A stretch buys an eighth and a twelfth of a turn

The crystallographic restriction is a statement about turning a lattice onto itself. A rotation that carries every lattice point to a lattice point is an integer matrix in a basis of the lattice, its trace is an integer, and the trace of a rotation by θ in the plane is $2\cos\theta$. So $2\cos\theta$ is one of $-2, -1, 0, 1, 2$, and the turn is a half, a third, a quarter, a sixth or nothing. Every finite list in the subject descends from that line, and a second proof by a shortest vector reaches the same five without a trace in sight.

The same argument has a quieter relative that crystallographers use every day without calling it a restriction. Surface scientists write the superstructure of an adsorbed layer as $(\sqrt{2} \times \sqrt{2})\,R45°$ or $(\sqrt{3} \times \sqrt{3})\,R30°$: a cell that is the substrate’s own cell turned by an angle and stretched by a factor. The sublattices that are the same shape worked out which stretches the square and hexagonal lattices allow, and found Fermat underneath. What it did not ask is the question the restriction asks. A turn combined with a stretch is not a symmetry of the lattice. It carries the lattice onto a part of itself. Which fractions of a full turn can such a map make?

The answer is short and comes in three parts. The plane gains exactly two orders, the eighth and the twelfth, and no others, at the price of a stretch of at least $\sqrt2$ and $\sqrt3$. Every lattice whose metric is rational gains a quarter turn, whatever its shape. And space gives the eighth and the twelfth back: no map of a lattice in three dimensions that turns and stretches it can turn by either.

What is being turned, and what counts #

A similarity of a lattice $L$ is a linear map $\sigma$ that sends every lattice point to a lattice point and that is a scalar $r$ times an orthogonal map $R$. The scalar is the stretch and $R$ is the turn. In a basis of the lattice, $\sigma$ is a matrix of integers, because it sends basis vectors to lattice vectors. The turn has finite order $m$ exactly when some power of $\sigma$ is a multiple of the identity, $\sigma^m = r^m I$, and that is a statement that can be checked in integers without knowing $r$ or $R$ at all.

The converse holds too, and it is what lets a census of matrices stand in for a census of lattices. Take any integer matrix whose $m$-th power is a positive multiple $s$ of the identity. Divide it by $s^{1/m}$ and the result has finite order, so it generates a finite group, and averaging any inner product over that group gives an inner product the divided matrix preserves. In that inner product, $\sigma$ is a stretch times an orthogonal map. So every integer matrix with a positive scalar power is a similarity of some lattice, and the question about turns becomes a question about integer matrices, as the restriction itself was.

Two conventions are fixed here because the counts depend on them. The order of a similarity is the order of its turn $R$, not of $\sigma$, which has infinite order whenever the stretch exceeds one. And $R$ may be a reflection. In the plane a reflection is always of order two and adds nothing, so the interesting orders are those of rotations, and the tables below list them.

Two integers are enough #

Write $t$ for the trace of $\sigma$ and $d$ for its determinant, both integers. If the turn is a rotation by θ and the stretch is $r$, the two eigenvalues of $\sigma$ are $r e^{\pm i\theta}$, so

$$d = r^2, \qquad t = 2r\cos\theta, \qquad \cos^2\theta = \frac{t^2}{4d}.$$

The squared cosine is a ratio of integers. Doubling the angle gives $\cos 2\theta = 2\cos^2\theta - 1$, which is rational too. But $2\theta$ is a rational fraction of a turn whenever θ is, and the only rational fractions of a turn with a rational cosine are those with cosine $0$, $\pm\frac12$ or $\pm1$, because $2\cos$ of a rational angle is an algebraic integer and a rational algebraic integer is a whole number. So $2\theta$ is a multiple of a quarter or of a sixth of a turn, and θ is a multiple of an eighth or of a twelfth. The orders that survive are

$$1, \; 2, \; 3, \; 4, \; 6, \; 8, \; 12.$$

The argument is the restriction’s own argument, run one step later. The restriction asks for $2\cos\theta$ to be an integer. A similarity only manages to make $\cos^2\theta$ rational, because the stretch enters the trace once and the determinant twice, and squaring the cosine is exactly what cancels it. Halving the angle at the end is what lets two new orders in.

