Ken Baker has posted some pictures of the one tetrahedron triangulation of the solid torus to his blog. This is a surprisingly hard construction to visualize, given how simple it is. (Ken showed me the 3D models at the Georgia topology conference and after ten minutes of looking at them from different angles, I still couldn’t quite see it.) Start with a tetrahedron and choose two of the sides (it doesn’t matter which two.) There are three orientation reversing ways to glue the two sides together (an orientation preserving gluing would produce a non-orientable manifold) – You can just fold them over the common edge, or you can glue them by a 2pi/3 rotation clockwise or counter-clockwise. The first gluing gives you a 3-ball. The other two produce a solid torus, and one of these is shown in Ken’s pictures. (The blue triangles in the second picture show a Mobius band formed by the two glued triangles.)
This triangulation has a single vertex, which is in the boundary of the solid torus. The boundary is triangulated with two triangles (the two that aren’t glued to anything). One can produce other triangulations by taking another tetrahedron and gluing two of its faces to the two faces in the boundary. This produces a new solid torus with a different triangulation of its boundary. The process can be repeated to produce any desired triangulation of the boundary. The resulting triangulation of the solid torus is called a layered triangulation. Gluing the last two faces to each other produces a lens space, and the induced triangulation, also called a layered triangulation, is studied in a paper by Jaco, Rubinstein and Tillman that just hit the arXiv a few days ago . They show that for a certain family of lens spaces, the minimal layered triangulation is in fact minimal among all triangulations of the lens space. They conjecture that this is true for all lens spaces, so I’m adding that to the open problem list as a question.