Given the electromagnetic field strength $F^{\mu\nu}$, and its dual $$\tilde{F}^{\mu\nu} =\dfrac{1}{2}\varepsilon^{\mu\nu\alpha\beta}F_{\alpha\beta},$$ how can I show that $$\tilde{F}^{\mu\nu}F_{\nu\rho} = -\dfrac{1}{4}\delta^\mu_\rho \tilde{F}^{\alpha\beta}F_{\alpha\beta} ??? $$
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A possible proof can be:
From lorentz invariance you have
$ \tilde{F}^{\mu \nu} F_{\rho \nu}= C \delta_{\rho}^{\mu} , \quad C \in \mathbb{R} $
now taking the trace both side you obtain $C$
$ \tilde{F}^{\alpha \nu} F_{\alpha \nu}= C \delta_{\alpha}^{\alpha}= 4 C $ so
$ \tilde{F}^{\mu \nu} F_{\rho \nu}= \frac{1}{4} \delta_{\rho}^{\mu} \tilde{F}^{\alpha \beta} F_{\alpha \beta} $ finally switching the index ($\rho \leftrightarrow \nu$) and using $F_{\rho \nu}=-F_{\nu \rho}$ you obtain
$ \tilde{F}^{\mu \nu} F_{\nu \rho}= -\frac{1}{4} \delta_{\rho}^{\mu} \tilde{F}^{\alpha \beta} F_{\alpha \beta} $
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$\begingroup$ Thank you! I do not understand the nature of your ansatz. Would not be necessary to verify the symmetry in $\mu$ and $\rho$ before proposing the delta on the ansatz? $\endgroup$ Commented Oct 24, 2022 at 16:08
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1$\begingroup$ No, it's the only Lorentz invariant symbol. Lorentz group have only $\epsilon^{\mu \nu \rho \gamma}$ and $ \delta_{\mu}^{\nu}$ like invariant symbols. See some reference on Lorentz group for details. $\endgroup$ Commented Oct 24, 2022 at 16:25
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1$\begingroup$ I don't see why $\tilde{F}^{\mu \nu} F_{\rho \nu}$ must be proportional to the identity tensor. Your proof says that if it is, it must be proportional with a factor of $- \frac{1}{4} \tilde{F} F$, but why couldn't it be proportional to some other tensor of rank (1,1)? $\endgroup$ Commented Oct 25, 2022 at 12:15