The search behind the table is a check on the argument rather than the argument. Every integer $2 \times 2$ matrix with entries from $-4$ to $4$ was raised to successive powers, in exact integers, until one came out a positive multiple of the identity or the forty-eighth power was reached. Of the 6,561, 904 have such a power. Their turns have orders $1$, $2$, $3$, $4$, $6$, $8$ and $12$ and nothing else. The smallest eighth is the matrix with rows $(-1, -1)$ and $(1, -1)$, determinant two, whose square is $2$ times a quarter turn. The smallest twelfth has determinant three. A stretch below $\sqrt2$ buys nothing new, because a determinant of one is a rotation and the restriction already applies to it.

The angles in a superstructure label #

The angle in a label like $(\sqrt{5} \times \sqrt{5})\,R26.6°$ is the argument of a Gaussian integer, here $2 + i$, and the stretch is its modulus. The essay on similar sublattices derived the stretches. Their angles are the other half of the same data. For the square lattice they are the arguments of the Gaussian integers, and for the hexagonal lattice those of the Eisenstein integers $a + b\omega$, with ω a primitive cube root of unity.

The figure takes every similar sublattice up to index 400, 313 of the square lattice and 242 of the hexagonal one once the lattice’s own rotations are folded out, and places each by its angle. The angles spread over the whole range. Index five gives 26.57° and its mirror 63.43°, index thirteen gives 33.69° and 56.31°, and the hexagonal lattice’s index seven gives the familiar $R19.1°$. None of them is a fraction of a turn. The only ones that are lie on a single vertical line in each panel: 45° for the square lattice, at indices two, eight, eighteen and every other twice a square, and 30° for the hexagonal lattice, at three times a square.

That is the restriction for similarities read off a label. A superstructure cell turned by a fraction of a turn is turned by 45° on a square substrate or by 30° on a hexagonal one. Any other angle that appears in a label is a number like $\arctan\frac12$, and no power of that turn ever returns to the start. Since $\arctan\frac12$ is not a rational multiple of a right angle, the $\sqrt5$ cell never lines up with the substrate however often it is stacked. The cells at 45° and 30° do, after eight and twelve steps. The same Gaussian integers fix the angles at which two copies of a lattice share a sublattice, the coincidence site lattice of a grain boundary, and those angles are fractions of a turn just as rarely. And the same $\sqrt2$ is the irrational number behind the eight-fold tilings, which keep the eighth of a turn as an exact symmetry and give up the lattice to do it, where a similarity keeps the lattice and gives up being a symmetry.

Which lattices can do it #

The proof says which orders are possible and not which lattices have them. The answer to the second question is cleaner than the restriction’s and stranger.

A similarity that is not just a multiple of the identity makes the lattice a module over a ring of complex numbers. Write the lattice points as complex numbers. Then σ acts as multiplication by $\lambda = re^{i\theta}$, an algebraic integer of degree two, and the lattice is closed under multiplication by it. The field λ generates is an imaginary quadratic field $\mathbb{Q}(\sqrt{-D})$, and the turn is visible in it: λ divided by its complex conjugate is $e^{2i\theta}$, a number of modulus one in that field. For the turn to have finite order, $e^{2i\theta}$ must be a root of unity, and the roots of unity in an imaginary quadratic field are ±1 in every field but two. $\mathbb{Q}(i)$ also has $\pm i$, and $\mathbb{Q}(\sqrt{-3})$ has the sixth roots of unity.

So the double angle $2\theta$ is a half turn or nothing in general, a quarter turn in the Gaussian field, a sixth in the Eisenstein field, and the orders follow at once. A lattice that is a module over some imaginary quadratic ring has turns of order $1$, $2$ and $4$. One whose field is $\mathbb{Q}(i)$ adds $8$. One whose field is $\mathbb{Q}(\sqrt{-3})$ adds $3$, $6$ and $12$. A lattice whose metric is not rational up to a scale is a module over no such ring, and has nothing but $\pm r$.

