11
$\begingroup$

(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 $\pi$.

Improve this question
$\endgroup$
11
  • $\begingroup$ 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. $\endgroup$
    – RavenclawPrefect
    Commented Mar 1 at 0:28
  • $\begingroup$ 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. $\endgroup$
    – RavenclawPrefect
    Commented Mar 1 at 20:53
  • $\begingroup$ 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. $\endgroup$
    – Timothy Chow
    Commented Mar 2 at 13:27
  • $\begingroup$ @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. $\endgroup$
    – RavenclawPrefect
    Commented Mar 2 at 20:18
  • 1
    $\begingroup$ Related by duality: mathoverflow.net/questions/422290 $\endgroup$
    – Ilya Bogdanov
    Commented Mar 5 at 14:12

1 Answer 1

Reset to default
7
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ “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)… $\endgroup$
    – Ilya Bogdanov
    Commented Mar 18 at 6:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .