A whole star needs a cubic box

Crystallography in a box worked out what a calculation does when it replaces a crystal by a finite block with its opposite faces glued. In the plane, a box of $N$ cells in each direction holds exactly the wavevectors whose denominators divide $N$, and no others. The missing ones are absent from the calculation altogether rather than approximated. A box is a sublattice of the crystal’s translations, and choosing one is choosing which wavevectors the calculation can see. That essay ended on space, where “a box commensurate with one modulation is usually incommensurate with another.”

This essay does the accounting in space, for the primitive cubic lattice, and finds that the plane’s rule hides a second condition that matters more in three dimensions. A wavevector in a cubic crystal never comes alone. The cube’s 48 operations carry it to a star of equivalent wavevectors, and a modulation with one of them is as likely as a modulation with any other. A box can hold one arm of a star and not the rest. One arm of the face centre X fits in a box of two cells; all three arms need eight. One arm of a third of the way along an axis fits in three cells; the whole star needs twenty-seven. The smallest box that holds an entire star turns out always to keep the cube’s full symmetry. So the price of not breaking the cube is exactly the price of a whole star, and a cheaper box breaks it.

A box is a sublattice, and it holds its dual #

A box of the primitive cubic lattice $\mathbb{Z}^3$ is any sublattice $L'$ of finite index $N$, the number of cells the box contains. Gluing the box’s faces identifies two points exactly when they differ by a vector of $L'$, so a function on the box is a function on the crystal that repeats under $L'$. It can contain a plane wave $e^{2\pi i k \cdot x}$ only when $k \cdot v$ is a whole number for every $v$ in $L'$. Those wavevectors form the dual of $L'$, and modulo the reciprocal lattice there are exactly $N$ of them.

The boxes of a given size are counted and enumerated exactly. Every sublattice of index $N$ has a unique Hermite normal form, as how many sublattices used in the plane, three rows $(a, b, c)$, $(0, d, e)$ and $(0, 0, f)$ with $adf = N$ and the off-diagonal entries reduced, and there are $\sum_{adf = N} d f^2$ of them: 7 boxes of two cells, 13 of three, 35 of four, 155 of eight. The three that stay cubic counted them the same way, and the shapes a lattice in space can thin to asked which of them keep the parent’s shape. Everything below searches all of them, up to 216 cells, rather than the handful of rectangular boxes a calculation usually tries.

The figure shows four small boxes and what each sees of the zone. The two that have two cells make the point of the whole essay at once. The obvious one, two cells along one axis, holds the zone centre and one face centre, a single arm of X. The other two-cell box is the sublattice of vectors whose coordinates add up to an even number, a face-centred lattice. It holds the zone centre and the corner R, the wavevector of rock-salt ordering, in which every site differs from its six neighbours. The first box is tetragonal: it has lost the two axes it did not double. The second keeps all 48 of the cube’s operations, because the face-centred lattice is a cubic lattice.

The other two panels show the body-centred box of four cells, which holds the centre and all three arms of M, and the two-by-two-by-two box of eight. Both keep the cube. The eight-cell box holds everything with halves in it: the centre, the corner, all three edge centres and all three face centres.

Stars, and the arms a box can miss #

The star of a wavevector is the set of wavevectors the point group carries it to. For the cube’s 48 operations the star of X, $(\tfrac12, 0, 0)$, has three arms, one along each axis. The star of M, $(\tfrac12, \tfrac12, 0)$, has three. The corner R is alone in its star, since every operation of the cube carries $(\tfrac12, \tfrac12, \tfrac12)$ to itself modulo the reciprocal lattice. A third of the way along an axis, $(\tfrac13, 0, 0)$, has a star of six, three arms up to sign; a third along the body diagonal has eight, four arms up to sign; a quarter along a face diagonal has twelve.

A wavevector and its negative describe one real modulation, so an arm here means a pair $\pm k$. The box test is unchanged by the sign, since $k \cdot v$ and $-k \cdot v$ are whole numbers together.

The picture at the head of this essay is the census. For each of nine wavevectors it gives the size of the smallest box holding one arm, found by testing every box in order of size, and the size of the smallest box holding every arm. The two are equal only for the zone centre and the corner, whose stars have a single arm. Everywhere else the whole star costs more, and the factor grows quickly. M goes from 2 cells to 4, X from 2 to 8, the thirds from 3 to 27, the quarters from 4 to 32 or 64.

The labels beside the filled dots say what a one-arm box keeps. Each is tetragonal, trigonal or orthorhombic, never cubic. A box that holds one arm of X has doubled one axis and not the others, and a box holding one arm of the body-diagonal third has picked out one of the cube’s four three-fold axes.

