Timeline for A.D. Alexandrov imbedding theorem for metrics with symmetry
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Aug 4, 2020 at 15:44 | comment | added | mr_e_man | Perhaps I should have searched more before asking. Wikipedia points to Proofs from THE BOOK, which shows that the index can't be $0$ or $2$. This completes the argument from my first link. | |
| Aug 3, 2020 at 4:29 | comment | added | mr_e_man | @IvanIzmestiev - "The rigidity of convex hyperbolic polyhedra can be proved". Can you expand on this? What is the classical argument? I found this, where the index argument looks like what you're referring to. It's after Equation (4.1). I guess $\omega_{ij}$ should instead be the change in dihedral angle, and $(\mathbf p_i-\mathbf p_j)$ should be an edge vector in the tangent space at $\mathbf p_i$. But Eq(4.1) still doesn't make sense, since the edge vectors may also change. How should the argument be modified? | |
| Aug 11, 2019 at 6:25 | comment | added | Ivan Izmestiev | Yes, it is not explicitly stated if $n=0$ is allowed, and the proofs are only indicated. | |
| Aug 10, 2019 at 11:56 | comment | added | asv | However it is not quite clear to me from the statement whether the case of closed surfaces is a special case of Theorem 4. In other words can one assume that the number of boundary components $n=0$? | |
| Aug 10, 2019 at 11:51 | comment | added | asv | Meantime I found a paper by Danelich (in Russian) where he announces Theorem 4) the uniquness result for hyperbolic space (no proof is presented in Russian tradition):mathnet.ru/links/e413e4a85fd1bec9e1d0c4013398bf1e/dan22149.pdf | |
| Aug 10, 2019 at 8:42 | comment | added | Ivan Izmestiev | Yes, from the last section of my article "Projective background of the infinitesimal rigidity of frameworks" or Jean-Marc Schlenker's "Hyperbolic manifolds with convex boundary". | |
| Aug 10, 2019 at 8:33 | comment | added | asv | Can I learn somewhere Pogorelov map? | |
| Aug 10, 2019 at 8:30 | comment | added | Ivan Izmestiev | There is a more artificial but elegant way to prove this: the Pogorelov map. It associates to any pair of isometric hyperbolic polyhedra a pair of isometric Euclidean polyhedra. | |
| Aug 10, 2019 at 8:28 | comment | added | Ivan Izmestiev | Probably this is contained as a remark in the Alexandrov book. But when stating this in an article I would just say that the classical argument in the Euclidean case carries over. | |
| Aug 10, 2019 at 8:26 | comment | added | asv | Thank you. I am wondering if there is a reference to the hyperbolic case. | |
| Aug 10, 2019 at 8:23 | comment | added | Ivan Izmestiev | The rigidity of convex hyperbolic polyhedra can be proved in exactly the same way as that of Euclidean polyhedra: compare the dihedral angles in two realizations, then there are at least four sign changes around every vertex, then this contradicts the Euler formula. | |
| Aug 10, 2019 at 8:00 | comment | added | asv | Why these polyhedral metrics have unique realization in hyperbolic space? In Euclidean space this is known, but in hyperbolic space I did not see a reference (but I am not a specialist). Thank you. | |
| Aug 10, 2019 at 7:55 | history | answered | Ivan Izmestiev | CC BY-SA 4.0 |