The third survives a rectangle

A winding does not dilute sorted the arrangements of square ice on a torus by their two windings. The windings are the amounts a height function on the faces gains going once round the torus in each direction. The height function exists because the ice rule is a conservation law: with two arrows in and two out everywhere, the arrow field has no sources, and a field with no sources is the rotated gradient of a height. An arrangement is a three-colouring of the faces, read three ways, exactly when both windings are divisible by three. Three colours on a chessboard had found that on even tori the colourings and the arrangements are the same number, and the winding census turned that into a sharper fact. On every even square torus, exactly one third of the arrangements have both windings divisible by three, as a ratio of integers: 990 of 2,970 at side four, and 967,031,356,014 of 2,901,094,068,042 at side eight.

The explanation offered was a limit. In the limit of a large torus the windings become independent Gaussians with a stiffness of $\pi/12$. The fraction of even windings divisible by three is then a ratio of two theta functions, which Jacobi’s identity sets to exactly $1/\sqrt3$, and two independent windings give a third. That essay also noticed that the finite tori are more exact than the limit explains, and proposed a test. On a torus with sides in the ratio two to one, the two windings have different stiffnesses, the Jacobi argument gives a different number, and the exact third should fail. If it does not, the finite identity is about something other than the Gaussian.

The test has now been run, and the third does not fail. This essay reports that result, and then the part that matters more. The Gaussian limit, worked through for a rectangle, predicts a third as well. The proposed test could not have come out the other way, and what a rectangle does show is something the square could not.

Counting on a rectangle #

The census is the same transfer matrix that counted the arrangements one rule allows, generalised to a torus $L_x$ across and $L_y$ down. Its state is the row of $L_x$ vertical arrows between one row of vertices and the next. The winding across is the number of those arrows pointing up less the number pointing down, fixed by the starting state. The winding down is accumulated one horizontal arrow per row. Every count is an exact integer.

The picture at the head of this essay lists twelve even tori, the four squares of the earlier census and eight rectangles: $2 \times 4$, $2 \times 6$, $2 \times 8$, $4 \times 6$, $4 \times 8$, $4 \times 12$, $6 \times 8$ and $6 \times 12$. At every one the arrangements with both windings divisible by three are exactly a third. On the $2 \times 4$ torus that is 38 of 114. On the $4 \times 8$ torus it is 1,166,990 of 3,500,970. On the $6 \times 12$ torus, with a hundred and ten trillion arrangements, it is 36,634,401,687,120 of 109,903,205,061,360. Aspect ratios of two and three make no difference, and neither does which side is longer.

Restated in the terms the question began with, the result is about colourings. A three-colouring of the faces is three times an arrangement whose windings are both divisible by three, so a third in the corner means the colourings and the arrangements are the same number, the equality three colours on a chessboard found on open boards. On every even rectangle the three-colourings of the faces are exactly as many as the arrangements of square ice. On the $4 \times 8$ torus both are 3,500,970. On the $6 \times 12$ torus both are 109,903,205,061,360. The equality three colours on a chessboard found on even squares was never a property of the square.

By the terms of the test as proposed, that verdict would mean the Gaussian explanation was the right answer for the wrong reason. Before accepting that, it is worth asking what the explanation actually predicts for a rectangle.

What the limit predicts for a rectangle #

On a square torus the two windings have the same stiffness, $\kappa = \pi/12$, because a height field on a square has no preferred direction. On a torus of aspect ratio $r$, with the long side $r$ times the short one, a winding round the short circumference is spread over fewer faces along the long direction, and it costs more. Its stiffness is $\kappa\,r$, and the other winding’s is $\kappa / r$. That is the standard behaviour of a Gaussian field on a torus, and it is the change the proposed test relied on.

The scaling can be seen without any theory of the lattice. A height field with stiffness $K$ costs energy $K\int |\nabla h|^2$ over the torus. A winding $w$ round the short circumference $L_x$ is a height gain of $w$ over that length, so a gradient of about $w / L_x$ spread over an area $L_x L_y$, and it costs $K w^2 L_y / L_x$. A winding round the long circumference costs $K w^2 L_x / L_y$. The two stiffnesses are the square’s multiplied by $r$ and divided by $r$, and their product is the square’s stiffness squared whatever the shape. That product is what the formula below turns into a third.

