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]