Most sheets roll into a tube that never repeats

The tube has a screw no lattice allows rolled one pattern. The honeycomb was wrapped round a cylinder so that a lattice vector became the circumference, and the tube that resulted turned and climbed with screws of order 14, 98 and 794 — orders the flat sheet could never have. Every question that essay asked was answered by two integers, and every answer was checked on the rolled atoms.

What it could not ask is whether the honeycomb is typical. A sheet of graphene is an unusually well-behaved thing to roll: its lattice is hexagonal, its symmetry is the largest a plane pattern can have, and it has mirrors in six directions. There are seventeen plane groups and five plane lattices, and rolling is defined for all of them. The table of what each one becomes is the subject here, and the first thing in it is a failure the honeycomb cannot show.

Rolling a general plane pattern usually produces a tube with no repeat at all. Out of 1,008 classes of rolling direction, taken over the seventeen groups, 567 give a tube that climbs forever without ever returning to the angle it started at. For the two groups on an oblique lattice the figure is every direction, without exception.

What rolling is, said precisely enough to be general #

Rolling is a covering map. The plane is the universal cover of a cylinder of circumference equal to the rolling vector’s length, and the deck transformation is the translation by that vector: two points of the plane land on the same point of the cylinder exactly when they differ by a whole number of copies of $C$.

That single sentence decides which symmetries survive. An operation of the plane pattern descends to the cylinder exactly when it carries the deck group to itself — when its linear part sends $C$ to $C$ or to $-C$ — and the group of the rolled pattern is the subgroup of survivors divided by the translations along $C$. Nothing is lost in the other direction either: every symmetry of the pattern on the cylinder lifts to the plane, because the plane is a simply connected cover. So the rolled group is the stabiliser modulo $\langle C \rangle$, exactly, and the classification is a computation on the plane group rather than a search on the tube.

A point of the cylinder has two coordinates, the distance $s$ round the circumference and the height $z$ up the axis, and any isometry of the cylinder acts on them by changing the sign of neither, of both, or of one, and then shifting. Four sign patterns, four kinds of survivor:

• a translation of the sheet keeps both signs and becomes a screw;
• a half-turn of the sheet reverses both and becomes a half-turn about an axis crossing the tube;
• a mirror or glide whose line is parallel to $C$ keeps $s$ and reverses $z$, giving a mirror across the tube or a rotoreflection;
• a mirror or glide whose line is perpendicular to $C$ reverses $s$ and keeps $z$, giving a mirror or glide containing the axis.

Every other operation of the plane group — the three-fold and four-fold and six-fold turns, and every reflection at any other angle — sends $C$ somewhere else and does not survive. That is the whole of the correspondence, and it holds for any plane group. It is also why the restriction on rotation orders has nothing to say about the result: the turns of the tube are about an axis with no lattice across it, and the restriction is a statement about a turn that preserves a lattice in the plane it turns in.

The repeat is a property of the lattice, not of the rolling #

A tube repeats when something carries it straight up its own axis with no turn. In the plane that is a lattice vector perpendicular to $C$. The honeycomb always has one, which is why the nanotube literature never states the condition; a general lattice need not.

The condition is one linear equation. Writing $Q$ for the lattice’s Gram matrix and $c$ for the rolling vector’s coordinates in the lattice basis, a lattice vector $v$ is perpendicular to $C$ when $v^{\mathsf T} Q c = 0$. That is a single equation on two whole numbers, and it has a nonzero whole solution exactly when the vector $Qc$ points along a rational direction — when the ratio of its two entries is a fraction.

Whether it does is decided by the lattice type, because the type decides how much freedom the metric has.

The square lattice’s Gram matrix is the identity, so $Qc$ is $c$ itself and is rational for every direction — which is the four-fold turn saying the same thing, since the vector at right angles to a lattice vector is again a lattice vector. The hexagonal Gram matrix has entries $1$ and $-\tfrac12$ and is rational as well, although here no rotation of the lattice carries $c$ to a perpendicular vector and the perpendicular has to be constructed: $(c_1 - 2c_2,\, 2c_1 - c_2)$, reduced by its own common factor.

