Thirteen sites obey Friedel's law without a centre

Three species hide a spot at the golden ratio counted the three-species superstructures of the square lattice with up to eight sites in the cell, extending the two-species argument of the spots say the cell and little else, and found something it could not explain. At a superlattice spot the structure factor is $a S_A + b S_B$, where $S_A$ and $S_B$ are the sums of the phase over the A sites and the B sites, and $a$ and $b$ are the scattering contrasts of A and B against the third species. The spot goes dark at some real ratio of contrasts exactly when the two sums are parallel. At five and seven sites the census found that this happens exactly in the structures with a centre of inversion. A centre makes every sum real about itself, so one direction of that statement takes a line to prove. The other direction was a pattern in two small tables, and the essay ended by asking whether it is a theorem about roots of unity of prime order.

It is not, and the first counterexample has thirteen sites. Finding it took a reformulation that turns out to say more than the question asked. The two sums are parallel at every spot exactly when the A–B pairs of atoms are as common at every separation forwards as backwards. That holds exactly when no choice of complex scattering factors produces a Bijvoet difference anywhere in the pattern. A structure with that property obeys Friedel’s law even when one of its species scatters anomalously. A centre is the obvious way to have it. At five, seven and eleven sites it is the only way. At thirteen sites one arrangement, six A, four B and three C with no symmetry of any kind, has it without a centre.

One spot decides every spot #

The earlier census reported that a structure with a dark ratio has one at every spot, never at some and not others. That was an observation, and it has a one-line explanation at a prime number of sites.

Put the structure on a cell of $p$ sites, so that the superlattice spots are the non-zero elements $k$ of the integers modulo $p$ and the phase of site $x$ at spot $k$ is $\zeta^{kx}$, with $\zeta = e^{2\pi i/p}$. The map that replaces $\zeta$ by $\zeta^k$ is an automorphism of the field the sums live in. It carries the sums at the first spot to the sums at the $k$-th, because it carries $\zeta^x$ to $\zeta^{kx}$ site by site. It also commutes with complex conjugation, since both are automorphisms of an abelian extension. So if $S_A \overline{S_B}$ is real at the first spot, its image at the $k$-th spot is real too. Parallel at one spot means parallel at all of them.

That is why the dark ratios came in full sets. It also explains their degree. The dark ratios of a structure are the images of one number under these automorphisms, and the spots at $k$ and $-k$ always share one, so a structure has at most $(p - 1)/2$ distinct dark ratios and they are conjugates of each other. The earlier census found every one of them of degree $(p - 1)/2$, which is the case where all of them are distinct.

At a composite size the argument fails, because the field generated by $\zeta$ then has automorphisms only for the $k$ prime to $n$. A six-site arrangement such as AABCBC has parallel sums at the half-way spot and not at the first. So from here on, parallel means parallel at every spot, which is what one spot guarantees at a prime and what has to be checked spot by spot elsewhere.

Parallel sums are pairs counted both ways #

Write $D(d)$ for the number of pairs of an A site and a B site with the A site $d$ places after the B site, counted round the cell. Expanding the product of the two sums gives

$$S_A(k)\,\overline{S_B(k)} = \sum_{a \in A}\sum_{b \in B} \zeta^{k(a-b)} = \sum_{d} D(d)\,\zeta^{kd}.$$

The product is real when it equals its own conjugate, and conjugation sends $\zeta^{kd}$ to $\zeta^{-kd}$. So the product is real at every spot exactly when $\sum_d \bigl(D(d) - D(-d)\bigr)\zeta^{kd}$ vanishes for every $k$. By Fourier inversion that means $D(d) = D(-d)$ for every separation. The two sums are parallel at every spot exactly when the A–B pair counts are symmetric, as many pairs with the B atom $d$ sites before the A atom as with it $d$ sites after.

The reformulation turns a statement about irrational numbers into one about whole numbers, and that makes the census exact. A floating-point test can say only that a product is within $10^{-7}$ of the real axis. The pair counts are integers, and the census confirms every candidate on them, so no rounding error can admit an arrangement or drop one.

It also says what a centre has to do with it. A centre maps the A sites to themselves and the B sites to themselves while reversing every separation, so it pairs each A–B pair at $+d$ with one at $-d$. The pair counts are then symmetric because the arrangement is. The open question becomes concrete: can the A–B pairs balance at every separation without a symmetry that balances them? The figure above shows the thirteen-site arrangement that does: one pair each way at separation one, two at two, three and four, three at five and two at six, twelve each way in all, which is six A atoms times four B. Exchanging two neighbouring sites leaves the composition alone and breaks the balance at separations three and six.

