Timeline for Reference request: embedding the hyperbolic triangulation in $\mathbb{R}^3$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 24, 2014 at 23:29 | comment | added | Ian Agol | yes, icosahedron, not dodecahedron! | |
| Feb 24, 2014 at 21:42 | answer | added | j.c. | timeline score: 2 | |
| Feb 24, 2014 at 21:04 | comment | added | Geoffrey Irving | @IanAgol: Thanks! I'm having trouble visualizing that, though. How do you attach an octahedron to a pentagonal face of a dodecahedron? Do you mean icosahedron? | |
| Feb 24, 2014 at 20:40 | comment | added | Ian Agol | There's an isometric immersion of $T_7$ into $R^3$, which may be obtained by taking a dodecahedron, and attaching 4 octahedra to non-adjacent faces, and repeating (one obtains an infinite diamond lattice, with dodecahedra representing atoms, and octahedra representing bonds). Here, I'm assuming you want the triangles to be Euclidean equilateral triangles. | |
| Feb 24, 2014 at 19:06 | comment | added | Geoffrey Irving | Comments hopefully addressed, let me know if you have other suggestions. I'm not sure why the images are so poorly arranged. | |
| Feb 24, 2014 at 19:05 | history | edited | Geoffrey Irving | CC BY-SA 3.0 |
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| Feb 24, 2014 at 18:59 | comment | added | Benoît Kloeckner | Maybe you could precise a few points: how exactly is defined $T_d$? I guess the triangle are regular, and that the valence is the valence of vertices. More important, I guess that the "isometric" embedding you seek is combinatorially isometric but realized with flat triangle, is that right? And of course, you do not mean isometric for the extrinsic distance, but that might be better to say so explicitly. | |
| Feb 24, 2014 at 18:54 | history | edited | Geoffrey Irving | CC BY-SA 3.0 |
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| Feb 24, 2014 at 18:22 | history | asked | Geoffrey Irving | CC BY-SA 3.0 |