A rhombohedral box saves a factor of two the hexagon never offered

A calculation on a crystal that wants to see a modulation, a frozen phonon or a charge wave with a wavevector away from the centre of the zone, has to run in a periodic box that the wavevector fits: a box whose every translation the wave sees as a whole number of wavelengths. Crystallography in a box set up the accounting, and a whole star needs a cubic box found the rule that governs it. The smallest box holding every arm of a star, the wavevectors the point group makes from one, is the star’s own lattice, and that lattice always keeps the full point group. A box that holds one arm is cheaper and lower in symmetry; a box that holds the star costs more and keeps everything.

The corner of a hexagon fits in three cells ran the same census on the primitive hexagonal lattice and found what the cube never offered: a box turned through thirty degrees, three cells, that holds the corner of the hexagonal zone where every regular grid needs nine. The saving was always exactly a factor of three, at stars a rule on numerators predicted in advance. And boxes keeping the six-fold axis but not the mirrors came in mirror-image pairs.

That essay ended on the other lattice of the hexagonal family. A rhombohedral crystal has a point group of twelve operations rather than twenty-four, a zone with points the tables call T, L and F, and two cells of its own, so it has two kinds of regular grid. Whether a star is ever held by a box that beats every grid by a factor other than three, and whether the boxes that lose the mirrors still come in pairs, was the same enumeration on a different lattice. This essay runs it. The answers are yes and yes, and one of the yeses comes with a size the hexagon could not produce.

Two cells, two grids #

The rhombohedral lattice is generated by three vectors of equal length at equal angles, and its three-fold axis is their sum. The point group $\bar3m$ acts on those three vectors by permuting them, and by permuting them and reversing all three: six permutations, six negatives, twelve operations. Those integer matrices are the same whatever the angle between the vectors, so the whole census needs no metric. A box is a sublattice in Hermite form, a wavevector is written in the dual basis, and a box holds a wavevector when every translation of the box has a whole-number product with it.

The convention that decides what counts as a saving is the choice of grid. Crystallographers describe a rhombohedral crystal on either of two cells. One is the rhombohedron itself, and repeating it $n$ times along each edge gives a box of $n^3$ cells, a rhombohedral grid. The other is the hexagonal cell, built from two in-plane differences of the rhombohedral vectors and their sum along the axis, three times the size of the rhombohedron; repeating it $n$ by $n$ in the plane and $m$ times along the axis gives a box of $3n^2 m$ cells, a hexagonal grid. A star is held more cheaply by its own box than by the grids when its own box is smaller than the smaller of the two grids that hold it.

The named points #

The figure at the head of this essay shows the census for eight wavevectors, and it repeats the cubic and hexagonal rule exactly. For every one, the smallest box holding the whole star is the star’s own lattice, the set of translations every arm sees as whole numbers of wavelengths, and that box keeps all twelve operations of $\bar3m$. The smallest box holding one arm is smaller wherever the star has more than one arm, and keeps only the operations that carry the arm to itself or its negative.

The axis end T, written $(\tfrac12, \tfrac12, \tfrac12)$, is a single arm, carried to itself by every operation, and the box that holds it is the two-cell lattice of translations whose three coordinates add to an even number. The face centre F, at $(\tfrac12, \tfrac12, 0)$, has three arms; one is held in two cells and the whole star in four, the lattice of translations whose three coordinates are all even or all odd. L, at $(\tfrac12, 0, 0)$, has three arms too, held one at a time in two cells and together only in eight, the lattice doubled along every edge. The halfway points along the axis, along a face and towards L, and the in-plane point a third of the way round the zone, follow the same pattern at four, thirty-two, sixty-four and nine cells.

Three of these numbers are already known, from a different lattice. T, L and F cost two, two-then-eight and two-then-four, exactly what the cube’s R, X and M cost in the cubic census. That is not a coincidence of small numbers. A rhombohedral lattice whose vectors meet at a right angle is the simple cubic lattice, and $\bar3m$ is the subgroup of the cube’s forty-eight operations that keeps one body diagonal. These three stars are not split when the cube’s group is cut down to that subgroup: the cube’s three arms of X are the rhombohedron’s three arms of L, and so on. The box arithmetic never sees the angle, so it gives the same numbers at every angle, and the cube is one of them.

One, two or three #

The census of every star with denominators up to eight has 154 stars in it, counting a star and its negative once, as one real modulation. For each, the star’s own box is computed directly as a lattice, and both grids are searched for the smallest that holds the whole star. The figure places each star by those two numbers on logarithmic axes, and every one of the 154 falls on one of three lines.

