A sheet with two faces rolls into two tubes

The centring that turns into a screw rolled all seventeen plane patterns into tubes and sorted the tubes into the thirteen families of line groups. Every sheet in that census was a flat pattern with nothing above or below it: a pattern of the plane, which has no faces. A real sheet has two. A single layer of molybdenum disulfide has sulfur above and below its metal atoms, a Janus layer of MoSSe has sulfur on one face and selenium on the other, and buckled silicene has half its atoms raised and half lowered. The symmetry group of such a sheet is not a plane group but one of the eighty layer groups, which eighty is seventeen, seventeen and forty-six derived as a plane group with a sign on each operation, plus one when the operation keeps the faces and minus one when it turns the sheet over.

The earlier essay ended by guessing what rolling does to those signs. Rolling puts one face inside and one outside, so every operation that turns the sheet over ought to die, and the guess was that only the side-keeping half of the group survives, so that a two-sided sheet rolls like a plane pattern and adds nothing. It also warned that the guess might miss operations that turn the sheet over and then turn it back in combination with something else, such as a half-turn about a line in the sheet composed with the mirror in the sheet’s own plane.

Both halves of that can be settled. The guess is right about the group. The combination it worried about keeps the faces, so it already belongs to the side-keeping half. What the guess misses is the count. A sheet with two faces does not make one tube. It makes two, one with each face outside, and the operations that turn the sheet over, which cannot be symmetries of either tube, turn out to be exactly the congruences between them.

The guess is right about the group #

Rolling is a covering map, as the tube has a screw no lattice allows set it up. Choose a lattice vector $C$ of the sheet, wrap the sheet round a cylinder whose circumference is $|C|$, and the plane becomes the cylinder’s universal cover, with the translations by multiples of $C$ as the deck group. An operation of the sheet descends to the tube exactly when it carries the deck group to itself, which for an operation with linear part $A$ in the plane means $AC = C$ or $AC = -C$. For a flat pattern that was the whole test.

A sheet with thickness adds one fact. A point at height $z$ above the sheet’s middle plane lands at distance $R + z$ from the axis, where $R$ is the radius of the middle plane. An isometry of space that carries a cylinder to itself keeps every point’s distance from the axis. An operation that turns the sheet over sends height $z$ to $-z$, so it would have to carry a point at radius $R + z$ to one at $R - z$, and no symmetry of the tube can do that. Every face-exchanging operation dies on the tube, whatever it does in the plane.

The face-keeping operations of a layer group form a subgroup, the kernel of the sign, and the sign is a homomorphism onto $\{\pm 1\}$. So the subgroup has index one, for a sheet with nothing turning it over, or two. Projected onto the middle plane it is a plane group, and the tube’s group is what that plane group rolls into, which the earlier census computed. The combination the earlier essay worried about is already in it. A half-turn about the $x$ axis lying in the sheet sends $(x, y, z)$ to $(x, -y, -z)$. The mirror in the sheet sends it to $(x, y, -z)$. Their product sends it to $(x, -y, z)$, a mirror perpendicular to the sheet that keeps both faces where they were. It is one of the face-keeping operations and adds nothing to the group.

A sheet makes two tubes #

The count is where the guess goes wrong.

Rolling a sheet along a given vector $C$ can be done in two ways, with face A bending outwards or with face B bending outwards. The picture shows both for a sheet whose faces carry different atoms. The two results, tube A and tube B, have the same group, since each tube’s group is the rolled side-keeping half of the same layer group. They need not be the same object. Whether either one repeats along its axis at all is a separate question, which most sheets roll into a tube that never repeats answered; the comparison here is between the two tubes, whatever their repeat.

For a Janus layer they plainly are not. MoSSe with sulfur outside has a sulfur surface and a selenium lining, and with selenium outside the reverse. The two tubes have different outer chemistry, different strain in each face, and different energies. Nothing about them is related by a motion of space. For a single layer of molybdenum disulfide, with sulfur on both faces and a mirror in the metal plane, the two tubes are one tube: exchanging the faces is a symmetry of the flat sheet that changes nothing, and rolling the sheet one way or the other makes the same object. For buckled silicene, with no mirror in its plane but a centre of inversion, the answer is less obvious, and it takes the rest of this essay to see that it is again one tube, related to itself in a different way.

