A four is not two twos

Square ice puts an arrow on every edge of a square lattice and asks for two in and two out at every vertex. A vertex can fail that rule in two different ways. It can have three arrows pointing in and one out, or the reverse, and then it carries a charge of plus or minus two — arrows in minus arrows out. Or it can have all four pointing the same way, and then it carries plus or minus four.

What a defect costs the count counted every arrangement on a small torus by how many vertices break the rule, and counted the two kinds together. It ended on two questions a single count could not answer: how rare is the charge of four, and does the dip that census found in the excess entropy belong to one kind of defect or the other?

Counting the two apart answers both, and the second answer is not the one the question expected. There is no second dip. There was never a first one in the sense it was read: the lowest point of that curve sits at exactly three defects on every torus, and three is the smallest number of broken vertices that cannot be made of twos alone.

Sixteen states at a vertex, and the arithmetic of charge #

A vertex has four edges and each edge’s arrow points in or out, so there are sixteen states. Six of them have two in and two out. Eight have three one way and one the other: four with three in, charge plus two, and four with three out, charge minus two. Two have all four the same way, one with charge plus four and one with minus four.

If the vertices were independent of one another — the assumption Pauling made in 1935 and the one every estimate of this kind starts from — a broken vertex would be a four with probability two in ten, one in five, since two of the ten broken states carry four. That is the number the census is measured against.

The first thing the charges do is constrain the counts, and they do it exactly. On a torus every arrow leaves one vertex and enters another, so the charges summed over all vertices come to nothing. The fours contribute a multiple of four to that sum, and so the twos must too; a sum of terms each plus or minus two is a multiple of four only when there is an even number of them. The number of charge-two vertices in any arrangement is even. An odd number of broken vertices is possible, but only with an odd number of fours among them — three defects must be two twos and one four, arranged as plus four, minus two, minus two or its reverse.

The census confirms it for every arrangement it counts. On the torus five across there are $2^{50}$ arrangements of arrows in all — about a thousand million million — and not one has an odd number of twos. The disc sizes grow with the logarithm of the count, so the grid reads as a landscape: the ice count alone at the corner, 143,224 arrangements, then a broad hill centred where independent vertices would put it, and a column of empties at every odd position along the bottom.

How the census is made #

The count is a transfer matrix run vertex by vertex. The state is the row of vertical arrows under the next vertices to be placed, the horizontal arrow arriving from the west, and the horizontal arrow the row began with, which the last vertex in the row must match so that the row closes round the torus. At each vertex both choices are made for the two arrows leaving it, the vertex’s charge is read off, and the running count is carried as a polynomial in two variables — one power for each vertex of charge two, one for each of charge four.

On tori up to five across the coefficients are carried as exact integers. At six across the total is $2^{72}$, which is past the sixteen significant figures an ordinary floating-point number holds, and there the counts are carried to that precision and said to be. Three tests guard every size: the coefficients must add to $4^N$, which is every assignment of arrows whatever; the constant term must be the ice count the row transfer matrix already gives by a different route; and every coefficient with an odd power of the first variable must be zero.

How rare is the charge of four #

Rarer than independence says, while the defects are few; exactly as common as independence says, once they are crowded.

With two broken vertices — one pair — the share of fours is 0.085 on the torus four across, 0.075 on the torus five across and 0.066 on the torus six across. Well under half of what independence predicts, and falling as the lattice grows. A pair of fours is a much less likely way of breaking the rule than a pair of twos, and the gap widens with the room the pair has to move in.

As the number of broken vertices rises, the share climbs and oscillates. The odd counts sit above the even ones, because an odd count must hold a four and an even count need not — at exactly three defects the share is one third on every torus, which is simply one four among three. By twelve broken vertices on the torus five across the share is within a thousandth of one in five, and it stays within a few thousandths of it all the way out to every vertex being broken.

That second half is worth a sentence of its own, because it is the claim that looks least like a result. When most vertices are broken the rule has nothing left to correlate: the arrows are nearly free, and a nearly free arrow assignment breaks each vertex into the ten broken states in the proportions the states’ own count gives. The correlations of square ice are a property of the rule’s being mostly obeyed, and they vanish from the defect statistics exactly when the rule is mostly broken.

A pair is squeezed, and the square of the charge says by how much #

Why a pair of fours is so much rarer is a question about pairs, and the census can ask it at every size it reaches.

Take the arrangements holding exactly one pair of opposite defects and nothing else. If the two defects could sit anywhere without affecting the arrows around them, the number of such arrangements would be the ice count, times the number of ways of choosing the two vertices, times a constant for the states each defect can take. Divide the measured count by the ice count and by the number of ways of choosing two vertices, and a free pair would give a number that does not change with the size of the torus.

