Two mirrors never make a crystal

Two operations make rock salt’s group, and Pmmm needs six counted, for every space group that its own operations can build, how few of them suffice. The operations allowed were a group’s own rotations, screws, mirrors, glides and rotoinversions, with no lattice translation handed over outright, and the answer was a number per group: two for sixty-four of the 212, up to six for Pmmm. That essay ended on a question its search could not answer. The search stopped at its first success, so it said how many operations a group needs and nothing about which kinds. Rock salt’s group $Fm\bar{3}m$ had turned up as a screw with a glide and as two rotoinversions. Could two rotations do it, or two mirrors?

This essay answers that question for all sixty-four two-generator groups and all fifteen pairs of kinds. There are five kinds: a rotation about a fixed axis, a screw, a mirror, a glide, and an improper rotation, which covers the inversion and the rotoinversions. Every one of the 960 cases is decided, either by a generating pair found or by a proof that none exists. A screw and a glide together make thirty-two of the sixty-four groups, more than any other pair. Two mirrors make none, and a rotation with a mirror makes none. Neither fact needs the computation, and the reasons are worth more than the zeros. Two plain rotations make exactly six groups, all of them cubic and none with a mirror or an inversion.

Three ways to decide a case #

A case is a group and a pair of kinds, say $Fm\bar{3}m$ with a screw and a glide. It can be settled three ways, and the picture at the head of this essay colours each pair of kinds by how its sixty-four cases were settled.

The first way is to find a pair. The group’s own operations of each kind are listed with their translates in a box of neighbouring cells, pairs are drawn from the two lists, and each is tested for whether it generates the whole group. The test is the one seven space groups cannot make their own translations introduced. The pair must reach every coset of the lattice, and the translations it produces, read off by Schreier’s lemma, must span the lattice itself rather than a sublattice. A pair that passes is a proof that the kinds suffice, and every one of the 211 pairs found is checked a second time from scratch.

The second way is to look only at the linear parts. Two operations generate a group whose point group is generated by their two linear parts, so if no linear part of the first kind and no linear part of the second together generate the point group, no pair of those kinds can generate the group. This settles 702 of the 960 cases at once. Two rotations, for instance, can never make a group with a mirror, since rotations compose only to rotations.

The third way handles what is left: pairs of kinds whose linear parts do reach the point group and for which no pair is found. Here the proof comes from a finite quotient, the method the previous count used for its fourteen exceptions. A pair that generates the group $G$ also generates every quotient of it, and in particular the finite group $G/mL$ of the operations modulo $m$ times the lattice. So if no pair of residues of the right kinds generates $G/mL$, no pair of operations generates $G$. That quotient is a finite group of a few hundred or a few thousand elements, and every pair of its residues can be tried. Forty-seven cases are settled this way, all at $m = 2$ or $m = 3$. In none was the search left without a verdict.

Two operations that undo themselves #

The two mirrors’ zero is a theorem about any group. A mirror undoes itself: applied twice it is the identity. So does a half-turn, and so does the inversion. And any two operations $a$ and $b$ that each undo themselves generate a dihedral group, the group of a regular polygon or its infinite relative. Every element of the group they generate is an alternating word $abab\ldots$ or $baba\ldots$, so the group is determined by the single product $ab$, and it consists of the powers of $ab$ together with those powers followed by $a$.

The picture shows what that means in space. Two parallel mirrors compose to a translation perpendicular to them, and every element of their group is a translation along that one direction or a reflection; the images of a point march along a single line. Two crossing mirrors compose to a rotation about the line where they meet, and every image of a point stays at the same distance from that line. Neither group has translations in three independent directions, so neither is a space group, and no choice of mirrors in any crystal changes that.

The same argument disposes of every pair drawn from half-turns, mirrors and the inversion. Two half-turns, a half-turn with a mirror, the inversion with either: none of these pairs makes a crystal, whatever the group. A pair of plain rotations or of rotoinversions can only succeed when at least one of the two has order three, four or six and so does not undo itself.

The plane already knew this. In six groups are made of glides mirrors generate four plane groups, but only with three or four mirrors at once, the walls of a triangle or a rectangle. Two mirrors in the plane make the same dihedral groups as here.

An axis and a plane always meet #

The other zero is a rotation with a mirror, and its reason is also geometric. A plain rotation leaves every point of its axis where it was, and a mirror leaves every point of its plane where it was. A line and a plane in space either meet or are parallel.

