Seventy-five is twenty-nine and forty-six

Eighty is seventeen, seventeen and forty-six accounted for the eighty layer groups in three lines. A layer’s operations all keep its middle plane, so its group projects onto a plane group, and the only thing the projection forgets is a sign on each operation, plus if the operation keeps the sheet’s two faces and minus if it exchanges them. The signs are either all plus, or doubled by a mirror in the sheet, or a two-colouring of the plane group. That gives seventeen, seventeen and forty-six.

Seventy-five ways to be a thread derived the seventy-five rod groups by a different route. It took each of the twenty-seven crystal classes that keep a direction, put that direction along the rod, attached translation parts along the axis on a grid of twelfths, and removed duplicates. The layer essay ended by asking whether the rods have an account of the layers’ shape: a group of the line, a crystal class that keeps a direction, and a rule attaching one to the other. The earlier census used all three without naming them.

They do, and the account is the layer account turned round. A layer is a group of the plane with a finite action on the one direction across it. A rod is a finite group of the plane across it with an infinite action on the one direction along it. Twenty-nine rods never turn their axis over and forty-six do. Built from that account alone, the seventy-five agree cell by cell with the ones counted class by class. The account also says what a screw is. A rod’s screw is the turn its smallest step up the axis gives the plane, and the part of the rod that leaves the axis untouched decides the screw’s subscript. That is how the eleven screws arise from a rod rather than being listed for one.

What fixes the axis, and what moves it #

Every operation of a rod group carries the rod’s axis to itself, so it acts on the axis as a motion of a line: $z \mapsto z + \tau$ or $z \mapsto -z + \tau$. Composition of operations composes these motions, so the rule sending an operation to its motion of the axis is a homomorphism. Its image is a discrete group of motions of a line containing a translation, and there are exactly two kinds. One is the translations alone, a copy of $\mathbb{Z}$. The other is the translations together with reversals about a row of points, the infinite dihedral group $D_\infty$.

The kernel of the homomorphism is everything that leaves every point of the axis where it is. An operation that fixes the axis pointwise is a rotation about the axis or a reflection in a plane containing it. So the kernel $K$ is a finite group acting on the plane across the rod, and since the rod is crystallographic it is one of the ten plane point groups: 1, 2, 3, 4, 6, m, 2mm, 3m, 4mm or 6mm, the ten left of the plane’s finite groups once the crystallographic restriction has had its say. That gives the three ingredients the layer essay asked for. The image is the group of the line. The whole rod’s point group is a crystal class keeping the axis. The rule is the extension of the line group by $K$.

For a layer the corresponding split is the other way up. A layer’s operations act on the direction across it by a sign, a homomorphism onto the finite group $\{\pm 1\}$, and its kernel is the operations keeping both faces, an infinite plane group. The layer is an infinite group extended by a finite one, and the rod is a finite group extended by an infinite one. The layer account counted signs. The rod account has to count something slightly larger, because the extension of an infinite line group by a finite plane group carries more information than a sign.

One step, or two reversal points #

When the line group is $\mathbb{Z}$, choose an operation $g$ that moves the axis by the smallest step $\tau$. Every operation of the rod is $g^n$ times something in the kernel, and since $\mathbb{Z}$ is free the rod is the semidirect product of $K$ with the infinite cyclic group $g$ generates. What $g$ does to the plane is its in-plane part $\rho$, a rotation or reflection that must carry $K$ to itself. Replacing $g$ by $g$ times a kernel element changes $\rho$ within the coset $\rho K$ and changes nothing else. So a rod over $\mathbb{Z}$ is fixed by the pair $(K, \rho K)$. Its repeat is $s\tau$, where $s$ is the number of steps before $\rho^s$ lands back in $K$.

The left of the picture is the simplest case with a screw. The kernel is trivial, the step turns the plane by a quarter, and four steps return to the kernel, so the repeat is four steps and the rod is the four-fold screw $4_1$. The screw of a rod over $\mathbb{Z}$ is its step’s turn. A step whose in-plane part is a reflection instead of a rotation is a glide plane containing the axis, and the same account covers it.

When the line group is $D_\infty$, the axis has reversal points half a step apart, and the rod is generated by its kernel and the operations at two neighbouring reversal points. Call them $a$ at height 0 and $b$ at height $\tau/2$. Each reverses the axis, and its in-plane part, $\alpha$ or $\beta$, must carry $K$ to itself and square into $K$, since $a^2$ and $b^2$ fix the axis pointwise. The rod is fixed by $(K, \alpha K, \beta K)$, and its smallest step is $ba$, with in-plane part $\beta\alpha$. The right of the picture shows one: kernel the half-turn about the axis, a centre of inversion and a mirror across the axis at every whole step, and a four-fold rotoinversion at every half step.

