Three colours, and why most patterns cannot have them

A counterchange pattern is a pattern with two colours in which some of the symmetries swap them. Every plane group but one admits at least one: seventy-four two-colourings across the seventeen, and only p3 has none.

Three colours is a different world. Across the same seventeen groups there are twenty-six three-colourings, they are concentrated in seven groups, and ten of the seventeen have none at all. p4m has none. pmm has none. p6m has none. The most symmetric patterns in the plane cannot be three-coloured in a way that makes the colour permutation part of the symmetry, and the reason is a single line of arithmetic about the number two.

What a colour symmetry is #

A coloured pattern is a pattern together with an assignment of colours to its parts. A symmetry of the coloured pattern is an operation of the underlying group together with a permutation of the colours, and the two have to be consistent: the operation must send every part of one colour to parts of one colour, and the permutation must record which.

For two colours the permutation group is ℤ₂ and the assignment is a homomorphism from the plane group onto it. Its kernel — the operations that preserve colour — is a subgroup of index two, and every subgroup of index two arises this way. That is why counting two-colourings and counting index-two subgroups are the same job.

For three colours the permutation group is ℤ₃ if the colours are permuted cyclically, and the same identification holds: a three-colouring is a homomorphism onto ℤ₃, and its kernel is a subgroup of index three whose quotient is cyclic. So the question becomes a question about subgroups, and the enumeration is the same one run at a different prime.

Why an operation of order two kills it #

Here is the argument, and it takes three sentences.

A homomorphism φ onto ℤ₃ sends an operation of order two to an element whose order divides two. The only element of ℤ₃ with that property is zero, since three is odd. So every operation of order two is in the kernel, which means every half-turn and every mirror preserves the colours.

That alone is not fatal — the kernel is allowed to contain things — but the consequences cascade. If a group’s point group is generated by elements of order two, the whole point group is in the kernel, and the homomorphism has to come from the translations alone. And there the second half of the argument bites: a half-turn conjugates a translation to its inverse, so φ(t) = φ(t⁻¹) = −φ(t), which forces 2φ(t) = 0, which in ℤ₃ forces φ(t) = 0.

So a group containing a half-turn has no three-colouring whatever. That deletes p2, pmm, pmg, pgg, cmm, p4, p4m, p4g and p6m in one stroke, and the mirror groups that survive do so because a mirror does not invert a translation parallel to it.

The seven that can #

The groups with a three-colouring are p1, pm, pg, cm, p3, p31m and p6. They fall into two families and the families are worth separating.

The ones with no rotation of even order and no rotation at all beyond a three-fold. p1 has only translations, and its three-colourings are the ways of choosing a sublattice of index three — of which there are four in a general lattice — with a colour for each coset. pm, pg and cm have a mirror, which fixes its own direction and so does not force the translation argument above; each has two.

The ones with a three-fold. p3 has eight, p6 has two, p31m has two. Here the colour permutation can be the rotation itself, and the result is the pattern everybody has seen: three colours arranged around each three-fold centre, cycling as the pattern turns.

p6 is instructive for having only two despite being large. Its six-fold rotation has order six, and a homomorphism onto ℤ₃ must send it to an element of order dividing six — but the square of a six-fold is a three-fold and its cube is a half-turn, and the half-turn is in the kernel, which constrains the six-fold to map to an element of order three or one. The two survivors are the two three-fold images, and the kernel of each is p2 — a case where the point group drops from six-fold to two-fold and the lattice is untouched.

Where the twenty-two come from #

Twenty-two of the twenty-six keep the point group and thin the lattice, and that number is a lattice fact before it is a group fact.

A klassengleiche three-colouring needs a sublattice of index three that the point group preserves, and the arithmetic of those is the sublattice question: a general lattice has four sublattices of index three, and how many survive the point group depends on what the point group is.

That is the whole reason p3 has eight and p4 has none: the hexagonal lattice has a sublattice of index three that its rotation preserves, and the square lattice does not have one of index three at all — three is not a sum of two squares, so no square sublattice of index three exists, and the four that do exist are permuted by the quarter-turn.

