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It could be that the preprint where I found this identity has a typo or that it is simply wrong, but I have been trying to see if this is true:

\begin{equation} \int \left(\nabla\times F_{\bf B}\right)C({\bf x}) \left(\nabla\times G_{\bf B}\right) d{\bf x} =-2\int{\langle \nabla,G_{\bf B}\times F_{\bf B}\rangle d{\bf x}}=0. \end{equation}

where $C({\bf x})$ is a $3\times 3$ matrix given by \begin{equation} C({\bf x}) = \left[ {\begin{array}{ccc} 0 & x_3 & -x_2 \\ -x_3 & 0 &x_1 \\ x_2 & -x_1 & 0 \end{array} } \right] \end{equation}

and $F_{\bf B}$ and $G_{\bf B}$ are smooth vector fields

I cannot prove it. Am I doing something wrong?

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  • $\begingroup$ can you disclose the source of the formula? $\endgroup$ Commented May 25, 2023 at 12:49
  • $\begingroup$ Actually no, it's a preprint of a collaborator and it's not published yet... neither inArxiv $\endgroup$ Commented May 25, 2023 at 12:50
  • $\begingroup$ Something like this could potentially hold if you know additional info about $F$ and $G$: you would need that they have compact support (or if the integration is over all of $\mathbb{R}^3$, decaying to zero sufficiently fast) so that you can integrate by parts with no boundary terms, and you also need that they satisfy certain first order differential equation (to get rid of some spurious second order terms). If these other conditions are available, please include them in the question. Else the general case this cannot be true. $\endgroup$ Commented May 25, 2023 at 20:39
  • $\begingroup$ Willie, I have imposed that they decay sufficiently fast. I have not imposed any differential equation... in principle they should be arbitrary but maybe I could impose some restriction... I wanna see the parallelism between this expression and E,B electromagnetic fields... and they are related thru first order Maxwell equations so it may work. I will look into that. Thank you very much. $\endgroup$ Commented May 27, 2023 at 8:19

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try $F=\{x_1,1,1\}$, $G=\{1,1,x_1\}$, then $$ \left(\nabla\times F\right)C({\bf x}) \left(\nabla\times G\right)=0,$$ because $\nabla\times F=0$, while $$ \langle \nabla,G\times F\rangle=-1,$$ contradicting the identity.

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