Five-fold centres stay few, and mirrors do not

A chain with no mirror has mirrors everywhere asked of the Fibonacci chain which of its windows read the same backwards. The answer was rigid. There is exactly one such window of every even length and exactly two of every odd length, however long the windows are taken, and the reason was a fact about a circle: the chain is read off a rotation of a circle, a reflection of the circle has two fixed points, and the palindromes are the windows that sit over them. That essay ended by asking the same question of a tiling of the plane. Does a two-dimensional quasiperiodic pattern have a rigid count of symmetric patches, the way the chain has its one-and-two?

The plane has more to ask about. A window of a chain can have one symmetry besides the identity, a reflection. A round patch of a Penrose tiling can have up to twenty, the rotations by multiples of a tenth of a turn and the reflections in ten lines through its centre, and they need not behave alike. This essay runs the census on all of them, and the answer is not one rigid count but three different kinds of behaviour. Patches with a five-fold centre stop at four, however large they are drawn. Patches with a mirror keep growing in number, by about fifty for each edge added to their radius. And no patch wider than a single tile keeps a half-turn at all. The one-and-two of the chain turns out to be the first of these three answers, and the other two are what the extra dimension adds.

The census, and the twenty operations it tries #

The tiling is the rhomb tiling, grown by the subdivision described in inflation and the golden ratio from a wheel of ten triangles, nine times over. That gives about 19,000 vertices in a disc 76 edges in radius; a tenth inflation gives about 51,000 in a disc 123 edges in radius. All distances are measured in edges.

A patch of radius $r$ about a centre is the set of vertices within $r$ of it, written in coordinates relative to the centre. Two patches are the same when those sets agree exactly, which is the definition how many patches of each size used to count them: equality up to translation, with rotations kept distinct. A patch and its copy turned by a tenth of a turn are two patches in this count, just as that essay counted them. Only centres far enough inside the disc that their patch is not cut by its rim are used. The total of distinct patches at each radius is the plane’s version of the chain’s complexity, the count that n plus one, and no fewer found to be as small as an aperiodic chain allows.

Every distinct patch is then tested against the twenty operations about its own centre: the ten rotations by multiples of $36^\circ$ and the ten reflections in lines at multiples of $18^\circ$. The test is the same exact comparison, the patch against its own image. The operations that hold form the patch’s local group, which is one of the subgroups of the decagonal group: $C_1$ for nothing, $D_1$ for a single mirror, $C_2$ or $D_2$ for a half-turn with or without mirrors, $C_5$ or $D_5$ for a five-fold centre, and so on up to $D_{10}$.

The radii are an edge and a twentieth, one and a half edges and a twentieth, two edges and a twentieth, and so on up to six. The twentieth is not decoration. A radius equal to a distance that occurs in the tiling, such as one edge exactly, puts vertices on the circle, and whether they count as inside is then decided by the last bit of a floating-point number. A census run at exactly one edge found 72 patches on a sample of 51,000 vertices and 62 on one of 8,000. The ten extra came from vertices on the circle, and the discrepancy vanished at an edge and a twentieth.

A chain has two kinds of place to centre a window: on a letter, for an odd length, and between two letters, for an even one. The plane has three, and all three are counted. There are the vertices; the midpoints of edges; and the centres of rhombs. A symmetry of a large patch carries the tiles in it to tiles, so its centre is a vertex fixed by it, or the midpoint of an edge carried onto itself, or the centre of a tile carried onto itself. No other place can be a centre.

Three behaviours at once #

About a vertex the census finds three local groups and never a fourth: $C_1$, $D_1$ and $D_5$. A vertex’s patch either has no symmetry, or exactly one mirror, or the full symmetry of a regular pentagon. It never has a five-fold rotation without the mirrors, never two mirrors without the rest, and never a half-turn.

The numbers are the argument. At radii of one, one and a half, two, three, four, five and six edges (each plus a twentieth), the patches with a five-fold centre number 2, 2, 4, 4, 4, 4 and 4. Those with exactly one mirror number 60, 70, 80, 140, 210, 260 and 310. Those with no symmetry number 0, 60, 100, 220, 520, 840 and 1,140. The first column stops. The second grows by about fifty for each edge of radius. The third roughly quadruples each time the radius doubles.

