The body-centred hypercube is the checkerboard lattice

Only two and three can centre a lattice derived the centring letters instead of listing them. A centred lattice $L'$ that becomes a sublattice of $L$ when multiplied by a prime $p$ is $L$ plus the lift of a subspace of $L/pL$. It keeps the crystal system’s holohedry $H$ exactly when $H$, reduced modulo $p$, keeps that subspace. Enumerating every invariant subspace for the seven holohedries and the primes up to seven gave exactly the fourteen Bravais lattices, with new ones only at 2 and 3. The reason was the crystallographic restriction one step removed. Every holohedry’s order is a product of twos and threes, and a prime that divides no group order sees only rescalings, by the theorem of Maschke.

That essay ended by pointing at four dimensions. No lattice carries all three twelves had computed the symmetry groups of six four-dimensional lattices whose groups are maximal, and one of those orders has a factor of five. $A_4$, the root lattice of the four-simplex, has 240 symmetries, because the four-dimensional restriction allows a five-fold rotation, as where five-fold becomes legal showed. The theorem then permits a centring by fifths.

This essay runs the census there, and it finds two things the three-dimensional census could not. Five does centre a lattice: $A_4$ centred by fifths is its dual, $A_4^*$, a new lattice with the same 240 symmetries. And the most familiar centring of all fails in the extra dimension. Centring the hypercubic lattice at its body centre does not make a centred hypercube. It makes $D_4$, which has three times as many symmetries, because in four dimensions the body centre lies exactly as far from a corner as the next corner along an edge.

Six groups and four primes #

The six lattices are the hypercubic lattice $\mathbb{Z}^4$ with 384 symmetries, the checkerboard lattice $D_4$ with 1,152, $A_4$ with 240, two hexagonal planes $A_2 \oplus A_2$ with 288, a hexagonal plane and a square one $A_2 \oplus \mathbb{Z}^2$ with 96, and a cube with a perpendicular line $\mathbb{Z}^3 \oplus \mathbb{Z}$, also with 96. Their groups were found by backtracking. The image of each basis vector must be a lattice vector of the same length, with the same inner products with the images already chosen.

The last two carry a free ratio of scales: how long the square is against the hexagon, how long the line is against the cube. The three-dimensional census learned to fix such ratios where nothing is a coincidence. Its first run used an orthorhombic cell with $c^2 = 4a^2$, and one centring accidentally produced a tetragonal lattice. So here the square part is taken at norm 7 beside the hexagon’s 2, and the line at norm 13 beside the cube’s 1, and the census checks that each still has exactly the symmetry group it is meant to have.

For each lattice and each prime $p$ of 2, 3, 5 and 7, the census lists every line, plane and three-dimensional subspace of $\mathbb{F}_p^4$. That is 65 subspaces for $p = 2$, 210 for 3, 1,118 for 5 and 3,650 for 7. It keeps the ones the group leaves invariant, tested against a handful of generating matrices of the group. Each kept subspace gives a centred lattice $L'$. Its basis comes from a Hermite reduction and a lattice reduction, its Gram matrix is scaled to a primitive integer matrix, and its own symmetry group is computed by the same backtracking.

The picture at the head of this essay is the census. Of the twenty-four lattice-and-prime pairs, twelve have no invariant subspace at all. The hypercubic lattice keeps nothing modulo 3, 5 or 7. $D_4$ and $A_2 \oplus A_2$ keep one subspace each, at 2 and 3 respectively, and it rescales them. $A_4$ keeps nothing modulo 2, 3 or 7 and exactly one line modulo 5. The two lattices with a free ratio keep a few subspaces at every prime, because each has rational invariant subspaces, its two parts, and halving or thirding one part is a centring that keeps the group. Only three cells hold anything new, and one of those holds something the three-dimensional census had no category for.

What each centred lattice is #

Deciding what a centred lattice is takes two different arguments, depending on the group.

