A rolled sheet is never one of a pair

Most sheets roll into a tube that never repeats took each of the seventeen plane groups, rolled it along every class of lattice vector, and said what kind of group of a line came out. It said so in families and orders, such as $C_4$, $D_{6h}$ and $S_8$, and it stopped there on purpose. A family and an order name a point group. A rod group needs two more things: the climb of the screw along the axis, and whether each reflection containing the axis is a true mirror or a glide sliding along it. Class 422 carries four rod groups, and saying that a rolling lands in 422 does not say which.

This essay finishes the job. Each crystallographic rolling is built as a group of motions of space and named as one of the seventy-five rod groups. The tally is short and it is not the one a reader would guess.

Fifty-four of the seventy-five are reached. That includes every one of the forty-three achiral rod groups, so every rod group containing a mirror, a glide, an inversion or a rotoinversion is some plane pattern rolled up. Only eleven of the thirty-two chiral rod groups are reached. The twenty-one that are missed fall into two families with nothing else in them. Sixteen are the rod groups whose screw has a mirror image different from itself: 3₁ and 3₂, 4₁ and 4₃, 6₁ and 6₅, 6₂ and 6₄, bare or with two-folds across. The other five are the bare axes that do not climb at all: p1, p112, p3, p4 and p6.

A name needs a form both lists can be put in #

The two lists being compared were built on different foundations. The seventy-five are written as integer matrices on each crystal class’s own lattice basis, so a hexagonal three-fold is a matrix with entries $0$, $1$ and $-1$ in a frame at 120°. A rolled group arrives as turns about an axis through multiples of $360°/N$ and reflections in vertical planes at whatever angles the plane group’s mirror lines happened to make with the rolling vector. Neither frame is the other’s, and comparing matrices entry by entry would find nothing in common.

So both are carried into one Cartesian frame and put into a canonical form. It is what is left after removing every freedom that does not change the rod group: turn the operations about the axis until the line of some vertical reflection, or of some half-turn crossing the axis, lies along a fixed direction; optionally apply a half-turn about that direction, which turns the rod end for end; and slide the origin along the axis to whichever position spells the operations smallest. Those three moves are exactly the changes of description the seventy-five are counted up to. A turn about a rod is legitimate by any angle, because nothing across a rod repeats and so nothing across it has to stay on a lattice. The half-turn across it reverses the rod’s direction and keeps its handedness. A reflection is not on the list, and that omission is the one that matters here, because a reflection is what turns 4₁ into 4₃.

Three things are checked before any rolling is named. The form must give seventy-five different answers for the seventy-five groups, or two of them could not be told apart and a rolling landing on either would be misnamed. It must give the same answer for a group turned through an arbitrary angle, slid a quarter of a repeat and given a half-turn across, since those are the same group. And it must give different answers for 4₁ and its reflection. All three hold.

One convention sits under the whole count and should be named rather than implied. The seventy-five keep enantiomorphs apart, so a left-handed and a right-handed screw of the same pitch are two groups. Merge the eight pairs, as two hundred and thirty becomes two hundred and nineteen for space groups, and the list is sixty-seven long; rolling then reaches fifty-four of sixty-seven, and the sixteen handed groups become eight unreached entries rather than sixteen. The finding below survives the change of convention. Only its arithmetic changes.

The list itself needed a correction before this comparison meant anything. The derivation of the seventy-five records how a count reaching the right total could still miss six groups and double-count six others, and among the missing six was p211, a chiral rod group this census reaches. Matching rolled groups against the uncorrected list would have reported a rolling that is no rod group at all.

Every improper class is full #

The first result is the one with no exceptions in it. Every achiral rod group is a rolled plane pattern, and most are reached several ways.

The distinctions that make those forty-three groups forty-three, rather than the eighteen improper classes they sit over, are all distinctions rolling can make. A mirror containing the axis comes from a mirror line of the sheet perpendicular to the rolling vector, and a glide containing the axis comes from a glide line in the same position: pm rolled six cells round its mirror direction gives p6mm, and pg rolled the same way gives the group with glides in place of mirrors. The climb of the principal screw comes from where the sheet’s cells sit up the tube, so cm, whose lattice is centred, rolls into the 6₃ versions of the same classes that pm reaches with a plain axis. And the operations a flat sheet cannot have at all come out of reflections that survive the rolling with a turn attached. pg rolled along its own glide at one, two and three cells gives the inversion 1̄, the four-fold rotoinversion 4̄ and the three-fold rotoinversion 3̄, three kinds of operation that do not exist in the plane.

The classes with two roles on a rod are filled in both. Class mm2 has five rod groups, three with its two-fold along the rod and two with the two-fold crossing it, and all five are reached: pm, pg and cm give the first three, rolled so that their reflections contain the axis, while pmm and pmg, rolled a single cell round, give the two in which the two-fold lies across the tube. That second pair is exactly the pair a count putting each class on the rod only one way cannot see. So the two roles are real groups, and ordinary sheets produce them.

