The tiling that points every way

Every pattern in this collection has had a finite list of directions in it.

A wallpaper pattern’s motif appears in as many orientations as the group has operations, and there are at most twelve. A Penrose tiling never repeats and is nevertheless just as restrained: its inflation rule turns each tile through a multiple of thirty-six degrees, thirty-six degrees closes up after ten, and the tiles of a Penrose tiling point in ten directions however far the rule is applied.

That finiteness is doing real work. A diffraction pattern’s symmetry is the symmetry of the set of directions in the structure, so a structure with ten orientations can scatter with ten-fold symmetry and a structure with four can scatter with four-fold. What would a structure with infinitely many orientations do?

The answer is that it could not have a rotational symmetry of any finite order, and it could have one of infinite order — a diffraction pattern that is the same at every angle. That is a stronger break with periodicity than a quasicrystal makes, and it is available from a rule short enough to state in two sentences.

One triangle, five copies #

Take the right triangle with legs 1 and 2. Its hypotenuse is √5, and it has a property no other right triangle has: the altitude from the right angle cuts off a copy of the whole triangle at exactly one over root five. That is not arranged; it follows from the leg ratio being one to two.

The piece left over is a copy of the same triangle at two over root five, and joining the midpoints of its sides cuts it into two corner triangles and a rectangle. Cut the rectangle along a diagonal and there are five pieces, every one of them the original at one over root five.

Apply that rule to every tile, over and over, and a patch of one shape and one size grows.

Nothing in that construction is random and nothing is chosen by eye. Each tile’s five children are determined by three points — the foot of an altitude and two midpoints — and the whole patch is the rule applied to itself.

Thirty-two directions at four steps, and it does not stop. Apply the rule again and there are more; there is no depth at which the list closes.

Why the triangle has to be this one #

The subdivision above used one property of the (1, 2, √5) triangle and it is worth isolating, because it is the only place the shape enters.

Drop the altitude from the right angle of any right triangle to its hypotenuse. The two pieces are always similar to the whole — that is a fact about right triangles and needs no special shape. What is special here is the ratio: the altitude cuts a triangle with legs a and b into pieces at scales b/c and a/c where c is the hypotenuse. For the pieces to be at 1/√5 and 2/√5, and so for one of them to be exactly a fifth of the area, the legs must be in the ratio one to two.

So this triangle is the unique one for which “one small copy plus four half-size copies of the large one” comes out to five equal pieces. Any other right triangle gives a small copy and a large one that do not divide into a whole number of equal pieces, and the substitution does not exist.

That the inflation factor is √5 and the turn has tangent 1/2 are the same fact, and the second is what matters. A leg ratio of 1 : 2 gives a hypotenuse direction with tangent 1/2, an angle whose cosine and sine are 2/√5 and 1/√5, and a complex number 2 + i whose norm is 5. Every argument below is about that Gaussian integer.

Which diagonal #

A rectangle has two diagonals, and both of them produce a valid subdivision of the triangle into five congruent copies. An exhaustive search over triangles with rational vertices finds exactly two subdivisions and no others, and they differ in precisely this.

The reason is one line of composition. If every child is a reflection through some angle −θ, then applying the rule twice composes two reflections — and the composition of two reflections is a rotation by the difference of their angles, which here is a multiple of a right angle. The turn cancels exactly. Two steps of the rule return every tile to a direction it had at the start.

The pinwheel escapes because its five children are not all of one kind. Two keep the parent’s handedness and three reverse it, and among them are turns of both +θ and −θ. Composing a +θ with a −θ gives 2θ, and 2θ is not going away.

This file’s first version drew the other diagonal, passed every check it had — the pieces added to the whole, every child was a copy at the right scale, the angle was proved irrational — and reported four directions at every depth with nothing complaining. A rule that produces the right shapes in the right places is not thereby the right rule.

The count is not the number of directions in the tiling. A patch grown four times has thirty-two of them; a patch grown five times has thirty-eight, and six gives forty-eight. Those are counts of what has appeared so far, and each is a lower bound on what the infinite construction contains. The upper bound is what the argument below removes: there is no ceiling, because a ceiling would require the turns to repeat.

It is also worth saying what the growth is not. The direction count does not grow like the tile count — five to the depth — because most children share a direction with some cousin. It grows roughly linearly in the depth, since the reachable turns after n steps are the multiples of θ up to n, and each of those with four quarter-turns added. Slow growth without a ceiling is still without a ceiling.

Why 2θ never closes #

The turn is the angle with tangent one half. Whether the directions can close up is the question of whether that angle is a rational part of a full turn, and the answer is an exercise in arithmetic rather than in geometry.

