Where the product is

Why it is a group makes the closure argument and stops one step short of the interesting part. Doing one symmetry after another lands on a third symmetry of the same pattern, always, and that is what separates a group from a list. It says nothing about where the third one is.

A symmetry of a plane pattern is a matrix and a translation. Its element — the point it turns about, the line it reflects across — is not written in the pair. It is computed from it, by solving for the fixed point or for the invariant line, and it is the only part of the operation a reader can see on the page. So the question this essay is about is a question about positions: given two operations whose elements are drawn, where is the element of their product?

The answer has three cases and each of them is a line of arithmetic. What the three cases add up to is not a line of arithmetic at all: the pattern of symmetry elements on a plate of a plane group is forced. Nothing about it was designed. Put two rotation centres somewhere and the rest of the plate follows, including the lattice.

Two half-turns make a translation, and it is twice as long as it looks #

Take the half-turn about a point p and the half-turn about a point q. Their composition has linear part (−I)(−I) = I, so it is a translation; and following a point through both motions gives the translation by 2(q − p).

The factor of two is the whole content. A pattern with two-fold centres a apart is periodic with period 2a, so the lattice of a p2 pattern is not chosen alongside its rotation centres — it is produced by them. And running the argument backwards: composing a half-turn with a repeat vector of the lattice gives a half-turn half that vector along, so one centre and the lattice force centres on the half lattice. That is why p2’s cell has four inequivalent two-fold centres and not one: the corner, the two edge midpoints and the cell centre, with nothing left over.

It is worth being exact about which pairs that argument applies to, because the loose version of it is false and looks true. Two centres a lattice vector apart have a centre halfway between them, by the composition just described. Two centres in general do not: the pair in the figure above are (½, ¼) apart, the group they generate has its centres strung along the line through them at intervals of that vector, and the point halfway between the two drawn is fixed by nothing at all. The half lattice of centres comes from the lattice, not from the pair.

The comparison in the figure is exact. The product is computed by multiplying the two operations as matrix-and-translation pairs; the prediction is computed as twice the difference of the two centres; and both are rationals with small denominators, so the assertion that they agree has no tolerance in it. It fails if the factor of two is dropped, which is the mistake the picture exists to make impossible to hold.

The canonical form had thrown the answer away #

Getting that figure to say anything required a change to the arithmetic behind it, and the change is worth stating because it is the kind of thing a convention gets right for one purpose and wrong for another.

A symmetry operation is normally stored with its translation reduced modulo the lattice, because the seventeen groups are classified modulo their translations and two operations differing by a lattice vector are the same coset. That is the correct convention for the classification and it deletes exactly the quantity this essay is about. The half-turn about (½, 0) has translation (1, 0), which reduces to (0, 0) — the half-turn about the origin. Compose two half-turns in the reduced form and the answer is the identity for every pair of centres. True modulo the lattice; useless as a picture.

So the compositions here are done on an unreduced pair, in the same rational arithmetic, and handed back to the same classifier. Nothing else changed. It is a small example of a general hazard: a canonical form is a decision about which differences do not matter, and a later question can be exactly about the differences it discarded.

The locator is one function and it is the same one every plate on this site is drawn from. For a rotation it solves (I − M)x = t; for a reflection it splits the translation into the part running along the invariant line, which is the glide, and the part running across it, which says how far from the origin the line sits — at half that part, because reflecting across a line offset by d moves a point at the origin by 2d. The factor of two appears there as well, and it is the same factor.

Two mirrors make a rotation through twice the angle #

Reflect across one line and then across another meeting it at angle θ, and the result is a rotation through 2θ about the crossing point. Reflect across two parallel lines a apart and the result is a translation by 2a. The second is the first with the crossing point sent to infinity, and both carry the same doubling.

This is the constraint that decides which mirror arrangements a plane group may have. The rotation a pair of mirrors produces has to be one of the orders a lattice permits — 1, 2, 3, 4 or 6 — so the angle between two mirrors of a plane group is 90°, 60°, 45° or 30° and nothing else. There is no group with mirrors at 50°, not because nobody drew one but because the pair would generate a rotation of order 3.6.

The doubling also explains a fact about counting that looks like an accident. p4m has eight mirror lines in a cell and p6m has twelve; each is twice the order of the principal rotation, because the mirrors through a centre of order n come in n families and the angle between neighbours is π/n. A plate of p6m is a fan of twelve lines through each six-fold centre, and the twelve is the doubling read backwards.