Nothing in the tube’s group distinguishes these three cases. Its group is computed from the face-keeping operations, and the question of whether tube A and tube B coincide is decided by the face-exchanging ones. Those are the operations the guess threw away.

An operation that dies becomes a congruence #

Take a face-exchanging operation $g$ of the sheet, with in-plane linear part $A$ and sign $-1$ on the normal. It carries the pattern at height $z$ onto the pattern at height $-z$. Suppose $A$ carries $C$ to $\pm C$. Then $g$ normalises the deck group, so it descends to the cylinder as a map of angles and heights. Follow an atom. In tube A, the atom at height $z$ on the sheet sits at radius $R + z$. The operation $g$ says the sheet has an identical atom at height $-z$ at the image position. In tube B that image atom sits at radius $R - (-z) = R + z$. So the map of angles and heights that $g$ induces, with radius left alone, carries every atom of tube A to an atom of tube B. The operation that could not be a symmetry of either tube is an isometry from one to the other.

Whether that isometry is a proper motion or a reflection is read off $A$. On the cylinder an in-plane map with $AC = \varepsilon C$ acts on the angle round the tube by the sign $\varepsilon$ and on the height along it by the sign $\det A \cdot \varepsilon$. This is the same four-way split that sorted the survivors in seven friezes round a cylinder. It is proper when the two signs agree, which is when $\det A = +1$, and a reflection when they differ.

The table lists the four cases. A face-exchanging operation with $A = I$ is a mirror in the sheet’s plane, or a glide in that plane, and on the cylinder it becomes a screw or a translation. Tube B is tube A moved along and round, which means it is tube A. A face-exchanging operation with $A = -I$ is a centre of inversion, and on the cylinder it becomes a half-turn about a line crossing the axis. Tube B is tube A turned end over end, and again it is the same tube. That is silicene’s case. Its sheet has no mirror in its plane, so no rolling of it is literally the same as the other rolling. But its inversion centres say that the tube with the raised atoms outside is the tube with the lowered atoms outside, flipped end for end.

The other two cases have $\det A = -1$. The in-plane part of such an operation is a reflection of the plane, and a face-exchanging operation whose in-plane part is a reflection is a two-fold rotation about a line lying in the sheet, turning the sheet over as it turns. If that line runs along $C$, the tube map is a mirror across the axis. If it runs across $C$, the tube map is a mirror containing the axis. Either way tube B is the mirror image of tube A.

A mirror image of a tube is the same tube when the tube has a reflection of its own, and a different tube, its enantiomer, when the tube is chiral. So three verdicts are possible for one rolling direction. If some face-exchanging operation fixing the line of $C$ is proper, the two tubes are one. If only improper ones exist, the two tubes are mirror images, which is one tube if it is achiral and a pair of hands if it is chiral. If none exists, the two tubes are different. That last verdict is exact and not merely the absence of a proof. Any isometry carrying tube A onto tube B lifts through the covering to an isometry of the sheet that exchanges its faces and keeps the line of $C$, and those are exactly the operations the test has already examined.

Eighty layers in four kinds #

The test depends on a layer only through the in-plane parts of its face-exchanging operations, and on a direction only through which of those parts fix it. So it can be run over all eighty layers and every rolling direction. The census takes each layer from its plane group and its sign, lists every primitive lattice vector that keeps the faces in the box of coordinates up to six, 3,400 layer-directions in all, and decides each one.

The picture at the head of this essay shows the result. It sorts the layers into four kinds, and the kind can be read from the linear parts alone before any direction is tried. The directions confirm it in every one of the 3,400 cases.

Seventeen layers are one-sided. They have no face-exchanging operation at all, and each is a plane group with every sign plus. Every rolling of them gives two different tubes: 816 layer-directions, all different. MoSSe is one of them, with the layer group of $p3m1$ one-sided. That is the symmetry form of what Janus tubes are known for: the two ways of rolling one Janus sheet are two materials.

Forty-six layers give one tube in every direction. Each has a face-exchanging operation with $A = I$ or $A = -I$, and those fix every line of the plane, so every rolling vector passes the test. Fifteen have a mirror or glide in the sheet and no inversion centre, and their two tubes are related by a screw. Nine have an inversion centre and nothing in the sheet’s plane, silicene’s layer among them, and their tubes are related by a half-turn. Twenty-two have both. The seventeen mirrored layers, a plane group doubled by the mirror in the sheet, are all here, and so are the twenty layers turned over by a translation. Those twenty are the glide planes lying in the sheet itself, which eighty is seventeen, seventeen and forty-six found were the twenty colourings an older count had missed.