A holohedry that keeps a single quadratic form up to scale is the easy case. $\mathbb{Z}^4$, $D_4$ and $A_4$ each keep one, as the census checks by solving the linear equations $M^{\mathsf{T}}XM = X$ for a symmetric $X$. For such a group, two lattices whose holohedries are conjugate by an integer change of basis must be similar. The conjugating matrix carries one lattice’s invariant form to a form the other group keeps, and there is only one such form up to scale. So for these three, a centred lattice with the same number of symmetries that is not similar to the original is a new Bravais type, proved and not merely suspected. Similarity is decided by searching for an explicit isometry between the two Gram matrices once both are scaled to primitive integer matrices. When a map is found the two are the same lattice, and when the exhaustive search finds none they are not.

The three lattices that keep two forms need the other argument used in three dimensions, an integral fingerprint of the symmetry group. For every element it records the order, the determinant, the trace and the elementary divisors of $M - I$, and it adds the coinvariants, the lattice modulo everything of the form $(M - I)v$. Different fingerprints prove different types. Equal fingerprints are read as the same type, which in every case here is a centring that rescales one part of a decomposable lattice, the same type by inspection.

The body centre is as far as an edge #

The hypercubic lattice modulo 2 has the same structure as the cube had. Its symmetries permute the four coordinates and change their signs, and signs disappear modulo 2, so the invariant subspaces of $\mathbb{F}_2^4$ are those the permutations keep: the line through $(1,1,1,1)$, the three-dimensional space of vectors with an even number of ones, and nothing else. In three dimensions the corresponding two subspaces were the body centring $\mathrm{I}$ and the face centring F. In four dimensions the census finds that both give the same lattice, and neither is a centred hypercube.

The reason is a length. The body centre of the unit hypercube is $\tfrac12(1,1,1,1)$, and its squared length is $4 \times \tfrac14 = 1$, exactly the squared length of an edge. In three dimensions the body centre $\tfrac12(1,1,1)$ has squared length $\tfrac34$, shorter than an edge, and the body-centred cubic lattice has eight shortest vectors, all to body centres. In four dimensions the sixteen vectors $\tfrac12(\pm1, \pm1, \pm1, \pm1)$ are as short as the eight edge vectors $\pm e_i$. The centred lattice has twenty-four shortest vectors, the kissing number of $D_4$, and they are the vertices of the 24-cell. In four dimensions it plays the part $E_8$ plays in eight. $E_8$ is proved to win the Gaussian sum at every width, and for $D_4$ the same is so far only measured. The picture counts the shells. The hypercube has 8, 24, 32 and 24 vectors in its first four shells. The centred lattice has 24, 24, 96 and 24, identical shell for shell with $D_4$, and the census finds an explicit isometry between the two.

Symmetries that mix edge vectors with half-diagonal vectors now exist, and the group grows from 384 to 1,152, the Weyl group of $F_4$. So the body-centred hypercube is not a centring of the hypercubic system. It belongs to a different crystal system altogether, the one $D_4$ defines. The even-weight centring, the four-dimensional analogue of F, gives $D_4$ again: it is $\tfrac12 D_4$, since $D_4$ is by definition the integer vectors with even coordinate sum. The hypercubic holohedry keeps two centrings by halves, and both leave its system. In three dimensions the corresponding statement was that a C-centred cube is not cubic, which forty-eight becomes sixteen computed as a loss of symmetry. Here $\mathrm{I}$ and F are not hypercubic, and the reason is the opposite, a gain of symmetry.

$D_4$ itself keeps one subspace modulo 2. It is the plane of half-vectors that makes the dual $D_4^*$, and $D_4^*$ is similar to $D_4$, so that centring only rescales. $D_4$ and $\mathbb{Z}^4$ are the two lattices where four dimensions differs most visibly from three. $\mathbb{Z}^4$’s centrings turn it into $D_4$, and $D_4$ is its own dual up to scale, which the three-dimensional centred cubic lattices are not: body-centred and face-centred cubic are each other’s duals.

