What forces a lattice

The five plane lattices, the seventeen groups, the two hundred and thirty: every enumeration that ends in one of those numbers begins by fixing a lattice and asking which motions it admits. Why there is a list at all names the theorem that licenses the start and then takes it as a definition — in two and three dimensions a crystallographic group can simply be defined as a lattice with operations attached. The same group means the same pattern calls the same theorem the engine of Bieberbach’s second result and moves on.

The definition hides a question. A pattern does not arrive with its translations marked; it arrives as a set of motions that leave it alone, and whether those motions include a lattice’s worth of translations is a question about the set. Bieberbach’s first theorem, of 1911, answers it: a group of motions of n-dimensional space that is discrete, and under which no point is far from the orbit of any other, contains n independent translations, and they form a subgroup of finite index. Nothing about repetition is assumed. The lattice is forced.

In the plane the proof has four steps, and each is a computation that can be drawn. In space the first step is false, and what replaces it is an inequality with a threshold at sixty degrees — which is also, exactly, the closest a symmetry of any lattice in any dimension can come to doing nothing.

A turn undone by another turn is a slide #

Take two rotations of the plane: g, a turn by θ about a point c, and h, a turn by φ about a point d. Perform h backwards, then g backwards, then h, then g. The motion that results is written $g h g^{-1} h^{-1}$, the commutator of g and h, and it would be the identity if the two turns could be performed in either order with the same effect.

They cannot, but they nearly can. A turn of the plane has two parts: a linear part that rotates every direction through the same angle, and a translation that records where the centre is. The linear parts of the two turns are rotations through θ and φ, and rotations about one point commute — the order of two turns about the same centre is immaterial. So in the commutator each rotation is cancelled by its own inverse, and what survives has linear part equal to the identity. A motion with no rotation left in it is a translation. Working the four compositions through, the translation is (I − A)(I − B)(c − d), where A and B are the two rotations, and its length is 4·sin(θ/2)·sin(φ/2)·|c − d|.

Three thousand random pairs of turns, with angles anywhere up to a half-turn either way and centres anywhere in a square six units across, were composed in exactly this way. The linear part of every commutator differed from the identity by at most 4.4 × 10⁻¹⁶, which is the rounding of the arithmetic, and the translation agreed with the closed form to 7.3 × 10⁻¹⁵.

Two consequences follow at once, and between them they are the first line of the proof. The formula vanishes only when one of the angles is zero or the two centres coincide, so a group of motions containing two turns about different centres contains a translation. And read the other way: if a group of orientation-preserving motions contains no translation other than the identity, every commutator in it is trivial, every two of its turns commute, and so every turn it contains shares a single centre. Both halves were checked on the same random pairs. Turns about one centre commuted to within 3.7 × 10⁻¹⁵, and no pair about centres at least a tenth of a unit apart, each turning by at least a degree, came closer to commuting than a translation of length 1.9 × 10⁻³.

A group that also contains mirrors or glides has an orientation-preserving part of index two, and every step here is applied to that part. Its translations are the whole group’s translations, so nothing is lost by setting the mirrors aside.

A turn a lattice forbids manufactures short translations #

Suppose the group does contain a translation, t. Conjugating t by a turn g — performing g backwards, then t, then g — gives a translation again, by the rotated vector At. So a group containing a turn and a translation contains every rotated copy of the translation, all of one length, all on one circle.

If the turn is through sixty degrees there are six copies and then they repeat. If it is through an angle that is not a rational fraction of a full turn, every copy is new, and the circle fills.

A circle with infinitely many points on it has points as close together as anybody cares to ask, and the difference of two translations in a group is itself a translation in the group. So a group containing a turn through an irrational angle and any translation at all contains translations arbitrarily close to zero, and a group of motions like that is not discrete. Discrete, or dense, and nothing between sorts the subgroups of the plane’s translations into their five kinds; here all that matters is that a discrete group cannot contain one of the dense ones.

The step does more than exclude irrational angles. Discreteness means that only finitely many translations of the group have any one length, so each turn’s linear part permutes a finite set of vectors. A rotation of the plane that fixes a single non-zero vector is the identity, so the linear parts act faithfully on that finite set, and the rotation parts form a finite group. That group is the point group, and its finiteness has now been deduced rather than assumed. Which finite groups it can be is the list of cyclic and dihedral groups, and which of those a lattice then admits is the crystallographic restriction — two arguments that start exactly where this one stops.

