Parametrizing the realization space of a polyhedron by its edges

Question

I alluded to this here, but at that point I hadn't really done enough work to know what I wanted to ask.

Call a polyhedron "trihedral" if three faces meet at each vertex. Each of the F faces can be varied according to three degrees of freedom; these can be written explicitly by fixing an origin not lying in the surface of the polyhedron, and representing each face F_j by the point v_j = (x_j,y_j,z_j) in its plane which is closest to the origin. However, of these 3F degrees of freedom, three correspond to translation and three to rotation; so there are E = 3F-6 degrees of freedom which determine the polyhedron's shape, where E is the number of edges. Also, E remains the "true" number of degrees of freedom even for non-trihedral polyhedra, since each degree of freedom is lost by contracting an edge to a point.

Now, each edge is characterized by its length L_i and dihedral angle θ_i, so to write the corresponding degree of freedom explicitly we need a fixed function f(L,θ). Together with the six values for translation and rotation, the values of f_i = f(L_i,θ_i) for each edge form a function ℝ^3F → ℝ^3F on the realization space of possible shapes, and the key criterion for f is that we want this function to be (at least locally) invertible for all shapes in the space.

Since f_i is a function of L_i and θ_i which in turn can be calculated explicitly from the (x_j,y_j,z_j), we can write the elements of the top E rows of the Jacobian J in the form ∂_Lf∂_xL + ∂_θf∂_xθ etc. So we want to solve |J| ≠ 0 ∀ (x_j,y_j,z_j) in terms of ∂_Lf and ∂_θf. But I've already encountered a problem here: I can't find ∂_xL etc. I've tried by using F_j = {p : p⋅v_j = |v_j|²} to find the co-ordinates of the vertices, and it's a mess. I think I'm missing something.

Any suggestions?

Thanks,

Robin

Answer 1

This is a nice problem is which related to a lot of nice mathematics. You are given a 3-dimensional polytope P and you would like to understand the space of $S(P)$ all gepmetric realization of P. The problem is of interest also in higher dimensions. (Two related miracles for 3-dimensional polytopes are the "Koebe-Abdreev-Thurston" circle packing problem and "Steinitz's theorem".) Indeed some proofs of Steinitz's theorem implies that $S(P)$ is a contractible space wose dimension is the number of edges.

While indeed the number of edges is the "right" dimension for this space, this is not obvious: for simplicial polytopes one can rely on "Cauchy's rigidity theorem". Connelly's flexible sphere demonstrates why the degree of freedom argument can fail. Works on rigidity of polyhedral graphs (Dehn, Alexandrov, and more modern works by Connelly, Whiteley, and many others can be of relevance).

The question is about an explicit parametrization of $S(P)$. I am not aware of an explicit parametization and description. The works of Sabitov and his school and collaborators are highly relevant. Sabitov's "bellow theorem" regarding the invariance of the volume for flexes of simplicial 2-sphere is related to the way the volume of the polytope can be algebraically described by the edge lengths. Sabitov, his students and partners have various additional results even closer to the question but I don't remember right now. (Try also this work by V. Alexander. http://www.springerlink.com/index/J4EXVR63M2QB95PP.pdf)

Answer 2

Just a quick comment: this question reminds me of Andreev's theorem (about the possibility of realizing some hyperbolic polyhedra using some conditions on the dihedral angles). May be the methods used there can help with your particular question?

For some recent accounts, see: "Andreev's theorem on hyperbolic polyhedra", by Roeder, Dunbar, Hubbard, and "a characterization of compact convex polyhedra in hyperbolic 3-space" by Rivin and Hodgson, based on several other papers by Rivin. The original article by Andreev contained apparently a mistake, found by Roeder.

A last remark: connected to the variation of dihedral angles, there is also a nice result that goes under the name of "Schl\"afli's formula" (see on arxiv an article by Rivin and Schlenker)...

Answer 3

In cases like this it might be practical to use the square of the lengths of the edges rather than the lengths. If $e_i$ is the edge between vertices $v_j$ and $v_k$ then the derivative of its square length can be written as $2\langle v_j-v_k, v'_j-v'_k\rangle$, which is quite simple.

Your problem is nice but probably difficult. If you consider only the dihedral angles and only convex polyhedra, you might want to look at a recent result by Mazzeo and Montcouquiol, http://arxiv.org/abs/0908.2981 They prove a rigidity result relevant to your question but their proof uses some real analysis.

Answer 4

Peter Schroeder et al use mean curvature half-density to navigate the shape space of Riemannian surfaces. They also define a discrete counterpart of this object. The latter might be helpful for the problem because, say, a discrete counterpart of total mean curvature is usually expressed through the edge lengths and dihedral angles.