Posted by: holdenlee | October 1, 2010

Hilbert’s Third Problem


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 V\otimes W as follows: First form the free product V*W, which has a basis consisting of v\otimes w where v\in V and w\in W. Then quotient out by the relations

(1)\; (av)\otimes w=a(v\otimes w)=v\otimes (aw)

(2)\; (v_1+v_2) \otimes w=v_1\otimes w+v_2\otimes w

(3)\; v \otimes (w_1+w_2)=v\otimes w_1+v\otimes w_2.

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

To any polyhedron define its Dehn invariant D(A) in V=\mathbb{R}\otimes_{\mathbb Q} (\mathbb R/\mathbb Q) to be

D(A)=\sum_{a\text{ edge}} l(a)\otimes \frac{\beta(a)}{\pi},

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

(A) Invariance under cuts

If the cut divides edge a into two edges b and c, then the term l(a)\otimes \frac{\beta(a)}{\pi} in D(A) is replaced by

l(b)\otimes \frac{\beta(b)}{\pi}+l(c)\otimes \frac{\beta(c)}{\pi}=l(a)\otimes \frac{\beta(a)}{\pi}

in D(B)+D(C) since l(a)=l(b)+l(c) and \beta(a)=\beta(b)=\beta(c).

If the plane of the cut goes through a, then the term l(a)\otimes \frac{\beta_A(a)}{\pi} in D(A) corresponds to the terms

l(a)\otimes \frac{\beta_B(a)}{\pi}+l(a)\otimes \frac{\beta_C(a)}{\pi}=l(a)\otimes \frac{\beta_A(a)}{\pi}

in D(B)+D(C). Equality follows since the angle measures in B and C add up to that in A: \beta_A(a)=\beta_B(a)+\beta_C(a).

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

l(a)\otimes \frac{\beta_B(a)}{\pi}+l(a)\otimes \frac{\beta_C(a)}{\pi}=l(a)\otimes \frac{\pi}{\pi}=0.

Hence if A can be cut into B and C by a straight cut, then D(A)=D(B)+D(C).

(B) \frac{\cos^{-1}(1/3)}{\pi} not rational

Suppose by way of contradiction that \frac{\cos^{-1}(1/3)}{\pi} is rational. Write \frac{\cos^{-1}(1/3)}{\pi}=\frac{2m}{n} with m,n\in\mathbb Z. We have

\cos\left(\frac{2m\pi}{n}\right)=\frac{1}{3}.

Then x=\text{cis}\left(\frac{2m\pi}{n}\right) is a nth root of unity with
x+x^{-1}=2\cos \left(\frac{2m\pi}{n}\right)=\frac{2}{3}
We show by induction that x^{k}+x^{-k} has denominator equal to 3^k when written in lowest terms. This is true for k=0,1. Suppose it true for k-1 and k-2; write x^{k-1}+x^{-(k-1)}=\frac{b}{3^{k-1}} and x^{k-2}+x^{-(k-2)}=\frac{c}{3^{k-2}} with 3\nmid b,c. Then

x^k+x^{-k}=(x^{k-1}+x^{-(k-1)})(x+x^{-1})-(x^{k-2}+x^{-(k-2)})=\frac{2b-9c}{3^k}.

Since 3\nmid 2b-9c, this completes the induction step. However, since x is a nth root of unity, x^n+x^{-n}=2, contradiction. Hence \frac{\cos^{-1}(1/3)}{\pi} cannot be rational.

An alternate solution is as follows: Suppose p and \cos(p\pi) are both rational. Let \omega=\cos (p\pi)+i\sin(p\pi). Then \omega is a root of unity and hence an algebraic integer; similarly \overline{\omega} is as well. Hence 2\cos(p\pi) =\omega + \overline{\omega} is an algebraic integer; since it is rational it is a rational integer. Hence we must have 2\cos(p\pi)=0,\pm 1,\pm 2; or \cos(p\pi)=0,\pm\frac{1}{2},\pm 1. Therefore \cos^{-1}\left(\frac{1}{3}\right) is not a rational multiple of \pi.

(C) Dehn invariant of regular tetrahedron and cube

The Dehn invariant of a cube is 0 because all angles between adjacent faces are \frac{\pi}{2}, which is a rational multiple of \pi.

Let s be the length of a side in a regular tetrahedron. The distance from the midpoint M of an edge to the centroid C of an adjacent face (the foot of the perpendicular from the opposite vertex V) is \frac{1}{3}\cdot\frac{\sqrt{3}}{2}s=\frac{\sqrt{3}}{6}s. The distance MV is \frac{\sqrt{3}}{2}s. Hence the angle between two faces is \cos^{-1}(MC/MV)=\cos^{-1}(1/3). The Dehn invariant is 6s\otimes \frac{\cos^{-1}(1/3)}{\pi}, 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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