MathLabs

Worked solution: Dehn's invariant: the tetrahedron and cube are not scissors-congruent (1900)

Step 7 of 7: Coda: Sydler's 1965 theorem completes the picture
In plain words

Dehn showed that equal volume and equal Dehn invariant are necessary for two solids to be scissors-congruent — but for over sixty years nobody knew if they were also enough. In 1965 the Swiss mathematician Jean-Pierre Sydler proved that they are: match the volume and match the Dehn invariant, and a dissection connecting the two solids is guaranteed to exist, even though Sydler's proof gives no practical recipe for actually finding the cuts.

vol⁡(P)=vol⁡(Q) ∧ Dehn⁡(P)=Dehn⁡(Q)  ⟺  P∼Q(Sydler, 1965)\operatorname{vol}(P) = \operatorname{vol}(Q) \ \wedge\ \operatorname{Dehn}(P) = \operatorname{Dehn}(Q) \;\Longleftrightarrow\; P \sim Q \quad \text{(Sydler, 1965)}
Detailed analysis

Dehn's 1900 work only establishes the 'only if' direction: scissors-congruent polyhedra have equal volume and equal Dehn⁡\operatorname{Dehn} invariant. Jean-Pierre Sydler proved the converse in 1965 (published in Commentarii Mathematici Helvetici): any two polyhedra P,Q⊂R3P, Q \subset \mathbb{R}^3 with vol⁡(P)=vol⁡(Q)\operatorname{vol}(P) = \operatorname{vol}(Q) and Dehn⁡(P)=Dehn⁡(Q)\operatorname{Dehn}(P) = \operatorname{Dehn}(Q) are in fact scissors-congruent. Together, Dehn's invariant and volume form a complete invariant for three-dimensional scissors-congruence — the precise three-dimensional analogue of how area alone classifies polygons in the Wallace–Bolyai–Gerwien theorem, just with one more number needed.

Sydler's proof is a highly technical existence argument and, unlike the Bolyai–Gerwien construction, does not supply an explicit cutting procedure; later work (Børge Jessen in 1968, extending the framework to four dimensions, and subsequent reinterpretations via algebraic K-theory and group homology by Dupont, Sah, and others) has re-derived and generalised Sydler's theorem, tying Hilbert's third problem to substantially more advanced areas of modern algebra than Dehn's original elementary construction required.

Terms in this step
complete invariant
A collection of invariants is complete for a classification problem if agreeing on all of them is not just necessary but also sufficient to guarantee the two objects are equivalent (here, scissors-congruent); volume alone is not complete in 3D, but volume together with the Dehn invariant is.
Knowledge used in this step