Three mirrors never make a crystal either

Two operations make rock salt’s group, and Pmmm needs six found, for each of the 212 space groups that their own operations can build, the fewest of them that suffice. Sixty-four need two, ninety-three need three, and the rest need four or more. Two mirrors never make a crystal then asked which pairs of kinds make the sixty-four. Every one of its 960 cases was decided, and two zeros needed no computation. Two mirrors make nothing, because two operations that undo themselves generate a dihedral group, and a rotation with a mirror makes nothing, because an axis and a plane always share a point or a direction.

It ended on the ninety-three, where the same question has thirty-five columns, one for each triple of kinds with repetition: rotation, screw, mirror, glide and rotoinversion, the last including the inversion. Some columns could be settled in advance. Three mirrors generate a group of reflections, and a group of reflections with translations in three directions needs at least four mirrors, so three mirrors should make nothing. But three fixed sets can avoid a common point where two cannot, so a rotation with two mirrors, or two rotations with a mirror, might succeed. Which ones do was left to the finite-quotient method that had decided every pair.

This essay runs it. Of 3,255 cases, 3,253 are decided. Three mirrors make none of the ninety-three, as predicted. A rotation with two mirrors makes exactly three, all of them cubic. Three plain rotations make thirteen, none with a mirror. A screw with two glides makes fifty-three, the most of any triple. One group, $Im\bar{3}m$, can be made thirty different ways, and three groups can be made only one way. The two cases left open are both in one trigonal group, $P\bar{3}m1$.

Three ways to decide a case, and a fourth #

The picture at the head of this essay is the result. The methods are those of the pair table, carried up by one generator.

The first is to find a triple. The group’s own operations of each kind, with their translates in a box of neighbouring cells, are drawn three at a time and tested for whether they generate the whole group: whether their products reach every coset of the lattice, and whether the translations they produce, read off by Schreier’s lemma, span the lattice rather than a sublattice. It is the same closure that what the closure adds watches to the end in a finite quotient. That finds 978 generating triples, and every one is checked a second time.

The second is the point group. Three operations generate a group whose point group is generated by their three linear parts, so if no three linear parts of the named kinds generate the point group, no three operations of those kinds generate the space group. This settles 2,114 cases without looking at a single translation. Three rotations never make a group with a mirror, three mirrors never make a group with a four-fold rotation, and so on.

The third handles what the first two leave. A triple that generates a group also generates every finite quotient of it, in particular the group of operations modulo $m$ times the lattice. If no three residues of the right kinds generate that finite group, no three operations do. That settles 133 cases modulo two and 27 modulo three. The fourth way was needed once. In $P\bar{6}2m$, a rotation, a rotation and a mirror survived the quotients modulo two, three and four and a search among operations up to two cells out, and were refuted modulo six, in a quotient of 2,592 elements where the three-fold and two-fold translations both appear.

That leaves two cases, both in $P\bar{3}m1$: two rotations with a glide, and two rotations with a rotoinversion. Neither was found among 200,000 triples drawn from operations up to two cells out, and neither is refuted modulo two, three, four or six. They are recorded as open, and they are the only gaps in the table.

Three mirrors #

The column of three mirrors is empty, and every one of its ninety-three cases is decided, 84 by the point group and nine in a quotient. That is the outcome Coxeter’s theory of reflection groups predicts. A group generated by reflections is determined by the angles between its mirrors, and the mirrors divide space into congruent cells, the group’s fundamental domains. For a group with translations in three independent directions those cells must be bounded, and a bounded polyhedron needs at least four faces. Three mirrors bound at best an infinite wedge or prism, and the group they generate is a point group or the group of a rod or a layer.

The plane had the same rule one dimension down. Four groups are made of mirrors found that three mirrors, the walls of a triangle, make the plane’s triangular reflection groups, and that two never do. In space it is four walls, a tetrahedron or more, and three is one short.

What the census adds is that the rule holds group by group for the three-generator groups, including the nine where the point group alone does not rule it out. Those nine are groups whose point group three mirror orientations do generate: $Cm$, $Imm2$, $I4mm$, $P6mm$, $P3m1$, $P\bar{6}m2$, $P\bar{6}2m$, $Pm\bar{3}m$ and $Im\bar{3}m$. In each the quotient shows that no choice of three of their mirrors closes up into the translations. The finite computation and the geometric theorem agree on all ninety-three.

