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I am working on the dual Lagrangian as given by $$\mathcal{L}_D=F_{\mu\nu}\tilde{F}^{\mu\nu}$$ In literature, this term is often written as $$\boxed{\mathcal{L}_D=2\partial^\mu(\varepsilon_{\mu\nu\rho\sigma}A^\nu\partial^\rho A^\sigma)}$$ I - with my limited knowledge on field theory - would like to reproduce this result. So far, I've computed $$\mathcal{L}_D=\frac{1}{2}(\varepsilon_{\mu\nu\rho\sigma}\partial^\mu A^\nu\partial^\rho A^\sigma-\varepsilon_{\mu\nu\rho\sigma}\partial^\mu A^\nu \partial^\sigma A^\rho - \varepsilon_{\mu\nu\rho\sigma} \partial^\nu A^\mu \partial^\rho A^\sigma + \varepsilon_{\mu\nu\rho\sigma}\partial^\nu A^\mu \partial^\sigma A^\rho)$$ By carefully evaluating a few combinations of the Levi-Civita symbol, I was able to deduce that $$\mathcal{L}_D=2\varepsilon_{\mu\nu\rho\sigma}(\partial^\mu A^\nu)(\partial^\rho A^\sigma)$$ To go from here to the boxed term, my reference states "One can write out the product rule for differentiation. All terms other than the boxed one are zero."

Here I do not really see how the product rule should be invoked and why terms drop out. Naively, I would say $$\mathcal{L}_D=2\varepsilon_{\mu\nu\rho\sigma}\left[\partial^\mu A^\nu \partial^\rho A^\sigma\right]=2\varepsilon_{\mu\nu\rho\sigma}\left[\partial^\mu(A^\nu)\partial^\rho A^\sigma + A^\nu \partial^\mu(\partial^\rho A^\sigma) \right]$$

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  • $\begingroup$ The partial derivatives in the last term commute, so you get two identical terms by swapping their indices. But the Levi-Civita tensor gives a minus sign for swapping indices, so you get pairs of identical terms with opposite signs. $\endgroup$
    – ragnar
    Commented Jun 24, 2019 at 16:25
  • $\begingroup$ That makes perfect sense, thank you! I was puzzled by the boxed equation as it still looks like something that needs to be expanded, but the extra term there will drop out as well then $\endgroup$
    – DaanMusic
    Commented Jun 24, 2019 at 16:27

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The question has been answered by ragnar yesterday: $\epsilon^{\mu\nu\rho\sigma}(\partial_\mu\partial_\nu(\cdots))=0$

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