A winding does not dilute

Colour the cells of a torus in three colours so that no two cells sharing an edge agree. Three colours on a chessboard showed that every such colouring is an arrangement of square ice — the difference of two neighbouring colours, read modulo three, is plus or minus one and becomes an arrow — and that on tori of even side the two counts are the same integer. On odd tori they are not. At side three there are 12 colourings against 148 ice arrangements, at side five 7,560 against 143,224, and the ratio was falling.

That essay ended by asking how it falls: as a power of the side, or exponentially? The obstruction is a condition on going round the torus, the reasoning ran, and a fixed obstruction set against a count that grows with the lattice should cost less and less.

It does not fall to nothing at all. Counted exactly out to side nine, the ratio goes 0.0811, 0.0528, 0.0444, 0.0410, each step smaller than the last, and it is settling on a number near 0.036 — about one arrangement in twenty-eight. The obstruction is not diluted by the lattice, because the thing it constrains is not local to any part of it. It is a winding, and a winding belongs to the whole surface.

What the colouring asks of an arrangement #

An arrangement of square ice has two arrows in and two out at every vertex, and that is exactly the condition for a height to exist on the faces. Cross an arrow and the height changes by one, up or down according to which way the arrow points; go once round any vertex and the four changes add to nothing, because two arrows point in and two out. So the heights are consistent round every vertex, which is the whole of the conservation law seen as a surface — the same surface that a pile of cubes in a corner makes visible, where the height is literally the height of the pile. A vertex that breaks the rule is a place where the height fails to close, and counting those by their charge is the other half of the same bookkeeping; here every vertex obeys the rule, and the only place the height can fail to close is round the torus itself.

On a torus the heights need not come back to themselves. Walk once round the torus across and the height gains some number — the winding across — and walking once round it down gains another. Each is a sum of plus and minus ones over the arrows crossed, one per column or one per row, so on a torus of side $L$ each winding has the parity of $L$ and lies between $-L$ and $L$.

A three-colouring is a height read modulo three. So an arrangement comes from a colouring exactly when both its windings are multiples of three — then the heights modulo three close up round the torus, and the colouring is determined by the colour of one cell, three ways. The number of colourings is three times the number of arrangements whose windings are both multiples of three. On the left of the figure the windings are one and one, and the arrangement is no colouring at all; on the right they are three and three, and it is three.

Counting by winding #

The census counts every arrangement on the torus and records its two windings as it goes. The transfer matrix adds one row of vertices at a time, and its state is the row of vertical arrows between one row and the next. The winding across is the number of those arrows pointing up less the number pointing down, and the ice rule carries it unchanged from row to row — so it is read off the starting state and never changes. The winding down is the sum of the horizontal arrows crossed by a vertical cut, one per row, and is accumulated as the rows are added.

Every count is an exact integer. At side nine there are 4,448,410,550,095,612 arrangements, and the census places each in one of a hundred sectors. Two checks guard it: the sectors must add to the ice count from the ordinary transfer matrix, and three times the sectors whose windings are both multiples of three must equal the number of colourings counted independently by a transfer matrix over colours — which they do at every side from two to six, where that second count runs.

The two tori look different in exactly the way the parity argument says. On the even torus every winding is even, the arrangements crowd the centre at nought and nought, and nought is a multiple of three; the solid disc at the centre is the largest there is. On the odd torus every winding is odd, there is no centre to crowd, and the arrangements sit at plus and minus one. The nearest multiples of three are plus and minus three, and the sectors at those corners hold 1.37 per cent of the arrangements between them.

The share settles, and the reason is a zero mode #

The odd shares are 0.0811 at side three, 0.0528 at five, 0.0444 at seven and 0.0410 at nine. The differences are 0.028, 0.008 and 0.003: they shrink by a factor of about three each step, which is a sequence converging on something positive rather than one decaying towards nought. A power law would lose a fixed fraction of its value at each doubling of the side and an exponential more than that; this loses less at every step, and the line it is approaching is drawn.

The reason is that the windings do not care how large the torus is. A torus has no boundary, so there is no edge whose share of the lattice shrinks as the lattice grows; what decides the winding is how the height behaves over the whole surface, and the whole surface grows with the lattice. The distribution of windings is therefore a property of the shape of the torus — here a square — and not of its size.

