Five groups can run every motion backwards

Every symmetry of a pattern can be undone. A quarter turn clockwise about a centre is undone by a quarter turn anticlockwise about the same centre, a translation by the opposite translation, a glide by the glide along the same line sliding the other way. That inverse is always in the group too, which is half of why it is a group and not a list. The question here is subtler. Is the inverse the same symmetry as the original, seen from somewhere else?

The same symmetry, somewhere else gave that phrase a precise meaning. Two operations $g$ and $h$ of a group are the same symmetry when some third operation $x$ of the group carries one onto the other, $h = xgx^{-1}$: do $x$ backwards, then $g$, then $x$. The question is then whether every $g$ is conjugate to its own inverse. A group in which it is, for every element, is called reversible. In such a group nothing distinguishes a motion from its reverse except where it is applied, and the group can run every motion backwards by looking at it from another position.

Two mirrors a coset cannot tell apart noticed one failure in passing: in $pg$ a glide sliding half a cell to the right is not conjugate to the same glide sliding left, its own inverse. Asked of every element of every plane group, the question has a sharp answer. Five of the seventeen are reversible: $p2$, $pmm$, $cmm$, $p4m$ and $p6m$. And the answer the point groups give, which is where the question would naturally be asked, is fourteen.

The question, made exact #

A finite group is reversible exactly when all of its characters are real numbers, a fact from representation theory, and it is easy to check from a character table. The point group of a plane group is finite, and it is the plane group with the lattice divided out. So the natural shortcut is to read the answer off the point group. Of the ten plane point groups, all are reversible except the three rotation groups $3$, $4$ and $6$, in which a turn and its reverse are different elements and no reflection exists to exchange them. On that reading, fourteen of the seventeen plane groups would be reversible.

The plane group is not its point group. It is infinite, and an element of it is a definite motion of the plane with a definite position: a mirror on one particular line, a half-turn about one particular point. Conjugacy in the infinite group is a stronger condition than conjugacy of the cosets, because a conjugating motion must carry the operation exactly onto its inverse, at its own position, not merely onto something in the same coset.

The essay on cosets made that decidable. An element is a pair $(A, a)$, an integer matrix and an exact translation, and whether $g$ and $h$ are conjugate reduces to finitely many linear parts $R$ and, for each, whether one vector lies in one lattice: $(I - B)r = b - Ra$ must have an integer solution $r$. Here the test is run on every element of each group in a window of five cells by five, with $h = g^{-1}$, translations of every length up to the window included. That is twenty-five elements in $p1$ and three hundred in $p6m$, each decided in integers.

The table sorts the failures by kind, and each kind fails for its own reason. There are three reasons, and between them they account for every one of the nine.

Translations need a half-turn #

A translation by a vector $v$ is conjugate only to translations by the vectors $Rv$, with $R$ in the point group, because conjugating a translation moves nothing and turns its vector. Its inverse is the translation by $-v$. So a translation is reversible exactly when some operation’s linear part sends $v$ to $-v$. A half-turn does that for every vector at once. Without one, a mirror reverses only the vectors perpendicular to it, and a rotation of order three reverses none.

So the seven groups without a half-turn fail on their translations: $p1$, $pm$, $pg$, $cm$, $p3$, $p3m1$ and $p31m$. In $pm$ the translations across the mirrors pass and every other translation fails, twenty of the twenty-four in the window. In $p3m1$ half pass. This is the case the point group gets wrong most simply. The point group of $pm$ has two elements, the identity and the reflection, and each is its own inverse, so it is reversible. But the point group has no translations to test, and translations are most of the plane group.

The physical reading is direct. A crystal whose group has no half-turn, no inversion centre in two dimensions, has directions that are not equivalent to their reverses. A translation along such a direction and the translation back are distinct symmetries, and no change of viewpoint within the pattern identifies them. The polar plane groups are exactly the groups in which this happens for some direction, and the census finds it is every direction except the few a mirror happens to reverse.

Rotations need a mirror through their centre #

A rotation by $\theta$ about a point $c$ has inverse the rotation by $-\theta$ about the same point. Conjugating by a rotation or translation keeps the angle and moves the centre. Only a reflection or a glide reverses the angle, and the conjugate is then the rotation by $-\theta$ about the image of $c$. For that to be the inverse, the image of $c$ must be $c$ itself. A glide fixes no point at all. So a rotation is reversible exactly when a mirror of the group passes through its centre. A half-turn is its own inverse and needs nothing.

