Three tiles keep a number that asks which way they face

One invariant for every chain that can be undone found a quantity that the Fibonacci chain cannot change, and then found that it was not the Fibonacci chain’s at all. Build the transfer matrices of the two words a substitution produces at any level, and the trace of their commutator is the same at every level and every energy, for every rule on two letters that can be run backwards. A theorem of Nielsen says why: such a rule sends the commutator of the two letters to a conjugate of itself or of its inverse. So the number belongs to the two tiles, and the chain has no say in it.

That essay ended on the obvious next case. Chains built from cubic inflation factors have three kinds of tile, and three matrices have no single commutator. The free group on three letters has automorphisms that move every commutator somewhere new, and the traces of three matrices form a space of six dimensions rather than three. The reasonable guess was that a three-letter chain conserves either a set of several quantities that again measure the tiles, or nothing.

Neither is what happens. Most three-letter rules do conserve nothing, up to the degree searched. The tribonacci rule, the simplest and most studied of them, conserves exactly one number. It does not measure how much the tiles differ, because tiles as different as anyone likes can have it equal to zero. It measures whether the tiles have a direction.

Looking for a conserved quantity without guessing one #

The earlier invariant was known before anyone computed it, since the commutator identity is a century older than the Fibonacci chain. Here there was no formula to test, so the search had to be one that could find a quantity nobody had written down.

Three matrices of determinant one have seven natural traces: one of each matrix, one of each of the three products of two, and one of the product of all three in a fixed order. Every trace of every word in them is a polynomial in those seven. A substitution acts on these numbers: it replaces the three matrices by three words, and the seven traces of the new matrices are polynomials in the old ones. That map is what is usually called a trace map, and a conserved quantity is a polynomial $P$ with $P$ of the new traces equal to $P$ of the old ones for every choice of matrices.

That condition is linear in the coefficients of $P$. So the search is a piece of linear algebra. Take every monomial up to some degree in the seven traces, of which there are 120 at degree three and 330 at degree four. Evaluate each at a large collection of random matrices over a large prime field, before and after the rule is applied. The polynomials that are conserved are exactly the ones whose difference vanishes at every sample, and counting them is a rank computation. Polynomials that vanish because of the one relation the seven traces obey are subtracted out by comparing with the rank of the evaluations themselves, so a relation is not reported as a discovery.

The first row is the control. A rule that does the Fibonacci thing to two letters and leaves the third alone has to find the Fricke quantity, plus the powers of the third letter’s own trace, which nothing touches. It finds them: four at degree three, namely three powers and the quantity from the earlier essay, without having been told about any of them. An instrument that can rediscover what is known may be believed about what is not.

The other rows are the result. The rule that sends $a$ to $ab$, $b$ to $ac$ and $c$ to $a$, which is the tribonacci substitution that Rauzy’s fractal window comes from, has one conserved polynomial at degree three and no more at degree four. Its neighbours among the cubic rules have none, with no exception up to that degree. The plastic rule, with the smallest of the inflation factors that exist, sends $a$ to $b$, $b$ to $c$ and $c$ to $ab$, and it has no conserved polynomial.

The number, once it has a name #

Reading the coefficients out of the null space gives a short answer:

$$J \;=\; 2\operatorname{tr}(ABC) \;-\; \operatorname{tr}A\operatorname{tr}(BC) \;-\; \operatorname{tr}B\operatorname{tr}(AC) \;-\; \operatorname{tr}C\operatorname{tr}(AB) \;+\; \operatorname{tr}A\operatorname{tr}B\operatorname{tr}C.$$

That looks like nothing in particular until one remembers a classical identity for three matrices of determinant one, due in substance to Fricke and Vogt: the sum of $\operatorname{tr}(ABC)$ and $\operatorname{tr}(ACB)$ equals the last four terms with their signs reversed. So the polynomial the search returned is

$$J \;=\; \operatorname{tr}(ABC) - \operatorname{tr}(ACB).$$

It is the difference between a product and the same product read backwards. For two matrices no such thing exists: $\operatorname{tr}(AB)$ and $\operatorname{tr}(BA)$ are always equal, which is why this quantity could not have been seen on two letters. Three is the first number of letters at which order around a cycle means anything, because two cyclic orders exist.

The picture at the head of this essay shows what conservation looks like. Exact integers are used for the reason the earlier essay learned: the traces grow to tens of digits and the conserved number is 24, and no floating-point computation can subtract two numbers of that size and keep a difference of two digits. Under tribonacci the value is −24 at every level. The plastic rule starts from the same number and has eleven digits by its ninth application. The rule for $x^3 = x^2 + 1$ meanders and at its sixth application it even passes through zero. A rule that cannot be undone behaves worst, as period-doubling did on two letters.