In the complex plane, the rotation carrying the parent onto a child is multiplication by (2 ± i)/5 up to a quarter turn — a number of modulus one over root five with the right argument. The composite turn 2θ is multiplication by

$$\frac{(2+i)^2}{5} \;=\; \frac{3+4i}{5}.$$

If that were a root of unity then some power of it would be one, so (3 + 4i)ⁿ would equal 5ⁿ.

It never does, and the reason is short enough to state. In the Gaussian integers 5 is not prime: it factorises as (2 + i)(2 − i), and 3 + 4i is (2 + i)². So (3 + 4i)ⁿ = 5ⁿ would read

$$(2+i)^{2n} = (2+i)^{n}\,(2-i)^{n},$$

which forces (2 + i)ⁿ = (2 − i)ⁿ. Those are two primes that are not associates — neither is the other times a unit — and unique factorisation forbids it. The angle is irrational, exactly, and the table above is a check on the argument rather than the argument itself.

That is the same shape of reasoning as the crystallographic restriction, and it is worth setting the two side by side. The restriction asks which rotations are integer matrices and finds five. This asks which rotations are roots of unity in a particular ring and finds that one particular rotation is not. Both are questions about arithmetic dressed as questions about shape, and both have answers a picture cannot supply.

There is a third member of that family already on this site and the comparison sharpens all three. The inflation factors a quasicrystal may have are decided by asking whether an algebraic integer’s conjugates are small — a condition on a number’s other embeddings rather than on the number itself. Here the condition is on a number’s order in a group, and there it is on a number’s size. Same subject, different question, and neither is reachable from the other.

One tempting shortcut is wrong and worth naming. It is easy to think the angle’s irrationality alone settles the matter — that any substitution turning tiles through an irrational angle must produce infinitely many directions. The other diagonal disproves it: its children are turned through exactly the same irrational angle, and its directions still close up after two steps, because each turn is undone by the reflection that comes with it. What has to be irrational is the composite, and which composites occur depends on how the children are placed and not on the angle each one is at.

Directions everywhere, and evenly #

Two claims have to be separated here, and only one of them is proved.

That the directions are infinite in number follows from the irrationality above: the turns 2θ, 4θ, 6θ … are all distinct modulo a full turn, because 2θ is not a rational part of one.

That the directions become evenly spread is a stronger statement and is Radin’s theorem rather than anything derived here. A set can be infinite and dense and still be lumpy at every scale. What this collection can do is measure it.

The consequence is what makes the tiling worth an essay. A structure whose parts point in every direction, evenly, has no preferred direction at all — so its diffraction has circular symmetry, a continuous rotational symmetry that no crystal and no quasicrystal may have. Sharp spots with ten-fold symmetry forced the definition of a crystal to be rewritten; a pattern with a continuous rotational symmetry is outside the rewritten definition too.

The tiles come in two hands and both occur. Three of the five children are reflected and two are not, so the tiling contains the triangle and its mirror image in comparable numbers. That is a difference from the Penrose tiling, whose two rhombs are each their own mirror image, and it is the reason the direction census here tracks handedness as well as angle: a tile and its reflection pointing the same way are two different tiles, and counting them as one would halve the answer for no reason.

And the handedness is not a detail of the drawing. A count of directions that ignored it would report the two hands of a triangle at one angle as a single direction, and would then be measuring the set of lines the tiles lie along rather than the set of orientations they take. The two are different sets and the smaller one is not what the theorem is about: a structure whose parts lie along infinitely many lines but in only two senses along each would still have a preferred sense, and the circular symmetry the essay ends on would not follow. So the census tracks the hand, and the wheel figure above draws the unreflected tiles more strongly for the same reason.

There is a broader point behind all of this, and it is the one that makes an aperiodic tiling worth the trouble at all. Order and repetition are different properties, and they were treated as one for a century because every ordered structure anybody had seen repeated. A patch of a Penrose tiling and a patch of a periodic one, seen through a small enough window, are indistinguishable: both are built by a rule, both have every local configuration occurring at a well-defined frequency, and neither betrays which it is until the window is large enough to look for a translation. The pinwheel takes that further than the Penrose tiling does — it is ordered, it repeats not at all, and it has no preferred direction either — which is why the definition of a crystal that Shechtman’s measurement forced open still does not reach it.

Where this sits among the site’s other aperiodic patterns #

The comparison with the Penrose tiling is worth making in numbers, because the two are separated by exactly the thing this essay is about and by nothing else.

The Penrose substitution has a two-by-two matrix and the pinwheel’s is one-by-one. Penrose has two tiles, thick and thin, and its matrix reads (2 1 · 1 1): its characteristic polynomial is λ² − 3λ + 1 and its dominant eigenvalue is φ² ≈ 2.618. That is what the tile counts grow by from one inflation to the next; the linear scale grows by the golden ratio itself, φ ≈ 1.618, which is the square root of it. The two numbers are easy to run together and this collection has done it, which is why both are stated here. The pinwheel has one tile, its matrix is the one-by-one matrix (5), its tile counts grow by five and its linear scale by root five.

