Eleven tilings, five groups

Eleven tilings by regular polygons have every vertex alike. Each of them is a plane pattern, and this collection has a machine for plane patterns: forget how the pattern was made, enumerate every operation its lattice could permit, and keep the ones that map the point set onto itself.

Point the machine at the eleven and the answer is startlingly coarse. The eleven tilings have five groups between them.

Asking the question properly #

Three things have to be got right before the answer means anything, and each of them is a place where an easier route gives a wrong number.

The lattice is found, not supplied. A detector needs to know which lattice the pattern repeats on, because that is what supplies the candidate rotations — the crystallographic restriction is a statement about integer matrices in a lattice basis. So the translations are discovered first: every vertex-to-vertex vector is a candidate, and a candidate survives only if it carries every polygon of the patch onto a polygon of the patch. The two shortest independent survivors are the cell.

That test is stricter than it sounds, and the strict version is necessary. A translation that carries the middle of a patch is easy to find and means nothing, because the vertex species does not always decide the tiling — a patch can be legal at every vertex and out of register with itself further out. Testing the core alone reported a cell of eight vertices for a tiling whose cell holds four.

The coordinates are exact and are not rational. Truncate a honeycomb so its twelve-gons and triangles share an edge length and the cut falls at 2 − √3 of the way along; a vertex of 4.8.8 sits at 1/(2 + √2) of its cell. Those are exact numbers and no fraction equals them.

So the detector was rewritten to work over ℚ(√d), where a number is a triple of integers (p + q√d)/r. Addition, multiplication and division are exact — division by the conjugate, since d is not a square — and two numbers are equal exactly when their reduced triples agree. The round trip survives the change of number system unaltered: it is still complete, still enumerates every operation the lattice permits, and still has no tolerance in it anywhere.

The search over operations is complete rather than clever. For each linear part the lattice permits, and each ordered pair of vertices, the translation carrying the first onto the second is a candidate — and every symmetry of the pattern must arise that way, because a symmetry sends some vertex to some vertex. Each candidate is then applied to every vertex and either lands the whole set on itself or does not. There is no search heuristic, no early exit on a near miss, and nothing that could quietly miss an operation; the cost is quadratic in a number that is never more than twelve.

Uniform is a stronger condition than “every vertex reads the same”. The eleven are the tilings whose group is transitive on vertices, so the machinery checks it: the group is detected, the orbit of one vertex under it is computed, and the orbit must be every vertex of the cell. The two tilings that admit non-periodic variants are exactly the ones where that check does work rather than merely passing.

What comes back #

Six of the eleven are p6m — the triangular, hexagonal and trihexagonal tilings, the truncated hexagonal, the rhombitrihexagonal and the truncated trihexagonal. Two are p4m: the square tiling and 4.8.8. The three that are left are the interesting ones.

3.3.3.3.6, the snub hexagonal tiling, is p6. Six operations, all rotations, no reflection anywhere. It is the only chiral member of the eleven, and its two mirror-image forms are genuinely different tilings that no motion of the plane relates — the same distinction Friedel’s law hides in a diffraction pattern and that separates the eleven enantiomorphous pairs of space groups.

3.3.4.3.4, the snub square tiling, is p4g — and it has mirrors, so “snub” is not a synonym for chiral. Its reflections are diagonal and its axial lines carry glides, which is the whole content of the difference between p4m and p4g.

That distinction very nearly went wrong here, and the way it did is instructive. An operation’s intrinsic translation — the part of it no choice of origin removes — is what separates a mirror from a glide, and it is computed by averaging the translation over one cycle of the linear part. Computed from a translation that has already been reduced into the cell, it is not well defined: reducing changes it by the projection of a lattice vector onto whatever the operation fixes, and for a diagonal mirror of a square lattice that projection is half the primitive translation along the mirror. So the truncated square tiling came back with one diagonal reflection a mirror and the other — its own conjugate under the four-fold — a glide, which no group can do. The ambiguity has to be quotiented out before an intrinsic part can name anything, and once it is, the real distinctions survive: p4g’s axial glides have no such ambiguity and stay glides. That p4m’s diagonals carry glide lines as well as mirror lines is not a defect in the arithmetic; it is a fact about p4m that the arithmetic was reporting.

3.3.3.4.4, the elongated triangular tiling, is cmm — four operations, on a centred rectangular lattice. It is by a long way the least symmetric of the eleven, and it is also one of the two whose vertex species does not determine the tiling. Those two facts are related: a tiling with room to be varied is a tiling with little symmetry to lose.

The lattice it comes back on is worth a second look, because the reduction found a basis of two different lengths — (1, 0) along a row of squares, and a vector to the next row — and the two are not each other’s mirror image. A centred rectangular lattice has to be handed over in its rhombic basis, the pair of equal-length vectors that a mirror exchanges, or the integer matrices that stand for its holohedry describe a different lattice. Handed the reduced basis instead, this tiling came back with two operations rather than four. The condition that decides it is not “the two vectors are the same length”: it is that one of them is equidistant from the origin and the other, so the perpendicular bisector between them is a mirror — a test that catches the elongated triangular tiling and the equal-length case together.

