# Prove Christoffel Symbol Identity

In a book I am reading, the following identity is claimed and then "left to the reader to prove." $g_{ij}$ is the metric tensor, and $\Gamma$ is the Christoffel symbol of the second kind with the appropriate indices.

$$\partial_k g_{ij} = g_{jl}\Gamma^{l}_{ki}+g_{il}\Gamma^{l}_{kj}$$

I have tried expanding the $g_{ij}$ term using its definition, $g_{ij}=\epsilon_{i}\cdot\epsilon_{j}$, but then I don't really know if a vector identity should be used. Moreover, I'm not even sure if that's even on the right track.

Could you possibly give me nudge in the right direction? Do I need to assume the covariant derivative of the metric tensor is zero?

At the most basic level, you can just use the definition of the Christoffel symbols in terms of the metric:

$\Gamma^i_{jk} = \frac{1}{2}g^{is} (\partial_j g_{sk} + \partial_k g_{sj} - \partial_s g_{jk})$.

Plugging this into the right-hand side of your expression will yield the left-hand side.

However, one can obtain your expression directly from one of the properties of the Christoffel symbols; namely, that they are the connection coefficients of a metric-compatible affine connection (i.e. they can be used to construct a covariant derivative operator $\nabla_i$ which satisfies $\nabla_i g_{jk} = 0$). Expanding the equation $0 = \nabla_i g_{jk}$ out explicitly, we obtain

$0 = \nabla_i g_{jk} = \partial_i g_{jk} - \Gamma^s_{ij} g_{sk} - \Gamma^s_{ik} g_{js}$,

which gives

$\partial_i g_{jk} = \Gamma^s_{ij} g_{sk} + \Gamma^s_{ik} g_{js}$.

This is precisely the equation you're after.