Every interesting thing about the pinwheel is invisible to that matrix. A one-tile substitution has the matrix (5) whichever diagonal of the small rectangle is drawn, so the matrix cannot tell the pinwheel from the subdivision that closes up after two steps — and the matrix is the whole of what the standard algebraic machinery reads. The counts, the growth rate, the inflation factor and the frequencies are all identical between a tiling with four directions and a tiling with infinitely many.

Which inflation factors exist answers a question about the scaling part of a substitution: the factor is an eigenvalue of an integer matrix, so it is an algebraic integer, and sharp diffraction requires its conjugates to be small. The pinwheel’s factor is √5, which passes that test comfortably.

So the two tilings are separated by nothing the earlier essay’s algebra can read. Both factors pass its test; both matrices are honest; and everything distinguishing a tiling with four directions from one with infinitely many lives in the rotations, which is why the argument here needed a different question and a different ring to answer it in.

What the pinwheel does not settle #

It is not a crystal and the machinery here decides nothing about it. This collection’s round trip works in a lattice basis, where a symmetry is an integer matrix. The pinwheel has no lattice, its tiles sit at angles that no integer matrix expresses, and every claim above about it is either an exact statement about Gaussian integers or a measurement of a finite patch. Nothing was detected and nothing was verified in the sense the rest of the site means.

And the diffraction is named rather than computed. That a pattern with evenly spread directions scatters with circular symmetry is stated here and derived nowhere; the chain’s spectrum is computed on this site because it is one-dimensional and a finite sum, and the pinwheel’s is not. What the essay establishes is the input to that claim — the directions, their growth and their evenness — and it says where the claim stops being its own.

Nor does anything here say the tiling exists in the plane. Every picture above is a finite patch grown from one triangle, and the step from an increasing sequence of patches to a tiling of the whole plane is a limiting argument this collection does not make. It is standard and it is not free: a substitution can produce arbitrarily large patches and fail to tile the plane, and the usual repair is to find a patch that reappears inside its own subdivision. The pinwheel has one, and this essay’s claim is about the patches.

The tiling is not unique either. A choice was made at every step of which subdivision goes where, and different choices give different tilings that no motion carries onto one another. They all have the same directions, the same patch statistics and the same diffraction, which is the freedom a crystal has not appearing in a third setting: a parameter that changes the structure and nothing measurable about it.

Finite complexity, but only up to a rotation #

There is a technical property every other pattern in this collection has and the pinwheel does not, and naming it says exactly what the infinitely many directions cost.

A tiling has finite local complexity when, for each radius, only finitely many different patches of that radius occur — up to translation. Every periodic pattern has it trivially, and so does a Penrose tiling: its tiles point in ten directions, its vertex configurations are a short list, and a patch of a given size is one of finitely many things.

The pinwheel does not. Its tiles point in infinitely many directions, so a patch of a given radius occurs in infinitely many orientations, and no two of them are related by a translation. Counting patches up to translation gives infinity at every radius above zero.

What it does have is finite local complexity up to rotation. Allow patches to be compared after turning as well as sliding, and the count is finite again — the same short list as any substitution tiling, because the substitution rule produces only finitely many local arrangements and the orientations are what the rotations remove.

That distinction is not a technicality; it is the whole content of the pattern. Every tool built for aperiodic tilings assumes finite local complexity in the translational sense, and the pinwheel is the standard example of what has to be redone when the assumption is dropped. The diffraction is the visible case: a pattern with finitely many directions scatters to a point set with the symmetry of those directions, and one with infinitely many, spread evenly, scatters to something with continuous rotational symmetry — rings rather than spots.

The same trick in space #

The construction generalises upwards, and the three-dimensional version is worth naming because its orientations are stranger than the plane’s.

Conway and Radin built a substitution on a tetrahedron whose children are turned through two rotations about different axes, and those two rotations generate a subgroup of the rotations of space that is dense: the orientations of the tiles are not merely infinite in number but come arbitrarily close to every possible orientation.

That is a stronger statement than the plane’s, and it is available only in three dimensions, because rotations of the plane commute and rotations of space do not. In the plane the orientations form a subgroup of a circle and their density is a statement about one irrational angle; in space they form a subgroup of a group with no such simple description, and two generic rotations generate a dense subgroup for reasons that have nothing to do with any single angle being irrational.

So the pinwheel’s oddity is the small case of something that gets stranger with dimension, rather than a curiosity confined to one triangle. What both cases share is the point this essay opens with: the finiteness of a pattern’s direction list is an assumption, it is doing work everywhere in the classification, and a substitution rule can break it without breaking anything else.