# Showing $\partial_{\mu}\tilde{F}^{\mu\nu}=0$ by the antisymmetric properties

The electromagnetic dual tensor is given by \begin{align} \tilde{F}^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\delta\rho}F_{\delta\rho} \end{align} Here, $$\epsilon^{\mu\nu\delta\rho}$$ is the Levi-Civita symbol. I want to show that $$\partial_{\mu}\tilde{F}^{\mu\nu}=0 ,$$ using the antisymmetric propeties of Levi-Civita.
I have exchanged indices $$\rho \to \delta$$, \begin{align} & \frac{1}{2}\partial_{\mu}\epsilon^{\mu\nu\rho\delta}F_{\rho\delta}\\ &=- \frac{1}{2}\partial_{\mu}\epsilon^{\mu\nu\delta\rho}F_{\rho\delta}\\ &=\frac{1}{2}\partial_{\mu}\epsilon^{\mu\nu\delta\rho}F_{\delta\rho} \end{align} In the last line, I've used $$F_{\mu\nu}=-F_{\nu\mu}$$. So, I can't use the property to prove this. In Ryder's QFT book he says that, because of the antisymmetry of $$\epsilon^{\mu\nu\delta\rho}$$ this will be true.

We know that $$F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu .$$ Now we get $$\partial_\mu \epsilon^{\mu\nu\sigma\rho} F_{\sigma \rho} = \partial_\mu \epsilon^{\mu\nu\sigma\rho} ( \partial_\sigma A_\rho - \partial_\rho A_\sigma ).$$ Now, from symmetry/antisymmetry we have $$\epsilon^{\mu\nu\sigma\rho} \partial_\mu \partial_\sigma A_\rho = 0$$ and similar for the other term. To see this last step, consider $$\epsilon^{\mu\nu\sigma\rho} \partial_\mu \partial_\sigma = [\textrm{rename variables } \sigma \leftrightarrow \mu] \\ = \epsilon^{\sigma\nu\mu\rho} \partial_\sigma \partial_\mu = [\textrm{use property of Levi-Civita } \epsilon^{\alpha \beta} = -\epsilon^{\beta \alpha}] \\ = - \epsilon^{\mu\nu\sigma\rho} \partial_\sigma \partial_\mu = [\textrm{use that } \partial_\alpha \partial_\beta = \partial_\beta \partial_\alpha] \\ = - \epsilon^{\mu\nu\sigma\rho} \partial_\mu \partial_\sigma .$$ So we have $$\epsilon^{\mu\nu\sigma\rho} \partial_\mu \partial_\sigma = - \epsilon^{\mu\nu\sigma\rho} \partial_\mu \partial_\sigma .$$ And if $$a = -a$$ then we must have $$a = 0$$.
In the parlance of differential forms, the identity (Abelian Bianchi identity) amounts to $$dF = 0.$$ This is a trivial fact in math: exact forms ($$F = dA$$) are closed $$dF = d^2A = 0,$$ because of nilpotency of $$d$$ ($$d^2 = 0$$, barring Nonholonomic coordinates).
Note that the opposite may not be true: closed forms ($$dF = 0$$) do not necessarily imply exact forms ($$F = dA$$). Case in point: non-trivial de Rham cohomology.