Sixteen candidates, ten groups

Take the point group with two perpendicular mirrors and the two-fold axis where they meet — the group crystallographers write mm2 and everyone else recognises as the symmetry of a brick with a top and a bottom. Put it on a primitive orthorhombic lattice. Ask how many space groups that gives.

The answer is ten, and this essay gets there by trying everything.

The search #

The group is generated by two mirrors: one perpendicular to a, one perpendicular to b. Their product is the two-fold along c, so fixing the two mirrors fixes everything.

Each mirror gets a translation. The translations are eighths, quarters or halves of the cell in principle, but a mirror’s square must be a lattice translation, which forces every component to be a half or a zero — so each generator has 2³ = 8 choices and there are 64 assignments in all. That is the search space, and it is small enough to walk.

Every assignment is closed: compose the two generators until nothing new appears, and keep the result if it has exactly four elements. In this class all sixty-four survive, because the consistency condition happens not to bite here — but the closure is run rather than skipped, because in other classes it does bite and the code should not know which case it is in.

Sixty-four groups, all genuinely different as sets of operations. Every one of them is a real symmetry group of a real periodic pattern. And there are ten space groups.

The first quotient: nobody chose the origin on purpose #

Most of the collapse is the origin, and it is the part that requires no judgement at all.

The mirror perpendicular to a has $M - I$ equal to diag(−2, 0, 0), so shifting the origin along a changes its translation’s a-component and nothing else. Shift by a quarter of the cell and the component changes by a half, which is enough to set it to zero. The same argument sets the b-component of the other mirror to zero.

So the a-component of the first mirror’s translation and the b-component of the second’s are not information — they are location rather than intrinsic — they say where somebody put the origin. Four of the six numbers survive: whether each mirror carries a half in each of the two directions lying in its own plane.

Sixteen. Each is a real, distinct group. And the Tables have ten.

The second quotient: which axis is called a #

Here is the point at which the counting stops being arithmetic.

The mirror perpendicular to a can carry a half along b, a half along c, both, or neither. Those four possibilities have names: m for neither, b for a half along b, c for a half along c, and n for both. The mirror perpendicular to b has the same four, with a in place of b. Sixteen combinations, sixteen symbols: Pmm2, Pmc2, Pma2, Pbn2, and so on.

But a and b are labels, in the sense that the cell is a choice and the labelling comes with it. Nothing in the crystal distinguishes them — the orthorhombic system has three mutually perpendicular axes of unequal length and which is called which is a convention. Swap them, and Pmc2₁ becomes Pcm2₁: the same crystal, described with the axes relabelled.

The Tables regard those as the same group in two settings, and count them once.

Working out which of the sixteen pair off is a small piece of combinatorics with a pleasant answer. Four of the sixteen are unchanged by the swap — the ones where both mirrors are doing the same thing to their own plane, which are Pmm2, Pcc2, Pba2 and Pnn2. The other twelve pair into six. Four plus six is ten.

The screw nobody put in #

That last observation is worth stopping on, because it is the clearest thing this enumeration shows and it is invisible from the symbols.

The two-fold along c was never given a translation. It is the product of the two mirrors, so its translation is whatever the product produces. Work it out: if the first mirror carries c₁ along c and the second carries c₂, the product carries c₁ + c₂ along c.

So when exactly one of the two mirrors has a half along c, their product is a 2₁ screw — an operation with an intrinsic translation, in a group where neither generator had one along that direction. Four of the ten groups get a screw axis this way, and the Tables record it in the symbol: Pmc2₁, Pca2₁, Pmn2₁, Pna2₁, where the subscript on the 2 was not chosen but computed.

The plane has exactly this, and this site has written about it: the group cm has a glide that nobody put in, produced by composing its mirror with the centring translation. The three-dimensional version is more common and less remarked on, and it is the reason the closure has to be run rather than the generators listed.

The same shape of argument, one dimension down #

There is a family resemblance here to something this site has already done, and it is worth naming because the resemblance is not a coincidence.

The seven frieze groups are derived on this site by enumerating sixteen candidates and watching nine of them collapse. Each of the four extra operations a strip can carry — a horizontal mirror, a vertical mirror, a half-turn, a glide — is either present or absent, giving 2⁴ = 16 subsets, and closing each subset under composition produces seven distinct groups.

Sixteen candidates, seven groups there; sixteen candidates, ten groups here. The numbers are a coincidence and the shape is not: in both cases a small set of independent binary choices is generated, and then a quotient is taken by something that does not change the group.

What differs is the kind of collapse, and comparing them is instructive. In the frieze case the collapse is renormalisation: composing a horizontal mirror with a glide gives a half translation, so the group generated by both has a lattice half the size of the one it started from, and against that smaller lattice the glide is just the mirror composed with a translation. The candidates collapse because two of them describe the same group on differently-sized lattices.

Here the lattice never changes size. The collapse is entirely about naming, and that is why the second quotient is the arguable one. Nothing about the crystal changed; only the label on an axis did.

Why not sixteen? A defence of the second quotient #

It is worth being honest that the drop from sixteen to ten is a convention in a way that the drop from sixty-four to sixteen is not.

Sixty-four to sixteen removes something nobody could defend keeping. Two descriptions differing only in where the origin sits are the same crystal by any reading; insisting they were different groups would make the count depend on a choice made for typographical convenience.

Sixteen to ten removes something a reasonable person could argue about, and it is the same kind of judgement that made the two-colour count come out at seventy-four here and forty-six in the literature. Pmc2₁ and Pcm2₁ are the same crystal seen with the axes renamed — but they are genuinely different subgroups of the group of all rigid motions, and a physicist working with an oriented sample in a magnetic field along a specific direction cares which is which. The Tables’ choice to count them once is a choice about what a space group is for: it is a classification of crystals, and a crystal does not know what its axes were called.

The number reported depends on that choice, and this site’s enumeration reports both because both mean something. What is not negotiable is that the choice must be stated: an enumeration that quietly quotients by axis relabelling and reports ten is giving a correct number for a question it did not ask out loud.

The change of basis everybody forgets #

The class m on a primitive lattice is where this bites, and it is where this site’s enumeration was wrong for an afternoon.

A single mirror perpendicular to b. Its glide vector lies in the a–c plane, so the candidates are a/2, c/2, (a+c)/2 and nothing: four groups, called Pm, Pa, Pc and Pn. Swapping the a and c axes — which keeps b unique, so the cell stays monoclinic — identifies Pa with Pc. That leaves three.

The Tables have two.

The missing identification is not a swap. It is a shear: replace c by a + c. That keeps b perpendicular to both, so the cell is still monoclinic and still conventional — the monoclinic system constrains only the two angles at the unique axis, and a shear in the plane perpendicular to it changes nothing that the system cares about. Under that change of basis, a glide vector of (a+c)/2 becomes c′/2. Pn is Pc.

That is why the Tables list Pn as a setting of Pc rather than as a group. It is also why the monoclinic system has the most confusing notation in crystallography, in a subject whose notation is already the barrier to entry: the same group appears in the literature as P2₁/c, P2₁/n and P2₁/a depending on which cell somebody chose, and the three symbols look like three groups.

The shear is available because the monoclinic cell has a free direction to shear along. The orthorhombic cell does not — all three axes are pinned perpendicular — which is why the mm2 class’s second quotient is only the axis swap, and why sixteen goes to ten rather than lower.

What the ten actually look like #

It is worth seeing the ten rather than only counting them, because the enumeration produces objects and not just a number.

The pattern across the ten is easy to state once the enumeration has been run. Of the sixteen candidates, four have neither mirror sliding along c, four have both, and eight have exactly one — and it is that last eight that produce a screw. Four of the ten groups therefore carry a 2₁ in the symbol, and it is the last character rather than the first, because it names the third direction rather than either of the mirrors.

Counting the glides gives the other useful summary: one of the ten has two plain mirrors, six have one mirror and one glide, and three have two glides. A crystal in this class with a mirror plane in it can hold a molecule lying flat on that plane, which is a real structural constraint and the reason the distinction matters outside the classification.

What the enumeration cannot do #

Three things, and the third is the largest.

It is conditional on the search grid. Translations are drawn from a grid of halves for the two-fold and mirror classes, thirds for the threefold, quarters for the fourfold — and the origin shifts from a grid computed from the generators rather than chosen. That last computation matters: a threefold’s $M - I$ has determinant three in the plane it turns, so an origin shift of a ninth is needed to reach a translation of a third, and a grid of sixths — which looks generous — reports nine groups in the class 3 where there are three. Every group in the nine is real, so the failure produces a plausible wrong answer with nothing to flag it.

It works one class at a time. Six classes are enumerated here, covering the classes the essays argue about. Running it on all seventy-three is not a matter of a longer loop: the second quotient needs the full normaliser of each point group in the integer matrices, and computing that correctly for each of the seventy-three is the content of the classification rather than an application of it.

