The three that stay cubic

How many ways there are to thin a lattice answers its question in the plane and gets a clean number: the sublattices of index n are counted by the sum of the divisors of n, and they can be written down one per Hermite normal form. Which of them keep the square symmetry answers the harder question, and gets Fermat: a square sublattice of the square lattice exists exactly at the indices that are sums of two squares.

Space is the same two questions with different arithmetic, and both answers change more than the extra coordinate suggests. The counting formula grows a factor, and the symmetric ones become rare — so rare that the entire list up to index eighty has nine entries in it, and the nine entries are the three cubic Bravais lattices.

What a third coordinate does to the count #

A sublattice is the column span of an integer matrix, two matrices span the same sublattice exactly when they differ by a change of basis, and every class has one representative in Hermite normal form — upper triangular, positive diagonal, off-diagonal entries reduced. That is what makes the sublattices of index n a list rather than a family.

In the plane a Hermite form is [[a, b], [0, d]] with ad = n and b below a, so the count is Σa over the factorisations, which is the sum of the divisors. In space it is a three by three form with two more free entries, and the count is the sum of a²d over the factorisations adf = n.

Both counts have a compact statement as a Dirichlet series: the sublattice-counting function of ℤ^d has series ζ(s)ζ(s−1)···ζ(s−d+1). One zeta in one dimension gives one sublattice at every index; two in the plane give the divisor sum; three in space give a longer sum. That formula and the enumeration are computed here by routines that share no code, and the build stops if any row disagrees.

The gap between the two dimensions is larger than an extra factor suggests. At index sixteen the plane has thirty-one sublattices and space has six hundred and fifty-one — twenty-one times as many, from one more coordinate — and the ratio grows.

The growth is worth a sentence because it is what makes the rest of this essay surprising. Summed over all indices up to N, the number of sublattices of a plane lattice grows like N², and of a space lattice like N³ — one power per dimension, with the zeta values as the constants. So by index sixteen there are already about two thousand sublattices of ℤ³ to sort through, and the number that keep the parent’s symmetry has reached four.

The reduction ranges, and a lattice that was invisible #

Building this census turned up an error in the enumeration that had been in the file since it was written, and the shape of it is worth recording because it is the kind that hides behind a correct total.

An off-diagonal entry of a column Hermite form is reduced against the pivot of the column to its left, not against the diagonal entry in its own column. The file had it the other way. Sum d·f² over the factorisations of n and sum a²·d over the same factorisations, and the two totals are equal — the same product relabelled — so the enumeration produced exactly as many matrices as the formula demanded and agreed with it at every index tried.

It produced the wrong matrices, though: duplicates and omissions in equal numbers. And the omission that mattered is exactly the object this essay is about. The face-centred cubic lattice sits inside the primitive one at index two, and its Hermite form is [[2,1,1],[0,1,0],[0,0,1]] — a 1 where the old ranges permitted only a 0. So the census of cubic sublattices returned the primitive ones and nothing else, and two of the three lattices in the title were simply not in the list being searched.

A count that agrees with its formula is not a list that is right. The check the file already had was the strongest one available at the time and it could not have caught this; what caught it was asking the list a question the count cannot answer.

Which of them keeps the symmetry #

The test is the same integer condition the plane uses, and it is worth saying again that it has no metric in it. A sublattice H·ℤ³ is carried onto itself by an operation M exactly when H⁻¹MH has integer entries. No distance, no angle, no tolerance — a division of integers that either comes out whole or does not.

Run it against all forty-eight operations of the cubic group and the answer is stark.

Nine indices in eighty. One sublattice at each — never two, never none-then-two. And when the shape of each is read off, the list is not nine things but three things at four sizes.

Reading the shape off the shortest vectors #

A sublattice arrives as a matrix, and a matrix is a basis rather than a lattice. Naming the shape needs a quantity that does not depend on which basis was written down, and the shell of shortest vectors is one: how many vectors have the shortest length in the lattice.

Six of them means a primitive cube — the six face neighbours. Twelve means face-centred, which is the twelve nearest neighbours of a close packing. Eight means body-centred, the eight corners of a cube seen from its middle. Those counts are properties of the lattice, so two different Hermite forms describing the same sublattice give the same answer and no basis is privileged.

