Going up costs the cell a parameter

The descent with no shortcut ends by naming the direction it does not go, and states the reason in the form everybody states it: a group has finitely many maximal subgroups and infinitely many minimal supergroups, so going up is the harder question.

Both halves of that sentence are false for a plane group, and the true asymmetry is somewhere else and is sharper.

Downwards, a plane group has infinitely many maximal subgroups. A sublattice of prime index carried onto itself by the point group gives a subgroup of prime index, which is maximal because nothing can sit between it and the parent, and there is no largest prime. Upwards, a superlattice of prime index gives a minimal supergroup by exactly the same argument. And the two infinities are the same infinity: the map sending a superlattice $L' \supset L$ of index $n$ to the sublattice $nL' \subset L$ of index $n$ is a bijection, and it preserves invariance under the point group. Counted index by index, the two directions match exactly.

What is genuinely asymmetric is the other kind of step — the one that changes the point group and leaves the lattice alone. Of the thirty-one such steps among the seventeen plane groups, seventeen cost the lattice a parameter going up and nothing going down.

One edge, read twice #

Take the step between p2 and p4. Going down, the quarter-turns are discarded and what is left is a group with a half-turn and the same translations. Nothing is asked of the lattice, because the lattice has not been touched: every p4 pattern contains a p2 pattern, on the same cell, without exception.

Going up, the quarter-turns have to be added, and an operation added to a plane group must carry that group’s lattice onto itself. A quarter-turn does that only if the cell has equal edges at a right angle. A general oblique cell has two parameters, the ratio of its edges and the angle between them, and being square uses both. So a p2 structure has a p4 supergroup exactly when its measured cell happens to be square — which is a question about the metric, not about the group.

The arithmetic is worth doing once in coordinates, because it is where the two parameters go. A lattice is fixed, up to size and orientation, by its Gram matrix — the three numbers $a^2$, $b^2$ and $a\cdot b$, of which only two ratios matter. A quarter-turn $R$ carries the lattice onto itself when $R^{\mathsf T} Q R = Q$, and writing that out for the matrix $R$ that sends the first basis vector to the second gives $a^2 = b^2$ and $a \cdot b = 0$: two equations, and the lattice had two free ratios. A half-turn is $-I$, and $(-I)^{\mathsf T} Q (-I) = Q$ identically — no equation at all, which is why that one step is free for every group.

That is the whole of the asymmetry, and it is worth saying what kind of statement it is. The list of edges is one list. Reading it downwards and reading it upwards give the same thirty-one relations, in the same numbers, group by group. The two directions differ in what each edge demands, and nothing about the count reveals it.

Three strata, and a step that only moves one way #

The space of plane lattices has a stratification, and it is short. Discount the overall size and an oblique cell has two parameters left. Fixing the angle at a right angle spends one and gives the rectangular cell; making the two primitive vectors equal instead spends one and gives the rhombic cell, which is a centred rectangle written on its own two vectors. Spending both gives the square cell, and the hexagonal condition — equal vectors at 120° — also spends both at once. So the strata have codimension 0, 1, 1, 2 and 2, and the five plane lattices are exactly those five.

Which stratum a measured crystal is on is itself a decision rather than an observation, and the cell that settles the argument is where that decision is made. Every plane group demands a stratum: p1 and p2 are content with a general cell, the five rectangular groups need a right angle, cm and cmm need equal primitive vectors, and the eight groups with a four-fold or a three-fold need the cell fully specialised.

A step downwards keeps the lattice and so never moves left across this picture: a subgroup’s own demand on the cell is at worst weaker than its parent’s. A step upwards may move right, and moving right is a condition on the measured cell.

The stratification also explains why the free edges are where they are. Fourteen of the thirty-one edges have the same lattice type at both ends, and every one of them either sits entirely inside the specialised strata — p3 inside p6, p6 inside p6m, p4 inside p4m — or adds a mirror to a group already on the stratum that mirror needs, as pm inside pmm does. A step is free exactly when the operation being added was already permitted by the cell the group had, and that is a statement about the holohedry of the lattice rather than about either group: the operations a cell will accept are its holohedry, and a supergroup is free when its point group is still inside that.

One more consequence of the stratification is worth extracting, because it reverses the usual intuition about which groups are awkward. A group of high symmetry has an easy time going up: p3 sits on a hexagonal cell, which has no parameters left, so all three of its minimal supergroups are free and the only question about each is whether the atoms cooperate. A group of low symmetry has a hard time, because it has the most to spend and every step spends some. The diagram is therefore hardest to climb exactly where a crystallographer most often wants to climb it — at the bottom, where a structure has been solved in p1 or p2 and the suspicion is that it has more symmetry than that.

The census, and where the shading is #

Thirty-one edges. Fourteen of them have the same lattice type at both ends and are free in both directions: pm inside pmm, p4 inside p4m, p3 inside p6, and the rest. Seventeen change the lattice type, and those are free downwards and priced upwards.

The distribution across the seventeen is not even, and the pattern in it is the useful part. The totals on the two sides are equal group by group only in aggregate — p6m has four maximal subgroups below it and nothing above, p1 has nothing below and five above — which is the diagram of containments read as a lattice with a top and a bottom. The groups at the top of the diagram — p4m, p4g, p6m — have no supergroups at all, so the question does not arise. The groups on a fully specialised lattice, the tetragonal and hexagonal ones, have supergroups that are all free, because there is nothing left to spend. The groups on a general oblique cell have almost none that are free.

p1 has five minimal translationengleiche supergroups, and exactly one of them is free. The free one is p2: a half-turn is multiplication by $-1$, which carries every lattice onto itself whatever its shape, so a pattern with no symmetry at all always has a supergroup with a half-turn. The other four ask the cell to become rectangular, rectangular, rhombic and hexagonal.

p2 has six, and not one is free. Every way of adding an operation to a half-turn adds a mirror or a turn of order three, four or six, and every one of those constrains the cell. So a structure measured with a general oblique cell and refined in p2 has no candidate supergroup at all until somebody notices a metric relation among the cell parameters.

The index is worth noticing too. Four of p1’s five steps have index 2 and the fifth, to p3, has index 3 — because a three-fold cannot be added by doubling anything. The same happens above p2, where the step to p6 is index 3 while the step to p4 is index 2, and it happens for the same reason: the index of a translationengleiche step is the index of the point groups, and a three-fold has to arrive whole. So a structure suspected of a hidden three-fold is suspected of a step that triples the number of operations, which is a larger claim about the atoms than doubling is, as well as a harder one about the cell.

That is the practical content of the whole essay, and it is the situation a crystallographer with a suspicious structure is actually in. Whether a cell has a symmetry it was not given is a question with a tolerance in it — an angle that refines to 90.03° either is or is not a right angle depending on the error bars — and the diagram above says that the tolerance is the only thing standing between the structure and every supergroup it might have. There is no group-theoretic screening to do first.

The other step, counted in both directions #

The step that enlarges or shrinks the cell behaves completely differently, and its symmetry is exact.

A sublattice of index $n$ is the column span of an integer matrix $K$ with determinant $n$, and it is carried onto itself by an operation $M$ exactly when $K^{-1}MK$ is integral — a test with no metric in it, which is the point the sublattice arithmetic rests on. A superlattice of index $n$ is the dual of a sublattice of index $n$: if $K$ spans a sublattice then $K^{-\mathsf T}$ spans a lattice containing the original at the same index, distinct sublattices give distinct superlattices, and the invariance test becomes $K^{-1}M^{\mathsf T}K$ — a different matrix, because conjugating the dual transposes the operation.

The oblique case is the calibration. An oblique lattice’s holohedry is the identity and the half-turn, and a half-turn carries every sublattice onto itself, so every sublattice of index $n$ is invariant and the count is the total number of sublattices of that index — the sum of the divisors of $n$. Measured against the divisor sum at every index to 24, it agrees: 1, 3, 4, 7, 6, 12 for the first six. That is the only row of the table with no arithmetic condition in it, and it is the row that checks the rest.

Counted with those two tests independently, at every index to 24 and for every one of the five holohedries, the numbers are equal. The dashes are as informative as the counts: the square holohedry has no invariant lattice of index 3 in either direction, and the hexagonal none of index 2, which is the same arithmetic that leaves a four-fold group with no subgroup of index three arriving from the lattice rather than from the permutations.