The remaining three carry free parameters. A rectangular lattice has $Q = \operatorname{diag}(1, \rho)$ with $\rho$ the squared ratio of the two edges, and $v_1 c_1 + \rho\, v_2 c_2 = 0$ can hold for all $\rho$ only if $v_1 c_1$ and $v_2 c_2$ vanish separately — so only when $c$ lies along a cell edge. A rhombic lattice, which is a centred rectangle described on its two equal primitive vectors, has $Q = \begin{pmatrix} 1 & k \\ k & 1\end{pmatrix}$, and the same argument leaves only the two diagonals, which are precisely its mirror lines. An oblique lattice has both parameters free, and the two conditions that result have no common solution at all.

So the criterion arrives twice over, from two unrelated directions. A perpendicular lattice vector exists whenever $C$ lies along a mirror of the lattice, because then $v - m(v)$ is a lattice vector perpendicular to $C$ for any $v$ at all; and it exists whenever the metric is rational, whatever $C$ is. The square and hexagonal lattices satisfy the second condition everywhere because their metrics are forced by their own rotations. The rectangular and rhombic lattices satisfy the first on two directions each and the second nowhere. The oblique lattice, which has neither a mirror nor a forced metric, satisfies neither.

Three questions and a number decide the rest #

Given that a tube repeats, what group it has is decided by very little. The number of turns in one repeat is $N = |c \wedge t|$, the number of cells of the sheet in the rectangle on $C$ and the perpendicular vector $T$, and the turns are exactly the multiples of $360°/N$ — one for each translation of the sheet modulo that rectangle. Beyond $N$, only three yes-or-no questions remain: does the plane group contain a half-turn, does it contain a mirror or glide parallel to $C$, and does it contain one perpendicular to $C$.

The answers name one of the seven families of groups with a single axis, the same seven the rolled friezes produced in Seven friezes round a cylinder, which are the axial families the classification of finite groups of motions reaches once the orders are not restricted: a bare axis, an axis with half-turns across it, an axis with mirrors along or across it, and the combinations. Only one of the three answers is subtle. An operation of the third kind — one that fixes the circumference and reverses the axis — has a turn that no choice of origin can remove, because rotating the origin about the axis commutes with it. When that turn is a multiple of $360°/N$ the operation is a mirror straight across the tube and the family is $C_{nh}$ or $D_{nh}$; when it sits half a step off, the same operation is a rotoreflection instead, and the family is $S_{2n}$ or $D_{nd}$. Every other offset is absorbed by moving the origin and decides nothing.

The rarest outcome is worth naming. Only pg produces the $S_{2n}$ family, and only along the direction of its own glide: 7 of the 441 repeating rollings, against 175 that give $D_n$. Rolled along its glide line at the shortest vector, pg gives $S_2$ — which is an inversion centre, and the plainest possible illustration that rolling manufactures operations rather than merely keeping them, since a plane pattern has no inversion in space at all.

The census, and what it says about the honeycomb #

The table splits three ways and the split is entirely by lattice. p4, p4m, p4g, p3, p3m1, p31m, p6 and p6m repeat in every direction. The six groups on a rectangular or rhombic lattice repeat in 14 classes out of 63, which are the classes along their two mirror directions. p1 and p2 repeat in none of their 112.

That is 441 repeating classes against 567 that do not, and the honeycomb sits in the first group by an accident of which lattice it has. The check that the arithmetic here is the same arithmetic as before is the honeycomb itself: rolled along the vector written $(4,2)$ in the notation the nanotube literature uses, the index comes out 28, and along $(9,0)$ it comes out 18 with the family $D_{18h}$ — the same numbers the search on the rolled atoms produced, reached this time without rolling anything.

Two of the eight always-repeating groups deserve a second look, because their answer is the same for a reason that is not the same. p4 has no mirror at all, so no rolling of it can produce one, and every rolling gives $D_n$ — an axis with half-turns across it and nothing else, which is a chiral group. p3 has no mirror and no half-turn either, so every rolling gives the bare axis $C_n$. A pattern with three-fold symmetry and nothing else rolls into a tube whose only symmetries are its own screws. Both outcomes are chiral in space, which is the rolling manufacturing a hand out of a sheet that had none — the same manufacture a layer’s handedness traces to what each operation does to the two sides of a sheet, arriving here through what each operation does to the two ends of an axis.