What a vertex’s own symmetry has to be #

Transitivity turns the count of vertices in a cell into a statement about each one. If the group has |G| operations per cell and carries one vertex onto all k of them, then the subgroup fixing a vertex — its site symmetry — has order |G|/k, and that is a number about the picture rather than about the algebra.

For 3.4.6.4 it is twelve over six, so every vertex of the rhombitrihexagonal tiling sits on a mirror and on nothing else, which is visible in the drawing once it is looked for: the line through a vertex bisecting its square and its hexagon. For 4.8.8 it is eight over four, again a mirror. For the trihexagonal tiling it is twelve over three, an order-four site symmetry — a two-fold rotation and two perpendicular mirrors, which is what a point at the midpoint of a lattice edge has. For the triangular tiling it is twelve over one: the vertex is a six-fold centre with mirrors, the most symmetric position a plane pattern has.

And for the snub hexagonal tiling it is six over six — site symmetry of order one. Its vertices are at general positions, which is the algebraic form of the observation that nothing about that tiling looks symmetric locally.

The twelve that never appear #

Twelve of the seventeen do not occur, and the absences have a pattern in them. Nothing without a rotation appears: p1, pm, pg and cm are all missing, so no uniform tiling gets by on translations and reflections alone. Ten of the eleven have a four-fold or a six-fold rotation, and the eleventh is the exception this census keeps producing — the elongated triangular tiling, whose group cmm has nothing above a two-fold. And — the absence that surprises — no three-fold group appears at all. There is no uniform tiling with group p3, p3m1 or p31m.

The cmm case is worth being precise about rather than waving at, because a rule with one exception in eleven is a rule that has to name it. The elongated triangular tiling is built from alternating rows: a row of squares, then a row of triangles, then squares again. Every row is like every other row of its kind, which gives mirrors along the rows and across them; but the two kinds of row are unlike each other, and no rotation of order three, four or six can carry a square onto a triangle. What survives is a two-fold, a pair of mirror directions and a centred rectangular lattice — the only one of the eleven that does not sit on a square or a hexagonal lattice.

The reason is that a regular polygon is more symmetric than the tiling needs it to be. A tiling whose group is p3 would have a three-fold centre where p6 has a six-fold one, and something at that point would have to distinguish the two halves of the turn. In a tiling by regular polygons at a common edge length there is nothing available to do it: the polygon sitting on a three-fold centre is regular, so it has the six-fold symmetry as well, and the group is p6 or larger.

That is an instance of the failure this collection watches for from the other side. Accidental symmetry is what happens when a motif is placed at a special position and quietly acquires operations the caption never claimed. Here it is not accidental and not avoidable: regularity is extra symmetry, and it propagates into the group.

The five that do occur are worth naming, because between them they say what kind of group a uniform tiling is forced to have. They are p4m, p4g, cmm, p6 and p6m, and the distribution is lopsided: six of the eleven tilings have p6m, two have p4m, and p4g, p6 and cmm have one each.

Four of the five carry a mirror, and the exception is p6. It has six-fold rotation and no reflection anywhere, and it belongs to exactly one tiling — the snub hexagonal, 3.3.3.3.6, which is therefore the only chiral member of the eleven: its mirror image is a different tiling, and no motion of the plane carries one onto the other. The other snub, the snub square tiling, is not chiral despite the name, because its group p4g contains reflections; the word snub describes how the tiling is constructed rather than what symmetry it ends up with.

The rest divide by lattice. p4m and p4g sit on a square lattice and are the two ways a four-fold group can carry mirrors: p4m puts them through the four-fold centres, p4g puts glides there instead and mirrors between. p6 and p6m sit on a hexagonal one, cmm on a centred rectangular one. So the eleven tilings use three of the five plane lattices, and the square and hexagonal ones between them account for ten of the eleven.

Four of them are a lattice with holes #

The four tilings with rational coordinates admit a description the other seven do not.

The honeycomb is a triangular lattice with one point in three removed. Its two vertices per cell generate a subgroup of order three; the missing coset is the centre of each hexagon.

The trihexagonal tiling is a triangular lattice with one point in four removed — three vertices in a subgroup of order four.

The snub hexagonal tiling is a triangular lattice with one point in seven removed. Six vertices, a subgroup of order seven, and a cell √7 across. Seven is a number this collection has already met as an index: it is the smallest sublattice index of a hexagonal lattice past three that keeps the six-fold symmetry, a Loeschian number 1² + 1·2 + 2². The snub hexagonal tiling is what is left of a triangular lattice when one coset of an index-seven sublattice is deleted, and its √7 cell is that sublattice’s cell.

For the remaining seven the question does not get a different answer — it does not arise. An irrational coordinate cannot lie in a finite subgroup of the torus at all, so those tilings are not any lattice with anything removed.

