# Infinitesimal lorentz transformation

I am stuck on a homework problem. The metric is an invariant tensor. $\Lambda^T \cdot g \cdot \Lambda=g$. Considering the infinitesimal lorentz transformation $\Lambda= 1+ \epsilon \Omega$, i have to show $\Omega_{\alpha \beta}=\Omega_{\beta \alpha}$.

I did the following expansion:

g$_{\alpha \beta} (1^\alpha_\mu+\epsilon\Omega^\alpha_\mu )(1^\beta_\nu+\epsilon \Omega^\beta_\nu)= (g_{\beta \mu}+\epsilon g_{\beta \mu})(1^\beta_\nu+\epsilon \Omega^\beta_\nu)$

g$_{\mu \nu}+ \epsilon g_{\mu \nu}+ \epsilon g_{\mu \nu}=g_{\mu \nu}$ neglecting $\epsilon^2 terms$

The infinitesimal Lorentz transformation takes the form $\Lambda^\mu{}_\nu = \delta^\mu{}_\nu + \epsilon\Omega^\mu{}_\nu$. So, neglecting terms of order $\epsilon^2$ we have $$g_{\alpha\beta}\Lambda^\alpha{}_\mu\Lambda^\beta{}_\nu = \left(g_{\mu\beta} + \epsilon\Omega_{\beta\mu}\right)\left(\delta^\beta{}_\nu + \epsilon\Omega^\beta{}_\nu\right) = g_{\mu\nu} + \epsilon\Omega_{\mu\nu} + \epsilon\Omega_{\nu\mu},$$ where $\Omega_{\mu\nu} = g_{\mu\alpha}\Omega^\alpha{}_{\nu}$. Since, as you say, $g_{\alpha\beta}\Lambda^\alpha{}_{\mu}\Lambda^\beta{}_{\nu} = g_{\mu\nu}$ we must have $\Omega_{\mu\nu} = -\Omega_{\nu\mu}$.
You stated that you should show that $\Omega_{\mu\nu} = \Omega_{\nu\mu}$, but the above result is the correct one.