Now repeat the Jacobi argument direction by direction. For even windings $w = 2n$ with weights $e^{-\kappa s w^2}$ at $s = r$ or $1/r$, the fraction divisible by three is

$$\frac{\sum_m e^{-3\pi s m^2}}{\sum_n e^{-\pi s n^2/3}} = \frac{\theta(3s)}{\theta(s/3)}, \qquad \theta(t) = \sum_n e^{-\pi t n^2}.$$

At $s = 1$ Jacobi’s identity $\theta(1/t) = \sqrt{t}\,\theta(t)$, the one-dimensional case of the sum that turns a lattice into its dual, turns this into $1/\sqrt3$. At $s = r \neq 1$ it does not simplify alone. But multiply the two directions together:

$$\frac{\theta(3r)}{\theta(r/3)} \cdot \frac{\theta(3/r)}{\theta(1/(3r))} = \frac{\theta(3r)}{\sqrt{3/r}\;\theta(3/r)} \cdot \frac{\theta(3/r)}{\sqrt{3r}\;\theta(3r)} = \frac13.$$

Each denominator is the other numerator’s dual, Jacobi’s identity turns one into the other, and the square roots multiply to three. The limit predicts a third on a torus of any shape.

The picture draws the two fractions as the aspect ratio runs from a quarter to four. At $r = 1$ both are $1/\sqrt3 \approx 0.577$. As the torus lengthens, the fraction for the stiff direction climbs towards one, since a stiff winding is almost always zero and zero is divisible by three. The fraction for the soft direction falls towards a third, since a soft winding is spread over many values and a third of them are multiples of three. Their product never moves from one third.

So the earlier essay’s test was misjudged, not its explanation. Stretching the torus changes each direction’s fraction, and it changes them in exactly compensating ways. What the Gaussian argument explains is not a coincidence of the square but a symmetry of the torus. A rectangle of aspect $r$ and one of aspect $1/r$ are the same torus turned, and the modular identity that relates them is the one Jacobi’s formula is the simplest case of. The same cancellation appears wherever a quantity on a torus is a product over its two directions. Exchanging the directions of a rectangle replaces its aspect ratio by the reciprocal, and in the Gaussian limit that exchange acts on each direction’s theta function exactly as Jacobi’s identity does. A quantity built symmetrically from both directions therefore cannot depend on the aspect ratio through the theta functions at all. The fraction in the corner is such a quantity. The fraction in one direction alone is not, and that is why the two marginals move with the shape while the corner does not. The limit could never have told the square from the rectangle. The finite census was always going to find a third on both.

What the rectangle does show #

The rectangle is not a wasted test. It separates two things the square kept together: the corner of the joint distribution, and the product of the two marginals.

On the $4 \times 8$ torus the fraction of arrangements whose winding across is divisible by three is 0.7767, and the fraction whose winding down is divisible by three is 0.4221. The limit’s values for aspect two are 0.802 and 0.416, so the finite marginals are near their limits and not at them. Their product is 0.3278, not a third, yet the corner is a third exactly. The two windings are not independent on a finite torus. They are correlated, and the correlation adds exactly what the product lacks. On the $6 \times 12$ torus the marginals are 0.790 and 0.419, their product 0.3307, and the corner is again a third exactly.

The full table of residues on the $4 \times 8$ torus shows the correlation directly. With both windings divisible by three the share is 0.3333. With the winding across divisible by three and the other not, each of the two cells holds 0.2217. With the winding across not divisible by three, the cells hold 0.0444 where the winding down is divisible and 0.0336 where it is not. So among arrangements whose winding across is divisible by three, the winding down is divisible by three 42.9 per cent of the time. Among the rest it is divisible 39.8 per cent of the time. A winding divisible by three in one direction makes one divisible in the other more likely. On a finite torus the two windings interact through the arrangements that realise them, and the interaction is exactly strong enough to put a third in the corner.

The marginals themselves approach the limit steadily. At aspect one they run 0.5663, 0.5714 and 0.5738 on the tori four, six and eight across, towards $1/\sqrt3 = 0.5774$. At aspect two the fraction across runs 0.7193, 0.7767 and 0.7902 on the $2 \times 4$, $4 \times 8$ and $6 \times 12$ tori, towards 0.8021. The correlation that holds the corner at a third fades as they converge, and the product of the marginals climbs to meet the corner from below, 0.3155, 0.3278 and 0.3307 at aspect two. The limit is reached by the marginals and by their product. The corner starts there.

The square hid this. On an even square torus the two marginals are equal by symmetry. The earlier census recorded them as close to $1/\sqrt3$ without being equal to it, 0.5738 at side eight. It could not tell whether the corner’s exactness was a property of the joint distribution or of each winding separately. The rectangle settles it. Neither marginal is exact, their product is not exact, and the joint count is.

The third as four twisted counts #

The exact third has an equivalent form that makes clear what kind of statement it is. Weight each arrangement by $\omega^{a w_x + b w_y}$, where $\omega$ is a cube root of unity and $(a, b)$ is a pair of residues modulo three. The weighted counts $Z(a, b)$ are the partition functions of the ice model with twisted boundary conditions. The fraction of arrangements with both windings divisible by three is the average of all nine of them divided by the untwisted one. A little algebra turns “exactly a third” into

$$Z(1,0) + Z(0,1) + Z(1,1) + Z(1,2) = Z(0,0),$$

one twisted count for each of the four lines through the origin of the plane over three elements.