The one-arm box keeps what the arm keeps #

The symmetry beside each filled dot is not arbitrary, and the census found a rule in it that holds in all nine cases. The smallest box holding one arm keeps exactly the operations of the cube that carry that arm to itself or to its negative, no more and no fewer. X’s arm along one axis is fixed up to sign by the sixteen operations that preserve that axis, and its two-cell box keeps those sixteen. The body-diagonal third is fixed up to sign by twelve, the operations of the trigonal group about that diagonal, and its three-cell box keeps twelve. The face-diagonal quarter and the off-centre face point S are each fixed up to sign by eight, and their boxes are orthorhombic with eight.

The reason is the same as for whole stars, applied to a single arm. The vectors $v$ with $k \cdot v$ a whole number form a lattice that every operation fixing $\pm k$ preserves, and that lattice is the smallest box holding the arm. It can keep nothing more, since an operation that moves the arm to a different arm moves the lattice to a different lattice. So the symmetry a one-arm box keeps is the arm’s own symmetry group up to sign, the group that matters at that wavevector, and the symmetry it loses is exactly the part of the cube that would have moved the arm.

Holding two arms can force a third #

Between one arm and the whole star there are boxes holding some arms and not others, and the figure counts them. Of the 155 boxes of eight cells, 70 hold no arm of X, 66 hold one, 18 hold two and one holds all three. Of the 1,210 boxes of twenty-seven cells, 858 hold no arm of the axial third and one holds the whole star.

Two of the bars are empty, and the gaps are forced. The wavevectors a box holds are closed under addition, because they are the dual of a lattice. Two arms of M add up, modulo the reciprocal lattice, to the third: $(\tfrac12, \tfrac12, 0) + (0, \tfrac12, \tfrac12)$ is $(\tfrac12, 1, \tfrac12)$, which is the third arm $(\tfrac12, 0, \tfrac12)$ shifted by a reciprocal lattice vector. So a box holding two arms of M holds all three, and none of the 35 boxes of four cells holds exactly two. The four arms of the body-diagonal third behave the same way, since any three of them combine with signs into the fourth, and none of the 1,210 boxes holds exactly three. X’s arms add up to M’s, not to each other, so a box can hold two of X’s three, and eighteen do.

That closure is the reason a modulation at M is cheaper to hold whole than one at X, four cells against eight, though both have three arms of halves. M’s arms already generate one another; X’s do not.

Why the smallest box for a star is a cube #

The open dots all fall on boxes with the cube’s whole symmetry, and the reason is two lines long.

The vectors $v$ with $k \cdot v$ a whole number for every arm $k$ of a star form a lattice, the lattice every box holding that star must sit inside. The cube’s operations permute the arms of the star among themselves, so they carry that lattice onto itself. The lattice is cubic, and it is itself a box: the smallest one holding the star. Every larger box holding the star is a sublattice of it.

The census confirms this rather than assuming it. For each of the nine wavevectors, the smallest box the enumeration finds holding the whole star has exactly the index of the star’s own lattice, computed separately by counting residues, and it keeps all 48 operations. And since the three that stay cubic showed that the only sublattices of the primitive cubic lattice keeping the cube are scaled copies of the primitive, face-centred and body-centred lattices, of $n^3$, $2n^3$ and $4n^3$ cells, every star’s smallest box is one of those, the three cubic types of the fourteen Bravais lattices appearing as sublattices of the primitive one. R’s is the face-centred lattice of two cells, M’s the body-centred lattice of four, X’s the doubled primitive lattice of eight, the thirds’ the tripled primitive lattice of 27, the face-diagonal quarter’s the doubled body-centred lattice of 32.

The converse matters as much. A box that keeps the cube holds a star whole or not at all, because its operations carry any arm it holds to every other. So the cubic boxes are the boxes in which no arm is privileged, and the cheaper boxes are the ones in which some arm is.

Which stars each cubic box holds #

Since only the cubic boxes hold whole stars with nothing privileged, the useful table for a calculation that must not break the cube is the one in the figure. Its rows are the nine cubic boxes up to 64 cells, in order of size, and its columns the nine wavevectors.

Some entries are expected. The $n \times n \times n$ boxes hold the stars whose denominators divide $n$, as in the plane. Others are not. The 27-cell box, three cells on every edge, does not hold R: the rock-salt ordering cannot be represented in a three-by-three-by-three box, because three is odd. The face-centred box of 54 cells, three times the two-cell face-centred lattice, holds R and both thirds. The body-centred box of 32 holds the face-diagonal quarter’s whole star, which no primitive box below 64 holds.

So the boxes that see the most, size for size, are often not the cubes of side $n$. That is the plane’s lesson, that a box of twelve cells sees more than one of sixteen, repeated with a twist. In space the box’s shape decides what it sees as much as its size does, and two of the three cubic shapes are boxes a calculation set up by hand as an $n \times n \times n$ grid would never try.

What a box that holds one arm does #

A box that holds one arm of a star and not its partners does more than miss some wavevectors. It turns an equivalence the crystal has into a distinction the box has.

