Timeline for Why do some uniform polyhedra have a "conjugate" partner?
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 17, 2018 at 19:05 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Details
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| Jan 16, 2018 at 23:14 | answer | added | Dr. Richard Klitzing | timeline score: -1 | |
| Dec 12, 2017 at 2:35 | vote | accept | Tito Piezas III | ||
| Dec 11, 2017 at 8:30 | answer | added | j.c. | timeline score: 10 | |
| Dec 11, 2017 at 5:28 | answer | added | Igor Pak | timeline score: 17 | |
| Dec 11, 2017 at 4:41 | comment | added | Tito Piezas III | @j.c.: I've added the list of 26 "conjugates" below. | |
| Dec 11, 2017 at 4:38 | answer | added | Tito Piezas III | timeline score: 3 | |
| Dec 10, 2017 at 19:25 | comment | added | Sylvain JULIEN | 26? As the number of sporadic finite simple groups? Is it a mere coincidence? | |
| Dec 10, 2017 at 17:44 | comment | added | j.c. | Can you post your list of 26 "conjugate" pairs of polyhedra, labeled by the Wythoff symbols and their ordering in the MathWorld site? There seem to be some interesting patterns in the two pairs you've shown and I'm curious to see what happens more generally. | |
| Dec 10, 2017 at 16:34 | comment | added | Noah Snyder | If all the vertices have coordinates in a real field K, then the K-linear combos of them will also lie in K, and that means a dense subset of every edge and every face are in K. So you can just apply conjugation to all those points and get a dense subset of the other polyhedron. | |
| Dec 10, 2017 at 15:42 | comment | added | Sylvain JULIEN | Would it be due to the fact that the isommetry group of those polyhedra are actually Galois groups of some field extension whose considered circumradii are elements? | |
| Dec 10, 2017 at 15:31 | comment | added | Tito Piezas III | @NoahSnyder: I've afraid I wouldn't know. Some pairs have the same convex hull, but the pair $U_{32}$ and $U_{72}$ do not. The former has a truncated icosahedron, while the latter has a truncated dodecahedron. | |
| Dec 10, 2017 at 15:22 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Pictures
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| Dec 10, 2017 at 14:41 | comment | added | Noah Snyder | Can they be positioned so that you just apply Galois conjugation to the vertices of one polyhedron to get the other? | |
| Dec 10, 2017 at 13:39 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Galois
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| Dec 10, 2017 at 13:32 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Fixed typo
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| Dec 10, 2017 at 13:28 | comment | added | Tito Piezas III | Because of this observation, I found two erroneous entries in Mathworld (for $U_{53}$ and $U_{61}$). Accurate ones can be found in David McCooey's Visual Polyhedra. | |
| Dec 10, 2017 at 13:19 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Another pair for more detail
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| Dec 10, 2017 at 12:49 | history | asked | Tito Piezas III | CC BY-SA 3.0 |