What the round trip checked, and how #

The claim under test is that a bare point set decides these groups. It could fail in the direction this collection worries about most: a point set has no edges, so an operation is free to carry a triangle’s corner onto a square’s, and the vertices might have symmetry the tiling does not.

They do not. For each tiling the detected operations were re-tested against the polygons — every tile mapped, matched by exact vertex keys modulo the lattice — and every operation survived. Eleven times out of eleven the vertices know where the edges are.

That is worth stating as a positive result rather than as a check that passed. A uniform tiling is recoverable from its vertex set, so the classification of these objects loses nothing by being carried out on points, which is the only form the detector accepts.

It is also not obvious in advance, and the reason it holds is worth having. The polygons of a uniform tiling are determined by the vertices because the edge length is fixed: a tile is a cycle of vertices at unit separation enclosing no other vertex, and that is a condition on the point set alone. Change the rule — allow two edge lengths, or allow a tile to contain a vertex in its interior — and the implication fails immediately, which is the same reason the near-symmetry essay’s tolerance argument cannot be run here: there is no continuum of nearly-right answers, only a point set that either has unit-separated cycles or does not.

The refusals are the other half. Displacing a single vertex of the trihexagonal tiling by a seventh of a cell takes its twelve operations to two — the detector is not reporting the symmetry a reader expects, it is reporting the symmetry that is there.

Where the exactness stops #

The detector is complete relative to the lattice, and that is not a limitation. It tries only the linear parts the lattice’s holohedry permits, which is every integer matrix of finite order preserving that lattice’s metric — and a symmetry of the pattern must preserve the lattice, so nothing can be missed. The completeness is inherited from the lattice being found correctly, which is why the translation test is the strict one.

“The group of the tiling” has been used here to mean the group of the plane pattern. A tiling also has a combinatorial automorphism group — the symmetries of its adjacency structure, ignoring geometry — and the two need not agree for a general tiling. For these eleven the geometric group is what is reported and the combinatorial one is not computed, because the question this collection asks throughout is which of the seventeen a pattern has.

Five groups is a count of these eleven, not a statement about tilings in general. Allow two kinds of vertex and far more groups become available; allow tiles that are not regular and all seventeen do. The coarseness here is a consequence of regularity, not of tiling.

The quadratic field is exactly wide enough and no wider. It holds these eleven because the surviving species use only triangles, squares, hexagons, octagons and twelve-gons. A tiling that needed a regular pentagon would need ℚ(√5) and the machinery would have to be told; a tiling that needed a heptagon would need a cubic field and this arithmetic would not reach it at all. That the classification stops exactly where the arithmetic does is a coincidence worth naming as one.

Who found them, and when #

Kepler drew all eleven in Harmonices Mundi in 1619 and did not have wallpaper groups to say anything about, since the seventeen were not classified until Fedorov in 1891. The two lists were made two and a half centuries apart for different purposes, and the map from one to the other has never had much to say — which is exactly what a five-for-eleven answer records. A tiling’s identity is carried mostly by information its symmetry group throws away.

The nineteenth-century ornament literature the Alhambra question belongs to made the same discovery in the other direction: pattern designers had far more than seventeen patterns and only seventeen groups, and the surplus was never a defect in the classification.

Where the ladder goes next #

Each of these eleven has one kind of vertex and several kinds of tile. Dualising swaps the two — a vertex for every tile, a tile for every vertex — and produces eleven tilings with one kind of tile and several kinds of vertex. The same eleven objects classify two different things, and the duals are worth drawing.

The coarseness is caused by the hypothesis #

Five groups out of seventeen looks like a fact about regular polygons. It is closer to a fact about the word uniform, and dropping that word restores most of the missing twelve immediately.

Vertex-transitivity is a demand for a large group. A tiling is uniform when its symmetry group carries any vertex to any other. The vertices are spread across the whole plane, so the group has to be big enough to reach all of them from one — and a group with few operations per cell can only do that by having a very small cell, which the polygons will not permit. The hypothesis therefore selects for large groups, and p6m is the largest group the hexagonal lattice carries.

Allow two orbits of vertices and the constraint relaxes at once. The 2-uniform tilings — every vertex still surrounded by regular polygons, but falling into two classes rather than one — number twenty, and they are drawn from a wider range of groups, because a group now only has to reach half the vertices from each of two starting points. Continue to three orbits and the count rises again, into the sixties.

So the right reading of the census is conditional. It says which groups are compatible with a very strong homogeneity assumption, not which groups regular polygons can produce. Regular polygons can produce a great deal more; they simply cannot produce it uniformly.

The same reading applies to the shape of the answer as well as its size. Six tilings answering p6m does not mean those six are alike — they have different vertex species, different cells, different numbers of tiles per cell. It means the detector was asked a question with a coarse range of answers, and coarseness in the answer is not evidence of sameness in the objects. That is the standing hazard of any classification: it reports the equivalence it was given.