The triangles that were never in the bands

A census of wrapped nets sorted seven nets by the third moment of their spectra, found that the sorting is complete and that neither dimension nor coordination does it, and then turned the third moment on continuously and found a crossover. Every one of those measurements held one thing fixed without meaning to: the coupling has always run along one of the bands’ own graphs.

That is not a small omission. The whole argument is that a first-order term is present or absent according to whether a graph has odd cycles, and there are two graphs in the problem — the one the bands hop on and the one the perturbation runs along. The attribution so far is to the first. With the coupling on a band’s own bonds, nobody could have seen the second.

Separated, the attribution is wrong.

Two kinds of triangle #

The composite system is one hopping matrix on twice as many sites, and its third moment counts closed three-step walks in it. Those walks come in exactly two kinds, and the counting is worth doing before any number is looked at.

A walk that stays inside one layer uses three band bonds. A walk that leaves its layer must come back, and the two layers bipartition the coupling edges, so it uses an even number of them — which in three steps means exactly two, plus one band bond. Three coupling bonds cannot close a walk of length three.

So the composite third moment is a sum of two counts: triangles inside a band, and triangles with one foot in each. Splitting them costs nothing — switch the coupling off and what remains is the first — and the split is what the census could not make, because with the coupling running along a band’s own bonds the two counts rise together.

Only one of them matters #

Nine combinations, and the gap-crossing count sorts all nine. Every row with none of them returns an exponent between −2.0066 and −2.1117 and a quotient that is nearly constant; every row with any of them returns one between −0.9846 and −1.1054 and a spread of 128 to 134 per cent.

The intra-band count sorts nothing, and it is not for want of range — it spans four decades across the census, from arithmetic noise to 7.296, which is more range than the crossing count has. A band that becomes a bell curve does so by its higher moments settling, and none of that is what is being read here. Among the rows at −2 it takes the values 0, 0, 1.296 and 7.296 — the largest third moment in the whole census — and the exponent does not move. That is the strongest form this kind of measurement takes, and it is the same form as two structures with identical second, third and fourth moments that still differ in binding: a quantity that sorts a table while a much larger quantity of the same kind sorts nothing.

The pair that says it outright is the one that holds the bands fixed.

Two triangular bands coupled along their own bonds: −1.1054, spread 128 per cent. Coupled along a square net instead: −1.0991, spread 130. Coupled along a matching — a graph in which every site has exactly one partner, so no site of the upper layer is joined to two of the lower and no triangle can cross — the constant comes back, at −2.1040 with a spread of 4.5 per cent.

Nothing about the bands changed between those rows. Their spectra are identical, their third moment is 7.296 throughout, and the seven-net census would have predicted −1 for all three.

The other direction is equally clean and was always available. Two square bands — no triangles of their own at all — coupled along a triangular set of bonds give −1.0732 and a spread of 128 per cent. The odd cycles are entirely in the perturbation, and they are enough.

There is a reading of that row worth pausing on, because it makes the original question sharper rather than merely correcting it. The question was whether the constant belongs to a net, and the first answer was that it belongs to a square net and not a triangular one. It does not belong to a net at all. It belongs to an arrangement — two bands and a way of joining them — and the same net appears on both sides of the answer depending on how it is joined. A property attributed to a material that turns out to be a property of how the material was probed is a familiar failure and it is not usually available to check; here it is, because the probe is a graph that can be written down and changed.

The matching is not a weaker coupling #

It is worth being careful about the matching, because the obvious objection is that it is simply a smaller perturbation and the constant returns because nothing much happened.

It is smaller — one coupling bond per site against a square net’s four — but that is a factor of four, and the effect being switched off is not reduced by a factor of four. It is zero: the gap-crossing third moment is $4\times10^{-13}$, which is the arithmetic noise of a diagonalisation and not a small quantity. The reason is combinatorial rather than quantitative, and it is the counting argument above: a matching gives every site one partner, and a triangle across the gap needs a site with two.