For 83 stars a grid is already the smallest box: nothing is saved by looking further. For 52 the better grid costs three times the star’s own box. And for 19 it costs twice. Nothing else occurs. No star with denominators up to eight is held by its own box more than three times more cheaply than by the better grid, and none by a factor that is not a whole number. The hexagonal lattice’s census gave one and three only, and three was its signature, the turned $\sqrt3 \times \sqrt3$ box. The rhombohedral lattice keeps the three and adds the two.

The two has a rule, and it was found in the data rather than stated in advance. Every one of the nineteen stars that save exactly a factor of two can be written with an even common denominator and numerators adding to an even number, and F’s star, with denominator two and numerators one, one and zero, is the first. Those are the wavevectors that see the lattice of translations whose three coordinates share one parity: a box that is the rhombohedral grid with half its translations removed. Not every star of that form saves two, since for some the hexagonal grid saves three and the larger saving wins, but every star that saves two is of that form.

What the box of one parity is #

The four-cell box that holds F’s whole star deserves a closer look, because it is the source of every saving of two. It consists of the translations whose three coordinates on the rhombohedral vectors are all even or all odd. Doubling the rhombohedron along each edge gives eight cells; adding back the translation that goes once along every edge, the sum of the three vectors, which lies along the three-fold axis, halves that to four. The result is a lattice with a translation through the middle of the doubled rhombohedron, a body-centred doubling of the cell, and it keeps all twelve operations of $\bar3m$ because every operation permutes or negates the three coordinates and so preserves their common parity.

Seen at the cube’s angle, this box is familiar. On the simple cubic lattice the translations with all three coordinates of one parity form a body-centred cubic lattice of twice the edge, one of the three sublattices that stay cubic. Its rhombohedral counterpart keeps less, since the rhombohedron has no four-fold axes to keep, but it is the same set of translations with the same parity rule, and it holds the same star. Where the cube had three cubic sublattices at each scale, the primitive, the body-centred and the face-centred, the rhombohedral lattice has the same three as boxes keeping $\bar3m$, and it has more, because twelve operations constrain a box less than forty-eight do. Fourteen lattices and no others counts the same freedom from the other side: the rhombohedral lattice is one Bravais type, and a cube is a rhombohedron that happens to have more operations.

The hexagonal grid cannot reach this box. The hexagonal cell is a sublattice of the rhombohedral lattice of index three, and a box built from it has a size divisible by three; four is not. The rhombohedral grid can reach the doubled cell of eight but not the four in between. So the parity box is exactly a box that neither familiar cell produces, and its existence is the whole reason the saving two appears on this lattice and not on the hexagonal one, where the corresponding box would have to keep a six-fold axis and cannot.

What such a box is for is concrete. A modulation at F, frozen into a crystal, produces new reflections at the positions of its star, the superlattice reflections that the reflections a superlattice adds describes, and a calculation that wants to compare the energies of the three arms’ combinations, one arm alone or all three together, needs a box holding all three. Four cells do it. And a calculation that samples the zone rather than freezing a wave, the reciprocal-space wedge of the domain in reciprocal space, gains from the same arithmetic: a sampling grid that is the dual of a star’s own box visits exactly the wavevectors that box holds and no others.

Two grids that miss in different ways #

The two grids are very different instruments, and the figure separates them. The hexagonal grid costs either the star’s own box or exactly three times it, for every one of the 154 stars: four stars at one and 150 at three. That is the hexagonal census found again from the other side. The hexagonal cell contains three rhombohedral cells, and a hexagonal grid can never be closer to a rhombohedral star’s own box than that factor allows unless the star happens to need the hexagonal cell’s own shape.

The rhombohedral grid misses by anything. A star whose arms lie along the three-fold axis fits a grid repeated equally along three edges very badly, and the grid can cost sixty-four times the star’s own box. A star whose arms lie in the rhombohedral directions fits it well, and the grid costs the box itself. In between, the cost runs through 2, 3, 4, 5, 6, 7, 8, 9, 12, 16, 18, 25, 36 and 49. The saving a star’s own box offers is the smaller of the two misses, and that is why it never exceeds three: whatever the rhombohedral grid does, the hexagonal one is never worse than three times the box.

A practical reading follows directly. A calculation that has to hold a whole star on a rhombohedral crystal can use the hexagonal grid and lose at most a factor of three, or the rhombohedral grid and lose anything up to sixty-four, or the star’s own box and lose nothing. The common habit of building supercells from the hexagonal cell is therefore a safer default than building them from the rhombohedron, and neither is necessary: the star’s own box is a lattice the arithmetic hands over directly.

