Three on a plane, and never four

A reflection in a plane, composed with a translation lying in that plane, is again a reflection in the same plane — a mirror becomes a glide, or one glide another. So a plane of a space group carries not one reflection operation but a family of them, one for each lattice vector in the plane. Counted modulo the conventional cell’s own translations, which the symbol’s letters are measured against, the family has as many members as there are cosets of the conventional mesh among the lattice vectors lying in the plane.

The plane that carries two glides is about planes with two, where a centring vector lies in the plane and the two operations it separates both have a claim on the symbol — the case for which the International Tables introduced the letter e in 1992. A line carries one screw showed that the same thing cannot happen on an axis. It ended by sharpening the plane’s question: a lattice with two independent centring vectors in one plane would give four operations, and a face-centred lattice has planes with exactly that. Does any of the two hundred and thirty have a plane carrying four?

No — and the answer needs no list of space groups at all. It is decided by the fourteen lattices, and running it turns up a case the question did not anticipate: every mirror plane of the rhombohedral lattice carries three.

Why the lattices decide it #

Every reflection operation of a space group has a linear part that is a symmetry of the group’s lattice: it is one of the mirrors of the lattice’s holohedry. So the planes that can carry reflections, in any space group on a given lattice, are the planes parallel to those mirrors, and nothing else. The operations a given plane carries differ from one another by lattice vectors lying in it, so their number modulo the conventional mesh is fixed by the lattice and the plane’s orientation — not by which space group it is, and not by where the plane sits.

That turns a question about two hundred and thirty groups into one about fourteen lattices and their mirrors. For each lattice the holohedry is found by searching for the integer matrices that preserve its metric and its centring, as the count of the fourteen does; for each mirror among them, the lattice vectors in its plane are listed, and the cosets of the conventional mesh among them are counted.

The primitive lattices carry one on every plane, as they must: every lattice vector is an integer combination of the cell edges, so no vector in a plane lies outside the plane’s conventional mesh. The centred lattices carry two on exactly the planes their centring vectors lie in. For the C-centred orthorhombic lattice that is one plane of three — the one containing the centred face. For the body-centred tetragonal and cubic lattices it is the diagonal planes, which contain the body diagonal; the body-centred orthorhombic lattice has no diagonal mirrors and carries one everywhere. For both face-centred lattices it is every mirror plane: each coordinate plane contains the centring of the face it is parallel to.

A centring vector out of the plane makes a new plane #

The count above is of vectors lying in a plane. A centring vector that does not lie in it has a different effect, and it is worth following, because it is where most glide planes come from.

Compose a reflection in the plane $z = 0$ with a translation $t$. The part of $t$ along the plane becomes a slide; the part perpendicular to it moves the plane, by half its length, because reflecting and then pushing by $t_\perp$ is the same as reflecting in a plane shifted by $t_\perp / 2$. So a centring vector with a perpendicular component of a half puts a new reflection plane at a quarter, carrying the reflection composed with whatever part of the vector lies along it.

Face centring shows both effects at once. Its three centring vectors are $(\tfrac12,\tfrac12,0)$, $(\tfrac12,0,\tfrac12)$ and $(0,\tfrac12,\tfrac12)$. The first lies in the plane $z = 0$, so a mirror there is joined by a glide with slide $(\tfrac12,\tfrac12,0)$ on the same plane — the second of its two operations, an $n$ glide coincident with the mirror. The other two have a perpendicular component of a half, so they build new planes at $z = \tfrac14$: one carrying a glide with slide $(\tfrac12,0,0)$ and the other with slide $(0,\tfrac12,0)$. Those two glides share the plane at a quarter, and that plane is a double glide plane of exactly the kind the letter e was invented for. In $Fm\bar3m$ every one of the three coordinate directions has a mirror at nought carrying a mirror and a diagonal glide, and a double glide plane at a quarter carrying two axial glides, all from the same three vectors.

So the question of how many operations a plane carries has two halves. The vectors lying in a plane set how many operations it carries; the vectors crossing it set how many other planes stand parallel to it, and where. The census counts the first half. The second is why the five glide planes need five letters and why most centred groups have glide planes that no generator names: they are mirrors displaced by half a centring vector, which is the closure doing to planes what it does to axes.

