The angle that is not a fraction of a turn

Two polygons of the same area can always be cut into finitely many pieces that rearrange into each other. That is not obvious and it is true — Wallace, Bolyai and Gerwien between them — and it is the reason area behaves the way anybody expects it to: a shape’s area is the only thing that decides what it can be cut into.

Hilbert asked whether the same holds for polyhedra, and it does not. Volume is not the only obstruction, and the other one is a sum over edges.

The difference between the plane and space is worth locating precisely before the machinery arrives, because it is not where intuition puts it. It is not that space is bigger or that polyhedra are more complicated. It is that a polygon has no invariant beyond area, and a polyhedron has one more, and the extra one lives on the edges — a feature polygons have too but which carries no information there, because a polygon’s “dihedral angles” are the angles of a flat figure and they can be traded freely. In space the angle at an edge is a genuine second measurement, and the whole of Hilbert’s third problem is that it cannot be traded away.

The invariant #

For a polyhedron, take every edge, multiply its length by its dihedral angle, and add. Written that way the quantity is useless: it changes under cutting, because cutting makes new edges. What fixes it is where the sum is taken. Form instead

Σ length ⊗ angle

in the tensor product of the real numbers with the reals modulo rational multiples of π — a group in which π, π/2, 2π/3 and every other rational part of a turn is zero, and in which lengths add. That is the Dehn invariant, and in that group cutting really does change nothing.

That is the whole of the machinery, and everything after it is arithmetic about angles.

Two features of that group are worth naming, because they are what make the whole thing work and both look like technicalities.

The first is that lengths add and angles do not. The invariant is linear in the length slot, so splitting an edge splits its contribution; it is linear in the angle slot too, so angles that meet around a new edge add. The two behaviours are what the two kinds of cut require, and the tensor product is simply the smallest object in which both hold at once.

The second is why the angles are taken modulo rational multiples of π rather than modulo π alone. Modulo π would kill a straight angle and nothing else, and then a cut through the interior of a face — which creates two angles of π/2 and π/2 on the new edge, or 2π/3 three times, depending on where it lands — would change the total. Killing every rational multiple is exactly enough to make every possible cut harmless, and killing more would destroy the invariant’s ability to distinguish anything.

Two polyhedra that cannot be cut into each other #

A cube’s dihedral angles are all right angles, and a right angle is π/2, which is zero in the group. So a cube’s invariant is zero, and so is that of anything cut from a cube.

A regular tetrahedron has six edges and all six dihedral angles are arccos(1/3), about 70.53°.

The claim doing all the work is that arccos(1/3) is not a rational multiple of π. If it were, six copies of it would be zero in the group and the tetrahedron’s invariant would vanish along with the cube’s, and the problem would have the answer Hilbert suspected it did not.

It is worth stating exactly what the invariant does and does not decide before going further. It is a necessary condition: two polyhedra that can be cut into each other have equal volume and equal invariant. It says nothing on its own about whether matching both is enough. So the argument here is an impossibility argument and only an impossibility argument, which is the useful direction — a construction can always be exhibited, and only an invariant can rule one out.

Before the arithmetic, one observation about which polyhedra this can possibly separate. The invariant is a sum over edges, so a polyhedron with no edges at all — there is none — or one whose every angle is a rational part of a turn has invariant zero and is indistinguishable by this test from a cube. That covers a great deal: every rectangular box, every prism on a regular polygon, the rhombic dodecahedron. So the invariant does not distinguish much, and the surprise is that it distinguishes anything, and that the thing it distinguishes is the most symmetric solid there is after the cube.

The claim is an integer computation #

Set cos θ = 1/3 and put bₙ = 3ⁿ · 2cos(nθ). The product formula for cosines gives 2cos((n+1)θ) = 2cos θ · 2cos(nθ) − 2cos((n−1)θ), and multiplying through by 3ⁿ⁺¹ turns that into

bₙ₊₁ = 2bₙ − 9bₙ₋₁

with b₀ = 2 and b₁ = 2. Every term is an integer, and the recurrence is exact.

There is nothing in that argument except divisibility. No estimate is made, no limit is taken, and no floating-point number appears; the computation is in exact integers because by the tenth term a double has lost the property being tested.