The pure rotation groups $p3$, $p4$ and $p6$ have no mirrors, and every rotation of order three or more fails, as the point group already said. Two other groups fail in a way the point group cannot say. In $p31m$ the three-fold centres come in two families, and one family lies on mirrors and the other does not: 32 of the 50 thirds in the window fail. In $p4g$ the four-fold centres lie at the crossings of glide lines and on no mirror at all, while its mirrors pass through its half-turn centres. Every quarter turn of $p4g$ fails, although its point group, $4mm$, has mirrors that exchange a quarter turn with its reverse.

A glide needs a half-turn on its line, or a mirror across it #

A glide reflects in a line and slides along it, and its inverse reflects in the same line and slides the other way. The slide survives conjugation in length. A conjugating operation must keep the line in place and reverse the direction of the slide. A translation keeps the slide, a mirror along the line keeps it, and a glide parallel to the line keeps it. A perpendicular glide reverses it but moves the line by its own slide. What remains is a half-turn about a point on the line, or a mirror across it, and a glide is reversible exactly when the group has one of them.

This is where $pgg$ fails. Its glide lines run in two perpendicular directions, its half-turn centres sit between them and on none of them, and it has no mirrors. All fifty glides in the window fail, while every translation and every half-turn passes.

The failure in $pmg$ is the most instructive in the census, because it is not in the glides anyone draws. $pmg$ has mirrors in one direction and glides perpendicular to them, and its true glides are reversed by those mirrors, which cross their lines. The mirrors have glides of their own. A mirror followed by a translation along its own line is a glide-reflection in that line, with a slide of a whole lattice vector. In the point group it is the same element as the mirror, since the slide is a lattice translation. In the plane group it is a glide, and its inverse slides the other way. To reverse it, the group needs a half-turn centre on the mirror line or a mirror across it, and $pmg$ has neither: its half-turn centres lie between its mirrors, and the operations across its mirrors are glides that move each mirror onto its neighbour. Twenty of the forty-five glide-reflections in the window fail, every one of them a mirror composed with a slide along itself.

The same thing happens in $pm$ and $cm$, where every glide-reflection along a mirror’s line fails for the same reason. In $pmm$, $cmm$, $p4m$ and $p6m$ it does not, because each mirror line is crossed by a mirror at right angles.

Three conditions and seventeen answers #

Put together, the three requirements are a criterion that can be read off a drawing of the group, with no element-by-element test.

• a half-turn in the point group;
• a mirror through every rotation centre of order three, four or six;
• on every line carrying a glide-reflection, including every mirror line, a half-turn centre or a mirror crossing it at right angles.

The criterion predicts all seventeen answers of the element-by-element test, and it explains why the point group cannot: the first condition is about translations, which the point group has divided out, and the other two are about where operations are, which the point group has forgotten. $pgg$ and $p4g$ pass the first condition and fail on position. $pm$, $pg$, $cm$ and $p3m1$ fail on translations. $pmg$ fails on its mirrors’ own glides. Of the five that pass, $p2$ passes almost vacuously, since every operation is a translation or a half-turn and the half-turns reverse everything. The other four pass for one shared reason. $pmm$, $cmm$, $p4m$ and $p6m$ have mirrors in at least two perpendicular directions, so every line carrying a reflection, whether a mirror or a glide parallel to one, is crossed at right angles by a mirror, which reverses every slide along it. $cmm$ is the instructive member: it has glides, halfway between its mirrors, and they would fail in a group like $pgg$, but the mirrors of the other direction cross them. The difference between $cmm$ and $pgg$, which share the point group $mm$, is exactly whether the operations across a glide line are mirrors or glides.

Two shortcuts, and how far each gets #

Two facts already established give quick partial answers, and seeing where they stop is a way of seeing what the census adds.

The first is the centre. The four groups with a centre found that a plane group has elements commuting with everything only when its point group leaves some translation alone, and that four of the seventeen do. An element of the centre is conjugate to nothing but itself, so it can be conjugate to its inverse only if it is its own inverse, and a non-zero translation never is. A group with a centre is never reversible. That disposes of four groups in one line. It says nothing about the other thirteen, since a group without a centre can still fail, as $p4$ and $pgg$ do.