Every mirror patch appears in ten orientations, since a tenth of a turn carries a patch with one mirror to a different patch with one mirror. Up to rotation, the mirror patches are 6, 7, 8, 14, 21, 26 and 31 shapes. The five-fold patches appear in two orientations each, so at two edges and beyond they are two shapes. At one edge they are one shape, five thick rhombs meeting at their sharp corners. The two shapes that separate at two edges share that first ring and differ in the ring beyond it.

The two samples, of nineteen and fifty-one thousand vertices, give identical counts at every radius and in every group. These are counts of the tiling, not of the sample, the distinction that how many patches of each size had to make with its own converged column. The growth that essay measured in its total is, at large radii, almost entirely the growth of the third column.

The picture at the head of this essay shows what the three columns mean on the tiling itself. In a region fifteen edges across, 98 vertices keep a mirror within three edges of themselves and 118 keep nothing. Seven have a five-fold centre. The mirrors point in only ten directions, those of the tiling’s edges and the lines halfway between them, and they are everywhere. That is the plane’s version of the chain’s finding that mirrors stand all along it. The five-fold centres are the rare thing, and even these seven are more than a larger radius would allow, since at four edges their share of the vertices falls below one per cent.

The patches themselves #

At three edges both five-fold shapes hold 31 vertices, and they differ in how the second ring is arranged about the first. Between them they are the patches of 1.2 per cent of all vertices, 0.81 per cent for one shape and 0.36 for the other. The two commonest shapes with one mirror are the patches of 0.66 and 0.51 per cent. A single five-fold shape is therefore about as common as a single mirror shape. The difference between the columns is not that five-fold patches are individually rare. It is that there are only two of them, while the mirror shapes keep multiplying.

Why the count behaves three ways #

The chain’s one-and-two was explained by reading the chain off a circle. The Penrose tiling is read off a higher-dimensional lattice in the same way, and the same reading explains all three columns.

Every edge of the tiling points along one of ten unit vectors, which are five vectors $e_0, \ldots, e_4$ and their negatives. So every vertex is $n_0 e_0 + \cdots + n_4 e_4$ for five integers $n_j$, which a walk along the edges from the centre of the wheel finds for every vertex. On both samples the walk reaches every vertex, and no two routes to one vertex ever disagree about its five integers. The same integers put each vertex at a second position, in a plane of its own called internal space, by replacing each $e_j$ with a different unit vector: the one at $144^\circ$ times $j$ rather than $72^\circ$ times $j$. Their sum modulo five sorts the vertices into four height classes. This is the construction of cut and project run backwards. There a lattice is sliced and projected to make a quasiperiodic pattern. Here the pattern is lifted back into the lattice, and which lattice points are vertices is decided by where they fall in internal space: inside a pentagon, one pentagon for each height class.

What a vertex’s patch looks like is decided by where the vertex sits in its pentagon. Two vertices whose internal positions are close have the same patch out to a large radius, since every neighbour of one has a counterpart near the corresponding neighbour of the other, on the same side of every pentagon’s edge. The pentagon is cut into regions, one for each distinct patch of radius $r$, and the regions get smaller as $r$ grows. Their number grows as the square of $r$, which is why the no-symmetry column grows that way.

A local symmetry is a condition on where the vertex sits. A reflection of the tiling in a line through a vertex acts on internal space as a reflection too, and the vertex’s patch keeps it only if the vertex’s internal position is close to that reflection’s mirror line, close enough that no neighbour’s position is carried across an edge of a pentagon. The allowed band narrows as the radius grows, because more neighbours have to be carried correctly. A rotation by a tenth of a turn acts on internal space as a rotation too, about the centre of the pentagons, and a five-fold patch needs its vertex close to that single point.

The picture shows the largest height class on the smaller sample. At two edges, 1,075 of its 2,490 vertices have a patch with a mirror, and they fill bands so wide that the five lines are hard to pick out. At four edges, 570 of 2,245 do, and they lie in strips about the pentagon’s five mirror lines, every one of them within 0.19 of its line, in units in which the five internal vectors have length one. At two edges the same measure is 0.33. The five-fold vertices shrink from a pentagonal blob of 65 at the centre to 5.