The correspondence is exact and the arithmetic behind it is short. Scaling the whole lattice by m gives a primitive sublattice of index m³. The face-centred lattice of cube edge m has four points per cube and therefore index m³/4 — but written as a sublattice of the unit cubic lattice it is the one with index 2m³. The body-centred one is 4m³. Those three arithmetic progressions are the whole of the nine-entry list, and no fourth progression appears.

Which is the fourteen Bravais lattices’ cubic row, obtained by a completely different question. The fourteen are reached by taking each holohedry, trying every centring on it, and discarding the candidates that are duplicates or that destroy the symmetry — an enumeration of cells. Here nothing is centred and no cell is chosen: sublattices of one lattice are listed, and the ones that keep its symmetry are kept. The two routes land on the same three, and there is still no C-centred cubic because there is still no fourth arithmetic progression.

Almost none of them keeps anything #

The complement of that list is where most of the census lives, and it says something about how special a symmetric sublattice is.

The first thing to notice is how quickly the cubic column empties. At index one, the lattice itself is the only sublattice and it keeps everything. At index two, one of seven keeps everything. At index four, one of thirty-five. At index sixteen, one sublattice of six hundred and fifty-one keeps all forty-eight operations. Twenty-four keep sixteen, which is the tetragonal amount; a hundred and twenty-six keep eight; three hundred and twenty-four keep four; and a hundred and sixty-eight keep two, which is the identity and the inversion — the least any lattice can keep, since every lattice is symmetric under negation.

The orders that occur are 48, 16, 12, 8, 6, 4 and 2, and they are the orders of subgroups of m3̅m, which is the least the table could satisfy and is checked rather than displayed.

One of them is worth reading twice. Order six is 3̅, and 3̅ is not a holohedry: no lattice in three dimensions has 3̅ as its full symmetry group. There is no contradiction, and the resolution is in what is being counted. The column reports the intersection of a sublattice’s symmetry with the parent’s, and a sublattice can have symmetry the lattice it sits in does not. A hexagonal sublattice built on the {111} planes of a cubic lattice has 6/mmm as its own holohedry; only twelve of those twenty-four operations lie in m3̅m, and a sublattice keeping fewer still reports six.

That is a distinction the plane’s version of this file never had to make, because in the plane the parent lattice’s holohedry is usually the ceiling. In space it is not, and reporting an intersection as a holohedry would be a wrong answer with a plausible-looking order attached to it.

The two that are nearly cubic #

The rows just to the right of the cubic column are the ones a crystallographer meets, and they are worth naming because they are how a cubic structure becomes something else.

Order sixteen is 4/mmm, the tetragonal holohedry, and those sublattices are the ones that keep one four-fold axis and lose the other two. There are twenty-four of them at index sixteen and six at index two — and the six at index two are the obvious thing: thin the lattice along one axis and the cube becomes a box twice as tall. Every ordered structure with a doubled axis, every tetragonal distortion of a cubic parent, is one of these.

Order twelve is 3̅m, the rhombohedral holohedry, and those keep one of the four body diagonals and lose the other three. There are four at index three, which is the count anyone who has thought about the {111} planes of a cubic lattice would predict: four diagonals, one sublattice each.

Between them those two orders account for most of the symmetric sublattices at low index, and the pattern is the one the descent of symmetry always has — a cubic parent loses symmetry by keeping one special direction, and which direction it keeps is the whole of the choice. That is the same accounting the descent of symmetry makes for point groups, appearing here for lattices, with the four body diagonals and the three axes as the objects being chosen among.

Where the two dimensions do not rhyme #

The plane’s version of this question has a beautiful answer, and space’s is beautiful in a different way, and the difference between them is instructive.

In the plane, square sublattices are common: they exist at every index that is a sum of two squares, which is roughly three indices in every four asymptotically thinned, and there can be several at one index — index five has two, index sixty-five has four. The counting is a divisor sum, so the number grows.

In space, cubic sublattices are rare and unique: nine indices in eighty, one at each. The counting is not a sum over anything; it is a question of whether n is a cube, twice a cube or four times a cube.

