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:

For more details, read this:

• 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.

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.


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!


Excitonium

10 December, 2017

In certain crystals you can knock an electron out of its favorite place and leave a hole: a place with a missing electron. Sometimes these holes can move around like particles. And naturally these holes attract electrons, since they are places an electron would want to be.

Since an electron and a hole attract each other, they can orbit each other. An orbiting electron-hole pair is a bit like a hydrogen atom, where an electron orbits a proton. All of this is quantum-mechanical, of course, so you should be imagining smeared-out wavefunctions, not little dots moving around. But imagine dots if it’s easier.

An orbiting electron-hole pair is called an exciton, because while it acts like a particle in its own right, it’s really just a special kind of ‘excited’ electron—an electron with extra energy, not in its lowest energy state where it wants to be.

An exciton usually doesn’t last long: the orbiting electron and hole spiral towards each other, the electron finds the hole it’s been seeking, and it settles down.

But excitons can last long enough to do interesting things. In 1978 the Russian physicist Abrikosov wrote a short and very creative paper in which he raised the possibility that excitons could form a crystal in their own right! He called this new state of matter excitonium.

In fact his reasoning was very simple.

Just as electrons have a mass, so do holes. That sounds odd, since a hole is just a vacant spot where an electron would like to be. But such a hole can move around. It has more energy when it moves faster, and it takes force to accelerate it—so it acts just like it has a mass! The precise mass of a hole depends on the nature of the substance we’re dealing with.

Now imagine a substance with very heavy holes.

When a hole is much heavier than an electron, it will stand almost still when an electron orbits it. So, they form an exciton that’s very similar to a hydrogen atom, where we have an electron orbiting a much heavier proton.

Hydrogen comes in different forms: gas, liquid, solid… and at extreme pressures, like in the core of Jupiter, hydrogen becomes metallic. So, we should expect that excitons can come in all these different forms too!

We should be able to create an exciton gas… an exciton liquid… an exciton solid…. and under the right circumstances, a metallic crystal of excitons. Abrikosov called this metallic excitonium.

People have been trying to create this stuff for a long time. Some claim to have succeeded. But a new paper claims to have found something else: a Bose–Einstein condensate of excitons:

• Anshul Kogar, Melinda S. Rak, Sean Vig, Ali A. Husain, Felix Flicker, Young Il Joe, Luc Venema, Greg J. MacDougall, Tai C. Chiang, Eduardo Fradkin, Jasper van Wezel and Peter Abbamonte, Signatures of exciton condensation in a transition metal dichalcogenide, Science 358 (2017), 1314–1317.

A lone electron acts like a fermion, so I guess a hole does do, and if so that means an exciton acts approximately like a boson. When it’s cold, a gas of bosons will ‘condense’, with a significant fraction of them settling into the lowest energy states available. I guess excitons have been seen to do this!

There’s a fairly good simplified explanation at the University of Illinois website:

• Siv Schwink, Physicists excited by discovery of new form of matter, excitonium, 7 December 2017.

However, the picture on this page, which I used above, shows domain walls moving through crystallized excitonium. I think that’s different than a Bose-Einstein condensate!

I urge you to look at Abrikosov’s paper. It’s short and beautiful:

• Alexei Alexeyevich Abrikosov, A possible mechanism of high temperature superconductivity, Journal of the Less Common Metals
62 (1978), 451–455.

(Cool journal title. Is there a journal of the more common metals?)

In this paper, Abrikoskov points out that previous authors had the idea of metallic excitonium. Maybe his new idea was that this might be a superconductor—and that this might explain high-temperature superconductivity. The reason for his guess is that metallic hydrogen, too, is widely suspected to be a superconductor.

Later, Abrikosov won the Nobel prize for some other ideas about superconductors. I think I should read more of his papers. He seems like one of those physicists with great intuitions.

Puzzle 1. If a crystal of excitons conducts electricity, what is actually going on? That is, which electrons are moving around, and how?

This is a fun puzzle because an exciton crystal is a kind of abstract crystal created by the motion of electrons in another, ordinary, crystal. And that leads me to another puzzle, that I don’t know the answer to:

Puzzle 2. Is it possible to create a hole in excitonium? If so, it possible to create an exciton in excitonium? If so, is it possible to create meta-excitonium: an crystal of excitons in excitonium?


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.


Information Processing in Chemical Networks (Part 1)

4 January, 2017

There’s a workshop this summer:

Dynamics, Thermodynamics and Information Processing in Chemical Networks, 13-16 June 2017, Complex Systems and Statistical Mechanics Group, University of Luxembourg. Organized by Massimiliano Esposito and Matteo Polettini.

They write, “The idea of the workshop is to bring in contact a small number of high-profile research groups working at the frontier between physics and biochemistry, with particular emphasis on the role of Chemical Networks.”

