A rolled sheet turns by a square root

A rolled sheet is never one of a pair found that no plane pattern rolled into a crystallographic tube lands on a handed screw. The seventy-five rod groups include eight pairs of screws that are mirror images of each other — 3₁ and 3₂, 4₁ and 4₃, 6₁ and 6₅, 6₂ and 6₄ — and the census of every rolling of the seventeen plane groups reached none of them. The reason was a count of cells. A tube with six turns or fewer in a repeat needs a primitive index of six or fewer, and on the lattices that repeat at all the only small values are one and two, which give climbs of nothing and of half a repeat. Neither of those is changed by a mirror.

That essay closed on the case it had set aside. Past six turns the primitive index is free, climbs of a fifth and a thirteenth appear, and handed tubes come with them. The obvious guess is that the restriction simply lifts: once the order is free, the screw is free too, and a tube of thirteen cells can turn by any of the twelve fractions of a circle a thirteen-fold screw can have. It cannot. A square sheet rolled into a thirteen-cell tube turns by five thirteenths or by eight thirteenths and never by anything else. Five squared plus one is twenty-six, eight squared plus one is sixty-five, and both are multiples of thirteen. No other number below thirteen has that property.

This essay shows that the coincidence is a theorem. The screw of a rolled tube is a square root of minus the sheet’s determinant, modulo the number of cells in its repeat. From that one congruence follow the orders at which handed tubes exist, the fact that they always come in both hands, the number of unrelated pairs at each order, and a way round the crystallographic gap that the previous census could not see.

One repeat, cut open #

Roll a sheet along a lattice vector $C$ with no common factor in its coordinates. If some lattice vector $T$ is perpendicular to $C$, the tube repeats after climbing $|T|$. Cut one repeat along its length and lay it flat, and it is the rectangle on $C$ and $T$, which holds some whole number $n_0$ of the sheet’s cells: the primitive index. Every lattice point of the sheet lands somewhere in that rectangle, at a turn round the tube and a climb up it.

The lattice points sit at exactly $n_0$ heights, equally spaced by $1/n_0$ of the repeat. The lowest point off the floor is the screw that climbs least, and every other point is one of its powers, since the tube’s symmetry is generated by it. So a rolled tube’s whole helix is described by one number: the turn $s/n_0$ of a circle that the least-climbing screw makes. The $(2, 1)$ tube has $s = 3$ and the $(3, 2)$ tube has $s = 5$. The tube built operation by operation in most sheets roll into a tube that never repeats records the same fact from the other end, as the climb $p/N$ of the screw that turns least. The two descriptions are inverse to each other: $s \cdot p \equiv 1$ modulo the number of turns in the tube, with the same sign on every one of 448 primitive rollings of the square and hexagonal sheets on which the two were compared.

Only the heights are forced by the count of cells. How far round the tube the point at each height sits is not obviously constrained at all, and on a picture of thirteen equally spaced heights any of twelve turns looks as plausible as any other.

Why the turn is a root of minus one #

The constraint is an identity about two vectors that is older than crystallography. For any two vectors in the plane,

$$(u \cdot w)^2 + (u \times w)^2 = |u|^2\, |w|^2 ,$$

the dot product squared plus the cross product squared is the product of the squared lengths. It is the Pythagorean theorem for the parallelogram the two vectors span, and it is exact for integer vectors.

Apply it to the rolling vector $C$ and a lattice vector $v$ that completes $C$ to a basis of the sheet, so that the parallelogram on $C$ and $v$ is exactly one cell. On a square sheet of unit spacing that cell has area one, so $v \times C = \pm 1$. The vector $v$ is the screw that climbs least: its height up the tube is its component along $T$, which is one cell’s area divided by the rectangle’s, $1/n_0$ of a repeat. Its turn round the tube is its component along $C$ divided by the circumference, which is $(v \cdot C)/|C|^2$ of a circle — and $|C|^2 = n_0$ on the square sheet, because the rectangle on $C$ and $T$ is a square of side $|C|$. So $s = v \cdot C$, and the identity reads

$$s^2 + 1 = n_0\, |v|^2 \qquad\Longrightarrow\qquad s^2 \equiv -1 \pmod{n_0}.$$

The turn of a square tube’s least-climbing screw is a square root of minus one modulo the number of cells in its repeat. The five-cell tube turns by three fifths, a root of $-1$ modulo five, and so is two. The thirteen-cell tube turns by five thirteenths, a root, and so is eight. Every other residue fails.

