On 12/22/08, Fred lunnon <fred.lunnon@gmail.com> wrote:
... Question: is it possible to actually prove this sort of result; or otherwise to bound the number of points in J(S)?
Third time lucky, I hope! Fred Lunnon
Hmm ... keep hoping! To those hanging on my every word regarding this comedy of errors, a quick update :--- The n-diamond comprises {X_ij | |i|+|j| <= n}, rather than ... < ... The dimension of the companion cone J(S) equals (f+1) - (e-1), so not 2 but 3 --- a result confirmed by another simpler argument, earlier discounted since it seemd meaningful only over continuous ground domains. As a consequence of the fact that J(S) is determined locally by any (sufficiently large) finite subset of the components, its point count is independent of the chosen vertex S; hence it can be computed by examining case of S the wall of the zero sequence (a single row of units), for which easily: The number of lines in J(S), all meeting in the vertex S, equals (p^2-1); The number of points (excluding S) equals (p^2-1)(p-1); The number of walls (properly) companion to S equals (p^2-1)(p-1)^2. Compare the J(S) (inclusive) point count p^3 - p^2 - p + 2, with the 3-flat point count p^3 + p^2 + p + 1; we expect the former to be somewhat reduced on account of the singularity at S. Might J(S) be something quite simple, such as a quadric cone? What about a parametric presentation for it? I have quite a few examples of these things now, but shan't bother posting them unless requested. Fred Lunnon