A homometric twin is the structure moved class by class

Two structures are homometric when they have the same interatomic vectors with the same multiplicities. Their Pattersons are then identical and so are their diffraction patterns in every intensity, and no measurement of intensities can tell them apart. Two structures, one Patterson found such pairs by exhaustive search on a ring of sites. Where the pairs come from explained every pair that search found by one construction. Write a structure as a polynomial with a term per atom. If it factors as a sumset $A = B \oplus C$, a union of translated copies of one set placed at the points of another, then $B \oplus (-C)$ has the same interatomic vectors, because both have the difference polynomial $b(x)b(x^{-1})c(x)c(x^{-1})$.

Two structures on a torus moved that search to a two-dimensional cell, and symmetry does not rescue a Patterson then imposed each of the seventeen plane groups on a $6 \times 6$ torus. Homometry survived in sixteen of them and grew worse, most of all in the three-fold groups, where a fifth to three tenths of all structures have a twin. It ended by asking whether those constrained pairs come from factor reversal too. A structure invariant under $p3$ that factors would have to factor into pieces the three-fold axis respects. That might be why the three-fold groups are worst, or the coincidences might be of a kind the factorisation never produces.

It is the second. This essay tests every homometric pair on the torus, for ten plane groups, in both directions. Factor reversal explains two pairs of 3,522. The other 3,520 come from a different construction, and it explains all of them. In that construction the partner is the structure itself, moved in Fourier space by a translation, a sign or an inversion, independently on each of the twenty Galois classes of the torus’s characters.

Six atoms cannot be a reversed factor #

The search takes each group of structures that share one Patterson and each pair within it. It looks for every way of writing the first structure as a sumset $B \oplus C$, with $|B| \cdot |C|$ atoms and both factors of at least two points. That means every set $C$ whose translates by the points of $B$ tile the structure exactly. It then asks whether $B \oplus (-C)$ is the second structure, up to a translation and an inversion, the two motions a Patterson cannot see. The search over $C$ is exhaustive. $C$ is anchored at a point of the structure and drawn from its other points, and the translates that fit are then covered exactly.

At six atoms the answer is fixed before any search is run. Six splits as a sumset only as two translates of three points or three translates of two. A set of two points is centrosymmetric: reflecting it through its midpoint gives it back. So reversing a two-point factor $C$ gives $-C = C + t$ for some $t$, and $B \oplus (-C)$ is the original structure translated. Reversing a two-point $B$ instead, by applying the construction the other way, gives the original inverted. Either way the partner is the structure seen by a Patterson in the same way it sees itself, not a second structure. No six-atom homometric pair on any torus can come from factor reversal. Yet the torus has 3,252 of them in the general position alone, 40 with a two-fold axis, 29 with the mirror-and-glide group $cm$, and three each for $p3$ and $p31m$.

The ring hid this because of where its search looked. Where the pairs come from examined nine points on a ring of thirteen, which factors as three times three. That is the smallest size at which both factors can be asymmetric and the construction can produce anything. The pairs it found were exactly the reversed factors because a prime ring of that size leaves no room for anything else.

What the table finds #

The picture at the head of this essay is the result, fourteen cells of group and number of atoms. Across them there are 3,522 homometric pairs. Factor reversal accounts for two: one $p3$ pair at twelve atoms and one $p31m$ pair at twelve atoms. Nothing else in the table is a reversed factor. That includes the twelve-atom pairs of $p3$, where the factorisations $3 \times 4$ and $2 \times 6$ are available and the three-fold group gives its structures every chance to factor symmetrically. Of forty-nine homometric pairs there, one is a reversed factor.

The two pairs factor reversal does explain are instructive in their own way. Both are twelve-atom structures that factor as four translates of a three-point set, and both are patchworks too. So where factor reversal happens on this torus it is one way of producing a patchwork, not a separate mechanism alongside it. That is a fact about this torus. On a ring of prime length, as the section below on the ring and the torus explains, the relation runs the other way.

So the answer to the question the constrained census asked is no. The three-fold groups are not worse because their structures factor. Their pairs are no more reversed factors than the general position’s are, and the general position’s are not reversed factors at all.

A Patterson sees moduli, class by class #

The mechanism that does explain the pairs comes from reading the Patterson in Fourier space. A structure $A$ on the torus has a Fourier transform $\hat A(\chi) = \sum_{a \in A} \chi(a)$ at each of the thirty-six characters $\chi$ of the group $\mathbb{Z}_6^2$. The Patterson is the transform of the interatomic vectors, $|\hat A(\chi)|^2$. Two structures are homometric exactly when their transforms have the same modulus at every character. That is the statement of the map that needs no phases on a finite group, and it leaves each phase free.