Take the two-cell box that doubles one axis. It can represent an ordering that alternates along that axis, one arm of X, and it cannot represent the same ordering along either of the other axes. A calculation of the cubic crystal in that box that finds an ordering at X will find it along the doubled axis, because there is nowhere else for it to be. It will then report a tetragonal structure. The tetragonal symmetry came from the box. The crystal had three equivalent ways to order and the box offered one.

The same thing happens to any quantity computed in the box. A soft vibration at X has three partners in the crystal, related by the cube; the box holds one of them, so a calculation sees one soft mode where the crystal has three. And the box’s lower symmetry lets the box itself relax tetragonally if its shape is allowed to change, which a cubic box cannot do without breaking a symmetry it holds exactly. None of these is an error in the arithmetic. Each is the arithmetic of a box answering a question about itself, as the plane essay put it, and only sometimes about the crystal.

The remedy is to pay for the star. For X that is a factor of four in cells, eight against two; for the thirds a factor of nine; for the off-axis quarters a factor of sixteen. The census says the remedy is never more than that, and never less.

Two modulations at once #

The plane essay’s closing worry was a box commensurate with one modulation and incommensurate with another, and in space the cost of avoiding that compounds with the cost of whole stars. The figure gives ten pairs. For each, it shows the smallest box holding one arm of each wavevector and the smallest holding both stars entire.

The extreme case is a structure that might order at the face centre X and might modulate with a period of three cells along an axis. One arm of each fits in six cells, a box six long in one direction. Both whole stars need a box six cells on every edge, 216 cells, thirty-six times as many. A box between those, holding the X star and one arm of the third, is still tetragonal and still privileges an axis.

Some pairs are much cheaper than the arithmetic of denominators suggests. R and either third need 54 cells for both stars, not 216, because the face-centred box three cells on an edge holds both. The two thirds together need 27, no more than either alone, since the body-diagonal third and the axial third are held by the same cubic box. These savings come entirely from the non-rectangular boxes. A search confined to $n \times n \times n$ grids would report 216 for R with a third.

What the census does not decide #

The census is for the primitive cubic lattice. A face-centred or body-centred crystal has a different lattice of translations, 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 each against the arms, and read the smallest off. The rule that the smallest box for a whole star is the star’s own lattice, and so keeps the full point group, holds for any lattice and any point group, by the same two-line argument.

The figures cannot show whether a calculation needs the whole star, or how many of its wavevectors it must then visit, which the irreducible wedge reduces again by the point group. A structure known in advance to order along one axis is well served by a box holding one arm, and the factor of four or nine is wasted. The census measures the price of not knowing, which is the usual situation when a calculation is looking for an instability rather than confirming one.

It also says nothing about the irrational wavevectors, which no box holds at all. For those the plane essay’s remedy applies unchanged: a twisted boundary condition on a small box, or a description of the structure in a space of one more dimension, where it repeats again.

Born, von Kármán, and the grids that followed #

Periodic boundary conditions were introduced by Max Born and Theodore von Kármán in 1912, for the vibrations of a crystal, as a device for making a finite count of modes come out right. The modes of the glued box are exactly the wavevectors of its dual lattice, and their paper took the box to be $N$ cells along each axis, so its wavevectors were the grid of fractions with denominator $N$.

The same grids returned when electronic-structure calculations needed to sample the zone. H. J. Monkhorst and J. D. Pack’s scheme of 1976 is a regular grid of wavevectors, chosen so that it is closed under the crystal’s point group, and the condition that a grid be closed under the point group is the reciprocal-space form of the condition here that a box keep it. Supercells built from the face-centred and body-centred sublattices are standard in practice. They are often chosen because they are more nearly spherical, and they also hold whole stars that the rectangular grids of the same size miss.

What the census must refuse #

The enumeration is checked against the divisor sum at every size from one to twelve, so a box missing from the search would show as a count one short. The cubic boxes it finds up to 64 cells must be exactly the nine at sizes $n^3$, $2n^3$ and $4n^3$ that the earlier census established. The smallest box for each star must coincide with the star’s own lattice, computed separately by counting residues. And the claim the essay refuses is refused: the two-cell box doubling one axis holds exactly one of X’s three arms.

Still open: the boxes of the hexagonal crystals #

The cube is the most symmetric case, and the factors of four and nine and sixteen come from its three-fold axes permuting the arms of each star. A hexagonal crystal has a six-fold axis and a unique direction, and its stars split differently. The edge centre M of the hexagonal zone has three arms in the plane, held one at a time by a box of two cells and together by the box two cells on each side of the hexagon. A point on the zone’s top face has arms that mix a direction in the plane with a halving along the axis, and a box can double the axis without touching the plane or the other way round. The rule that the smallest box for a whole star keeps the full point group still holds, by the same argument. How large those boxes are, and how often the hexagonal lattice’s own non-rectangular sublattices, the $\sqrt3 \times \sqrt3$ box among them, undercut the regular grids as the face-centred boxes do here, is the same census on a different lattice. It has not been run here.