The planes that could carry four #

The face-centred lattice’s three centring vectors — the centres of the three pairs of faces — are not coplanar, so no coordinate plane holds two of them. But two of them always span a plane, and that plane holds both.

Among all forty-nine planes through the origin with indices up to two, the census finds exactly four on which the face-centred lattice has four cosets of the conventional mesh: the four {111} planes, perpendicular to the body diagonals. Each holds two independent centring vectors, and a reflection in one of them would carry four operations. None of them is a mirror. The nine mirrors of the cubic holohedry are perpendicular to the three cube axes and the six face diagonals; the body diagonals are three-fold axes, and a mirror perpendicular to a three-fold axis would make it a six-fold rotoinversion, which no cubic lattice has.

The face-centred orthorhombic lattice has no mirrors but the three coordinate planes, so the same holds there. The planes that could carry four are planes no symmetry of the lattice reflects in, and the question the earlier essay posed has its answer in the geometry of one lattice: the centrings that could double up are placed exactly where no mirror can go.

The case the question did not expect #

The census has one row with a three in it. The rhombohedral lattice, described on the hexagonal axes the Tables use for every rhombohedral space group, has a holohedry of order twelve and three mirrors, and each of the three carries three operations.

The reason is in the centring vectors themselves. On hexagonal axes the rhombohedral lattice is the hexagonal one with two extra points in each cell, at $(\tfrac23, \tfrac13, \tfrac13)$ and $(\tfrac13, \tfrac23, \tfrac23)$. Neither lies in a mirror plane as written, but each is a lattice vector away from one that does: $(\tfrac23, \tfrac13, \tfrac13) - (1, 0, 0) = (-\tfrac13, \tfrac13, \tfrac13)$ lies in the plane $x = -y$ that contains the $c$ axis, and so does its double. So the plane holds the conventional mesh and two more cosets of it, displaced along the plane by thirds, and a reflection in that plane comes in three operations: in R3m a mirror and two glides whose slides are those vectors, in R3c three glides.

The letters have no name for that. The symbol R3m names the mirror, R3c names the $c$ glide, and the other two operations on each plane are left for the list of operations to carry. It is the same situation the e plane was invented to resolve, one operation more crowded — and it is a situation every rhombohedral group with a reflection is in, because it comes from the lattice and not from the group.

The count belongs to the cell #

There is a second reading of the rhombohedral row, and it is the more important one.

On the rhombohedral lattice’s own primitive axes — three equal vectors at equal angles, the cell the rhombohedral setting describes — every lattice vector is an integer combination of the axes, and the same three mirror planes carry one operation each. The census computes this rather than taking it on trust: the holohedry of the primitive rhombohedral metric has the same three mirrors, and each carries one. Nothing about the planes changed; the cell against which operations are counted did.

The same is true of every centred lattice. On a primitive cell there are no centring vectors, so there is nothing to lie in a plane, and every plane of every lattice carries one operation modulo the cell. The two operations of an e plane and the three of a rhombohedral plane are the conventional cell’s centring vectors, seen from inside the plane. That does not make them unreal — they are genuinely different operations, with different slides — but it makes the count a statement about a description. The Tables have a reason to prefer the hexagonal description of the rhombohedral lattice, and it is the same reason they prefer centred cells generally: on hexagonal axes the three-fold axis is the $c$ axis and the mirrors contain it, so every operation has a matrix with small entries aligned with the symmetry, and the rhombohedral groups sit beside the hexagonal ones in one coordinate system. The price is a cell three times the primitive volume, with two centring vectors, and those two vectors are the two extra operations on every mirror plane. A reader of R3m on hexagonal axes is reading a description in which each mirror plane carries three operations; a reader of the same group on rhombohedral axes is reading one in which it carries one. Both are correct, and only the first has a counting problem.

It is the same lesson the cell is a choice draws for the lattice as a whole, and the bigger cell wins because its axes follow the symmetry, at the price of centring vectors that reappear as extra operations on planes.

