Timeline for Parallelepiped is defined by the volumes of its faces
Current License: CC BY-SA 4.0
27 events
| when toggle format | what | by | license | comment | |
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| May 18, 2018 at 9:58 | comment | added | Liviu Nicolaescu | You're right. I've erased my speculations. | |
| May 18, 2018 at 9:57 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| May 17, 2018 at 20:43 | comment | added | erz | Another counterexample for the stronger version: for $n=k=3$ consider $v_i$ and $w_j$ so that $\|v_i\|=\|w_i\|=1$, $(v_i,v_j)=(w_1,w_2)=(w_2,w_3)=-(w_1,w_3)=\alpha$, where $\alpha$ is sufficiently small. Then you cannot find an orthogonal operator that moves $v_i$ into $w_i$, because the $3$-dimensional volumes are different. | |
| May 17, 2018 at 10:15 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| May 17, 2018 at 10:08 | vote | accept | erz | ||
| May 17, 2018 at 10:07 | comment | added | erz | Your explanation in your previous comment is rather satisfying (the reason why there is no way around considering those components is that they are exactly what is invariant). | |
| May 17, 2018 at 9:58 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| May 17, 2018 at 9:56 | comment | added | erz | I don't think the comment is correct: consider $k=n=3$ and consider $v_i$ forming the regular tetrahedron, while $w_i$ lie in one plane with angles between each pair equal to $\frac{\pi}{3}$. | |
| May 17, 2018 at 9:49 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| May 17, 2018 at 9:28 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| May 16, 2018 at 23:29 | comment | added | Liviu Nicolaescu | One seeks an element of the group of orthogonal transformations $T$ such that $Tv_i=\pm v_i$. The last remark in my proof shows that this group is isomorhic to $(\mathbb{Z}/2\mathbb{Z})^c$ where $c$ is the number of components of the graph $\Gamma$. Somehow this should be relevant in any approach. | |
| May 16, 2018 at 21:06 | comment | added | erz | It seems that your proof is more or less based on the same idea as mine, and is also the more directly computational proof from the article. Probably there is just no way around considering the components of non-orthogonality. | |
| May 16, 2018 at 10:08 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| May 16, 2018 at 9:43 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| May 16, 2018 at 9:34 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| May 16, 2018 at 9:22 | comment | added | Liviu Nicolaescu | I've fixed the argument. | |
| May 16, 2018 at 9:21 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| May 15, 2018 at 21:12 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| May 15, 2018 at 20:28 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| May 15, 2018 at 13:37 | comment | added | Liviu Nicolaescu | You are correct. I'll ttry to fix the argument. | |
| May 15, 2018 at 3:47 | comment | added | erz | I'm sorry, but I still don't understand: with the same example $[v_0]_3=e_1+e_2$, while $[w_0]_3=e_1-e_2$, and they are not co-linear. I think that the argument that you use several times about equality of $\eta_i$ is incorrect as $T$ in the example above is neither $Id$, nor $-Id$ when restricted to the span of $e_1$ and $e_2$. In my opinion, this is exactly the fundamental difficulty - we cannot really fix the transformation on the $n$ vectors and only tweak $n+1$-th (if I understand correctly what is happening). | |
| May 15, 2018 at 0:48 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| May 15, 2018 at 0:42 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| May 15, 2018 at 0:36 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| May 15, 2018 at 0:29 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| May 14, 2018 at 22:27 | comment | added | erz | I am still halfway through your answer, but your claim seems incorrect to me: consider the case when $n=3$ and $v_i=e_i$, $1\le i\le 3$ with $v_0=e_0+e_1+e_2$ and $w_0=e_0+e_1-e_2$ (only two dimensions matter here). Then $[v_0]_2=e_1+e_2$, while $[w_0]_2=e_1-e_2$. | |
| May 14, 2018 at 20:58 | history | answered | Liviu Nicolaescu | CC BY-SA 4.0 |