Hilbert's third problem

related topics
{math, number, function}
{math, energy, light}
{@card@, make, design}
{work, book, publish}
{school, student, university}
{son, year, death}

The third on Hilbert's list of mathematical problems, presented in 1900, is the easiest one. The problem is related to the following question: given any two polyhedra of equal volume, is it always possible to cut the first into finitely many polyhedral pieces which can be reassembled to yield the second? Based on earlier writings by Gauss, Hilbert conjectured that this is not always possible. This was confirmed within the year by his student Max Dehn, who proved that the answer in general is "no" by producing a counterexample.

The answer for the analogous question about polygons in 2 dimensions is "yes" and had been known for a long time; this is the Bolyai-Gerwien theorem.

Contents

History and motivation

The formula for the volume of a pyramid,

had been known to Euclid, but all proofs of it involve some form of limiting process or calculus. Similar formulas in plane geometry can be proven with more elementary means. Gauss regretted this defect in two of his letters. This was the motivation for Hilbert: is it possible to prove the equality of volume using elementary "cut-and-glue" methods? Because if not, then an elementary proof of Euclid's result is also impossible.

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,...,Pn, then D(P) = D(P1) + ... + D(Pn)

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.

Full article ▸

related documents
Euler's identity
Elias gamma coding
Linear function
Parse tree
Regular graph
Inner automorphism
Earley parser
Injective function
Axiom of power set
Unitary matrix
Connectedness
Unit interval
Subring
Kleene star
Profinite group
Hausdorff maximal principle
Row and column spaces
Subset
Disjunctive normal form
Identity matrix
Random sequence
Class (set theory)
Specification language
Inverse transform sampling
Additive function
Inequation
Discrete probability distribution
Urysohn's lemma
Just another Perl hacker
Sharp-P