The tube has a screw no lattice allows

Seven friezes round a cylinder rolled a strip into a cylinder and found the seven families of point groups with a single axis. A strip has one direction of repeat, and the rolling used it up: every translation of the strip became a turn about the axis, and nothing was left to climb with. The essay ended on the case it could not reach. A plane pattern has two directions of repeat, and rolling it along one of them keeps the other.

That is not an abstraction. It is the construction of a carbon nanotube, which is a single sheet of graphene’s honeycomb rolled so that one lattice vector of the sheet becomes the circumference. The vector is written $C = n a_1 + m a_2$, with $a_1$ and $a_2$ two lattice vectors of the sheet sixty degrees apart, and the pair (n, m) is how every nanotube is named.

The tube that results has symmetries the sheet cannot have. Its turns come in steps of 360°/N, with N = 14 for the (2, 1) tube, 98 for (5, 3) and 794 for (12, 11), and the crystallographic restriction, which confines the sheet to turns of order 1, 2, 3, 4 and 6, has nothing to say about any of them. The rolling also discards most of the sheet’s mirrors, and keeps any at all only when C lies along one of two directions.

What the rolling keeps, and what it spends #

The shortest lattice vector of the sheet perpendicular to C is $T = \big((2m + n) a_1 - (2n + m) a_2\big)/d_R$, where $d_R$ is the greatest common divisor of 2n + m and 2m + n. The rectangle with sides C and T is one repeat of the tube: roll it so that its two long edges meet, and stacking copies of it along T builds the whole tube. It holds $N = 2(n^2 + nm + m^2)/d_R$ hexagons and twice as many atoms.

A motion of the sheet becomes a motion of the tube only if it respects the rolling — only if it carries the circumference C onto itself or onto its reverse. Four kinds of motion do.

• Every translation of the sheet does, and it becomes a screw: its component along C turns the tube about its axis, and its component along T moves it up the axis.
• Every half-turn of the sheet reverses C, and becomes a half-turn about an axis crossing the tube at right angles.
• A mirror whose line is parallel to C becomes a mirror across the tube, perpendicular to its axis.
• A mirror whose line is perpendicular to C becomes a mirror containing the axis.

The sheet’s glides join the mirrors of their direction. Everything else — the three-fold and six-fold turns, and every mirror at any other angle — would carry C to a vector in a different direction, and does not survive.

That is a prediction about which operations the tube has, and it was checked against the tube rather than argued from the sheet. Every isometry that maps a cylinder to itself acts on the two coordinates of its surface — distance round the circumference and height up the axis — by changing the sign of neither, of both, or of one, and then shifting. So for each of the four sign patterns, every shift that carries one atom of the rolled tube onto another was tried against all the atoms, and the shifts that carried every atom onto an atom were counted.

For the (4, 2) tube, whose repeat holds fifty-six atoms, the search finds twenty-eight screws in a repeat, two of which climb not at all and are pure turns, twenty-eight half-turns, and no mirror or glide of either kind. The screw that climbs least rises 0.3273 per step, in units of the sheet’s lattice spacing, while turning through 64.29°.

A screw of order fourteen from a sheet that allows six #

Where the turns come from is a one-line calculation. A translation v of the sheet turns the tube through $2\pi\,(v \cdot C)/|C|^2$, the fraction of the circumference it covers. For $v = i a_1 + j a_2$ the dot product $v \cdot C$ is $\big(i(2n + m) + j(2m + n)\big)/2$, and as i and j run over the integers that takes exactly the multiples of $d_R/2$. So every turn of the tube is a multiple of 360°/N, and every multiple occurs, one per translation of the sheet modulo the repeat.

