Three species hide a spot at the golden ratio

The spots say the cell and little else sorted the two-species superstructures of the square lattice by what their diffraction patterns record, and found that at a prime number of sites in the cell no superlattice spot is ever missing. At a superlattice position the structure factor is a sum of the phases of the A atoms. When the cell has a prime number $p$ of sites, those phases are distinct $p$-th roots of unity. The polynomial $1 + z + \dots + z^{p-1}$ is irreducible over the rationals, so a sum of distinct $p$-th roots of unity vanishes only when it takes all of them or none. A structure with both species present lights every spot its cell allows, and its pattern says which cell and nothing more.

That essay ended with the case an alloy designer usually has, three species instead of two. The structure factor then weighs two sums of roots of unity against each other with two different scattering contrasts. A vanishing combination no longer needs all the roots or none, and whether a prime still forbids a missing spot depends on what the contrasts are.

This essay settles it for the square lattice at every cell size from three to eight sites. At a prime number of sites no spot is dark for every choice of contrasts, and none is dark at a rational ratio of them. But a spot can be dark at one irrational ratio. At five sites the ratio is a power of the golden ratio, the number inflation and the Fibonacci chain are built on, and at seven a root of a cubic. The census also finds exactly which structures can do it. They are the ones with a centre of symmetry, and in those every spot has its own dark ratio.

Two sums and two contrasts #

Put species A, B and C on the sites of a superstructure cell, with scattering factors $f_A$, $f_B$ and $f_C$. At a superlattice position $k$ the structure factor is the sum of $f$ times the phase over the sites. The phases of all $n$ sites sum to zero at every non-zero $k$, which is the statement that $k$ is a superlattice position and not a fundamental one. So the third species can be eliminated:

$$F(k) = f_A S_A + f_B S_B + f_C S_C = a\,S_A + b\,S_B, \qquad a = f_A - f_C,\quad b = f_B - f_C,$$

where $S_A$ and $S_B$ are the sums of the phases over the A sites and the B sites. The two numbers $a$ and $b$ are the contrasts of A and B against C. Only they matter, which is why a pure crystal and a pure alloy of equal scatterers show no superlattice spots at all.

For real contrasts, as scattering factors are away from an absorption edge, $F(k) = 0$ requires $a S_A = -b S_B$. If both sums vanish, the spot is dark for every contrast. If exactly one vanishes, the spot is dark only when the other contrast is zero, which means two species scatter identically and the structure is really a two-species one. That case is set aside as degenerate. If neither vanishes, $a S_A + b S_B = 0$ has a real solution only when $S_A$ and $S_B$ point along the same line in the complex plane. It then has exactly one, the ratio $a/b = -S_B/S_A$. So every spot of every three-species structure is dark for all contrasts, dark at one ratio of contrasts, or never dark.

The census #

The census builds every three-species structure of the square lattice with three to eight sites in its primitive cell, one per structure. It takes every sublattice of the right index up to the parent’s point group, as every superstructure counted once did. It colours the cosets with three species, all three present, and discards colourings that repeat on a smaller cell. The rest are identified under the parent’s translations and the operations of $4mm$ that keep the cell. At each of the $n - 1$ superlattice positions of each structure it computes $S_A$ and $S_B$ and sorts the spot into one of the three kinds.

The picture at the head of this essay is the result. There are 2 structures at three sites, 24 at four, 45 at five, 275 at six, 441 at seven and 2,478 at eight. At the prime sizes, three, five and seven, no spot is ever dark for every contrast, over all 2,830 spots of those 488 structures. At six and eight sites, twenty and 411 spots are. The prime argument survives the third species in that form. A spot dark for every contrast needs both $S_A$ and $S_B$ to vanish, and each is a sum of distinct $p$-th roots, which vanishes only if it is empty or everything. With all three species present neither can be.

The second kind is where the third species changes things. At five sites, 60 of the 180 spots are dark at some ratio of contrasts, and at seven sites 648 of 2,646. The number of distinct ratios at which this happens is six at five sites and twenty-four at seven. At three sites there are none at all, and the two three-site structures light both their spots for every choice of contrasts.

None of the ratios is rational #

The ratios at a prime size have a property that the census checks and a short argument explains. None of them is rational. Suppose $a/b$ were rational and $a S_A + b S_B = 0$. Clearing denominators gives a vanishing combination of the $p$ distinct $p$-th roots of unity with whole-number coefficients: $a$ on the A sites, $b$ on the B sites and zero on the C sites. The irreducibility that settled the two-species case says such a combination vanishes only if every coefficient is the same. That would need $a = b = 0$, which is two species scattering like the third. So at a prime size a rational ratio of contrasts, $-1$, $\tfrac12$, $2$ or any other, never darkens a spot.

