Hilbert's third problem and Dehn invariant -- an elementarization with spherical triangles (Q827537)

To avoid the tensor product of Abelian groups involved in the Dehn invariant, which is considered non-elementary, the author introduces another invariant of a polyhedron \(P\) in \(\mathbb{R}^3\), defined by \(L(P)=\sum_{e\in P} \epsilon_e\), where, for every vertex \(e\) of \(P\), \(\epsilon_e=\sum_{k\in K_e} \alpha(k)-(|K_e|-2)\cdot \pi\), where \(\alpha(k)\) denotes the dihedral angle of edge \(k\), and \(K_e\) the set of all edges emanating from \(e\). Computing the \(L\)-invariant of a cube and of a regular tetrahedron and showing that: (*) ``if two polyhedrons \(P\) have \(Q\) of equal volume have each equal dihedral angles, and those of the former are rational multiples of \(\pi\), whereas those of the latter are irrational multiples of \(\pi\), then \(P\) and \(Q\) are not equidecomposable'' solves Hilbert's third problem. Using (*), one can also provide two tetrahedra with the same basis and congruent altitudes which are not equidecomposable. The Bricard condition can also be stated using (*) in a Dehn-Hadwiger type theorem as: Let \(P\) and \(Q\) be two polyhedra with dihedral angles of the edges \(\alpha_1, \ldots \alpha_p\) and \(\beta_1, \ldots \beta_q\), respectively. If \(P\) and \(Q\) are equidecomposable, then there exists an integer \(n_{\pi}\) and there are natural numbers \(m_1, \ldots, m_p\) and \(n_1, \ldots, n_q\), so that \[\sum_{i=1}^pm_i\alpha_i-\sum_{j=1}^qn_j\beta_j = n_{\pi}\cdot \pi\]