / \Gamma \cong \{ (V,E,F) \in \mathbb{C}^3 \colon \; V^5 + E^2 + F^3 = 0 \} . }$

This complex surface is smooth except at $V = E = F = 0,$ where it has a singularity. And hiding in this singularity is $\mathrm{E}_8$!

To see this, we need to ‘resolve’ the singularity. Roughly, this means that we find a smooth complex surface $S$ and an onto map

that is one-to-one away from the singularity. (More precisely, if $X$ is an algebraic variety with singular points $X_{\mathrm{sing}} \subset X,$ $\pi \colon S \to X$ is a resolution of $X$ if $S$ is smooth, $\pi$ is proper, $\pi^{-1}(X - X_{\textrm{sing}})$ is dense in $S,$ and $\pi$ is an isomorphism between $\pi^{-1}(X - X_{\mathrm{sing}})$ and $X - X_{\mathrm{sing}}.$ For more details see Lamotke’s book [L].)

There are many such resolutions, but one minimal resolution, meaning that all others factor uniquely through this one:

What sits above the singularity in this minimal resolution? Eight copies of the Riemann sphere $\mathbb{C}\mathrm{P}^1,$ one for each dot here:

Two of these $\mathbb{C}\mathrm{P}^1$s intersect in a point if their dots are connected by an edge: otherwise they are disjoint.

This amazing fact was discovered by Patrick Du Val in 1934 [DV]. Why is it true? Alas, there is not enough room in the margin, or even in the entire blog article, to explain this. The books by Kirillov [Ki] and Lamotke [L] fill in the details. But here is a clue. The $\mathrm{E}_8$ Dynkin diagram has ‘legs’ of lengths $5, 2$ and $3$:

On the other hand,

$\displaystyle{ \mathrm{A}_5 \cong \langle v, e, f | v^5 = e^2 = f^3 = v e f = 1 \rangle }$

where in terms of the rotational symmetries of the icosahedron:

$v$ is a $1/5$ turn around some vertex of the icosahedron,

$e$ is a $1/2$ turn around the center of an edge touching that vertex,

$f$ is a $1/3$ turn around the center of a face touching that vertex,

and we must choose the sense of these rotations correctly to obtain $vef = 1.$ To get a presentation of the binary icosahedral group we drop one relation:

$\displaystyle{ \Gamma \cong \langle v, e, f | v^5 = e^2 = f^3 = vef \rangle }$

The dots in the $\mathrm{E}_8$ Dynkin diagram correspond naturally to conjugacy classes in $\Gamma,$ not counting the conjugacy class of the central element $-1 \in \Gamma.$ Each of these conjugacy classes, in turn, gives a copy of $\mathbb{C}\mathrm{P}^1$ in the minimal resolution of $\mathbb{C}^2/\Gamma.$

Not only the $\mathrm{E}_8$ Dynkin diagram, but also the $\mathrm{E}_8$ lattice, can be found in the minimal resolution of $\mathbb{C}^2/\Gamma.$ Topologically, this space is a 4-dimensional manifold. Its real second homology group is an 8-dimensional vector space with an inner product given by the intersection pairing. The integral second homology is a lattice in this vector space spanned by the 8 copies of $\mathbb{C}P^1$ we have just seen—and it is a copy of the $\mathrm{E}_8$ lattice [KS].

But let us turn to a more basic question: what is $\mathbb{C}^2/\Gamma$ like as a topological space? To tackle this, first note that we can identify a pair of complex numbers with a single quaternion, and this gives a homeomorphism

$\mathbb{C}^2/\Gamma \cong \mathbb{H}/\Gamma$

where we let $\Gamma$ act by right multiplication on $\mathbb{H}.$ So, it suffices to understand $\mathbb{H}/\Gamma.$

Next, note that sitting inside $\mathbb{H}/\Gamma$ are the points coming from the unit sphere in $\mathbb{H}.$ These points form the 3-dimensional manifold $\mathrm{SU}(2)/\Gamma,$ which is called the Poincaré homology 3-sphere [KS]. This is a wonderful thing in its own right: Poincaré discovered it as a counterexample to his guess that any compact 3-manifold with the same homology as a 3-sphere is actually diffeomorphic to the 3-sphere, and it is deeply connected to $\mathrm{E}_8.$ But for our purposes, what matters is that we can think of this manifold in another way, since we have a diffeomorphism

$\mathrm{SU}(2)/\Gamma \cong \mathrm{SO}(3)/\mathrm{A}_5.$

The latter is just the space of all icosahedra inscribed in the unit sphere in 3d space, where we count two as the same if they differ by a rotational symmetry.