A rotation and two mirrors #

The pair table found that a rotation and a mirror make nothing, because a line and a plane always share a point or a direction, and whatever they share the group keeps. With a second mirror the obstacle can be avoided: two planes meet in a line, and a rotation axis can be skew to that line and miss both planes’ common points. The census finds that it works for exactly three groups, $Pm\bar{3}$, $Pm\bar{3}m$ and $Im\bar{3}m$, and all three are cubic.

The generating triples the search found show how. In $Pm\bar{3}$ the two mirrors are parallel, at different heights, so together they make a translation across them, and the rotation is a three-fold one about a body diagonal, which carries that translation onto the other two axes. In $Pm\bar{3}m$ the rotation is a four-fold one and the mirrors are a coordinate mirror and a diagonal one; in $Im\bar{3}m$ it is a two-fold rotation with two mirrors of different orientations. What the three share is a rotation that carries one mirror’s orientation to another, so that the group the three generate has mirrors, and products of mirrors, in more directions than the two it was given. In the cubic groups the rotations have the room to do that in every direction at once. The census finds no non-cubic group made by a rotation and two mirrors, and it does not supply a one-line reason for that, as the pair table had for its zeros.

The neighbouring columns show how much a glide or a rotoinversion changes. Two mirrors with a glide make ten groups, from $Imm2$ to $Im\bar{3}m$, because a glide carries its own translation and supplies a direction the two mirrors cannot. Two mirrors with a rotoinversion make six, among them $P\bar{6}2m$ and the cubic groups with a centre.

Three rotations #

Three plain rotations make thirteen groups. None of them has a mirror, a glide or an inversion, because rotations compose only to rotations: the point group check rules out all seventy-one types with a mirror or a glide, and every type with an inversion. The thirteen are Sohncke groups, the groups a single hand may sit in, and they reach well beyond the cubic system. Two plain rotations made only six groups, all cubic. Three make tetragonal, trigonal, hexagonal and orthorhombic groups too: $I222$, $P422$, $P321$, $P622$, $P6_222$, $P6_322$, $R32$ and others, as well as the cubic $P23$ and $F23$.

The reason two were confined to the cubic system and three are not is the pair table’s geometry again. Two plain rotations must have skew axes of different directions, and outside the cubic system every rotation of order three or more is parallel to one direction, so a pair can only be a rotation of high order and a perpendicular two-fold one, which share a point or fail to reach the third direction. With three, two perpendicular two-fold rotations about skew axes and a rotation along the principal axis can avoid every common point and every common direction. $P6_222$ is the striking case. Its name advertises a screw, and three of its plain rotations generate it anyway: the screw is in the group, and it is a product of the rotations.

Three screws make eighteen groups, and twelve of them are in the three-rotation list too. The other six are monoclinic, orthorhombic and tetragonal groups with screw axes, $P2_1$ and $P4_2$ among them, where the screws reach what the plain rotations cannot. Three rotoinversions make twelve, all of them with an inversion or a rotoinversion in their names: $P\bar{4}$, $P4/m$, $P\bar{3}$, $P6/m$ and the cubic groups with a centre.

Where the translations come from #

A space group’s translations have to be made by its generators, and the kinds divide by whether they can make one on their own. A screw and a glide each carry a translation in their square, a slide along the axis or along the plane, so a set of them brings its own translations with it. A plain rotation, a mirror and a rotoinversion each fix a point, and they can make a translation only by being placed apart, so that the product of two of them about different centres moves everything.

The table separates the ninety-three by that. Eighty-three can be made by three operations none of which fixes a point, screws and glides alone. Fifty-six can be made by three that each fix a point, rotations, mirrors and rotoinversions alone, placed so that no point and no direction is fixed by all three. The other thirty-seven cannot be made without at least one screw or glide among the three. They include every three-generator group whose only non-translational operations are screws or glides, such as $P2_1$ and $Pc$, but also groups like $Pmm2$ types and $Imm2$ types, which have plenty of mirrors and still need a glide to finish the translations. Eleven screws and no others and the five glide planes list the slides a lattice allows; here those slides turn out to be what a third of the three-generator groups cannot be built without.

