# What is a quasicrystal?

Posted by in math

This post was originally a talk given in Pizza Seminar on October 15, 2014. Most of the mathematical content is from Wikipedia and this paper by Niizeki.

## Abstract

The crystallographic restriction theorem says a finite-order automorphism of $$\mathbb{Z}^3$$ (in chemistry language, a symmetry of a crystal) has order 1, 2, 3, 4, or 6. But $$\mathbb{Z}^3$$ is small potatoes; so let’s go to $$\mathbb{Z}^n$$, do some Galois theory in cyclotomic fields, and get us some icosahedral symmetry! (Icosahedrite is a real thing, say serious scientists.) Pretty pictures will be provided, including demonstrations of self-similarity. Self-similarity is always cool, right?

## The crystallographic restriction theorem

A crystal is a regular grid (like $$\mathbb{Z}^3$$) of atoms. There is a remarkable theorem about what symmetries a crystal can admit:

Theorem (crystallographic restriction theorem)
If $$A \in \operatorname{Aut} \mathbb{Z}^3$$ has finite order, then the order is 1, 2, 3, 4, or 6.
Proof
Properties of $$A$$ imply properties of its eigenvalues:
$$A$$ eigenvalues
finite order on the unit circle $$S^1$$
real in complex conjugate pairs (so one is $$\pm 1$$)
integer their sum, $$\operatorname{tr} A$$, is an integer

So if $$e^{i\theta}$$ is an eigenvalue of $$A$$, then $\operatorname{tr} A = e^{i\theta} + e^{-i\theta} \pm 1 = 2 \cos \theta \pm 1.$ Therefore $$\cos \theta \in \frac{1}{2} \mathbb{Z}$$. The eigenvalues $$e^{i\theta}$$ satisfying this are of order 1, 2, 3, 4, and 6.

## X-ray crystallography: peeking inside real crystals

### Setup

Here’s how scientists figure out the internal structure of a crystal. To start, shine an x-ray on it and record the scattered radiation on film. For our purposes, an x-ray is a function $$f$$ satisfying $f(x+?) = e^{i k \cdot x} f(?)$ for some vector $$k$$ that depends on the ray direction and wavelength. This lets us calculate the factors by which $$f$$ changes from source $$x_1$$ to atom $$x$$ and from atom to film $$x_2$$.

The total change in $$f$$ is by a factor of $$e^{i[(k_1 - k_2) \cdot x - k_1 \cdot x_1 + k_2 \cdot x_2]} .$$ Integrating this over all atom positions $$x$$ in the crystal yields the total radiation bounced in the direction of $$k_2$$. So…

Upshot
The photo records (part of) the Fourier transform of the distribution of atoms.1

### A troublesome photo

Since rotational symmetries of a lattice are inherited by its Fourier transform, photos like the one below would seem to violate the crystallographic restriction theorem.

### Demo: Making waves

To get a picture of what kind of crystal we’re looking at, let’s approximate the inverse transform by adding waves at $$n$$ regularly spaced angles. Try it for 5, 8, 12, and 7.

You can animate these too; and Jason Davies has a WebGL adaptation of the animation.

## Slicing higher-dimensional lattices

### The problem: locating atoms

The last demo showed you the graph of $$f(x) = \displaystyle\sum_{i=1}^n \cos (k_i \cdot x)$$ in $$\mathbb{R}^2$$ for evenly spaced $$k_i$$ on the unit circle. This is all very nice and smooth, but we want to actually place atoms—discrete points—in a grid. Let’s put them in the hotspots—i.e. where $$f$$ hits some threshold.

To turn this into a discrete process, use the following trick: realize the waves $$\cos (k_i \cdot x)$$ as 2-D slices of independent waves in $$\mathbb{R}^n$$—then hotspots are where the slice passes near lattice points.

### The explicit recipe

Let $$f$$ be the restriction of the function $\mathbb{R}^n \ni x \mapsto \sum_{i=1}^n \cos (e_i \cdot x)$ to a 2-plane on which the basis vectors $$e_i$$ project to $$k_i$$. To choose the 2-plane, let $$A \in \operatorname{GL}_n \mathbb{Z}$$ act by cycling the basis vectors $$e_i$$. Diagonalizing over $$\mathbb{C}$$ gets you the matrix $\begin{pmatrix} \zeta & & & & \\ & \zeta^{-1} & & & \\ & & \zeta^2 & & \\ & & & \zeta^{-2} & \\ & & & & \ddots \end{pmatrix}$ where $$\zeta = e^{2 \pi i / n}$$. Over $$\mathbb{R}$$, the eigenvalues pair into 2×2 blocks, corresponding to 2-planes on which $$A$$ acts as rotation by a multiple of $$\frac{2\pi}{n}$$. Pick any one of them.

Upshot
• The 2-plane is a 2-dimensional subrepresentation $$V$$ of $$\langle A \rangle \curvearrowright \mathbb{R}^n$$.
• Atoms (hotspots) are points of $$V$$ that lie kinda near points of $$\mathbb{Z}^n$$.

### Demo: Atoms in 2-D quasicrystals

With this in hand, a computer can be convinced2 to find the atoms for you, drawing larger dots for better hotspots. Try it for several values of $$n$$; I suggest 5, 7, 8, 6, 2, 12, and 11.

Remark
An alternative method is to slice fundamental domains ($$n$$-cubes) with the 2-plane, which yields pictures like the Penrose tiling below. Jason Davies has an applet you can play with, based on an older one by Greg Egan.

## Self-similarity

At this point, we could go home—we have a description of quasicrystals from which we can produce both interactive computer renderings and theorems (like quasicrystals not actually having translational symmetry, due to $$V$$ containing no integer points besides 0).

Of course, there is more—the “quasi”-ness of quasicrystals allows a sort of self-similarity that honest crystals can’t have. To see it in action, go back to that atoms demo and try zooming in or out for a while.

To explain this self-similarity—the apparent indistinguishability of different zoom levels—we will need a bit of number theory.

### A number-theoretic model of quasicrystals

Theorem (crystallographic restriction in arbitrary dimensions)
$$\operatorname{GL}_m \mathbb{Z}$$ contains a matrix $$A$$ which has $$\zeta_n = e^{2\pi i/n}$$ as an eigenvalue (so it rotates some 2-plane by $$\frac{2 \pi}{n}$$) if and only if $$m \geq \varphi(n)$$, where $$\varphi$$ is Euler’s totient function.
Proof
The Galois conjugates of $$\zeta_n$$ must also be eigenvalues of $$A$$.
• Since $$\zeta_n^n = 1$$, they’re $$n$$th roots of 1.
• Since $$\zeta_n$$ generates all $$n$$th roots of 1, they’re primitive—i.e. $$\zeta_n^k$$ where $$k$$ and $$n$$ are coprime.

The number of such $$k$$ is $$\varphi(n)$$. To prove that $$\varphi(n)$$ works, observe that $\mathbb{Z}[\zeta_n] \hookrightarrow \mathbb{Q}(\zeta_n) \cong_{\mathbb{Q}\text{-vect}} \mathbb{Q}^{\varphi(n)} \subset \mathbb{R}^{\varphi(n)}$ as a discrete subgroup, since the field norms of the leftmost terms are in $$\mathbb{Z}$$.

Remark
If $$p$$ and $$q$$ are large enough primes, you can produce an integer matrix $$A$$ of order $$pq$$ and size $$\varphi(p) + \varphi(q) < \varphi(pq)$$, but it won’t rotate any 2-plane by $$\frac{2\pi}{pq}$$.

### A mathematical description of zooming

Here’s how we implement3 zooming: scale the 2-plane $$V$$ by some factor and the unseen $$m-2$$ dimensions by some other factor to preserve $$m$$-dimensional volume. That way the total visual density of dots remains about the same. Here’s how we interpret zooming and self-similarity:

• Zooming acts by scalars on $$V$$ and the other irreducible sub-representations of $$\langle A \rangle \curvearrowright \mathbb{R}^{\varphi(n)}$$, so zooming is a 1-parameter subgroup $$\gamma$$ in $$\operatorname{SAut}_A \mathbb{R}^{\varphi(n)}$$ (volume preserving automorphisms).
• Visually, transforming by anything in $$\operatorname{Aut} \mathbb{Z}[\zeta_n]$$ isn’t detectable.
• Self-similarity means that the image of $$\gamma$$ in $$\operatorname{SAut}_A \mathbb{R}^{\varphi(n)} / \operatorname{Aut} \mathbb{Z}[\zeta_n]$$ accumulates on itself (or is periodic).

The existence of self-similarity rests on one key fact from number theory.

Theorem (Dirichlet’s Unit Theorem)
Suppose a number field has $$r$$ embeddings in $$\mathbb{R}$$ and $$c$$ embeddings in $$\mathbb{C}$$. Then its group of units has rank $r + \frac{c}{2} - 1 .$
Corollary
$$\operatorname{SAut}_A \mathbb{R}^{\varphi(n)} / \operatorname{Aut} \mathbb{Z}[\zeta_n]$$ is compact.
Proof
Since automorphisms act by scalars on all irreducible subrepresentations,

\begin{align*} \operatorname{Aut}_A \mathbb{R}^{\varphi(n)} &\cong (\mathbb{C}^\times)^{\frac{\varphi(n)}{2}} \cong (\mathbb{R} \times S^1)^{\frac{\varphi(n)}{2}} \\ \operatorname{SAut}_A \mathbb{R}^{\varphi(n)} &\cong \mathbb{R}^{\frac{\varphi(n)}{2} - 1} \times (S^1)^{\frac{\varphi(n)}{2}} . \end{align*}

Since $$\mathbb{Z}[\zeta_n] = \mathcal{O}_{\mathbb{Q}(\zeta_n)}$$, Dirichlet’s Unit Theorem implies that $$\operatorname{Aut} \mathbb{Z}[\zeta_n]$$ contains a copy of $$\mathbb{Z}^{\frac{\varphi(n)}{2} - 1}$$. Then $$\operatorname{Aut} \mathbb{Z}[\zeta_n]$$, being discrete (it’s conjugate to a subgroup of $$\operatorname{SL}(\varphi(n), \mathbb{Z})$$), is cocompact.

Self-similarity now follows, having been reduced to the statement that a 1-parameter subgroup of a torus is recurrent.

Remark
The image of $$\gamma$$ is periodic when (and only when) $$\varphi(n) = 2$$—so when $$n$$ is 5, 8, 10, or 12. This makes the self-similarity exact—i.e. different zoom levels can look identical (not just suspiciously similar)—and allows one to build games like Richard Schwartz’s Lucy and Lily based on facts like $(2 + \sqrt{5})(2 - \sqrt{5}) = -1.$
Remark
Michael Rule’s collection of quasicrystals graphics includes an alternative description of zooming that sidesteps the number theory, instead using Shepard Tones to produce self-similarity.

## 3-D quasicrystals

### The n-dimensional viewpoint

A quasicrystal with irreducible symmetry $$G$$ is an irreducible-over-$$\mathbb{R}$$ subrepresentation $$V$$ of an irreducible-over-$$\mathbb{Z}$$ representation $$G \hookrightarrow \operatorname{SL}(m,\mathbb{Z})$$. The atoms are points of $$\mathbb{Z}^m$$ near $$V$$, and the Galois conjugates of $$V$$ (the other subrepresentations over $$\mathbb{R}$$) allow the possibility for self-similarity.

### The search for 3-D quasicrystals

To look for 3-dimensional quasicrystals, one first finds a list of the seven irreducible finite subgroups of $$\operatorname{SL}(3,\mathbb{R})$$. Most of these preserve an ordinary 3-D lattice—i.e. the representation can be defined over $$\mathbb{Z}$$. The two exceptions are the two icosahedral groups—the alternating group $$A_5$$ and the product $$A_5 \times \{\pm 1\}$$—so up to finite index subgroups, there’s only one irreducible 3-D quasicrystal, which you can explore in the following demo.

The standard representation of $$A_5$$ has one Galois conjugate, which can be found in a list of representations of $$A_5$$. The two representations can be described explicitly as follows: in the standard representation, $$A_5$$ acts by rotation on the icosahedron with vertices $$\pm(\sqrt{5}, 0)$$ and $$\pm(1, 2\zeta_5^k)$$ in $$\mathbb{R} \times \mathbb{C}$$. To obtain the other representation, merely conjugate the action on the set of vertices by the permutation induced by $$\zeta_5 \mapsto \zeta_5^2$$.

Oh, and—yes, they’re real.

## Footnotes

1. The photo loses phase information, the recovery of which is a nontrivial problem

2. For efficiency reasons, the demo actually uses the number-theoretic model from the section after.

3. Self-similarity is what makes zooming possible (i.e. not unbearably slow) in the atoms demo—the atom finder can get away with exploring a 2-dimensional slice instead of the entire $$m$$-dimensional space.