The same argument works for any sheet whose Gram matrix $Q$ has whole-number entries, and the cross product is what changes. The squared area of the cell is $\det Q$, so the identity becomes $(v^{\mathsf T} Q C)^2 - (C^{\mathsf T} Q C)(v^{\mathsf T} Q v) = -\det Q$. When the vector $QC$ has no common factor, the turn satisfies $s^2 \equiv -\det Q$ modulo $n_0$. The square sheet has determinant one. The hexagonal sheet, written with the Gram matrix whose entries are two and minus one, has determinant three, and a rectangle whose squared axial ratio is a whole number $\rho$ has determinant $\rho$.

The picture at the head of this essay is the congruence drawn. Its columns are the possible primitive indices, and a column is occupied only where minus one, or minus three, has a square root. Where it is occupied it holds two turns or four, never the dozen a column of that height has room for. The census found every turn by rolling along every primitive vector out to sixteen cells in each coordinate, independently of the identity, and then compared: at every index, the turns rolling produced and the roots found by trying every residue are the same list.

The table is the identity checked from both sides. The column found by rolling cannot contain a residue that is not a root, since the identity forbids it. The column of roots could, in principle, contain one that no rolling produces, and it never does. That half is the converse: for every root $s$ of $-1$ modulo $n_0$ there is a primitive vector of length squared $n_0$ whose tube turns by exactly $s$. It is the square lattice’s version of a classical fact about sums of two squares, and it is where the argument meets the sublattices that stay square, which found that a square sublattice exists at an index exactly when that index is a sum of two squares. A square sublattice of index $n_0$ is spanned by $C$ and $T$, and a sum of two squares is the length squared of $C$. The rectangle that one repeat of the tube cuts out is the same square the sublattice essay counted.

Fermat decides which orders have a hand #

A handed screw is one whose turn is not its own negative, so neither nought nor a half. On the square sheet that needs a root of $-1$ other than those two, and the table says where they exist. Minus one has a square root modulo an odd prime exactly when the prime is one more than a multiple of four, which is Fermat’s condition for the prime to be a sum of two squares. It has none modulo four, so a factor of four in the index is fatal, and modulo two it has the single root one, which is a half-turn. So the square sheet makes a handed tube only when the number of cells in its repeat is a product of primes of the form $4k + 1$, possibly doubled: five, ten, thirteen, seventeen, twenty-five, twenty-six, twenty-nine.

A tube’s number of turns in a repeat is the primitive index times the common factor of the rolling vector, $N = g\,n_0$, and a vector with a common factor makes a tube whose screw is $g$ times the primitive one. So the orders with a handed tube are the multiples of the handed indices, and the figure lists them. Seven, nine, eleven and twelve never appear, from either sheet. A square or hexagonal sheet can be rolled into a tube of seven turns — along seven cell edges, for instance — but the tube’s screws are then plain turns with no climb at all, because seven is prime and not one more than a multiple of four, and the only way to get seven turns is $g = 7$, $n_0 = 1$.

The hexagonal sheet runs on the same arithmetic with $-3$ in place of $-1$, and with one complication: the vector $QC$ has a common factor of three whenever three divides the rolling vector’s squared length. The two kinds of direction obey two congruences.

Every hexagonal index is even, which most sheets roll into a tube that never repeats proved from the parity of the lattice’s cell and checked on six and a half thousand directions. Half the index is then a product of primes of the form $6k + 1$, the primes that are Loeschian — seven, thirteen, nineteen, thirty-one — and fourteen is the first index with a handed turn. A hexagonal sheet cannot make a handed tube with fewer than fourteen turns in its repeat, where a square one manages five.

Two hands, always #

A congruence $s^2 \equiv -\det Q$ has its roots in pairs: if $s$ is one, so is $n_0 - s$. The two are the turns of a tube and of its mirror image, since a reflection in a plane containing the tube’s axis reverses every turn and keeps every climb. So wherever a sheet makes a handed tube, the congruence at least permits the other hand. The census confirms that the sheet actually makes it.