It changes, for both kinds, and it falls. The measured ratio for a pair of twos goes 1.068, 1.020, 0.969, 0.925 on tori three to six across; for a pair of fours it goes 0.115, 0.095, 0.078, 0.065. On logarithmic axes those are nearly straight lines, which is what a power law in the side looks like, and the slopes are the whole of the finding: between successive sizes the fours fall 4.16, 3.85 and 3.74 times as steeply as the twos.

Four is the square of the ratio of the charges. That is the arithmetic of electrostatics in two dimensions. The arrangements between two defects are constrained by both, the constraint grows with the distance between them as a logarithm — exactly as the potential between two charges in a plane does — and the strength of a logarithmic interaction between charges $q$ and $-q$ goes as $q^2$. A pair of fours is a pair of doubled charges and pays four times over. The ice rule is a conservation law made the field of arrows divergence-free; a defect is where that fails, and a count with nothing but arrows in it has turned out to know Coulomb’s law.

The measured ratio is drifting, and the sizes do not settle where it ends. Four tori is a short run, and the two slopes are not yet the straight lines they would be on a large lattice. What the census establishes is that the fours’ squeeze is several times the twos’ at every size it reaches and close to four throughout; what it does not establish is the limit, which is the kind of number a field theory supplies and a finite count only approaches.

The correlations cost a fixed amount, not a cost per defect #

The same census separates what the correlations cost each kind of defect when there are many of them, which is the question the single curve could only answer as an average.

The quantity drawn is the logarithm of the count with only one kind of defect present, less the ice count, less the exact logarithm of the number of ways of choosing where the defects sit, less what independent vertices would give each defect. What is left is what the arrows’ correlations cost, and for twos it has a shape that no per-defect account predicts. It falls over the first few pairs and then stops: on the torus five across it reaches −1.04 by eighteen twos and does not move from there to twenty-four. On the torus six across it levels at −1.33.

So the cost is a fixed sum, paid while the defects are dilute, and not a cost per defect. Once the twos are dense enough that the regions each would constrain overlap, adding another constrains nothing new.

The level it reaches has an exact explanation, and it is the one place in this census where the answer is not a measurement. Asking every vertex to have an odd number of arrows in is a condition on the arrows’ parity, and parity conditions are linear over the integers modulo two: $N$ conditions, one per vertex, on $2N$ arrows, with exactly one dependency among them, because every arrow enters exactly one vertex and so the arrows in, summed over all vertices, are $2N$. So the number of arrangements with every vertex a two is $2^{2N-N+1} = 2^{N+1}$ exactly, whenever $N$ is even. Pauling’s independent vertices give $4^N (8/16)^N = 2^N$. The independent estimate is wrong by a factor of exactly two at full density, the census confirms it — 131,072 against 65,536 on the torus four across — and the factor is the one dependency that treating the vertices separately cannot see.

The line for fours has no such plateau within reach, and no dip either. It drops at the first pair and holds roughly level while the fours remain few; on the smallest torus it turns up past eight fours, because sixteen vertices do not have room for many more without the fours crowding one another, and the drawing stops it there.

The dip was a count of three #

Which leaves the question the earlier census asked: does the dip it found belong to one kind of defect or the other? Neither line has a dip. So where did it come from?

It came from two places, neither of them the screening cloud it was read as.

The fall from nought defects to two is Stirling’s approximation. The earlier curve took out the freedom to choose which vertices break by subtracting the familiar mixing term, $-\rho \log \rho - (1-\rho)\log(1-\rho)$ per vertex. That expression is the large-lattice limit of the logarithm of a binomial coefficient, and at a count of two on sixteen vertices it overstates the binomial by more than a unit — the error in Stirling’s formula at small arguments, which is exactly where the first defects live. With the exact binomial in its place, the first pair adds entropy on the torus four across, a tenth of a unit, rather than subtracting more than one.

The lowest point is the forced four. Three is the first count of broken vertices that cannot be made without a four, and a four is expensive — the census has just measured how expensive. So the curve drops at three, recovers at four where twos alone are possible again, and climbs from there. The earlier account placed the dip “near a fifth” of the vertices, and on the torus four across three defects is 0.19 of sixteen; on the torus five across the lowest point is still three defects, now 0.12 of twenty-five. A feature that stays at the same count while the density it is quoted at halves is a feature of the count.

