Dehn invariants are the principal tool used (by Max Dehn, and others after him) to solve Hilbert's 3rd problem. Their use is a beautiful example of how Algebra can help us solve problems in Geometry. Note that many impossibility proofs in Geometry use Algebra (trisection of the angle and squaring the circle are the classic examples). Geometry itself is quite weak when it comes to showing its own failures; it is interesting that Algebra developed for other purposes proves so useful here. There are also other (similar) applications of Dehn invariants.

Let B⊂R3 be some polyhedron. B has a set EB of edges. For any edge e∈EB of B, we let le be the length of e, and αe be the dihedral angle formed between the two faces of B which meet along e. For "convenience", we express all angles in radians; thus, π is 180o. Dehn's proof makes heavy use of the fact that the angles αe needn't all be rational multiples of π.

Let AB={α1,...,αn}={αe: e∈EB} be the set of all dihedral angles αe which appear in B. We pretend that π is a dihedral angle of B (just form a spurious edge of B flat across one of its faces), so WLOG π∈B. We form the vector space

MB = {r1α1+...+rnαn: r1,...,rnQ}
of all rational combinations of angles of B. MB is a vector space over the field of rational numbers Q of dimension ≤ n. In practice, though, MB will have much lower dimension that n. MB precisely expresses the rational relationships between the angles of B. It turns out (see Dehn invariants are "additive" for a sneak preview) that this is very pertinent to equidecomposability!

We still haven't seen a Dehn invariant, though! Before we do, we need one more "let". So let f:MBQ be a Q-linear functional which satisfies f(π)=. Of course, this in itself means that f(2π)=0, f(π/2)=0, and indeed f(any rational fraction of a right angle)=0. Define the Dehn invariant of B with respect to f to be

Df(B) = e∈EB lef(αe).
It's probably called an "invariant" because the value doesn't depend on the orientation or position of B in space -- just on the lengths of the edges of B and the angles which they form. Congruent transformations of B preserve the value of the Dehn invariant.

The requirement that f(π)=0 is somewhat obscure. In fact, it makes perfect sense. Suppose we decide to add another edge e' to B, by splitting one of its faces. Of course, we'll keep the dihedral angle at the edge flat -- αe'=π -- so B doesn't change. The requirement that f(αe')=f(π)=0 means that the value of Df(B) isn't changed by such meaningless tomfoolery. In fact, it turns out that this requirement is key to the remainder of the proof. The seemingly arbitrary requirement is in fact deeply connected with the structure of R3!

Probably the most interesting thing about Dehn invariants is that we can use them to show that two polyhedra are not equidecomposable! This follows immediate from the fact that Dehn invariants are "additive".

Log in or register to write something here or to contact authors.