Forty-eight becomes sixteen

Three of the four ways to centre a cube keep it cubic and one does not. This is one row of the Bravais enumeration, taken slowly, because it is the row where the enumeration’s method is easiest to see. Which one is not obvious from looking, and the difference between the survivors and the casualty is a number.

What the cube’s forty-eight are #

Before asking what centring destroys, it is worth having the thing being destroyed clearly in view.

A cube has forty-eight symmetries and they factor neatly. There are twenty-four rotations: the identity; three fourfold axes through opposite face centres, giving nine non-identity rotations; four threefold axes through opposite corners, giving eight; and six twofold axes through opposite edge midpoints, giving six. One plus nine plus eight plus six is twenty-four. Every one of them can then be composed with the inversion, doubling the count to forty-eight.

In the lattice basis these are the signed permutation matrices — computed here rather than listed, by the same bounded search that produces every holohedry on this site: the six ways to permute three coordinates, times the eight ways to attach signs. Six times eight is forty-eight, and that is the cubic holohedry written as arithmetic. The fourfold axes are the permutations of order four with a sign; the threefolds are the cyclic permutations, which send a to b to c and back.

The threefolds are the whole of what makes a lattice cubic. A lattice with three mutually perpendicular fourfold axes and nothing else does not exist; put a fourfold on c and a fourfold on a and their composition generates the threefolds automatically. Conversely, take away the threefolds and the fourfolds cannot all survive — which is exactly what the C-centring does.

The computation #

The four candidates in the cubic system, with the order of each resulting lattice’s automorphism group:

P — the plain cube. 48. I — a point at the body centre. 48. F — a point at the centre of each face. 48. C — a point at the centre of one pair of opposite faces. 16.

Sixteen is the tetragonal number. So a C-centred cubic cell describes a tetragonal lattice, and tetragonal lattices are already on the list; the candidate is not a fifteenth Bravais lattice because the thing it produces is not new and is not cubic.

Why the threefolds go #

The argument in words takes one sentence and it is worth having, because the number alone does not explain anything.

A threefold axis along a body diagonal cycles the three coordinate axes: a → b → c → a. For that to be a symmetry of the lattice, the lattice must look the same along all three directions.

C-centring puts a point at (½, ½, 0). Apply the threefold: it goes to (0, ½, ½). Is that a lattice point? Only if the lattice is also A-centred, and it is not — C-centring centres one pair of faces and leaves the other two pairs alone. So the threefold maps a lattice point to a non-lattice point, and it is not a symmetry.

The same argument passes for body centring and for face centring, and the difference is exactly why.

I-centring puts a point at (½, ½, ½), which is symmetric in the three coordinates. The threefold cycles it to (½, ½, ½) — itself. Nothing to check.

F-centring puts points at (½, ½, 0), (½, 0, ½) and (0, ½, ½), which is the whole orbit of (½, ½, 0) under the threefold. The threefold permutes the three centring vectors among themselves, so the set is preserved.

So the rule is that the centring vectors have to form a set the point group permutes. That is a general statement, it applies to every system, and it is the reason the surviving centrings are the ones that look symmetric in the coordinates: the body centre, which is fixed, and the full set of face centres, which is an orbit. A single face centre is neither.

The general rule, and the other two casualties #

The observation above generalises and it is the cleanest statement of the whole Bravais enumeration:

A centring survives when its set of centring vectors is closed under the point group.

That is one sentence, it is checkable by hand for every candidate, and it accounts for all three failures in the enumeration and every one of the survivors.

The two hexagonal casualties fall to it immediately. A hexagonal cell’s point group contains the threefold that cycles a → b → −(a+b). Apply it to the C-centring vector (½, ½, 0): the result is (−½, 0, 0) plus a lattice vector, which is (½, 0, 0) — not a lattice point of a C-centred hexagonal cell. So the threefold fails and the twenty-four drops to eight.

I-centring in a hexagonal cell fails for the same reason with an extra step: (½, ½, ½) goes to (½, 0, ½), which is again not present.

Applied to the survivors the rule is equally quick. I in any system puts a point at (½, ½, ½), which every point group of every system fixes, because it is the “all coordinates equal” point and the point groups act by permutations and sign changes that fix it modulo the lattice. F in the cubic system is the orbit of a face centre, and in the orthorhombic system every centring survives because the orthorhombic point group only changes signs and never permutes — so it fixes each of (½,½,0), (½,0,½) and (0,½,½) individually, and any subset of them is closed.

That last observation explains why orthorhombic contributes four of the fourteen and no other system contributes more than three: it is the system whose point group does the least, so it destroys the least.

The B and A cases, which are the same case #

A reader who has been counting will notice that the enumeration asks about four centrings and there are six single-face centrings available: A, B and C each centre one pair of faces, and each could in principle be tried.

