Timeline for Parametrizing the realization space of a polyhedron by its edges
Current License: CC BY-SA 4.0
11 events
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| Oct 30 at 13:31 | history | edited | M. Winter | CC BY-SA 4.0 |
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| S Apr 17, 2023 at 7:33 | history | suggested | The Amplitwist | CC BY-SA 4.0 |
fixed broken link to emis.ams.org; added full citation in tooltip
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| Apr 17, 2023 at 5:34 | comment | added | The Amplitwist | Reposting a link mentioned in a previous comment so that it appears in the "Linked" questions list: Degree of freedom restricted by inequalities | |
| Apr 17, 2023 at 5:32 | comment | added | The Amplitwist |
The link to springerlink.com is broken. I'm also unable to find any copy saved on the Wayback Machine.
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| Apr 17, 2023 at 5:31 | review | Suggested edits | |||
| S Apr 17, 2023 at 7:33 | |||||
| Jan 24, 2013 at 17:05 | comment | added | Gil Kalai | Dear Hans, I believe the count e already took into account translation and rotations. Namely for the tetrahedron you have 3 times 4 degrees of freedom to locate the vertices and when you subtract 6 for translations and rotations you are left with 6. | |
| Jan 23, 2013 at 17:11 | comment | added | Hans-Peter Stricker | Dear Gil: You say it is not obvious, so where do I find a calculation of $e$ (the number of edges) as the "right" dimension of $S(P)$? If I substract translations and rotations I come up with $e-6$ (as in Richter-Gebert's REALIZATION SPACES OF POLYTOPES, p. 14, if I did understand him correctly). But what does this mean in the case of the tetrahedron with $e=6$? I for myself came up with $e-2$ (+ 7 degrees of freedom for translations, rotations, and scaling), i.e. $e+5$, but that was only a guess. See here mathoverflow.net/questions/119607/… | |
| Jun 30, 2011 at 9:59 | history | edited | Gil Kalai | CC BY-SA 3.0 |
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| Jun 30, 2011 at 9:39 | history | edited | Gil Kalai | CC BY-SA 3.0 |
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| Jun 28, 2011 at 2:12 | history | edited | Gil Kalai | CC BY-SA 3.0 |
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| Jun 27, 2011 at 9:30 | history | answered | Gil Kalai | CC BY-SA 3.0 |