How short, and at which copy #

How fast the translations shrink under a turn of one radian is worth measuring, because the answer turns out to be arithmetic in an unexpected place.

The shortest translation made from the first J copies does not fall smoothly. It falls in steps, and each step comes at a copy whose angle — a whole number of radians — happens to land unusually close to a whole number of turns. Six radians is a little short of one turn and leaves a translation of length 0.282. Nineteen radians against three turns leaves 0.150; twenty-five against four leaves 0.133; forty-four against seven leaves 0.0177; three hundred and thirty-three against fifty-three leaves 0.0088. Seven hundred and ten radians against a hundred and thirteen turns leaves 0.00006, and nothing shorter appears before two thousand copies.

The pairs are 6/1, 19/3, 25/4, 44/7, 333/53 and 710/113, and they are the successive best rational approximations to 2π — the convergents of its continued fraction, which begins [6; 3, 1, 1, 7, 2, 146]. The last of them is twice 355/113, the value of π Zu Chongzhi gave in the fifth century, and the reason the shortest translation then stalls for so long is the same reason his fraction is so good: the next term of the expansion is 146, so no better approximation arrives for a very long time.

The same expansion decides how evenly the copies spread round the circle, where at every J the gaps between neighbouring copies take at most three lengths. So the difference between a lattice and a dense group of translations is visible in a single number. For a lattice the shortest translation is fixed at the first copy and never moves; for the dense group it obeys a law written in the continued fraction of 2π, and it goes to zero.

No translations, or translations in one direction only #

The first two steps show that a discrete group containing turns and at least one translation has a finite group of rotation parts. Neither shows that a translation exists, nor that translations point in two directions, and both can fail. What excludes the failures is the second hypothesis: that the group leaves no point of the plane far from the orbit of a chosen point, so that there is a distance R within which every point has an image of the chosen one. The smallest such R is the orbit’s covering radius, and the hypothesis is that it is finite — which is what it means for the plane divided by the group to be compact.

The two failures fail in the same way. A group with no translations has, by the first step, all its turns about one centre, so the orbit of any point is a finite set on one circle; points far from the centre are far from all of it, and the distance measured in a disc of radius ten is 9.78. A group whose translations all lie along one line — a frieze — has, because its rotation parts are finite, an orbit made of finitely many rows of points along that line, all inside a strip; points far from the strip are far from the orbit, and the measurement reaches 9.90. The group p6 on the hexagonal lattice has an orbit that the measurement never finds more than 0.323 from any point, however large the disc.

So a finite covering radius forces translations in two independent directions, and the discreteness of the group makes them a discrete subgroup of rank two, which is a lattice. Its index in the whole group is the order of the point group, because two elements with the same rotation part differ by a translation. That is the whole theorem in the plane, and it fits on four lines:

1. the commutator of two turns is a translation;
2. a rotated translation is a translation, so discreteness makes the rotation parts a finite group;
3. no translation at all puts every turn on one centre, and the covering radius is unbounded;
4. translations along one line only leave the covering radius unbounded too.

The lattice underneath every plane pattern is a consequence of two conditions on a set of motions, and neither condition mentions repetition.

In space, turns do not commute #

Every one of those steps survives the move to three dimensions except the first, and the first fails outright. Two rotations of space about axes that are not parallel do not commute, and their commutator is not a translation but another rotation. The plane had a free cancellation that space does not provide, which is the same fact every motion of space is a screw runs into from the other side.

What replaces the cancellation is a statement about size. Measure how far a rotation A is from doing nothing by ‖A − I‖, the most it moves any unit vector. For a turn through ψ that is the chord 2 sin(ψ/2): a sixth of a turn sits at 1, a quarter-turn at $\sqrt{2}$, a half-turn at 2. For orthogonal matrices A and B in any dimension,

$$[A, B] - I = \big((A - I)(B - I) - (B - I)(A - I)\big)\, A^{-1}B^{-1},$$

and the last factor changes no lengths, so ‖[A, B] − I‖ ≤ 2 ‖A − I‖ · ‖B − I‖. The commutator of two small rotations is much smaller than either of them.

