Aperiodic is two words in space

One tile, and no period is about the hat, a single shape that tiles the plane and never does so periodically. That essay, and every essay here that follows the hat further, read “periodically” one way: a tiling is periodic when some translation carries it onto itself. There is a second reading in the literature, and it is not a pedantic variant. On it, a tiling is periodic in the relevant sense when it has infinitely many symmetries, of whatever kind.

For a tile of the plane the two readings ask the same question, and that is a theorem with a two-line proof, not a convention. For a tile of space they ask different questions. The difference matters to the history of the problem. A single shape that tiles space with no translation in any of its tilings was found in 1988, thirty-five years before the hat. It was never counted as the answer, and the reason is the gap between the two words.

What the two readings ask of a tiling #

Call a tiling translation-free when no translation other than doing nothing carries it onto itself, and finitely symmetric when its whole group of symmetries is finite. A symmetry here is any rigid motion carrying the tiling onto itself: a turn, a reflection, a glide or a screw as well as a translation.

The second condition implies the first everywhere. A translation that carries a tiling to itself can be repeated, and its powers are infinitely many different symmetries. So a finitely symmetric tiling is always translation-free. The question is the other direction: whether a tiling with no translation can still have infinitely many symmetries.

Both readings assume the symmetry group is discrete, and for a tiling by copies of one bounded tile it is. A motion close enough to doing nothing that still carries the tiling onto itself would have to carry each tile onto itself, since it moves no tile far enough to land on another. A bounded tile has only finitely many symmetries of its own, so only finitely many motions can come that close. The symmetries of a tiling therefore stay a definite distance apart, and every argument below uses that.

In the plane, anything infinite makes a translation #

Suppose a discrete group of motions of the plane contains no translation. Every motion of the plane is a translation, a turn about some point, a reflection in some line, or a glide along some line, and the other three kinds each produce a translation almost at once.

A glide applied twice is a translation, so the group has no glides. Two turns about different centres produce a translation too: turn about the first, turn about the second, undo the first, undo the second, and the result has no turn left in it but has moved the plane. The figure computes it for two quarter-turns a unit apart, and the translation is two units long. So every turn in the group has the same centre. Two reflections in parallel lines compose to a translation by twice the distance between the lines. Two reflections in lines crossing at a point compose to a turn about that point, which must be the one centre. So every mirror passes through the centre as well.

That leaves a group of turns about one point and mirrors through it. A discrete group of that kind is finite, since turns about one point by angles that never repeat would come arbitrarily close to doing nothing, and the finite ones are the cyclic and dihedral groups. So a translation-free tiling of the plane is finitely symmetric. The two readings are one condition in two dimensions, and a tile whose every tiling is translation-free, such as the hat, meets both.

The argument is the same one what forces a lattice uses to show that a discrete group of plane motions containing a translation has a finite point group, run in the other direction. There, a turn and a translation together manufacture translations as short as anybody likes. Here, any two operations that do not share a centre manufacture a translation where there was none.

In space, a screw is infinite and makes nothing #

Space has a kind of motion the plane lacks. A screw turns about an axis and slides along it at the same time, and every motion of space that keeps handedness is one. Repeat a screw and the turns add up and the slides add up. If the turn is a rational part of a whole turn, some power of the screw has turned a whole number of times and only the slide is left: a translation. The screw axes of crystallography are all of that kind, with turns of a half, a third, a quarter or a sixth.

If the turn is not a rational part of a whole turn, no power of the screw ever comes back to zero turn, and no power is a translation. The slides still add up, so each power sits one step further along the axis than the last, and the powers never crowd together. The group the screw generates is infinite, discrete and translation-free: exactly the combination the plane forbids.

Why the plane has no screw is worth saying precisely, because it is the whole difference. A motion of the plane that turns and also slides is not a new kind of motion. The slide lies in the same plane the turn acts on, and a turn followed by a slide in that plane is simply a turn about a different centre: the slide has been absorbed into where the turn happens. There is nowhere for the slide to go that the turn does not reach. A motion that reverses the plane and slides is a glide, and a glide applied twice leaves only the slide, which is a translation.

