The Minkowski metric transforms under Lorentz transformations as
\begin{align*}\eta_{\rho\sigma} = \eta_{\mu\nu}\Lambda^\mu_{\ \ \ \rho} \Lambda^\nu_{\ \ \ \sigma} \end{align*}
I want to show that under a infinitesimal transformation $\Lambda^\mu_{\ \ \ \nu}=\delta^\mu_{\ \ \ \nu} + \omega^\mu_{{\ \ \ \nu}}$, that $\omega_{\mu\nu} = -\omega_{\nu\mu}$.
I tried expanding myself:
\begin{align*}
\eta_{\rho\sigma} &= \eta_{\mu\nu}\left(\delta^\mu_{\ \ \ \rho} + \omega^\mu_{{\ \ \ \rho}}\right)\left(\delta^\nu_{\ \ \ \sigma} + \omega^\nu_{{\ \ \ \sigma}}\right) \\
&= (\delta_{\nu\rho}+\omega_{\nu\rho})\left(\delta^\nu_{\ \ \ \sigma} + \omega^\nu_{{\ \ \ \sigma}}\right) \\
&= \delta_{\rho\sigma}+\omega^\rho_{\ \ \ \sigma}+\omega_{\sigma\rho}+\omega_{\nu\rho} \omega^\nu_{{\ \ \ \sigma}}
\end{align*}
Been a long time since I've dealt with tensors so I don't know how to proceed.
Best Answer
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.