The census #

The search runs over every arrangement of three species on a cyclic cell of $n$ sites with all three species present, for $n$ from four to nineteen. That is $3^n - 3 \cdot 2^n + 3$ arrangements at each size, from 36 at four sites to 1,160,688,606 at nineteen. Both the property and the centre are unchanged by renaming the species, so only arrangements with site 0 holding an A atom are visited and the counts are multiplied by three. At a prime the two sums at the first spot are kept up to date as the arrangement changes one site at a time, and only the arrangements whose product falls within $10^{-7}$ of the real axis are passed to the integer test. At a composite size the integer test is run on every arrangement. Each arrangement that passes is tested for a centre, every map $x \mapsto t - x$ against every site.

The picture at the head of this essay is the result. At five, seven and eleven sites every arrangement with parallel sums has a centre: 30 of 30, 252 of 252, and 5,940 of 5,940. At thirteen sites 24,414 arrangements have parallel sums and 23,478 have a centre, so 936 do not. At seventeen sites the exceptions are 27,744 of 336,294, and at nineteen 97,128 of 1,160,748. As a share of the arrangements with parallel sums the exceptions are 3.8 per cent at thirteen sites, 8.2 at seventeen and 8.4 at nineteen. A centre is the commonest reason for the balance at every size, and from thirteen sites it is not the only one.

The last column sorts the exceptions into kinds. Two arrangements count as the same kind when one becomes the other by a translation, by the inversion $x \mapsto -x$, by multiplying every position by a number prime to $n$, or by renaming the species. The multiplication needs a word. Multiplying positions by $u$ changes the arrangement, but its sums at spot $k$ are the old sums at spot $uk$, so it shuffles the spots and keeps every intensity. The property is therefore preserved, and the equivalence that every superstructure counted once used, an operation of the square lattice that keeps the cell, is one of these maps. Under all of them the 936 exceptions at thirteen sites form a single kind. The orbit has $12 \times 13 \times 6 = 936$ members, twelve multipliers times thirteen translations times six renamings, with no coincidences. Nothing but the identity fixes the arrangement, so it has no symmetry of any kind. At seventeen sites the exceptions make seventeen kinds and at nineteen forty-eight.

The arrangement at thirteen sites #

Read round the ring from site 0, the arrangement is AAAABAACBCBBC. The A atoms sit at sites 0, 1, 2, 3, 5 and 6, the B atoms at 4, 8, 10 and 11, and the C atoms at 7, 9 and 12. Every renaming and every multiplier gives an equivalent one, so the species counts six, four and three are the only part of the description that is not a choice.

The figure sets it beside its inverse. If the arrangement had a centre, some turn of the right-hand ring would bring it into register with the left-hand one, and none does. Six consecutive A sites on one side are interrupted by a B on the left and by nothing comparable on the right. Yet the two rings have exactly the same numbers of A–A, B–B, C–C, A–B, A–C and B–C pairs at every separation. The self-pairs always agree, since a pair counted from either end is the same pair. The cross-pairs agree here because the arrangement’s pair counts are symmetric. The arrangement and its inverse are homometric species by species: every partial Patterson function, the map of vectors between atoms of two named species, is the same for both.

That puts the exception in a known family with a new member. Two structures, one Patterson showed that on a ring of eight sites two genuinely different arrangements of four atoms can share every interatomic vector. A homometric twin is the structure moved class by class found where such twins come from on a torus. A single-species arrangement is always homometric to its own inverse, because its Patterson function is symmetric by construction. With three species that stops being automatic, since an A–B vector and a B–A vector are different things, and the thirteen-site arrangement is the smallest prime cell where it happens anyway.

The ring is an abstraction of a crystal, and the figure above lays it back on the square lattice. Thirteen is a sum of two squares, $13 = 3^2 + 2^2$, so by the theorem of Fermat’s that decides which sublattices stay square the square lattice has a square sublattice of index thirteen. It is spanned by $(3, 2)$ and $(-2, 3)$, and the site at $(x, y)$ lies in coset $x + 5y$ modulo 13. A quarter turn of that cell multiplies every coset by 5, since $5^2 \equiv -1$. So on this cell one of the multipliers in the census’s equivalence is literally a rotation of the crystal, and the half turn, multiplication by $-1$, is the inversion that the arrangement lacks. A two-dimensional crystal built this way has no half-turn centre anywhere. The same ring laid along a row of a cubic crystal, with thirteen-layer repeats, would give a three-dimensional structure with no centre of inversion.