The ratio that does darken one lies in a definite field. $S_A$ and $S_B$ are sums of $p$-th roots of unity, and a real ratio of two such sums lies in the real subfield of the $p$-th cyclotomic field. That field is generated by $\cos(2\pi/p)$ and has degree $(p-1)/2$ over the rationals. The census computes each ratio’s minimal polynomial from its conjugates, the values it takes when the phases are replaced by their powers, and finds every one of full degree. At three sites the degree is one, so the real subfield is the rationals and no ratio is possible at all. That is why three sites never shows a dark spot. At five sites the degree is two, at seven three.

Five sites, and the golden ratio #

The five-site ratios are the most concrete, because the real subfield of the fifth cyclotomic field is generated by $\sqrt5$, and so by the golden ratio $\varphi = (1 + \sqrt5)/2$. The census finds six distinct ratios at five sites, and they are $-\varphi$, $-1/\varphi$, $1/\varphi^2$, $1/\varphi$, $\varphi$ and $\varphi^2$. Their minimal polynomials are $x^2 + x - 1$, $x^2 - x - 1$ and $x^2 - 3x + 1$, each ratio taken with its conjugate.

The picture shows how one of them arises. The structure puts B atoms on two sites, A atoms on two, and C on the fifth. At the superlattice spot drawn, the phases put the B atoms at the roots $\zeta^0$ and $\zeta^2$ and the A atoms at $\zeta^3$ and $\zeta^4$. The two sums point in exactly opposite directions along one line, and the structure factor $a S_A + b S_B$ is zero when $a/b$ is $0.3820$, which is $1/\varphi^2$. The cancellation is the geometry of the regular pentagon, whose diagonal and side stand in the golden ratio. It is the same fact that makes a five-fold rotation impossible in a lattice and possible in four dimensions, as where five-fold becomes legal showed.

The two dark ratios of this structure can be computed by hand, and doing so shows where the golden ratio comes from. The structure has a centre at its C site. Measured from there, the two B atoms sit at positions $\pm1$ and the two A atoms at $\pm2$ along the cell’s repeat. At the first superlattice spot the phases are the fifth roots of unity $\zeta^{\pm1}$ for the B atoms and $\zeta^{\pm2}$ for the A atoms. Each pair sums to twice a cosine:

$$S_B = 2\cos 72^\circ = \frac{1}{\varphi}, \qquad S_A = 2\cos 144^\circ = -\varphi.$$

The structure factor $a S_A + b S_B$ vanishes when $a/b = -S_B/S_A = 1/\varphi^2$, the value the census reports. At the second spot every phase is squared, so the B atoms move to $\zeta^{\pm2}$ and the A atoms to $\zeta^{\pm4} = \zeta^{\mp1}$. The two cosines exchange roles, and the dark ratio becomes $\varphi / (1/\varphi) = \varphi^2$. The regular pentagon’s two cosines, doubled, are $2\cos72^\circ = 1/\varphi$ and $2\cos144^\circ = -\varphi$, and every dark ratio at five sites is a ratio of sums of them and of the one contributed by a species at the centre. That is why all six are powers of $\varphi$ up to sign.

The intensity of a spot as the contrasts vary is $|a S_A + b S_B|^2$, a quadratic in $a/b$ that is positive except where the sums are parallel. For this structure all four spots have a dark ratio, two at $1/\varphi^2$ and two at $\varphi^2$, since the spots at $k$ and $-k$ always share one. At every other ratio, including every rational one, all four spots are lit.

Seven sites and the cubics #

At seven sites the real subfield has degree three and is generated by $2\cos(2\pi/7)$, a root of $x^3 + x^2 - 2x - 1$. The census finds twenty-four distinct dark ratios, the roots of eight cubics. One is that minimal polynomial itself, and the others are its relatives: $x^3 - x^2 - 2x + 1$, $x^3 + 3x^2 - 4x + 1$, $x^3 - 5x^2 + 6x - 1$ and four more. Each cubic contributes three real roots, the three conjugates of one ratio.

The number line shows the difference between prime and composite sizes at a glance. At four sites the dark ratios are $-1$, $\tfrac12$, $1$ and $2$, all rational, because the fourth roots of unity include $-1$ and a vanishing combination can have unequal whole-number coefficients. At six sites there are ten ratios, again all rational. At eight sites rational ratios sit beside irrational ones built from $\sqrt2$, which generates the real subfield of the eighth cyclotomic field. Only at the primes are the ratios all irrational, and only there is none of them a simple fraction.

Only a centre makes the sums parallel #

Which structures have dark ratios at all? At five sites, fifteen of the forty-five do, and at seven sites 108 of 441. In each, every spot has one. That all-or-nothing pattern points to a property of the structure rather than of a spot, and the census identifies it.

