Combinatorily a polyhedron can be just thought of as a decomposition of the sphere into geodesic polygons.
If you expand the definition to include other compact surfaces besides the sphere, there are many lovely examples of regular polyhedra.
For instance, the 3-holed torus can be tiled in a regular way with 8 hexagon, with each vertex surrounded by 4 of them. Resulting in a symmetry group of size 96. (Determined by: one hexagon going to any of the 8, in any of 12 ways.)
But if you try to tile the sphere with hexagons, you find it won't work. Suppose there are N hexagons averaging q >= 3 around each vertex.
The number of vertices must be V = 6N/q.
Then the number of edges must be E = 6N/2 = 3N.
The number of faces is F = N.
Since the Euler characteristic X(S^2) = 2, this means
2 = X(S^2) = V - E + F = 6N/q - 3N + N = N(6/q - 2).
But since q >= 3, this is impossible.
—Dan
Colin writ:
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Forgive me if this is obvious and well-known.
Because of Euler we know that for polyhedra without holes we have
V+F=E+2. As a result we can't have a polyhedron made entirely of
hexagons, and need at least 12 pentagons or an equivalent set of
shapes. Etc.
But things change if we allow intersecting faces. Clearly we have
a lot of scope for variations, but we can always retain limits to
make things reasonable.
But ...
Can we make a "polyhedron" consisting only of hexagonal "faces"?
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