Hilbert's Third Problem - Dehn's Answer

Dehn's Answer

Dehn's proof is an instance in which abstract algebra is used to prove an impossibility result in geometry. Other examples are doubling the cube and trisecting the angle.

We call two polyhedra scissors-congruent if the first can be cut into finitely many polyhedral pieces which can be reassembled to yield the second. Obviously, any two scissors-congruent polyhedra have the same volume. Hilbert asks about the converse.

For every polyhedron P, Dehn defines a value, now known as the Dehn invariant D(P), with the following property:

• If P is cut into two polyhedral pieces P₁ and P₂ with one plane cut, then D(P) = D(P₁) + D(P₂).

From this it follows

• If P is cut into n polyhedral pieces P₁,...,P_n, then D(P) = D(P₁) + ... + D(P_n)

and in particular

• If two polyhedra are scissors-congruent, then they have the same Dehn invariant.

He then shows that every cube has Dehn invariant zero while every regular tetrahedron has non-zero Dehn invariant. This settles the matter.

A polyhedron's invariant is defined based on the lengths of its edges and the angles between its faces. Note that if a polyhedron is cut into two, some edges are cut into two, and the corresponding contributions to the Dehn invariants should therefore be additive in the edge lengths. Similarly, if a polyhedron is cut along an edge, the corresponding angle is cut into two. However, normally cutting a polyhedron introduces new edges and angles; we need to make sure that the contributions of these cancel out. The two angles introduced will always add up to π; we therefore define our Dehn invariant so that multiples of angles of π give a net contribution of zero.

All of the above requirements can be met if we define D(P) as an element of the tensor product of the real numbers R and the quotient space R/(Qπ) in which all rational multiples of π are zero. For the present purposes, it suffices to consider this as a tensor product of Z-modules (or equivalently of abelian groups). However, the more difficult proof of the converse (see below) makes use of the vector space structure: Since both of the factors are vector spaces over Q, the tensor product can be taken over Q.

Let ℓ(e) be the length of the edge e and θ(e) be the dihedral angle between the two faces meeting at e, measured in radians. The Dehn invariant is then defined as

where the sum is taken over all edges e of the polyhedron P.