What one looks like when it is drawn #

The arithmetic above is about kernels; the picture it produces is about where the colours sit, and the translation between them is worth making explicit.

Take p3 with its colour permutation equal to the three-fold rotation. Around every three-fold centre there are three copies of the motif, one for each power of the rotation, and each gets a different colour — so the picture is a plane of three-petalled pinwheels, each petal a different colour, and turning the picture by a third of a turn advances every colour by one. The kernel is p1: the only operations that leave the colours alone are the translations.

Now take one of the six that thin the lattice instead. Every rotation preserves its colour and the translations do the permuting: a cell and its neighbour are different colours, and the colouring repeats over the √3 × √3 cell while the pattern repeats over the small one. The picture is a plane of pinwheels all of one colour per cell, tiled in three colours.

The consistency condition then decides whether a given drawing can carry the colouring: a point fixed by a colour-changing operation would have to be several colours at once, so a motif sitting exactly on a three-fold centre kills the first colouring above and not the second. The motif has to be asymmetric for the same reason it has to be asymmetric everywhere on this site, and here the penalty for getting it wrong is a colouring that is not a function.

Where ornament got there first #

Three-colour symmetry has a much longer history in workshops than in journals, and the workshops chose the same seven groups without knowing why.

Islamic ornament is full of three-colour and six-colour work built on a three-fold or six-fold framework — the Alhambra among many others — and effectively none built on a square framework, which the argument above explains: a square pattern’s half-turns force every three-colouring to collapse. Weaving has the same bias for a different reason, since the groups a woven cloth can realise are constrained by the grid before colour is considered at all.

The interesting part is that a craftsman meeting the obstruction meets it as a practical failure — an attempt to three-colour a square repeat produces a pattern whose colouring does not close, and the only fixes are to enlarge the repeat or to give up one of the symmetries. Both fixes are in the enumeration above. Enlarging the repeat is the klassengleiche case; giving up a symmetry is the translationengleiche one.

More colours, and the point at which this stops being a subgroup question #

Three is the last easy case, and the reason is that four is not prime.

For a prime p, a colouring in which the colours are permuted cyclically is a homomorphism onto ℤ_p and its kernel has index p — so counting colourings is counting subgroups, and the enumeration above works unchanged at five, seven and eleven. For four colours the permutation group could be ℤ₄ or it could be the Klein four-group, and those are different classifications with different counts; for six it could be ℤ₆ or S₃. The general object is a transitive permutation group on the colours together with a homomorphism onto it, and the subgroup language only covers the cyclic cases.

So the twenty-six above is a complete count of one thing and a partial count of another: complete for cyclic three-colourings, which is what “three-colour symmetry” usually means, and silent about colourings whose permutation group is not cyclic — of which, at three colours, there are none other than S₃ acting through its quotient, since ℤ₃ is the only transitive group on three points that a plane group can map onto without losing the transitivity.

The general count, and where the number of colourings comes from #

The two-colour census produced numbers one less than a power of two, and the three-colour one produces numbers that look unrelated. They are the same formula.

The colourings with cyclic permutation of p colours are the surjective homomorphisms onto ℤ_p, and those form a vector space over the field of p elements once the trivial one is added back — so if there are r independent choices, the number of colourings is

$$\frac{p^{r} - 1}{p - 1},$$

which is the count of one-dimensional subspaces rather than of vectors, because two homomorphisms differing by a relabelling of the colours give the same colouring. At p = 2 the denominator is one and the formula collapses to 2^r − 1, which is the earlier result; at p = 3 it gives 1, 4, 13 and so on, and p3’s eight is not on that list because eight is a count of homomorphisms up to nothing at all rather than of subgroups.

The useful part is r, which is the rank of the group’s abelianisation modulo p — how many independent generators survive being reduced mod p. That single number decides the whole count, and it is why a group with a half-turn scores zero at p = 3: reducing mod three leaves the order-two element with nowhere to go but the identity, and it drags the translations with it.

Which primes, and why the answer is about factorisation #

For p = 3 the point group does most of the work. For a larger prime it does none, and what is left is a question about the lattice that turns out to be number theory.

