It's worth noting that a polyhedron need not be a subset of 3-space. In fact the existence of a polyhedron with its intrinsic properties should be considered a separate issue from whether it can be embedded in 3-space (or n-space, for that matter). To define such a polyhedron abstractly, it's necessary only to specify the faces, and which pairs of faces share an edge. For example, identifying antipodal points of the dodecahedron gives a tiling of the projective plane P^2 by 6 pentagons. This is particularly nice since each pentagon is surrounded by the other 5 of them, realizing the chromatic number 6 of P^2. For any regular p-gon Q in the Euclidean plane R^2 (p ≥ 3), there is an interesting example of an infinite polyhedron X_Q whose faces are copies of Q. The interior angle of Q is (p-2)π/p. Start by placing copies of Q around a vertex of Q until one of the copies finally shares an edge with Q again. (The number of copies will be the smallest multiple of (p-2)/p that is an even integer.) Now imagine doing the same for each copy of Q that is created in this way, until all edges are shared by 2 copies of Q. Of course in most cases there will be rampant overlapping of the copies. But by interpreting each copy of Q as an abstract p-gon, sharing only its edges with other copies, one gets an abstract polyhedron composed of countably many p-gons. This polyhedron is certainly not embedded in the plane, except when Q is a triangle, square, or hexagon. The surface constructed in this way is an example of a orientable infinite regular polyhedron, since mapping Q to any copy of Q (including Q itself) in any of the 2p dihedral ways can be extended to an isometry of the entire constant-curvature surface to itself. If p = 3, 4, or 6, then clearly the surface will just be the plane. Assuming p = 5 or p ≥ 7, the topology of this infinite polyhedron will be that of the unique orientable surface with a countably infinite number of handles and one "end".* This may be thought of as a kind of limit of an n-holed torus stretching to the right, as n –> oo if more tori are repeatedly added by connected sum to the right end. (This surface has been affectionately nicknamed "the Loch Ness monster", though I prefer "Nessie" since it has fewer syllables.) Still assuming p = 5 or p ≥ 7, the surface can be given constant negative curvature of K = -1. By its construction, it has a highly symmetric map onto the Euclidean plane, and it's easy to check that this map preserves angles away from vertices — where the angle is multiplied by some positive integer ≥ 2. This shows that its map to the plane can be interpreted as a highly symmetric holomorphic map of Riemann surfaces. —Dan _____ * An "end" of a non-compact surface is, roughly speaking, a distinct way that it can "go off to infinity". For any closed subset X of the Cantor set K, including K itself, there is a surface of any genus g (including g = oo) that has X as its space of ends. In fact, the genus and the space of ends together determine the topological type of any orientable surface. See <https://en.wikipedia.org/wiki/End_(topology)>. Veit Elser wrote: ----- I think your negative F, E, V counts come from insisting the surface has Euler characteristic 2. But I completely agree that there are more regular polyhedra than the usual five! Who said they had to be finite? We all know the Greeks had issues with infinity. Your "polyhedral foam based on the cubic lattice” should be the 6th regular polyhedron. It has a dual (analogous to the duality between the cube and octahedron) where all faces are regular hexagons and four meet at every vertex (forming a “saddle”). Think of gluing together truncated octahedra by their square faces so only the hexagon faces remain, forming an infinite foam. If you take the quotient by the translations you get a closed surface with F=8 faces, E=24 edges, and V=12 vertices, so F-E+V = -4 = 2-2(3) and the genus is 3 (the three pairs of square faces of the truncated octahedron that are identified form three handles on the sphere). Just think, if Kepler had known about the two infinite regular polyhedra he probably would not have proposed that silly model of the solar system. -----