Infinitesimal Lorentz transformation is antisymmetric 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.
 A: 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 
$$\omega_{\rho\sigma}+\omega_{\sigma\rho}=0\Rightarrow\omega_{\rho\sigma}=-\omega_{\sigma\rho}$$
A: 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.