Five centres a lattice #

$A_4$ behaves differently at every prime. Its order, 240, is $2^4 \times 3 \times 5$, so the theorem allows a new lattice at 2, 3 or 5. The census finds nothing modulo 2 or 3. The symmetric group on five letters acts on $A_4$ modulo 2 and modulo 3 with no invariant line or plane at all. Modulo 5 it keeps exactly one line, the one spanned by $(1, 2, 3, 4)$ in the basis of simple roots. That is the familiar fact that $A_4^*/A_4$ has order 5 and is generated by one fundamental weight. Centring by it gives $A_4^*$, the dual, which the dual lattice defines as every vector whose inner product with the lattice is whole.

The census confirms that $A_4^*$ has the same 240 symmetries, and that the exhaustive isometry search finds no map between it and $A_4$ at any scale. Since $A_4$’s group keeps a single form, that makes $A_4^*$ a new Bravais type. In four dimensions a lattice can be centred by fifths, and that is the only centring $A_4$ admits. The three-dimensional census found that its fourteen lattices were made by halves and thirds because every holohedry’s order was a product of twos and threes. Four dimensions allow a five-fold rotation and so a holohedry whose order has a five, and five then does exactly what the theorem allowed it to.

Why five, and only five, for $A_4$ #

The census’s answer for $A_4$ has a reason that is easy to see once $A_4$ is written in five coordinates. $A_4$ is the set of integer vectors in $\mathbb{Z}^5$ whose coordinates sum to zero, and its symmetries are the permutations of the five coordinates, with a sign. Modulo a prime $p$ that makes $A_4/pA_4$ the space of vectors in $\mathbb{F}_p^5$ with coordinate sum zero, permuted by the symmetric group on five letters. A line that every permutation keeps must be spanned by a vector that every permutation fixes up to sign. The only such vectors are the multiples of $(1, 1, 1, 1, 1)$. That vector lies in the sum-zero space exactly when $5 \equiv 0$ modulo $p$, which means $p = 5$.

For every other prime the sum-zero space has no invariant line, and it has no invariant plane or three-space either. This is the standard fact that the permutation module splits off its sum-zero part cleanly, with nothing further inside, when $p$ does not divide the number of coordinates. The census checks it at 2, 3 and 7. At 5 the kept line is exactly $(1,1,1,1,1)$. In the basis of simple roots, $e_1 - e_2$ through $e_4 - e_5$, the census reports it as $(1, 2, 3, 4)$, and $(1,2,3,4)$ in that basis is $(1, 1, 1, 1, -4)$ in coordinates, which is $(1,1,1,1,1)$ modulo 5.

The same argument runs for every root lattice of this kind, and it makes sense of pairs already met. $A_n$ sits in $n + 1$ coordinates and is centred into its dual by the primes dividing $n + 1$. The hexagonal net is $A_2$, with three coordinates, and the three-dimensional census found its one invariant line modulo 3, the centring that turns the net into itself rotated. Face-centred cubic is $A_3$, with four coordinates, and its dual, body-centred cubic, is reached from it by two centrings by halves, since $A_3^*/A_3$ is cyclic of order four. $A_4$, with five coordinates, is the first case in which the prime is one no three-dimensional crystal can use.

That $A_4$ and $A_4^*$ are not similar is a four-dimensional fact that the plane does not share. In the plane every lattice is similar to its dual, which every plane lattice is its own dual proved. In three dimensions, body-centred and face-centred cubic are each other’s duals and are different lattices, a pair made by 2. In four dimensions $A_4$ and $A_4^*$ are such a pair made by 5, and the two lattices where a four-dimensional descent stops short found them as the two places a search for good packings can get stuck. The pair is a genuinely separate pair of lattices with one symmetry group between them.

The cube and a line #

The lattice $\mathbb{Z}^3 \oplus \mathbb{Z}$ is a cube with a line perpendicular to it, and its 96 symmetries are the cube’s 48 times the reversal of the line. Modulo 2 it keeps eight subspaces. Two rescale it. One halves the line and the other halves the cube, which gives the same type since the ratio of scales is free. The other six make three new Bravais types, two subspaces each, and the fingerprints separate all three from each other and from the original.

