Three colours on a chessboard

The residual entropy of ice is a physical quantity. A calorimeter reads it: cool a lump of ice towards absolute zero, integrate the heat capacity, and the number that does not go away is the logarithm of how many arrangements the protons can take. Pauling estimated the count in 1935 and Lieb solved the two-dimensional case exactly in 1967, and every count of it so far has treated it as a fact about arrows on a lattice.

It is also the answer to a question with no arrows in it.

Colour the cells of a board in three colours so that no two cells sharing an edge carry the same colour. Count the ways. On a torus of even side, that number is the number of square-ice arrangements on the same torus — not approximately, not in the growth rate, but the same integer to the last digit.

The difference on an edge is a direction #

The correspondence is one line long and the line is arithmetic rather than geometric.

Call the colours nought, one and two, and read them modulo three. Two cells sharing an edge carry different colours, so their difference is one or two modulo three — which is to say plus one or minus one. Never nought. So the difference is a sign, and a sign on an edge is an arrow.

Now walk round a corner of the lattice. Four cells meet there, and walking from one to the next to the next and back to the first accumulates four differences and returns to where it started, so they add to nothing modulo three. Each is plus or minus one, so their sum is four, two, nought, minus two or minus four; of those only nought is divisible by three. So exactly two of the four are plus and two are minus, and reading a plus as an arrow one way gives exactly two arrows in and two out.

The ice rule is not imposed anywhere in that argument. It is a consequence of colouring modulo three, and it could not have been anything else: four steps of plus or minus one that have to cancel modulo three have to cancel outright.

The counts are the same integer #

The correspondence argues that every colouring gives an arrangement. It does not argue that every arrangement gives a colouring, and the counts are where that is settled.

At side two there are 18 colourings and 18 arrangements. At side four, 2,970 and 2,970. At side six, 16,448,400 and 16,448,400. Three numbers running to eight figures, agreeing exactly, computed by two transfer matrices that share nothing — one walks a row of arrows choosing directions, the other walks a row of cells choosing colours.

At side three there are 12 colourings against 148 arrangements. At side five, 7,560 against 143,224. The odd sizes are not close.

Twelve is small enough to account for by hand. A three-by-three torus of cells has each cell adjacent to four others, and a proper colouring of it is very nearly forced: fixing one cell’s colour leaves the rest with almost no freedom, and the twelve are three choices of that first colour times four ways of arranging the rest. A four-by-four torus fixed the same way has hundreds. The difference is not that nine cells are fewer than sixteen; it is that an odd cycle of three cells can be coloured in fewer ways than an even cycle of four, and every row and every column of the torus is such a cycle.

That is the whole obstruction in miniature, and it is worth having before the general account, because the general account is about windings and windings are what a cycle of cells is.

Going round the torus is a condition, and an odd side makes it a real one #

The gap is not a failure of the correspondence at a vertex. Every local step of the argument above holds at any size; what fails is the closing up.

Recovering a colouring from an arrangement means picking a colour for one cell and walking outwards, adding the difference each arrow prescribes. On a plane that always works. On a torus the walk can come back to where it started by going round, and the colour it arrives with must be the colour it left with — so the differences accumulated round any loop that wraps must add to nothing modulo three.

A loop going once round a torus of side L crosses L edges. Each contributes plus or minus one, so the total has the same parity as L. With L even the total is even, and nought is even; with L odd the total is odd, and nought is not. An odd loop can still sum to nought modulo three — three itself is odd — but the arrangements for which it does are a small minority, and the count says how small: at side three, twelve of a hundred and forty-eight.

So the correspondence is exact where there is no obstruction and lossy where there is, and the obstruction is about windings rather than about vertices. That is the same kind of statement the boundary makes, arriving in a different disguise: the count depends on what happens at the edges, and a torus’s “edge” is its two ways of going round.

The growths per cell tell the same story with the sizes divided out. The even sides trace the ice sequence — 2.060, 1.648, 1.586 — and the odd ones run 1.32 and 1.41, rising. Both sequences are heading for the same limit from opposite sides, because a condition on windings costs a fixed amount however large the lattice is, and a fixed amount divided by the number of cells goes to nothing. An obstruction that is fatal at side three is negligible at side thirty.

A defect is a corner the colouring cannot get round #

The colouring picture has something to say about the conservation law itself, and it says it in one sentence.

A defect is a vertex with three arrows in and one out, or the reverse. In differences that is three of one sign and one of the other, which sums to plus or minus two — and two is not nothing modulo three. So the colouring does not close round a defect. Walking once round that corner and adding the differences lands on a colour two away from the one it started with.

That is exactly what a branch point is, and it is the same statement as the height function failing to be a function there. The arrow picture calls a defect a charge, the height picture calls it a dislocation, and the colouring picture calls it a place where the colours do not join up — three descriptions of one thing, and the colouring’s is the one in which the obstruction is a residue modulo three rather than a number.

It also explains the vertices with all four arrows the same way. Four differences of one sign sum to plus or minus four, which is plus or minus one modulo three — a different obstruction from the defect’s two, and a smaller one in the sense that three of those cancel where three of the others do not. The census of what a defect costs counts the two kinds together and the colouring is where they would most naturally be told apart.

Three colours on the edges, where the lattice does the work #

There is a second three-colouring in this neighbourhood and it is not the same one, which is worth saying because the two are easy to run together.

Colouring the cells of the square lattice is what this page is about, and the three colours are labels on faces. Colouring the edges of a lattice where three meet at every vertex, so that the three at each vertex are all different, is a different problem — and it is the one that belongs to the accounting of curvature, where three edges meet at every atom and the whole accounting of curvature depends on it.