So the tribonacci chain has something the other cubic chains lack, and the question is why it, of all of them.

Why a single site cannot see it #

Before asking which rules keep it, there is a more basic thing to ask about the number itself, because a physicist would compute it first and the answer is alarming. Take the standard model of the earlier essays, in which each tile is one site with a site energy, and whose transfer matrix has $E - V$ in one corner. For any three such matrices, at any energy and any site energies, $J$ is zero.

That is not a coincidence of the numbers. The transpose of a single-site matrix is its conjugate by one fixed diagonal matrix with entries 1 and −1. So the trace of $ABC$ equals the trace of its transpose, which is the trace of $C^{T}B^{T}A^{T}$, which by that conjugation equals the trace of $CBA$, which cyclically is the trace of $ACB$. The conserved number of the most-studied chain in the subject vanishes identically on the most-studied tiles.

The same argument shows when it does not vanish. It used only that each tile is reversal-symmetric. Let a tile have an inside, two sites in a particular order, so that it has a left end and a right end, and the argument fails.

For the tiles drawn, with site energies 1 then 0, 0 then 2, and 3 then 1, it is a parabola in energy. So this differs from the earlier invariant in a second way, since that one was the same number at every energy, and the difference is honest: for tiles with insides the conserved number is a property of the tiles at that energy. Reversing the inside of every tile reverses the sign of $J$ exactly, and leaves it otherwise alone. Three-site tiles that read the same both ways give zero, however different they are from each other. Exchanging two of the tile labels reverses the sign as well.

This is where the expected reading fails. Fricke’s quantity on two letters came out as the square of the difference in site energies, so it was natural to expect a three-letter version measuring how far the three tiles are from one another. $J$ does nothing of the sort. Tiles that differ enormously, with site energies 1, 2 and 5, have $J$ zero. Tiles with the same set of site energies in a different internal order, which any measure of difference would call identical, have $J$ of opposite sign. It is much closer to a handedness than to a distance. It reads the direction of the tiles and the orientation of the cycle $A \to B \to C$, in the same way that a tile with no reflection is a question about hand rather than about shape, and that the law that hides handedness is about a quantity an ordinary measurement cannot reach.

Why tribonacci, and the test that needs no matrices #

The reason a rule keeps this number can be stated about words alone, and the argument is short enough to give. Let $R$ be reversal, the map that writes every word in the free group backwards. It is an anti-automorphism, and it swaps $\operatorname{tr}(ABC)$ with $\operatorname{tr}(ACB)$ and leaves the other six traces alone. So every polynomial in the traces splits into a part $R$ fixes and a part $R$ negates, and the negated part is $J$ times something $R$ fixes.

Suppose a rule commutes with reversal, up to a conjugation. Reversing the image of a word then gives a conjugate of the image of the reversed word, and conjugate matrices have equal traces. So applying the rule to $J$ gives something $R$ still negates, which must be $J$ times a function $f$. And since the rule has an inverse, $f$ must have an inverse too in the ring of polynomials on this space, which means $f$ is a constant, and for integer reasons it is plus or minus one. Any invertible rule that commutes with reversal keeps $J$, up to sign.

The condition can be read directly. A rule commutes with reversal up to conjugation when one word $x$ carries each of the three images onto its own reverse, by the same conjugation.

Tribonacci passes at sight. Its images are $ab$, $ac$ and $a$, and all three begin with $a$. Conjugating by $a$ moves that first letter to the end, turning $ab$ into $ba$ and $ac$ into $ca$ and leaving $a$ as it is, which are precisely the three reverses. The plastic rule’s images $b$, $c$ and $ab$ cannot pass. A word that fixes both $b$ and $c$ under conjugation must be trivial, and then $ab$ does not become $ba$.

This also explains the two-letter case in a way the earlier essay did not need. On two letters the three kinds of elementary move all pass the test. The move sending $a$ to $ab$ and fixing $b$ has images $ab$ and $b$, and conjugating by $b$ reverses both, since it fixes $b$ itself. On three letters the same move has images $ab$, $b$ and $c$, and now the same conjugation must also fix $c$, and nothing does that but the identity. The elementary move that builds every reversible rule on two letters fails the test on three. With a third letter present, $c$ stands by and prevents the conjugation. That is what a passive letter does: the row of the table for the Fibonacci rule with a passive $c$ fails, which is why that rule does not conserve $J$ although it conserves Fricke’s quantity.

Tests from words and from matrices are entirely different computations, so they can be compared.

