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(Originally on MSE.)

Suppose $P$ and $Q$ are combinatorially equivalent non-self-intersecting polyhedra in $\mathbb{R}^3$, with $f$ a map from edges of $P$ to edges of $Q$ under said combinatorial equivalence. Let $\theta(e)$ be the dihedral angle of edge $e$.

If it is the case that for all $e\in D$, $\theta(e)\le \theta(f(e))$, must we have $\theta(e)=\theta(f(e))$ for all $e$?

That is, is it possible to distort the geometry of $P$ such that no dihedral angle decreases, and at least one dihedral angle increases?

My intuition is that this should not be possible, that there is some conserved notion of "total curvature" which would be violated by such an operation. But I can't seem to prove any such invariant (the sum certainly isn't one, for instance).

I'm also interested in the restriction to the convex case, ie where every dihedral angle is less than $\pi$.

RavenclawPrefect
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  • One approach to this problem that seems promising is to consider the spherical graph in which vertices are the normal vectors to faces, and edges are between any two adjacent faces. Then we're asking about a distortion of this graph in which every edge length increases. Frustratingly, there do exist graphs of polyhedra on the sphere which can be distorted so every edge length increases - it's just that the ones I've found so far aren't valid adjacency graphs after the distortion, because the faces have moved far enough apart that they don't share an edge on the resulting polyhedron anymore. – RavenclawPrefect Mar 01 '24 at 00:28
  • In any such transformation, all the solid angles at the vertices of the polyhedron will also have to increase, but that is possible in general - compare the tetrahedron whose vertices are at $(\pm 1, 0, \epsilon)$ and $(0,\pm 1, -\epsilon)$ with the regular tetrahedron. – RavenclawPrefect Mar 01 '24 at 20:53
  • I think this follows from the discrete Gauss-Bonnet theorem (not sure of the best reference for discrete Gauss-Bonnet, but Google suggests this or this). The angular defect of a vertex is usually defined in terms of the interior angles of the polygonal faces meeting at the vertex, but it's also related to the dihedral angles between the faces meeting at the vertex. – Timothy Chow Mar 02 '24 at 13:27
  • @TimothyChow: This seems related to the approach I mentioned above, using the spherical graph of normal vectors. I confess I'm still not seeing the route to a proof here - discrete Gauss-Bonnet prevents the areas of the faces of this graph from all growing simultaneously, but we care about its edge lengths, and it's possible for all edge lengths in a spherical planar graph to increase. – RavenclawPrefect Mar 02 '24 at 20:18
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    Related by duality: https://mathoverflow.net/questions/422290 – Ilya Bogdanov Mar 05 '24 at 14:12
  • @TimothyChow : According to one of my high-school teachers, that is "Descartes's theorem." The defect of vertex is the amount by which the sum of the angles between edges at that vertex falls short of a full circle. It is negative if the sum exceeds a full circle. The theorem says that in a polyhedron homeomorphic to a sphere, the sum of the defects is two full circles. – Michael Hardy Mar 05 '24 at 15:59
  • @TimothyChow : Can you explain the connection between$ \uparrow $that and the dihedral angles? – Michael Hardy Mar 05 '24 at 16:00
  • @TimothyChow imagine a cube with the top square deleted. Put a new vertex at the center of the cube and let edges connect it with the four upper vertices of the cube, getting a polyhedron with five square faces and four triangular faces. Contrast that with a cube with a square pyramid replacing the upper square face. Note that the intrinsic geometries of these two surfaces are exactly the same: if distances are measure along the surface, then there is an isometry between these two. But the dihedral angles differ. – Michael Hardy Mar 05 '24 at 16:04
  • @TimothyChow : (And in that first-describe polyhedron, with one vertex at the center of the cube, each vertex where two squares and two triangles meet has a positive defect, even though a vertex with that shape can occur only in a non-convex polyhedron. A counterexample to an obvious naive guess.) – Michael Hardy Mar 05 '24 at 16:08

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From Igor Pak's Lectures on Discrete and Polyhedral Geometry (link to PDF, table of contents here) we have

Theorem 28.3 (Schläfli formula). For a continuous family of [combinatorially identical] polyhedra $\{P_t:t\in [0,1]\}$ preserving the faces, we have: $$\sum_{e\in E}\ell_e(t)\cdot \theta'_e(t) =0$$ where $\ell_e(t)$ are edge lengths and $\theta_e(t)$ are dihedral angles.

Since all edge lengths are positive this implies the desired result for any local infinitesimal transformation of a polyhedron. I don't see how to extend this to the general case of any two combinatorially equivalent $P$ and $Q$, but it's substantial progress.

RavenclawPrefect
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  • “A substantial progress” —- I hope in the same way as Csikos’ result was a progress towards Kneser—Poulsen (which was much later solved by Bezdek and Connelly i; the plane)… – Ilya Bogdanov Mar 18 '24 at 06:56