The phases are not free independently. $\hat A$ is a sum of sixth roots of unity with whole-number coefficients, so its values at characters that generate the same cyclic subgroup are Galois conjugates of one another. Knowing it at one such character fixes it at the others. Those characters form a Galois class, and $\mathbb{Z}_6^2$ has twenty classes: four single characters of order one or two, and sixteen pairs $\{\chi, \chi^5\}$ of characters of order three or six. A change to $\hat A$ that keeps every modulus and still gives the transform of a set of integer points must act on a whole class at once. The simplest such changes use rational multipliers. The first is multiplying the class by $\chi(t)$ for some translation $t$, which is what translating the structure does. The second is multiplying by $-1$. The third is replacing every value by its complex conjugate, which is what inverting the structure does.

Applied to the whole transform at once, those moves give the structure translated, inverted or negated, and a Patterson cannot see any of them. The question is what happens when different classes are moved differently.

Every pair is a patchwork #

Call a pair a patchwork when the partner’s transform is the structure’s, class by class, under one of those moves: some translation phase, a sign, and possibly conjugation, chosen separately for each class. The test tries all $36 \times 2 \times 2$ moves on each of the twenty classes. The picture shows the $p3$ pair drawn above. On thirteen classes the second structure’s transform is the first’s with one common translation. On three the first structure’s transform is zero and there is nothing to move. On the remaining four classes it is moved differently, by another translation, a sign or an inversion. The second structure is the first on thirteen classes and a different image of the first on four.

The same pair can be read in the positions of its atoms, and the two readings agree. Each structure is two orbits of $p3$ of three atoms each. The first structure is the orbit through $(0,1)$, $(1,0)$ and $(5,5)$ together with the orbit through the origin, $(2,4)$ and $(4,2)$. The second keeps the first orbit and slides the second by $(0,2)$ along the torus. Sliding one orbit and not the other changes the vectors between the two orbits. A Patterson records every such vector, so it would normally notice. It does not here, because the slide happens to carry the set of cross-vectors onto itself. In Fourier space the slide is exactly the patch in the picture. The transform is kept on thirteen classes. On the three classes of characters of order six, $(1,1)$, $(1,4)$ and $(2,5)$, it is inverted and shifted by $(0,2)$, and on one class of order three it is shifted by $(0,1)$.

All 3,522 homometric pairs in the table are patchworks. That covers every group, every number of atoms and both factor-reversal pairs. The census makes no claim that a patchwork must be homometric in the other direction as a theorem. It checks that direction as well: a patchwork keeps every $|\hat A(\chi)|$ by construction, and the test declines every non-homometric pair it is offered.

The mechanism accounts for what factor reversal could not. A six-atom structure has no asymmetric factors, but its transform has twenty classes, and a partner can differ from it on some of them. What limits the partners is not the algebra of sumsets but the requirement that the patched transform still be the transform of a set of six points with weight one each. Most patchworks of a transform are not. They give fractional or negative occupations when transformed back. The homometric pairs are the rare patchworks that come back as sets.

How much of the transform a twin changes #

A patchwork that moved one class alone would change the transform at one or two characters. The census finds that no homometric pair does so little. In the general position at six atoms, the number of classes on which the partner departs from the structure’s most common shift is never fewer than four. It runs from 4 to 17, most often 8, and the other groups show the same floor of four. The classes where the partner departs are not a detail on a structure that is otherwise the same. They are a quarter or more of the transform, redistributed so that it still comes back as a set of points.

Why symmetry helps twins, partly #

The patchwork suggests a reason for the earlier finding that symmetry does not rescue a Patterson. A patchwork can only move a class on which the transform is non-zero, and moving it independently is the freedom that makes a twin. So structures whose transforms vanish on many classes have less room to have twins. The census checks the prediction. In every row but one the structures with a partner have fewer vanishing classes than those without, and in that one, $p6$ at twelve atoms, the two averages are equal. In the general position at six atoms the averages are 1.6 against 2.5. In the two-fold group $p2$ they are 1.6 against 5.8, and in $pmm$, the quietest group of all, structures without a partner average 6.7 vanishing classes.

