Timeline for A question on vectors in $\mathbb{R}^4$
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 11, 2021 at 14:37 | comment | added | Michael Engelhardt | @WillieWong - Yes, that explains nicely why we have to do what we have to do. Thank you! | |
| Oct 11, 2021 at 14:04 | comment | added | Willie Wong | This is not an issue for the cross product as $u\wedge v$ is preserved under $SO(2)$ rotations, it also having an even number of factors. | |
| Oct 11, 2021 at 14:03 | comment | added | Willie Wong | Incidentally: any linear combination of the four 3-vectors $ u\wedge v\wedge Mu, u\wedge v\wedge Mv, u\wedge Mu\wedge Mv, v\wedge Mu\wedge Mv$ will necessarily vanish under the action $(u,v)\mapsto (\rho u,\rho v)$ with $\rho\in SO(2)$ some rotation in the plane of $u\wedge v$. This is because $-I\in SO(2)$ and each of the four 3 vectors swap signs under negation of $u$ and $v$, and hence by the intermediate value theorem (the space of 3-vectors is one dimensional) for some $\rho$ you hit zero. Hence a "simple multilinear construction" is not possible. | |
| Oct 11, 2021 at 4:28 | comment | added | Michael Engelhardt | @WillieWong - Ah, you are right, I hadn't considered that loophole. I've fixed it with an unfortunately somewhat ugly kludge that constructs a linear combination of $M\mathbf{u} $ and $M\mathbf{v} $ which does not lie in the span of $\mathbf{u} $, $\mathbf{v} $. | |
| Oct 11, 2021 at 4:12 | history | edited | Michael Engelhardt | CC BY-SA 4.0 |
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| Oct 11, 2021 at 3:51 | history | edited | Michael Engelhardt | CC BY-SA 4.0 |
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| Oct 11, 2021 at 2:19 | history | edited | Michael Engelhardt | CC BY-SA 4.0 |
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| Oct 11, 2021 at 1:39 | comment | added | Willie Wong | Probably you meant that the kernel of $\mathcal{M}^T$ is the linear span of $\{\epsilon(u,v,Mu), \epsilon(u,v,Mv)\}\}$? | |
| Oct 11, 2021 at 1:32 | comment | added | Willie Wong | For example, set $M = \begin{pmatrix} 1 & 1\\ -1 & 1 \\ & & 1 \\ -1 & 1 & & 1\end{pmatrix}$. If you set $u = (1,0,0,0)$ and $v = (0,1,0,0)$, then each of $Mu$ and $Mv$ is linearly independent of $u$ and $v$. And $\{ u,v, Mu, Mv\}$ is linearly dependent. But $Mu + Mv = 2u$ and so your $w$ as constructed is $0$. | |
| Oct 11, 2021 at 1:29 | comment | added | Willie Wong | I am not sure about your final equality. Can it not be the case that $M(u+v)$ is in the span of $u$ and $v$? | |
| Oct 10, 2021 at 22:00 | comment | added | Michael Engelhardt | $\epsilon_{ijkl} $ is the totally antisymmetric (Levi-Civita) tensor. Sorry, I'm more fluent in this notation than in producing the appropriate combination of wedges. | |
| Oct 10, 2021 at 21:59 | comment | added | LSpice | What does $\epsilon_{i j k l}$ mean? | |
| Oct 10, 2021 at 21:56 | history | edited | Michael Engelhardt | CC BY-SA 4.0 |
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| Oct 10, 2021 at 21:51 | history | answered | Michael Engelhardt | CC BY-SA 4.0 |