Levi-Civita connection and Christoffel symbols

[WendysRules]

Homework Statement

Show that $$g(d \sigma ^k, \sigma _p \wedge \sigma _q) = \Gamma _{ipq} - \Gamma _{iqp}$$

Homework Equations

Given $$\omega_{ij}=\hat e_i \cdot d \hat e _j = \Gamma_{ijk} \sigma^k$$, we can also say that $$d \hat e_j = \omega^i_j \hat e_i$$. Where $$\sigma^k, \sigma_p, \sigma_q$$ are to be one forms.

The Attempt at a Solution

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! (I've posted this on another forum, but no replies).

 

[mighty2000]

Thank you for your post. Your solution looks correct to me. You have correctly used the definition of the connection form and the metric compatibility to arrive at the desired result. Your justification for saying that $d\sigma^k = \omega_{ji} \sigma^j$ also seems valid. Keep up the good work, and don't hesitate to ask if you have any further questions.