Brian Bi
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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.