What a tube with no repeat actually is #

The cases that do not repeat are not degenerate. The rolled pattern is still a perfectly good discrete set of points on a cylinder, and it still has symmetries; what it lacks is a translation.

The translations of the sheet, divided by $\langle C \rangle$, form an infinite cyclic group generated by one screw. Its climb is the cell’s area divided by $|C|$, always, and its climbs stack up to a discrete set of heights. Its turn is $v \cdot C / |C|^2$ for a lattice vector $v$ completing $C$ to a basis, and that number is a fraction exactly when the perpendicular exists. Where it is irrational, the powers of the screw take angles that are dense in the circle: every angle is approached and none is reached twice.

A structure generated by a single screw with an irrational turn is a helix and not a crystal, and this is where the classification of rolled patterns meets the structures with a wavevector that is no fraction of a reciprocal lattice vector. Both are one-dimensional periodicities whose relation to a second length is incommensurate; both have sharp diffraction and no cell. The difference is that a modulated crystal is aperiodic because two lengths in it do not divide, while a rolled oblique sheet is aperiodic because a direction is not a lattice direction, and what does not divide is an angle. The line between discrete order and dense disorder runs between the two halves of this census, and the lattice type is what puts a rolling on one side of it.

The protein $\alpha$-helix is the standard example of a helix whose turn per residue is not a fraction, and the seventy-five rod groups closes on exactly that point — helices that are not on its list. A rolled oblique sheet is another, and it arrives from a different place: not from a chain whose repeat happens to be awkward, but from a perfectly ordinary periodic pattern rolled in a perfectly ordinary direction.

The complete reach, and the four cells with no crystal class at all #

A rolled group can only be a rod group if its number of turns in a repeat is one of the five orders a lattice permits, and since $N$ grows with the length of the rolling vector that is a finite list which can be written out completely.

Having one of the five orders is necessary and it is not sufficient, which is the trap in the paragraph above and worth being caught by. The family $S_{2n}$ at $n = 4$ is $S_8$: its rotations are the four turns of $C_4$, perfectly crystallographic, and its other four operations are eight-fold rotoreflections. An eight-fold rotoreflection has matrix trace $\sqrt2 - 1$, which no integer matrix has, so $S_8$ is not one of the thirty-two crystal classes and a rolled group in that cell is not a rod group. The same goes for $S_{12}$, $D_{4d}$ and $D_{6d}$.

Deciding it is a computation on the matrices rather than a lookup: each family at each order is built as a set of rotations and reflections about the axis, and its multiset of determinants and traces is compared with the same multiset taken over each of the twenty-seven axial classes. Thirty-one of the thirty-five cells match a class, and between them they name all twenty-seven — four of the classes twice over, because at order one the distinctions collapse: $C_{1h}$ and $C_{1v}$ are both the mirror, $D_1$ is the two-fold, $D_{1d}$ is $2/m$ and $D_{1h}$ is $mm2$.

All four of the non-crystallographic cells are reached. pg rolled along its own glide at the fourth multiple gives $S_8$, and a tube with an eight-fold rotoreflection is exactly the essay’s own theme arriving one level up: the restriction bounds a turn that preserves a lattice in the plane it turns in, a tube has no such plane, and an improper axis is as free as a proper one.

So the count that answers the question asked is twenty-nine: twenty-nine of the thirty-five cells are both a crystal class and reached by some rolling. Thirty-three of the thirty-five are reached. The two that are not are $C_1$ and $C_3$ — a bare axis of order one or three, with no half-turn and no mirror of any kind — and the reason is a single arithmetic fact.

A bare axis requires a plane group with no half-turn and with no mirror or glide in any direction a rolling can catch, and only four qualify: p1, p3, p3m1 and p31m. The first is oblique and never repeats. The other three are hexagonal, and a hexagonal rolling index is always even. Writing the index out,

N = 2(c₁² − c₁c₂ + c₂²) / gcd(2c₂ − c₁, 2c₁ − c₂)

the divisor is odd whenever either coordinate is odd, since then one of the two arguments is odd; and when both coordinates are even the whole expression halves to the same shape again. So the factor of two never cancels. Checked over 6,560 hexagonal directions, no index is odd — which is a strong test of a claim proved in two lines, and it is the kind of claim where a proof and an enumeration are worth having together.

