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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?

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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.

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  • $\begingroup$ 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? $\endgroup$ – M. Winter Mar 10 '19 at 14:23
  • $\begingroup$ 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. $\endgroup$ – Gerhard Paseman Mar 10 '19 at 15:52
  • $\begingroup$ @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. $\endgroup$ – M. Winter Mar 10 '19 at 16:28
  • $\begingroup$ 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. $\endgroup$ – Gerhard Paseman Mar 10 '19 at 17:14
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    $\begingroup$ @M.Winter but the facets are fixed as labelled polytopes, do not they? $\endgroup$ – Fedor Petrov Sep 22 '20 at 10:12

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