The picture sets the four side by side for the even tori and three odd ones. On the even tori the four pieces vary a great deal with shape. The twist across dominates a long thin torus, 0.78 of the whole on the $2 \times 6$ torus, while on a square the two single twists are equal. Their sum is always one. On the $3 \times 3$, $4 \times 5$ and $5 \times 5$ tori they add to $-0.38$, $-0.11$ and $-0.42$. This form is closer to what a proof would have to show. Four traces of one transfer matrix with twisted boundaries add up to the plain trace on every even torus, whatever its shape.

An odd side breaks it #

With one side odd, the winding across that side is odd, since it is a sum of an odd number of plus and minus ones. It can still be divisible by three, three or minus three for instance. But the census finds no third on any of the eight odd tori it counts. The fractions run from 0.005 on the $3 \times 6$ torus to 0.086 on the $4 \times 5$, and none is a simple fraction. The Jacobi argument does not reach this case. For odd windings the sum runs over $w = 2n + 1$, a theta function with a shifted argument, and the identity that made the even case self-dual does not apply to it in the same way. The identity needs both sides even, as it did on squares.

What the census does and does not settle #

The finite exactness is still unexplained. Twelve even tori, eight of them rectangles, give a third exactly as integers. The limit explains why the answer is a third, now for every shape. It does not explain why each finite torus already has it exactly, when neither its marginals nor their product do. The twisted-count form says what a finite proof would establish. It gives no hint of the mechanism, whether a bijection between colourings and arrangements on even tori or a symmetry of the twisted transfer matrices.

The stiffness is quoted. The value $\pi/12$ and the scaling of the stiffness with aspect ratio come from the theory of the height field, not from this census. The rhombus tilings of a hexagon have a height field of the same kind, and it is what makes the average pile in a facet with no energy in it flat at its corners and rounded in the middle. The census confirms that the finite marginals approach the resulting fractions, 0.790 against 0.802 and 0.419 against 0.416 at the largest rectangle, without deriving the stiffness.

No single twisted count is fixed. The identity constrains the sum of the four twisted counts, not any one of them. Each depends on the shape: the twist across is a third of the whole on the $2 \times 2$ torus and 0.78 of it on the $2 \times 6$, and the mixed twists shrink as the torus lengthens. A statement about one twisted count alone, such as the one the earlier essay made about the twist in one direction plus the twist in both, holds on squares because of the square’s symmetry, and it would be false on a rectangle. Only the four together are shape-free.

The rectangles are small. The widest torus has eight arrows across, because the transfer matrix has $2^{L_x}$ states. Longer tori are cheap and wider ones are not. None of the conclusions depends on size, but the approach of the marginals to their limits is seen over a short range. The torus is also a choice of boundary. The count that depends on the edge found that for square ice a boundary can change even the entropy per vertex, and the windings exist at all only because a torus has no edge.

Who worked out the pieces #

The six-vertex model and its residual entropy are Lieb’s, from 1967, and the correspondence between ice and three-colourings of the square lattice is Lenard’s, reported by Lieb. The Gaussian description of the height field goes back to the work of van Beijeren and of Nienhuis on height models. The modular behaviour of partition functions on tori under exchange of the two directions, of which the product formula here is the simplest example, was developed for such models by Cardy. Jacobi’s identity for the theta function is from 1829.

What the census must refuse #

The second test is the correction of the earlier prediction, and it is a computation, not an argument. It evaluates the two theta ratios at six aspect ratios from a quarter to four and checks their product against a third to twelve decimal places. The third test is the new information. A torus on which the marginals multiplied to a third would be one where the windings behaved independently, and it would say nothing about whether the exactness is joint. The $4 \times 8$ torus shows it is joint.

Still open: a bijection on even tori #

On every even torus the arrangements of square ice and the three-colourings of its faces are equinumerous, and the census has now checked this on twelve shapes. A bijection would explain it in one stroke. It would need a map from arrangements with arbitrary windings to arrangements with windings divisible by three, three to one, or equivalently a way of reading a colouring from any arrangement. The twisted-count identity says the map would have to respect all four lines of residues at once. The first place to look is the sublattice structure an even torus has and an odd one lacks. An even torus’s faces split into two classes like the squares of a chessboard, and a map that shifts the heights on one class by one modulo three changes both windings by amounts the parity of each side controls. Whether such a shift, combined with the arrangement’s own arrows, gives a three-to-one map is a finite question on each torus. It has not been asked.