Brian Bi

## Section 2.13. Hilbert's third problem

Problem 2.13.1

1. A straight cut can do any or all of the following, any number of times:

1. Cut an edge transversely, cutting it into two shorter edges.
2. Cut an edge longitudinally, creating two edges with the same length as the original.
3. Cut a face, producing a new edge.

In case 1, if the edge had length $$\ell$$ and dihedral angle $$\theta$$, then it is divided into two edges with lengths $$\ell_1, \ell_2$$ and the same dihedral angle $$\theta$$, where $$\ell_1 + \ell_2 = \ell$$. The cut therefore removes the contribution $$\ell \otimes \theta/\pi$$ from the Dehn invariant and adds $$\ell_1 \otimes \theta/\pi$$ and $$\ell_2 \otimes \theta/\pi$$, leaving it unchanged.

Case 2 is similar, but here the length remains the same but the dihedral angle is divided.

In case 3, the two new edges would contribute $$\ell \times \theta/\pi$$ and $$\ell \times (\pi - \theta)/\pi$$, for a sum of $$\ell \times 1$$. Since $$1 \in \mathbb{Q}$$, this vanishes, and the Dehn invariant is left unchanged again.

2. Lemma: If $$x + x^{-1} = 2/3$$, then $$x^k + x^{-k}$$ has denominator $$3^k$$ in lowest terms, where $$k$$ is a positive integer

Proof: Obviously this is true for $$k = 0$$ and $$k = 1$$. Proceed by induction. For $$k \geq 2$$, we have $$x^k + x^{-k} = (x + x^{-1})(x^{k-1} + x^{-(k-1)}) - (x^{k-2} + x^{-(k-2)})$$. By the inductive hypothesis, the first term on the RHS has denominator $$3^k$$ and the second $$3^{k-2}$$, so the result has denominator $$3^k$$.

Following the Hint, suppose $$\alpha = 2m/n$$ for integers $$m, n$$, so that $$\cos(2\pi m/n) = 1/3$$. Consider the equation $$x + x^{-1} = 2/3$$. If we put $$x = \cos(2\pi m/n) + i \sin(2\pi m/n)$$, then since $$|x| = 1$$, it follows that $$x + x^{-1} = 2\Re(x) = 2 \cos(2\pi m/n) = 2/3$$. By the Lemma, $$x^n + x^{-n}$$ has denominator $$3^n$$, but that's a contradiction since $$x^n + x^{-n} = 2$$.

3. The cube has zero Dehn invariant since each dihedral angle is $$\pi/2$$. For the regular tetrahedron, let AB be one of the edges and M the midpoint of CD. Then the dihedral angle is the angle $$\angle AMB$$. If $$|AB| = 1$$, then $$|AM| = |BM| = \sqrt{3}/2$$. By the law of cosines, $$1 = 3/4 + 3/4 - 2(3/4) \cos \angle AMB$$, so $$\angle AMB = \cos^{-1}(1/3)$$. The Dehn invariant for the regular tetrahedron is thus $$6l \otimes \cos^{-1}(1/3)/\pi$$. The second term is nonzero by the previous part, so the regular tetrahedron has nonzero Dehn invariant. We conclude that the cube and regular tetrahedron are not interdissectable.