For rotations of space the constant is better than two. Write a rotation as a unit quaternion a + u, with u a vector of length sin(ψ/2), so that its chord is 2|u|. Two such quaternions satisfy pq − qp = 2 u × v, and from that one line the commutator’s chord is at most the product of the two chords times the sine of the angle between the axes. In space the constant is one, and the sine accounts for the plane: two turns of the plane are turns about parallel axes, the sine is zero, and the commutator’s linear part is exactly the identity — which is the first line of the plane’s proof, recovered as a special case.

Twenty thousand random pairs of rotations, with random axes and turns up to a half-turn, put that ratio between zero and one, and the largest value found was 0.99999. The quaternion identity itself was checked on another twenty thousand pairs and held to within 8.9 × 10⁻¹⁶, and the bound with the sine in it was not exceeded once. The histogram leans towards one because two random axes are more often nearly perpendicular than nearly parallel — the sine of the angle between them passed 0.9 for 44% of the pairs — and every one of the 4,361 ratios above 0.9 came from such a pair, where the sine in the bound is close to one.

Commutators that shrink #

The inequality earns its place by being used repeatedly. Fix a rotation A and take any rotation B; replace B by [A, B], then by [A, [A, B]], and so on. Each step multiplies the distance from the identity by at most ‖A − I‖, so whenever A turns through less than sixty degrees the nested commutators shrink geometrically towards the identity.

The measured factors do more than respect the bound; they converge to it. Against a turn of ten degrees each step multiplies the distance by 0.174, against twenty by 0.347, against forty by 0.684 — the chords of those three angles, to three places. The reason is that the axis of the nested commutator swings round until it is perpendicular to A’s, where the sine in the bound is one. At sixty degrees the factor allowed is exactly one and the distance only creeps, from 0.68 to 0.41 in twelve steps. At ninety degrees it grows and settles near 1.73.

In a discrete group the shrinking cannot continue for ever. Only finitely many elements of a discrete group of motions lie close to the identity with translation parts of bounded length, and a companion inequality for the translation parts keeps nested commutators from wandering off; so a sequence of nested commutators approaching the identity has to arrive. That is the lever the general proof pulls. Elements close to the identity have commutators that eventually vanish, and from there Bieberbach’s argument, in the shorter geometric form Peter Buser gave it, works outwards from the elements near the identity to a subgroup of translations that spans the space. Those later steps are quoted here and not reproduced, and the translation-part inequality is not drawn.

The case that fails the hypothesis is instructive. Two ten-degree turns about perpendicular axes generate a group of rotations with no translations in it at all. Eight nested commutators come out each nearer the identity than the last, the eighth at 1.5 × 10⁻⁷, and none of them equal to it. The group they belong to is not discrete, the sequence has nowhere to stop, and nothing in it is the symmetry of a crystal.

The gap every lattice leaves round the identity #

Once the lattice exists, the elements the proof worked with have a very simple description, and the reason is a small piece of number theory.

With a lattice in hand, every rotation part of the group is an integer matrix of finite order in a basis of that lattice. The average that makes it finite builds a metric the matrix preserves by averaging over the group; in that metric the matrix is orthogonal, and its distance from the identity is the largest |λ − 1| over its eigenvalues λ. Those eigenvalues are roots of unity, and because the characteristic polynomial has integer coefficients, an eigenvalue that is a primitive m-th root of unity brings every other primitive m-th root with it.

A complete set of primitive m-th roots adds up to the Möbius function of m, which is −1, 0 or 1. If every root in the set lay within sixty degrees of 1, each would have real part greater than a half, and the φ(m) of them together would add up to more than φ(m)/2. That exceeds one unless φ(m) is at most two, which leaves the orders 3, 4 and 6 — and of those, only the sixth roots lie as close as sixty degrees, and they lie at exactly sixty. So no symmetry of a lattice, in any dimension, other than the identity, is closer than 1 to the identity, and only a six-fold rotation is that close.

The statement was checked two ways. For every order from 2 to 200 the largest chord among the primitive roots is at least 1, with equality at six alone; the nearest competitors are order 4 at 1.414, order 10 at 1.618 and order 3 at 1.732. For every order up to 30 the integer matrix itself was built — the companion matrix of the cyclotomic polynomial, in dimensions up to 28 — its metric was averaged, and its distance from the identity computed as an operator norm rather than read off the eigenvalues. The two agreed to 3.3 × 10⁻¹⁶, and the sums of primitive roots matched the Möbius function at every order to 200.