Sixteen layers relate their tubes only along special directions. Their face-exchanging operations are two-fold axes lying in the sheet, and nothing else. A two-fold axis in the sheet has an in-plane part that is a reflection, and a reflection fixes exactly two lines, its own and the one perpendicular to it. So these layers pass the test along the directions of their axes and their perpendiculars, and nowhere else. That is two, four or six directions, depending on how many axes the layer has.

One layer is none of these. It is taken up after the sixteen.

Along the axes, and only as a pair of hands #

The sixteen special layers can be drawn direction by direction.

The layer over $p3m1$ whose mirrors, lifted into space, turn the sheet over is the layer group with three two-fold axes lying in the sheet and three-fold axes normal to it. Along the six directions of its axes and their perpendiculars its two tubes are mirror images. Along all the other rolling directions drawn, they are different tubes. The census contains 48 such directions in the box for this layer, of which six are related and 42 are not. For a real sheet with this symmetry, almost every tube rolled from it can be rolled in two genuinely different ways, and the only exceptions come in handed pairs.

That last point holds for all sixteen. At every special direction of every special layer the tube is chiral, and the two tubes are enantiomers. That sits beside what a rolled sheet is never one of a pair found about the groups: rolling reaches eleven chiral rod groups but none of the sixteen that come in left- and right-handed pairs. Here the pair of hands is carried by the two tubes, not by their group. There is no special direction where the two rollings give one tube.

The census finds it in 48 of 48 special layer-directions. The argument is short and explains why. Suppose some special layer, along the line of some rolling vector $C$, gave an achiral tube. Then the tube has a reflection, and a reflection of the tube comes from a face-keeping operation of the sheet whose in-plane part $K$ is a reflection fixing the line of $C$. The special direction also has its two-fold axis, a face-exchanging operation whose in-plane part $S$ is a reflection fixing the same line. The two reflections each have their mirror line along $C$ or across it, so their product $SK$ is $I$ if the lines agree and $-I$ if they do not. And $SK$ belongs to a face-exchanging operation, since one factor exchanges the faces and the other does not. A face-exchanging operation with in-plane part $\pm I$ is a mirror or glide in the sheet or an inversion centre, which would put the layer in the forty-six, not the sixteen. So along a special direction of a special layer the tube is always chiral. The one sheet then rolls into both hands of it, and which hand depends only on which face goes out.

Faces exchanged, and never the same tube #

The one layer that fits none of the three kinds is the layer group $p\bar{4}$. It is the layer over the square group $p4$ in which the quarter-turns turn the sheet over and the half-turns do not. Its face-exchanging operations are four-fold rotoinversions about lines normal to the sheet: turn by a quarter and reflect through the sheet’s plane. The two faces of a $p\bar{4}$ sheet are equivalent. Every atom on face A has a partner on face B that the group carries it to, and a flat $p\bar{4}$ sheet looks the same from above as from below, turned by a quarter.

Rolled, the equivalence does nothing. The face-exchanging operations have in-plane part a quarter-turn, clockwise or anticlockwise, and a quarter-turn carries no vector of the plane to itself or its negative. So no face-exchanging operation fixes the line of any rolling vector, and the two tubes are different in every direction, all 48 in the box. The group guarantees that the face-A-out tube and the face-B-out tube are made of the same atoms in the same local arrangements. It does not make them the same tube, because the operation that matches one face to the other also turns the rolling direction through a right angle, and a tube rolled along $C$ is not a tube rolled along a vector perpendicular to it.

What the rotoinversion does give can be stated exactly. Rolling a $p\bar{4}$ sheet face B out along $C$ gives the same tube as rolling it face A out along the vector $C$ turned by a quarter. The two rollings of one direction are the one rolling of two directions. So the tubes of $p\bar{4}$ come in pairs as the tubes of a Janus sheet do, but they are paired across directions rather than within one.

