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Let's consider 3-dimensional convex polyhedra $P\subset\Bbb R^3$. The angular defect at a vertex $v$ is $2\pi$ minus the sum of the interior angles of the incident faces at $v$.

Question:
Is a convex polyhedron determined (up to isometry) by its edge graph, edge lengths and angular defects?

If Yes, this would imply Alexandrov's uniqueness theorem ("a convex polyhedron is determined by the metric on its surface"), because from the metric we can read off the edge lengths and angular defects. (it was pointed out to me by Ivan Izmestiev that this would not be as straight-forward because we don't know the edge graph from the metric)

A potential approach is to reduce it to Alexandrov's theorem. Can we reconstruct the metric from the edge lengths and angular defects?

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  • $\begingroup$ I assume you want the edge lengths and angular defects to come with associated combinatorial incidence data (ie which edges connect which vertices), rather than determining the polyhedron solely by the multiset of such lengths and defects? $\endgroup$
    – RavenclawPrefect
    Commented Mar 5 at 20:49
  • $\begingroup$ @RavenclawPrefect Yes, you can assume that the combinatorics of the polyhedron is given. $\endgroup$
    – M. Winter
    Commented Mar 5 at 20:50
  • $\begingroup$ Such a polyhedron with E edges has 2E angles to be determined. Sums of angles inside a face and sums of angles around a vertex are known.These make F+V=E+2 linear equations, which don't seem enough for determining 2E unknowns. Are there other relations that the angles must satisfy? $\endgroup$
    – Pietro Majer
    Commented Mar 5 at 23:03
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    $\begingroup$ @PietroMajer: One also knows the edge lengths, which impose additional constraints. For polyhedra with non-triangular faces, the planarity of the faces is an additional constraint not compatible with some assignments of edge lengths and face angles. (By Cauchy's rigidity theorem for convex polyhedra, we know the answer to the question in the post must be yes for triangular-faced polyhedra, even if we only have the edge lengths and not the defects!) $\endgroup$
    – RavenclawPrefect
    Commented Mar 5 at 23:07
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    $\begingroup$ @PietroMajer The realization space of a 3-dimensional convex polyhedron is a manifold of dimension $|E|+6$ (Steinitz). Thus, in a perfect world, after factoring out the six trivial motions, specifying $|E|$-many parameters should suffice -- i.e. edge lengths. Unfortunately the world is not perfect and edge lengths are not sufficient (think of the cube). Still, from the perspective of counting DOFs, specifying the defects on all vertices feels like an overkill. But it is still not clear whether it suffices. $\endgroup$
    – M. Winter
    Commented Mar 5 at 23:25

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