The centring that turns into a screw

A rolled sheet turns by a square root named the screw of every rolled tube by one congruence. The screw is only part of the group. A tube can also have half-turns across its axis, mirrors containing it, a mirror across it, glide planes, and rotoreflections, and which of these it has decides what the tube is as surely as its screw does. Up to order six those possibilities are sorted by the seventy-five rod groups. Past order six the crystallographic list stops and the tube does not. A tube of sixty-five turns has a sixty-five-fold screw, and no table of crystal symmetry has a name for its group.

The groups of motions of a line that contain translations along it form thirteen infinite families, each running over every order $n$ of pure rotation. That is the classification the nanotube literature uses, in the numbering of Milan Damnjanović and Ivanka Milošević, and it is to the unbounded orders what the seventy-five are to orders two, three, four and six. The question the square-root essay left is which of the thirteen the seventeen plane patterns reach when they are rolled. This essay runs the census and answers it: all of them.

The more useful finding is the reason the families come out as they do. Twelve of the thirteen are pinned down by their translations, and for a rolled tube the translations are decided by a single integer. Every tube that keeps any reflection holds one or two cells of its sheet in a repeat, never more. It holds two exactly when the sheet’s lattice is centred along the rolling direction. The centring vector of the flat lattice becomes the half-step screw of the tube. That is why a centred pattern and a primitive one with the same mirrors roll into different families, and why every zigzag and armchair carbon tube is in the same family as a rolled centred rectangle.

Thirteen families, and the four kinds of translation #

A group of motions of a line has an isogonal point group: its linear parts, with every climb along the axis forgotten. Seven friezes round a cylinder showed that these fall into seven families of groups with one axis. They are $C_n$, $S_{2n}$, $C_{nh}$, $C_{nv}$, $D_n$, $D_{nd}$ and $D_{nh}$, according to which of the half-turns across the axis, mirrors across it, mirrors containing it and rotoreflections are present. The tube has a screw no lattice allows placed every carbon tube in one of two of them, $D_N$ or $D_{Nh}$. Most sheets roll into a tube that never repeats did the same for all seventeen plane groups.

The line group is that point group together with its translations, and the translations come in four kinds. A general screw turns by any fraction and climbs by any fraction. It is allowed only in the two families with no reflection at all, family 1 over $C_n$ and family 5 over $D_n$, and those are the chiral families. Pure translations climb a whole period without turning, beside the $n$ pure rotations. Glide planes contain the axis, reflect across themselves and climb half a period. The half-step screw turns by half the rotation step, $1/(2n)$ of a circle, and climbs half a period. A group with the half-step screw has an isogonal point group of order $2n$, twice its pure rotations, because forgetting the climb turns the screw into a rotation by $1/(2n)$.

The thirteen are these combinations with the impossible ones removed. $C_n$ and $D_n$ take a screw of any kind, and each makes one family, 1 and 5. $S_{2n}$ takes only pure translations and makes family 2. $C_{nh}$ takes pure translations or the half-step screw, families 3 and 4. $C_{nv}$ takes all three achiral kinds, families 6, 7 and 8, and so does $D_{nh}$, families 11, 12 and 13. $D_{nd}$ takes pure translations or glides, families 9 and 10. The table lists them with the rolled patterns that produce each, and it is the result of the census that follows, not its premise.

The census, with the order left free #

The census takes each of the seventeen plane groups on its generic lattice and rolls it along every lattice vector out to nine steps in each coordinate. It keeps the rollings that repeat, 666 of them, and sorts each tube into a family. The isogonal point group comes from which of the sheet’s operations survive, as before: an operation of the sheet survives when it carries the rolling vector $C$ to itself or to its negative. The translational part comes from two whole numbers. One is $n$, the common factor of the rolling vector’s coordinates, which is the order of the tube’s pure rotations. The other is $n_0$, the number of the sheet’s cells in one repeat of the primitive tube. If $n_0 = 1$, some lattice vector climbs one period with no turn, and the translations are pure, or glides when the planes containing the axis only glide. If $n_0 = 2$, the least-climbing screw turns by half a step and climbs half a period, which is the half-step screw. Anything larger is a general screw.

The picture at the head of this essay is the census. Every one of the thirteen columns has a mark, so rolling a plane pattern reaches every family of line groups there is. Each is reached at every order: the common factor of the rolling vector is free, so a rolling along $g$ times a primitive vector has $g$ pure rotations, and every family appears at every $n$ from one to nine in the census and, by the same multiplication, at every $n$ beyond.

