Two mirrors a coset cannot tell apart

The same symmetry, somewhere else gives the question “are these two symmetries the same one?” an exact meaning: two operations are the same symmetry when some operation of the group carries one onto the other. It then computes the answer for the seventeen wallpaper groups modulo their lattices — treating every translation as nothing, so that each group becomes finite, four operations for p2, twelve for p6m, and the classes can be listed by composing every pair.

That convention is convenient and it answers a coarser question than it appears to. The same group means the same pattern points at the finer one. Of Dehn’s three decision problems for a group — whether two words are equal, whether two presentations give the same group, whether two elements are conjugate — the first two have been settled here for the plane groups, the word problem by a normal form and the isomorphism problem by Bieberbach, and the third was left with a sentence. It is settled below, and settling it shows exactly what the quotient loses.

One coset, two families of mirrors #

The group pm has mirrors in one direction, a cell’s width apart and half a cell’s width apart: a mirror on the line y = 0, another on y = ½, another on y = 1, and so on across the plane.

Modulo the lattice, a mirror is its linear part together with its translation reduced to lie between nothing and one cell. The mirror on y = 0 is y ↦ −y. The mirror on y = ½ is y ↦ −y + 1, and reducing its translation by the whole cell leaves y ↦ −y again. So in the quotient they are one element, and nothing computed there can ask whether they are the same symmetry, because it cannot tell that there are two.

In pm itself they are not the same symmetry. The operations of pm are translations and mirrors of that one direction. A translation by whole cells carries the mirror on y = 0 to the mirrors on y = 1, 2, 3 and never to one on a half-integer line. A mirror of pm carries the line y = 0 to itself or to another whole-integer line. Nothing in the group moves a mirror by half a cell, so pm has two conjugacy classes of mirror, and its quotient has one. The picture below is the same fact for p2: four families of half-turn centre, which are four classes in p2 and one element in its quotient.

The earlier essay’s convention is not wrong about what it computes. Its classes are classes of cosets — each a whole family of parallel mirrors, or of half-turns a lattice vector apart — and it said so. What it could not do is tell two families inside one coset apart, and in every group but p1 some of the distinctions live exactly there.

What conjugate means, written out #

Write an element of a plane group as (A, a): the motion x ↦ Ax + a, with A an integer matrix in the lattice’s basis and a a vector of fractions, never reduced. Two elements g = (A, a) and h = (B, b) are conjugate when some element x = (R, r) of the group satisfies $x g x^{-1} = h$. Composing the three motions, that is two conditions:

• $R A R^{-1} = B$, a condition on linear parts only;
• $(I - B)\, r = b - R a$, a condition on translations.

The first leaves finitely many candidate linear parts R, because a plane group has at most twelve. For each, the element’s translation r is not free: it is one particular vector $r_0$ plus any lattice vector t. Substituting, the second condition becomes $(I - B)\, t = b - R a - (I - B)\, r_0$. The right-hand side is one known vector w; the left-hand side ranges over the lattice (I − B)ℤ² as t ranges over the integer vectors.

So g and h are conjugate exactly when, for some one of finitely many R, the vector w lies in the lattice (I − B)ℤ². That is a question with an integer answer, and when the answer is yes the lattice vector t that solves it gives the conjugator. Every conjugator the decision produces is composed back and checked to carry g onto h.

The lattice a rotation is moved through #

For a rotation, I − B is invertible, and (I − B)ℤ² is a sublattice of the integer lattice with an index equal to the size of the determinant of I − B.

For a half-turn, I − B is twice the identity and the sublattice is every second point in both directions, index four. For a rotation through a third of a turn the index is three, for a quarter-turn two, and for a sixth of a turn one. Those four numbers are the determinant of I − B, which is 2 − 2cos θ for a rotation through θ, and they are what translations alone can do: conjugating a rotation by the lattice’s translations moves its centre by lattice vectors, and leaves exactly that many families of centre apart — four half-turn centres to a cell, three centres of a third of a turn, two of a quarter, one of a sixth. The other operations of the group can merge some of those families; they cannot split them.

The index is the same arithmetic that the crystallographic restriction reads off a trace. There the trace 2cos θ had to be an integer; here 2 − 2cos θ, the determinant of I − B, counts how many inequivalent places a rotation of that order can sit, and a sixth of a turn can sit in only one.