Three sums on one line, and a line that wanders #

At every spot the three sums lie on one line through the origin. That follows from the pair counts, and with three species present it covers all three sums at once: away from the origin of reciprocal space the phases of all thirteen sites sum to zero, so $S_C = -S_A - S_B$, and a vector that is minus the sum of two parallel vectors is parallel to them.

The figure shows the first six spots; the other six are their mirror images in the real axis. The line is at 75.0° at the first spot, 23.4° at the second, 102.2°, 106.8°, 33.2° and 100.2° at the rest. For a structure with a centre those angles are forced. Measured from the centre every sum is real, so the line is the real axis, and measured from any other site of a thirteen-site cell the line at spot $k$ turns by $k$ times a fixed multiple of $360°/13$. So for a centred arrangement every line falls on a multiple of $180°/13 \approx 13.85°$, and the angles step evenly from spot to spot. The exception’s lines fall on no multiple of $13.85°$, and they do not step evenly. Every spot has its own line and nothing relates the lines to each other except that all three species agree on each one. No choice of origin makes the sums real, and they are parallel anyway.

Each spot still has a dark ratio, the value of $a/b$ at which $a S_A + b S_B$ vanishes. At the six spots it is 0.485, 1.565, 0.332, 1.880, $-0.410$ and 5.148. The six numbers are the images of one another under the automorphisms of the first section, so they are the roots of a single polynomial,

$$x^6 - 9x^5 + 24x^4 - 22x^3 + 2x^2 + 4x - 1,$$

of the full degree six that the real subfield of the thirteenth cyclotomic field allows. That polynomial has whole-number coefficients and leading coefficient one, so every dark ratio of this arrangement is an algebraic integer. In the structures with a centre the dark ratios were ratios of sums of cosines, read off from where each species sits relative to the centre. Here there is no centre to read them from, and the polynomial is the shortest description available.

What an anomalous scatterer sees #

The reason to care about parallel sums goes beyond dark spots. Friedel’s law says that the intensities at $k$ and $-k$ are equal, and it holds for real scattering factors whatever the structure, as the law that hides handedness explains. Near an absorption edge one species acquires an imaginary part, and in a structure without a centre the law then breaks. The difference between the two intensities is the Bijvoet difference, and it is how an absolute structure is determined at all. With complex scattering factors $f_A$, $f_B$ and $f_C$,

$$|F(k)|^2 - |F(-k)|^2 = -4 \sum_{X<Y} \operatorname{Im}\bigl(f_X \overline{f_Y}\bigr)\, \operatorname{Im}\bigl(S_X \overline{S_Y}\bigr).$$

The scattering factors enter only through the first imaginary part in each term, and the structure only through the second. When all three sums are parallel, every $\operatorname{Im}(S_X \overline{S_Y})$ is zero and the difference vanishes, whatever the scattering factors are. So an arrangement has parallel sums at every spot exactly when no anomalous scatterer, placed on any of its species at any strength, produces a Bijvoet difference anywhere in its pattern. The three imaginary parts $\operatorname{Im}(S_X \overline{S_Y})$ are equal up to sign, because $S_C = -S_A - S_B$, so the converse holds too: if the sums are not parallel, one absorbing species is enough to produce a difference, unless the other two scatter identically.

The two patterns above are drawn with $f_A = 1 + 0.6i$, $f_B = 0.4$ and $f_C = 0$, so that only A absorbs. The left-hand pattern is the exception’s, and every spot matches the spot opposite it through the origin exactly. The right-hand pattern belongs to the arrangement with sites 3 and 4 exchanged, which has the same composition and no centre either. There opposite spots differ, by up to 1.59 in the units of the figure, which is what anomalous scattering does to an ordinary structure without a centre. In a diffraction experiment the left-hand crystal would be reported as centrosymmetric. Worse, its absolute structure is undefined rather than merely uncertain: how much of it is the other hand estimates the fraction of a specimen that is the inverted structure from exactly these differences, and here every fraction from nought to one predicts the same intensities. The absence of Bijvoet differences is the test that anomalous scattering offers for the absence of a centre, and this structure passes it falsely at every wavelength. One experiment gives the cosine, the other gives the sine showed that the anomalous difference is what fixes the sine of a phase, and here it is zero at every spot, so the measurement it rests on has nothing to measure.

The intensities themselves do not settle the matter either. Without anomalous scattering the left-hand pattern is centrosymmetric for every structure, and whether there is a centre is a statistic of the intensity distribution, which a cell of thirteen sites is far too small to supply. Imposing a space group would not have rescued it either, as symmetry does not rescue a Patterson found for single-species pairs. The structure is told from its inverse only by a measurement that is not a sum over pairs, such as a phase determined some other way or a direct image of the arrangement.

