The corner of a hexagon fits in three cells

A whole star needs a cubic box asked what a periodic box costs a calculation on a cubic crystal. A box is a sublattice of the crystal’s translations, as crystallography in a box first set it out in the plane, where a box of twelve cells turned out to see more than one of sixteen. Gluing its faces lets it hold exactly the wavevectors of its own dual lattice, and a wavevector never comes alone, because the crystal’s operations carry it to a star of equivalent ones. The census found that one arm of a star is cheap and the whole star costs more, and that the smallest box holding a whole star always keeps the cube’s full symmetry. It also found that the boxes doing best for their size were the face-centred and body-centred sublattices, which no calculation set up as an $n \times n \times n$ grid would try.

The essay ended on the hexagonal crystals, where a six-fold axis and a unique direction split the stars differently and where the lattice has a sublattice of its own that a grid never reaches: the box $\sqrt3$ cells on a side, turned through thirty degrees. This essay runs the census there. The rule about whole stars carries over unchanged, as its two-line proof promised. What the hexagonal lattice adds is that the turned boxes win by a factor that is always three, at stars a rule can name in advance, and that boxes turned through other angles exist too and keep the six-fold axis while losing every mirror.

The same accounting on a different lattice #

The primitive hexagonal lattice is written on two vectors of equal length at a hundred and twenty degrees in the plane and a third along the axis. On that basis every translation has integer coordinates, so a box is again a sublattice of the integer lattice in three dimensions, written in Hermite form exactly as how many sublattices wrote them, and the enumeration of the cubic census can be reused without a change. Every box of every size up to a hundred and forty-four cells is in it.

What changes is the point group. The hexagonal holohedry 6/mmm has twenty-four operations, generated by a sixth of a turn about the axis, a mirror exchanging the two in-plane vectors, the mirror across the axis and the inversion, and on this basis every one of them is an integer matrix. Each was checked to preserve the hexagonal metric rather than taken to. A wavevector is written in the dual basis, where an operation acts by the inverse of its transpose, and a box holds a wavevector exactly when its pairing with every vector of the box is a whole number. That test needs no lengths or angles at all, which is why the same function decides it on both lattices.

The named points of the hexagonal zone are the ones the tables use. Γ is the centre. A is the centre of the top face, half a reciprocal vector up the axis. M is the centre of a side face, half a reciprocal vector in the plane, and it has three arms. K is a corner of the hexagon. L is M raised to the top face, H is K raised to it. Between them sit Δ, a third of the way up the axis, Σ halfway from Γ to M, T halfway from Γ to K, and U halfway from M to L.

A star keeps the hexagon, an arm does not #

The figure at the head of this essay is the census. For M, one arm fits in two cells, a box doubled along one in-plane direction, and the whole star needs four, the box two cells on each side of the hexagon. For L the step is from two cells to eight. For Σ and U it is from four to sixteen, for T from six to twelve. Γ, A, K, H and Δ have a single arm up to sign, so for them one arm and the whole star are the same box.

The two rules the cubic census found hold here in every case, and both are checked rather than assumed. The smallest box for a whole star is the star’s own lattice, the set of translations every arm pairs with to a whole number: its size, found by searching boxes in order, equals the index of that lattice, counted separately by residues. And that lattice keeps all twenty-four operations, since the operations permute the star’s arms and so carry the lattice onto itself. In the other direction, the smallest box for one arm keeps exactly the operations that carry the arm to itself or to its negative. M’s arm is fixed up to sign by eight operations, and the two-cell box holding it keeps those eight and loses the other sixteen. A calculation in that box is orthorhombic, whatever the crystal is.

The star of a wavevector is the orbit the point group gives it, and the group that keeps one arm is the arm’s little group up to sign; the one-arm box keeps that little group and nothing more. So the price of not breaking the hexagon is the price of a whole star, as the price of not breaking the cube was. A factor of two for M, of four for L, Σ and U, of two for T. What a box that holds one arm does, as the cubic essay put it, is turn an equivalence the crystal has into a distinction the box has, and here the distinction is between three in-plane directions that the six-fold axis says are the same. A calculation in the two-cell box that finds an instability at M will find it along the doubled direction, because there is nowhere else for it to be, and report an orthorhombic structure; the crystal had three equivalent ways to order and the box offered one. Folding the zone into the smaller one of a bigger cell does the same thing to the bands: the two-cell box folds one arm of M onto the centre and leaves the other two where they were, so degeneracies the hexagon guarantees appear split.

The corner, in three cells #

The corner K is where the hexagonal lattice parts company with the cube. Its star under 6/mmm has two members, K and $K'$, and $K'$ is K’s negative, so as a real modulation the star is a single arm. A wavevector with a single arm needs no symmetry broken to hold it. The question is only how small a box can do it.