Four dimensions supply the fourth #

The question has a clean answer in three dimensions for a reason that is specific to three dimensions: a lattice’s centring vectors that lie in a common plane with a mirror are few, because the mirrors are few and the centrings are few. Adding a dimension changes both.

Take the four-dimensional hypercubic lattice and centre every pair of coordinates — add the midpoints $(\tfrac12, \tfrac12, 0, 0)$ and its permutations, which brings in $(\tfrac12, \tfrac12, \tfrac12, \tfrac12)$ as well. That is the four-dimensional analogue of face centring. Every mirror of the hyperoctahedral group keeps it, and the census checks all sixteen of those mirrors: each hyperplane holds four cosets of its conventional mesh. The hyperplane $x_4 = 0$, for instance, holds exactly the three-dimensional face-centred lattice, whose three face-centring vectors all lie in it. So in four dimensions a mirror carries four reflection operations, and the obstruction that saved three dimensions — face-centring vectors doubling up only in planes no mirror can occupy — is gone, because the hyperplanes that hold them are mirrors.

What the letters can name #

A symbol’s letter for a reflection plane was designed to name one operation: a mirror, or a glide by half an axis, half a face diagonal or a quarter of a body diagonal. Planes with two operations exposed that design, because when both are glides neither has a better claim on the letter, and reading a symbol before 1992 meant knowing which one a table had chosen. A mirror and a glide on one plane never had that problem: the mirror wins. Three operations on a plane have no letter and no convention beyond “name one”, and the census says that is the whole of the problem: there is no fourth case to design for, in any of the two hundred and thirty.

What the census depends on, and cannot show #

It counts operations modulo the conventional cell. That is the convention the letters use, and the only one under which the question has a non-trivial answer; modulo the full lattice every plane carries one, and the count is about how much larger the conventional cell is than a primitive one, seen from inside a plane.

It counts what a lattice permits, not what a group uses. A plane of a face-centred lattice that is a reflection plane in some group carries two operations in that group, because the second is the first composed with a lattice translation — so permitted and used coincide here. What the census cannot say is which of the space groups on each lattice have reflection planes at all; that is a count over the two hundred and thirty, and it is not needed for the answer.

The search for planes is bounded, at indices up to two for the face-centred planes and at small boxes of lattice vectors for the rest. For the mirrors of the fourteen lattices the boxes are complete, because every mirror plane there is spanned by vectors with small entries; for the survey of all planes it is a survey, and a plane of index three or more holding four cosets would not have been found. None can be a mirror of a face-centred lattice, which is all the answer needs.

No figure here shows an operation. The planes are drawn as their lattice points, and an operation is a reflection composed with a translation, which a set of points can only suggest.

Planes, lines and the dimension that separates them #

The line carries one screw because a line has one dimension for a slide to be ambiguous in, and the translations along a line are all whole multiples of one shortest vector. A plane has two, so centring vectors can lie in it and create ambiguity — two operations, or three. Neither count depends on the space group, and both depend on the lattice alone: on a line, on the fact that a one-dimensional lattice has no centring; on a plane, on which centring vectors lie in which mirrors.

That makes the list of cases short and closed. Screw axes, eleven of them, one per line. Reflection planes, one, two or three operations per plane, with the three confined to one lattice and the two to the planes its centring vectors occupy. The rest of the variety in the two hundred and thirty — what the closure adds, which elements sit where — is built from these pieces, and the pieces are fixed by the fourteen lattices before any group is chosen.

Still open: the census over the two hundred and thirty #

The lattice census bounds what any group can do; the census over the groups would say what they do. Which of the space groups on a face-centred or body-centred lattice actually have a reflection plane in a doubly occupied orientation — every one with a mirror there, since the second operation comes free — and which of the rhombohedral groups carry three reflections on a plane is a count over the full list, and it is the census the double planes essay ran over forty-five groups and could not run over all.

The four-dimensional case deserves its own census. Four-dimensional space groups number in the thousands, and a hyperplane carrying four operations is the least of what a fourth dimension adds; whether a hyperplane can carry eight, and whether a letter system for four dimensions would need names for four and for eight, is the same computation run on the four-dimensional Bravais lattices, of which there are sixty-four.