That is the whole explanation of the three columns. The distinct patches with a given symmetry are the regions that meet the set the symmetry leaves fixed. A point meets a bounded number of regions, however finely the pentagon is cut. A line meets a number that grows as fast as the regions shrink, in proportion to $r$. The whole pentagon meets all of them. The five-fold column is bounded because a rotation of internal space fixes a point. The mirror column grows linearly because a reflection fixes a line. The no-symmetry column grows as the square of the radius because doing nothing fixes everything.

And this is the chain’s one-and-two again. The Fibonacci chain’s internal space is a circle, one-dimensional. A reflection of a circle fixes two points and no line, and so the chain’s palindromes are bounded, one or two of each length, for the same reason the plane’s five-fold patches are bounded at four. The chain had no operation whose fixed set was a line, because a circle has no room for one. The plane’s mirrors are the first case of a symmetry whose fixed set has a dimension, and the dimension is exactly the exponent with which its patches grow.

How often a vertex keeps a symmetry #

The same picture predicts how the share of vertices with a symmetric patch falls, not only how many distinct patches there are. The share with a mirror is the area of the strips, which is their length times a width shrinking like $1/r$. The share with a five-fold centre is the area of a small disc about one point, shrinking like $1/r^2$.

At one edge every vertex has a mirror, since every vertex star of the rhomb tiling is symmetric. The share with a mirror falls to 64 per cent at one and a half edges, 53 at two, 45 at three, 32 at four, 29 at five and 25 at six. Multiplied by the radius, those shares run between about one and one and a half, rather than falling by a factor of four. The five-fold share holds at 5.7 per cent to two edges, then falls to 2.1, 0.87, 0.89 and 0.34. Multiplied by the square of the radius, from two edges on it stays between about an eighth and a quarter. Neither product is constant. The regions of internal space shrink in steps, whenever a new neighbour enters the patch, and a step can leave the share unchanged over a whole edge of radius, as the five-fold share does between four and five. But the shares follow the prediction to within a factor of two, over a range in which they themselves change by factors of four and seventeen.

So most of the tiling is locally mirror-symmetric at small scales and a shrinking fraction of it at large ones, and yet the number of distinct mirror-symmetric neighbourhoods keeps growing. Those are two different measurements, the first of how often and the second of how many kinds. Neither makes the symmetric patches scarce in the sense that matters to a reader of the tiling: each of them, like every patch of an aperiodic chain, comes back within a bounded distance of anywhere. How often each patch occurs found each patch’s frequency as an entry of an eigenvector, and the shares here are sums of such frequencies over the patches in one column.

The half-turn does not survive the first ring #

The table extends the census to the other two kinds of centre. About the midpoint of an edge there are patches with one mirror, 20 at one edge and 160 at six, growing like the vertex column. The mirror is the perpendicular bisector of the edge or the line through it. About the centre of a rhomb there are also mirror patches, 50 at one and a half edges and 180 at six. And at one edge there are ten patches with the group $D_2$, a half-turn and two mirrors. They are the rhombs themselves, thick and thin in five orientations each, with nothing else within an edge of their centre. A rhomb, as a shape, has a half-turn.

At one and a half edges those ten are gone, and no patch about any vertex, edge midpoint or rhomb centre, of any radius from there to six, has a half-turn. By the argument above, a centre is the only place a large patch could keep one. The half-turn is not merely bounded, like the five-fold rotation. It is absent.

The picture shows where it fails. Widen the patch about a rhomb centre by half an edge and the first vertices of the neighbouring tiles come in. In the rhomb drawn, two of them have no partner at the point opposite, so the half-turn fails at the first ring of neighbours. The table says it fails there about every rhomb in the tiling.

