WHAT'S THE "SIMPLEST" POLYHEDRAL g-HOLED TORUS?
If you wish to make a g-holed torus with V vertices, E edges, and
F triangular faces, then comparing boundaries gives 3F=2E while
Euler's formula is 2-2g = V-E+F so that E=3(V-2+2g). On the other hand,
with V vertices, the largest number of edges you could draw would be
to connect every possible pair, giving E <= ( V choose 2 ), so there is
an inequality relating V and g: V^2-7V +12(1-g) >= 0, i.e. V is
larger than the positive root V >= (7/2) + sqrt(48g-1)/2.
It can be shown that the minimal number of vertices is integral, that is,
using the minimum number of vertices requires joining every pair of them, iff
g is of one of these forms: [proof: write 48g+1=x^2; find x=1,7,17,23 mod 24]
g=12k^2+k, V=12k+4 (V,g) = (4,0), (16, 13), ...
g=12k^2+7k+1, V=12k+7 (V,g) = (7,1), (19,20), ...
g=12k^2-7k+1, V=12k (V,g) = [(0,1),] (12,6), (24, 35),...
g=12k^2-k, V= 12k+3 (V,g) = [(3,0),] (15,11), (27,46),...
So nicest genera are g=0, 1, 6, 11, 13, 20, ...
The case g=0 is topologically a sphere; the standard tetrahedon shows
it is possible to build one with the minimum number of vertices (4).
The case g=1 is discussed in the other files in this directory. As far
as I know it is unknown whether it is possible to glue 44 triangles
onto the 66 edges joining 12 points in R^3 in such a way that
the triangles do not intersect (except at the edges); this would realize
the minimal configuration for g=6.
TABLES FOR SMALL GENUS:
For small g we compute the minimal V, the corresponding number of
total edges (V choose 2) and edges to be used E=3(V-2+2g), with the
difference (=possible edges not used):
g: 0 1 2 3 4 5 6
Vmin 4 7 9 10 11 12 12
Eavail 6 21 36 45 55 66 66
Eused 6 21 33 42 51 60 66
Eunused 0 0 3 3 4 6 0