Composite cells, for comparison #

The census includes composite sizes because the prime is doing work in the argument, and the composite rows show what that work is. At four, six and nine sites every arrangement with parallel sums at every spot has a centre. At eight sites 48 do not, all one kind, and at ten sites 240 in two kinds. From twelve sites on the exceptions are numerous: 4,320 in 30 kinds at twelve, 66,864 in 151 at sixteen.

The eight-site exception, AABACCBC, has a stabiliser of four maps where the thirteen-site one has only the identity. A composite cell has partial structures of its own, cosets of its subgroups, and an arrangement can be balanced on each of them separately, which is how vanishing sums of roots of unity of composite order come apart into smaller ones. A prime cell has no such pieces. At a prime the balance has to be achieved by the whole arrangement at once, and the thirteen-site exception is not built from anything smaller. That the first prime exception arrives later than the first composite one, at thirteen sites rather than eight, is what the absence of those pieces costs.

Why thirteen, and what the census does not explain #

The census says that eleven sites have no exception and thirteen have exactly one kind. It does not say why. The thirteen-site arrangement is fixed by no map but the identity, so it is not explained by a hidden symmetry. Nor is it a product of a centrosymmetric pattern with a smaller set of shifts, the construction that produces homometric pairs on a ring: because the three species together fill every site, such a product would tile the cell with translates of the smaller set, and a cell of prime size cannot be tiled that way. Its pair counts balance by what looks like arithmetic coincidence, and as the cells grow there is more room for coincidence. The number of arrangements grows as $3^n$, the balance imposes $(n - 1)/2$ equalities between whole numbers, and the share of balanced arrangements that are exceptions grows from 3.8 per cent at thirteen sites to 8.4 per cent at nineteen.

A coincidence that the census finds is still a fact about the integers modulo 13. A proof that eleven sites have none, or a construction that produces the thirteen-site arrangement and the seventeen kinds at seventeen sites from one rule, would say more, and neither is attempted here. The eleven-site result rests on the enumeration, all 171,006 arrangements tested.

What the pictures cannot show #

A finite cell. Every count here is for a cyclic cell of $n$ sites with one atom on each, a model of an ordered alloy rather than a molecular crystal with atoms at general positions. In a real structure the sites have coordinates, the phases are not roots of unity, and the irreducibility argument that ties one spot to all of them does not apply. What carries over is the pair-count criterion: a structure shows no Bijvoet difference anywhere exactly when every partial Patterson function is symmetric, which for continuous coordinates is a condition that a centre satisfies and that the census shows can be satisfied otherwise.

Real occupancy. Each site holds one species. A structure with mixed or partial occupancy has weights in place of indicators, and the balance can then be met by continuous adjustment, which makes exceptions easy and uninteresting. The interesting statement is the integer one.

The pattern, not the experiment. The figures draw exact intensities. An experiment measures them with errors, and a Bijvoet difference of a few per cent is easily lost in noise. The claim made here is the strong one: the exception’s differences are zero, not small, at every wavelength and for any choice of which species absorbs.

Who worked out the pieces #

Friedel stated the law that bears his name in 1913. Bijvoet, Peerdeman and van Bommel used its failure near an absorption edge to determine an absolute configuration in 1951. Patterson introduced the function of interatomic vectors in 1934 and, with homometric structures in 1939 and 1944, the problem that two structures can share it. The description of homometric sets on cyclic groups by factoring and reversing is due to Rosenblatt and Seymour in 1982. The census here is an enumeration over every arrangement, with the pair-count test applied exactly.

What the census must refuse #

The first refusal guards the statement that the exception is special rather than typical. If any arrangement of six A, four B and three C balanced its pairs, the result would be about composition and not about arrangement. Exchanging two neighbouring sites gives a Bijvoet difference of 1.59 at once. The second refusal guards the Galois argument, whose whole content is the prime. Six sites are enough to break it.

Still open: a rule that produces the exceptions #

The census settles the question the earlier essay asked, and replaces it with one it cannot answer. The exceptions at seventeen and nineteen sites come in seventeen and forty-eight kinds, some fixed by a multiplier of order three, and nothing here organises them. A construction would be a rule that takes a prime and produces arrangements whose cross-pairs balance without a centre. The partial Pattersons of such an arrangement agree with those of its inverse, so any such rule is a source of homometric pairs in which the two partners are mirror images, and the Galois classes of a torus are the natural place to look for it. Whether eleven sites have none for a reason, and whether a square cell of a two-dimensional crystal can carry such an arrangement in an alloy that actually orders, are the questions the counts leave.