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How can we prove that the christoffel symbol is \[ \Gamma^k_{ij} = \frac 12 \sum_{l=1}^2 g^{kl} \left(\frac{\partial g_{il}}{\partial u^j} + \frac{\partial g_{jl}}{\partial u^i} - \frac{\partial g_{ij}}{\partial u^l}\right) \] I can think of some substitution using the first and second fundamental forms, but can't see how to really fit it in.


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$\def\+#1{\frac{\partial}{\partial u^{#1}}}$We have, as the metric has vanishing covariant derivative \[ 0 = \nabla_{\+j} g_{ik} = \+j g_{ik} - \sum_m (g_{im}\Gamma^m_{kj} + g_{mk}\Gamma^m_{ij}) \] or \[ \+jg_{ik} = \sum_m (g_{im}\Gamma^m_{kj} + g_{mk}\Gamma^m_{ij})\] That is \begin{align*} \frac 12 \sum_l g^{kl}\left( \+j g_{il} + \+ig_{jl} - \+l g_{ij}\right) &= \frac 12 \sum_{m,l} g^{kl} \left(g_{im}\Gamma^m_{lj} + g_{ml}\Gamma^m_{ij} + g_{jm}\Gamma^m_{li} + g_{ml}\Gamma^m_{ji} - g_{im}\Gamma^m_{jl} - g_{mj}\Gamma^m_{il}\right)\\ &= \frac 12 \sum_{m,l} \left(g^{kl}g_{im}\Gamma^m_{jl} - g^{kl}g_{im}\Gamma^m_{lj}\right) + \frac 12(\Gamma_{ij}^k + \Gamma^k_{ji})\\ &= \Gamma^k_{ij} \quad\text{by symmetry} \end{align*}

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One approach is to use the fact that the Levi-Civita connection is metric compatible: $$ 0 =\nabla_ig_{jk}=\frac{\partial g_{jk}}{\partial u^i}-\Gamma^l_{ji}g_{lk}-\Gamma^l_{ki}g_{jl},$$ and then permute indices, add and multiply by the inverse metric to isolate $\Gamma^i_{jk}$. Remember that by definition, the Christoffel symbols are symmetric: $\Gamma^i_{jk}=\Gamma^i_{kj}$. (A slight cheat: knowing the answer gives a guide as to which permutation to consider.)

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