In the cubic system they are the same question. The three axes of a cube are interchangeable — the point group permutes them — so centring the ab faces, the bc faces or the ac faces produces three lattices related by relabelling, and testing one tests all three. The enumeration asks about C and takes A and B as the same row, which is a use of exactly the argument that makes sixteen candidates into ten groups elsewhere in this collection.

In the monoclinic system they are not the same question, and that is where the letters earn their existence. A monoclinic cell has one distinguished axis and two ordinary ones, so centring the face containing the unique axis is a different operation from centring the face perpendicular to it. The Tables’ convention picks C, and A and B appear in the literature as alternative settings of the same lattice — which is why monoclinic groups have three or four symbols each and orthorhombic ones have one.

What the sixteen actually is #

Saying “sixteen is the tetragonal number” leaves something out, and it is worth chasing down because it makes the whole thing concrete.

A C-centred cubic cell with edge a — points at the corners and at the centres of the top and bottom faces. Take a new cell with a′ = (a + b)/2, b′ = (−a + b)/2, c′ = c. The first two are half-diagonals of the centred face, which are lattice vectors precisely because the face is centred.

The new cell has a′ and b′ of length a/√2 and c′ of length a. Two equal edges and one different, all perpendicular: a primitive tetragonal cell with a c/a ratio of √2.

That figure is worth reading twice, because it separates two things the word “cubic” runs together. The shape of a cell and the symmetry of a lattice are different questions. The primitive cell drawn inside the C-centred cube has a tetragonal metric and the lattice has sixteen symmetries, and those two facts agree. They do not always: the primitive cell of a face-centred cubic lattice is a rhombohedron, its metric has rhombohedral shape, and the lattice has forty-eight symmetries. A cell shaped like one system can perfectly well describe a lattice belonging to another, which is why nothing in the enumeration is settled by looking at a box.

So the C-centred cube is not merely “tetragonal, somehow”. It is a specific tetragonal lattice — the one with c/a = √2 — and it is the same lattice-type as any other primitive tetragonal lattice, because tetragonal does not constrain the ratio.

The plane’s version, which is smaller and clearer #

The plane has this too and it is worth the comparison, because there the whole thing fits in one’s head.

Four plane systems and one centring operation. Centring an oblique cell gives an oblique lattice, so nothing is gained. Centring a square cell gives a square lattice turned through 45° and shrunk — the same transformation as above, one dimension down — so again nothing. Centring a hexagonal cell breaks the threefold, for the same reason C-centring breaks a cube’s: the centring vector is not in an orbit of the point group.

Centring a rectangular cell is the one case that gives something new, and it is the centred rectangular lattice this site has already met. The result has the rectangular lattice’s four symmetries still, and no change of cell makes it primitive rectangular, so it is a fifth lattice — and its existence is the reason cm and cmm are groups and the plane has seventeen rather than thirteen.

Four systems, one survivor: five plane lattices. Seven systems, seven survivors: fourteen. The pattern is that centring almost never gives anything new, and the exceptions are the interesting rows.

The one thing this does not explain #

The rule accounts for which centrings survive. It does not, on its own, account for the duplicates — the eight candidates that keep the symmetry and turn out to be a lattice already counted — and it is worth being clear that these are two different arguments.

Tetragonal C keeps all sixteen symmetries. Its centring vector (½, ½, 0) is fixed by every tetragonal operation, because the fourfold along c sends (½, ½, 0) to (−½, ½, 0), which is (½, ½, 0) plus the lattice vector (−1, 0, 0). So the closure rule passes and the candidate survives it.

It is nonetheless not a new lattice, and finding that out requires the change of cell rather than a symmetry count. The 45° turn produces a primitive tetragonal cell, and no amount of examining what the point group does to the centring vectors would have revealed it.

So the enumeration needs both halves: a computation of what survives, and an exhibition of what duplicates. Three failures by the first, eight by the second, fourteen left.

Why the picture cannot show this #

The four cubic candidates in the first figure are drawn correctly and no amount of looking at them distinguishes the one with sixteen symmetries from the three with forty-eight.

That is not a failure of the drawing. It is a property of the thing being drawn. Symmetry is a statement about the whole infinite lattice, and a picture shows one cell; the missing threefolds are missing because a distant point fails to map correctly, and no cell-sized picture contains the evidence.

The nearest a drawing gets is to show the failure directly: mark the point (½, ½, 0), mark where the threefold sends it, and show that the destination is empty. That is a picture of one operation failing rather than of a symmetry group, and it is a different kind of figure from the ones in this collection — which are pictures of things that are true.

This is the case this site was built for. Every claim here is decidable, so the honest response to it cannot be seen is to compute it, print the number under the picture, and let the number carry the content. The four cubes above look alike and read differently, and the reading is where the argument is.

