Closing the plane from two centres

Where the product is establishes the object: composing two symmetries lands on a third, and the third one is somewhere. Two half-turns make a translation by twice the distance between their centres; a rotation and a rotation make a rotation at a computed place. It ends by naming three questions it leaves, the first of which is

the fundamental theorem of the classification: the seventeen fall out of asking which sets of located elements are consistent, and the constraint that makes the list finite is exactly the doubling law meeting the crystallographic restriction.

Every route to the seventeen taken so far starts by fixing a lattice. The classification proof cuts on rotation orders and mirror positions with the lattice already there; the orbifold argument prices the features of a folded pattern; the cocycle count attaches translations to a point group acting on a lattice. A lattice is an input in all three.

It need not be. Put two rotation centres on an empty plane, close under composition, and see what happens.

Two rotations generate rotations, and the angles add #

The first thing to establish is what the closure can contain, and it is short.

A rotation by θ about any point has linear part the rotation by θ, whatever the point is; composing two of them multiplies the linear parts, so the product’s linear part is the rotation by θ + φ. The centres move and the angles add. So the set of linear parts a closure reaches is the group generated by two rotations of the circle, which is cyclic, and its order is the least common multiple of the two orders.

That is a statement about a group of two-by-two matrices with no lattice in it, and the computation confirms it at every pair tried: two centres of orders three and three reach three linear parts, two and three reach six, three and four reach twelve, five and five reach five.

Now the crystallographic restriction says a lattice admits rotations of order one, two, three, four and six and nothing else. So a closure that is to be a plane group must have a least common multiple among those five, and that is a condition on the pair of orders before any lattice is mentioned.

Eight pairs pass and four fail, of the twelve tried: (2,2), (2,3), (2,4), (2,6), (3,3), (3,6), (4,4) and (6,6) pass; (3,4) and (4,6) reach twelve, (2,5) reaches ten and (5,5) reaches five.

The translations settle, or they do not #

The condition on the linear parts is necessary. What makes it sufficient is the other half of the closure: the translations.

Where the product is computes what two turns make. The commutator of a turn by θ about c and a turn by φ about d is a translation of length 4 sin(θ/2) sin(φ/2) |c − d| — a translation, out of two rotations, with a length the two centres decide. So a closure containing two centres contains a translation immediately, and then it contains every rotated copy of that translation, and their differences, and so on.

For two four-fold centres a unit apart the shortest translation is the square root of two and stays there however far the words run. The closure is p4 on a square lattice whose cell is that translation and its quarter-turn — and nobody put a lattice down. It arrived as the set of translations the two centres force.

For a three-fold against a four-fold the shortest translation falls, and the factor it falls by is not approximate. A group holding a translation $t$ and a rotation of order $L$ holds the difference of $t$ with its own rotated copy, which has length $2\sin(k\pi/L)\,|t|$, and it holds the sum, which has length $2|\cos(k\pi/L)|\,|t|$. The smallest of those over every $k$ is what the group forces on its own shortest translation, and for $L = 12$ it is $2\sin(\pi/12) = 0.5176$. The search finds exactly that and nothing else: $2.4495$ at word four, $1.2679$ at word seven, and then forty thousand elements are not enough to reach the word length that would show the next one. The step applies to its own result, so the sequence it starts has no floor.

And that factor is the restriction. It is below one at $L = 5$, where the sum rather than the difference gives $2|\cos(2\pi/5)| = 0.6180$, and below one for every $L$ of seven or more, since $2\sin(\pi/L)$ is then under $0.87$. It is one or more at $L = 1, 2, 3, 4$ and $6$ and nowhere else — the five orders a lattice admits, arriving here as the orders that cannot shrink a translation. A group with translations as short as anybody cares to ask is not discrete, and is not the symmetry of anything.

The third centre, and where it is #

The closure’s first non-trivial content is the third centre, and it is the object the classification’s case analysis is really about.

