Given $\omega_{ij}=\hat e_i \cdot d \hat e _j = \Gamma_{ijk} \sigma^k$, we can also say that $d \hat e_j = \omega^i_j \hat e_i$.
Show that $g(d\sigma^k, \sigma_p \wedge \sigma_q) = \Gamma_{ipq} - \Gamma_{iqp}$
Where $\sigma^k, \sigma_p, \sigma_q$ are to be one forms. I'm self -studying this material, and I just want to make sure I'm making the right connections!
So what I did was to say, $g(d\sigma^k, \sigma_p \wedge \sigma_q) =g(\omega_{ji} \sigma^j, \sigma_p \wedge \sigma_q) = g(\Gamma_{jik} \sigma^k \wedge \sigma^j, \sigma_p \wedge \sigma_q) = g(-\Gamma_{ijk} \sigma^k \wedge \sigma^j, \sigma_p \wedge \sigma_q)$
$= -\Gamma_{ijk}g(\sigma^k \wedge \sigma^j, \sigma_p \wedge \sigma_q) = -\Gamma_{ijk}[\delta^k_p \delta^j_q - \delta^k_q \delta^j_p] = -\Gamma_{ijk}\delta^k_p \delta^j_q + \Gamma_{ijk} \delta^k_q \delta^j_p = -\Gamma_{iqp}+\Gamma_{ipq} = \Gamma_{ipq} - \Gamma_{iqp}$
Cool, that's what we wanted, but I'm not sure if everything I did was valid, so I'm just seeing if it's fine what I did. My justification for saying $d\sigma^k = \omega_{ji} \sigma^j $ is that we see that $d \hat e_j = \omega^i_j \hat e_i$ since $\hat e_j$ is a basis vector, and $\sigma_i$ is a one form that can also form a basis, they should have the same connection. Thus, $d\sigma_i = \omega^k_i \sigma_k = \omega^k_i g_{kj} \sigma^j = \omega_{ji} \sigma^j$
We know $\Gamma_{jik} = - \Gamma_{ijk}$ by metric compatibility.
Sorry for the long post, just need some pointers, or reassurance that this is all right!