What the three casualties have in common #

Set the three failures side by side and there is a pattern that says something about which systems are fragile.

Cubic C, hexagonal C, hexagonal I. All three are failures of a threefold axis, and there is no failure anywhere in the enumeration of a fourfold, a sixfold or a mirror.

The proportions are worth noticing. A threefold axis is not one operation but a factor: removing it from the cubic holohedry takes forty-eight to sixteen, and from the hexagonal holohedry twenty-four to eight. In both cases the survivor is a third of what there was, and in both cases the third that survives is the part of the group that never mixed the three axes in the first place.

The reason is that a threefold axis is the only crystallographic operation that genuinely permutes coordinate directions rather than reflecting or reversing them. A fourfold along c swaps a and b with a sign; a mirror negates one coordinate; the inversion negates all three. Every one of those maps a set of half-integer vectors to another set of half-integer vectors in the same directions, so a centring built from halves survives them almost automatically.

A threefold sends a to b to c — three different directions — and a centring that treats those directions unequally cannot survive it. C-centring treats one pair of faces differently from the other two, which is exactly the inequality a threefold cannot tolerate.

So the two systems with threefolds are the two that lose candidates, and the fragility is not about how much symmetry a system has but about what kind. Hexagonal has twenty-four operations and loses two candidates; orthorhombic has eight and loses none.

Where it stops #

Two limits.

The verdict is about the lattice, not the crystal. A crystal whose lattice is the C-centred-cubic-that-is-really-tetragonal will have tetragonal symmetry at best, and the gap between what a lattice permits and what a crystal has is the space group’s whole subject, and might have less — the lattice’s symmetry is an upper bound on the crystal’s, never a guarantee. Everything in this essay is about the point set of the lattice and says nothing about what sits on it.

The verdict is about this system’s target, not about lattices in general. A candidate that fails here may be the right answer somewhere else. C-centred cubic fails as a cubic lattice and succeeds as a tetragonal one, and hexagonal C fails as hexagonal and succeeds as orthorhombic. Every failure in this enumeration is a success in another row, which is what “already on the list” means and why the total comes out at fourteen rather than eleven.

“Destroyed” is a word about a target, not about a lattice. The C-centred cubic lattice is not damaged; it is a perfectly good tetragonal lattice with sixteen symmetries. What failed was the attempt to describe it as cubic, and calling that a destruction is shorthand for “did not meet the criterion the row was testing”. A reader who took the word literally would conclude that some lattices are broken, and none are.

The next rung is the one case in the whole enumeration where a centring lowers a system’s symmetry and the result is still a new lattice — which is R, and it is the reason the fourteen include a lattice whose conventional description is on somebody else’s axes.

The enumeration, as two tests in order #

The essay’s closing observation — that the classification needs both halves — is worth turning into the procedure, because the procedure is short and it is what produces fourteen.

For each system and each candidate centring, ask two questions in order.

First, is the centring vector set closed under the point group? If not, the resulting lattice has a smaller automorphism group than the system’s, so it belongs to a different system and the candidate is not a lattice of this one. That is the test this essay is about, it is a finite check over at most forty-eight matrices, and it disposes of C-centred cubic, of the hexagonal casualties, and of nothing else.

Second, is the surviving lattice new? Reduce it — take the shortest basis, compute the reduced metric — and compare against the lattices already on the list. If the reduced form matches one, the candidate is that lattice described on different axes and contributes nothing. That test disposes of face-centred tetragonal, of C-centred cubic’s fate as a tetragonal lattice, and of several others.

Neither test alone gives fourteen. The first is about symmetry and returns the system; the second is about descriptions and returns whether anything new has been produced. A candidate can pass the first and fail the second — tetragonal F is the standing example, keeping every one of its sixteen operations and being a body-centred tetragonal lattice on smaller axes.

Frankenheim’s fifteen #

The two-test structure is the reason the classification was got wrong the first time, and the mistake is instructive enough to be the standard cautionary tale of the subject.

Moritz Frankenheim enumerated the lattices in 1842 and found fifteen. His list was not careless: every entry was a genuine lattice, every symmetry count was right, and the argument was the one above’s first half run thoroughly.

Two of his fifteen were the same lattice. They differed by a change of cell — one described on axes the other did not use — and no amount of counting symmetries reveals that, because the two descriptions have the same symmetry by construction. The first test cannot see a duplicate, and Frankenheim had not run the second.

Bravais found the duplication in 1848 and the number has been fourteen since, which is why the lattices carry his name and not Frankenheim’s.

The moral is the one this collection keeps arriving at. A classification needs an invariant to separate and a construction to identify, and Frankenheim had the first and not the second. A count of symmetries separates lattices that differ and says nothing about two that agree — and the reduced cell, which is the construction that settles it, was a century away.