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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 Fj by the point vj = (xj,yj,zj) 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 Li and dihedral angle θi, 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 fi = f(Lii) for each edge form a function ℝ3F → ℝ3F 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 fi is a function of Li and θi which in turn can be calculated explicitly from the (xj,yj,zj), 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 ∀ (xj,yj,zj) 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 Fj = {p : pvj = |vj|2} to find the co-ordinates of the vertices, and it's a mess. I think I'm missing something.

Any suggestions?

Thanks,

Robin

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Do you want to characterize all admissible $f$ or do you want just to find one that works for all polytopes? –  fedja Jun 24 '11 at 11:59
    
The latter, although the former would be nice since I might want to put some extra constraints on it later. –  Robin Saunders Jun 24 '11 at 12:57
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4 Answers

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 relies on "Cauchy's rigidity theorem". Connelly's flexible sphere demonstrates why the degree of freedom argument can fail. Works on rigiditi 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" ( http://www.emis.ams.org/journals/BAG/vol.38/no.1/1.html ) regarding the invariance of the volume for flexes of simplicial 2-sphere is related to the way the colume 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 dont remember right now. (Try also this work by V. Alexander. http://www.springerlink.com/index/J4EXVR63M2QB95PP.pdf /)

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Dear Gil: You say it is not obvious, so where do I find a calculation of $e$ (the number of edges) as the "right" dimension of $S(P)$? If I substract translations and rotations I come up with $e-6$ (as in Richter-Gebert's REALIZATION SPACES OF POLYTOPES, p. 14, if I did understand him correctly). But what does this mean in the case of the tetrahedron with $e=6$? I for myself came up with $e-2$ (+ 7 degrees of freedom for translations, rotations, and scaling), i.e. $e+5$, but that was only a guess. See here mathoverflow.net/questions/119607/… –  Hans Stricker Jan 23 '13 at 17:11
    
Dear Hans, I believe the count e already took into account translation and rotations. Namely for the tetrahedron you have 3 times 4 degrees of freedom to locate the vertices and when you subtract 6 for translations and rotations you are left with 6. –  Gil Kalai Jan 24 '13 at 17:05
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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)...

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Wait, I have to mention that one: "Shapes of polyhedra and triangulations of the sphere" by W. Thurston, who covers Andreev's theorem in his Notes (msri.org/publications/books/gt3m ) –  Sylvain Bonnot Jun 24 '11 at 14:48
    
Thanks for your response. I've seen Thurston's paper before, in fact. It has in common with other literature I've found that it deals specifically with convex triangulated polyhedra. These are "nice" in the sense that they are determined completely by their edge-lengths, and each vertex has three degrees of freedom; but somehow they are the exact opposite of what I am studying. Trihedral polyhedra are dual to triangulated ones. That in itself might not be an issue, but I have no canonical surjection from triangulated to trihedral polyhedra. Also, I don't want to demand convexity. –  Robin Saunders Jun 25 '11 at 17:25
    
Oh I see, you allow non convex ones... There is one interesting point happening then: imagine one such surface with a protruding tetrahedron,and suddenly you push this tetrahedron inside (from a bump it switched to a dent), then lengths and dihedral angles are the same. How would such a phenomenon be reflected in the parameter space? Perhaps the whole parameter space could be thought as a covering space of the space recording the lengths and dihedral angles, and perhaps this operation of "creating a dent" corresponds to a jump from one sheet of the covering space to another? –  Sylvain Bonnot Jun 25 '11 at 20:40
    
The inner dihedral angles are not the same: they are complementary. –  fedja Jun 25 '11 at 23:16
    
The dihedral angles change sign, so as long as f is not locally an even function of θ everything should be fine. Even if the polyhedron is non-orientable so that there is no global sign for θ, it is still meaningful to say that it changes sign locally, so that the Jacobian at the transition can distinguish between the two directions. –  Robin Saunders Jun 25 '11 at 23:19
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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.

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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.

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