# Massaging an infinitesimal Lorentz transformation into a divergence

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?

Ok, never mind. I did indeed miss something completely trivial. In my scenario, the $$\omega^\mu_{\;\nu}$$'s are constant; I've just been doing far too much general relativity lately, where everything depends on position ;)
Anyway, remembering that $$\partial_\sigma \omega^\mu_{\;\nu}=0$$, the proof is indeed trivial:
$$\partial_\mu(\omega^\mu_{\;\nu} x^\nu)=\omega^\mu_{\;\nu}\partial_\mu x^\nu=\omega^{\mu\sigma}\eta_{\sigma\nu}\delta^\nu_{\;\mu}=\omega^{\mu\sigma}\eta_{\sigma\mu}=0$$ since $$\eta_{\sigma\mu}$$ is symmetric. My bad.