The same fullness has a more practical reading. A tube built by rolling a sheet with any reflection in it — a strip of paper printed with a mirror-symmetric motif, a sheet of boron nitride, a layer of a rectangular crystal — has one of these forty-three groups, and which one is decided by three questions about the sheet and one integer. Nothing in the achiral half of the classification is out of reach of that construction.

A rolled lattice climbs nothing or half a repeat #

The chiral half is where rolling runs out, and the reason lies in the sheet’s translations rather than in its symmetry.

A translation of the sheet becomes a screw of the tube: it turns by its component along the rolling vector $C$ and climbs by its component along the perpendicular repeat $T$. Cut one repeat of the tube open and every lattice point of the sheet lands somewhere in a rectangle, at a turn between nothing and a whole circumference and a climb between nothing and a whole repeat.

In the first panel every point sits on the floor, so the tube’s screws are plain turns. In the other two, half the points sit exactly halfway up, so the shortest screw climbs half a repeat for its turn. What never appears, in any crystallographic rolling of any of the seventeen plane groups, is a point a third or a quarter of the way up.

The reason is a count of cells. Write the rolling vector as $C = g\,C_0$, a whole number $g$ of copies of a shortest lattice vector $C_0$ in the same direction. The rectangle on $C_0$ and the perpendicular repeat $T$ contains some whole number of the sheet’s cells, the primitive index $n_0$, and the number of turns in one repeat of the tube is $N = g\,n_0$. The climbs of the sheet’s translations are multiples of $1/n_0$ of a repeat. So a primitive index of one gives climbs of nothing and a primitive index of two gives climbs of nothing and a half.

A crystallographic tube has $N$ equal to 1, 2, 3, 4 or 6, so it needs $n_0 \le 6$, and the lattice decides which small values exist.

On a square lattice with $C_0 = i\,a + j\,b$, the perpendicular $-j\,a + i\,b$ is also a lattice vector and the rectangle holds $n_0 = i^2 + j^2$ cells, which runs 1, 2, 5, 10, 13 and upward. On a hexagonal lattice the index is twice the Loeschian norm $i^2 - ij + j^2$, divided by three when three divides the norm, and a hexagonal rolling index is always even: it takes 2, then 14, 26 and larger. A rectangular lattice repeats only along its edges, index one, and a rhombic lattice only along its diagonals, index two. An oblique lattice never repeats.

So the only small primitive index other than one and two is five, on the square lattice, and five times any whole number is never 1, 2, 3, 4 or 6. Every crystallographic rolling therefore has $n_0$ equal to one or two, and its translations climb nothing or half a repeat. That is checked on all one hundred and six rollings rather than trusted to the argument: each has primitive index one or two, and each generating screw climbs zero or $N/2$ steps of $N$.

A climb of nothing or a half is its own mirror image #

A rod group’s handedness lives in its principal screw. An axis of order $n$ with climb $k/n$ is the screw $n_k$, and reflecting it in a plane containing the axis keeps the climb and reverses the turn, which carries $n_k$ to $n_{n-k}$. The screw is its own mirror image exactly when $k = n - k$ modulo $n$, which means $k = 0$ or $k = n/2$. The eleven screw axes split into pairs and singletons by that one congruence.

The two statements are the same statement. The screws of a rolled tube come only from the sheet’s translations, because a translation is the only operation of a plane group that keeps both the rolling vector and the perpendicular direction pointing the way they pointed. The turns of a three-fold or six-fold plane pattern, for instance, do not survive. So a tube’s screws climb nothing or half a repeat, and those are the two climbs a mirror leaves alone. A rolled tube with an axis of order six or less can have a hand, but it cannot be one of a pair. Its mirror image has the same rod group as the tube itself, even when the tube, with its half-turns and everything else, contains no reflection at all.

This is a sharper form of something the previous census of rolled tubes noticed for bare axes, where only 2₁, 4₂ and 6₃ were reached. Here it holds across every family, with half-turns across the tube or without them, and it accounts for sixteen of the twenty-one unreached groups at once: p3₁ and p3₂, p4₁ and p4₃, p6₁, p6₂, p6₄ and p6₅, and the same eight screws with two-folds crossing the axis.

The consequence runs against what carbon nanotubes suggest, and the conflict is only apparent. A rolled honeycomb turns with screws of order 14, 98 and 794, and almost every nanotube comes in a left-handed and a right-handed form. Those tubes are handed in exactly the sense this essay says rolling cannot reach. They escape it because their orders are not crystallographic: fourteen turns in a repeat is a primitive index of fourteen, and at fourteen the climbs are fourteenths, most of which are not their own mirror image. Every handed tube lives outside the seventy-five. Inside them, where a lattice could also hold the axis, rolling produces nothing with a mirror-image partner.