One decision, worked #

The smallest case shows the whole method.

The half-turn g about the origin is x ↦ −x. The half-turn h about the point one cell along is x ↦ −x + (2, 0), and the one h′ about the point half a cell along is x ↦ −x + (1, 0). For R the identity, w is (2, 0) for h and (1, 0) for h′; for R the half-turn, the same. The lattice 2ℤ² contains (2, 0) and does not contain (1, 0). So g is conjugate to h, by a translation of one cell, and not to h′, by anything. Modulo the lattice, all three are the same element.

That decision took two candidate linear parts and two membership tests. The same procedure applied to every pair of small elements of five groups — p2, pmg, p4g, p31m and cmm, 1,120 pairs in all — agreed with a direct search for a conjugator among the group’s elements with small translations, on every pair.

The classes are where the positions are #

Colouring every rotation centre and mirror line of a group by its class makes the result visible, and one group with everything in it is enough to see the pattern.

In p4m the quarter-turns are in two classes, at the cell corners and at the cell centres, and the quotient has one. The half-turns are in three: at the corners, at the centres, and at the midpoints of the edges; the quotient has one. The mirrors are in three: the ones along the cell edges through the corners, the parallel ones half a cell over through the centres, and the diagonal ones; the quotient has two.

Those are not arbitrary families. A crystallographer reading the International Tables entry for p4m finds its special positions listed in exactly this shape — a position at the corner with symmetry 4mm, one at the centre with 4mm, one at the edge midpoints with 2mm, and lines of mirror symmetry along the edges, half an edge over, and along the diagonals. The table of positions is a table of conjugacy classes in the infinite group: two points share a position when an operation carries one to the other, which is the same as the rotations or mirrors fixing them being conjugate. The quotient cannot compute that table, because it cannot see the difference between a corner and a centre; the same site under two names is about when even the table’s own entries can be exchanged.

All seventeen #

Of the forty entries, twenty-eight grow when the lattice is kept. The largest growth is in the groups with the least to merge with: p2’s half-turns go from one class to four, and p3’s rotations from two to six, because in both groups nothing but the rotations themselves and the translations acts, and the index of (I − B)ℤ² stands unreduced. pmm’s half-turns go from one to four and its mirrors from two to four. p4’s quarter-turns go from two to four and its half-turns from one to three. In the other direction, p6m’s six-fold rotation stays one class, because the index for a sixth of a turn is already one, and so do its two families of mirror.

Every rotation count was obtained twice. Once through the lattice (I − B)ℤ², with its classes merged under every operation whose linear part commutes with B; and once as orbits of rotation centres under the same operations, which never mentions (I − B)ℤ². The two agreed for all twenty-four rotation linear parts across the seventeen groups.

Why twelve entries do not grow #

A class of the quotient splits in the group only if nothing in the group can carry one of its families onto another, and in twelve of the forty entries something can. There are three ways it happens, and each is visible in the census.

The index is already one. A rotation through a sixth of a turn has I − B with determinant one, so its lattice (I − B)ℤ² is the whole integer lattice and translations alone reach every centre. p6’s six-fold rotations stay at two classes — one for each sense — and p6m’s at one, because its mirrors reverse the sense.

A translation of the lattice already moves by half the spacing. cmm’s mirrors in one direction are half a conventional cell apart, and in a centred lattice the half-diagonal of that cell is itself a lattice translation. Asked whether the mirror through the origin and its neighbour are conjugate, the decision returns a conjugator whose linear part is the identity — a translation by one step of the lattice. p4g’s parallel diagonal mirrors are joined the same way. Nothing splits, because the quotient’s coset was never hiding two families there.

Another operation exchanges the families. pmg has mirrors on the lines x = ¼ and x = ¾, and no translation of pmg moves a mirror by half a cell. But pmg has half-turns, and the half-turn about the point midway between the two lines carries one onto the other: the conjugator the decision returns for that pair has linear part minus the identity. So pmg’s mirrors are one class in the group as in the quotient, while its half-turns — which nothing exchanges — go from one class to two.

So growth is the ordinary case and non-growth needs a reason, and the three reasons are the three things a plane group can have: a lattice fine enough, or an operation placed between the families. The quotient cannot see any of them, because every one of them is a statement about where operations are rather than about what they are.