Two rods are the same type when they are carried onto each other by a motion of space. That allows a turn of the plane about the axis by any angle, which rotates mirror lines by any amount. It also allows a half-turn about a line across the axis, which reverses the axis and conjugates the in-plane parts by a reflection, and for $D_\infty$ it allows moving along the axis to the next pair of reversal points. It does not allow a reflection of space. The two hands of a screw are different types, as they are in the International Tables’ count of seventy-five.

Counting in a group of order twenty-four #

Every rotation and reflection a crystallographic rod can use lies in one finite group: rotations by multiples of 30° and reflections in lines at multiples of 15°, twenty-four elements in all. A subgroup of it is crystallographic exactly when it contains no rotation of order twelve. That is the crystallographic restriction in one line, since twelve is the only order of rotation in this group that is not 1, 2, 3, 4 or 6. So the count is a finite enumeration. Take every crystallographic subgroup as a kernel. For $\mathbb{Z}$ take every element normalising it as a step, and for $D_\infty$ every pair of elements normalising it and squaring into it. Keep the data whose whole group is crystallographic, and identify them under the motions just listed.

The picture at the head of this essay is the result. Twenty-nine rods never reverse their axis, forty-six do, seventy-five in all. The twenty-nine are exactly the rods that are polar along their axis, since a rod whose operations never turn the axis over has a direction up the rod that nothing reverses. The distribution by kernel is uneven in an informative way. The trivial kernel carries twenty-two rods, nine over $\mathbb{Z}$ and thirteen over $D_\infty$. The kernel with a two-fold rotation carries fourteen, and the kernels with a four-fold or six-fold rotation, alone or with mirrors, carry two to five each. A large kernel leaves little room for the step to add anything. A four-fold kernel allows only a plain translation or a glide containing the axis, and a 6mm kernel allows only the translation.

The step’s turn, and the kernel’s subscript #

The twenty-nine can be read off one at a time. Over the trivial kernel there are nine. A plain translation. A half-turn step, which is the screw $2_1$, or a reflection step, which is a glide plane containing the axis with the slide half the repeat. And two each of the steps that turn by a third, a quarter or a sixth: $3_1$ and $3_2$, $4_1$ and $4_3$, $6_1$ and $6_5$, each pair being the two hands of one screw. Over the two-fold kernel there are five: the translation, a quarter-turn step that makes $4_2$, a glide step, and two third-turn steps that make $6_2$ and $6_4$. Over the three-fold kernel there are three: the translation, a sixth-turn step that makes $6_3$, and a glide.

That table contains the rule for screw subscripts. A screw $n_m$ turns by $1/n$ of a circle and climbs $m/n$ of the repeat, and it is standard that its powers include a pure rotation of order $\gcd(n, m)$, so $4_2$ contains a two-fold rotation and $6_3$ a three-fold one. In the account that pure rotation is the kernel, and the subscript is decided by the kernel. A step that turns by a quarter makes $4_1$ or $4_3$ over a trivial kernel and $4_2$ over a two-fold kernel, because four steps must return to the kernel and two steps already do when the kernel holds the half-turn. The two hands disappear exactly when the kernel is large enough to absorb them: $4_2$, $6_3$ and the translation have no mirror image distinct from themselves. $6_2$ and $6_4$ still come as a pair, because a two-fold kernel absorbs a half-turn and not a third.

The steps the account allows are also the whole of the crystallographic restriction for rods. A step’s order modulo its kernel is 1, 2, 3, 4 or 6, and the kernels with four-fold and six-fold rotations have no steps of order greater than two. They cannot, since a step turning by a further twelfth would put a twelve-fold rotation in the rod.

Forty-six that turn their axis over #

The forty-six are read the same way, with the operations at two neighbouring reversal points in place of one step. Over the trivial kernel there are thirteen. The three with one step to the repeat have the same kind of operation at every reversal point: all mirrors across the axis, all centres of inversion, or all two-fold axes across the rod. The four with two steps alternate two kinds, or alternate two-fold axes in two different directions. The six with three, four or six steps have two-fold axes across the rod at every reversal point, turning by a third, a quarter or a sixth between one point and the next. They are the screws $3_1$, $4_1$ and $6_1$ with two-fold axes added across them, $3_121$, $4_122$ and $6_122$, and like those screws they come in pairs of hands.

Over larger kernels a reversal point holds several operations at once, because the kernel’s elements compose with the reversal. With the two-fold kernel, a point holding a centre of inversion also holds a mirror across the axis, since the half-turn about the axis times the inversion is that mirror. A point holding a four-fold rotoinversion holds only that, since the square of a four-fold rotoinversion is the half-turn about the axis already in the kernel.

