Timeline for Is there a 4-polytope without 3-gonal and 4-gonal faces, other than the 120-cell?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jul 21, 2020 at 8:28 | vote | accept | M. Winter | ||
| Jul 20, 2020 at 22:21 | vote | accept | M. Winter | ||
| Jul 20, 2020 at 22:49 | |||||
| Jul 20, 2020 at 22:15 | answer | added | M. Winter | timeline score: 0 | |
| May 24, 2020 at 0:25 | history | became hot network question | |||
| May 23, 2020 at 18:17 | comment | added | M. Winter | I found this highly relevant question with an equivalently relevant answer. | |
| May 23, 2020 at 16:57 | comment | added | M. Winter | @BrianHopkins I still don't have a source, but I realized the following: one can show (via a standard double counting arguments) that a planar graph has a vertex of degree 5 or smaller, or equivalently (considering its dual), a 3-gonal, 4-gonal or 5-gonal face. Since the edge-graph of a (3-dimensional) polyhedron is a planar graph, this proves it in dimension three. This then carries over to higher dimensions by considering the 3-faces of the polytopes. | |
| May 23, 2020 at 16:54 | comment | added | Brian Hopkins | Did you find who proved that the only possible faces are 3-, 4-, or 5-gonal? (An earlier version requested a citation for that result.) | |
| May 23, 2020 at 16:54 | answer | added | Dmitri Panov | timeline score: 8 | |
| May 23, 2020 at 16:50 | history | edited | M. Winter | CC BY-SA 4.0 |
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| May 23, 2020 at 16:41 | comment | added | M. Winter | @YCor Yes, thanks. I edited that into the question. | |
| May 23, 2020 at 16:41 | history | edited | M. Winter | CC BY-SA 4.0 |
added 105 characters in body
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| May 23, 2020 at 16:38 | comment | added | YCor | "other than" means: not isomorphic as polyhedral complex? (this is a reasonable isomorphism notion; an a priori stronger one would be being isotopic, i.e., have a continuous deformation from one to another) | |
| May 23, 2020 at 16:25 | history | asked | M. Winter | CC BY-SA 4.0 |