This is a nice description of the points of $\mathbb{H}/\Gamma$ coming from points in the unit sphere of $\mathbb{H}.$ But every quaternion lies in some sphere centered at the origin of $\mathbb{H},$ of possibly zero radius. It follows that $\mathbb{C}^2/\Gamma \cong \mathbb{H}/\Gamma$ is the space of all icosahedra centered at the origin of 3d space—of arbitrary size, including a degenerate icosahedron of zero size. This degenerate icosahedron is the singular point in $\mathbb{C}^2/\Gamma.$ This is where $\mathrm{E}_8$ is hiding.

Clearly much has been left unexplained in this brief account. Most of the missing details can be found in the references. But it remains unknown—at least to me—how the two constructions of $\mathrm{E}_8$ from the icosahedron fit together in a unified picture.

Recall what we did. First we took the binary icosahedral group $\Gamma \subset \mathbb{H},$ took integer linear combinations of its elements, thought of these as forming a lattice in an 8-dimensional rational vector space with a natural norm, and discovered that this lattice is a copy of the $\mathrm{E}_8$ lattice. Then we took $\mathbb{C}^2/\Gamma \cong \mathbb{H}/\Gamma,$ took its minimal resolution, and found that the integral 2nd homology of this space, equipped with its natural inner product, is a copy of the $\mathrm{E}_8$ lattice. From the same ingredients we built the same lattice in two very different ways! How are these constructions connected? This puzzle deserves a nice solution.

#### Acknowledgements

I thank Tong Yang for inviting me to speak on this topic at the Annual General Meeting of the Hong Kong Mathematical Society on May 20, 2017, and Guowu Meng for hosting me at the HKUST while I prepared that talk. I also thank the many people, too numerous to accurately list, who have helped me understand these topics over the years.

#### Bibliography

[CS] J. H. Conway and N. J. A. Sloane, Sphere Packings, Lattices and Groups, Springer, Berlin, 2013.

[DV] P. du Val, On isolated singularities of surfaces which do not affect the conditions of adjunction, I, II and III, Proc. Camb. Phil. Soc. 30, 453–459, 460–465, 483–491.

[KS] R. Kirby and M. Scharlemann, Eight faces of the Poincaré homology 3-sphere, Usp. Mat. Nauk. 37 (1982), 139–159. Available at https://tinyurl.com/ybrn4pjq

[Ki] A. Kirillov, Quiver Representations and Quiver Varieties, AMS, Providence, Rhode Island, 2016.

[Kl] F. Klein, Lectures on the Ikosahedron and the Solution of Equations of the Fifth Degree, Trüubner & Co., London, 1888. Available at https://archive.org/details/cu31924059413439

[L] K. Lamotke, Regular Solids and Isolated Singularities, Vieweg & Sohn, Braunschweig, 1986.

[N] O. Nash, On Klein’s icosahedral solution of the quintic. Available at https://arxiv.org/abs/1308.0955

[Sh] J. Shurman, Geometry of the Quintic, Wiley, New York, 1997. Available at http://people.reed.edu/~jerry/Quintic/quintic.html

[Sl] P. Slodowy, Platonic solids, Kleinian singularities, and Lie groups, in Algebraic Geometry, Lecture Notes in Mathematics 1008, Springer, Berlin, 1983, pp. 102–138.

[VH] J. van Hoboken, Platonic Solids, Binary Polyhedral Groups, Kleinian Singularities and Lie Algebras of Type A, D, E, Master’s Thesis, University of Amsterdam, 2002. Available at http://math.ucr.edu/home/baez/joris_van_hoboken_platonic.pdf

[V] M. Viazovska, The sphere packing problem in dimension 8, Ann. Math. 185 (2017), 991–1015. Available at https://arxiv.org/abs/1603.04246

## Wigner Crystals

7 December, 2017

I’d like to explain a conjecture about Wigner crystals, which we came up with in a discussion on Google+. It’s a purely mathematical conjecture that’s pretty simple to state, motivated by the picture above. But let me start at the beginning.

Electrons repel each other, so they don’t usually form crystals. But if you trap a bunch of electrons in a small space, and cool them down a lot, they will try to get as far away from each other as possible—and they can do this by forming a crystal!