In space the slide has a direction the turn does not touch. A turn about a vertical axis moves nothing vertically, so a vertical slide commutes with it, and the two stay separate however often the screw is repeated: the turns accumulate in the horizontal plane and the slides accumulate along the axis, and neither can cancel the other. That separation is what lets the turns wander round the circle forever without the slides ever being pulled back to zero. One more dimension is exactly enough room for a motion that is infinite in one coordinate and never repeats in the others. It is the same extra room that lets a rolled sheet have screws of order fourteen and ninety-eight: a tube’s turns are about an axis with no lattice across it, so nothing forces them into the short list a plane lattice allows.

The angle used throughout is the one with cosine $\tfrac35$ and sine $\tfrac45$, about 53.13°. It is not a rational part of a turn, by a theorem of Niven’s: the only rational parts of a turn whose cosine is a rational number are those with cosine $0$, $\pm\tfrac12$ or $\pm1$. That can be checked without appealing to the theorem. A turn through this angle is multiplication by $(3+4i)/5$ in the complex plane, and its $n$-th power is a whole number of turns exactly when $(3+4i)^n$ is a positive real number. Now $3+4i = (2+i)^2$, and $2+i$ and $2-i$ are different Gaussian primes. A real number contains them to equal powers, and $(2+i)^{2n}$ contains only the first, so no power is real. The check is also run in exact integer arithmetic over the first four hundred powers, and none is real.

The near misses are worth looking at, because they show why the plane’s version of this fails. After 393 steps the screw has turned within about two thousandths of a radian of a whole number of turns. In the plane, a group containing such a turn and any translation would contain the corrected turn itself, a turn about a point by two thousandths of a radian, and then turns smaller still, which is the loss of discreteness what forces a lattice measures. In space the correction is not available, because the screw’s power has also climbed 393 steps and there is no translation in the group to bring it back down. Add any translation along the axis and the group stops being discrete; leave it out and the screw is safe. It works because it never pays its climb back.

The screw is also the only way. By Bieberbach’s theorem in its general form, which covers every discrete group of motions and not only the crystallographic ones, an infinite discrete group of motions of space has an element of infinite order. A turn or rotoreflection of infinite order about a fixed point is not discrete, and a glide reflection, or a screw with a rational turn, has a translation among its powers. So an infinite discrete group of space with no translation contains a screw through an irrational part of a turn.

A stack of layers that share less and less #

A screw of this kind is the symmetry of a particular kind of structure: a stack of identical periodic layers, each turned from the one below through the same angle.

Carrying layer $n$ onto layer $n+1$ is the screw: turn through the angle and climb one layer. Repeating it carries the whole stack onto itself, however many layers there are, so the stack has infinitely many symmetries. Whether it also has a translation is a separate question, and it has two parts. A vertical translation would carry a layer onto a layer above it with the same orientation, and no layer above has the same orientation. That holds even allowing for the square lattice’s own quarter-turns, because $(3+4i)^n$ would then be $5^n$ times a power of $i$, which the Gaussian primes forbid by the same argument as before. A horizontal translation would have to be a period of every layer at once.

Two layers at this angle share a sublattice, and it is the coincidence site lattice of index five that grain boundaries are described by: the angle with cosine $\tfrac35$ is the $\Sigma 5$ rotation of a square lattice. A third layer, turned again, cuts the shared periods down by another factor of five, and so on up the stack. In Gaussian integers the pattern is exact. A vector $v$ is a period of layer $j$ when $v \cdot \big((3-4i)/5\big)^j$ is again a Gaussian integer, and the vectors that pass this test for every $j$ from $0$ to $k$ are precisely the multiples of $(2+i)^k$, a sublattice of index $5^k$. The figure’s filled points are found by testing every vector in the window and agree with that description for every $k$ shown. Five layers share nothing in the window except zero, and an unending stack shares nothing at all, since a nonzero vector is not divisible by every power of $2+i$.

So the stack has no translation and infinitely many symmetries. Every layer is as periodic as a layer can be, the whole is carried onto itself by one screw, and nothing moves it without turning it. On the reading “no translation”, such a structure is aperiodic. On the reading “finitely many symmetries”, it is not.