That accounts for the half-turn groups being quiet. An inversion centre makes a structure’s transform real up to one phase, and many of its classes then cancel. It does not by itself account for the three-fold groups being loud. Their structures average two or three vanishing classes, close to the general position, and their rates of homometry are two to three times the general position’s. What makes the three-fold groups worse has not been isolated. The patchwork has turned the question into one about which classes a three-fold axis couples, since a three-fold rotation permutes characters and its structures’ transforms are constant on the resulting orbits, and that question has not been answered here.

Why the ring and the torus differ #

The two constructions divide the ground between them according to how many Galois classes a group has. A ring of prime length $p$ has two classes: the trivial character, and all $p - 1$ others, which are conjugate to one another. A patchwork there can only move the whole non-trivial transform at once, which gives back the structure translated or inverted and never a genuine twin. So on a prime ring every homometric pair must come from something the rational moves cannot express, and factor reversal is exactly that. Its multiplier $\overline{\hat C}/\hat C$ has modulus one and is in general not a root of unity. The $6 \times 6$ torus is at the other extreme. Its characters have orders 1, 2, 3 and 6, the cyclotomic fields they live in are the rationals and the field of cube roots of unity, and its thirty-six characters fall into twenty classes, most of them pairs. There are twenty independent places to apply a rational move, and the census finds that they are enough for every pair it contains.

Real crystals sit between these. A structure’s cell has many atoms, and its reciprocal lattice has reflections of every order the lattice allows, but the classes of a finite torus are an artefact of making the cell periodic on a coarse grid. What survives from the torus to a real Patterson is the principle, not the count. The intensities fix each reflection’s modulus, which is the whole of the phase problem in one clause, conjugate reflections are tied together, and a twin is any other assignment of phases, class by class, that still comes back as a set of atoms.

What the census does and does not settle #

The patchwork is observed, not proved to be the only mechanism. Every one of 3,522 pairs is one, on one torus. The moves tried are the rational ones. A multiplier of modulus one that is not a root of unity is conceivable in principle. Every sixth-root value of a transform lives in the field of cube roots of unity, which has many elements of modulus one that are not roots of unity. None is needed for any pair here. On a ring of prime length there is only one non-trivial class, rational moves give nothing, and the pairs factor reversal finds there use exactly such an irrational multiplier. The two constructions are complementary, and the torus, with twenty small classes, is where the rational one does all the work.

The factor search is exhaustive and the translation box is not a limit. Every factor $C$ was drawn from the structure’s own points and every exact cover checked, so a missed factorisation is impossible. On a torus there are no translates beyond the torus.

Point atoms only. Every structure here has identical atoms. When the atoms are not all the same found the ambiguity growing with unequal scattering powers, and there a structure’s transform is a weighted sum over its atoms. With rational weights the Galois classes still act on it in the same way, so the patchwork test applies unchanged; whether those twins pass it has not been run.

One torus, ten groups. The census uses the $6 \times 6$ torus of the earlier essay, the smallest on which all seventeen plane groups act, and ten of the groups. Its conclusions about the rational moves may change on a torus whose side has more prime factors, where the Galois classes are larger and fewer.

Who worked out the pieces #

The word homometric is Patterson’s, from his papers of 1939 and 1944 on the ambiguity his map leaves. The theorem that every homometric pair of finite sets of integers comes from factor reversal is due to Joseph Rosenblatt and Paul Seymour, in 1982, and it is the line’s version of what where the pairs come from found on the ring. That it fails in finite groups other than the integers has been known in the phase-retrieval literature since. The Galois grouping of characters is the standard decomposition of the group ring of a finite abelian group into cyclotomic fields.

What the test must refuse #

The refusal guards the one place the factor test could cheat. Reversing a factor of a structure can give back the structure’s own inversion, which is homometric to it for the dull reason the law that hides handedness gives. A test that counted those would find factor reversal everywhere. The test compares the partner with the structure up to translation and inversion first and declines it, so every pair it credits to factor reversal is a genuine second structure.

Still open: what a three-fold axis couples #

A three-fold rotation of the torus permutes its characters, and a structure invariant under it has a transform constant on the resulting orbits. So a $p3$ structure’s twenty Galois classes fall into fewer independent blocks, and a patchwork of such a structure must move whole blocks if the partner is to be $p3$-invariant too. Fewer blocks means fewer independent moves, which would predict fewer twins, and the three-fold groups have more. The resolution must lie in how often a patched transform of a symmetric structure comes back as a set. The symmetric transforms take values in a smaller set, and a coincidence among a few values is likelier than among many. Counting, group by group, how many of the patchworks of each structure come back as sets would measure that directly. It is one more loop over the moves already enumerated, and it has not been run.