$p\bar{4}$ is the only layer where this happens, and the reason it is alone is short. A layer’s face-exchanging operations form a coset of the face-keeping half. If one of them has in-plane part a reflection, the layer is special or better. If one has in-plane part $\pm I$, the layer is in the forty-six. What is left is a coset whose in-plane parts are all rotations other than the half-turn, and the half-turn must then keep the faces. Among plane point groups the only coset of that kind is the quarter-turns of the four-fold group, since a three-fold or six-fold rotation that exchanged the faces would have a cube that exchanged them too, and the cube of a three-fold rotation is the identity and of a six-fold rotation the half-turn. That leaves $p4$ with its quarter-turns reversing the normal, which is $p\bar{4}$.

What the verdicts do and do not say #

Two tubes are compared as idealised coverings. The construction treats rolling as wrapping a sheet of fixed shape round a cylinder, with each atom keeping its height above the middle plane as a radial offset. A real tube relaxes. Its inner face is compressed and its outer face stretched, and the two faces of a Janus sheet respond differently, which is why a Janus sheet is expected to prefer one direction of curvature. None of that changes a verdict. A relaxation that respects the tube’s group cannot make two tubes congruent that were not, or separate two that were, because the congruence, where it exists, is an operation of the sheet and carries any symmetric relaxation of one tube to a symmetric relaxation of the other.

The census is finite and the kinds are not. The 3,400 layer-directions confirm a classification that was read from the linear parts before any direction was tried. The kind of a layer depends only on which in-plane parts its face-exchanging operations have, and the argument above settles every direction of every length. The box is a check that nothing was misread, not the source of the answer.

Congruence is not energy. For a one-sided sheet the two tubes are different objects, but the census says nothing about which one forms. That is a question of bending stiffness, of the strain each face can take, and of how the sheet was made, and symmetry has no opinion on any of them. The same holds for the sixteen special layers. Symmetry says that along an axis of a special layer the two rollings make the two hands of one tube, which means they have equal energies and a synthesis with no reason to prefer a face gives both. It says nothing about the directions where the tubes differ.

The seam is refused rather than examined. Twenty layers have a translation that turns the sheet over, a glide in the sheet’s plane. Rolling along such a translation would glue face A to face B at the seam. A point going once round the tube would come back on the other face, which no tube with an inside can do. The census rolls only along vectors that keep the faces.

Who worked the pieces out #

The layer groups were first tabulated in full by Elizabeth Wood in 1964, and the eighty now appear with their subgroup relations in volume E of the International Tables, edited by Vojtěch Kopský and Daniel Litvin. The line groups of tubes, and the practice of reading a nanotube’s symmetry off the sheet it was rolled from, are due largely to Milan Damnjanović and his collaborators from the late 1990s. Janus layers of MoSSe were first made in 2017, and tubes rolled from them have been studied theoretically since. The reading of a face-exchanging operation as a congruence between two tubes follows from the covering construction already used for flat sheets, together with the fact that an isometry preserving a cylinder preserves distance from its axis.

What the census must refuse #

The second test is the one that carries weight. The four kinds are defined from the in-plane parts of the face-exchanging operations, and the verdicts are computed separately, direction by direction, by applying each operation to each rolling vector. If a layer’s operations had been listed wrongly, with a missing glide or a sign attached to the wrong coset, the two readings would disagree, and they agree for all eighty. The fourth test is a refusal. The census must decline to roll a sheet along a vector that turns it over, and it does, on the first layer with a glide in its own plane.

The last test states the finding that motivated the census. It asks for a layer that exchanges its faces and still gives two different tubes in every direction, and finds exactly one. Cutting a sheet can give it a hand found the same kind of loss at an edge, where a cut keeps only the operations that fix one direction. Rolling keeps only the operations that fix one line, and in both cases an operation that keeps nothing of the direction concerned is lost, however much of the rest of the sheet it preserves.

Still open: a sheet rolled into a scroll #

A tube closes on itself. A sheet can also be rolled into a scroll, a spiral of several turns with the edge left free, and carbon and dichalcogenide scrolls are both known. In a scroll each turn lies on the previous one, so face A of one turn faces face B of the turn beneath. A scroll has no closed circumference and no deck group, and whatever symmetry it keeps is a group of motions of a spiral rather than of a cylinder. For a one-sided sheet a scroll wound face A out and one wound face B out are two different objects, as tubes were. For a mirrored sheet they are probably one, and for the sixteen special layers perhaps a pair of hands again. But the operation that relates them would have to preserve a spiral of fixed pitch, and whether the special directions survive that requirement, or whether the extra constraint of the next turn removes them, has not been worked out here.