Understanding Christoffel Identity and its Application in Differential Geometry

[etotheipi]

Homework Statement

We need to show that $\Gamma_{cb}^e = \frac{1}{2}g^{ea}(\partial_c g_{ab} + \partial_b g_{ac} - \partial_a g_{cb})$

Relevant Equations

N/A

We use $g_{\alpha \beta} = \vec{e}_{\alpha} \cdot \vec{e}_{\beta}$ to show that$$\partial_c g_{ab} = \partial_c (\vec{e}_a \cdot \vec{e}_b) = \vec{e}_a \cdot \partial_c \vec{e}_b + \vec{e}_b \cdot \partial_c \vec{e}_a$$then because $\partial_{\alpha} \vec{e}_{\beta} := \Gamma_{\alpha \beta}^{\gamma} \vec{e}_{\gamma}$, we get$$\begin{align*}\partial_c g_{ab} = \vec{e}_a \cdot \Gamma_{cb}^{d} \vec{e}_{d} + \vec{e}_b \cdot \Gamma_{ca}^{d} \vec{e}_{d}

&= \Gamma_{cb}^{d} \vec{e}_a \cdot \vec{e}_d + \Gamma_{ca}^{d} \vec{e}_b \cdot \vec{e}_d \\

&= \Gamma_{cb}^{d} g_{ad} + \Gamma_{ca}^{d} g_{bd}

\end{align*}$$Now I am uncertain as to how to proceed. I wonder if someone can give me a hint? I try contracting both sides with $g^{ea}$ but I can't see how that helps. Thanks!

 

[Ibix]

I'm not certain what you know about Christoffel symbols at this point. If you are allowed to use that $\Gamma^a_{bc}=\Gamma^a_{cb}$ then you can permute the indices in your last expression and use that symmetry to get three equations in three terms like $\Gamma^\cdot_{\cdot\cdot}g_{\cdot\cdot}$. That ought to get you where you're going.

 

[etotheipi]

Thanks! I'll write out my solution then, using this fact about symmetry in the downstairs indices of the Christoffel symbols,$$\begin{align*}
\partial_c g_{ab} &= \Gamma_{bc}^d g_{ad} + \Gamma_{ca}^d g_{bd} \\
\partial_b g_{ca} &= \Gamma_{ab}^d g_{cd} + \Gamma_{bc}^d g_{ad} \\
\partial_a g_{bc} &= \Gamma_{ca}^d g_{bd} + \Gamma_{ab}^d g_{cd}
\end{align*}$$It follows that$$\partial_c g_{ab} + \partial_b g_{ca} - \partial_a g_{bc} = 2\Gamma_{bc}^d g_{ad}$$Now I halve both sides, and contract both sides with $g^{ea}$, making use of the identity $g^{\alpha \beta}g_{\beta \gamma} = \delta^{\alpha}_{\gamma}$,$$\frac{1}{2} g^{ea} \left(\partial_c g_{ab} + \partial_b g_{ca} - \partial_a g_{bc} \right) = \Gamma_{bc}^d g^{ea} g_{ad} = \Gamma^e_{bc}$$Nice! Muchas gracias señor Ibix!