The eight pairs of hands in the whole list are four over each line group, and in every one of them the axis carries a single hand, since a line carries one screw: $3_1$, $4_1$, $6_1$ and $6_2$ with their mirror images over $\mathbb{Z}$, and the same four with two-fold axes across them over $D_\infty$. The account finds them without being told. Allowing a reflection of space as an equivalence, which identifies each rod with its mirror image, reduces seventy-five to sixty-seven, eight fewer.

Two derivations of one number #

The account was built without consulting the earlier census, and the earlier census was built without mentioning kernels. The comparison is therefore a real check. For each of the seventy-five rods found class by class, the check reads off the kernel, the operations that fix the axis pointwise, and the line group, whether any operation reverses the axis. It also reads off the number of distinct climbs among the operations that keep the axis’s direction, which is the number of steps to the repeat. It tallies those into cells and sets them beside the account’s own tallies. Forty-four cells hold at least one rod and all forty-four agree.

The total alone would not be worth much, and seventy-five ways to be a thread found out why. Its first version matched seventy-five with two errors of six groups each that cancelled. Six were missing where a class has a second orientation on the rod, and six were counted twice where a turn about the rod identifies them. A cell-by-cell agreement over forty-four cells is a much stronger statement. The two derivations would have to make matching errors in the same kernel, line group and step.

The first run of the account did not agree. It came to seventy-one, four short, and the four missing rods were one hand each of the screws over $D_\infty$: $3_121$, $4_122$, $6_122$ and $6_222$ had each merged with its mirror image. The cause was an equivalence allowed that should not have been. The count had let the two reversal points $a$ and $b$ be exchanged freely. Exchanging them puts the second point below the first, which is a reversal of the axis, and so reverses the screw. A half-turn across the rod exchanges them legitimately, because it reverses the axis and conjugates by a reflection at the same time. With the free exchange removed the count is seventy-five and every cell agrees.

What the account does and does not say #

It counts types, not rods. Each datum fixes a group up to the motions that identify rod types. The step length $\tau$ and the positions of reversal points are not part of the datum, because a rescaling along the axis and a shift of the origin are among those motions. Two rods with the same datum and different repeats are the same type.

The two line groups are a complete list. A discrete group of motions of a line containing a translation is $\mathbb{Z}$ or $D_\infty$. That is the one-dimensional case of the classification, the two groups of the line, and the account inherits its completeness.

The restriction is imposed, not derived. The account asks for crystallographic kernels and steps by excluding rotations of order twelve from a group that contains no other forbidden orders. It does not prove that a rod repeating in a crystal cannot have a five-fold axis. That is the crystallographic restriction, and rods inherit it from the three-dimensional lattice they are cut from. With the restriction removed the same enumeration finds 110 data. They are rods with twelve-fold rotations and screws, the start of the unbounded families that the centring that turns into a screw met as the thirteen families of line groups.

Who worked the pieces out #

The rod groups have been enumerated several times since the first half of the twentieth century, and they appear with the layer and frieze groups in volume E of the International Tables, edited by Vojtěch Kopský and Daniel Litvin. Describing a group of motions of a line as a finite group of the cross-section extended by a group of the axis is the approach Milan Damnjanović and Ivanka Milošević took to line groups without a restriction, where it produces thirteen infinite families. The account here is that approach restricted to the crystallographic case and counted. A space group is an extension is the same idea for the two hundred and thirty, with the lattice as the kernel instead of the image.

What the account must refuse #

The test that carries weight is the second, the cell-by-cell agreement. The third is the sharpest check on the equivalences. A count that allowed one equivalence too many would merge pairs of hands and come in under seventy-five, as the first run did. A count that allowed one too few would split rods that are the same and come in over. Allowing reflections must remove exactly the eight enantiomorphic pairs, and it does.

Still open: why forty-six twice #

The layers came to seventeen, seventeen and forty-six, and the rods to twenty-nine and forty-six. The two forty-sixes have nothing obvious in common. One counts two-colourings of the plane groups up to the affine normaliser. The other counts extensions of the infinite dihedral group of a line by the plane’s finite groups. The first is a count of index-two subgroups of infinite groups, and the second of finite data in a group of order twenty-four. Whether the equality is a coincidence of small numbers or the shadow of a correspondence, a way of reading a two-coloured plane pattern as a rod that turns its axis over, has not been examined. A correspondence would have to send the seventeen plane groups’ colourings to kernels and reversal points. The first thing to check is whether the forty-six colourings and the forty-six rods have the same distribution over any natural invariant, such as the order of the point group. That is one table beside another, and it has not been drawn.