At five and seven sites the structures with dark ratios are exactly the centrosymmetric ones. Every structure with a centre of inversion has a dark ratio at every spot, and no structure without one has a dark ratio at any spot. A centre is usually read from diffraction by a statistic of the intensities; here it shows as the existence of a contrast at which a spot goes dark. One direction is simple. Measured from a centre of inversion, every species’ sites come in pairs $x$ and $-x$ whose phases are complex conjugates, so every sum $S_X$ is real. Measured from any other origin, all the sums are multiplied by the same phase. So $S_A$ and $S_B$ are real multiples of one complex number, parallel at every spot, and each spot has a dark ratio.

The same argument says which ratio each spot gets. From a centre, each species contributes a sum of cosines, $2\cos(2\pi k x/p)$ for each pair of its sites at $\pm x$, plus one for a species at the centre itself. A spot’s dark ratio is the ratio of two such sums. So a centrosymmetric structure’s dark ratios can be read off from where its species sit relative to the centre, and changing which species occupies the centre changes them all at once.

The converse, that a spot’s sums are parallel only when the whole structure has a centre, is what the census finds at five and seven sites and does not prove. At composite sizes it fails. Structures without a centre have dark ratios at four sites (three of them), six (ninety-five) and eight (1,232), where sums can be parallel because some partial sums vanish on their own. The prime condition that forbade the missing spot with two species reappears here as a condition that ties a missing spot to a symmetry of the arrangement.

What a measurement can do with this #

A diffraction experiment does not choose its contrasts freely, but it moves them. X-ray scattering factors fall off with angle at different rates for different elements, and near an absorption edge one element’s factor changes sharply with wavelength while the others stay put. The two most anomalous wavelengths are the wrong pair made use of exactly that. So the ratio $a/b$ is not one number for a given sample. It traces a path as the reflection’s angle or the wavelength changes. For a five-site centrosymmetric structure, a spot can then vanish at one angle or one wavelength and reappear on either side of it. An absence of that kind is not a symmetry extinction. It is not in the tables of systematic absences, and the census says when to expect one: only in a centrosymmetric structure at a prime size, and only at an irrational ratio of contrasts. For a structure without a centre at a prime size, the census says that no choice of wavelength darkens any spot.

What the census assumes #

Real contrasts. Near an absorption edge a scattering factor acquires an imaginary part, and then $a$ and $b$ are complex. A complex ratio can match any two non-zero sums, so every spot of every structure could in principle be darkened at some complex contrast. The statements here are for real contrasts, which is the case away from edges and for neutrons, whose scattering lengths are real for almost every nucleus.

A two-dimensional parent. The census uses the square lattice, because that is what the two-species enumeration used. The prime argument does not depend on the dimension, only on the cell’s group of superlattice positions being cyclic of prime order. It carries to a face-centred cubic parent unchanged, but the counts do not.

Up to eight sites. The census stops at eight. At eleven sites the ratios would have degree five and the counts would be larger, and nothing here was computed there.

Who worked out the pieces #

The irreducibility of $1 + z + \dots + z^{p-1}$ is Gauss’s. The description of all vanishing sums of roots of unity with whole-number coefficients, which governs the composite sizes, is due to Rédei, de Bruijn and Schoenberg in the 1950s and 1960s, and to Lam and Leung, who determined for which $n$ a vanishing sum of $n$-th roots of unity with $k$ terms exists. Enumerating derivative superstructures up to symmetry is the method of Hart and Forcade, used in alloy theory to list every ordering of a parent lattice.

What the census must refuse #

The refusal is the tempting extension. At six sites the argument that forbade dark spots does not apply, and the census finds twenty spots dark for every contrast and ten rational dark ratios. A rule stated for all sizes would be false at the first composite one that allows three species room to cancel. The degree test is the check on the whole method. A ratio computed with a rounding error would have a minimal polynomial of the wrong degree, or none with small integer coefficients. Every ratio at five and seven sites has one of exactly the degree the field requires.

Still open: whether the converse is a theorem #

At five and seven sites the census finds that a spot’s two sums are parallel only when the whole structure has a centre. That is a claim about sums of roots of unity of prime order. If $S_A / S_B$ is real, is the colouring of the $p$ sites necessarily symmetric under some reflection $x \mapsto t - x$? The forward direction is the one-line argument above. The converse would have to use the prime again, perhaps through the fact that a real ratio is fixed by complex conjugation, which acts on the $p$-th roots as the reflection $x \mapsto -x$. Proving it for every prime, or finding the first prime where it fails, is a question about roots of unity that the census raises and does not answer. The census at eleven sites would be the first place to look for a counterexample.