Every primitive rule with an inverse and images of total length at most eight is listed, 8,880 of them. Tested on matrices, 168 keep the number. Tested on words, 168 have images that one conjugation reverses. The two lists are identical, with nothing in either off-diagonal cell. That the second implies the first was argued above. That the first implies the second was not, and across eight thousand eight hundred and eighty rules it held with no exception, which is a measurement and not a theorem. Keeping $J$ is unusual, at one rule in fifty, and being a tribonacci-type rule is not required: rules with inflation factors of 2.2, 2.8 and 3.4 appear on the list.

What this does not show #

A conserved number on six dimensions does not confine anything. The reason the Fibonacci chain’s invariant has such consequences is that its trace map lives in three dimensions, so a conserved quantity pins an orbit to a surface, and the shape of that surface decides what a spectrum can be, which is what a spectrum that is a Cantor set drew. Here one number removes one dimension out of six. The orbit is free on a five-dimensional hypersurface, and nothing has been shown about whether it stays bounded or what its boundedness would say. No spectrum was computed in the course of this, and nothing is claimed about a tribonacci chain’s gaps.

The search bound is a bound. Plastic has no conserved polynomial up to degree four. A conserved quantity of degree twenty would not be seen. A periodic orbit of the map, or a conserved quantity of the rule’s square rather than the rule, would need a separate search, and only for tribonacci and its relatives is the structural reason available. The two search results that matter here, the Fricke quantity recovered and $J$ found, are positive and exact, since $J$’s conservation was then checked in exact integers. The negative results are the weaker half.

A tile with a direction is a modelling choice. The two-site tiles here are the simplest thing with a direction, a chain whose repeating unit is a pair of different atoms in a set order, and the numbers drawn belong to the three arbitrary pairs chosen. Reversal symmetry at the level of tiles is often exact in a real material, in which case $J$ is simply zero. Nothing here identifies a material with this structure.

The picture of a conserved line hides the reason for it. A flat line on a plot of a number against level looks like a property of the trajectory. It is really the rule’s relationship to reversal, which appears in no plot of traces at all. The figure of words carries that part of the argument and the figure of levels is only its consequence.

Where it comes from #

The two-letter identity is Fricke’s, of the late nineteenth century, and the three-matrix relation is in Vogt’s and Fricke’s work of the same period. Nielsen’s theorem on the commutator dates from the 1910s. The trace map itself entered physics with Kohmoto, Kadanoff and Tang in 1983, for the Fibonacci chain. Rauzy introduced the tribonacci substitution in 1982 for reasons that had nothing to do with chains. Automorphisms that commute with reversal have been studied as a subject of their own, under the name of palindromic automorphisms, and three-letter trace maps have a literature of their own. Whether $J$ has been noted for tribonacci before is likely, though no specific source is offered for it here; the search that found it was run without one.

How it was checked, and what it had to refuse #

The search evaluates monomials at random matrices modulo a prime of about $2^{26}$, chosen so that a product fits in a double exactly, and it counts ranks over that field. The polynomial it returns is checked once more as an identity in exact integers, on tiles with actual energies. The level-by-level values use arbitrary-precision integers throughout.

The search has to find what is already known. With two letters active and a passive third it must return Fricke’s quantity, and it does, with the right count for the powers of the passive trace, or nothing else it says about three letters would mean anything. The result must not be an artefact of the prime or the sample. Tribonacci’s count is the same at degrees three and four, and the exact-integer run is an independent check on the positive result.

A rule with an inverse must not be assumed to keep it. Plastic has an inverse and loses $J$ at its first application. And the natural reading must be refused. If $J$ measured only how much tiles differ, as the two-letter quantity does, reversing each tile’s interior could not change it, because the set of energies in each tile is unchanged. It reverses the sign instead, in every case tested. And the test that decides all of this is conjugation once more, the test of sameness that every question about a group’s description comes back to.

Still open: the other half of the converse #

The list of 168 is a list of agreements, and only one direction is proved. Whether keeping $J$ forces a rule to commute with reversal is a statement about the automorphism group of the free group on three letters, and a counterexample would need images longer than eight. Beyond that lies four letters, where the traces of words have more than one such difference, since four things can be cycled in three different ways up to reversal. A rule on four letters might keep several numbers, or the single reversal might leave one over. The word test generalises at once and costs nothing to run. And the physical question this essay stops short of is whether the tribonacci chain’s spectrum at $J \neq 0$ differs from its spectrum at $J = 0$ in any way a calculation would notice, given tiles with insides. It is one transfer matrix computation away, and it has not been made.