Suppose we define a slightly generalized type of regular "spherical" polyhedron (RSP) as follows. (They are a kind of metric and regular definition of a branched covering.) ---------------------------------------------------------- Definition: An RSP is an abstract metric space X that is a finite union of identical *copies* of spherical polygons called faces {P_k | 1 <= k <= N} (on say the unit sphere), equipped with a mapping F: X -> S^2 that is an isometry when restricted to any one P_k. Note that the P_k are not necessarily either disjoint or distinct, and they may have interior angles > pi. Note that the polygons P_k are subsets of X but not of S^2. X must satisfy: * For any vertex v of any one polygon, the sum of the angles at v, over all the polygons having v as a vertex, is an integer multiple of 2pi. * For any edge e of any one polygon, there are exactly two polygons that have e as a full edge. Two such polygons sharing exactly one full edge are called "adjacent". No two adjacent polygons have the same image under F. * For any two polygons, there is a sequence of polygons starting with one and ending with the other, with any two consecutive polygons in the sequence being adjacent. * (Regularity): For any triple (v,e,f) with vertex v in edge e in face f, and any other such (v',e',f'), there is an isometry I: X -> X that carries v |-> v', e|-> e', f |-> f'. * Any two RSP's are called "equivalent" if there is an isometry of one onto the other that carries each vertex, edge or polygon of one to a vertex, edge, or face, respectively, of the other. ---------------------------------------------------------- I know the complete classification of RSP's, but I wonder if anyone has seen this definition before, or such a classification. Examples of nonstandard polyhedra that are RSP's include: * Taking any of the Platonic solids projected radially to the sphere S^2, and replacing each face of the projection with the closure of its complement in S^2. (E.g., starting with a cube, the complement of a projected face has angles = 4pi/3.) * Project the regular tetrahedron to the sphere and choose any two disjoint edges e, e' of it. Now take two identical copies T_1, T_2 of this spherical tetrahedron and "slit" each of them them along each edge corresponding to e and e', creating 8 loose edges. Now identify these in 4 pairs (as when creating a Riemann surface) so that one side of e on T_1 is identified to the other side of e on T_2, etc., so that no loose edges remain. (In this case the topology of X is that of a torus.) --Dan ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx