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

Question

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.

Answer 1

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.

Answer 2

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 $a = 0$.