The count has a convention in it. A rotation and its inverse are different elements, and they are conjugate only if the group contains something reversing orientation. p3 has three centres of three-fold rotation to a cell and no mirrors, so its rotations by a third of a turn form three classes and its rotations by two thirds form three more: six. p3m1 has the same three centres and mirrors that reverse every rotation’s sense, so its six collapse to three.

A glide is not conjugate to its own inverse in pg #

Glides show a different effect, and it is the one the quotient hides most completely.

A glide reflects in a line and slides along it, and the slide survives conjugation. Conjugating by a translation moves the line and keeps the slide; conjugating by a mirror or glide parallel to it keeps the slide; conjugating by a half-turn reverses it. So two glides with slides of different lengths are never conjugate, and in pg, which has no half-turn, a glide sliding half a cell to the right is not even conjugate to the same glide sliding half a cell to the left, which is its own inverse.

The census for pg’s glides reads accordingly. The glides that slide half a cell fall into four classes: two families of glide line — the lines y = 0 and y = ½, which no operation of pg exchanges — each with two senses of slide. The glides that slide one and a half cells fall into four more classes, and so on without end. In the quotient all of them are one element. The group has infinitely many conjugacy classes, as every infinite group of motions with a translation must — a translation by v is conjugate only to translations by the rotated copies of v — and a finite list of classes is only possible after deciding not to look.

What makes this decidable, and what would not be #

The conjugacy problem is usually the hardest of Dehn’s three. There are finitely presented groups in which the word problem is solvable and the conjugacy problem is not, so knowing how to tell whether two words are the same element does not tell whether two elements are the same symmetry.

For the plane groups it is decidable for the reason telling two words apart found the word problem to be: every element is a small integer matrix and a vector, the group has a lattice of translations of finite index, and every question about the group becomes a question about that lattice. The conjugacy question becomes a membership question for one sublattice, and a membership question for a lattice is a division. The same argument works in any dimension, with three-by-three matrices and (I − B)ℤ³, and that is the structural reason the space groups’ positions can be tabulated at all.

What the decision does not settle #

The census is of plane groups in this collection’s settings. Each group is written on a conventional cell with integer linear parts, and a class count does not depend on the cell chosen; which lines are called y = 0 and y = ½ does.

The glide classes are counted for one slide at a time. There are infinitely many classes of glide in every group with glides, indexed by the slide; the census reports the shortest slide that is not itself a lattice vector.

The space groups are not computed. The decision is the same with three-dimensional matrices, and a screw axis’s slide plays the part of a glide’s; nothing about it has been run.

The checks, and what they refuse #

The pm refusal is the one that matters, because it is the case the quotient answers wrongly: both mirrors reduce to the same coset, and the vector the decision tests, (0, 1), is not in the lattice (I − B)ℤ² = 0 × 2ℤ, while the mirror on y = 1 gives (0, 2), which is. The second refusal is about the question rather than the answer. A motion that is not an element of pm — the mirror x ↦ −x, when pm’s mirrors are y ↦ −y — has no conjugacy class in pm, and a procedure that returned “not conjugate” for it would be answering a question nobody asked. It returns nothing.

Who asked, and who tabulated #

Max Dehn posed the word, conjugacy and isomorphism problems in 1911 and 1912, in papers about the fundamental groups of surfaces — groups that are, for the torus and the Klein bottle, two of the plane groups on this list. The special positions of the space groups were first tabulated systematically by Ralph Wyckoff in 1922, and the positions that now carry his name are listed in the International Tables for every group. That the list of positions is a list of conjugacy classes of stabilisers in the infinite group is part of how those tables are constructed, and it is the reason a table computed from the finite quotient alone would come out too short.

Where this goes: the same test in space #

The step up is direct and has not been taken. In a space group a rotation’s I − B has rank two and a screw axis’s slide along the axis behaves as a glide’s does here, so the classes of each kind of axis are counted by a sublattice of ℤ³ and a slide; the four-fold screw axes of a group with 4₁ and 4₃ axes, and the question of when they are conjugate in the group but not in a normaliser, are the natural place to see whether the plane’s pattern — the quotient too coarse, the table of positions exactly right — carries over unchanged.