Equal area is enough, and equal volume is not

The operation set is a pair of scissors and a hand: cut along straight lines, pick the pieces up, put them down somewhere else, do not overlap and do not leave gaps. The question is the one this field always asks — what can that reach?

In the plane the answer is everything of the right area, which is a strong statement and an old one. In space the answer is not, and the obstruction is the first problem on Hilbert’s 1900 list to be solved — within the year, by his own student.

The chain #

The plane result is proved by reduction, in three steps, each of which is a picture.

Any polygon into triangles. Cut along diagonals until nothing is left but triangles. The count is not in doubt — a polygon of $n$ sides falls into exactly $n - 2$ triangles however the diagonals are chosen, which is the same count that Euler’s formula produces for a planar subdivision. For a convex polygon this is a fan from one corner and needs no argument; for a general polygon it needs the observation that every polygon has a diagonal lying entirely inside it, which is true and slightly fiddly.

Any triangle into a rectangle. That is the hero figure: slice at half the height, cut the small triangle down its own altitude, and swing each half about the end of the slice it stands on. Three pieces, and the rectangle has the triangle’s area because the pieces are the same pieces.

Any rectangle into a square. This is the step that takes real work in general, because a rectangle a thousand times as long as it is wide cannot be squared in a few pieces. When the sides are in a convenient ratio a single staircase cut does it.

Put the three together and any polygon becomes a stack of squares; make the squares the same size, put two together into one, and repeat. Two polygons of equal area both reduce to the same square, and a dissection is reversible, so each reduces to the other.

Why the theorem is about an operation set #

It is worth saying why this belongs beside straightedge-and-compass constructions rather than beside geometry generally, because the two questions have the same shape and it is not a coincidence.

What two points can build fixes a set of moves — draw a line through two known points, draw a circle about a known point through another — and asks which points are reachable. The answer is a field, and the impossibilities follow from a degree count. Here the moves are cut along a line and place a piece by a rigid motion, and the question is which shapes are reachable from which.

In both cases the reach is characterised by an invariant: something preserved by every move, whose value therefore cannot change, so two objects with different values are not connected. For constructions the invariant is the degree of a number over the rationals; for dissections in the plane it is the area, and in space it is the area together with Dehn’s quantity.

And in both cases the operation set can be enlarged and the reach grows. Add a marked ruler to the compass and cube roots come within reach; allow a hinged dissection, where the pieces stay connected and swing, and the plane theorem survives — which is a theorem of Abbott and others from 2008, and was open for a century. Allow pieces that are not measurable and, as the section below records, everything is reachable and the subject collapses.

What the theorem does not promise #

The theorem says finitely many and says nothing else about the number. The chain as written is wasteful — the figure counts its pieces and the count is large — and the question of the fewest pieces for a given pair of shapes is a separate subject with almost no general theory in it.

The most famous instance is Dudeney’s dissection of an equilateral triangle into a square in four pieces, from 1902, which is beautiful and which nobody has proved to be optimal. That is the standing situation: particular dissections are found by ingenuity, lower bounds are rare, and the general theorem guarantees existence and no more.

There is a second thing the theorem does not promise, and it matters for what follows. Nothing above uses reflections. Every piece in the figures is moved by a translation or a rotation, never turned over, and the theorem holds with that restriction — which is not obvious and is worth noticing, because it means the answer would not change if the pieces were physical objects with a top and a bottom.

The obstruction in space #

Hilbert asked, as the third of his problems, whether the same holds for solids: given two polyhedra of equal volume, can one be cut into finitely many polyhedral pieces and reassembled into the other?

He suspected not, and said so, and the reason he suspected it is worth stating because it is the reason the plane proof does not carry over. Every argument for the plane theorem eventually uses the fact that a triangle can be halved into two triangles similar to nothing in particular and rearranged freely. The three-dimensional analogue would need a tetrahedron to be cut into pieces that reassemble into a prism, and nobody could do it.

Max Dehn found the obstruction in 1900. Attach to a polyhedron a quantity built from its edges: for each edge, the length times the dihedral angle along it, added up — but with the angles taken in a way that ignores rational multiples of $\pi$. Cutting a polyhedron and reassembling the pieces leaves that quantity unchanged, because every cut either creates two new edges whose angles add to the old one, or creates edges whose angles add to $\pi$ or $2\pi$, and both are invisible to a quantity that ignores rational multiples of $\pi$.

For a cube every dihedral angle is $\pi/2$, a rational part of a turn, so the quantity is zero. For a regular tetrahedron every dihedral angle is $\arccos(1/3)$, and the whole question is whether that is a rational part of a turn.

The angle that is not a rational part of a turn #

It is not, and the proof is short enough to be a figure.

Write $\theta = \arccos(1/3)$ and consider $\cos(n\theta)$. The double-angle recurrence gives $\cos((n+1)\theta) = 2\cos\theta\cos(n\theta) - \cos((n-1)\theta)$, and with $\cos\theta = 1/3$ that becomes, writing $\cos(n\theta) = a_n/3^n$,

$$a_{n+1} = 2a_n - 9a_{n-1}.$$

Start from $a_0 = 1$ and $a_1 = 1$. Modulo three, the recurrence is $a_{n+1} \equiv 2a_n$, so no term is ever divisible by three, since none of the first two is.

