# Prove $F_{\mu\rho} \tilde F_\nu^{\phantom{\nu}\rho} = \frac14 \eta_{\mu\nu} F_{\rho\sigma} \tilde{F}^{\rho \sigma}$ using Schouten identity [Done] [closed]

How to prove \begin{align} F_{\mu\rho} \tilde{F}_{\nu}^{\phantom{\nu}\rho} = \frac{1}{4} \eta_{\mu\nu} F_{\rho\sigma} \tilde{F}^{\rho \sigma} \end{align} using Schouten identity \begin{align} 0 = 5 \delta_{\mu}^{[\nu } \epsilon^{\rho\sigma\tau\lambda]} = \delta_{\mu}^{\phantom{\mu}\nu} \epsilon^{\rho\sigma\tau\lambda} +\delta_{\mu}^{\phantom{\mu}\rho}\epsilon^{\sigma\tau\lambda\nu} +\delta_{\mu}^{\phantom{\mu}\sigma}\epsilon^{\tau\lambda\nu\rho} +\delta_{\mu}^{\phantom{\mu}\tau}\epsilon^{\lambda\nu\rho\sigma} +\delta_{\mu}^{\phantom{\mu}\lambda} \epsilon^{\nu\rho\sigma\tau} \end{align} Where the dual tensor is defined in this way \begin{align} \tilde{F}^{\mu\nu} = -\frac{1}{2} i \epsilon^{\mu\nu\alpha\beta} F_{\alpha\beta} \end{align}

My guess \begin{align} &R.H.S=\frac{1}{4} \eta_{\mu\nu} F_{\rho\sigma} \tilde{F}^{\rho \sigma} = -\frac{1}{8} i \eta_{\mu\nu} \epsilon^{\rho\sigma\alpha\beta} F_{\rho\sigma} F_{\alpha\beta} \\ &L.H.S=F_{\mu\rho} \tilde{F}_{\nu}^{\phantom{\nu}\rho} = F_{\mu\rho} \tilde{F}^{x\rho}\eta_{x \nu} = -\frac{1}{2} i \epsilon^{x \rho \alpha \beta} \eta_{x \nu} F_{\mu \rho} F_{\alpha\beta} \\ & \eta_{\mu\nu} \epsilon^{\rho\sigma\alpha\beta} F_{\rho\sigma} F_{\alpha\beta} = 4\epsilon^{x \rho \alpha \beta} \eta_{x \nu} F_{\mu \rho} F_{\alpha\beta} \quad (?) \end{align} To prove the last line i think Schouten identity is needed.

Due to below answer (@Noiralef) Explicit computations are following

Rewrite the schouten identity by pulling down $\nu$ \begin{align} 0 = \eta_{\mu\nu} \epsilon^{\rho\sigma\tau\lambda} +\delta_{\mu}^{\phantom{\mu}\rho}\eta_{x\nu}\epsilon^{\sigma\tau\lambda x} +\delta_{\mu}^{\phantom{\mu}\sigma}\eta_{x\nu}\epsilon^{\tau\lambda x\rho} +\delta_{\mu}^{\phantom{\mu}\tau}\eta_{x\nu}\epsilon^{\lambda x\rho\sigma} +\delta_{\mu}^{\phantom{\mu}\lambda}\eta_{x\nu} \epsilon^{x \rho\sigma\tau} \end{align} Then \begin{align} &\left( \eta_{\mu\nu} \epsilon^{\rho\sigma\tau\lambda} +\delta_{\mu}^{\phantom{\mu}\rho}\eta_{x\nu}\epsilon^{\sigma\tau\lambda x} +\delta_{\mu}^{\phantom{\mu}\sigma}\eta_{x\nu}\epsilon^{\tau\lambda x\rho} +\delta_{\mu}^{\phantom{\mu}\tau}\eta_{x\nu}\epsilon^{\lambda x\rho\sigma} +\delta_{\mu}^{\phantom{\mu}\lambda}\eta_{x\nu} \epsilon^{x \rho\sigma\tau} \right) F_{\rho\sigma}F_{\tau\lambda} \\ &=\eta_{\mu\nu} \epsilon^{\rho\sigma\tau\lambda}F_{\rho\sigma}F_{\tau\lambda} + \epsilon^{\sigma\tau\lambda x} \eta_{x\nu} F_{\mu\sigma}F_{\tau\lambda} + \epsilon^{\tau\lambda x\rho} \eta_{x\nu} F_{\rho\mu}F_{\tau\lambda} + \epsilon^{\lambda x \rho\sigma}\eta_{x\nu} F_{\rho\sigma}F_{\mu\lambda}+ \epsilon^{x \rho\sigma\tau} \eta_{x\nu}F_{\rho\sigma} F_{\tau\mu} \\ & = \eta_{\mu\nu} \epsilon^{\rho\sigma\tau\lambda}F_{\rho\sigma}F_{\tau\lambda} + 4\epsilon^{\sigma\tau\lambda x} \eta_{x\nu} F_{\mu\sigma}F_{\tau\lambda} =0 \\ & \quad \Rightarrow \quad \eta_{\mu\nu} \epsilon^{\rho\sigma\tau\lambda}F_{\rho\sigma}F_{\tau\lambda} =- 4\epsilon^{\sigma\tau\lambda x} \eta_{x\nu} F_{\mu\sigma}F_{\tau\lambda} = 4\epsilon^{x \sigma\tau\lambda } \eta_{x\nu} F_{\mu\sigma}F_{\tau\lambda} \end{align} Here we used \begin{align} \epsilon^{\tau\lambda x \rho} F_{\rho\mu} F_{\tau\lambda} = \epsilon^{\tau\lambda x \sigma} F_{\sigma\mu} F_{\tau\lambda} = (-1)^3 \epsilon^{\sigma\tau\lambda x} F_{\sigma\mu} F_{\tau\lambda} = \epsilon^{\sigma\tau\lambda x} F_{\mu\sigma} F_{\tau\lambda} \end{align}

## closed as off-topic by JamalS, ACuriousMind♦, Danu, Kyle Kanos, MartinMay 4 '15 at 13:45

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Rewrite the Shouten identity by pulling down $\nu$: $$0 = \eta_{\mu\nu} \epsilon^{\rho\sigma\tau\lambda} + \delta_\mu^\rho \eta_{\chi\nu} \epsilon^{\sigma\tau\lambda\chi} + \cdots$$ Then plug this into the left hand side of the equation you marked with (?). You will get four terms, all having the correct structure of the right hand side (one free index at $\eta$ and one at one of the $F$'s, all other lower indices are contracted with the $\epsilon$).