And every row goes on forever, which is what makes the usual statement of the asymmetry wrong in its first half as well as its second. The lattice scaled by any whole number $n$ is carried onto itself by everything the parent is carried onto itself by, and it has index $n^2$. So every plane holohedry has an invariant sublattice at every square index, every plane group has infinitely many maximal subgroups of the klassengleiche kind, and the duality gives it infinitely many minimal supergroups to match.

Why the mistake is a natural one #

The received statement is not arbitrary, and tracking down where it comes from says something about what the two directions are for.

A subgroup is found by deleting and a supergroup is found by searching. Given a plane group, its maximal translationengleiche subgroups can be written down from the maximal subgroups of its point group, mechanically, with no further information — which is what sorting the index-two subgroups does and what a Bärnighausen tree records. Going up there is nothing to delete: the supergroup contains operations that the group does not have and that nothing in the group’s own data names. They have to be looked for, and the looking has to consult the lattice.

So the practical difficulty is real and the arithmetic reason given for it is wrong. The difficulty is that the answer depends on data outside the group — the cell parameters, with their error bars — rather than on there being more answers. Saying “infinitely many supergroups” puts the difficulty in the wrong place, and it suggests that a longer computation would settle it. A longer computation would not. A better measurement might.

The infinity people have in mind may also be a three-dimensional one. A space group’s minimal supergroups are harder than a plane group’s in a way that is not only a matter of size: there are more strata, the relations among cell parameters are more varied, and a single group can sit inside a supergroup in several inequivalent positions. Whatever is true there, importing the sentence into two dimensions imports a claim that the arithmetic above refutes, and the refutation takes a paragraph.

And the count that is genuinely infinite is infinite in both directions and is not usually what anybody means. A crystallographer asking for the supergroups of a structure is not asking about superlattices; a supergroup on a cell three times smaller is a different structure, not a symmetry the present one was hiding. The question is about the translationengleiche step, the step this essay prices, and there the list is short — at most six — and every entry is conditional.

What the pricing does not say #

A free edge is not a guarantee that the supergroup exists. Every p1 pattern has a p2 supergroup in the sense that a group containing it exists on the same lattice; whether a given structure has that symmetry is a question about the atoms, not the cell. The pricing here is a necessary condition, and it is only ever necessary. The sufficient half is the ordinary one — the motif has to be carried onto itself — and a motif drawn carelessly acquires symmetry it was never given, which is the failure in the opposite direction, and the reason a pattern meant to illustrate a group has to be grown from a motif with no symmetry of its own.

Nor does a priced edge mean the supergroup is rare. Cells become special for reasons: a structure built from rigid octahedra sharing corners tends to a cubic cell because the octahedra are cubic, and a layer stacked on itself tends to a right angle because stacking is perpendicular. The conditions this essay counts are conditions on a continuum, and real cells are not drawn from that continuum at random. What the count does say is that the condition has to hold rather than be nearly held, and that the group supplies no reason for it.

A cost of zero and a cost of two are not two points on one scale. The strata are nested and the codimensions add along a path, so a step from oblique to square costs two conditions and a step from rectangular to square costs one. But a rectangular cell that satisfies one more condition is square, while an oblique cell satisfying one condition might be rectangular or rhombic and the two are different specialisations. The number counts conditions rather than naming them.

And the whole account is about the plane. In three dimensions there are fourteen Bravais types over six metric strata rather than five over three, and the stratification is not a chain — so the same argument runs but the arithmetic is longer, and whether the received statement about infinitely many supergroups becomes true somewhere in that longer arithmetic is a question this does not answer. What is safe to carry across is the shape: the edge is priced in metric, and the price is the codimension.

Still open: what the tolerance turns the diagram into #

Every conditional edge above becomes a decision when a real cell is measured, and the decision has a tolerance in it. An oblique cell whose angle refines to 90.02(4)° is rectangular or it is not, and the diagram gives no advice about which.

What the diagram could give is a ranking, and it does not yet. Two conditional edges are not equally close to being met: a cell may be a tenth of a degree from rectangular and ten per cent from having equal edges, and the supergroup needing only the first is a live possibility while the one needing both is not. A distance from each stratum, measured in the same units the cell’s own standard uncertainties are in, would turn a list of candidates into an ordered one — and the ordering would be the thing a structure report ought to carry and does not. The measure exists in the literature for cells; putting it on the edges of this diagram is a computation that nothing above has done.