This is sometimes called an electron crystal. It’s also called a Wigner crystal, because the great physicist Eugene Wigner predicted in 1934 that this would happen.

Only since the late 1980s have we been able to make electron crystals in the lab. Such a crystal can only form if the electron density is low enough. The reasons is that even at absolute zero, a gas of electrons has kinetic energy. At absolute zero the gas will minimize its energy. But it can’t do this by having all the electrons in a state with zero momentum, since you can’t put two electrons in the same state, thanks to the Pauli exclusion principle. So, higher momentum states need to be occupied, and this means there’s kinetic energy. And it has more if its density is high: if there’s less room in position space, the electrons are forced to occupy more room in momentum space.

When the density is high, this prevents the formation of a crystal: instead, we have lots of electrons whose wavefunctions are ‘sitting almost on top of each other’ in position space, but with different momenta. They’ll have lots of kinetic energy, so minimizing kinetic energy becomes more important than minimizing potential energy.

When the density is low, this effect becomes unimportant, and the electrons mainly try to minimize potential energy. So, they form a crystal with each electron avoiding the rest. It turns out they form a body-centered cubic: a crystal lattice formed of cubes, with an extra electron in the middle of each cube.

To know whether a uniform electron gas at zero temperature forms a crystal or not, you need to work out its so-called Wigner-Seitz radius. This is the average inter-particle spacing measured in units of the Bohr radius. The Bohr radius is the unit of length you can cook up from the electron mass, the electron charge and Planck’s constant:

$\displaystyle{ a_0=\frac{\hbar^2}{m_e e^2} }$

It’s mainly famous as the average distance between the electron and a proton in a hydrogen atom in its lowest energy state.

Simulations show that a 3-dimensional uniform electron gas crystallizes when the Wigner–Seitz radius is at least 106. The picture, however, shows an electron crystal in 2 dimensions, formed by electrons trapped on a thin film shaped like a disk. In 2 dimensions, Wigner crystals form when the Wigner–Seitz radius is at least 31. In the picture, the density is so low that we can visualize the electrons as points with well-defined positions.

So, the picture simply shows a bunch of points $x_i$ trying to minimize the potential energy, which is proportional to

$\displaystyle{ \sum_{i \ne j} \frac{1}{\|x_i - x_j\|} }$

The lines between the dots are just to help you see what’s going on. They’re showing the Delauney triangulation, where we draw a graph that divides the plane into regions closer to one electron than all the rest, and then take the dual of that graph.

Thanks to energy minimization, this triangulation wants to be a lattice of equilateral triangles. But since such a triangular lattice doesn’t fit neatly into a disk, we also see some ‘defects’:

Most electrons have 6 neighbors. But there are also some red defects, which are electrons with 5 neighbors, and blue defects, which are electrons with 7 neighbors.

Note that there are 6 clusters of defects. In each cluster there is one more red defect than blue defect. I think this is not a coincidence.

Conjecture. When we choose a sufficiently large number of points $x_i$ on a disk in such a way that

$\displaystyle{ \sum_{i \ne j} \frac{1}{\|x_i - x_j\|} }$

is minimized, and draw the Delauney triangulation, there will be 6 more vertices with 5 neighbors than vertices with 7 neighbors.

Here’s a bit of evidence for this, which is not at all conclusive. Take a sphere and triangulate it in such a way that each vertex has 5, 6 or 7 neighbors. Then here’s a cool fact: there must be 12 more vertices with 5 neighbors than vertices with 7 neighbors.

Puzzle. Prove this fact.

If we think of the picture above as the top half of a triangulated sphere, then each vertex in this triangulated sphere has 5, 6 or 7 neighbors. So, there must be 12 more vertices on the sphere with 5 neighbors than with 7 neighbors. So, it makes some sense that the top half of the sphere will contain 6 more vertices with 5 neighbors than with 7 neighbors. But this is not a proof.

I have a feeling this energy minimization problem has been studied with various numbers of points. So, there either be a lot of evidence for my conjecture, or some counterexamples that will force me to refine it. The picture shows what happens with 600 points on the disk. Maybe something dramatically different happens with 599! Maybe someone has even proved theorems about this. I just haven’t had time to look for such work.

The picture here was drawn by Arunas.rv and placed on Wikicommons on a Creative Commons Attribution-Share Alike 3.0 Unported license.