# How prove that the contraction of electromagnetic tensor and its dual, $F_{\mu\nu}\tilde F^{\mu\nu}$, is an total derivative? [closed]

I have tried to solve this exercise from Supergravity-Freedman and Van Proeyen (2012),

Excercise 4.10 Show that the quantity $F_{\mu\nu}\tilde F^{\mu\nu}$ is a total derivative, i.e. $$F_{\mu\nu}\tilde F^{\mu\nu}=-i\partial_\mu(\epsilon^{\mu\nu\rho\sigma }A_\nu F_{\rho\sigma}).$$

$F_{\mu\nu}\equiv\partial_\mu A_\nu-\partial_\nu A_\mu$ is the Electromagnetic Tensor, $\tilde F_{\mu\nu}\equiv -\frac i2 \epsilon_{\mu\nu\rho\sigma}F^{\rho\sigma}$ its dual tensor, $\epsilon^{\mu\nu\rho\sigma }$ is the Levi-Civita symbol and $A_\mu$ is an $c$-vector field (the $c$-potential).

I've tried this:

By definition $$F_{\mu\nu}\tilde F^{\mu\nu}=-\frac i2 \epsilon^{\mu\nu\rho\sigma}(\partial_\mu A_\nu -\partial_\nu A_\mu)F_{\rho\sigma}.$$ Using the Antisymmetry property of $\epsilon$ we have $$F_{\mu\nu}\tilde F^{\mu\nu}=-i \epsilon^{\mu\nu\rho\sigma}(\partial_\mu A_\nu)F_{\rho\sigma}.$$ With the product rule, $$F_{\mu\nu}\tilde F^{\mu\nu}=-i \epsilon^{\mu\nu\rho\sigma}\left(\partial_\mu ( A_\nu F_{\rho\sigma})-A_\nu\partial_\mu F_{\rho\sigma}\vphantom{\frac yy}\right).$$

But I don't kown how cancel the second term in the brackets. Can yo help me?

## closed as off-topic by AccidentalFourierTransform, Kyle Kanos, John Rennie, stafusa, Emilio PisantyAug 12 '18 at 15:37

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• $\epsilon^{abcd}\partial_a\partial_b(\text{anything})\equiv0$. – AccidentalFourierTransform Aug 11 '18 at 17:24
• Tanks!! I'ts Easy. – Cristian Rodríguez Aug 11 '18 at 18:36

I'm glad the OP understood the hint, but in case a future reader interested in this problem doesn't, here it is spelled out: the final term is $iA_\nu (\epsilon^{\mu\nu\rho\sigma}\partial_\mu\partial_\rho A_\sigma-\epsilon^{\mu\nu\rho\sigma}\partial_\mu\partial_\sigma A_\rho)$. Each term then vanishes by contracting a pair of indices in which the Levi-Civita symbol is antisymmetric with the indices of a symmetric second-order derivative.