If they meet, the point where they meet is fixed by both operations, so it is fixed by everything they generate. A group fixing a point is finite, the point group of a molecule, and has no translations at all.

If they are parallel, the axis runs along the mirror’s plane. Then neither operation moves anything along the axis’s direction: the rotation turns about it and the mirror reflects across a plane containing it. So nothing they generate does either, and the group has no translation with a component along that direction. Two dimensions of translation at most, and again no space group.

So a rotation and a mirror never generate a space group. Fifty-six of the sixty-four cases are settled by the point group already. The other eight have linear parts that do reach the point group, $R3m$ among them, and a finite quotient refutes them in the census. The two-line argument covers them too: what the quotient detects is the shared fixed point.

The same argument says when two plain rotations can work. Their axes must not meet, or they share a point, and must not be parallel, or they share a direction. They must be skew. In every crystal system but the cubic, all rotations of order three or more are parallel to one axis, and the census finds that skew pairs of plain rotations never make a group outside the cubic system.

What a quotient sees #

The forty-seven cases the finite quotients settle are the ones where geometry gives no quick answer and the point group gives none either. Each is a small exhaustive computation, and two of them show how the method works.

$P31m$ with a mirror and a glide is the clearest. Its point group is generated by one of its mirrors and one of its glides’ linear parts, so the point group raises no objection. Modulo twice the lattice the group has 48 elements, and among the residues of its mirrors and glides there are pairs that generate all 48, so the quotient modulo two proves nothing. Modulo three times the lattice it has 162 elements. There are 486 pairs of a mirror’s residue and a glide’s residue whose linear parts generate the point group, and every one of them generates a proper subgroup. A mirror and a glide that generated $P31m$ would generate the quotient too, so none does. A pair can pass one quotient and fail another; the proof needs only one failure, and here the failure is in the threes, where the three-fold axes put their translations.

Rock salt’s group with a screw and a mirror is the other kind. Modulo twice its lattice it has 384 elements, there are 192 pairs of a screw’s residue and a mirror’s residue that reach the point group, and none generates the quotient. In this case the computation says no and gives no reason, and the reason has not been looked for. What the proof does show is that no cleverer choice of screw and mirror, further out in the crystal or along some other line, can rescue the pair. Every screw and every mirror of the group falls into one of the residues tried, so the whole infinite family of candidate pairs is covered by 192 finite cases.

That is what makes the table complete rather than merely large. A search can only find pairs, and a group has infinitely many pairs of each kind, so a search that finds nothing proves nothing. The quotient turns each infinite family into a finite list. What the closure adds made the same move in the other direction: closing a few operations produces every operation the group has, and the finite quotient is where that closure can be watched to the end.

Six groups from two plain rotations #

Two plain rotations make six groups: $P432$, $P4_132$, $P2_13$, $I2_13$, $I432$ and $F432$. All six are cubic, and all six are Sohncke groups, with no mirror, no glide and no inversion. That last fact is forced, since rotations compose only to rotations. Every one of them is among the sixty-five groups a single hand may sit in.

In each, the two rotations have skew axes of different directions, typically a three-fold along a body diagonal and a four-fold or two-fold along an axis, placed so that neither crosses the other. Two of the six, $P4_132$ and $P2_13$, have screw axes in their names, and yet plain rotations generate them. A screw is an element of the group, and so is a plain rotation about a different line. Each group has both, and the plain ones suffice.

The six are not the whole cubic Sohncke list. $P23$ itself, the simplest cubic group, needs three operations of any kind, so it is not in the table at all. $I23$ is in it, and two plain rotations are refuted for it modulo two: every pair of its plain rotations whose linear parts reach the point group generates, modulo twice the lattice, a proper subgroup.

Groups made one way only #

Eight groups are made by exactly one pair of kinds, and the list is lopsided. Five need two screws: $P2_12_12_1$, $P4_1$, $P3_1$, $P6_1$ and $P4_232$. Two need two glides: $Cc$ and $P31m$. One, $P\bar{6}$, needs two rotoinversions.

The screw groups are the natural ones to be found here. $P4_1$ has only four-fold screws, two-fold screws, and translations, and the only own operations that can produce a translation along the axis are the screws themselves. Two four-fold screws about different parallel axes generate the group, and nothing else of its own can. $P2_12_12_1$ has nothing but screws, along three perpendicular directions, so two of its screws are the only own operations available at all. The census confirms that two do suffice for it, as the previous count found.