If $\theta$ were a rational multiple of $\pi$, then some multiple $n\theta$ would be a whole number of half turns, and $\cos(n\theta)$ would be $\pm 1$ — which needs $a_n = \pm 3^n$, a multiple of three for every $n$ above zero. It never is. So $\theta$ is not a rational part of a turn, the tetrahedron’s Dehn quantity is not zero, the cube’s is, and no number of straight cuts turns one into the other.

The figure runs the recurrence in exact whole numbers and checks each term against the cosine it is supposed to be, so the arithmetic and the trigonometry are compared rather than one being trusted.

Why this was the first of the twenty-three to fall #

Hilbert’s third problem was answered by Dehn within the year, which makes it the fastest resolved of the list and by a wide margin. That speed is worth a comment.

The reason is that the problem is the one on the list where the right kind of object was already available. An invariant that is preserved by cutting and reassembling is exactly what is needed, and by 1900 the idea of attaching an algebraic quantity to a geometric object and watching it survive a transformation was thoroughly familiar. What Dehn supplied was the specific quantity, and its cleverness is in the quotient: taking angles modulo rational multiples of $\pi$ is what makes the quantity blind to the cuts and sensitive to the shape.

The comparison with the plane case is instructive. In the plane the Dehn quantity is always zero — every polygon’s angles are irrelevant because a polygon has no edges with dihedral angles — and there is nothing to obstruct. The theorem holds in two dimensions precisely because the obstruction is empty there.

Sydler proved in 1965 that volume and the Dehn quantity together are sufficient in three dimensions: two polyhedra of equal volume and equal Dehn quantity are scissors-congruent. So the classification is complete, and the answer to the original question is that one number was not enough and two are.

The invariant, in the shape every invariant has #

Dehn’s quantity is the third invariant this site has built, and the three are worth putting side by side, because the construction is the same each time and only the moves change.

The linking number is a signed count that a deformation cannot change, because a deformation creates and destroys crossings in pairs. The sign of a permutation is a parity that a redrawing cannot change, because a redrawing adds and removes crossings in pairs. Dehn’s quantity is a sum over edges that a cut cannot change, because a cut either splits an angle into two that add back to it or produces angles that add to a straight one.

In all three the recipe is identical: find something the move alters in a way that cancels, and quotient out exactly what does not cancel. The quotient is the clever part every time — halving in the first case, reading modulo two in the second, ignoring rational multiples of $\pi$ in the third — and it is chosen so that the residue after a move is zero.

The test of whether the quotient is right is that the invariant still distinguishes something. Quotient too little and the moves change it; quotient too much and it is zero for everything. Dehn’s quantity sits exactly between: rational angles are quotiented away because a cut can produce them, and irrational ones survive because a cut cannot.

Where it fails, and what it needs #

Finitely many pieces is doing all the work. Allow infinitely many pieces, drop the requirement that they be reasonable sets, and equal volume stops mattering at all: the Banach–Tarski construction cuts a ball into five pieces and reassembles them into two balls of the same size. The pieces are not measurable, so they have no volume, so there is nothing to conserve. Everything above assumes pieces that are polygons or polyhedra, which have area and volume, and the whole subject depends on that.

The plane theorem needs straight cuts. Curved cuts change nothing in the plane, since a dissection with curved pieces can be refined to one with straight pieces, but the statement is about polygons and its proof uses them throughout.

The Dehn quantity is not a number. It lives in a tensor product of the reals with the reals modulo rational multiples of $\pi$, which is an infinite-dimensional space over the rationals and cannot be written down as a single quantity. It can be shown non-zero, which is all that is ever needed, and the figure shows it by exhibiting an angle that is not a rational part of a turn.

And equal volume with equal Dehn is a theorem for three dimensions only. In four dimensions and above, whether volume and the Dehn quantity classify scissors congruence is open.

What the pictures cannot show #

The dissections drawn here are of specific shapes with convenient numbers. The staircase cut works because the sides are in the ratio of two squares; the general rectangle needs a different construction and more pieces, and the figure draws the easy case rather than the general one.

The piece count in the polygon figure is the chain’s, and the chain is not efficient. Nothing in the figure suggests how far from optimal it is, because nobody knows.

And the impossibility is not drawn at all. The figure shows a cube, a tetrahedron, and a table of integers proving that one angle is not a rational part of a turn. That the quantity built from those angles is preserved by cutting is a theorem, quoted here, and the picture of a thing that cannot be done is — as always in this field — the search that failed rather than the failure itself.

The ladder from here #

Below: the Pythagorean theorem by rearrangement, which is a dissection whose pieces are the argument, and what two points can build, where an operation set’s reach is asked about for the first time. Sideways: the cube that will not double, an impossibility settled by a different invariant of a different operation set, and two right angles and a box, where a plane theorem is carried into space and survives. Above: Sydler’s theorem, hinged dissections, and the scissors-congruence groups of spherical and hyperbolic space.

What is worth carrying away #

A single number classified the plane case, and the temptation is to expect a single number in space. What happened instead is the usual thing: the number that worked in low dimensions was one of two, and the second was invisible because it is identically zero where there is nothing for it to measure.

The way to find such a thing is the way Dehn found it — look for a quantity that the operation cannot change, rather than a quantity that describes the object. Volume is preserved by cutting because cutting preserves volume, and so is the angle sum, once the angles are read in a way that forgets exactly what a cut can produce. An invariant is built against the moves, not against the shapes.