Across all ninety tubes with n up to twelve, the screws found on the rolled atoms numbered N in every case, and their turns sat at the multiples of 360°/N. The values run from 2 to 794. The orders a lattice allows — 1, 2, 3, 4 and 6 — occur for exactly six tubes: (1, 0), (1, 1), (2, 0), (2, 2), (3, 0) and (3, 3). Every one of those is about four tenths of a nanometre across or less, at the very edge of what exists — tubes that thin have been reported only grown inside the channels of other materials — so a screw a crystal could have belongs to the thinnest tubes alone, and every tube of ordinary size has one no crystal could.

That is not a violation of the crystallographic restriction, and saying why is the point. The restriction is a statement about a turn that carries a lattice onto itself, in the plane perpendicular to the axis of the turn. A tube repeats in one direction only, along its axis, and a turn about that axis leaves the axis exactly where it was. There is no lattice across the axis for the turn to preserve, so the restriction has nothing to act on. The sheet’s lattice has not been violated either; it has been rolled up, and its translations along C have become turns.

Seventy-five ways to be a thread is the classification of the rod groups that sit inside space groups, and there the screws are restricted, because a rod group that is part of a crystal has the crystal’s lattice around it. Its closing section names the helices that are not on its list, the protein helix among them. A nanotube is another, and it is off the list for a reason that can be computed in advance from two integers.

How many pure turns, and why #

Among the N screws of a repeat, the ones that do not climb are the translations of the sheet lying along C itself. The shortest lattice vector in that direction is C divided by the greatest common divisor of n and m, so there are gcd(n, m) pure turns, and the tube has a rotation axis of that order. The search confirmed it for all forty-four tubes with n up to eight.

The difference between the two numbers is the whole character of a nanotube. The zigzag tube (9, 0) has nine pure turns and eighteen screws; its atoms sit in rings, and a turn by forty degrees carries each ring onto itself. The chiral tube (6, 5) has one pure turn, the identity, and 182 screws. Nothing turns it onto itself without also moving it along its length, and every one of its atoms lies on one helix.

Mirrors in two directions only #

The honeycomb has mirrors in six directions, thirty degrees apart. A mirror of the tube needs one of them to be parallel or perpendicular to C, and since the mirrors come in perpendicular pairs, that happens exactly when C itself lies along a mirror of the sheet — along $a_1$, which gives the tubes called zigzag, or along $a_1 + a_2$, thirty degrees away, which gives the armchair tubes.

Among the 168 rolling vectors with n and m between zero and twelve, 36 lie on a mirror direction and 132 do not. The search on the rolled atoms agreed tube by tube: every zigzag and armchair tube has N mirrors across its axis and N mirrors or glides containing it, and every other tube has none.

So almost every nanotube is chiral. The sheet it is rolled from is not — it has mirrors in its own plane and a mirror in the plane of the sheet itself, and whether a sheet has a hand in space depends on what its operations do to its two sides, which for graphene is nothing. The hand comes from the rolling: it is the choice of a direction for C that no mirror of the sheet contains.

Forget the climb, and a family of point groups comes back #

A screw has two parts, a turn and a climb, and a tube’s operations with their climbs set to zero are a finite group of motions about a point on the axis. For a strip, that forgetting was the whole construction, and it produced the seven families of point groups with one axis. For a tube it can be done afterwards, and it tells which of those families the tube belongs to.

The turns of every tube, forgotten, are the N multiples of 360°/N, a cyclic group of order N. A chiral carbon tube adds its N half-turns across the axis and nothing else, so its point group is $D_N$, the dihedral group of order 2N: $D_{14}$ for the (2, 1) tube, $D_{76}$ for (6, 4), $D_{794}$ for (12, 11). A zigzag or armchair tube adds mirrors of both kinds, and among its mirrors across the axis there is always one that turns through nothing, so its point group is $D_{Nh}$: $D_{18h}$ for (9, 0), $D_{10h}$ for (5, 5). Both families are ones the rolled friezes reached — the spinning hop gives $D_n$ and the spinning jump $D_{nh}$ — and the orders are ones that essay’s census never came near, because it stopped at the orders a lattice around the axis would allow. Every one of the forty-four tubes with n up to eight was checked against those two names, and every one fitted.

