I have seen many times on different sources that the equation:
$$ \epsilon^{\mu\nu\rho\sigma}\partial_{\nu}F_{\rho\sigma}=0$$
Is true identically following from it being a product of an anti-symmetric (levi-cevita) and a symmetric tensor, but I don't understand which symmetric tensor do they refer to? is it the $\partial_{\nu}\partial_{\rho}$ due to partial derivative interchangeability? Is this equation true even without heeding the fact that it represents two of maxwell equations?
Best Answer
Yes you are right, $\partial_\nu F_{\rho \sigma}$ has mixed symmetry, but the Levi-Civita symbol is fully anti-symmetric. This means that their contraction is identically 0. In fact, one can decompose this equation as follow: \begin{equation} I^\mu = \epsilon^{\mu \nu \rho \sigma}\partial_\nu \partial_\rho A_\sigma-\epsilon^{\mu \nu \rho \sigma}\partial_\nu \partial_\sigma A_\rho \end{equation} One can see that the two terms vanish identically because of the same symmetry argument.
In differential forms, this equation is still an identity because it is the double exterior derivative of a 1 form. As long as the one form belongs to the $\Omega^1(\mathcal{M}) \equiv \Gamma(\wedge^1 T^\ast \mathcal{M})$, where $\mathcal{M}$ is our manifold, this equation can be written as: \begin{equation} \underbrace{d\circ d}_{\stackrel{!}{=}0} A=0 \end{equation} So this is true even if one doesn't take into account the maxwell equations.