On a grid of $n$ by $n$ cells, written on the hexagonal vectors, K is held when three divides $n$, so the smallest grid is three by three, nine cells. The smallest box is three cells: the sublattice whose vectors are the hexagonal lattice turned through thirty degrees and stretched by $\sqrt3$. It keeps all twenty-four operations, and the figure shows what it holds: the centre and the two corners, nothing else. This is the box in which every third atom of a honeycomb or a triangular lattice can be distinguished from its neighbours, which is the pattern of the Kekulé distortion of graphene and of the $\sqrt3 \times \sqrt3$ arrangements adsorbed atoms make on many hexagonal surfaces. A calculation on a three-by-three grid holds it too, at three times the cost. What the order does to a diffraction pattern is the same in either box: the reflections a superlattice adds appear at K and $K'$, two new reflections for each of the parent’s, because the box has three cells and so holds three wavevectors.

The saving is not a curiosity of small numbers. A calculation of the electronic or vibrational structure of a hexagonal crystal costs, at the least, in proportion to the cells in its box and often to their cube, so a factor of three in cells is a factor of up to twenty-seven in time. And the larger box is not only slower. A three-by-three grid holds nine wavevectors where the turned box holds three, and six of the extra ones are arms of other stars that the order being looked for does not need; any instability at those arms, real or numerical, now has room to appear in the result as well.

The same holds a floor up. H, the corner raised to the top face, is a single arm held by the turned box doubled along the axis, six cells, against eighteen for the smallest grid. And T, halfway from the centre to K, has three arms whose whole star the turned box doubled in the plane holds in twelve cells, where the smallest grid holding it is six by six, thirty-six.

Always three, and always the same stars #

The three examples have the same ratio, and the question is whether that is a pattern or a coincidence of small numbers.

The census answers it by brute force over every star of in-plane wavevectors whose coordinates have a common denominator up to twelve: sixty-eight stars. For each, the smallest whole-star box is found by counting its lattice, and the smallest $n$ by $n$ grid by trying $n$ in turn. The ratio is one at sixty stars and three at eight, and nothing else ever occurs.

A rule was stated before the search and tested by it. Write the wavevector as $(a, b)/n$ with $n$ the common denominator in lowest terms. The turned box wins exactly when three divides $n$ and $a$ and $b$ are congruent modulo three. For those stars the star’s own lattice is a turned lattice, a multiple of the $\sqrt3$ box, rather than a multiple of the original. The rule predicts all sixty-eight verdicts.

The ratio is three and never more because the hexagonal lattice’s sublattices that keep its full symmetry come in exactly two shapes in the plane: $n$ by $n$, of $n^2$ cells, and $n\sqrt3$ by $n\sqrt3$ turned through thirty degrees, of $3n^2$ cells. That is what the census finds for every such box up to forty-eight cells, and it is the hexagonal counterpart of the three that stay cubic, where the cube had three shapes and the face-centred and body-centred ones beat a grid by factors of four and two. A whole star’s box is one of the two shapes. If it is a grid, the grid is the answer. If it is a turned box of $3n^2$ cells, the smallest grid containing it has $9n^2$, three times as many.

Arms that close and arms that do not #

The cubic census found that the wavevectors a box holds are closed under addition, since they form a lattice, and that this forces some counts to be impossible. The hexagonal zone has one case of each kind. M’s three arms add in pairs to the third: half of one in-plane reciprocal vector plus half of the other is half of their sum, which is the third arm. So a box holding two arms of M holds all three, and none of the thirty-five boxes of four cells holds exactly two.

L’s three arms do not close. Each is an M arm with a half added along the axis, and two of them add to an M arm with a whole reciprocal vector along the axis, which is an arm of M and not of L. So a box can hold exactly two of L’s arms, and eighteen of the hundred and fifty-five boxes of eight cells do. That is why L costs eight cells for its star when M costs four: the two have the same number of arms and the same denominators, and M’s arms generate one another while L’s do not. The cubic zone had the same pair, M closing and X not, in a different place.

Two modulations at once #

A crystal that might order at one wavevector and modulate at another needs a box holding both, and if the calculation must not prejudge which arm, both whole stars. The table gives fifteen pairs. Wherever K or H is in the pair the turned box wins by exactly three: M with K needs the $2\sqrt3$ by $2\sqrt3$ box of twelve cells against thirty-six for a grid, L with H twenty-four against seventy-two, K with Δ nine against twenty-seven. Wherever neither corner is involved, the grid is already the smallest box, and pairs such as M with L, held by the two-by-two-by-two box of eight cells, cost no more than the harder of the two.