The claim the dip was offered for is still true, and the census now says how. The first pair’s exact excess on tori four, five and six across is +0.109, +0.046 and −0.010: positive on the smaller tori, negative on the largest, and falling by a steady amount per size. That fall is the logarithmic squeeze of the previous section, and on a large enough lattice it makes the first defects subtract entropy rather than add it — which is what the dip was taken to show, arriving by a different road and with the size dependence that is its signature. The account of the dip has been corrected to say this.

What the pictures cannot show, and what the counts depend on #

The size is small and the limits are not here. Tori of three to six vertices across are what an exact count reaches in reasonable time, and the power laws above are read off four points. The census establishes the evenness of the twos, the parity count at full density and every individual count exactly; the ratio near four, the level of each plateau and the falling share of fours at the first pair are measurements on those sizes. The field theory that would give the limiting exponents is quoted by name only.

The boundary is a torus, and it matters. A torus has no boundary, so it constrains no arrow and makes every vertex equivalent; that is why its counts are the ones that compare cleanly across sizes. A different boundary gives a different count, sixteen per cent apart for the ice arrangements alone, and nothing here says how the defect statistics would move under a domain wall.

The mixing term is the exact multinomial throughout. Every excess quoted on this page subtracts the logarithm of the number of ways of choosing which vertices carry twos and which carry fours, computed exactly, and never its Stirling approximation. That is the convention the dip depended on, and it is named here because leaving it implicit is how the dip came to be read as physics.

And no picture here shows a defect’s cloud. The grids and lines are counts over all arrangements at once, and a count cannot say where in a given arrangement the constraint around a defect lies. The squeeze is inferred from how the counts change with size; it is consistent with a logarithmic attraction and it does not draw one.

The tests include the refusal on purpose. A census that reproduced the earlier curve and nothing else would have passed with the artefact still in it, so one test requires the old curve to be reproduced to nine decimal places and another requires the Stirling term to be rejected at exactly the point where it misleads.

Where charge and counting meet #

Nothing in the rule mentions charge. It is two arrows in and two out, a local condition on sixteen states, and the census is a count of arrangements with no energy anywhere in it. Yet the defects it counts behave as charges in a plane: they must cancel, they attract with a strength that grows as the square of their size, and a doubled charge is suppressed as a doubled charge in electrostatics is.

The reason is the height. An arrow field with two in and two out everywhere is the rotated gradient of a height on the faces — the construction behind the pile of cubes and behind the colourings of three colours on a chessboard — and a defect is a place where going once round changes the height by two or by four instead of by nothing. The arrangements near such a place are the ones that accommodate a height mismatch, and the number of ways of accommodating it falls with the logarithm of the distance to the defect that cancels it, exactly as the energy of a dislocation pair does in a crystal. A charge of four is a mismatch of four and pays for twice the mismatch at every radius, which is four times the logarithm’s coefficient.

That is the same accounting that makes a point defect’s charge a topological number read on a loop, arriving here as a count rather than as a winding. And it is the reason a calorimeter and a diffractometer disagree about what they see: the average arrangement is as symmetric as the lattice at every defect density, so every one of these statistics is invisible in the sharp scattering and lives only in the counts and in the diffuse.

Who counted what #

The six-vertex model and its ice rule are Pauling’s, Slater’s and Lieb’s; the exact solution on the square lattice is Lieb’s of 1967. The description of its long-wavelength behaviour as a height field — a Gaussian surface whose defects interact logarithmically, with a strength set by one stiffness — is the Coulomb-gas picture developed through the 1970s and 1980s by Nienhuis and others, and the statement that a charge’s cost goes as its square belongs to that picture rather than to this census.

What the census adds is the pair of things a field theory takes for granted and a finite count can check: that the charge of two and the charge of four really are separate species with separate statistics on a lattice small enough to count exactly, and that the dip a single-variable count produced is a count of three rather than a density.

Still open: how the fours are placed #

The census counts arrangements by how many fours they hold and not by where. Whether a four prefers to sit next to a two of opposite sign — whether it is, in the height picture, two twos that have fallen together — is the natural next measurement, and it needs the positions of the defects carried through the transfer matrix rather than only their number. A four adjacent to a pair of twos is a different arrangement from a four far from them, and the census as it stands adds the two together.

The other question left is the exponent. The two slopes drift, and a field theory predicts where they go; checking a prediction against four sizes is weak, and the transfer matrix at seven across is within reach for the ice count and beyond it for the two-variable polynomial with exact coefficients. A census that carried the pair’s separation instead of the defect counts would reach much larger tori, because one pair is a small thing to carry, and it would turn the drifting ratio into a measured one.