If you want to define a covariant derivative which transforms correctly, you need to define it as ##\nabla_i f_j = \partial_i f_j - f_k \Gamma^k_{ij}##, where ##\Gamma^k_{ij}## has the transformation property(adsbygoogle = window.adsbygoogle || []).push({});

##\bar{\Gamma}^k_{ij} = \frac{\partial \bar{x}_k}{\partial x_c}\frac{\partial x_a}{\partial \bar{x}_i}\frac{\partial x_b}{\partial \bar{x}_j}\Gamma^c_{ab} + \frac{\partial \bar{x}_k}{\partial x_c}\frac{\partial x_a}{\partial \bar{x}_i}\frac{\partial^2 x_c}{\partial \bar{x}_a\partial \bar{x}_j}##

This definition doesn't yet assume that ##\Gamma^k_{ij}## is the Levi-Civita connection. Any affine connection needs to have this transformation law in order to make the covariant derivative transform properly.

Now, if we want a torsion-free connection which preserves the metric (i.e. ##\Gamma^k_{ij} = \Gamma^k_{ji}## and ##\nabla_k g_{ij} = 0##), then it's straightforward to show that

##\Gamma^k_{ij} = \frac{1}{2}g^{kl}(\partial_i g_{lj} + \partial_j g_{il} - \partial_l g_{ij})##

However, this only shows that if a metric-compatible connection exists, it would have to take the form of that equation. It doesn't necessarily prove that this definition of ##\Gamma^k_{ij}## actually transforms according to the above equation.

I'd like to show this explicitly. I started by substituting in the transformation laws for each component of this definition, and tried to massage it into the form of the above transformation law, but I can't seem to wade my way through the sea of indicies to get everything to line up. I'm guessing that I'm missing some kind of trick for working with tensor transformation laws, but I'm not quite sure what it would be. Can somebody sketch out the method by which you would show this correspondence?

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# A Transformation properties of the Christoffel symbols

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