Some of these pairs have names in the literature. The kagome metals of the family $A\mathrm{V_3Sb_5}$ have a charge order reported at the three arms of M and L together, which this census says needs the two-by-two-by-two box of eight cells and nothing smaller. A calculation of such a crystal that holds one arm of M in two cells, or the star of M in four without the doubling along the axis, cannot represent the order the experiments see; the box that can is four times the first and twice the second.

Turned through other angles #

The turned box of thirty degrees keeps every mirror of the hexagon. There are boxes turned through other angles, and they do not.

The boxes in the plane that a sixth of a turn carries onto itself are the ideals of the Eisenstein integers, the multiples of a fixed complex number in the ring generated by a sixth root of unity, and their sizes are the numbers of the form $a^2 + ab + b^2$: one, three, four, seven, nine, twelve, thirteen and so on, as the sublattices that are the same shape found when it asked which sublattices of the hexagonal lattice are hexagonal. The census here asks the further question of which of them keep the mirrors. At one, three, four, nine, twelve, sixteen, twenty-five and twenty-seven cells there is one such box and it keeps all twenty-four operations; those are the $n$ by $n$ and the turned $n\sqrt3$ boxes. At seven, thirteen, nineteen, twenty-one and twenty-eight cells there are two, each keeping only twelve: the six-fold axis, the inversion and the mirror across the axis, but none of the mirrors containing the axis. The two are mirror images of each other, turned through equal and opposite angles.

Such a box holds a whole star of the group 6/m, three arms in the case of seven cells, and only half a star of 6/mmm. The other half is held by its mirror image, and both together need forty-nine cells, the box and its partner’s intersection. So for a crystal whose symmetry includes the vertical mirrors, a turned box of seven cells privileges a sense of rotation that the crystal does not have, in the way a two-cell box privileges a direction. For a crystal whose symmetry is 6/m or lower, the same box is a legitimate whole-star box, and the cheapest.

The best-known example is the commensurate charge order of tantalum disulphide in its 1T form, whose superstructure cell is $\sqrt{13} \times \sqrt{13}$, one of the thirteen-cell boxes here. The census predicts that it comes in two mirror-image orientations, and domains of both have been reported. The box is not wrong for that crystal. It is a statement that the order has chosen a sense of rotation that the parent lattice did not.

What the census must refuse #

Each test could fail. The smallest box for each star is found by search and compared with an index counted by residues. The shapes of the full-symmetry boxes are checked against the two allowed forms rather than listed. The rule for the turned boxes was stated before the sixty-eight stars were examined. And the refused claim, that a box holding one arm of L holds its star, fails by the factor of four the figures show.

Where the census stops #

It is a census of the primitive hexagonal lattice. A rhombohedral crystal described on hexagonal axes has a different lattice, the R-centred one, a different zone and different stars, and every number here would change. The method would not: enumerate the sublattices of the crystal’s own lattice, test them against the arms, and read the smallest off.

It says nothing about irrational wavevectors, which no box holds and which a description in a space of one more dimension makes periodic again, nor about how many of a box’s wavevectors a calculation must actually visit, which the domain in reciprocal space reduces by the point group. It measures the cost of not knowing in advance which arm a crystal will choose.

And a box that keeps the full symmetry is not automatically the right box. A crystal that has already broken a mirror, or that orders along one arm, is well served by a cheaper box, and the factor saved is the census’s ratio. What a box may not do is decide the question for the crystal by having room for only one answer.

Born, von Kármán, and the turned grids #

The periodic box of Born and von Kármán in 1912 was $N$ cells along each axis, and so was the grid of wavevectors Monkhorst and Pack chose in 1976 for sampling a zone. For a hexagonal crystal both are the $n$ by $n$ grids of this census. The turned boxes entered practice from the other side, from experiment: low-energy electron diffraction on hexagonal surfaces kept finding superstructures written $(\sqrt3 \times \sqrt3)R30°$ and $(\sqrt7 \times \sqrt7)R19.1°$, in Elizabeth Wood’s notation of 1964, which names exactly the turned boxes this census enumerates. The census adds the accounting: which turned boxes keep the crystal’s mirrors, which stars each holds whole, and that against a grid the saving is three or nothing.

Still open: the rhombohedral lattice #

The rhombohedral lattice is the other hexagonal-family lattice, and its sublattices include the primitive hexagonal one at index three, so its census contains this one and more. Its zone has the points the tables call T, L and F, its point group $\bar3m$ has twelve operations rather than twenty-four, and the boxes that keep $\bar3m$ need not keep a six-fold axis at all. Whether some of its stars are held by boxes that beat every grid by a factor other than three, and whether the turned boxes that lose the mirrors still come in mirror pairs when the mirrors were fewer to begin with, is the same enumeration on a different lattice with a different group. It has not been run here.