Subject: Re: two on proj plane
Date: Mon, 16 Sep 2002 15:57:20 PST
From: Kevin Iga <kiga@pepvax.pepperdine.edu>

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