# Proving that the Christoffel connection transforms like a connection

In Sean Carroll's intro to GR, he shows that a connection transforms as follows: $$\Gamma^{\upsilon'}_{\mu'\lambda'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\frac{\partial x^\lambda}{\partial x^{\lambda'}}\frac{\partial x^\upsilon}{\partial x^{\upsilon'}} \Gamma^\upsilon _{\mu\lambda} + \frac{\partial x^\mu}{\partial x^{\mu'}}\frac{\partial x^\lambda}{\partial x^{\lambda'}}\frac{\partial^2x^{\upsilon'}}{\partial x^\mu \partial x^\lambda}$$

I remember reading somewhere on here that this textbook might have a typo, and there is actually a minus sign instead of a sum here, but I haven't actually performed that calculation to check, for now. Later on in the book, after introducing some properties the covariant derivative must have in order to be of use in GR, he shows that the Christoffel connection takes the form: $$\Gamma^\sigma _{\mu\upsilon} = \frac 12 g^{\sigma\rho} (\partial _\mu g _{\upsilon\rho} + \partial _\upsilon g _{\rho\mu} - \partial _\rho g _{\mu\upsilon})$$

I want to prove this myself, so I started with an arbitrary transformed Christoffel connection and replaced all the transformed metric forms with their transformation laws:$$\Gamma^{\sigma'} _{\mu'\upsilon'} = \frac {1}{2} \frac{\partial x^{\sigma'}}{\partial x^\sigma} g^{\sigma\rho}\left( \frac {\partial}{\partial x^{\mu'}}\frac{\partial x^\upsilon}{\partial x^{\upsilon'}} g _{\upsilon\rho} + \frac {\partial}{\partial x^{\upsilon'}}\frac{\partial x^\mu}{\partial x^{\mu'}} g _{\rho\mu} - \frac {\partial}{\partial x^{\rho}}\frac{\partial x^\mu}{\partial x^{\mu'}}\frac{\partial x^\upsilon}{\partial x^{\upsilon'}} g _{\mu\upsilon} \right)$$

I didn't transform the $$\rho$$ index because it is a dummy index that should fall out of the equation if everything goes to plan (typing this out now, I realize this might be a mistake, because partial derivatives of tensors generally don't transform as tensors, but if it really is a mistake please let me know). After this, I used the Leibniz rule on the transformed metrics (I broke the last term into the metric itself and the transformation matrices by themselves) and managed to get the first term of the connection transformation law. My result was: $$\Gamma^{\sigma'} _{\mu'\upsilon'} = first term + \frac {1}{2} \frac{\partial x^{\sigma'}}{\partial x^\sigma} g ^{\sigma\rho}\left( g _{\upsilon\rho}\frac{\partial x^\mu}{\partial x^{\mu'}}\frac{\partial}{\partial x^\mu} \frac{\partial x^\upsilon}{\partial x^{\upsilon'}} + g _{\rho\mu}\frac{\partial x^\upsilon}{\partial x^{\upsilon'}}\frac{\partial}{\partial x^\upsilon} \frac{\partial x^\mu}{\partial x^{\mu'}} - g _{\mu\upsilon}\frac{\partial}{\partial x^{\rho}}\frac{\partial x^\mu}{\partial x^{\mu'}} \frac{\partial x^\upsilon}{\partial x^{\upsilon'}}\right)$$

I broke up the last term using Leibniz rule again, and I summed up the terms I had by noticing that they only differed by index name (for example, the first two terms within the brackets in the above equation are the same and only differ by renaming $$\mu \to \upsilon$$), and I ended up with: $$\Gamma^{\sigma'} _{\mu'\upsilon'} = first term + \frac{\partial x^{\sigma'}}{\partial x^\sigma} g ^{\sigma\rho}\left(g _{\rho\mu}\frac{\partial x^\upsilon}{\partial x^{\upsilon'}}\frac{\partial}{\partial x^\upsilon} \frac{\partial x^\mu}{\partial x^{\mu'}} - g _{\mu\upsilon} \frac{\partial x^\mu}{\partial x^{\mu'}}\frac{\partial}{\partial x^\rho} \frac{\partial x^\upsilon}{\partial x^{\upsilon'}} \right)$$

Finally, in order to get the $$\frac {\partial x^{\sigma'}}{\partial x^\sigma}$$ term under the $$\frac {\partial}{\partial x^\upsilon}$$ derivative, I used the Leibniz rule one last time, and managed to get a term very similar to the second term in the connection transformation law, along with some extra terms. The result: $$\Gamma^{\sigma'} _{\mu' \upsilon'} = first term - \delta^\sigma _\mu \frac{\partial x^\mu}{\partial x^{\mu'}}\frac{\partial x^\upsilon}{\partial x^{\upsilon'}}\frac{\partial}{\partial x^{\upsilon}}\frac{\partial x^\sigma}{\partial x^{\sigma'}} + g ^{\sigma\rho}g _{\mu\rho}\frac {\partial x^\upsilon}{\partial x^{\upsilon'}}\frac {\partial}{\partial x^\upsilon}\left( \frac {\partial x^\mu}{\partial x^{\mu'}} \frac {\partial x^{\sigma'}}{\partial x^\sigma}\right) - g ^{\sigma\rho} g _{\mu\upsilon}\frac {\partial x^{\sigma'}}{\partial x^\sigma}\frac {\partial x^\mu}{\partial x^{\mu'}} \frac {\partial}{\partial x^\rho}\frac {\partial x^\upsilon}{\partial x^{\upsilon'}}$$

I can see that if the Kronecker delta works on transformation matrices the same way it works on tensors, then the 2nd term in the above equation matches the second term in the connection transformation law exactly, except for the minus sign (like I mentioned, the plus sign from the book might be a typo; please let me know which version is correct). The terms that have me stumped are the last two; if the Kronecker delta works on transformation matrices, I do not know how to reduce them to zero. The last term especially, the $$g ^{\sigma\rho} g_{\mu\upsilon}$$, is difficult to deal with because I cannot contract that term to a delta, because the indices don't match.

Am I on the right path in proving this assertion? If yes, how do I deal with the last two terms in the last equation? And if not, how do you go about proving that the Christoffel connection transforms like a connection?