Two of the three are the cube’s own centrings with the line carried along: the body-centred cube beside a line, and the face-centred cube beside a line. They are the three-dimensional $\mathrm{I}$ and F one dimension up. The third is new to four dimensions. It is the centring by $\tfrac12(1, 1, 1, 1)$, which mixes the cube’s body diagonal with the line. With the line at a generic length it is not the $D_4$ accident. It is a lattice with exactly the 96 symmetries of the cube and the line, and a fingerprint of its own. Modulo 3, 5 and 7 the lattice keeps only the two rational subspaces, the cube and the line, and centring by either rescales that part.

A prime must divide the group, and need not help #

The census confirms the theorem’s direction everywhere. At every prime that does not divide a group’s order, every invariant centring gives the type being centred. Seven divides none of the six orders, and nothing modulo 7 is new. Five divides only 240, and the only centring by fifths is $A_4$’s.

The converse fails often, and the table shows how. Two and three divide every one of the six orders, 384, 1,152, 240, 288, 96 and 96. Yet two makes new types only for the cube and a line, and three makes none at all among these six. In three dimensions three made the rhombohedral lattice, by centring the hexagonal lattice under its subgroup $\bar{3}m$ rather than under its full holohedry. The census here uses only the six maximal groups. The four-dimensional analogues of the rhombohedral lattice, if there are any, would be found under subgroups of these groups and are not in this count.

What the census does and does not cover #

It covers six groups, not every four-dimensional holohedry. Four-dimensional lattices have many more crystal systems than the six whose groups are maximal. The complete classification, by Brown, Bülow, Neubüser, Wondratschek and Zassenhaus in 1978, is far larger than the list examined here. A census over every four-dimensional holohedry would need those groups built first, and it would find the centrings that only a subgroup keeps, which is where three dimensions found R.

Two ways of deciding type, one of them a proof. For $\mathbb{Z}^4$, $D_4$ and $A_4$ a centred lattice is identified by an explicit isometry or shown to be new by the failure of an exhaustive one. With a single invariant form, that settles the Bravais type. For the three decomposable lattices, different fingerprints prove types different, but equal fingerprints are only evidence of the same type. In every such case here the centring rescales one of the parts, and the sameness can be seen without the fingerprint.

The primes stop at seven. Past seven no prime divides any of the six orders, so the theorem already says every invariant centring rescales. The census checks this for 5 and 7 where it applies.

Who worked out the pieces #

The four-dimensional crystallographic groups, including the Bravais types, were classified by Harold Brown, Rolf Bülow, Joachim Neubüser, Hans Wondratschek and Hans Zassenhaus and published in 1978. The lattices $D_4$ and $A_4$ with their duals are standard in the theory of root systems and sphere packings, and Conway and Sloane’s Sphere Packings, Lattices and Groups treats both. That $\mathbb{Z}^4$ with its body centres added is the $F_4$ root lattice, $D_4$ rescaled, is classical, and it is why the 24-cell exists as a regular polytope with no analogue in three or five dimensions. The coprime-order theorem is Maschke’s.

What the census has to refuse #

The refused claim is one the construction cannot produce. A centring by a subspace the holohedry keeps gives a lattice that the whole holohedry still maps to itself, so its symmetry group can only be the same or larger. The census checks that none of the invariant centrings, at any prime, has fewer symmetries than the group it was built with. A failure there would mean the invariance test or the symmetry search was wrong, and nothing in the table would reveal it.

Still open: the subgroups, and the fifths #

The census found $A_4^*$ as the only centring by fifths among six maximal groups. Five also divides the orders of subgroups of $A_4$’s group, among them the cyclic group of order five generated by one five-fold rotation, and the dihedral groups containing it. A lattice whose holohedry is one of those subgroups would have its own centrings modulo 5, and nothing here says whether any of them is new. That is the question of four-dimensional lattices with a five-fold rotation that are not $A_4$ or $A_4^*$, the kind of lattice on which the four-dimensional description of a Penrose tiling is built. Running the census over the subgroups of the six groups would answer it, and would reach the four-dimensional rhombohedral analogues at the same time.