Two things in the figure are worth stopping on, because neither is what the restriction would lead anyone to expect.

Every rational lattice turns by a quarter. The oblique lattice with Gram matrix rows $(2, 1)$ and $(1, 3)$ has no symmetry beyond $\pm1$, and it nevertheless has a similarity that turns it by exactly a right angle, with stretch $\sqrt5$, five being its determinant. The reason is a known theorem: every plane lattice is its own dual, turned through a right angle and scaled. For a lattice with an integral metric the dual contains the lattice, and the lattice contains the dual multiplied by the determinant. Composing gives a quarter turn that lands inside, with stretch the square root of the determinant. The 1 : $\sqrt2$ rectangle does it with stretch $\sqrt2$, and so does every integral binary form, at no cost but its own determinant.

The eighth does not need a square lattice, only a square lattice inside. The rectangle with sides in the ratio 1 : 2 is not square and has no four-fold symmetry. It contains the square lattice of twice its short side, with index two, and that is enough: it turns by 45° with stretch $2\sqrt2$, landing on a sublattice of index eight. The rectangle with sides 1 : $\sqrt3$ is the orthohexagonal cell, the C-centred rectangular setting crystallographers use to describe hexagonal crystals with orthogonal axes, and it contains a hexagonal lattice of index two. It turns by a twelfth with stretch $2\sqrt3$. Whether a lattice can make the extra turns is a question about the lattices it contains, which is a question about its field, and not about its own point group.

Space gives both back #

In three dimensions rotations keep the plane’s five orders, and the analogous search for similarities finds nothing new at all. Every monic integer cubic with coefficients from $-9$ to $9$, taken as a companion matrix, and every block sum of a $1 \times 1$ and a $2 \times 2$ block with the same stretch was tested. The orders of turn that occur are $1$, $2$, $3$, $4$ and $6$, exactly the restriction’s. The eighth and the twelfth are gone. The reason takes three lines, and the axis is what does it.

A turn in space keeps a line, so σ has a real eigenvalue, $\pm r$. Either the characteristic cubic of σ has that root among the rationals, or it does not.

If it does, $r$ is a whole number. The line the turn keeps is stretched by an integer, and the plane across it is turned with stretch $r$ as well. On that plane σ has a quadratic factor $x^2 - 2r\cos\theta\, x + r^2$ with integer coefficients, so $\cos\theta$ itself is rational, not just its square, and the restriction applies in full: the orders are $1$, $2$, $3$, $4$ and $6$. The eighth would need $r^2 = 2$ on the plane and the twelfth $r^2 = 3$, and neither is the square of the whole number the axis demands.

If it does not, the cubic is irreducible, and its three roots have the same modulus $r$. A short argument with the Galois group then forces the roots to differ by cube roots of unity. If ρ is the real root and λ a complex one, then $\lambda\bar\lambda = \rho^2$. Some automorphism carries ρ to λ, and applying it to that equation gives $\rho\bar\lambda = \lambda^2$, so $\lambda^3 = \rho\lambda\bar\lambda = \rho^3$. The turn is a third, or a sixth if the sign of the real root is reversed.

That second case is real and not a curiosity of the algebra. The companion matrix of $x^3 - 2$ is a similarity of some lattice in space with stretch $\sqrt[3]{2}$, an irrational number, and the search found fourteen irreducible cubic blocks of this kind, all turning by a third or a sixth. Their lattices are not rational: the metric is scaled by $2^{2/3}$ under σ, so no multiple of it has rational entries. The shapes a lattice in space can thin to found the index of a similar sublattice of a cubic lattice to be a perfect cube. That essay was about lattices with a rational metric, for which the stretch is a whole number. A lattice that is not rational can thin to a similar sublattice of index two, and only by turning a third or a sixth of the way round.

The plane and space therefore differ for a reason that has nothing to do with room. The plane has one more order than its rotations allow because a stretch can hide in a square root. Space cannot hide it, because any turn there keeps a line, and the line has to be stretched by the same factor as everything else. A square substrate supports a $(\sqrt2 \times \sqrt2)R45°$ surface cell, but the cubic crystal underneath has no bulk cell that is its own cell turned by 45° and stretched. The 45° turn of the surface cell is a fact about the surface alone.

The counts, and what they rest on #

Nothing in the essay’s claims rests on a search. The plane’s seven orders, the field criterion for which lattice has which, and the three cases in space are proved above, each in a few lines. The searches are there to catch a slip in the proofs, and they would have caught one: a missing order or an extra one would have shown up as a matrix.

They were made in exact integer arithmetic throughout. A matrix counts as a similarity when a power of it is a positive multiple of the identity, compared entry by entry in arbitrary-precision integers, so a power that comes close is not mistaken for one that arrives. The plane search takes every integer matrix with entries from $-4$ to $4$ and powers to the forty-eighth. The lattice table tests every matrix with entries to $\pm6$ against each lattice’s own metric, through the identity $\sigma^{\mathsf T} G \sigma = |\det\sigma|\, G$. The space search takes every cubic with coefficients to $\pm9$ and every block sum with a common stretch, and the Wood angles come from every Gaussian and Eisenstein integer of norm up to 400.

Each test in the figure could have failed. The first would fail on an integer matrix in the box whose turn had order five or ten, which the trace argument forbids and the search would have found if the argument were wrong. The third would fail on a lattice that turns by an eighth without holding a square lattice, or that holds one and cannot. The orthohexagonal cell and the 1 : 2 rectangle were chosen for that test because neither has the symmetry its turn seems to require. The last would fail on an irrational stretch with a turn of order four, which the Galois argument rules out and which nothing but that argument would have made unlikely.

What the pictures cannot show #

Every figure here shows a lattice, and a similarity is not a symmetry of a crystal. A crystal’s pattern, laid over its own image turned by 45° and stretched by $\sqrt2$, does not coincide with it. The similarity maps the lattice onto a sublattice and says nothing about where the atoms go. What it supplies is the possibility of a superstructure cell of that shape, which an ordering of atoms on the substrate may or may not adopt. Whether a real adsorbed layer chooses $(\sqrt2 \times \sqrt2)R45°$ is chemistry, and the restriction only says that if a layer’s cell is turned by a fraction of a turn, it is turned by 45° or 30°.

Nor do the pictures show the absences that carry most of the argument. That no plane lattice turns by a fifth with any stretch is a claim about all integer matrices, proved by the trace and checked in a box. A drawing of a lattice turned by 72° and stretched shows points that miss each other, and it cannot show that no stretch would make them meet. The Wood figure draws angles that occur and cannot draw the angles that never do.

Where the pieces come from #

The notation for superstructure cells is E. A. Wood’s, from 1964, and the angles in it have been computed from Gaussian and Eisenstein integers for as long as the notation has existed. Similar sublattices were classified systematically by Conway, Rains and Sloane in 1999, and by Baake and Moody for the modules of quasicrystals. That a plane lattice with a non-trivial similarity is a module over an imaginary quadratic order, and that its turns are governed by the roots of unity in that field, is the classical theory of complex multiplication. What is assembled here is the restriction read for similarities: seven orders in the plane, one per lattice field, and five in space.

Still open: what four dimensions allows #

Three dimensions loses the two orders the plane gained, and the reason was specific: a turn in space keeps a line. In four dimensions a turn can keep nothing, rotating two planes at once, and a lattice there can be stretched without any axis to object. The space argument has nothing to say about it, and the plane argument generalises only partly. The determinant of a similarity is $\pm r^n$, so $r^n$ is a whole number, and the $n$-th power of the turn is a rational matrix of finite order, which must therefore be one of the orders an integer matrix can have in $n$ dimensions. That bound is a ceiling and nothing more. Which orders actually occur in four dimensions, whether the stretch buys more there than in the plane, and whether the ceiling is reached, is worked out in four dimensions adds a sixteenth and a twenty-fourth. The question was already in view when five-fold rotation became legal in four dimensions without any stretch.