From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: How do you make a polyhedral 2-holed torus
Date: 28 Feb 1995 16:33:59 GMT
rusin@olympus.math.niu.edu (Dave Rusin) writes:
>
>What is the minimal number of vertices in a polyhedron in R^3 which
>is homeomorphic to the 2-holed torus? (It's either 9 or 10).
LAURAHELEN@DELPHI.COM wrote:
>Sheesh, how do you manage it with 10 vertices?
I found (by accident one day) a 7-vertex form of the 1-holed torus in
A. Cs\'asz\'ar, "A polyhedron without diagonals",
Acta Sci Math Szeged 13 (1949) 140-142,
adjusted it lightly (he had a "15" where I use "5"), and glued two together
to make a 10-vertex 2-holed torus, below.
If you make a polyhedron with F triangles, it will have E = 3F/2 edges
(assuming every edge is the meet of precisely two faces). Of course the
number of edges is at most (V choose 2), where V is the number of
vertices. On the other hand, Euler's formula is V - E + F = 2-2g, where
g is the genus (=# of holes). Then E = 3(V-2+2g) and E<= V(V-1)/2 force
V >= (7+sqrt(1+48 g))/2.
For g=0 (topologically a sphere) this says V>=4. A tetrahedron
shows V=4 is possible. Then E = 3(V-2+2g) = 6 and F= (2/3)E = 4.
For g=1 (topologically a torus) this says V>=7. Csaszar 's
construction shows V=7 is possible. Then E=3(V-2+2g)=21, F = (2/3)E=14.
(If you want to make a model of this one but can't figure out how to
cut the cardboard, write me for details.) This construction, by the way,
shows you need at least 7 colors to color a graph on a torus, since all
7 vertices are joined by edges to each other.
For g=2 (the 2-holed torus) we need V>=8.4..., so at minimum we
will need 9 vertices. If V=9 is possible, it will require
E=3(V-2+2g)=33 edges and F=(2/3)E=22 faces. On the other hand, V=10 is
sufficient as I will show below. Here E=3(V-2+2g)=36 and F=(2/3)E=24.
For g=3,4,5,6 the minimum number of vertices is V=10,11,12,12. It
seems that the minimum is not going up very fast, so I should think it
unlikely that the corresponding polyhedra can be constructed. But then,
my intuition was that V=10 was impossible for g=2, too. At g=6, we have
a perfect square for 1+48g, so the minimal configuration would once again
have every pair among the 12 vertices joined by one of 66 edges.
dave
2-holed torus with 10 vertices:
Save the data below to a file "2torus", fire up Mathematica, and
type " <<2torus " as your first (and only) input. You'll see 2 stills and a
a spinning version of the 2-holed torus. The input should be comprehensible
to a non-machine, too.
A1={ 0, 0, 5}
A2={ -1, -2, 2}
A3={ 1, 2, 2}
A4={ -3, -3, 1}
A5={ 3, 3, 1}
A6={ -3, 3, 0}
A7={ 3, -3, 0}
A8={ -1, 2, -1}
A9={ 1, -2, -1}
A10={ 0, 0, -4}
P1=Polygon[{ A6, A4, A3}]
P2=Polygon[{ A4, A2, A5}]
P3=Polygon[{ A2, A3, A7}]
P4=Polygon[{ A3, A5, A1}]
P5=Polygon[{ A7, A1, A4}]
P6=Polygon[{ A1, A6, A2}]
P7=Polygon[{ A6, A2, A3}]
P8=Polygon[{ A4, A3, A5}]
P9=Polygon[{ A2, A5, A7}]
P10=Polygon[{ A3, A7, A1}]
P11=Polygon[{ A5, A1, A6}]
P12=Polygon[{ A1, A4, A2}]
P13=Polygon[{ A4, A6, A9}]
P14=Polygon[{ A6, A8, A7}]
P15=Polygon[{ A8, A9, A5}]
P16=Polygon[{ A9, A7, A10}]
P17=Polygon[{ A5, A10, A6}]
P18=Polygon[{ A10, A4, A8}]
P19=Polygon[{ A4, A8, A9}]
P20=Polygon[{ A6, A9, A7}]
P21=Polygon[{ A8, A7, A5}]
P22=Polygon[{ A9, A5, A10}]
P23=Polygon[{ A7, A10, A4}]
P24=Polygon[{ A10, A6, A8}]
T={P1, P2, P3, P4, P5, P6, P7, P8, P9 , P10, P11, P12, P13, P14, P15, P16, P17, P18, P19, P20, P21, P22, P23, P24}
H:=Graphics3D[T]
Show[H, ViewPoint ->{ 1.5, 1.3, 0}]
Show[H, ViewPoint ->{ 1.5,-1.3, 0}]
<{ 0, 0, 0}, Frames ->100]