From: Nick Halloway Newsgroups: sci.math.research,sci.math Subject: Re: Flexheptagons, flexoctagons, etc. (123) Date: Wed, 17 Sep 1997 23:59:59 -0700 I asked earlier whether it is possible to embed in 3-space a finite triangulation of a (many-holed) torus, that has the same number of triangles meeting at each vertex, with each face an equilateral triangle. You can if you don't mind triangles lying flat against each other -- triangulations of many-holed tori with 8 triangles meeting at a vertex can sit in 3-space. I'm not sure yet if it's possible to make the torus enclose volume. You can make a 1-holed torus triangulated with 6 equilateral triangles at a vertex enclose volume -- just make a circular chain of tetrahedra, each with 4 triangles to a face; join the edge 12 of the first tetrahedron to the edge 34 of the second, edge 12 of the second to edge 34 of the third ... edge 12 of the 2n'th to edge 34 of the first. Here's how you can make the flattened triangulations with 8 triangles meeting at a vertex. Squares can be arranged in space with 3, 4, 5, 6 triangles at a vertex (maybe more). With 3 you get a cube; with 4 you get a plane. With 5 you get an infinite lattice. The lattice looks like this: Put a cube with sides of length 1 over each integer point in the plane. Then take away the cubes that are over (x,y) with x and y even. With 6 cubes at a vertex you get a 3-dimensional infinite sponge, similar to the infinite planar lattice. Now, cut off the corners of each square to get an octagon. The vertices are converted into triangular holes if it's a cube, square holes if it's planar, pentagonal holes for the infinite lattice and hexagonal holes for the infinite sponge. Now divide the octagons into 8 triangles each by adding a vertex at the center of each octagon. Four triangles meet at each vertex of a hole. Now warp the polyhedra by converting the isoceles triangles to equilateral triangles. This creates an angle surplus at the center of each octagonal face and subtracts angle at the holes, but doesn't change the total curvature of the polyhedra. So now one can see whether these warped polyhedra can sit in 3-space. For the infinite polyhedra, one just uses the smallest repeating unit. The warped cube may sit in 3-space, I'm not sure. It forms a tangle :) The warped plane makes a nice torus -- you can use 8 octagons to make a 4-octagon long flexible tube, which wraps around to meet itself. This has 8 square holes. The warped infinite lattice forms a tangle which I'm not sure embeds into 3-space. It has 10 octagons, 8 pentagonal holes. The warped infinite sponge makes a nice 2-holed torus with 4 hexagonal holes. This uses 6 octagons. Each hole is only one edge away from other holes. So, we can't just seal up the holes by glueing two copies of the polyhedra at the holes, because we'd get a degenerate triangulation. But, the graphs of the vertices of the original polyhedra that were made with squares can be 2-colored. So, we can seal up the holes by glueing four sheets of the polyhedra at the holes. Say the holes are labelled red and blue. Then, at the red holes glue sheets 1 and 2 together, and sheets 3 and 4 together. At the blue holes, glue sheets 2 and 3 together, and sheets 1 and 4 together. The sheets can separate somewhat, at least in the case of the warped plane. But they don't seem to be able to enclose volume, I wonder if there is a general principle behind this. There are some very symmetric combinatorial polyhedra with 12 vertices and 7 triangles meeting at a vertex. This one was especially nice: it has vertices 0, ..., 11 and around vertex 0, the vertices 1,5,6,11,4,8,7. You get the vertices around vertex n by adding n to each of the vertices around 0, mod 12. This, and one other polyhedron with 12 vertices, 7 triangles at a vertex, can be assembled in 3D with equilateral triangles, with 2 cuts. However, they aren't flexible and I don't think they can be used as building blocks to make a finite closed surface. I wonder if it's possible to make polyhedra with 7 triangles at a vertex so that they form a flexible branched tube, which might bend around to meet itself. A convex triangular framework is rigid, so if the tube is a union of convex frameworks, it's rigid.