Compose a turn by θ about c with a turn by φ about d and the result turns by θ + φ about a third point e, and the triangle c d e has angles θ/2 and φ/2 at two of its corners. For the three centres to be centres of orders n, m and k the three half-turns must add to a straight angle:

$$\frac{\pi}{n} + \frac{\pi}{m} + \frac{\pi}{k} = \pi, \qquad \text{that is} \qquad \frac{1}{n} + \frac{1}{m} + \frac{1}{k} = 1.$$

The whole-number solutions are (3, 3, 3), (2, 4, 4) and (2, 3, 6), and the degenerate case (2, 2, ∞) where the third element is a translation rather than a rotation. Four ways, and they are p3, p4, p6 and p2.

That equation is where the finiteness comes from, and it is worth seeing that it is the same equation twice. Read as a condition on a triangle it is a statement about angles; read as a condition on a closure it is the statement that the least common multiple is permitted. The four solutions of the first are exactly the pairs whose multiple is two, three, four or six — since a pair with multiple twelve would need a third angle the triangle has no room for.

Why a forbidden multiple manufactures short translations #

The measurement shows the translations falling and it is worth having the mechanism, because it is the same mechanism the proof of the restriction uses and it is three lines.

Suppose the closure contains a translation t and a rotation R of order N. Then it contains R t, the translation by the rotated vector, and their difference, which is a translation of length 2|t| sin(π/N). For N greater than six that factor is less than one: at N = 12 it is 2 sin 15°, which is 0.518. So the difference is shorter than the translation it came from, and repeating the step shortens it again, without limit.

At N = 6 the factor is exactly one, at N = 4 it is the square root of two and at N = 3 it is the square root of three — all at least one, so the step produces nothing shorter and the closure has somewhere to stop. Six is the largest order for which the factor reaches one, which is the crystallographic restriction in the form this measurement sees it.

N = 5 is the case worth watching, because the factor is 2 sin 36° = 1.176, which is greater than one — so a single step lengthens rather than shortens. The shortening comes two steps later: the five rotated copies of a translation include pairs whose difference is 2 sin 72° apart and pairs 2 sin 36° apart, and combining them produces a vector shorter than either by the golden ratio. The measurement records it as a shortest translation of 0.449 that goes on falling, which is the golden ratio doing what it does in the geometric proof of the restriction.

The lattice as an output #

What the closure produces is a group with translations in it, and those translations are a lattice rather than being assumed to be one.

For two four-fold centres the translations are generated by the commutator and its quarter-turns, which is a square lattice. For (2, 3, 6) the commutator and its sixth-turns give a hexagonal one. For two three-fold centres the same, and for (2, 2) a rectangular lattice whose two directions are twice the vector between the centres and whatever second centre is added.

So the lattice type is a consequence of the rotation orders, and the five plane lattices appear as five answers to a question about angles. That is the reverse of the usual order, in which the lattice types are enumerated first — five of them, by their own symmetry — and the groups are then built on each.

Neither order is better and the reverse one says something the usual one does not. In the usual order the crystallographic restriction is a theorem about lattices that has to be proved before the classification can start. Here it is the condition that a closure terminates, and the proof is that a rotation of a forbidden order manufactures translations shorter than the ones it was given — which is the argument Bieberbach’s first theorem uses, run forwards instead of backwards.

The same closure, one dimension down and one up #

The closure is a general procedure and running it in other dimensions says which parts of the answer are about the plane.

On a line there is nothing to close. The only rotation is the half-turn, two of them compose to a translation by twice the distance between them, and the closure is the infinite dihedral group at every separation. No condition arises because there is no second order to take a multiple with.

On a sphere the closure always terminates, because there are no translations at all and the linear parts are the whole group. The condition 1/n + 1/m + 1/k = 1 becomes greater than one, its solutions are the finite rotation groups, and five families come out rather than four. The closure is finite whatever the orders, so the measurement this page makes has nothing to measure.

In space the closure of two rotations is much harder, and the reason is what happens to axes: two rotations about skew axes compose to a screw, so the closure contains motions with a translation along their own axis and the bookkeeping of “where the product is” needs a line rather than a point. Whether a pair of located axes closes discretely is not settled by their orders alone.

So the plane is the case where the closure is finite to decide and infinite to contain, which is exactly the case in which a condition on the orders is the whole answer.

What two centres do not reach #

The closure above generates the rotation groups and nothing else, and it is worth being exact about the gap.

