Let $\{e_\mu\}$ be a base for vectors and $\{\omega^\mu\}$ for dual vectors that we write
$e_\mu=e_\mu^i\frac{\partial}{\partial x^i}$ and $\omega^\mu=\omega_\mu^id x^i$ with $e_\mu^i\omega^\nu_i=\delta^\nu_\mu$ (and $i$, $j$ coordinate components).
Further we have the commutator rule $[e_\mu,e_\nu]=\gamma_{\mu\nu}^\rho e_\rho$ with $\gamma_{\mu\nu}^\rho$ the commutator components.
My questions is
What are these $\gamma_{\mu\nu}^\rho$?
I want to prove the identity $$e_\mu^ie_\nu^j\frac{\partial\omega^\sigma_j}{\partial x^i}-e_\nu^ie_\mu^j\frac{\partial\omega^\sigma_j}{\partial x^i}=-\gamma_{\mu\nu}^\sigma,$$ but I don't know where to start if I don't know what these $\gamma$ are.
Starting on the left, $$ e_{\mu}^i e_{\nu}^j \frac{\partial \omega^{\sigma}_j}{\partial x^i} - e_{\nu}^i e_{\mu}^j \frac{\partial \omega^{\sigma}_j}{\partial x^i} = e_{\mu}^i \frac{\partial }{\partial x^i} (e_{\nu}^j\omega^{\sigma}_j) - \omega^{\sigma}_j e_{\mu}^i \frac{\partial e_{\nu}^j }{\partial x^i} - e_{\nu}^i \frac{\partial }{\partial x^i} (e_{\mu}^j\omega^{\sigma}_j) + \omega^{\sigma}_j e_{\nu}^i \frac{\partial e_{\mu}^j}{\partial x^i} \\ = 0 - \omega^{\sigma}_j \left( e_{\mu}^i \frac{\partial e_{\nu}^j }{\partial x^i} - e_{\nu}^i \frac{\partial e_{\mu}^j}{\partial x^i} \right), $$ using that $e_{\nu}^j \omega^{\sigma}_j = \delta^{\sigma}_{\nu}$ is constant, and the bracket is the coordinate expression of the Lie bracket, $[e_{\mu},e_{\nu}]^j$.