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?


1 Answer 1


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.


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