The reason is short. Reflect the sheet in a line along the rolling vector. The result is a pattern with the same lattice and a plane group of the same type, since the mirror image of a four-fold pattern is a four-fold pattern with its turns the other way. Rolling it along the same vector gives the mirror image of the original tube. For the square and hexagonal lattices the reflected pattern is one of the patterns the census already rolls, and the reflected vector is a lattice vector of the original sheet. So each handed tube’s partner is another rolling of the same sheet, along the mirror-image direction.

Across the seventeen plane groups, eight make handed tubes at all, and all eight make every handed tube they reach in both hands. The census found no unpaired screw. The other nine make none, because their lattices repeat only along cell edges and diagonals, where the primitive index is one or two.

The eight divide in a way the reflections decide. The square groups p4, p4m and p4g and the hexagonal groups p6 and p6m all contain a half-turn of the sheet, which survives rolling in every direction as a half-turn across the tube, so their handed tubes have a principal axis with two-folds crossing it. The three groups p3, p3m1 and p31m have no half-turn, and their handed tubes are bare screws. The mirrors of p4m, p6m and the two trigonal groups with mirrors survive only along the few directions that lie in them or across them, and those are exactly the directions with a primitive index of one or two. Every handed tube is made by rolling in a direction that meets no mirror, so a mirror in the sheet is no obstacle to a handed tube. It only makes the mirrored direction the tube’s partner rather than a separate pattern’s.

The contrast with the crystallographic census is complete. There, no handed pair is reached at all. Past six turns, every handed screw that is reached is reached in both hands. The crystallographic restriction did not bias rolling towards one hand. It removed the handed screws entirely, and when they come back they come back in pairs.

Sixty-five turns and two unrelated pairs #

The number of roots of $-1$ modulo $n_0$ doubles with each distinct prime factor of the form $4k + 1$, so an index with $\omega$ such primes has $2^{\omega}$ roots and $2^{\omega - 1}$ handed pairs. Thirteen has one pair. Sixty-five, which is five times thirteen, has two: the turns $8$, $18$, $47$ and $57$ sixty-fifths.

The two pairs have nothing to do with each other geometrically. Sixty-five is a sum of two squares in two genuinely different ways, $64 + 1$ and $49 + 16$. The vector $(8, 1)$ rolls into a tube turning by fifty-seven sixty-fifths, and its mirror image by eight. The vector $(7, 4)$ rolls into one turning by forty-seven, and its mirror image by eighteen. Both tubes have sixty-five turns in a repeat, the same circumference and the same repeat length, and screws that are not related by any motion of space or by any reflection. The number of unrelated handed tubes of one circumference is the number of ways its squared length is a sum of two squares, which the congruence counts as roots and Fermat counted as representations.

It is the first place where rolling produces two chiral tubes of exactly the same size that are not mirror images of each other. A thirteen-cell tube has one partner and no alternative. A sixty-five-cell tube has one partner and one alternative, with its own partner. The figure of orders counts four pairs at sixty-five, because two more come from the five-cell and thirteen-cell tubes wound round thirteen and five times, whose screws are the primitive ones multiplied by the winding.

The gap was the determinant #

The crystallographic census can now be read from the other end. A handed screw of order three, four or six needs a handed root modulo three, four or six. Minus one has no square root modulo three or four, and modulo six the only candidates fail too. Minus three is no better. The generic sheets have determinants one and three, fixed by their symmetry, and those two numbers have no roots where the crystallographic handed screws would need them. The gap the rod-group census found is not a property of rolling. It is a property of the two determinants that a four-fold and a three-fold lattice are forced to have.

A sheet with a different determinant escapes it. A rectangular lattice’s squared axial ratio $\rho$ is free, which is why its rollings do not generally repeat. If the ratio happens to be the square root of a whole number, the Gram matrix has whole entries and determinant $\rho$, and every direction repeats.

A rectangle whose sides are in the ratio one to the square root of two, rolled along its diagonal, has three cells in a repeat and turns by a third: minus two is one modulo three, which is a square. The tube is the rod group p3₁12, and the other diagonal gives p3₂12. The same sheet rolled along twice the diagonal gives the six-fold screws 6₂ and 6₄. A ratio of root three gives a four-cell repeat and 4₁ and 4₃, and root five a six-cell repeat and 6₁ and 6₅. Every one of the eight handed pairs is reached, each named against the seventy-five by the same canonical form the earlier census used, and each in both hands.

