Timeline for Are there polytopes with precisely two realizations?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 27 at 22:01 | comment | added | M. Winter | @RavenclawPrefect That sounds like an affine transformation. A projective transformation is more general. | |
| Aug 27 at 1:54 | answer | added | Karim Adiprasito | timeline score: 1 | |
| Aug 26 at 22:59 | comment | added | RavenclawPrefect | I believe so? 3 DOFs for the location of the origin, then 9 more for the 3x3 matrix specifying a linear transformation, makes 12. | |
| Aug 26 at 22:37 | comment | added | M. Winter | @RavenclawPrefect Are you sure you counting correctly the DOFs of projective transformations including translations? | |
| Aug 26 at 21:05 | comment | added | RavenclawPrefect | Whoops - I miscounted and should have $15$ DOFs for both the prism and the triangular bipyramid, but this doesn't resolve my confusion. (The triangular bipyramid is easy to count DOFs for: it is combinatorially equivalent under small perturbations of any of its vertices, so we have 3 DOFs per vertex times 5 vertices.) I don't think the triangular bipyramid has a unique realization either, though? I don't think there are projective transformations which fix 4 out of 5 vertices and let the fifth move freely, at least in the generic case. | |
| Aug 26 at 17:48 | comment | added | M. Winter | @RavenclawPrefect The easiest way for me to see that it is projectively unique is to use that it is dual to the triangular bipyramid which has five vertices which can be arranged arbitrarily by projective transformations. The realization space of a polytope and its dual are "the same". But I am not exactly sure where the mistake lies. How do you count DOFs for the prism? | |
| Aug 26 at 16:35 | comment | added | RavenclawPrefect | Why is the triangular prism projectively unique? It seems like it has $14$ degrees of freedom, compared with the $12$ offered by projective transformations. (Concretely, I think the prism with vertices $\{(0,0,0),(1,0,0),(0,1,0),(0,0,1),(1,0,1),(0,1,1)\}$ and the prism where we replace the $(0,1,1)$ vertex with $(0,1,2)$ are combinatorially but not projectively equivalent.) Am I misunderstanding the definition? | |
| Aug 26 at 10:50 | history | asked | M. Winter | CC BY-SA 4.0 |