$Cc$ is the monoclinic group whose only operations besides translations are glides, so two glides are forced. $P31m$ is less obvious. It has mirrors, glides and three-fold rotations, and the only pair of kinds that generates it is two glides. Two mirrors cannot, by the dihedral argument. A rotation with a mirror cannot, by the fixed-point argument. A rotation with a glide is refuted modulo two, and a mirror with a glide modulo three.

$P\bar{6}$ is made by two six-fold rotoinversions and by nothing else. Its other own operations are three-fold rotations and the mirror perpendicular to the axis, and every pair involving them either shares a fixed point or is refuted in a quotient.

How many ways each group is made #

At the other end, most cubic groups can be made in many ways. Rock salt’s group $Fm\bar{3}m$, the question the previous essay ended on, is made by six pairs: a rotation with a glide, a rotation with a rotoinversion, a screw with a glide, a screw with a rotoinversion, a glide with a rotoinversion, and two rotoinversions. Two rotations cannot make it, because it has mirrors, and two mirrors cannot make it, because no group can be made by two mirrors. Another type in the same arithmetic class is made by seven pairs, the most of any group.

The full table is in the grid, one row per group. Read down a column and the pattern of the kinds appears. The screw-and-glide column is the fullest, because a screw and a glide each carry a translation of their own, so between them they can produce translations along independent directions without any help. The mirror columns are the emptiest. A mirror is a fixed plane, and pairing it with anything that also has fixed points runs into the arguments above. The improper column is fuller than the rotation column for the same reason: a four-fold or six-fold rotoinversion fixes only a single point, much less than an axis.

What the table rests on #

The verdicts rest on the definitions of the kinds, and the definitions are a convention. A rotation here is a plain rotation, an operation with a fixed axis and no slide along it. A screw slides along its axis by a fraction of a lattice vector, so that it fixes nothing; eleven screws and no others lists the fractions a lattice allows. An operation whose slide is a whole lattice vector is neither: it is a plain rotation composed with a translation, and since translations are not allowed as generators, it is excluded. The same holds for mirrors and glides. The kind is judged on the operation itself, not on its coset, since a coset contains both a plain rotation and screws only when the group has both.

The table is also only for the sixty-four groups that two operations of any kinds can make. A group that needs three cannot be made by any pair, and asking which triples of kinds suffice is a larger question: thirty-five triples of five kinds, for ninety-three groups.

And the names. The 64 types are identified by their arithmetic class and, where a standard symbol could be read off the operations themselves, by that symbol: a four-fold screw axis climbing a quarter of a cell makes a type $P4_1$ rather than $P4_2$. Where no symbol is read off, a type is named by its class and its index among that class’s types. The verdicts do not depend on the names.

Coxeter, and the groups made of reflections #

That a group generated by reflections is determined by the angles between them is H. S. M. Coxeter’s theorem of 1934, and it is the reason two mirrors give only dihedral groups. A discrete group generated by reflections in space that has translations in three directions needs at least four mirrors, the walls of a tetrahedron, and more for a prism or a box. That is the three-dimensional version of what four groups are made of mirrors found for the plane, where three or four walls are needed. The question of how few operations of each kind generate a crystallographic group is younger, and the finite-quotient method that settles the negative cases here is the one the previous essay borrowed from computational group theory, where it is a standard way to prove that a set does not generate.

What the table must refuse #

Each verdict of “yes” is re-checked independently: the two operations’ kinds are recomputed and the pair is handed to the generation test again, and all 211 pass. A case that the point group should settle at once is checked to be settled that way: two rotations are refused for $Pmm2$ because its point group has mirrors. And the dihedral argument is checked against an example. Two inversions about different points generate translations along the line joining the points and nothing else, so they do not make $P\bar{1}$, though $P\bar{1}$ is made of nothing but inversions and translations and needs four of its own inversions.

Still open: the three-generator groups #

Ninety-three types need three of their own operations, and the same table for them has thirty-five columns, one for each choice of three kinds with repetition. Some columns are settled in advance. Three mirrors generate a reflection group, and a reflection group with translations in three directions needs at least four walls, so three mirrors make nothing. A rotation with two mirrors, or two rotations with a mirror, can still share a fixed point and fail for the reason here, but they need not: three fixed sets can avoid a common point where two cannot. The columns that mix kinds with fixed points and kinds without are the ones no argument here settles, and the finite-quotient method would decide each case as it decided these.