The mirrors lost, and always in pairs #

The second half of the lead asked about boxes that keep the three-fold axis but not the mirrors. A box the three-fold axis carries onto itself, and the inversion with it, keeps the six operations of $\bar3$; if it also keeps one of the operations that swap two of the rhombohedral vectors, it keeps all twelve of $\bar3m$. The census finds both kinds at every size up to twenty-seven, and it finds the second kind, $\bar3$ without the mirrors, at exactly nine sizes: 7, 9, 13, 14, 18, 19, 21, 26 and 27 cells.

At each of those sizes there are exactly two such boxes, and a swap of two rhombohedral vectors, one of the mirror operations of $\bar3m$, carries each onto the other. The boxes that lose the mirrors still come in mirror pairs, as on the hexagonal lattice, and the pairing is checked box by box rather than inferred from the count. A mirror that is not a symmetry of a box can only carry it to a different box of the same size that keeps the same three-fold axis, and the census shows that different box is always there and always unique.

The sizes are not the hexagonal ones. There the turned pairs sat at 7, 13, 19, 21 and 28 cells, the norms of the Eisenstein integers that are neither squares nor three times squares, the sizes the sublattices that are the same shape found for turned copies of the hexagonal net. Here 7, 13, 19 and 21 come back, with a turned cross-section and a straight stack. And 9, 18 and 27 appear, sizes at which the hexagonal lattice’s turned boxes all kept their mirrors. The pairs at nine cells contain three axis periods but not one, so they are not a turned box in the plane stacked straight up. The three-fold axis acts on the rhombohedral lattice by permuting three vectors, which mixes the plane and the axis, and a box can be turned and stacked at once in a way the hexagonal lattice, which keeps its axis separate, never allows.

What the census must refuse #

The claim refused is the economical one: that a regular grid, either of the two a crystallographer would draw, already holds every star as cheaply as any box can. Seventy-one of the 154 stars are held more cheaply by their own box. Every number the essay quotes is held to a test that would fail if the enumeration were wrong: the box counts against the divisor sum at every size, the cubic twins against the cubic census computed separately, and the mirror pairs against the swap applied box by box.

What the pictures cannot show is the box itself. A sublattice of three-dimensional space is three vectors, and a figure that drew them in the rhombohedral cell would show a skewed parallelepiped from one angle and hide the property that matters, which of the zone’s wavevectors it holds. The figures therefore show what the boxes cost and which symmetries they keep, and leave their shapes to the Hermite rows the census records. Nor do they show the angle of the rhombohedron, because nothing here depends on it, which is also why the census applies equally to calcite, to corundum, to bismuth and to any other rhombohedral crystal at whatever angle it has.

Where the census stops #

The denominators stop at eight. The rule for a saving of two and the bound of three hold over all 154 stars to that size, and both have structural reasons: the hexagonal grid’s factor of three is fixed by the hexagonal cell’s size, and the parity lattice is fixed by the three-fold permutation. But a denominator of nine, twelve or sixteen could in principle reach a star whose own box beats both grids by more, and only a larger census or a proof would close that.

The mirror pairs stop at twenty-seven cells. Their sizes up to that point are the norms of ideals in the ring generated by the three-fold permutation, an arithmetic richer than the Eisenstein integers because it carries the axis as well as the plane, and the census gives no reason to expect the pairs to stop. Nor has the census looked at boxes that keep less than $\bar3$: the boxes that keep the three-fold axis but not the inversion, $3$ alone, would be the rhombohedral lattice’s chiral supercells, and they are the obvious next count.

Born, von Kármán, and two cells for one crystal #

Periodic boundary conditions entered crystal physics with Born and von Kármán in 1912, and the question of which wavevectors a finite periodic crystal contains has been asked ever since. The habit of choosing supercells from a crystal’s conventional cell is older than any census of the alternatives, and for the rhombohedral crystals the conventional cell is ambiguous by design: the International Tables give every rhombohedral space group on both axes. What this census adds is the measured cost of each choice, star by star, and the observation that a third choice, the star’s own lattice, is never worse and up to three times better.

Still open: the chiral boxes of a three-fold axis #

The census sorted boxes keeping the three-fold axis into those that keep $\bar3m$ and those that keep $\bar3$. A crystal of class $3$ or $32$, quartz among them, has no inversion to keep, and a box that keeps its three-fold axis need not keep the inversion either. Such boxes come in enantiomorphic pairs rather than mirror pairs, and a calculation on a chiral crystal that uses one of them builds in a handedness the crystal may or may not have. How many there are at each size, whether they include the sizes the mirror pairs occupy or new ones, and whether a whole star of class $32$ is ever held more cheaply by a chiral box than by any box that keeps the inversion, is the same enumeration with three operations fewer. It has not been run here.