This matters beyond ice, because a torus is how almost every lattice calculation is made: a finite block with its opposite edges glued is what stands in for an infinite crystal, and the gluing is chosen precisely because it adds no boundary. It adds loops instead. Any quantity that depends on a loop — a winding, a flux, a phase carried round the box — is a property of the box that no amount of enlarging it removes, and the colouring count is a clean case of one.

The census measures that directly. The fraction of arrangements whose winding across is plus or minus three is 0.1086 on the torus five across, 0.1090 at seven and 0.1092 at nine; the fraction at plus or minus one is 0.891, 0.890 and 0.890. The torus grew from twenty-five vertices to eighty-one and the distribution moved in its fourth decimal place. The winding of five is still growing, because on the torus five across a winding of five means every arrow in the row points the same way, and that is the edge of what the torus permits rather than a sample of the distribution.

That is what a zero mode is. A surface with a stiffness has fluctuations at every wavelength, and the ones at short wavelengths are local — they are what the per-cell entropy counts, and they grow in number with the lattice. The windings are the one mode that is not a fluctuation at any wavelength: the height’s overall tilt, the part of it that goes once round the torus. There is one such mode each way however large the torus is, and its weight is set by the stiffness alone.

The stiffness, read two ways #

If the height is a Gaussian surface with a stiffness, the windings are weighted by $e^{-\kappa w^2}$ for some coefficient $\kappa$ that does not depend on the size. That is a claim the census can test, and it can test it twice, independently.

On an odd torus, the ratio of the weight at winding one to the weight at winding three is $e^{8\kappa}$, so $\kappa$ is an eighth of its logarithm: 0.2630 at side five, 0.2625 at seven, 0.2623 at nine. On an even torus the ratio of winding nought to winding two gives $e^{4\kappa}$: 0.2463 at side four, 0.2537 at six, 0.2570 at eight. The odd estimates are coming down and the even ones going up, and the number between them is $\pi/12 = 0.2618$, which is the stiffness the standard Coulomb-gas treatment of square ice assigns to its height field. Two sequences from disjoint sets of tori bracketing one number from opposite sides is better evidence than either converging alone, and the census does not derive $\pi/12$; it lands on either side of it.

With the coefficient in hand the limit follows by summing a Gaussian. On odd windings, the weight at the multiples of three is 0.1095 of the whole; both windings must be multiples, so the fraction of arrangements that are colourings tends to the square of that, and the share of colourings to three times the square — 0.0359. The census at side nine is at 0.0410 and heading there.

The exact third on even tori #

The even tori have a finding of their own, and it is the sharper one.

At every even side the census reaches — two, four, six and eight — exactly one third of the ice arrangements have both windings divisible by three. Not approximately: 6 of 18, 990 of 2,970, 5,482,800 of 16,448,400, and 967,031,356,014 of 2,901,094,068,042, as integers. That is the reason the colourings and the ice arrangements are the same number on an even torus — three colourings each for a third of the arrangements — and it says that the identity the chessboard essay found is not a coincidence of formulas but a statement about windings.

The table for the even torus has a shape worth looking at. The corner is a third; the four cells where one winding is a multiple of three and the other is not hold 0.1202 each; the four where neither is hold 0.0464 each. Those are not thirds and are not simple, and they drift with the size — at side four they were 0.1165 and 0.0502. Only the corner is fixed, which is what makes it an identity rather than a pattern.

Where the third comes from: a theta function that is its own dual #

In the Gaussian limit the third has a reason, and the reason is the identity that turns a lattice into its dual.

Take the even windings, $w = 2n$, with weights $e^{-\kappa w^2}$ at $\kappa = \pi/12$. Their total is $\sum_n e^{-\pi n^2/3}$ — a theta function evaluated at a third, the one-dimensional case of the sum that counts a lattice’s vectors by length. The multiples of three among them are $w = 6m$, and their total is $\sum_m e^{-3\pi m^2}$ — the same theta function evaluated at three. Jacobi’s identity says that a theta function at $t$ and at $1/t$ differ by exactly a factor of $\sqrt{t}$, which is the one-dimensional case of the sum over a lattice equalling the sum over its dual. At $t = 3$ that factor is $\sqrt{3}$. So the multiples of six carry exactly $1/\sqrt{3}$ of the even windings’ weight, and two independent windings carry $1/3$.