Five axes that never climb #

The five remaining misses are chiral and are their own mirror images, so the climb argument does not touch them. They are missed for a reason about the lattice rather than the screw.

p1, p112, p3, p4 and p6 are bare axes of order 1, 2, 3, 4 and 6 whose screws do not climb. A tube with such a group needs every translation of its sheet at climb zero, so its primitive index must be one. It also needs the plane group to keep nothing but translations under the rolling: no half-turn, since that would cross the tube, and no reflection line along or across the rolling vector.

Index one needs a rectangular primitive cell with one edge along $C$, which only the square and rectangular lattices have, and only along their cell edges. There are eight plane groups on those two lattices, and each brings something to every index-one rolling. p4, p4m, p4g, pmm, pmg and pgg contain a half-turn, and a half-turn always survives rolling, since it reverses $C$ and $T$ together. pm and pg have no half-turn, but their reflection lines run along one cell edge and across the other, so the reflection survives rolling along either. No lattice that could roll a resting axis comes without something extra, and the five bare resting axes are unreachable for that reason.

The hexagonal pattern that comes closest, p3, has no half-turn and no reflection, and fails on the other condition. A hexagonal index is always even, so its tubes always carry a translation halfway up, and a bare axis from p3 is 2₁, 4₂ or 6₃ rather than 2, 4 or 6. The unreached five and the reached three differ by exactly half a repeat.

The eleven chiral tubes, and where they come from #

The chiral tubes all come from the three plane groups that have no reflection and a lattice that repeats. That follows from what survives rolling. The directions giving six turns or fewer are cell edges and diagonals, and every plane group with a mirror or glide has one along or across each of those directions, where it survives as an improper operation of the tube. p1 and p2 are on an oblique lattice and never repeat at all.

Two of the eleven are worth looking at, because each shows rolling building a group that has nothing obviously to do with the sheet it came from. p4 rolled three cells round a cell edge gives p312: a three-fold screw axis with two-folds crossing it, made from a pattern with four-fold symmetry. The four-fold turn of the sheet does not survive; its square, the half-turn, survives as the two-folds across the tube; and the three-fold comes from the three cells, not from anything in the pattern. p3 rolled along a shortest lattice vector gives p112₁, a two-fold screw and nothing else, from a pattern with a three-fold axis and no two-fold at all. In both cases the axis order is set by the choice of rolling vector, and only the decoration comes from the plane group.

And the chirality is manufactured. p4, p3 and p6 are not chiral objects in any sense a plane pattern can be, since every plane pattern is carried to itself by turning the sheet over. Rolled, they become tubes with no improper operation, which is the same manufacture of a hand that a sheet undergoes when its two faces are allowed to differ. What rolling cannot manufacture is a hand with a partner of a different kind. The mirror image of every chiral rolled tube has the same rod group as the tube, although the group contains no reflection.

What the naming cannot show #

The search is finite and it is complete for this question. Rolling vectors are taken out to coordinates of seven, which yields one hundred and six crystallographic rollings. A longer vector along a new direction has a larger primitive index, and on every lattice the index grows past six before the coordinates reach seven, so no crystallographic rolling lies outside the range. A longer vector along an old direction multiplies $N$, which past six is no longer crystallographic.

The group named is the group of the pattern, not of any particular decoration of it. A real sheet has atoms at definite positions, and a sheet whose atoms have less symmetry than its lattice has a smaller plane group, which is one of the seventeen and is covered. What the census does not model is a sheet that is strained as it rolls, where the inner surface is compressed and the outer stretched. A strained tube can lose symmetry the ideal rolled pattern has, and that is a question about elasticity rather than about groups.

Reachable is not realised. Fifty-four rod groups are possible for a rolled sheet. Which of them any real tube, fibre or scroll actually has is a survey question, and what a thread scatters is where a measurement would read it off. Nor does the census say anything about how handed a tube is. A group that is its own mirror image can still describe a tube that is strongly twisted in one sense, and a continuous measure of chirality answers that where the group gives only one bit.

And the equivalence carries the result. Under a convention that merged enantiomorphs, the finding would read “rolling reaches every rod-group type up to reflection except eight pairs and five resting axes”, which is the same fact in a form that hides its most striking half. The sixteen separate entries are what make visible that rolling never produces a handed pair.

Still open: a sheet with two faces, and the orders past six #

A plane pattern has no inside and outside, but a real sheet usually does. A layer with different atoms on its two faces has one of the eighty layer groups, and rolling it puts one face inside the tube and the other outside. No symmetry of the tube can exchange its inside and its outside, so every operation of the layer that turns the sheet over dies in the rolling. The obvious guess is that only the side-keeping half of the layer group survives, which is a plane group, so the reach cannot grow. Whether a turned-over operation composed with something else can survive, landing on a rod group this census does not reach, has not been computed.

The other question is where every handed tube actually lives. Past order six, the primitive index is free, climbs of a third, a fifth and a fourteenth all appear, and handed pairs appear with them. Line groups with unbounded orders form a small number of infinite families, and a rolled sheet lands in some of them. Which families, at which orders, and whether rolling reaches both members of every pair it reaches at all, is the same census this essay took, with the crystallographic restriction removed from its last step.