The same argument runs on any cos θ = p/q in lowest terms with q odd and above one: modulo q the recurrence gives bₙ ≡ (2p)ⁿ, which is never zero, while ±2qⁿ always is.

There is a small subtlety in the scan and it is worth flagging rather than hiding. The divisibility witness needs a prime of the denominator that does not divide twice the numerator, which every odd denominator above one supplies. An even denominator can fail to — cos θ = 1/4 has only the prime two to work with, and two divides 2p — so those rows are reported as undecided by this witness rather than as rational. Nothing in the argument here depends on them, and the honest presentation is to say which cases the method settles rather than to let a silent default answer for them.

Three exceptions is a small number and it is worth noticing which three. cos θ = 0 gives a right angle, cos θ = ±1/2 gives sixty or a hundred and twenty degrees, and cos θ = ±1 gives nothing or a straight angle. Those are exactly the angles a crystallographer draws — and every other rational cosine, however innocent, belongs to an angle no lattice can be built out of at a rational fraction of a turn.

The generalisation is worth a sentence because it is where the crystallography enters. The dihedral angles a crystal actually shows are not arbitrary: they are angles between lattice planes, and the cosine of such an angle is a ratio of integers divided by a product of square roots. When the metric is cubic those cosines are rational, and the test above applies directly. So for a cubic crystal the question “is this dihedral angle a rational part of a turn” is decidable by an integer computation, every time, with no approximation — which is unusual, since most questions about angles are not decidable at all.

What it forbids #

A shape that tiles space fills a large box, and a box’s invariant is zero. Cutting and rearranging cannot change the invariant, so a shape that tiles space must have invariant zero as well.

The regular tetrahedron’s is 6α and α is not zero in that group. So a regular tetrahedron does not tile space — not by translations, not with rotations, not in any arrangement at all. That is a strong statement got from one angle, and it is the honest version of a fact this collection has met from the other side: five copies, and the gap they leave measures the 7.36° that five tetrahedra fail to close by, and treats the failure as a quantity to be absorbed. Here the failure is not a quantity but a prohibition.

The two readings are compatible and the difference between them is worth stating. The 7.36° deficit says that a particular arrangement of five tetrahedra does not close. The Dehn invariant says that no arrangement of any number of them does, and it says it without examining any arrangement.

There is a second consequence, and it is about what the tetrahedron can do. Two regular tetrahedra and an octahedron do fill space between them — the tetrahedral–octahedral honeycomb is a standard structure and it is the arrangement of close packing itself. The Dehn invariant explains that too: the octahedron’s dihedral angle is π − α, so its invariant is 12ℓ ⊗ (−α) while two tetrahedra give 12ℓ ⊗ α, and the two cancel. A shape with a non-zero invariant can tile space if something with the opposite invariant tiles alongside it, and here that is exactly what happens. Two stackings, one density describes the same honeycomb from the packing side; the invariant is why the two shapes need each other.

There is a converse worth stating carefully because it is easy to overstate. The invariant being zero does not make a shape a tile. A regular octahedron has invariant 12ℓ ⊗ (−α), which is not zero, and it does not tile alone — consistent. A regular dodecahedron has invariant non-zero too and does not tile. But a shape with invariant zero may still fail to tile for reasons of shape rather than of angle: which shapes tile by themselves shows in the plane how far a tiling question can be from an arithmetic one, and nothing about the Dehn invariant closes that gap in space.

The cells that do fill space #

Turning the argument round, every space-filling cell this collection knows about must have invariant zero, and checking that is a real test of the machinery rather than a formality.

The truncated octahedron is the interesting row and it is the body-centred cubic lattice’s cell, which is to say it is the cell nobody chose for the commonest arrangement of metal atoms there is. Its dihedral angles are 109.471° and 125.264°, and neither is a rational part of a turn: the first is π − α and the second is π/2 + α/2.

That cancellation is exact and it is not a coincidence, in the sense that it had to happen: the truncated octahedron tiles space, so its invariant is zero, so whatever its angles are they must arrange themselves this way. What is worth seeing is the mechanism — the equality of all thirty-six edge lengths is what makes the twelve and the twenty-four balance, and a truncated octahedron with unequal edges would not tile.