The speakers may include John Baez, Sophie de Buyl, Massimiliano Esposito, Arren Bar-Even, Christoff Flamm, Ronan Fleming, Christian Gaspard, Daniel Merkle, Philippe Nge, Thomas Ouldridge, Luca Peliti, Matteo Polettini, Hong Qian, Stefan Schuster, Alexander Skupin, Pieter Rein ten Wolde. I believe attendance is by invitation only, so I’ll endeavor to make some of the ideas presented available here at this blog.

Some of the people involved

I’m looking forward to this, in part because there will be a mix of speakers I’ve met, speakers I know but haven’t met, and speakers I don’t know yet. I feel like reminiscing a bit, and I hope you’ll forgive me these reminiscences, since if you try the links you’ll get an introduction to the interface between computation and chemical reaction networks.

In part 25 of the network theory series here, I imagined an arbitrary chemical reaction network and said:

We could try to use these reactions to build a ‘chemical computer’. But how powerful can such a computer be? I don’t know the answer.

Luca Cardelli answered my question in part 26. This was just my first introduction to the wonderful world of chemical computing. Erik Winfree has a DNA and Natural Algorithms Group at Caltech, practically next door to Riverside, and the people there do a lot of great work on this subject. David Soloveichik, now at U. T. Austin, is an alumnus of this group.

In 2014 I met all three of these folks, and many other cool people working on these theme, at a workshop I tried to summarize here:

Programming with chemical reaction networks, Azimuth, 23 March 2014.

The computational power of chemical reaction networks, 10 June 2014.

Chemical reaction network talks, 26 June 2014.

I met Matteo Polettini about a year later, at a really big workshop on chemical reaction networks run by Elisenda Feliu and Carsten Wiuf:

Trends in reaction network theory (part 1), Azimuth, 27 January 2015.

Trends in reaction network theory (part 2), Azimuth, 1 July 2015.

Polettini has his own blog, very much worth visiting. For example, you can see his view of the same workshop here:

• Matteo Polettini, Mathematical trends in reaction network theory: part 1 and part 2, Out of Equilibrium, 1 July 2015.

Finally, I met Massimiliano Esposito and Christoph Flamm recently at the Santa Fe Institute, at a workshop summarized here:

Information processing and biology, Azimuth, 7 November 2016.

So, I’ve gradually become educated in this area, and I hope that by June I’ll be ready to say something interesting about the semantics of chemical reaction networks. Blake Pollard and I are writing a paper about this now.


Semantics for Physicists

7 December, 2016

I once complained that my student Brendan Fong said ‘semantics’ too much. You see, I’m in a math department, but he was actually in the computer science department at Oxford: I was his informal supervisor. Theoretical computer scientists love talking about syntax versus semantics—that is, written expressions versus what those expressions actually mean, or programs versus what those programs actually do. So Brendan was very comfortable with that distinction. But my other grad students, coming from a math department didn’t understand it… and he was mentioning it in practically ever other sentence.

In 1963, in his PhD thesis, Bill Lawvere figured out a way to talk about syntax versus semantics that even mathematicians—well, even category theorists—could understand. It’s called ‘functorial semantics’. The idea is that things you write are morphisms in some category X, while their meanings are morphisms in some other category Y. There’s a functor F \colon X \to Y which sends things you write to their meanings. This functor sends syntax to semantics!

But physicists may not enjoy this idea unless they see it at work in physics. In physics, too, the distinction is important! But it takes a while to understand. I hope Prakash Panangaden’s talk at the start of the Simons Institute workshop on compositionality is helpful. Check it out:


Jarzynksi on Non-Equilibrium Statistical Mechanics

18 November, 2016

santa_fe_institute

Here at the Santa Fe Institute we’re having a workshop on Statistical Physics, Information Processing and Biology. Unfortunately the talks are not being videotaped, so it’s up to me to spread the news of what’s going on here.

Christopher Jarzynski is famous for discovering the Jarzynski equality. It says

\displaystyle{ e^ { -\Delta F / k T} = \langle e^{ -W/kT } \rangle }

where k is Boltzmann’s consstant and T is the temperature of a system that’s in equilibrium before some work is done on it. \Delta F is the change in free energy, W is the amount of work, and the angle brackets represent an average over the possible options for what takes place—this sort of process is typically nondeterministic.

We’ve seen a good quick explanation of this equation here on Azimuth:

• Eric Downes, Crooks’ Fluctuation Theorem, Azimuth, 30 April 2011.

We’ve also gotten a proof, where it was called the ‘integral fluctuation theorem’:

• Matteo Smerlak, The mathematical origin of irreversibility, Azimuth, 8 October 2012.

It’s a fundamental result in nonequilibrium statistical mechanics—a subject where inequalities are so common that this equation is called an ‘equality’.

Two days ago, Jarzynski gave an incredibly clear hour-long tutorial on this subject, starting with the basics of thermodynamics and zipping forward to modern work. With his permission, you can see the slides here:

• Christopher Jarzynski, A brief introduction to the delights of non-equilibrium statistical physics.

Also try this review article:

• Christopher Jarzynski, Equalities and inequalities: irreversibility and the Second Law of thermodynamics at the nanoscale, Séminaire Poincaré XV Le Temps (2010), 77–102.