Timeline for Is the dodecahedron flexible (as a polytope with fixed edge-lengths)?
Current License: CC BY-SA 4.0
18 events
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| Aug 19 at 3:57 | comment | added | Ian Agol | If convexity is removed, then the endo-dodecahedron is an example of an equilateral dodecahedron. mathworld.wolfram.com/Endododecahedron.html I’m guessing if non-Platonic convex equilateral dodecahedra exist, they would have been discovered by now. Nevertheless it is a nice question. | |
| Jan 14, 2023 at 11:26 | comment | added | Ivan Izmestiev | Yes, it should be added in my previous comment that P and Q are simplicial. The dodecahedron becomes "simplicial" after a subdivision of faces by diagonals. Your requirement that the faces remain planar can be translated as the angles at the new edges staying equal to $2\pi$. | |
| Jan 14, 2023 at 8:42 | comment | added | M. Winter | @IvanIzmestiev Thank you Ivan for this interesting comment. When you say that Cauchy solves the "length only" case, you mean for simplicial polytopes, right? The Oberwolfach question is about general polytopes? One should note that polytopes with known edge length can still be deformed beyond affine equivalence, such as zonotopes. Perhaps of interest: this article from 2022 claims to have solved the general Stoker's conjecture across all polytopes in all dimensions. | |
| Jan 12, 2023 at 17:33 | comment | added | Ivan Izmestiev | This is a special case of the following question from a question session in Oberwolfach about 12 years ago. Let P and Q be two combinatorially equivalent convex polyhedra. Assume that for any pair of corresponding edges either their lenghts are equal or the dihedral angles are equal. Does this imply that the polyhedra are congruent (or similar, if there are no equal lengths conditions, only equal angles). The pure length condition is the (Cauchy) rigidity, the pure angle condition is known as the Stoker conjecture, its infinitesimal version was solved 2009 by Mazzeeo and Montcouquiol. | |
| Nov 18, 2022 at 18:19 | comment | added | Wlodek Kuperberg | Nice question. Bob Connelly <[email protected]> may know the answer or at least have some idea of how to approach your question. | |
| Nov 18, 2022 at 10:55 | comment | added | M. Winter | @RavenclawPrefect This seems reasonable. I have seen the tetartoid which looks like its edges are of unit length, but I don't know. There is also this deformation, which is suggestive but does not work. Maybe some symmetry can be retained, but I don't care too much. | |
| Nov 18, 2022 at 8:25 | comment | added | RavenclawPrefect | As some evidence against, I believe this fails if you try to retain $S_3$ symmetry about a particular vertex (whereas in the cube case one can deform things in a way that preserves bilateral symmetry of all faces and keeps things symmetric relative to a particular vertex projection). | |
| Nov 17, 2022 at 20:41 | comment | added | M. Winter | @DanielAsimov It was certainly my intention to keep them planar and I believe that the listed constraints capture this: if the vertices of a face become non-coplanar then their convex hull has more edges now, which changes the edge-graph. I could have stated this more explicitly though. | |
| Nov 17, 2022 at 18:57 | comment | added | Daniel Asimov | A priori it seems that the pentagonal faces are not required to remain planar, which is not explicitly one of the constraints. Must they remain planar anyway? | |
| Nov 17, 2022 at 15:37 | comment | added | M. Winter | @TheoJohnson-Freyd I haven't, and I think it is tricky for exactly the flatness reason. :) I might try a physics simulation. But currently I am still hoping for a more systematic approach. I admit, the dodecahedron is just a placeholder for any polytope that feels like "it should have a flex" and I hope to get some broader understanding already from this discussion. | |
| Nov 17, 2022 at 15:28 | comment | added | Theo Johnson-Freyd | Have you tried building a physical model? I mean with actual sticks and elastics? Ah, you want the faces to stay flat. So that’s a harder engineering puzzle. But I still think you should try to build a toy and test it. | |
| Nov 17, 2022 at 15:23 | comment | added | Joseph O'Rourke | Apologies for misunderstanding. | |
| Nov 17, 2022 at 15:16 | comment | added | Wojowu | @JosephO'Rourke That theorem only applies if you require the faces to keep their shape. I don't think this is what OP wants. I think what they are asking about is whether there is something similar to how you can deform a cuboid into parallelepipeds. | |
| Nov 17, 2022 at 15:09 | comment | added | Joseph O'Rourke | Well then Cauchy's rigidity theorem applies: all convex polyhedra are rigid. I must still be misunderstanding... | |
| Nov 17, 2022 at 14:52 | history | edited | M. Winter | CC BY-SA 4.0 |
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| Nov 17, 2022 at 13:43 | comment | added | M. Winter | @Joseph Note that this is not rigidity of frameworks. You are forgetting to count the constraints that keep the faces flat. If my calculations are correct then you get exactly as many constraints as DOFs (for every 3-polytope). | |
| Nov 17, 2022 at 13:39 | comment | added | Joseph O'Rourke | Rigidity requires the graph to have at least $3n-6$ edges. Here $n=20$ and the dodecahedron has $30$ edges. So I think there are not enough edges to make it rigid. | |
| Nov 17, 2022 at 13:21 | history | asked | M. Winter | CC BY-SA 4.0 |
