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Let's say $P\subset\Bbb R^d$ is some convex polytope with the following two properties:

  1. all vertices are on a common sphere.
  2. all edges are of the same length.

I suspect that such a polytope is already rigid, i.e. there is (up to scaling and rotation) only a single way to realize it geometrically. Is this true? What if we only look at uniform polytopes?

M. Winter
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I guess you also fix the combinatorial structure. Then yes. Induct in dimension with obvious base $d\leqslant 2$. By induction proposition the facets are fixed. By Cauchy - - Alexandrov rigidity theorem the whole polytope also is fixed.

Fedor Petrov
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  • Thank you very much. Do you have any source for the rigidity theorem in dimension $d>3$? Also, is there anything obvious to say if I do not fix the combinatorial type, but just the edge-graph? – M. Winter Mar 10 '19 at 14:23
  • Maybe I am missing something, but what makes a (say) triangular prism rigid, especially if it is twisted slightly? Gerhard "Maybe The Definitions Are Tweaked?" Paseman, 2019.03.10. – Gerhard Paseman Mar 10 '19 at 15:52
  • @GerhardPaseman if you "twist" the triangular prism, the square faces are "folded" into two triangular faces each. This is what Fedor meant with fixing the combinatorial structure. Also, you create new edges which are now probably not of the same length as the others. – M. Winter Mar 10 '19 at 16:28
  • Even if you fix the sphere, I have two ways of embedding the prism (assuming the edges are hinged). Is a continuous motion on the sphere surface desired? Gerhard "Trying To See It Mentally" Paseman, 2019.03.10. – Gerhard Paseman Mar 10 '19 at 17:14
  • @GerhardPaseman No, a continuous motion is not required. Which two ways do you see? – M. Winter Mar 10 '19 at 20:18
  • Right hand twist and left hand twist. Gerhard "Baby Let's Do The Twist!" Paseman, 2019.03.10. – Gerhard Paseman Mar 10 '19 at 20:29
  • Wiki page of it seems to have a reference to Alexandrov's work – Fedor Petrov Mar 10 '19 at 22:52
  • @FedorPetrov Not really (or I can't see it). The relevant sentence in Wikipedia has no link to a source, and all the other sources in the end of the article seem to only talk about polyhedra, not polytopes (especially Alexandrov's book). – M. Winter May 20 '19 at 11:26
  • Hm. What is the difference? – Fedor Petrov May 20 '19 at 18:07
  • @FedorPetrov This answer to another question of mine made me think that there might be some details missing in this answer that are not obvious to fix. Cauchy's theorem seems to need more than just congruent faces, and I have no idea what a higher dimensional analogue might need. – M. Winter Sep 22 '20 at 08:12
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    @M.Winter but the facets are fixed as labelled polytopes, do not they? – Fedor Petrov Sep 22 '20 at 10:12
  • @FedorPetrov I don't think that we get a labeling out of my question straight away. Anyway, I still think this approach works after I considered an extended version of Cauchy's theorem (see here, especially the last theorem almost at the bottom of the question). – M. Winter Oct 12 '20 at 10:26