Take an irrational slice through a lattice and a quasicrystal falls out
There is a second way to make a pattern that never repeats, and it does not involve substitution, matching rules, or any cleverness about shapes. Take a lattice — perfectly periodic. Take a line, or a plane, through it at an irrational angle. Keep the lattice points near it and project them down.
Nothing in that recipe is aperiodic. The shadow is, and the reason is a single sentence: a repeat in the shadow would need the irrational slope to be rational.
Start small enough to see the whole thing. The lattice is the integer grid, the line has slope 1/φ, and the strip around it is exactly one unit cell wide in the perpendicular direction — the width that makes the projection a genuine tiling of the line rather than something with gaps or overlaps.
The strip is exactly one unit cell wide in the perpendicular direction, which is the choice that makes the projection a tiling of the line: any narrower and there would be gaps, any wider and points would land on top of each other. At that width exactly one of an accepted point’s two neighbours is itself accepted, so the walk along the strip is forced. This slope is irrational, and that alone is what stops the word repeating.
Try it: The accepted points form a staircase, and because the strip is exactly one cell wide, the staircase steps right or up but never both. So the projection is a sequence of two lengths, in the ratio φ. Then switch the slope to 1/2 and watch the word fall into an obvious repeat — that one change is the whole difference.
The sequence that falls out of the strip has another description with no geometry in it at all: start with L, and repeatedly rewrite every L as LS and every S as L. That is a substitution, in one dimension, on two letters.
The two constructions agree, letter for letter. That is the bridge between the two halves of this subject — slicing a periodic thing and substituting forever are two descriptions of one object.
Each rewrite multiplies the length by roughly φ, so the letter counts are consecutive Fibonacci numbers and their ratio walks to φ = 1.618033989. Both routes are ways of writing down an irrational number: the strip does it by slope, the rewriting by the ratio of successive lengths, and it is the same irrational number either way. The projection is started 1/φ of the way across the window, which is what puts it at the same letter of the sequence the rewriting starts from; another offset gives the same word read from elsewhere.
Try it: Push the rewrites up and watch the words stay green all the way along. The letter counts are consecutive Fibonacci numbers, so the ratio of long to short marches to φ — the same φ as the slope, arrived at without ever drawing a line.
Count the distinct blocks of each length appearing in the sequence. A word with period p can contain at most p distinct blocks of any given length — past p positions you are seeing the same blocks come round again. So a word whose block count grows without bound has no period, and not even an eventual one.
The Fibonacci word has exactly n + 1 blocks of length n. Words with that complexity are called Sturmian, and n + 1 is the smallest a non-periodic word can manage — these are the one-dimensional quasicrystals, and there is nothing between them and periodicity.
| block length n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| distinct blocks | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
| n + 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
This is the whole argument for the one-dimensional case, and it is worth sitting with, because the hat proof is the same shape: find a structural quantity that a periodic object is forced to keep bounded or rational, then show this object does not.
Try it: Switch to a rational slope. The block count climbs for a while and then stops dead, and a period appears. This is the same shape of argument as the hat proof — find something a periodic object is forced to keep bounded or rational, and show this object does not.
Now the real thing. In 1981 N. G. de Bruijn showed that Penrose tilings are the dual of a pentagrid: five families of parallel lines at 72° to each other, with one rhomb per crossing. He also showed these are exactly the tilings the five-dimensional cut and project produces — which explains where the five edge directions on the previous page come from. They are the shadows of five lattice axes.
Family k is the set of lines where z · e_k plus an offset is a whole number, with the five offsets summing to zero. A crossing of family j with family k becomes a rhomb spanned by e_j and e_k, so |j − k| of 1 or 4 gives the fat rhomb and 2 or 3 the thin one. Grow the patch and the perpendicular radius stays bounded while the tile count does not — which is the definition of a cut-and-project set, arrived at from a construction that never mentions five dimensions.
Try it: Turn the grid on and find a crossing, then find its rhomb. Grow the patch and watch the widest perpendicular image: the tile count runs away and that radius does not. Staying inside a bounded window is the definition of a cut-and-project set, and it is being measured here on a construction that never mentions five dimensions.
Cut and project explains why quasicrystals diffract. A diffraction pattern is a Fourier transform, and the transform of a slice of a lattice is a projection of the reciprocal lattice — sharp spots, in impossible symmetries. That is what Shechtman photographed in 1982. What the method does not do is produce a tile: it makes point sets and tilings, not shapes with a rule attached. The einstein problem is about a shape, and the next pages go back to that.