# Christoffel identity

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!

## Answers and Replies

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!