Only two and three can centre a lattice

Fourteen lattices and no others derived the Bravais lattices by asking twenty-five candidate cells what symmetry they have. Each crystal system’s conventional cell was centred in each of the usual ways: P for none, A, B or C for a pair of opposite faces, $\mathrm{I}$ for the body, F for all faces, R for the rhombohedral centring of a hexagonal cell. Each candidate was then kept, found to destroy the system’s symmetry, or found to be a lattice already listed. The answer was fourteen, computed rather than quoted. The candidates themselves were a list, though. Nothing in that essay says why the letters are those seven, or whether centring by a third, a fifth or a seventh of a lattice vector might make a lattice the list does not have.

Centring, counted as a sublattice already had the idea that settles it, in the plane: a centred lattice contains the lattice of its conventional cell as a sublattice. This essay runs that idea backwards in space, for every prime at once. The centrings turn out to be the subspaces of a small vector space over a finite field, the ones a holohedry keeps can be listed completely, and the list of letters comes out rather than going in.

A centring is a subspace modulo a prime #

Let $L$ be the lattice of a conventional cell, and $L'$ a centred lattice containing it. If every vector of $L'$ becomes a vector of $L$ when multiplied by a prime $p$, then $L \subset L' \subset \tfrac1p L$. The quotient $\tfrac1p L / L$ is a three-dimensional vector space over the field of $p$ elements, and $L'$ is $L$ together with the lift of a subspace $V$ of it. A body centring is the line through $(1,1,1)$ modulo 2, the vector $\tfrac12(a + b + c)$. A C centring is the line through $(1,1,0)$. An F centring is the plane of vectors whose coordinates sum to nothing modulo 2, which holds all three face centrings at once.

The crystal system’s symmetry is its holohedry $H$, a finite group of integer matrices acting on $L$. The centred lattice keeps that symmetry when every operation of $H$ carries $L'$ to itself, which is exactly when $V$ is carried to itself by $H$ reduced modulo $p$. So the candidates for a system and a prime are the subspaces of the three-dimensional space over $\mathbb{F}_p$ that $H$ modulo $p$ leaves invariant. There are $p^2 + p + 1$ lines and as many planes, finitely many, and every one can be tested.

The seven points over two #

Modulo 2 the space has seven non-zero vectors and seven planes, the points and lines of the smallest projective plane, the Fano plane. The picture names every point. Three are the halvings $a/2$, $b/2$ and $c/2$, which give a lattice of the same kind as $L$ with one axis shortened. Three are the face centrings A, B and C, and one is the body centring I. The lines are the centrings by two independent half-vectors, three non-zero vectors each, and one of them, drawn as the circle through A, B and C, is F. The letters of the International Tables are the invariant points and one invariant line of this plane. They are not a convention.

The cube’s holohedry acts on the Fano plane by permuting the three axes, and signs disappear modulo 2. The only point every permutation fixes is $(1,1,1)$, which is I. The only line every permutation keeps, other than the ones made of halvings that no permutation keeps at all, is the circle, which is F. So the cubic holohedry allows exactly two centrings by halves, and they are $\mathrm{I}$ and F. The absence of a C-centred cube, which why there is no C-centred cubic computes as a loss of symmetry, is the statement that the point C is not fixed by the permutation that exchanges the first and third axes.

The census #

The same test runs for every system and every prime. For the rhombohedral system the holohedry is taken on the hexagonal lattice, as its twelve operations $\bar{3}m$ among the hexagonal twenty-four, so that the centring R is found where the letters put it.

The picture at the head of this essay counts them: 340 invariant subspaces over the seven holohedries and four primes. The triclinic holohedry, whose only operation besides the identity is $-I$, keeps every subspace at every prime, since $-I$ reverses every vector and keeps every line. The cubic holohedry keeps only two subspaces modulo 2 and none at all modulo 3, 5 or 7. The others lie between.

Each invariant centring still has to be identified, and a count of symmetries is not enough to do it. Body-centred and face-centred cubic lattices both have 48 symmetries, of the same kinds. The identification here uses an integral fingerprint of the centred lattice’s symmetry group, computed from its matrices in a primitive basis: every operation’s order and determinant, and the elementary divisors of $M - I$, which record how an operation acts on the lattice rather than on space. That still leaves $\mathrm{I}$ and F together, because they are dual lattices and their groups match element by element. The last ingredient separates them: the coinvariants, the lattice modulo everything of the form $(M - I)v$. For body-centred cubic that quotient has order 2 and for face-centred cubic it is trivial. The fingerprints of the fourteen Bravais lattices, taken from the independent construction, are pairwise distinct, and every one of the 340 candidates matches one of them.

