Given any two polyhedra with the same volume, is it always possible to cut one of them with a finite number of straight cuts and assemble it to form the other?

In the corresponding problem for polygons, the answer is affirmative, but for polyhedra, the answer is NO. We will show that a cube cannot be cut with straight cuts and reassembled into a tetrahedron with the same volume.

We will assign each polyhedron with an invariant, that doesn’t change when we cut it with a straight cut. This invariant should have something to do with side lengths, and something to do with angles between faces. Indeed, a cut may split a side into two, an angle into two. So we want to do something with adding up side lengths, or adding up angles.

But since there are two pieces of information, not just one, we encapsulate the information about the angles and the side lengths in a* tensor*.

Given two vector spaces V and W, we form their tensor product as follows: First form the free product , which has a basis consisting of where and . Then quotient out by the relations

That is, the elements in are sums of elements of the form , and they are required to follow the rules above (and only those rules). In our case will represent lengths and will represent angles.

To any polyhedron define its **Dehn invariant** in to be

where is the length of and is the dihedral angle between the two planes meeting at . We’re considering the vector spaces over , that is in the tensor product we’re allowed to “shuffle” elements of in (1), but not arbitrary elements in . The second space is is modded out by ; this is just saying that we want angles differing by rational multiples of to be considered the “same”.

**(A) Invariance under cuts**

If the cut divides edge into two edges and , then the term in is replaced by

in since and .

If the plane of the cut goes through , then the term in corresponds to the terms

in . Equality follows since the angle measures in and add up to that in : .

Finally, any new edge created is an edge in both and . The angles at in and add up to a straight angle so these terms contribute

Hence if can be cut into and by a straight cut, then .

**(B)** **not rational**

Suppose by way of contradiction that is rational. Write with . We have

Then is a th root of unity with

We show by induction that has denominator equal to when written in lowest terms. This is true for . Suppose it true for and ; write and with . Then

Since , this completes the induction step. However, since is a th root of unity, , contradiction. Hence cannot be rational.

An alternate solution is as follows: Suppose and are both rational. Let . Then is a root of unity and hence an algebraic integer; similarly is as well. Hence is an algebraic integer; since it is rational it is a rational integer. Hence we must have ; or . Therefore is not a rational multiple of .

**(C) Dehn invariant of regular tetrahedron and cube**

The Dehn invariant of a cube is because all angles between adjacent faces are , which is a rational multiple of .

Let be the length of a side in a regular tetrahedron. The distance from the midpoint of an edge to the centroid of an adjacent face (the foot of the perpendicular from the opposite vertex ) is . The distance is . Hence the angle between two faces is . The Dehn invariant is , nonzero by (B).

A regular tetrahedron and cube of the same volume do not have the same Dehn invariant. By (A), if a polyhedron is cut with finitely many straight cuts, the sum of the Dehn invariant of the pieces stays the same. Thus a tetrahedron cannot be cut with finitely many cuts and be reassembled to form a cube.

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