Two rotations force a third centre, and the triangle has the half-angles #

The general case is Euler’s, and it is the one that makes the classification feel inevitable. Rotate by α about A and then by β about B and the result is a rotation by α + β about a third point C, where the triangle ABC has angle α/2 at A and β/2 at B. When α + β is a whole turn there is no third centre and the product is a translation, which is the half-turn case seen from further away.

For a plane group this pins everything down. In p3 the three-fold centres come in three inequivalent kinds per cell, and their positions are not a choice: one centre and the two lattice translations produce the others by exactly this construction. In p6 the same argument produces the two-fold and three-fold centres from the six-fold one. A plate of a plane group is a consequence of two or three generators, and this is the machinery that produces it.

There is a second way to read the same law, and it is the one the site’s generators use. Growing an orbit is composing repeatedly: apply a generator, apply it again, apply the other one, and stop when nothing new appears. The orbit closes because the products fall back into a finite set of cosets, and the positions of the images are where the located products put them. So the pattern and the plate of elements are two views of one calculation, and a figure that drew the pattern from one and the marks from the other would be two calculations with nothing requiring them to agree.

A mirror and a translation, which is the fourth case #

The three cases above compose two operations of the same kind. The mixed case is where the fourth motion comes from, and it is short enough to do here rather than take on faith.

Reflect, then slide along the mirror. A reflection across the line $y = 0$ sends $(x, y)$ to $(x, -y)$; follow it with a translation by $(a, 0)$ and the result sends $(x, y)$ to $(x + a, -y)$. That has the same matrix as the mirror and a translation that no change of origin removes, because an origin shift changes the translation by $(M - I)\mathbf{s}$, which for this mirror is $(0, -2s_y)$ — able to alter the second coordinate and nothing else. It is a glide, and the slide is what was added.

Reflect, then slide across the mirror. Translate instead by $(0, b)$ and the result sends $(x, y)$ to $(x, -y + b)$, which fixes the line $y = b/2$ pointwise. It is a mirror, moved by half the translation — the same halving as two half-turns, arriving in a different place.

A general translation does both, since it splits into a part along the mirror and a part across it: the across-part moves the line by half, and the along-part becomes a glide of that amount. So one mirror and one translation in a pattern force a whole family of mirrors and glides, alternating, at half-steps of the lattice.

That derivation settles a fact that is usually presented as a table entry. Take a mirror across $y = 0$ and the centring translation $(\tfrac12, \tfrac12)$ of a centred rectangular lattice. The composite sends $(x, y)$ to $(x + \tfrac12, -y + \tfrac12)$: half a cell along the mirror direction, and half a cell across. The across-half puts the line at $y = \tfrac14$, and the along-half stays as a slide. The product is a glide at quarter height — which is why cm has glide lines between its mirrors and pm does not, and why the two are different groups on lattices that look alike. Nobody put the glide there. The centring vector and the mirror did, by composition.

Where the half-turn centres are, all of them #

The doubling in the first case has a consequence for the picture of a pattern rather than for its arithmetic, and it is the fastest route to a fact that is otherwise memorised.

Compose a half-turn with a lattice translation. If $h_p$ is the half-turn about $p$ and $t$ is a translation by $\mathbf{v}$, then $t \circ h_p$ is a half-turn about $p + \tfrac12\mathbf{v}$ — read straight off the first result, which said that the translation between two half-turns is twice the vector between their centres.

So one half-turn plus the lattice gives half-turns on the half-lattice. Every point of the form $p + \tfrac12\mathbf{v}$, for $\mathbf{v}$ a lattice vector, is a half-turn centre. The centres therefore sit on a lattice with half the spacing in each direction, which contains four times as many points per cell as the original.

Four of them are inequivalent. Modulo the translations there are exactly four classes — at the cell corner, at each of the two edge midpoints, and at the cell centre — because the half-lattice contains the lattice with index four. A pattern in p2 has four distinct kinds of two-fold centre, no two of them related by any symmetry of the pattern, and they can be occupied independently.

That is a statement a reader can check by eye and would not guess. The obvious expectation is one centre per cell, or perhaps one kind of centre appearing four times; the truth is four kinds, each appearing once, and it follows from a factor of two in a composition rule.

