On Section 3.2 of Carroll's Spacetime and Geometry, he works on the proof of how Christoffel symbols transform under a change of coordinate system, in order to the covariant derivative of a tensor continue to be a tensor. On equation (3.9), he writes:
$$\Gamma^{\nu'}_{\mu' \lambda'}\dfrac{\partial x^{\lambda'}}{\partial x^\lambda}V^\lambda + \dfrac{\partial x^\mu}{\partial x^{\mu'}}V^\lambda\dfrac{\partial}{\partial x^\mu}\dfrac{\partial x^{\nu'}}{\partial x^\lambda} = \dfrac{\partial x^\mu}{\partial x^{\mu'}}\dfrac{\partial x^{\nu'}}{\partial x^\nu}\Gamma^{\nu}_{\mu \lambda}V^\lambda. \tag{3.9}$$
This equation must be true for any vector $V^\lambda$, so we can eliminate that on both sides. Then the connection coefficients in the primed coordinates may be isolated by multiplying by $\dfrac{\partial x^\lambda}{\partial x^{\sigma'}}$ and relabeling $\sigma' \rightarrow \lambda'$. The result is
$$\Gamma^{\nu'}_{\mu' \lambda'} = \dfrac{\partial x^\mu}{\partial x^{\mu'}}\dfrac{\partial x^\lambda}{\partial x^{\lambda'}}\dfrac{\partial x^{\nu'}}{\partial x^\nu}\Gamma^{\nu}_{\mu \lambda} + \dfrac{\partial x^\mu}{\partial x^{\mu'}}\dfrac{\partial x^\lambda}{\partial x^{\lambda'}}\dfrac{\partial^2 x^{\nu'}}{\partial x^\mu\partial x^\lambda}.\tag{3.10}$$
However, as far as I can see, there should be a minus sign in the middle of the expression, not a plus. Was that a typo by Carroll?
Besides, if I choose to isolate the unprimed Christoffel symbol instead of the primed one, I obtain
$$\Gamma^{\nu}_{\mu \lambda} = \dfrac{\partial x^{\mu'}}{\partial x^{\mu}}\dfrac{\partial x^{\lambda'}}{\partial x^{\lambda}}\dfrac{\partial x^{\nu}}{\partial x^{\nu'}}\Gamma^{\nu'}_{\mu' \lambda'} + \dfrac{\partial x^\nu}{\partial x^{\nu'}}\dfrac{\partial^2 x^{\nu'}}{\partial x^\mu\partial x^\lambda}$$
Shouldn't there be some kind of symmetry between the transformation formulas? In the sense that if in this last equation I just prime what is unprimed and unprime what is primed, shouldn't I obtain the transformation law from the unprimed Christoffel symbols to the primed ones? (I really don't know if that's true, but intuitively it seems so, because the indices shouldn't matter).
Best Answer
Yes. You are right. The sign should be minus.
It is interesting that in the original version published by Addison Wesley in 2004, the sign in (3.10) is right. But in the following versions reissued by Pearson in 2014, and Cambridge at 2019, the signs are wrong.
So, the correct one is $$ \Gamma^{\nu'}_{\mu' \lambda'} = \dfrac{\partial x^\mu}{\partial x^{\mu'}}\dfrac{\partial x^\lambda}{\partial x^{\lambda'}}\dfrac{\partial x^{\nu'}}{\partial x^\nu}\Gamma^{\nu}_{\mu \lambda} - \dfrac{\partial x^\mu}{\partial x^{\mu'}}\dfrac{\partial x^\lambda}{\partial x^{\lambda'}}\dfrac{\partial^2 x^{\nu'}}{\partial x^\mu\partial x^\lambda}.\tag{3.10} $$ It can also be rewritten as $$ \begin{split} \Gamma^{\nu'}_{\mu' \lambda'} &= \dfrac{\partial x^\mu}{\partial x^{\mu'}}\dfrac{\partial x^\lambda}{\partial x^{\lambda'}}\dfrac{\partial x^{\nu'}}{\partial x^\nu}\Gamma^{\nu}_{\mu \lambda} - \dfrac{\partial x^\mu}{\partial x^{\mu'}}\dfrac{\partial x^\lambda}{\partial x^{\lambda'}}\dfrac{\partial^2 x^{\nu'}}{\partial x^\mu\partial x^\lambda} \\ &= \dfrac{\partial x^\mu}{\partial x^{\mu'}}\dfrac{\partial x^\lambda}{\partial x^{\lambda'}}\dfrac{\partial x^{\nu'}}{\partial x^\nu}\Gamma^{\nu}_{\mu \lambda} - \dfrac{\partial x^\mu}{\partial x^{\mu'}}\dfrac{\partial x^\lambda}{\partial x^{\lambda'}}\dfrac{\partial}{\partial x^\lambda}\left(\dfrac{\partial x^{\nu'}}{\partial x^\mu}\right) \\ &= \dfrac{\partial x^\mu}{\partial x^{\mu'}}\dfrac{\partial x^\lambda}{\partial x^{\lambda'}}\dfrac{\partial x^{\nu'}}{\partial x^\nu}\Gamma^{\nu}_{\mu \lambda} - \dfrac{\partial x^\mu}{\partial x^{\mu'}}\dfrac{\partial}{\partial x^{\lambda'}}\left(\dfrac{\partial x^{\nu'}}{\partial x^\mu}\right) \\ &= \dfrac{\partial x^\mu}{\partial x^{\mu'}}\dfrac{\partial x^\lambda}{\partial x^{\lambda'}}\dfrac{\partial x^{\nu'}}{\partial x^\nu}\Gamma^{\nu}_{\mu \lambda} - \dfrac{\partial}{\partial x^{\lambda'}}\left(\dfrac{\partial x^\mu}{\partial x^{\mu'}}\dfrac{\partial x^{\nu'}}{\partial x^\mu}\right) + \dfrac{\partial^2 x^\mu}{\partial x^{\lambda'}\partial x^{\mu'}}\dfrac{\partial x^{\nu'}}{\partial x^{\mu}} \\ &= \dfrac{\partial x^\mu}{\partial x^{\mu'}}\dfrac{\partial x^\lambda}{\partial x^{\lambda'}}\dfrac{\partial x^{\nu'}}{\partial x^\nu}\Gamma^{\nu}_{\mu \lambda} + \dfrac{\partial^2 x^\nu}{\partial x^{\mu'}\partial x^{\lambda'}}\dfrac{\partial x^{\nu'}}{\partial x^{\nu}}. \end{split} $$