A few comments: The three simply connected surfaces M of constant curvature 1, 0, -1 resp. are the sphere M=S^2, the (Euclidean) plane M=R^2, and the hyperbolic plane M=H^2. There is a tiling of M by regular p-gons, q per vertex according as (p-2)(q-2) is <, =, or > 4, respectively. By taking the dual tiling, we get the tiling by regular q-gons, p per vertex. This is proved by focusing on the existence of a regular polygon of p sides and interior angle 2*pi/q. There are then various ways of quotienting out the simply connected space to obtain a surface of constant curvature, covered by the appropriate M above, with a tiling of the same description. It's a theorem of Hilbert that no surface of constant Gaussian curvature = -1 is a subset of 3-space R^3. So the Coxeter surface with 6 squares per vertex can't be modified *in 3-space* so as to have its ideal constant curvature = -1. There's also the Gauss-Bonnet theorem, which says that for a closed surface (1/2π) * Integral K dA = X(M) where K is Gaussian curvature and X(M) denotes Euler characteristic of M. I haven't seen this mentioned anywhere, but one could define "curvarea" to be the area of a piece of a constant-curvature surface to be its area multiplied by its curvature (or the integral of this product for non-constant curvature). Then Gareth's idea of a negative polyhedron makes sense. —Dan _____ * A closed surface is one that has finite area and empty boundary. Topologically the orientable ones are the sphere, the torus, and the connected sum of n tori, n > 1. The non-orientable ones are the projective plane, the Klein bottle, and the connected sum of n projective planes, n > 2. Gareth McCaughan wrote: ----- In another place, someone asked (approximately) the following question: [question begins] A regular polyhedron in which each {vertex, face} has {a,b} neighbours has vertex, edge and face counts satisfying av = bf = 2e (count pairs (vertex, neighbouring edge) for av=2e and count pairs (face, neighbouring edge) for bf=2e) and of course v-e+f = 2. So (2/a)e-e+(2/b)e = 2 so e = 2ab/(2a+2b-ab) v = 4b/(2a+2b-ab) f = 4a/(2a+2b-ab) and in addition to the "usual" solutions like a=3,b=5 for the regular dodecahedron there are some illegal ones where v,e,f are _negative_. For instance, a=4, b=5 yields e=-20, f=-8, v=-10. Is it in any way meaningful to think of this thing as a polyhedron with -8 pentagonal faces, four of which meet at each of the -10 vertices, joined by -20 edges? [question ends] ... and I was annoyed to find I didn't have a good answer. Purely numerologically, these "virtual" things of genus 0 / characteristic 2 correspond to not-virtual things of genus 2 / characteristic -2, because (-v)-(-e)-(-f) = -2. So this minus-8-faced thing corresponds to some thing that we might hope is a quotient of the hyperbolic plane by some suitable group, having 8 faces, 10 vertices, and 20 edges. There's a tessellation of the hyperbolic plane by right-angled pentagons, which (like our virtual polyhedron) has four pentagons meeting at each vertex; the quotient is an orbifold whose Euler characteristic is -1/4, so it makes some kind of sense that maybe we can put 8 of these together to get a thing of Euler characteristic -2. It's probably obvious to those skilled in the art how one does this, but I was never very skilled in that particular art and have forgotten almost everything now. I can imagine that there might be some more precise way to think of this sort of hyperbolic-plane thing as a "negative" spherical polyhedron, on the super-handwavy grounds that we can think of the faces, edges and vertices of a polyhedron as places where in the sphere we have "excess angle"; e.g., if you take a regular dodecahedron and project it out onto the sphere, then at each vertex you have three spherical pentagons with pi/3 angles meeting neatly, whereas the dodecahedron has three euclidean pentagons with 3pi/5 angles. Something something Gauss-Bonnet something. And, just as the sphere has uniform positive curvature, the hyperbolic plane has uniform negative curvature. So maaaaybe there's some way to think of faces, edges and vertices in the hyperbolic plane as "negative" compared with the sphere because they have angle _defects_ instead of angle _excesses_. But all the details elude me. I bet they're well known to those who know such things, and perhaps some such people are reading this. - Is it true that there's some sort of hyperbolic thing with 8 pentagonal faces meeting four to a vertex? What do you need to quotient the hyperbolic plane by to get it? What does the fundamental region look like? (And does something similar hold for all the negative solutions to the equations above?) - In the case a=6,b=4 there's a famous _infinite_ thing in euclidean 3-space, a sort of polyhedral foam based on the cubic lattice. The equations above suggest that there "should" be a "negative polyhedron" with a=6, b=4, e=-12, v=-4, f=-6. This seems suspiciously similar to the cube, with e=12, v=8, f=6. Is there some sort of quotient of the "mucube" foam that collapses it down to one cube with ?opposite? pairs of vertices identified? Is either the mucube itself or this hypothetical quotient, again, a quotient of the hyperbolic plane by a reflection group? - Is there anything more than numerology to the idea that we can think of these as spherical things with negative numbers of vertices/edges/faces and Euler characteristic +2, rather than (so I assume) hyperbolic things with positive numbers of vertices/edges/faces and Euler characteristic -2? -----