# Christoffel symbol transformation law

It is known that the transformation rule when you change coordinate frames of the Christoffel symbol is:

$$\tilde \Gamma^{\mu}_{\nu\kappa} = {\partial \tilde x^\mu \over \partial x^\alpha} \left [ \Gamma^\alpha_{\beta \gamma}{\partial x^\beta \over \partial \tilde x^\nu}{\partial x^\gamma \over \partial \tilde x^\kappa} + {\partial ^2 x^\alpha \over \partial \tilde x^\nu \partial \tilde x^\kappa} \right ]$$

Is there any way to prove this rule using only the definition of the Christoffel via the metric tensor? That is, using:

$$\Gamma^\mu _{\nu\kappa} = \frac{1}{2}g^{\mu\lambda}\left(g_{\lambda\kappa,\nu}+g_{\nu\lambda,\kappa}-g_{\nu\kappa,\lambda} \right)$$

All proofs have I've seen of the transformation law involve another method.

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This is very straightforward, just substitute the transformation rules and collect the terms.

Here are some details.

The inverse metric transforms, as we know, by the rule: $$g^{\mu \lambda} = \frac{\partial{\bar{x}}^\mu}{\partial{x}^\alpha} \frac{\partial{\bar{x}}^\lambda}{\partial{x}^\delta} g^{\alpha \delta}$$

The partial derivatives need some calculations that can be presented as \begin{align*} g_{\lambda \kappa , \nu} & = \frac{\partial}{\partial{\bar{x}^\nu}} \Big( \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} g_{\delta \gamma} \Big) \\ &= \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} \frac{\partial{x^\beta}}{\partial{\bar{x}^\nu}} g_{\delta \gamma , \beta} + g_{\delta \gamma} \frac{\partial}{\partial{\bar{x}^\nu}} \Big( \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} \Big) \end{align*}

Similarly, $$g_{\nu \lambda , \kappa} = \frac{\partial{x^\beta}}{\partial{\bar{x}^\nu}} \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} g_{\beta \delta , \gamma} + g_{\beta \delta} \frac{\partial}{\partial{\bar{x}^\kappa}} \Big( \frac{\partial{x^\beta}}{\partial{\bar{x}^\nu}} \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} \Big)$$ and $$g_{\nu \kappa , \lambda} = \frac{\partial{x^\beta}}{\partial{\bar{x}^\nu}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} g_{\beta \gamma , \delta} + g_{\beta \gamma} \frac{\partial}{\partial{\bar{x}^\lambda}} \Big( \frac{\partial{x^\beta}}{\partial{\bar{x}^\nu}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} \Big)$$

Substituting these identities into your "definition" $$\Gamma^\mu _{\nu\kappa} = \frac{1}{2}g^{\mu\lambda}\left(g_{\lambda\kappa,\nu}+g_{\nu\lambda,\kappa}-g_{\nu\kappa,\lambda} \right)$$ and taking into account that $$\Gamma^\alpha _{\beta \gamma} = \frac{1}{2}g^{\alpha \delta}\left(g_{\delta \gamma , \beta}+g_{\beta \delta , \gamma} - g_{\beta \gamma , \delta} \right)$$ it is not difficult now to show the required transformation rule for the Christoffel symbols.

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