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