One more thing worth extracting from that row. The truncated octahedron’s angles are π − α and π/2 + α/2, and the α in both is the tetrahedral angle — the same number the regular tetrahedron is made of. That is not an accident either: the body-centred cubic Voronoi cell is built from the same close-packing geometry, and its angles are what that geometry allows. A single irrational number runs through the tetrahedron, the octahedron, the truncated octahedron and the tetrahedral bond angle of every carbon atom, and the reason it recurs is that all of them are consequences of four directions from a point at equal angles.

What this is doing in a collection about crystals #

Three things, and the third is the one that changes how a reader should hear the earlier essays.

The first is that it settles a question the tiling essays raise and cannot answer with their own tools. Which shapes tile by themselves works in the plane, where the Dehn invariant does not exist and equal area really is the whole story; the difference between the plane and space is exactly this invariant, and it is why a plane question about tiles and a space question about tiles are not the same kind of question.

The second is that it is a genuine obstruction rather than a search that failed. Most impossibility results in this collection come from a group being finite or a congruence having no solution — why five-fold is impossible is the standing example. This one comes from a quantity being conserved, which is a different kind of argument and reaches things the group-theoretic ones do not: nothing about the symmetry of a regular tetrahedron forbids it from tiling, and it is forbidden anyway.

The third is about angles in general. A crystallographer meets 90°, 120°, 60° and 109.47° constantly, and the first three feel like ordinary numbers while the fourth feels like an awkward decimal. The scan above says the awkwardness is real and structural: 109.47° is π − arccos(1/3), it is not any fraction of a turn, and no amount of choosing a better cell will make it one. The tetrahedral angle is irrational for the same reason the tetrahedron does not tile, and the two facts are one fact.

The third of those is the one that keeps the argument honest. A divisibility test that refused everything would prove nothing at all, and arccos(1/2) is the case it must accept — the sixty-degree angle every hexagonal lattice is built from. Running the test on it and getting the answer rational is what shows the test is measuring something.

A last remark about scale, because a reader may reasonably ask whether any of this matters below the level of a theorem. It does, in one place: it says that the volume of an atom’s Voronoi cell is not the only thing that decides what shapes can be assembled from such cells. Two arrangements with the same total volume are not interchangeable, and a rearrangement that preserves volume can be geometrically impossible for a reason that has nothing to do with packing efficiency. That is a constraint on structure that never shows up in a density.

One more consequence is worth drawing because it connects to how this collection treats angles elsewhere. The crystallographic restriction forbids five-fold rotation because 2cos(2π/5) is not an integer, and the argument here forbids a tetrahedral dissection because 2cos(nθ) never lands on ±2. Both are statements that a trigonometric quantity fails to be an integer, both are settled by an exact computation on a recurrence, and neither has anything to say about the other. The recurrence is the same object in both — the Chebyshev relation 2cos((n+1)x) = 2cos x · 2cos(nx) − 2cos((n−1)x) — used once to bound what a trace can be and once to show a sequence misses a residue.

That is a small piece of unity worth pointing at, since the two prohibitions look nothing alike. One is about which rotations a lattice permits and is proved in a line; the other is about which solids can be cut into which and took a century to state. Both come down to the same three-term recurrence, and in both cases the whole force of the argument is that integers cannot be nudged.

Where this stops #

Two things this does not do.

It does not decide scissors congruence. Dehn’s invariant and volume together are known to be a complete pair of invariants for polyhedra in Euclidean three-space — Sydler’s theorem, forty years after Dehn — so two polyhedra with the same volume and the same invariant really can be cut into each other. Sydler’s proof is not a construction and this collection does not carry it; what is here is the obstruction, which is the half that does the forbidding.

And it does not settle which polyhedra tile space. Zero invariant is necessary and it is very far from sufficient: a shape can have every angle a right angle and still fail to tile for reasons of shape alone. Five parallelohedra and no others settles the question for the convex shapes that tile by translations, and it does so by an argument about symmetry rather than about angles. The two arguments do not overlap, and neither one implies the other.