The same argument runs at every order. Composing a quarter turn with the lattice produces quarter turns at the corner and the cell centre, and half turns at the edge midpoints — which is why p4 has two inequivalent four-fold positions and one two-fold position rather than four of anything. The rule is always the same: the fixed point of a product sits at a fraction of the translation, the fraction is decided by the rotation, and the census of positions falls out of the arithmetic instead of out of a diagram.

What the round trip checked, and how #

Every claim above is computed twice.

The product is computed by multiplication and by construction, and the two are compared. For the half-turns the construction is 2(q − p) and the comparison is between rationals — exact, no tolerance. For Euler’s case the product is obtained by multiplying the pairs and the construction is checked against it, which is the direction that can fail: the arithmetic does not know about triangles.

The mirror case is the one place a tolerance appears, and it is worth being precise about where. The angle between two axes is a real number, measured in the plane rather than in the lattice basis, so the comparison between “twice the angle between the axes” and “the turn of the rotation the matrices produce” is made to within 10⁻⁹ radians. The rotation order is exact — it comes from the trace of an integer matrix — and only the angle is measured. Nothing about the classification depends on the tolerance; the figure would draw the same picture without it.

The refusals are three, and each is a way of getting the doubling wrong. Two half-turns about the same centre compose to the identity rather than to a translation. A four-fold rotation about a point that is not a centre of p4 is not an operation of p4 at all, which is what “where” means. And the translation two half-turns produce is required not to equal the vector between the centres, so a figure that dropped the factor of two would fail rather than draw a plausible arrow.

The same forcing has a shape as well as a plate. A fundamental domain is bounded by mirror lines and cornered on rotation centres, and both of those are located objects — so the domain’s shape is decided by the composition law and not by taste. Two different-looking domains of one group are related by a motion of the group, and the count of copies filling a cell is the group’s order modulo translations, which is the orbit-stabiliser statement seen through positions.

Where the exactness stops #

The arithmetic is exact and its reach is narrow, and the boundary is worth marking.

Positions of elements are exact. A rotation centre is the solution of (I − M)x = t, and the determinant of I − M is a small integer for every permitted order, so the centre is a rational point with a small denominator. The same holds for the offset of a mirror line. Nothing here is measured.

Angles between elements are not, and are not needed. The doubling law is stated in angles because that is how it reads, and it is used through the rotation order, which is an integer. A pattern is never classified by measuring an angle on it.

Nothing here decides anything about a pattern’s contents. The composition law is a fact about the group. Two patterns with the same group have the same plate of elements and can look nothing alike, which is the standing separation between the lattice underneath and the motif on it.

The three-dimensional version is the same law with one more case #

In space the law grows a term rather than a case. Two half-turns about parallel axes compose to a translation by twice the vector between them, exactly as in the plane. Two half-turns about intersecting axes compose to a rotation about the common perpendicular. And two reflections in intersecting planes compose to a rotation about the line where they meet, through twice the dihedral angle.

The new case is the one that has no plane analogue: composing a rotation with a translation along its own axis gives a screw, and composing a reflection with a translation in its own plane gives a glide. Both are operations whose translation part cannot be removed by any choice of origin, and both are invisible to the plane version of this argument because a plane has no room for the direction they need. That is the content of what the closure adds: the same composition law, in a space with one more direction, produces operations the plane cannot hold.

Who found it, and when #

The composition rules are older than the classification they produce. Euler published the construction for composing two rotations about different points in 1776, as a result about spherical motions; the plane version was in wide use through the nineteenth century, and it is the argument by which Fedorov and Schoenflies, working independently and finishing within a year of each other around 1891, could be confident their lists of space groups were complete. The counting only closes because the positions are forced. An enumeration of point groups plus lattices gives the arithmetic classes; what turns those into the seventeen and the two hundred and thirty is the question of which translation parts survive a change of origin, and that question is this composition law asked repeatedly.

The International Tables’ plates are the modern form of the same fact. Every mark on them is derivable from a generating set, which is why the Tables list generators at all rather than only diagrams.

Where the ladder goes next #

This rung establishes the object: composition with positions kept. Three rungs sit above it.

The first 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. The second is the extension problem — given a point group and a lattice, which assignments of translations to the generators give genuinely different groups — which is a space group is an extension and which the plane version of answers with the four non-symmorphic groups. The third is the normaliser, which asks which motions carry the whole plate of elements onto itself rather than one element to another, and which is the same pattern described twice.

All three are the same question at different scales: not what the operations are, but where.