- #1

- 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!

&= \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!