The two sixty degrees are the same sixty degrees. In space, nested commutators shrink against any turn under sixty degrees. A lattice admits no symmetry that turns every direction through less than sixty degrees, in any dimension — a twelve-fold symmetry of a four-dimensional lattice turns one plane through thirty degrees and pays for it by turning another through a hundred and fifty. The proof needs a neighbourhood of the identity in which commutators shrink, and a finished crystallographic group has no rotation in that neighbourhood at all: its elements there are exactly its translations. The six-fold rotation sits on the boundary from both sides. It is the one lattice symmetry at distance exactly 1, and 1 is where the shrinking stops.

What is proved, what is measured, and what is quoted #

Proved, in the plane. The four steps are a complete proof, and every computation above checks one of them rather than replacing it. The checks are finite — three thousand random pairs, one disc of radius ten, one irrational angle out to two thousand copies — and a finite check of a statement about all groups is evidence, not proof.

Measured, in space. The inequality with constant two has the one-line proof written out above; the constant one rests on the quaternion identity, whose algebra is two lines and whose check is twenty thousand pairs. That the nested commutators’ factors converge to the chord is observed at three angles and not proved.

Quoted. In three dimensions and more, the passage from shrinking commutators to translations in every direction is Bieberbach’s and Buser’s and is not reproduced, and neither is the inequality for the translation parts.

The convention every distance depends on. Each distance from the identity is an operator norm in a metric the matrices preserve: the Euclidean metric for rotations, the averaged metric for integer matrices. Without a preserved metric the inequality is simply false. A stretch that doubles one axis and halves the other, and a shear, each sit at distance 1 from the identity, and their commutator sits at distance 3 — a ratio of three against a bound of two. The metric is part of the statement, not a detail of the proof.

The checks, and the two cases they turn away #

Everything quoted in the sections above comes from a set of tests that could fail, and two of them exist to fail.

The first refusal is the non-discrete group, recognised by a sequence of nested commutators that keeps approaching the identity without reaching it. The second is the inequality used where it does not apply, and it is the more useful of the two, because the mistake it catches is a natural one: an integer matrix written in a lattice’s own basis looks like the obvious object to measure, and in that basis it is not orthogonal and the bound does not hold.

Hilbert’s question, and the proofs that answered it #

Bieberbach proved the theorem in 1911, as the first of three results answering the part of Hilbert’s eighteenth problem that asks whether each dimension has only finitely many crystallographic groups. Georg Frobenius gave a second proof in the same year, by a different route. The argument through nested commutators turned out to be larger than its subject: Hans Zassenhaus showed that every Lie group has a neighbourhood of the identity in which the elements of a discrete subgroup generate a nilpotent group, and the Margulis lemma carries the same idea into hyperbolic geometry, where it describes the thin parts of a hyperbolic manifold. Peter Buser’s paper of 1985 wrote the crystallographic case as a short geometric argument, and it is the version most accounts now follow.

The quaternion identity is older than all of it — Hamilton had the product in 1843 — and the continued fraction of 2π older still. What the theorem adds to both is a reason a crystallographer should care about them.

Where this goes: the groups that keep one hypothesis #

Each hypothesis has been dropped here one at a time, and each failure is a subject of its own.

Keep discreteness and give up the bounded orbit, and the groups with translations in fewer directions than the space has appear: the layer and rod groups and the friezes. Seventy-five ways to be a thread is the case of one direction of repeat in space, classified by the same means once the missing directions are admitted to be missing. Keep a bounded orbit and give up discreteness, and the symmetries of a quasicrystal live there, where a dense group of translations seen through a window is what makes a structure ordered without repeating.

What neither failure touches is the arithmetic in between. The gap of 1 is a fact about integer matrices, true in every dimension whether or not any crystal is involved. Which finite orders a lattice of a given dimension admits is a question about cyclotomic degrees, and the gap adds a limit that no dimension lifts: however many orders become legal, none of them brings a symmetry closer to doing nothing than a six-fold turn does.