# Infinitesimal Lorentz transformation is antisymmetric

Note that if you lower an index of the Kronecker delta, it becomes the metric:

$\eta_{\mu\nu}\delta^{\mu}_{\rho}=\delta_{\nu\rho}=\eta_{\nu\rho}$

And in your last step you got a wrong index. It should be $\omega_{\rho\sigma}$, not $\omega^{\rho}_{\sigma}$.

Then, the metric terms cancel and you neglect cuadratic terms.

That should be enough to solve it.

Since the Lorentz transformation is valid for any $x\in M_{4}$, it can be rewritten as $\Lambda_{\rho}^{\mu}\eta_{\mu\nu}\Lambda_{\sigma}^{\nu}=\eta_{\rho\sigma}$. Substituting the infinitesimal form of the Lorentz transformation into the previous formula we get

$$(\delta_{\rho}^{\mu}+\omega_{\rho}^{\mu})\eta_{\mu\nu}(\delta_{\sigma}^{\nu}+\omega_{\sigma}^{\nu})+o(\omega^{2})=\eta_{\rho\sigma}$$

after expanding

$$\eta_{\rho\sigma}+\omega_{\rho}^{\mu}\eta_{\mu\nu}\delta_{\sigma}^{\nu}+\omega_{\sigma}^{\nu}\eta_{\mu\nu}\delta_{\rho}^{\mu}+o(\omega^2)=\eta_{\rho\sigma}$$

and from this we can see that

$$\omega_{\rho\sigma}+\omega_{\sigma\rho}=0\Rightarrow\omega_{\rho\sigma}=-\omega_{\sigma\rho}$$