The two are related and the relation runs through the same arithmetic. A proper three-edge-colouring of a trivalent net assigns to each edge one of three labels; reading the labels as residues modulo three and asking that the three at a vertex be distinct is asking that they sum to nought modulo three, which is the same condition in the same modulus. What differs is which objects carry the labels and how many meet.

And the counts are unrelated. The number of proper three-colourings of the square lattice’s cells grows as 1.5396 a cell; the number of three-edge-colourings of a honeycomb grows at a different rate, and neither is an approximation to the other. Two problems can share a modulus and an argument and have nothing to say about each other’s answers — which is the opposite of what this page’s own identity shows, and is the reason the identity had to be checked rather than read off the shape of the argument.

What an entropy is when it is two things #

The identity is worth stating carefully, because “the same count” is doing three different amounts of work in three places.

In the limit, the two problems have the same entropy per site, and that is a statement about the substance as much as about the arithmetic: the residual entropy Pauling was estimating is 1.5 log units per proton in his approximation and 0.4315 in Lieb’s exact one, and the same 0.4315 is the growth of proper three-colourings per cell. Nothing about colouring is an approximation to ice or the other way round; they are one problem in two vocabularies.

At even finite sizes the totals agree exactly, which is much stronger and is a fact about the torus rather than about the limit. Two enumerations with no common step producing the same eight-figure integer three times is the kind of agreement that is either an identity or a mistake, and it is checked here at three sizes and fails at none.

At odd sizes they do not agree, and that failure is the useful part, because it says what the identity is made of. A correspondence that held at every size would look like a coincidence of formulas; one that holds exactly at even sizes and breaks at odd ones has a mechanism, and the mechanism is a loop closing up.

This is the opposite of what colour usually does in crystallography. Three-colourings of a plane group are colourings a symmetry permutes, and counting them is counting subgroups of index three — a question about the group, in which most plane groups have no three-colouring at all. Here nothing permutes anything: the colours are labels with an arithmetic on them, and the count is a count of labellings rather than of orbits. The two uses of the word share nothing but the number three, and setting them side by side is the only way to keep them apart.

Where the exactness stops #

Computed here. Proper three-colourings of the cells of an L × L torus for L from two to six, by transfer matrix over rows, with the states being the cyclic colourings of a row and two rows adjacent when they differ in every position. The ice counts on the same tori, by the transfer matrix the residual-entropy count uses. And the ratio of the two at every size, which is one exactly at even sizes and is not at odd ones.

The colouring counts are of a torus’s cells, not of a chessboard’s squares. A chessboard has edges, and a finite board with free edges has many more proper colourings than a torus of the same size, for the reason the boundary count is entirely about: a boundary that constrains nothing leaves the cells near it freer than the ones inside. The title’s board is the picture and the torus is what is counted.

Six is where it stops and the reason is the ice count. The colouring transfer matrix is cheap — sixty-six row states at side six — and the ice count at side seven runs past the exactness of a double-precision integer.

The limit is quoted. That the growth per cell is 1.5396 for both problems is Lieb’s result for ice and Baxter’s for the three-colouring, and nothing on this page derives either. What is derived is that the finite counts agree, which is consistent with the limits agreeing and is not the same statement.

And the correspondence is stated, not proved in both directions. That a colouring gives an arrangement is the argument above and is complete. That an arrangement with no winding obstruction gives exactly three colourings — one for each choice of the first cell’s colour — is the standard converse and is not written out; what stands in for it is the measured equality of the totals, and the measured inequality where the obstruction bites.

The tests include the odd sizes, and that is deliberate. An identity between two counts is worth nothing unless it is checked where it should fail as well as where it should hold, and a check that only ever confirmed the even sizes would pass equally well for a bug that computed the ice count twice.

Who found the colouring #

The three-colouring model on the square lattice is Rodney Baxter’s, from 1970, and he solved it by the Bethe ansatz and found the entropy per site to be the same as Lieb’s for square ice. The equivalence with the six-vertex model at its ice point is the step that makes the two results one result, and it is the one written out above: a colouring modulo three has differences of plus or minus one on every edge, and four of them round a plaquette must cancel.

The same construction goes on being rediscovered because it is the natural way to describe a divergence-free field with a height behind it. The height function of the conservation law is a colouring with infinitely many colours; reducing it modulo three loses the magnitude and keeps exactly the information the rule constrains. Three colours is the smallest alphabet in which the ice rule can be written down as a proper colouring, and that is why the number is three rather than something chosen.

The odd-torus obstruction is standard in the same literature under the name of a topological or winding sector, and the numbers above are a small instance of it.

Still open: colouring under a boundary rather than a torus #

The identity is about a torus, and what a boundary does is a separate count, so the obvious next question is what a colouring does inside a domain wall.

A domain-wall arrangement is a colouring of a square with its edge colours fixed, since the boundary arrows prescribe the differences all round the rim. The corner-freezing that costs the domain wall sixteen per cent of its entropy should therefore be visible in the colouring as a region where the colours are forced — a fixed staircase of colours near each corner — and the colouring picture might make it easier to see than the arrow picture does. Whether the alternating sign matrix numbers have a natural colouring statement is not something this page can say.

And the odd-torus count is its own sequence. Twelve at side three and seven thousand five hundred and sixty at side five are the first two terms of something, and nothing here says what. The ratio to the ice count — 0.081 and 0.053 — is falling, which is what a fixed obstruction divided by a growing count does, and whether it falls as a power of the side or exponentially would say how many winding sectors an odd torus has and how unequal they are.