Suppose P is a compact convex polyhedron in 3-space, topologically equivalent to the sphere, with finitely many convex planar faces each bounded by a convex polygon with finitely many straight sides. Then every point x of P that is not a vertex has a neighborhood that is isometric to a neighborhood in the plane. And each vertex is locally isometric to a neighborhood of the vertex of a cone, with total angle around the vertex* equal to some theta in the open interval (0, pi). For the jth vertex, call this number theta_j. If this cone is approximated by a smooth surface then its unit normals translated to the origin will cover an area of the unit sphere equal to 2pi - theta. So if we assign the number 2pi - theta_j to the jth vertex, the Gauss-Bonnet theorem tells us that Sum (2pi - theta_j) = 4pi or equivalently Sum theta_j = 2pi(V-2) if V is the number of vertices.** Now suppose we have a finite number of regions P_j in the plane, each bounded by a polygon with straight sides, and that we are given a pairing of all the edges with edges of equal length so that, if the resulting abstract surface has V vertices, E edges, and F faces, the following two conditions are satisfied: 1) V - E + F = 2 (The Euler characteristic of the resulting surface is 2, from which we know that it must be a sphere.) and 2) The total angle about vertex j is theta_j in the interval (0, pi), such that Sum theta_j = 2pi(V-2). Question: --------- Does there then necessarily exist a convex polyhedron in 3-space that is isometric to the resulting abstract surface? —Dan ————— * The total angle of a cone can be seen if you slit the cone from the base to the vertex and flatten it out, it will fill a certain fraction f of a disk. Then its total angle is 2pi*f. ** E.g., for a cube V = 8 and each theta_j = 3pi/2, and indeed 8(3pi/2) = 2pi(8-2).