The marks are spread very unevenly. Six families are reached by exactly one plane group each, and all six belong to the rectangular groups. Family 2, the bare rotoreflection, comes only from pg. Family 3, the mirror across the axis with pure translations, comes only from pm. Family 6 also comes only from pm, and family 7, glide planes containing the axis, only from pg. Families 9 and 12 come only from pmg, which has a mirror, a glide and a half-turn and rolls into different families along its two directions. A rectangular lattice repeats only along its two mirror directions, so each of these groups has exactly two directions to be rolled in, and they are spent on six families that nothing else reaches.

The square and hexagonal groups reach fewer families, but along every direction. Every square and hexagonal pattern with a half-turn reaches family 5, the chiral $D_n$ family with a general screw, along every direction that meets no mirror. The trigonal groups without a half-turn, p3, p3m1 and p31m, reach family 1, the bare screw. Those two families are the only ones with a general screw, and they are the only ones that the square and hexagonal lattices reach off their mirror directions.

A tube with a reflection holds one cell or two #

A general screw is available only to a chiral tube, and the census shows something stronger than that rule requires. Any family other than 1 and 5 would permit a tube with a reflection whose primitive index was large, provided its screw were a pure translation or a half-step in disguise. None occurs.

Every one of the 216 achiral rollings has primitive index one or two, split evenly at 108 each, while the 450 chiral ones spread up to 434. The reason is in which directions can keep a reflection at all. An operation of the sheet survives rolling only if it carries $C$ to $\pm C$, and a reflection does that only when its line is parallel or perpendicular to $C$. So an achiral tube is always rolled along a mirror or glide direction of its sheet, and along such a direction the lattice has a vector perpendicular to $C$ with the shortest possible repeat. The rectangle on $C$ and that perpendicular vector is a conventional cell of the lattice in that orientation, and a conventional cell of a plane lattice holds one lattice point or two. It holds one if the lattice is primitive in that orientation and two if it is centred.

That turns a claim about screws into a claim about centring, and it can be tested without reference to the screws. For each of the 216 achiral rollings the census takes the primitive $C$ and its perpendicular $T$ and asks whether their midpoint $(C + T)/2$ is a lattice vector. That test knows nothing about tubes. It agrees with the primitive index in all 216: the midpoint is a lattice vector exactly when $n_0 = 2$.

The centring becomes the screw #

The cut-open repeat shows what that midpoint does once the sheet is rolled.

In the rectangular repeat the only lattice points are the corners. Rolled, they give pure rotations round the circumference and pure translations up the axis, and pm rolled along its mirror is family 6, $T C_{nv}$. In the rhombic repeat of cm a fifth point sits at the centre, half a rotation step round and half a period up, and on the tube that point is a screw. So cm rolled along the same kind of mirror is family 8, $T^1_{2n} C_{nv}$. The two tubes have mirrors containing the axis and nothing else, and forgetting the climbs sorts them both into the family $C_{nv}$. They are not the same group. The one rolled from the centred sheet has a screw the other lacks, which alternates its mirrors with glides and doubles the order of its isogonal point group.

Centring, and why cm is not pm made this distinction in the flat plane, where the centred group’s extra translation puts glides between its mirrors. Rolling carries the distinction onto the tube unchanged: the translation that made cm differ from pm in the plane is the screw that makes family 8 differ from family 6. The same happens across the axis, where pm rolled the other way is family 3 and cm is family 4, $T^1_{2n} C_{nh}$.

The dictionary between sheet and tube is short, and the table spells it out. A reflection line of the sheet parallel to $C$ runs round the tube, so it reflects the axis end for end: a mirror across the axis, or, if the line glides, a rotoreflection, since the slide along $C$ becomes a turn. A reflection line perpendicular to $C$ runs up the tube: a mirror containing the axis, or a glide plane if it glides. A half-turn of the sheet becomes a half-turn across the axis. The only item not visible in the operations themselves is the centring, which appears as the half-step screw. pmg shows two of the rarer outcomes. Rolled one way its glide runs round the tube and becomes a rotoreflection between mirrors, family 9, $T D_{nd}$. Rolled the other way its glide runs up the tube and becomes a glide plane, family 12.

Why every carbon tube with a mirror is family 13 #

The hexagonal lattice is the case that matters in practice, and the centring explains it. Along any of its mirror directions the hexagonal lattice is a centred rectangle. Its conventional rectangular cell has sides in the ratio one to $\sqrt{3}$ and a lattice point at its centre. So every hexagonal pattern rolled along a mirror has $n_0 = 2$ and the half-step screw, and a honeycomb rolled that way has mirrors both across and containing the axis as well: family 13, $T^1_{2n} D_{nh}$.

The triangle is the familiar one from the tube literature, and the families on it are the literature’s answer: family 13 along both edges and family 5 inside. What the census adds is the reason for the thirteen. The tube has a screw no lattice allows observed that the zigzag tube $(9, 0)$ has nine pure turns and eighteen screws, and it read the doubling off the rolled atoms. The doubling is the hexagonal lattice’s centring. The eighteen screws are the nine pure turns together with nine half-step screws, each a turn of one eighteenth of a circle combined with a climb of half the period, and that half-step screw is the midpoint of the centred rectangle that one repeat of a zigzag tube cuts from the sheet.

The check here does not trust the rolling. For all sixty-five tubes with indices up to ten, the carbon atoms are placed on the cylinder one by one. The order of the tube’s isogonal point group is then found by searching for operations that carry the atoms onto themselves, and it agrees with the order the rolled lattice predicts, twice the pure rotations on both edges, in every case.

What rolling cannot reach, and what it cannot say #

The census has an edge, and it is set by the convention that the lattice is generic. A rectangular sheet rolled in a direction that is not a mirror direction does not repeat, because the perpendicular lattice direction is irrational for a general axial ratio. That is why the rectangular groups reach so few families. With an axial ratio that happened to be rational, or the square root of a whole number, the sheet would repeat in other directions too. A rolled sheet turns by a square root used exactly that accident to reach the handed crystallographic screws. Those directions meet no mirror, so the extra tubes would all be chiral, in families 1 and 5. They cannot add a family to the census, which already reaches all thirteen; they could only add plane groups to the first and fifth columns. The oblique groups p1 and p2 stay empty for the same reason, since an oblique lattice has no direction with a perpendicular lattice vector.

Two limits come from the pictures rather than from the arithmetic. The cut-open repeats show lattice points and not patterns, so they cannot show the motif that makes a mirror a mirror. The claim that pm rolled along its mirror has mirrors containing the axis rests on the sheet’s operations, not on anything drawn. The honeycomb triangle shows families and not the screws inside family 5. Those differ from tube to tube: every chiral tube in the triangle is in family 5, and no two with different $(n, m)$ have the same screw. The family is the coarser statement, and the screws are what the square-root congruence decides.

The census also takes a one-sided sheet, a flat pattern with no inside and outside, as everything in this series has. A real sheet with different atoms on its two faces has one of the eighty layer groups instead of one of the seventeen. Its operations that turn the sheet over cannot survive rolling, since no motion of a tube exchanges its inside and its outside. Whether the operations that survive can put such a tube in a family the flat patterns reach through different generators, or in the same families for different reasons, is not something the seventeen can answer.

What the census has to refuse #

The first refused claim is that an achiral tube can climb by a general screw. The families would permit one if their translations could be disguised. The census finds no achiral rolling with an index above two, and the centring argument says none can exist. The second refused claim is the one a point-group table invites: that two tubes with the same point-group family are the same line group. pm and cm rolled along their mirrors both forget their climbs down to $C_{nv}$, and they are families 6 and 8.

The tests that pass are the ones built to catch an error that would otherwise pass silently. The isogonal order of every tube in the census is compared with the family’s prediction: twice the pure rotations for families 4, 8 and 13, and exactly the pure rotations for the pure and glide families. A mistake in reading the index, or in deciding between a mirror and a glide, would change a family label without changing anything visible, and this test would catch it. The centring test and the atom-by-atom test each check the rolled lattice against a computation that does not use it.

Still open: turning a two-sided sheet into a tube #

The census covers every flat pattern. The next question is the two-sided sheet. A layer with different atoms above and below has one of the eighty layer groups. Rolling it puts one face inside and one outside, so every operation that turns the layer over dies, and the obvious guess is that only its side-keeping half survives, a plane group, so that the two-sided census adds nothing. That guess ignores operations that turn the sheet over and then turn it back in combination with something else. A half-turn about a line lying in the sheet swaps the faces, and so does a mirror in the sheet’s own plane, but their product keeps the faces and is a half-turn about the sheet’s normal. Which layer groups roll into tubes whose families differ from those of their side-keeping halves, if any do, is the computation this census would take next. It needs the eighty in the form the layer groups derived them, and it has not been run.