I asked this question in a comment to this question, but got no response. I thought that perhaps it needed more exposure, so I made it a question in itself.

Define a set of general position vectors $v_i$ given in the following way:

Consider a convex $n$-polytope, $P_n$, with $f_i$ the set of $(n-1)$-polytopes "faces".

Associate the vector $v_i$ to the face $f_i$ such that $\left|\left|v_i\right|\right|=V_{f_i}^{n-1}$ the $(n-1)$-volume of $f_i$. Each vector $v_i$ has basepoint given by the center of mass of the associated face. $f_i$ and is normal to that face.

I have two questions:

  1. Is this notation strange? Are these notions defined elsewhere that I don't know about and perhaps I should reference?
  2. Now consider a triangulation of $P_n$ into $n$-tetrahedron, given by $T^j$. Each of these tetrahedron has a set of general position vectors $\lbrace v^j_i\rbrace$. Can we determine these general position vectors from the initial general position vectors? In particular can we characterize triangulations of this type by relations on the general position vectors of the tetrahedron?

Note: In case you are wondering the motivation of this question, let me make it a little more clear. I was looking at this paper by John Baez, following the work of Barrett-Crane on the same subject, and in particular the construction of constraints for face-bivectors of tetrahedron on page 4 of the latter.


To the question 1: If your vectors are perpendicular to the correspondent faces, then you have a situation from the Minkowski theorem about polyhedra(you can find it in the Lectures on Discrete and Polyhedral Geometry by Igor Pak, theorem 36.2):

Let $u_1, . . . , u_n \in R^d$ be vectors which span $R^d$. Then $u_1 + . . . + u_n = 0$ if and only if there exists a unique convex polytope $P ⊂ R^d$ with n facets $F_1, . . . , F_n$, such that $area(F_i) = |u_i|$ and the normal to $F_i$ is $u_i$, for all $1\le i \le n$.

  • $\begingroup$ Yes, thank you! Somehow I missed the normal vector condition from the original posting. I have updated the question to reflect this change. Also, thank you for this reference. This does indeed answer my first question and possibly the second, but I need to think a little first. $\endgroup$ – B. Bischof Dec 9 '09 at 2:46

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