The quotient’s own size says the same thing from the other side. If the matching were merely a feeble perturbation, its quotient — the excess gap over the shape departure — would be some small residue. It is 1.068 for the triangular bands, larger than the triangular-coupled value of 0.105 by a factor of ten, and the square-band matching gives 4.742 against the square-coupled 2.442. The response is not smaller. It is a different response, with a different power in it.

What the matching does cost is precision. The quotient’s spread across the six separations is 2.8 per cent for square bands and 4.5 for triangular, against 0.08 per cent when the coupling runs along a square net, and 13.7 per cent for the one mixed case. So the constant is a constant to a few per cent rather than to a fiftieth of one. That is worth reporting rather than smoothing over: the claim being made is that these rows are in the −2 family and not that they are as clean as the square-net case.

The crossover, produced without touching a band #

Turning the third moment on continuously makes the exponent slide. That family moved the bands. The same slide can be produced with the bands held exactly fixed.

Two square bands, coupled along a square net whose diagonals carry a strength $\lambda$: the exponent runs from −2.0118 at $\lambda = 0$ through −2.0020, −1.9231, −1.7365, −1.5313, −1.3677, −1.2317, −1.1169 to −1.0732 at $\lambda = 1$. Every band in that family is the same band. The intra-band third moment is exactly zero at every setting.

So the crossover is not a property of the bands at all. It is entirely a property of the perturbation, and the continuous family produced it by moving a variable that happened to move the coupling as a side effect.

The shape of the two slides is worth comparing before the next section makes anything of it. The coupling family is half-way to −1 at λ = 0.05 and the band family at λ = 0.02, so on the λ axis they are a factor of two and a half apart — which looks like agreement and is a coincidence, because the two λ’s mean different things. In the band family a diagonal at λ carries the full band strength; in the coupling family it carries the coupling strength, which is sixteen times smaller. The variable that has to be compared is the one both families compute, and that is what the next figure does.

What the split buys, and what it does not #

Having two families that produce the same slide by different means is an opportunity, because the two can be plotted against a common variable and the variable can be tested.

Against the total composite third moment the two families are a factor of twenty-seven apart: the band family reaches −1.76 at a total of 0.037 and the coupling family reaches it at 0.0014. Against the gap-crossing part alone they lie within about a factor of two.

That is a large improvement and it is not a collapse. So the honest statement is the one in the refutations above: the gap-crossing third moment is the right predicate — it is zero exactly for the rows that keep the constant, in all nine combinations — and it is not yet the right coordinate, because the two families’ crossovers do not fall on top of one another when it is used as one.

The residual factor of two is a real thing and it is not explained here. The most likely candidate is that a gap-crossing triangle in the band family carries one band bond of strength 1 and two coupling bonds of strength 0.06, while in the coupling family the same triangle carries a weak diagonal in one of its coupling bonds instead of in its band bond — so the two are triangles of the same count and different composition, and the moment weights them identically while the energy denominators do not. Testing that means a third family in which the composition is varied at fixed count, which is one more construction.

What was computed, and how #

Every net is a wrapped graph, and the composite is one $2m \times 2m$ hopping matrix diagonalised exactly. The bands are separated by a site energy $\pm\Delta/2$ and hop at 1 and 0.6; the coupling is 0.06 on whatever bonds the coupling graph names.

The shape departure is the change in the upper band’s kurtosis against its own isolated value, and the excess gap is the measured gap minus the four band edges written down from the two isolated spectra — never from a second calculation with the coupling switched off, so that the reference is exact rather than computed. The exponent is a least-squares slope on logarithms over six separations from 12 to 48, and the quotient is the excess gap divided by the shape departure and by the separation, averaged over the same six.

The split of the third moment is made by running the composite twice, once at the stated coupling and once at zero, on the hopping matrix alone with the site energies removed. The site energies are what separate the bands and they have nothing to do with the graph’s cycles, so leaving them in would make every moment a function of $\Delta$ and the count would not be a count.