What the halves make #

Modulo 2 the halvings always give a primitive lattice of the same system, a cell with one or two axes shortened, and they account for the triclinic holohedry’s fourteen invariant subspaces and most of the others. The genuinely new lattices come from the letters. For the monoclinic holohedry, with its two-fold axis along $b$, the face centrings A and C, the body centring $\mathrm{I}$ and the full centring F all give the same new lattice, the one conventionally called C. B, the centring of the face perpendicular to the two-fold axis, gives a primitive lattice again. That is the duplication fourteen lattices and no others settled by exhibiting changes of cell, found here as a match of fingerprints. For the orthorhombic holohedry, which fixes every point and line of the Fano plane because it only changes signs, all fourteen subspaces are invariant. Six give a primitive lattice: the three halvings and the three planes made of two halvings. Six give the C type: the three face centrings A, B and C, and the three planes that combine one face centring with the halving of the axis inside that face. One gives $\mathrm{I}$ and one, the circle, gives F. A, B and C are three different lattices of one type, which the fingerprint recognises and the letters record by convention as C, since relabelling an orthorhombic cell’s axes is a change of description and not of lattice. So for the orthorhombic holohedry C, $\mathrm{I}$ and F are three different lattices. For the tetragonal holohedry C gives a primitive lattice rotated by forty-five degrees and F gives I, leaving P and I. For the cubic holohedry the halves give $\mathrm{I}$ and F.

That is ten lattices from the halves: monoclinic C, orthorhombic C, $\mathrm{I}$ and F, tetragonal I, cubic $\mathrm{I}$ and F, with the triclinic, monoclinic, orthorhombic, tetragonal and cubic primitive lattices as the uncentred ones. The hexagonal holohedry keeps only two subspaces modulo 2, the halving of $c$ and the halving of the whole hexagonal net, and both give the hexagonal lattice back.

What the thirds make #

Modulo 3 the full hexagonal holohedry keeps four subspaces, and all four give the hexagonal lattice again. One of them is the line through $(1, 2, 0)$, the centring of the hexagonal net at the centres of its triangles, which gives a hexagonal net turned by thirty degrees. That the hexagonal lattice has exactly one index-three sublattice with its whole six-fold symmetry is the arithmetic three colours, and why most patterns cannot have them traced to the prime three ramifying in the Eisenstein integers. Here the same fact appears from the other side, as the one invariant line of the net modulo three.

Under the rhombohedral holohedry $\bar{3}m$ two more lines are invariant, through $(2,1,1)$ and $(1,2,1)$ modulo 3. These are the obverse and reverse R centrings, and each gives the rhombohedral lattice with its twelve symmetries. The six-fold rotation exchanges the two lines, so the full hexagonal holohedry keeps neither, and that is the whole reason R is a lattice of a lower system. Modulo 3 nothing else in any system is new. The fourteenth lattice is the only one made by thirds.

Why nothing is made by fives or sevens #

Modulo 5 and modulo 7 the lower holohedries still keep subspaces: 62 and 114 for the triclinic, 14 and 18 for the monoclinic, 6 for the orthorhombic, 2 for the tetragonal and hexagonal. Every one of them gives back a lattice of the kind being centred. The cubic holohedry keeps none.

The reason is a theorem about finite groups, which is quoted here and whose consequence the census checks for these two primes. When a prime $p$ does not divide the order of a finite group, the group’s action modulo $p$ is as well behaved as its action over the rationals: every invariant subspace has an invariant complement, which is Maschke’s theorem, and the invariant lattices between $L$ and $\tfrac1p L$ are, piece by piece, rescalings of $L$ along the group’s own invariant directions and planes. For a holohedry those are its symmetry directions and planes, the two-fold axis of a monoclinic cell or the basal plane of a tetragonal one. Centring by such a subspace shortens the lattice along that direction or in that plane, and the lattice keeps its type. A new lattice needs an invariant subspace modulo $p$ that is not a rescaling of this kind, and that can happen only when $p$ divides the group’s order. The body centring of a cube is the example at 2: the line through $(1,1,1)$ modulo 2 is invariant, while no rational line through $(1,1,1)$ is invariant under the cube’s sign changes.

