From: snowe@rain.org (Laura Helen)
Newsgroups: sci.math
Subject: minimal polyhedra
Date: 12 Mar 1995 16:56:31 -0800
I have been looking at the minimal number of vertices that you need to
form a polyhedron which is topologically a compact connected 2-manifold.
It was rather surprising that you do not need many extra vertices to
get extra holes in your donuts.
One constraint on the number of vertices needed comes from the equation
for the Euler characteristic:
V - E + F = EC
Since E = 3/2 * F, you have
E = (V-EC) * 3
but also E <= V * (V-1)/2, which gives a lower bound on the number of
vertices needed to get a given Euler characteristic.
For more negative Euler characteristics the minimal number of
vertices often suffices. You use more vertices; it seems that with more
vertices there is more latitude to form the topology you want, within the
constraints mentioned above.
I like the 6-vertex projective plane below. It looks a lot like a Boy
surface. :)
1 2 3
1 2 4
1 3 5
1 4 6
1 5 6
2 3 6
2 4 5
2 5 6
3 4 5
3 4 6
Here are the minimal numbers of vertices needed:
Euler characteristic Orientable Non-orientable
2 4
1 6
0 7 8
-1 9
-2 10 9
-3 9
-4 10 10 !Surprise! A
3-hole torus with
10 vertices.
-5 10
-6 11 11
-7 11
-8 12 12
-9 12
-10 12 12
-11 13
-12 13 13