Subject: Re: two on proj plane Date: Mon, 16 Sep 2002 15:57:20 PST From: Kevin Iga Another comment about the projective plane: One approach to triangulating the projective plane is to triangulate the two-sphere, and consider the projective plane as the two-sphere, with opposite points identified. There are many triangulations of the sphere, but the most famous are the platonic solids (using the ones with only triangles): the tetrahedron, the octahedron, and the icosahedron. The tetrahedron will not serve our purpose since the point on the opposite side of a vertex is not a vertex, but lies in the middle of a face. Thus, we cannot identify opposite points on a tetrahedron and meaningfully retain the triangulation. The octahedron also will not work, since identifying opposite points would yield triangles that intersect in a vertex and an edge, which, as David Green mentioned, is not allowed by definition. This is what you would get if you merged the four inner triangles into a single triangle, and if you merged pairs of triangles that would then be suggested by the edges of this inner triangle: 5 and 10, 6 and 7, 8 and 9. This is perhaps the "triangulation" that you were imagining, but as David Green pointed out earlier, it doesn't fit the definition of a polyhedron. The icosahedron does work, and identifying opposite points on it will give rise to the diagram in Alexandroff. Kevin Iga