Timeline for Parallelepiped is defined by the volumes of its faces
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 17, 2018 at 10:08 | vote | accept | erz | ||
| May 15, 2018 at 18:43 | answer | added | erz | timeline score: 1 | |
| May 14, 2018 at 20:58 | answer | added | Liviu Nicolaescu | timeline score: 2 | |
| May 14, 2018 at 8:29 | history | edited | erz | CC BY-SA 4.0 |
added 25 characters in body
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| May 14, 2018 at 8:26 | comment | added | erz | @AndrejBauer I was indeed wrong about the vector of translation: it should be just $v_k$. As for the second claim, the parallelepiped has $2^n$ vertices (e.g. $n$-dimensional cube). Also, we really need to use all the information, I'll provide a counterexample if needed. However, I did try something along the lines that you suggest, and I really hope it is possible to fins a proof like that. | |
| May 14, 2018 at 8:03 | comment | added | erz | @RichardStanley so do I understand correctly that you suggest to consider all $2n$ vectors $\pm 1v_i$ and apply Minkowski's theorem to them? But still, how do we know that there is an orthogonal operator that takes the set $\{v_i, -v_i\}$ into the set $\{w_i, -w_i\}$? We do know that $|\left<v_i,v_j\right>|=|\left<w_i,w_j\right>|$, but I don't see how to finish the proof. | |
| May 13, 2018 at 14:36 | comment | added | Richard Stanley | A theorem of Minkowski is applicable. See Theorem 36.2 of math.ucla.edu/~pak/geompol8.pdf. | |
| May 13, 2018 at 8:40 | comment | added | Andrej Bauer | Here's a thought: if we look just at the $1$-dimensional cases, i.e, the $W(\{i_1, i_2\}) = V(\{i_1, i_2\})$ then we see that the two shapes have congruent sides and all the diagonals. Since two triangles are congruent if they have the same lengths of sides, we should be able to get a proof from this observation. Or is my sense of rigidity betraying me in higher dimensions? | |
| May 13, 2018 at 8:37 | comment | added | Andrej Bauer | Is $P(v_1, \ldots, -v_k, \ldots, v_n)$ really a translation by $-2 v_k$? I would have expected $P(v_1 - v_k, \ldots, v_k - 2 v_k, \ldots, v_n - v_k)$. As for the claim that $P(\pm v_1, \ldots, \pm v_k)$ correspond to the vertices, that looks odd as well. There are $n$ vertices but $2^n$ combinations of $\pm$'s. | |
| May 13, 2018 at 7:53 | history | asked | erz | CC BY-SA 4.0 |