The same forgetting is what makes the order N visible in an experiment. A tube’s diffraction pattern is the pattern of a helix, and what a thread scatters is confined to layer lines by a selection rule written in the helix’s own turn and climb. The turn and climb of the tube’s screws are in that rule, which is why the (n, m) of a single tube can be read off its electron diffraction rather than guessed from its width.

(n, m) and (m, n) are the two hands #

A vector and its reflection in the armchair line, $n a_1 + m a_2$ and $m a_1 + n a_2$, are the same length and make the same angle with the nearest mirror on opposite sides of it. The reflection is a symmetry of the honeycomb, so it carries the rectangle for one tube onto the rectangle for the other, and the two tubes are mirror images.

Four such pairs were rolled and searched. In each, the screw with the smallest climb had the same climb and the same turn in both tubes, with the sign of the turn reversed: 77.14° for (4, 1) against −77.14° for (1, 4), and likewise for (5, 2), (6, 5) and (8, 3). Neither tube of a pair has a mirror, and no motion of space carries one onto the other except a reflection, which is the definition of an enantiomorphic pair.

The convention that makes this bookkeeping work is worth naming. Nanotubes are listed with 0 ≤ m ≤ n, which puts C between the zigzag direction and the armchair direction, and the other hand of every chiral tube is the one not listed. A table of tubes in that convention is a table of pairs, and a synthesis that grows a given (n, m) grows both hands unless something chiral is present to prefer one.

One width, two groups #

The circumference of a tube is $|C| = \sqrt{n^2 + nm + m^2}$, and the integers that can appear under that square root are the ones of the form $n^2 + nm + m^2$. They are the same numbers that index a hexagonal lattice’s sublattices of the same shape — the indices at which a hexagonal group sits inside itself — and the same numbers Caspar and Klug used to count the triangles of an icosahedral virus shell. A nanotube’s width, a sublattice’s index and a virus’s size are one quadratic form read three ways, because all three are the squared length of a vector in the hexagonal lattice.

A number can have more than one such vector, and then two different tubes have the same width.

Seven squared is 49, and 49 is 7² + 7·0 + 0² and also 5² + 5·3 + 3². The (7, 0) tube repeats every 1.732 and has fourteen screws in a repeat, with mirrors across and along its axis; the (5, 3) tube repeats every 12.124 and has ninety-eight screws and no mirror. A measurement that reads only the diameter cannot tell them apart, and yet one has fourteen screws in its repeat and the other ninety-eight, one has fifty-six operations in a repeat and the other a hundred and ninety-six, and only one of them has a hand.

Among the ninety tubes with n up to twelve there are three other widths shared in this way: 91, taken by (6, 5) and (9, 1); 133, by (9, 4) and (11, 1); and 147, by (7, 7) and (11, 2). The first two pairs happen to share their N as well. The last is the armchair case of the (7, 0) and (5, 3) story, fourteen screws against ninety-eight.

Two kinds of atom #

Boron nitride makes the same honeycomb with the two sites of each cell occupied by different atoms, boron on one and nitrogen on the other. An operation of a boron nitride tube must now carry boron onto boron, and the search was run again with that requirement added.

Every half-turn of the carbon sheet exchanges the two sites, so all of them go. The mirrors split: the honeycomb’s six mirror directions are two families of three, one family containing the bonds and one crossing them, and only the family containing the bonds survives when the two ends of a bond are different atoms. The zigzag tube keeps its mirrors containing the axis and loses the ones across it; the armchair tube keeps the mirrors across and loses the ones containing the axis; the chiral tube had neither and keeps its screws alone. The orders fall from 72 to 36 for (9, 0), from 40 to 20 for (5, 5), and from 152 to 76 for (6, 4).