It says nothing about which of the ten a crystal will choose. All ten are equally legal and they are wildly unequal in the wild. Pna2₁ and Pca2₁ are common; several of the others are rare enough that a structure in one is worth a remark. Why is a question about packing and about hydrogen bonds and about what molecules are shaped like, and this machinery cannot see any of it.

Why this class and not another #

Six arithmetic classes are enumerated in this field and mm2 is the one that got the essay, which is worth defending.

The class 2 on a primitive lattice gives two groups and the class m gives two. Those are the right places to start, because they are small enough to hold in the head, and both appear as figures here. But two is not enough of a number to show anything: with two answers, the difference between “the enumeration produced this” and “somebody listed the two possibilities” is invisible.

The class 222 gives four, and the class 4 gives four, and both are better. The fourfold class is particularly clean, because its four groups are P4, P4₁, P4₂ and P4₃ — the four possible rises of a fourfold axis, and nothing else — which makes the enumeration look like a statement about a single number rather than about a structure.

mm2 earns the essay because it is the smallest class in which both quotients do work, and in which the second quotient’s answer is arguable. Sixty-four to sixteen is the origin doing what the origin does; sixteen to ten is a convention being applied, and a reader can see exactly which rows it merged and decide whether they agree. A class where the second quotient does nothing hides the question.

The next rung takes the “what counts as the same group” question one step further, to the one place where the Tables’ answer and the abstract algebraist’s answer differ by eleven.

The ten, by name #

An enumeration that arrives at a number is worth checking against the answer somebody else got, and this one can be checked completely, because the ten have names.

They are Pmm2, Pmc2₁, Pcc2, Pma2, Pca2₁, Pnc2, Pmn2₁, Pba2, Pna2₁ and Pnn2 — and in the International Tables they are numbers 25 through 34, consecutive, with nothing else between them. The Tables order space groups by class, so a class occupying an unbroken run of ten numbers is exactly the statement this essay’s search makes.

Reading the symbols back against the search is the more useful check. The three letters after the lattice are the two mirrors and the axis: m for a mirror with no intrinsic translation, c or a or b for a glide with a half along that axis, n for a glide with halves along two. So Pmm2 is the assignment where both mirrors got nothing; Pnn2 is the one where both got two halves; Pcc2 is the one where both got a half along c.

The subscript is the screw nobody put in. Four of the ten are written with a 2₁ rather than a 2 — Pmc2₁, Pca2₁, Pmn2₁ and Pna2₁ — and in each of them exactly one mirror carries a half along c, which is precisely the condition the previous section derives. The symbol is not recording a decision anybody made; it is recording what the product of the two generators turned out to be.

One of the ten is symmorphic and nine are not. Pmm2 is the assignment with no intrinsic translation anywhere, so its point group sits inside it; the other nine each contain at least one glide or screw and contain no copy of mm2 at all. That ratio — one out of ten — is a fair sample of the whole classification, where seventy-three of two hundred and thirty are symmorphic.

What the same search costs for everything else #

This class was chosen because the search is small enough to state, and it is worth saying how the full classification scales, because the scaling is what decided how the two hundred and thirty were actually found.

The unit of the search is not the point group. It is the point group together with a lattice type — a pair sometimes described as an arithmetic crystal class, and there are seventy-three of them. The class enumerated here is mm2 on a primitive lattice; mm2 on a C-centred lattice, on an A-centred one, on a face-centred one and on a body-centred one are four further searches, each with its own count, and together the five account for every space group of this class.

Each search has the same shape and a different size. Attach a translation to every generator, close, discard the assignments that fail consistency, quotient by origin shifts, then quotient by the relabellings the lattice permits. The last step is where the arithmetic stops and judgement begins — here it was which axis is called a, and in a cubic class it is a much larger group of relabellings.

Adding the seventy-three results gives two hundred and nineteen, and the familiar two hundred and thirty appears only when eleven of them are counted twice because they come in mirror-image pairs that no proper motion relates. Which number is right depends on what is being asked, and that question has its own essay.

The reason to do one class by hand is that the machinery is invisible in the total. Two hundred and thirty is a number to be looked up. Sixty-four, then sixteen, then ten, with a reason for each collapse, is a computation — and the two collapses are of quite different kinds, one forced by arithmetic and one chosen by convention. No amount of staring at the final figure recovers that distinction.