The reason for the difference is that the plane’s square lattice has a rotation available — multiplication by i in the Gaussian integers — and a sublattice closed under it is an ideal of ℤ[i], of which there are as many as there are factorisations. The cubic group has no such single generator: it is not the multiplicative structure of a ring acting on the lattice, and a cubic sublattice cannot be built by multiplying by anything. So the plane’s count is arithmetic and space’s is geometric, and the two questions look identical and are not.

Why 1, 2 and 4 and no other multiplier #

The three arithmetic progressions can be read off the answer, and it is worth saying where they come from, since the search reports them and does not explain them.

A lattice invariant under the full cubic group is a cubic lattice, and there are exactly three cubic lattices up to scale. That much is the fourteen and is not in question here. What this essay adds is at which indices each of them sits inside a primitive cubic lattice, and that is a determinant.

Take the primitive cubic lattice of edge one as the parent. A primitive cubic sublattice of edge m has determinant m³, so index m³. A face-centred cubic lattice whose conventional cube has edge m has four lattice points per cube, so its primitive cell has volume m³/4 — but it is only inside the parent when m is even, and writing m = 2j gives volume 2j³, so index 2j³. The body-centred one has two points per cube, primitive volume m³/2, again needing m even, giving index 4j³.

So the multipliers 1, 2 and 4 are the three primitive-cell volumes of the three cubic lattices at the smallest size each can have inside the parent, and the cubes are the scalings. There is no fourth multiplier because there is no fourth cubic lattice, which is the statement the enumeration up to eighty is a check on rather than a proof of.

What the census refuses #

The one worth naming is the third. There must be indices at which no cubic sublattice exists — and there are eleven of them below sixteen. A construction that reported a cubic sublattice at every index would be one whose integrality test had a bug in the permissive direction, and every other check on this page would still pass. Asserting that something is impossible is the only way to catch that, and it is the same shape of check as the plane’s assertion that indices 3, 7 and 11 have no square sublattice.

Where the exactness stops #

Computed here: every sublattice of index up to sixteen in three dimensions as a Hermite normal form; the same counts from the Dirichlet coefficient; the cubic operations each one keeps, by integer division; the shape of each fully symmetric sublattice from its own shell of shortest vectors; and the list of indices up to eighty at which one exists.

One parent lattice, and it is the cubic one. Everything above thins ℤ³ with the cubic metric. The same machinery runs on any lattice — the integrality test does not know what metric is being used, because it does not use one — but the shapes that come out and the indices at which they occur are facts about the cubic lattice. A hexagonal parent has a different list and it is not computed here.

Shortest-vector counts name three shapes and would not name fourteen. Six, twelve and eight separate the primitive, face-centred and body-centred cubic lattices because those are the three cubic lattices and nothing else is in the running once all forty-eight operations are known to survive. Used as a general classifier the count would fail: two lattices of different systems can have the same number of shortest vectors. It is a legitimate label here and would not be one for the whole census.

A sublattice is not a structure. Everything counted here is a set of points, and what a crystallographer usually wants to know is what happens to a pattern when its lattice is thinned — which atoms become inequivalent, which reflections appear. That is a different question with a different answer, and it is the two kinds of subgroup rather than this one; the lattice census is the part of it that has no motif in it.

The bound is a bound. The list of indices runs to eighty because that is where the enumeration stops being cheap — index eighty has tens of thousands of Hermite forms. The pattern m³, 2m³, 4m³ continues, and a proof that it continues is not in this file: what is here is a complete search up to a stated bound, which is what this collection means by an enumeration rather than an argument.

Where the ladder goes next #

Back, to the plane’s version and the arithmetic it turns on: how many ways there are to thin a lattice, and the sublattices that stay square, where Fermat decides the answer.

Sideways, to the same three lattices reached by enumerating centrings instead of sublattices: the fourteen, and the reason the list has no C-centred cubic in it.

And onward, to what a thinned lattice does to a pattern rather than to a lattice: every way down, and no way round, where the two kinds of subgroup a plane group has are separated, and the reflections a superlattice adds, which is what a diffraction pattern makes of one.