The prism that tiles space this way #

In 1988 Peter Schmitt found a single polyhedron that tiles space in which no tiling has a translation. John Conway modified it, and Ludwig Danzer gave a convex version, now called the Schmitt–Conway–Danzer prism or SCD prototile: a biprism combinatorially like the gyrobifastigium, with parallelogram and irregular triangular faces. Its tilings are built from layers, and the tile’s shape forces the layers to turn relative to one another through an angle that is an irrational part of a turn. Some of its tilings have exactly the symmetry computed above, a screw with an irrational turn and no translation.

That forcing is quoted here, not derived. What is derived is the consequence: a tiling with that structure is translation-free and has infinitely many symmetries, so the SCD prism meets the first reading of aperiodic and fails the second. The pictures above use square layers and the angle with cosine three fifths because every step is then exact integer arithmetic. The SCD prism’s own layers and angle are different, and nothing here depends on which lattice or which irrational angle is used.

This is why the prism was not accepted as a solution to the problem the hat later solved. Chaim Goodman-Strauss proposed, in a survey of open questions in tiling written in 2000, that a tile set be called strongly aperiodic if none of its tilings admits an infinite cyclic group of symmetries, and weakly aperiodic otherwise. The SCD prism is weakly aperiodic. The hat, by the theorem above, is strongly aperiodic as soon as it is aperiodic at all, because in the plane there is no third option. Before 2023 the plane’s only single aperiodic tile was the Socolar–Taylor tile of 2010, which is disconnected. The hat is a connected shape with no rules added to it, and for it the two properties are one.

The same split appears in the hyperbolic plane, for a different reason. The binary tiling of the hyperbolic plane uses one tile and has no tiling with a two-dimensional family of symmetries, yet its tilings can have a one-dimensional family, so it is weakly aperiodic in Goodman-Strauss’s sense. The Euclidean plane is the case where the words coincide because its groups of motions are small, not because tiling problems are easy there.

What the argument cannot show #

The plane theorem is proved; the tile facts are quoted. That a translation-free tiling of the plane has a finite symmetry group is argued completely above, and each of its three constructions is computed. That the hat has no periodic tiling is due to Smith, Myers, Kaplan and Goodman-Strauss and is discussed in the essay on the search that finds it. That the SCD prism has no translation in any tiling, and that some of its tilings have a screw with an irrational turn, is due to Schmitt, Conway and Danzer. Neither tile’s forcing is rederived here.

The stack drawn is a model of a symmetry, not of the prism. Square layers turned by $\Sigma 5$ have the same kind of symmetry group as an SCD tiling with a screw, and the computations are about that group. They say nothing about which shapes force such a stack, and a finite search could not say it either, since no procedure decides whether a shape tiles at all.

Finitely many layers always share a period. The common sublattice of $k$ layers has index $5^k$, so a sample with a hundred layers has shared periods, spaced about $5^{50}$ lattice spacings apart. The claim that the stack has no translation is about the unending stack, and for a real material it becomes a statement about length scales, the same distinction that separates order from periodicity wherever an aperiodic structure meets a finite sample.

And the turned stack is not exotic. A twisted bilayer, two sheets of one lattice laid at an angle, is the two-layer case of this picture, and whether it has a common supercell depends on whether the angle is a coincidence angle, exactly as here. A stack in which every layer turns the same way is the many-layer case. The difference between a helical stack of that kind and a crystal is the difference between the two readings of aperiodic, and it is measurable in the plane of the layers. The in-plane diffraction of such a stack superposes every layer’s pattern, each turned by the step angle, and by the same arithmetic as the shared periods no in-plane spot except the centre is common to all of them. The layers are still evenly spaced, so the spots along the stacking direction are as sharp as a crystal’s.

Still open: a strongly aperiodic tile of space #

The plane now has a single tile that is strongly aperiodic, and space has one that is weakly aperiodic. The obvious question is whether space has a single tile that is strongly aperiodic, one whose every tiling has a finite symmetry group. Thickening the hat into a prism does not give one, since stacking prisms of hats straight up gives a tiling with a vertical translation.

The argument above shows what such a tile would have to avoid. In space an infinite discrete group with no translation contains a screw through an irrational part of a turn. So a strongly aperiodic tile of space is a weakly aperiodic one that also rules out every tiling carried onto itself by such a screw. Whether a known aperiodic tile of space already does this, or whether one exists at all, is a question of forcing that no computation here can reach.