The reason is the arrows. A Penrose tiling is a tiling by rhombs whose edges carry markings, single and double arrows, and the rule that makes it aperiodic is that two tiles sharing an edge must agree on its marking and on the direction of its arrow. Now suppose a patch had a half-turn about the midpoint of an edge. The half-turn exchanges the two tiles on either side of that edge, and it reverses the edge. It carries the first tile’s arrow, pointing one way along the edge, to an arrow on the second tile pointing the other way. The two tiles would disagree about their shared edge, which the rule forbids. About the centre of a rhomb the same thing happens to the rhomb’s own edges, since no marked rhomb keeps its markings under a half-turn. About a vertex the census found no half-turn even at one edge: no vertex star of the tiling has one.

That argument is about marked tiles, and the census never saw a marking. It looked at bare vertex positions. What the census shows is that the positions carry the arrows, at least as far as the half-turn is concerned: by one and a half edges, the vertices near a rhomb already say which way its edges point.

The internal-space picture gives the same verdict by another route, and more cautiously. A half-turn of the tiling acts on internal space as a half-turn, fixing a point, so by the argument for the five-fold rotation its patches should be bounded. Bounded includes zero. The census says zero, and the arrows say why.

What the picture cannot show #

The census is a finite count on two finite samples. It establishes the numbers to a radius of six edges, and that the counts agree between the two samples at every radius. A patch that keeps a mirror to six edges may lose it at seven, which is the same limit how much pattern is enough measured for periodic patterns: a finite window can fail to refute a symmetry the whole pattern lacks. It does not prove that the five-fold column stays at four for ever, or that the mirror column grows linearly without limit. Those are what the internal-space argument predicts, and the argument is standard, but it was checked here only by the narrowing of the strips and by the shares, not proved.

The census also depends on a convention it does not hide. Patches are compared up to translation, so a patch and its rotation are two patches. Counted up to rotation instead, every column except the five-fold one shrinks by a factor of ten and the five-fold one by two. The three kinds of growth, bounded, linear and quadratic, do not depend on the choice. And the samples were grown from a wheel with a five-fold centre of its own, so they are a special Penrose tiling, one with a global five-fold symmetry about the middle. Every patch in them occurs in every Penrose tiling, which is why a count of distinct patches is not affected. The positions of the symmetric vertices near the middle are, which is why the field in the first picture was drawn well away from it.

Nor does the arrow argument come from the census. It explains why the half-turn must die, given that the vertex positions carry the arrows. That they carry them within an edge and a half is what the census found, not something it assumed.

Where the ideas come from #

The rhomb tiling and its arrows are Penrose’s, from the 1970s, and they became widely known through Gardner’s column of 1977. The description of the tiling as a slice of a five-dimensional lattice, with pentagons as the windows, is de Bruijn’s, from 1981. The counting of palindromes in Sturmian words, whose one-and-two this essay generalises, goes back to Droubay and Pirillo in 1999. The link between local symmetries of a quasicrystal and fixed sets in internal space is part of the general theory of cut-and-project sets. The census on this page is a direct count.

What the census must refuse #

The refused claim is the one this essay set out to test, in the form the chain suggested. Had the plane behaved like the chain, the mirror column would have settled at some small number and stayed there, and the tiling’s local mirrors would have been as rigid as the chain’s palindromes. The census finds 80 mirror patches at two edges and 210 at four, and the test that refuses a bounded count is run on that difference.

The lift test belongs in the list because the explanation depends on it. The internal-space argument assumes every vertex has one well-defined position in the five-dimensional lattice, and the walk over the edges checks that assumption on every vertex of the sample, by finding the same five integers for every vertex whichever route reaches it.

Still open: a count that depends on a dimension #

The three columns say that a local symmetry’s patches grow with the dimension of the set it fixes in internal space. The Penrose tiling has a two-dimensional internal space, and the only fixed sets available are a point, a line and the whole plane. The icosahedral quasicrystals are sliced from a six-dimensional lattice and have a three-dimensional internal space, and there every operation of the icosahedral group fixes a point, a line or a plane of internal space: the inversion and the rotoinversions a point, every rotation its axis, every mirror a plane. The same argument predicts four columns there, bounded, linear, quadratic and cubic, in the radius of a three-dimensional patch, and it predicts that a five-fold axis, which is a point’s worth of symmetry in the plane, becomes a line’s worth in space. Six integers and the lattice that holds them builds that lattice, and the census would need its slice rather than an inflation. Whether the prediction holds there too is a count nobody here has made.