- #1

WendysRules

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## Homework Statement

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

## 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).