The second is the abelianisation. Conjugate elements become equal once the order of multiplication is forgotten, so in a reversible group every element becomes equal to its own inverse, and the abelianisation can contain only elements of order two. What is left when order is forgotten computed it for all seventeen. $p31m$ abelianises to a cyclic group of order six, which has elements of order three, and so cannot be reversible. But $p3m1$ abelianises to a group of order two, and $pmg$ to three factors of two, exactly as $p2$, $cmm$ and $p4m$ do. Both pass the abelian test and both fail the census. The test is necessary and not sufficient, and it fails for the same reason the point group fails: it divides out part of the group, here the commutators, and the failures of $p3m1$ and $pmg$ live in what it divides out.

Nor is reversibility the same as having mirrors. The census’s five include $p2$, which has no reflection of any kind, and exclude $pm$, $pg$, $cm$, $pmg$, $pgg$, $p3m1$, $p31m$ and $p4g$, all of which have reflections. The groups one hand may sit in are the five with no reflection, $p1$, $p2$, $p3$, $p4$ and $p6$, and only one of them is reversible. A pattern can be chiral and still run every motion backwards, provided every one of its motions is reversed by its half-turns. What a group needs is enough operations of order two in the right places, and a mirror is only one kind.

An involution always suffices #

A stronger property holds in all five. When $g$ is conjugate to its inverse, the conjugating operation can always be chosen to have order two, a mirror or a half-turn: $xgx = g^{-1}$ with $x^2 = 1$. The census checked that for every element of each reversible group in a smaller window, against every operation of order two in a larger one. Such a group is called strongly reversible.

Strong reversibility has a consequence that connects it to an older fact. If $xgx = g^{-1}$ with $x$ an involution, an operation of order two, then $xg$ is an involution as well, since $(xg)^2 = xgxg = g^{-1}g = 1$. So $g = x \cdot (xg)$ is a product of two involutions. In the five reversible groups, every operation is a product of two operations of order two, each a mirror or a half-turn of the group itself. Three reflections, and never four showed that every motion of the plane is a product of at most three mirrors. The plane’s motions can be built from reflections freely, but a pattern’s own operations can be built from its own involutions only in these five groups.

What the census rests on, and what it does not show #

The test is exact and the window is finite. Every element in a window of 5 × 5 cells was decided, and the window includes translations up to two cells long in each direction and operations at every position in the cell. Conjugacy depends only on an operation’s linear part and its position modulo a lattice that is visible within a cell or two, so the window sees every class. The criterion, which has no window in it, agrees with the test on all seventeen. That agreement is the evidence the window was large enough.

The refused claim is the shortcut, that a plane group is reversible when its point group is, and it is refused nine times over. The tests that pass are each independent of the others: the criterion is read off the geometry and the census off the arithmetic of conjugacy, and the two agree on all seventeen groups.

The figures show the operations of a pattern, never its reversibility. Whether a quarter turn of $p4g$ can be reversed is a statement about every operation of the group, and the drawing shows the centre and the mirrors around it and leaves the reader to see that none passes through. The absence is the whole of the argument and the figure can only make it easy to look for.

The convention matters too. Conjugacy here is within the group itself. Allow conjugation by a larger group, the group’s normaliser among all motions of the plane, and more pairs become equivalent. In $p4$, a reflection of the whole pattern exchanges its two senses of turn, which is why $p4$ looks the same in a mirror. Within $p4$ itself the two remain different symmetries.

Who established which part #

Reversibility is an old notion for finite groups, where it is the same as having only real characters, and for groups of motions it has been studied under the name reversing symmetry in dynamics, where a map conjugate to its inverse is one whose orbits can be run backwards. O’Farrell and Short’s book of 2015 surveys it for groups of isometries among others. The census of the seventeen plane groups, the three-part criterion, the failure of $pmg$ on its mirrors’ glides, and the comparison with the point groups are computed here, on the conjugacy test of the essay on cosets.

Still open: the space groups, where a screw needs a perpendicular axis #

In space, a screw and its inverse turn the same way about their axis and advance the opposite way, so the inverse of a right-handed screw is right-handed too. A mirror containing the axis reverses the turn and keeps the advance, which turns the screw into one of the other hand. An inversion centre on the axis does the same. Neither yields the inverse. Only a two-fold axis perpendicular to the screw’s axis and meeting it reverses both turn and advance, and so the space groups should divide differently from the plane groups, with inversion, the three-dimensional half-turn’s analogue, no longer enough. Where the plane had places, space has pitches decided conjugacy among screws in all 219 types. Running the same test with $h = g^{-1}$ there would say which space groups are reversible, and whether any centrosymmetric group fails while some group without a centre passes. It has not been run.