The counts at 5 and 7 can be predicted from that alone, and they come out right. The monoclinic holohedry keeps its two-fold axis, the plane perpendicular to it, every line in that plane, since the two-fold rotation reverses every vector in it, and every plane containing the axis. That is $1 + 1 + (p + 1) + (p + 1) = 2p + 4$ subspaces, 14 at five and 18 at seven, as found. The orthorhombic holohedry keeps its three axes and its three coordinate planes, six at every odd prime. The tetragonal and hexagonal holohedries keep their principal axis and the plane perpendicular to it, two. The cubic holohedry keeps no rational line or plane at all, and none modulo any odd prime. The one exception to the prediction is the hexagonal holohedry modulo 3, which keeps four subspaces rather than two, and three divides its order.

The orders of the seven holohedries are 2, 4, 8, 16, 12, 24 and 48, every one a product of twos and threes. That is the crystallographic restriction at one remove. A lattice’s rotations have orders 1, 2, 3, 4 and 6, so its symmetry groups are built from twos and threes, so only the primes 2 and 3 can centre it. The same five numbers that forbid a five-fold axis forbid a centring by fifths.

The same letters in the diffraction pattern #

A centring leaves its mark on a diffraction pattern as a pattern of absences, and the census has a reciprocal reading. The reciprocal of a centred lattice is a primitive lattice with some points removed, and systematic absences are the points removed. A centring by the subspace $V$ removes exactly the reflections $h$ with $h \cdot v$ not a whole number for some $v$ in $V$, so the absences of a body centring are the reflections with $h + k + l$ odd, and those of F the reflections with mixed parities. The reciprocal of body-centred cubic is face-centred cubic and the reverse, which is why the two are a dual pair and why the fingerprint needed the coinvariants to tell them apart. In the reciprocal lattice’s terms the census is a census of invariant absence patterns, and the same two primes govern it: a crystal’s absence conditions from centring involve only parities and residues modulo three.

What the census does and does not show #

The primes stop at seven. The theorem covers every prime beyond three, and the census checks two of them. A centring whose index is not a prime, a quarter or a sixth, is a chain of prime centrings, and each step is one of those computed here.

The identification leans on the independent fourteen. The fingerprints are matched against the Bravais lattices of the earlier construction, which established them by symmetry and changes of cell. The census shows that every invariant centring is one of those fourteen and that all fourteen occur. It does not re-derive that the fourteen are distinct lattices. That the fingerprints differ is consistent with it, and the earlier essay proves it.

The metrics are generic by construction. Each holohedry is paired with a metric whose lengths have no small rational ratios, so that a centring cannot acquire extra symmetry by coincidence. The first run of this census used the metrics of the earlier construction, and its orthorhombic metric has $c^2 = 4a^2$. Halving $c$ then produced a lattice with square cross-sections, a tetragonal lattice by accident, and the census reported one among the orthorhombic centrings. A metric with $a^2$, $b^2$ and $c^2$ of 13, 29 and 71 has no such coincidences for any prime up to seven.

Who found it #

The fourteen lattices are Bravais’s, from 1848, and the letters are the International Tables’ convention. Describing centrings as invariant subgroups of $\tfrac1p L / L$ belongs to the arithmetic theory of crystallographic groups developed by Zassenhaus and his school, in which the Bravais types are the integral classes of the holohedries. The coprime-order theorem is Maschke’s, from 1898, in the form that allows coefficients modulo a prime. The Fano plane is Fano’s, from 1892, and its appearance as the geometry of face and body centrings is simply what the space of half-vectors is.

What the census has to refuse #

The test that makes the rest possible is the first, the fourteen fingerprints being pairwise distinct. Without the coinvariants it fails, for two pairs: body- and face-centred cubic, and body- and face-centred orthorhombic. Each pair consists of a lattice and one related to it by duality, and their symmetry groups cannot be told apart element by element. A census built on the weaker fingerprint would have matched every F centring to I, or the reverse, and the count would have come out twelve.

Still open: four dimensions #

The same construction runs in four dimensions, where no lattice carries all three twelves computed the symmetry groups of the six lattices whose groups are maximal. Their orders are 384, 1,152, 240, 288, 96 and 96. Every one is again a product of twos and threes, except $A_4$’s 240, which has a factor of five, because the four-dimensional restriction allows five-fold rotations. The theorem then permits centrings by fifths of $A_4$’s holohedry, and one is known: $A_4$ and its dual differ by exactly such a centring. How many four-dimensional Bravais types arise from centrings by 2, 3 and 5, and whether the census reproduces the sixty-four types of the complete enumeration, is the same calculation with four coordinates and one more prime.