The plane had the same division in miniature. Six groups are made of glides found the glide doing work that mirrors could not, and the mirror columns are the emptiest here for the same reason as there. A mirror’s fixed set is a whole plane, and two of them always share a line or a direction; only a third operation of a different kind, placed off that line, can break the deadlock.

How many ways each group is made #

The spread is wide. Three groups are made by one triple only, five by two, and at the other end $Im\bar{3}m$ is made by thirty of the thirty-five triples. The cubic groups with mirrors and a centre are the most versatile, because they contain every kind of operation in many orientations, and almost any three kinds can be placed so as to reach everything.

The three made one way are all monoclinic, and two of them are decided before any search. They appear in the figure under names that give their arithmetic class and their index within it, since no standard symbol was read off their operations. $P2$ type 1 is $P2_1$: apart from translations its operations are all two-fold screws, so three screws are the only triple available, and the census finds that they suffice. $Pm$ type 1 is $Pc$, whose operations are all glides, and three glides make it. $Cm$ has both mirrors and glides, and of the triples drawn from them only a mirror with two glides makes it; two mirrors with a glide do not, and neither do three of either.

Three generators are needed for these groups and not two because of what two operations make rock salt’s group measured. The quotient of the group modulo twice its lattice needs three generators, so no pair can reach it.

The cases that took the most #

The quotient modulo two settles most of the cases the point group leaves, 133 of 161. The modulus matters for the same reason as in the pair table: a group with three-fold axes puts their translations in thirds of a cell, and only a quotient by three times the lattice sees them. The twenty-seven cases settled modulo three are all in trigonal and hexagonal groups.

The single case settled modulo six is the one where both kinds of translation had to be seen at once. In $P\bar{6}2m$ a rotation, a rotation and a mirror pass the quotients by two, three and four times the lattice. Modulo six the group has 2,592 elements, and every triple of residues of the right kinds whose linear parts reach the point group generates a proper subgroup. So no triple of the group’s own rotations, rotation and mirror generates it.

The two open cases in $P\bar{3}m1$ are the table’s honest gaps. A search can only find triples, and a group has infinitely many of each kind; a quotient can only refute. Neither has worked here, and the next attempt would be a quotient modulo twelve, with $12 \times 12^3 = 20{,}736$ elements and correspondingly more residue triples to close.

What the table rests on #

The kinds are the pair table’s. A rotation is a plain rotation, with a fixed axis and no slide along it; a screw slides by a fraction of a lattice vector; mirrors and glides likewise; and the improper kind collects the inversion and the rotoinversions. An operation whose slide is a whole lattice vector is a plain rotation or mirror composed with a translation and is excluded. The verdicts depend on that convention and on nothing else.

The table covers only the ninety-three types that need three operations, and it says nothing about the sixty-four that need two except that a pair of kinds that makes them would be one triple’s worth of help. Nor does it cover the types that need four or more, of which $Pmmm$, needing six, is the extreme.

Who worked out the pieces #

Coxeter’s classification of the discrete groups generated by reflections, from 1934, is the reason three mirrors make nothing: the fundamental domain of a crystallographic reflection group in space is a polyhedron, and a polyhedron needs four faces. The method of proving that a set does not generate a group by exhibiting a finite quotient it fails to generate is standard in computational group theory. Generating sets for every space group are listed in the International Tables, but as generators of the group’s coset structure rather than as the fewest of its own operations of named kinds, which is the question here.

What the table must refuse #

Each verdict of “yes” is re-checked from scratch, the three operations’ kinds recomputed and the triple handed to the generation test again. The three-mirror column is recomputed in full on every check, because it is the column with a theorem behind it, and a table that found a crystal made of three mirrors would be wrong about Coxeter before it was wrong about anything else. The refusal is the one the point group should decide before any search: three rotations for any of the seventy-one types with a mirror or a glide.

Still open: two cases in one trigonal group, and the four-generator groups #

The table has two gaps, both in $P\bar{3}m1$, a group that has two-fold rotations across its three-fold axis, mirrors, glides and the inversion. Whether two of its plain rotations and a glide, or two rotations and a rotoinversion, can generate it is undecided. A quotient modulo twelve, or a proof that every triple of those kinds shares a fixed point or a fixed direction, would close it. Beyond the table lie the fifty-five types that need four operations or more, for which the columns number seventy, one for each choice of four kinds, and the same three methods would decide them.