That changes what the tube is allowed to do, and not only how many operations it has. A vector along the axis is reversed by a half-turn crossing it and by a mirror across it, and left alone by screws and by mirrors containing the axis. So the zigzag and chiral boron nitride tubes have no operation reversing their axis and are polar along their length, able to carry a dipole from one end to the other, while the armchair boron nitride tube is not, and no carbon tube is. Forgetting the climbs says the same thing in the language of point groups: boron nitride’s zigzag tube (9, 0) has $C_{18v}$, its armchair tube (5, 5) $C_{10h}$ and its chiral tube (6, 4) $C_{76}$, and $C_N$ and $C_{Nv}$ are the polar families while $C_{Nh}$, with its mirror across the axis, is not. The group permits the dipole; how large it is, if it is there at all, is a question about the electrons that no symmetry argument answers.

What rolling does not settle #

The positions are the ideal ones. Each tube here is the honeycomb wrapped onto a cylinder with every bond keeping its length along the surface. A real tube relaxes: its bonds across the circumference and along the axis come out slightly different, and a narrow tube strains noticeably. A relaxation that respects the tube’s operations leaves the group unchanged, and the counts above are the counts of that group; a relaxation that broke a symmetry would be a transition, and nothing computed here rules one out.

The census is finite and the arithmetic is not. The formulas for N, for the pure turns and for which tubes have mirrors are proved by the dot-product calculation and the mirror argument, for every n and m. The search on the atoms confirms them for the ninety tubes with n up to twelve, and the checks on half-turns and hands for the smaller ranges named above.

The steepest screw is a choice. A screw and the same screw combined with a whole turn are the same operation, and the helix drawn for each tube is the one whose turn is smallest in size, which is the one a reader sees. Its handedness is a property of the tube; its particular angle is a property of the convention.

And only the honeycomb has been rolled. The seventeen plane groups have their mirrors and glides in other arrangements, and which tube groups each of them rolls into, along which vectors, is a larger census than this one.

The checks, and the two cases they turn away #

Every count above comes from a test that could fail, run on the rolled atoms.

The first refusal is the seam. A rolling vector that is not a lattice vector, such as $4.5\,a_1 + a_2$, carries atoms on one edge of the rectangle onto hexagon centres on the other, so the rolled sheet does not close. The second is the mirror that is not there: on the (5, 3) tube every candidate mirror and glide the search could construct, 392 of them, failed to carry the atoms onto atoms.

A helix in the title, in 1991 #

Sumio Iijima’s report of 1991 that introduced carbon nanotubes to most of the scientific world was titled Helical microtubules of graphitic carbon, and the helix was in the electron diffraction from the start. Single-walled tubes followed in 1993. The (n, m) description was set out in 1992 by Noriaki Hamada, Shin-ichi Sawada and Atsushi Oshiyama, who also showed that the pair decides whether a tube conducts like a metal or like a semiconductor. The tubes’ full symmetry groups, as groups of motions of a line rather than of a lattice, were worked out later in the decade by Milan Damnjanović and his collaborators.

The quadratic form under the circumference is older than any of it. Eisenstein’s integers are the points of the hexagonal lattice, and the norm of the integer $n + m\omega$, with $\omega$ a primitive cube root of unity, is $n^2 - nm + m^2$ — which is $n^2 + nm + m^2$ once $a_2$ is taken at sixty degrees rather than at a hundred and twenty.

Where this goes: the seventeen, rolled #

The honeycomb is one pattern with its mirrors in one arrangement. A plane pattern with glides rolls into a tube with glides or rotoreflections in their place; one with mirrors in only one family, like boron nitride’s, loses the other family whatever the rolling vector; one with no two-fold centres has no half-turns crossing its tubes. Each of the seventeen plane groups rolls, along each class of lattice vector, into a definite group of motions of a line, and the question left is the whole table — which of those groups each plane group reaches, along which vectors, and which groups of a line no rolled pattern reaches at all.