That is a striking thing for a count of ice arrangements to contain. The stiffness is $\pi/12$ for a reason in the model, the point where the height’s Gaussian weight becomes self-dual under the change of scale by three is $\pi/12$ for a reason in number theory, and nothing obvious connects the two. The coincidence of those two numbers is what makes even-sided tori exactly as colourable as they are arrangeable.

And the finite tori are more exact than the limit explains. At side eight the fraction of arrangements whose winding across is a multiple of three is 0.5738, not $1/\sqrt{3} = 0.5774$; the windings are not quite Gaussian and not quite independent at that size. Yet their joint distribution puts exactly a third in the corner, as an integer identity, at every even size. The Gaussian limit explains why the answer is a third; it does not explain why every even torus already knows it. That is a finite identity with no proof here, measured at four sizes.

What the counts depend on #

The torus is square. A torus with sides in a ratio other than one weights its two windings differently — the Gaussian coefficient for a winding round the long way is larger — and both the odd limit and the even third depend on the square shape. Nothing here counts a rectangular torus.

The limits are measurements. The census is exact to side nine; the odd share of 0.0359, the coefficient π/12 and the statement that the windings become independent Gaussians are what the numbers approach, with π/12 quoted from the theory of the height field rather than derived. What is established outright is the colouring identity at every size, the exact third at the four even sizes, and the sector counts themselves.

No figure here shows a winding on a large torus. The height picture is drawn on the torus three across, where a winding can be read off by eye; on the torus nine across the same quantity is a sum of nine arrows along a cut, and the sector plot shows only how many arrangements have each value. A reader has the arithmetic and the count, not a picture of the tilt.

The refusal matters because the extrapolation is the one the earlier account invited. A shrinking ratio with three data points fits a power law and an exponential equally well, and both go to nothing; what rules them out is not the ratio but the distribution underneath it, which is not shrinking at all.

A count that belongs to the surface #

The per-cell entropy of square ice is a local quantity. The arrangements a crystal keeps at absolute zero is about it, and it is what a calorimeter measures: so much entropy a proton, whatever the shape of the lump. The windings are the opposite kind of quantity. They are one pair of numbers for the whole torus, and what they decide — whether the arrangement is a colouring — is decided for the whole torus at once.

That is the same distinction the count that depends on the edge found in a different form. A square with domain walls has a different entropy per vertex from a torus, and the difference lived in the corners — a region that grows with the square rather than one confined to its edge. A torus has no edge, and what plays the edge’s role is the pair of loops round it; they are not a region at all, and what they constrain is not diluted by making the torus larger.

And on a plane there is no winding. A three-colouring of a large square region with free edges is an ice arrangement and every ice arrangement is three colourings, because there is no loop for the height to fail to close round. The whole gap between the two counts is topological, which is why it can be fixed exactly at even sides and fixed at a positive fraction at odd ones, and why neither fact changes as the lattice grows.

Who found what #

The three-colouring model on the square lattice is Baxter’s, from 1970, and its equivalence with square ice at the ice point is standard. The description of the six-vertex model’s long-range behaviour as a Gaussian height field with a single stiffness, and the treatment of windings on a torus as that field’s zero modes, belong to the Coulomb-gas picture developed through the 1970s and 1980s; the value $\pi/12$ for square ice is that picture’s, quoted here and bracketed by the census rather than derived. Jacobi’s transformation of the theta function is from 1828.

What the census adds is the finite statement: exact sector counts to side nine, the odd share settling rather than vanishing, and an exact third on every even torus it reaches — an identity whose limit is explained and whose finite form is not.

Still open: why the finite third is exact #

The exact third wants a proof. It holds at sides two, four, six and eight as an integer identity, while the marginal distributions it is built from are not yet the Gaussian ones. An identity that holds exactly at every finite size usually has a finite reason — a bijection, or a symmetry of the transfer matrix that the twisted counts respect — and the census has only the evidence. The equivalent statement in the census’s own terms is that the count weighted by a cube root of unity for the winding in one direction, added to the count weighted by it for both windings, is exactly half the unweighted count at every even side; that is a fact about three traces of one transfer matrix, and it is where a proof would start.

And a rectangle would test the explanation. On a torus with sides in the ratio two to one, the two windings have different Gaussian coefficients, the Jacobi argument gives a different number for the corner, and the exact third should fail. If it does not, the finite identity is about something other than the Gaussian, and the limit’s explanation was the right answer for the wrong reason.