So, I was watching this video https://www.youtube.com/watch?v=cj-QpsZsDDY&t=3968s , where at some point the prof explains that a divergence can be ontained by "exploiting the antisymmetry" of an infinitesimal Lorentz transformation.
What it boils down to is this: Let $x^\mu$ be some inertial coordinate system in flat 4D spacetime with Minkowstki metric $\eta_{\mu\nu}$. Consider the infinitesimal Lorentz transformation $$ x'^\mu=\Lambda^\mu_{\;\nu}x^\nu=(\delta^\mu_{\;\nu}+\omega^\mu_{\;\nu})x^\nu $$ with $|\omega^\mu_{\;\nu}| \ll 1$. It is easy to show (see e.g. this question) that this requires that $\omega^{\mu\sigma}=\eta^{\sigma\nu}\omega^\mu_{\;\nu}$ be antisymmetric.
Now here's my question: apparently this antisymmetry allows one to conclude, that $$ \partial_\mu(\omega^\mu_{\;\nu} x^\nu)=0 $$ Yet, I can't manage to prove that. Here's what I came up with sofar $$ \partial_\mu(\omega^\mu_{\;\nu} x^\nu)=\partial_\mu(\omega^{\mu\sigma}\eta_{\sigma\nu}x^\nu) \\ =\frac{1}{2}\eta_{\sigma\nu}\partial_\mu(\omega^{\mu\sigma}x^\nu)-\frac{1}{2}\eta_{\sigma\nu}\partial_\mu(\omega^{\sigma\mu}x^\nu) \\ =\frac{1}{2}\eta_{\sigma\nu}\partial_\mu(\omega^{\mu\sigma}x^\nu)-\frac{1}{2}\eta_{\mu\nu}\partial_\sigma(\omega^{\mu\sigma}x^\nu) $$
none of which seems to bring me any closer though... Am a missing something trivial?