Mirrors are not reachable from rotations. Two rotations compose to a rotation or a translation, both of which preserve handedness, so no closure of rotations contains a reflection. The eleven plane groups with a mirror or a glide need a reflection among the generators, and three reflections and never four is what supplies it: every motion of the plane is a product of at most three mirrors, and the parity of the number decides handedness.

So the full classification from located elements takes mirrors as a second kind of generator and asks the same closure question of them — two mirrors at an angle generate a rotation by twice that angle, so the permitted angles between mirrors are the halves of the permitted rotations. The doubling law meeting the restriction is exactly that sentence, and it is why the mirror angles in the seventeen are 90°, 60°, 45° and 30° and nothing else.

That half is not computed here. What is computed is the rotation half, where the closure is a closure of orientation-preserving motions and the condition is one arithmetic statement.

Where the exactness stops #

Computed here. For twelve pairs of rotation orders, the group generated by a centre of each placed one unit apart, closed out to words of six letters: the number of distinct linear parts reached, the shortest translation among the elements found, and how that shortest behaves as the words lengthen. Two checks that can fail: the linear parts must be exactly the least common multiple at every pair, and every permitted pair’s shortest translation must settle.

A closure to six letters is not a closure. For a discrete group the set of elements is infinite and the measurement is of a finite part of it; for a dense group there is no finishing at all. What is measured is the behaviour of the shortest translation as the search deepens, which is the quantity discreteness is about, and a finite search of a statement about all words is evidence rather than proof.

One separation, and the answer depends on it only up to scale. The two centres are placed one unit apart throughout. Moving them scales every translation by the same factor and changes nothing about which pairs settle, which is why the separation is not a parameter of the result.

Rotations only. Nothing in the closure is a reflection, so the eleven plane groups with a mirror or a glide are outside what is measured. The condition on the least common multiple is about the rotation subgroup of a plane group and applies to all seventeen through it; what it does not do is produce the eleven.

And the triangle equation is quoted rather than derived here. That the three centres form a triangle with the half-angles at its corners is the located product’s computation, and the figures above are that computation at two further pairs of orders rather than a new argument.

The refusal is a three-fold and a four-fold centre offered as generating a plane group. It is the case a reader is most likely to think works, because both orders are permitted individually and both appear in the seventeen — and what rules it out is a property of the pair rather than of either.

Who put the centres down first #

Deriving the seventeen from the configurations of rotation centres is the nineteenth-century route. Camille Jordan attempted a classification of the groups of motions in 1868 and missed some; Evgraf Fedorov completed the plane case in 1891, and Arthur Schoenflies and William Barlow the space groups in the same decade, all working with located elements rather than with lattices, because the language of lattices as modules over rings came later.

The triangle equation is older than any of it. The condition 1/n + 1/m + 1/k = 1 and its solutions (3,3,3), (2,4,4), (2,3,6) are the Euclidean cases of the triangle groups, and the same equation with the sum greater than one gives the finite groups on the sphere and less than one the hyperbolic ones — which is the list past two dollars in a different notation. That one equation separates three geometries is the reason it is worth deriving from composition rather than quoting.

What the computation adds to the history is the negative half. Fedorov’s route establishes that certain configurations work; it does not display what goes wrong with the others, because in 1891 the thing that goes wrong had no convenient name. A translation that keeps getting shorter is the name, and it is a measurement rather than an argument by cases.

Still open: closing with mirrors in the generators #

Two things would complete the derivation from located elements and neither is here.

The mirror closure. Two mirrors at an angle α generate a rotation by 2α about their intersection, so a closure containing mirrors reaches rotations of order π/α — and the condition becomes that α is one of 90°, 60°, 45° or 30°. Adding a mirror to a rotation closure and asking which of the seventeen come out is the same computation with reflections among the generators, and the shortest-translation measurement would work unchanged. What is new is that the closure can produce a glide, which is neither a mirror nor a rotation and has no fixed point, so the measurement of “where the product is” needs a second kind of answer.

And the count. Nothing above counts the groups. The closure of a pair of centres is one group; getting seventeen means asking which sets of located elements are consistent and identifying the ones that differ only by where the origin was put, which is a quotient the closure does not take. The route from “these pairs close” to “there are seventeen” passes through that quotient, and it is the step the cocycle count performs by a different means.