The consequence is that the unreachable groups are the smallest ones. It would be reasonable to guess the other way — that rolling produces small groups easily and large ones with difficulty — and the truth is that $D_{794}$ is reached by a tube two nanometres across while $C_3$ is reached by nothing at all.

Three screws of sixteen #

The family names an axis and its decorations but not the screw. An axis of order $n$ admits $n$ screws counting the plain rotation, since the $n$-th power of the operation has to be a whole repeat, so the five permitted orders carry sixteen screws between them — the sixteen that the enumeration of rod groups finds on the classes that are nothing but an axis.

Three of the sixteen are reached, and they are $2_1$, $4_2$ and $6_3$ — each the screw that climbs exactly half a repeat for every turn. This is the parity fact again, one level down: a bare rolled axis can only come from a hexagonal pattern, and the generating screw of a hexagonal rolling always has a half-repeat climb, because the perpendicular vector and the rolling vector span a rectangle containing exactly two cells for every turn.

An axis of order 4 with a quarter-repeat climb, $4_1$, is an entirely ordinary thing to find in a crystal — the screws that turn while climbing enumerate it along with the other ten. No rolling of a plane pattern produces one.

What is checked, and what the checking cannot reach #

The counterfactual is the test that matters most, because the central claim is negative and a negative claim computed at one numerical shape is worth nothing. “A general oblique lattice has no perpendicular lattice vector” cannot be checked by picking an oblique lattice and looking, since a badly chosen one would have a rational metric by accident and the check would report the opposite of the truth. So the free parameters are kept as symbols and a coefficient has to vanish identically in them; and then, to show that the symbols are doing work, the parameters are fixed at the fractions $k = \tfrac14$ and $\rho = \tfrac32$ and the repeat comes back, with a perpendicular vector $(-7, 5)$ and an index of 12.

That is the convention the whole census depends on and it should be stated rather than assumed. “Oblique” here means a lattice with no relation among its parameters, which is what a crystallographer means by a triclinic cell and is not what any particular measured cell is. Any real oblique lattice has measured parameters, and measured parameters are rational to the precision they were measured at, so some very long perpendicular vector exists for every direction. The distinction is the ordinary one between a symmetry and an accident: the repeat that a rational metric provides is not stable under the smallest change of the cell, and the repeat a square lattice provides is stable under every change that keeps it square. What the picture cannot show is where the boundary lies in practice — how long a repeat has to be before it stops being a repeat for any purpose a reader cares about — and that is a question about coherence lengths and diffraction line widths rather than about groups.

Two further limits are worth naming. The census counts classes of rolling vector out to a bounded length, and the fraction that repeat is exact for that range rather than a limiting density; for the groups on a rectangular or rhombic lattice the repeating fraction falls towards zero as the range grows, since the number of directions grows and the number of mirror directions does not. And the family is not the group: two rollings with the same family, the same order and the same screw can still differ in whether a mirror containing the axis is a mirror or a glide, and that distinction has not been pushed through to the seventy-five names.

Still open: which of the seventy-five, by name #

The reach above is stated in families and orders, which is enough to say which combinations occur, which are crystal classes and which two are unreachable. It is not enough to name the rod groups: twenty-seven axial classes carry seventy-five rod groups between them, and knowing that a rolling lands in the class 422 does not say which of that class’s four rod groups it is. Doing that needs the glide content of the operations containing the axis carried through the quotient as well as the screw pitch, and then a match against the seventy-five derived from the axial classes — a comparison between two enumerations built on different bases, which is the point at which this stops being arithmetic about a plane group and becomes bookkeeping about symbols.

The other question left is the one the split itself raises. Rolling divides the seventeen into eight groups that always repeat, seven that repeat on two directions and two that never do, and the division is by lattice alone. A layer group, which is what a rolled sheet becomes if it is wrapped the other way, has eighty types against the rod groups’ seventy-five, and the layer groups are built from the same twenty-seven axial classes. Whether the reach of the layer construction has holes in it of the same kind — small groups unreachable while large ones are easy — is a question this census’s method answers and this census has not asked.