Two things had to hold, and both do. The intra-band moment must take both a large and a vanishing value among the rows at −2, so that a census in which the two counts happened to agree would fail rather than pass. And the matching’s gap-crossing moment must be zero to machine precision rather than merely small — if it were small, the predicate would be a threshold and the counting argument would be decoration.

The matching used here pairs each site with its neighbour along one axis, which is one perfect matching of many. Nothing in the argument depends on which: the property being used is that every site has degree one, and every perfect matching has it.

Where the model stops #

This is a tight-binding hopping matrix and no repulsion, and the standing caution is sharpened here rather than relaxed: the bipartite behaviour is fragile because a small second-neighbour hop puts a third moment back. What is added here is that the hop that matters is not the one inside the band — it is any coupling between the two bands that joins one orbital to two neighbouring orbitals of the other. In a real two-band problem that is the ordinary case rather than the exception: a $\sigma$ orbital overlaps more than one $\pi$ orbital of its neighbours, which is the same geometry that makes two bands of one chain interact at all.

So the −2 behaviour is rarer than it first appeared. It requires a coupling of a shape that no ordinary overlap has, and the census’s five clean rows include two that were constructed for the purpose.

The exponents at the matching rows are −2.03, −2.10 and −2.11 rather than −2.00, and the quotient spreads are a few per cent rather than a fraction of one. Both are worse than the square-coupled reference, and the reason is presumably that a matching’s coupling is structured rather than uniform — it distinguishes the two sites of each pair, which the shape departure can see. What is claimed is the family, not the precision.

And nothing here is a derivation. The counting argument for the two kinds of triangle is exact; the claim that the crossing kind is the one that enters the excess gap at first order is an inference from nine measurements, and the factor of two between the families says it is not yet the whole of the mechanism.

The generalisation #

A variable that has never been varied is not a variable that has been controlled for. The census measured a real effect, attributed it correctly to a third moment, and attributed it to the wrong object’s third moment — not by carelessness but because the experiment available held the two objects equal. The tell was there: the confound could be named before it was tested. The same shape appears wherever a quantity is computed from a system whose parts have not been separated — an underdetermined force field is the standing example, where the fit is excellent and the constants are not determined by it.

And a perturbation has a shape, not just a size. The instinct that a smaller coupling is a weaker version of the same thing is what makes a matching look like a quarter of a square net. It is not: it is a coupling with a different combinatorial structure, and the structure switches a term off entirely rather than reducing it. The same distinction is what separates a symmetry-forbidden overlap from a small one, and it is why a count is a better description of a perturbation than a strength wherever the count can be made. It is also why a second moment that is a count of neighbours is a stronger statement than a bandwidth: the count is exact and the width is a consequence.

Who found it, and when #

The moment expansion for a coupled two-band tight-binding problem is standard, and so is the observation that the odd moments of a bipartite graph vanish. That the relevant graph for a perturbative term is the one the perturbation runs on rather than the one the unperturbed states live on is elementary once stated — the term is a product of matrix elements along a path, and the path uses whatever bonds it uses.

What made it easy to miss here is that the natural construction couples two bands along the bonds one of them already has, which is what the earlier measurements did, and what most treatments do for the same reason. The matching is an unnatural coupling and that is exactly why it separates the two.

Still open: a triangle’s composition, and the coupling strength #

The obvious open question is the composition of a triangle rather than the count of them. The two families cross over a factor of two apart against the crossing moment, and the candidate explanation is that their triangles carry the weak bond in different places — one band bond and two coupling bonds are not interchangeable, because only the coupling bonds cross the gap and pick up a denominator. A family in which the count is held and the composition is moved would say so directly, and it is one more construction of the kind used here.

The nearer question is the coupling strength. Everything above is at 0.06, and the crossover between a first-order and a second-order term should sit where the two are equal — which makes its position in the crossing moment a computable function of the coupling. Three couplings would test that, and a crossover whose position can be predicted is one that can be used to measure a coupling rather than merely reported alongside it.