If p does not divide the order of the point group, then every element of the point group maps to the identity — the argument that killed the half-turn at p = 3, applied generally — so every index-p subgroup is klassengleiche. Colouring in five or seven or eleven colours is therefore purely a matter of choosing a sublattice of that index which the point group preserves, whatever the group is.

A lattice has p + 1 sublattices of prime index p, and which of them the point group preserves is decided by arithmetic in the ring the lattice belongs to: the Gaussian integers for the square lattice, whose point group is multiplication by i, and the Eisenstein integers for the hexagonal one. A prime that splits in that ring leaves two sublattices fixed; one that stays prime leaves none; and one that ramifies leaves exactly one.

That is where the hexagonal lattice’s answer at index three comes from. Three ramifies in the Eisenstein integers, so there is exactly one sublattice of index three carrying the full six-fold symmetry — which is the one the figure above picks out, and the reason the three-colourings of the three-fold groups are as few as they are. The square lattice at index three has none, because three stays prime among the Gaussian integers, and that is a second and quite independent reason p4 cannot be three-coloured.

Two obstructions, from two different halves of the group, agreeing. The point group’s half-turn kills the translationengleiche colourings and the arithmetic of three in the Gaussian integers kills the klassengleiche ones, and neither argument needs the other.

What the count is not #

Two disclaimers, and the first is the same one the two-colour essay had to make.

These are subgroups, not colourings up to equivalence. The classical enumerations of coloured plane groups identify colourings related by a change of basis or a relabelling of the colours, and that equivalence is finer than this counting. The two-colour case makes the size of the difference visible: this site enumerates 74 index-two subgroups where the classical count of two-colour plane groups is 46. The same gap will be there at three, in an amount this site does not compute, because the equivalence needs a normaliser rather than a subgroup search.

And a colouring has to be drawable. A colouring is inconsistent if some point of the pattern is fixed by a colour-changing operation, since that point would have to be two colours at once. Whether that happens depends on where the motif sits, not on the group — so a group can admit a colouring that a particular pattern cannot show. The colour figures on this site test for it while they draw and refuse rather than produce a picture whose colours are undefined.

There is one more asymmetry between the two cases worth stating, because it explains why two colours feel natural and three feel special. A two-colouring needs a subgroup of index two, and index-two subgroups are automatically normal — every group of even order has them in abundance and no compatibility question arises. A three-colouring needs a subgroup of index three with cyclic quotient, which is a genuine condition: the quotient has to be ℤ₃ rather than merely a set of three cosets, and that is what fails in ten of the seventeen. Two colours ask for a split; three ask for a rotation of the colours, and a rotation of order three cannot be carried by an operation of order two.

Who counted them, and in what order #

Colour symmetry arrived from ornament rather than from mathematics. Woven and printed textiles have been counterchanged for as long as there have been two dyes, and the mathematics caught up in the 1930s with Heesch and Shubnikov, who formalised antisymmetry — the two-colour case — as a group with an operation of order two attached to it.

The generalisation to more colours came later, and messily: several enumerations of the three-colour plane groups were published in the 1960s and 1970s with different answers, because the authors were using different equivalences and not always saying which. That is the same difficulty this page ends on and it is not a criticism of anybody — the objects are genuinely different depending on whether a relabelling of colours counts as a difference.

What is stable across all of the accounts is the structural statement: a colouring is a homomorphism, colour-preserving operations are its kernel, and the number of colours is the index. Everything else is a choice of equivalence, and the sensible thing is to state which is meant, which is what the disclaimers above are for.

Where the ladder goes next #

The natural next rung is the case where the colours are not permuted cyclically — four colours with the Klein group, six with S₃ — where the object stops being a subgroup and becomes a homomorphism onto a specified permutation group. The count is larger, the argument is the same shape, and the deletions are decided by which elements of the permutation group have which orders, exactly as the number two decided everything above.

Both of those are worth keeping together, because they say the obstruction is not one obstruction. A prime dividing the point group’s order is stopped or admitted by the point group; a prime that does not is decided entirely by the lattice, and by arithmetic that predates crystallography by two centuries.