The convention this depends on should be said plainly. The census of the seventeen rolled treated the free parameters of a lattice as independent of every rational number, which is what a crystal class means by an oblique or rectangular lattice, and under that convention the gap is exact. A real crystal’s axial ratio is a measured number, and no measured number is exactly the square root of two. So the accident is a construction and not a prediction about any material. What it shows is where the gap comes from: from a specific pair of determinants and not from any general law about turning a sheet into a tube.

Two lattices share the roots between them #

The congruence names a turn from a determinant, and a determinant does not name a lattice. How many lattices share a determinant counted the inequivalent lattices of each determinant, and at five there are two: the rectangle $x^2 + 5y^2$ and the oblique form $2x^2 + 2xy + 3y^2$, which no change of basis turns into the rectangle. Both give tubes whose turns are roots of $-5$. The question is how the roots are divided between them.

They are divided exactly. Every root of $-5$ at every index up to fifty coprime to five is the turn of a tube rolled from exactly one of the two sheets, with none left over and none shared. At a prime index the roots go together: twenty-nine and forty-one, which leave one or nine on division by twenty, belong to the rectangle; three, seven, twenty-three, forty-three and forty-seven, which leave three or seven, to the oblique sheet. At twenty-one, three times seven, both factors belong to the oblique sheet and the product belongs to the rectangle. That multiplication rule, in which two classes that are not the identity compose to the identity, is Gauss’s composition of forms. The genus that lattices that agree at every prime sets beside the class number is what the remainder modulo twenty detects here.

So a pair of rolled sheets of the same area can be told apart by the set of screws their tubes make, and the two sets are complementary. The square sheet never faces this, because determinant one has a single class of form and it collects every root there is. That is why the square sheet’s table above has no gaps.

What the census has to refuse #

Every statement above is a comparison between two computations that do not know about each other: rolling along vectors, which knows nothing about congruences, and trying residues, which knows nothing about vectors.

The first refusal is the guess this essay began with. With the order free, a thirteen-cell tube has twelve handed screws available on paper, and the claim that rolling can make any of them is refused by the square sheet making exactly two. The second is a single wrong number offered as a turn: a $(3, 2)$ tube turning by four thirteenths of a circle. Four squared plus one is seventeen, which thirteen does not divide, and the tube in fact turns by five. A check that accepted any residue would have accepted it.

The identity is also checked where it could go wrong silently. For 1,840 rollings over five forms, the completed basis must have a cross product of exactly one and the Lagrange expression must come out at exactly minus the determinant, in integers. A rolling vector with a common factor, or a completion that is not a basis, would break it at once, and a bug in the completion step would otherwise surface only as a turn slightly wrong in a few directions.

Still open: the group, not only the screw #

The congruence names a tube’s screw, and a rod group is more than its screw. Whether the tube has half-turns across it, mirrors along it or a mirror across it is decided by which reflections and half-turns of the sheet survive the rolling, and that is the business of the seven families with one axis rather than of any congruence. Past six turns those families do not stop at the thirty-two crystal classes. A tube of sixty-five turns from p4 has two-folds crossing a sixty-five-fold screw, and the groups of a line with unbounded orders fall into a small number of infinite families that the crystallographic classification does not contain. Which of those families each plane group reaches, at each order and each root, is the census the seventy-five did at the crystallographic orders, run with no bound on the order and a name for every family. It has not been run here.

The other direction is physical. The argument treats a sheet as a lattice that rolls without distortion. A real sheet strains as it rolls, differently on its inner and outer faces, and a strained sheet’s metric is no longer the one its symmetry fixes. Rolling a square sheet into a tube of a few dozen cells is therefore rolling a sheet whose determinant is not quite one. Whether the screw of such a tube is still the root the ideal congruence predicts depends on how the strain moves the lattice points, and it has not been computed. The screw of a real tube is measured from its diffraction, as what a thread scatters describes, and the comparison would test whether the integer arithmetic survives in real tubes.