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This question is about pages 95 and 96 of Carroll's book: Spacetime and Geometry.
We have the formula for the covariant derivate: $$\nabla _\mu V^\nu=\partial _\mu V^\nu + \Gamma _{\mu\lambda}^\nu V^\lambda$$ We want the covariant derivate to be a good tensor operator, so we impose that the right hand side of this equation is a tensor. This means that the right side should transform like a tensor.
After a lot of tedious math shenanigans we arrive at the following equation: $$\Gamma ^{\nu'}_{\mu ' \lambda '}\frac{\partial x^{\lambda '}}{\partial x^\lambda}V^\lambda+\frac{\partial x^\mu}{\partial x^{\mu '}}V^\lambda\frac{\partial}{\partial x^\mu}\frac{\partial x^{\nu '}}{\partial x^\lambda}=\frac{\partial x^\mu}{\partial x^{\mu '}}\frac{\partial x^{\nu '}}{\partial x^\nu}\Gamma ^{\nu}_{\mu \lambda }V^\lambda \ \ \ \ \ \ \ \ \ (1)$$ this equation must be true if the right side transforms like a tensor. Wonderful, so now from this one we can discover how the connection coefficients $\Gamma ^{\nu}_{\mu \lambda }$ have to transform.
Carroll now states that we can eliminate $V^\lambda$ from both sides, and then the connection coefficients in the primed coordinates may be isolated by multiplying by $$\frac{\partial x^\lambda}{\partial x^{\sigma '}}$$ and relabeling $\sigma ' \to \lambda '$.
At last Carrol states that the result is: $$\Gamma ^{\nu'}_{\mu ' \lambda '}=\frac{\partial x^\mu}{\partial x^{\mu '}}\frac{\partial x^\lambda}{\partial x^{\lambda '}}\frac{\partial x^{\nu '}}{\partial x^\nu}\Gamma ^{\nu}_{\mu \lambda }+\frac{\partial x^\mu}{\partial x^{\mu '}}\frac{\partial x^\lambda}{\partial x^{\lambda '}}\frac{\partial ^2 x^{\nu '}}{\partial x^\mu \partial x^\lambda} \ \ \ \ \ \ \ \ \ (2)$$ Here I have a problem.

My questions are:

  1. In (1) we have two terms on the left, in equation (2) we have two terms on the right; so we should get a minus in front of the second term on the right side in (2). Where did the minus sign go?
  2. Why multiply by $$\frac{\partial x^\lambda}{\partial x^{\sigma '}}$$ and then relabel? We could simply multiply by $$\frac{\partial x^\lambda}{\partial x^{\lambda '}}$$ right?
  3. On this wikipedia page are reported the transformation properties of the connection coefficients. But this properties are different from what we get by following Carroll. Why is that? (To be precise the wikipedia page is about Cristoffel Symbols, so a specific kind of connection coefficients, but anyway the transformation properties should be the same for every connection coefficients, so the problem remains)

Edit: This is what I have at page 96; indeed there is a plus sign.

enter image description here

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My copy of Carrol has the minus sign before the second term that you worry is missing in your eq 2.

carrol eqation

Now the Wikipedia page you cite has a plus in this location and so appears to be different --- but note that in Carrol the downstairs indices in the second derivative refer to the unprimed frame, while in the Wikipedia article they are for the primed frame. The Wiki article has taken the term from left to right therefore; hence the difference in minus signs.

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  • $\begingroup$ I have edited my question, check it out. $\endgroup$ – Noumeno Jul 11 at 14:29
  • $\begingroup$ Also you haven't commented about the strange relabeling. $\endgroup$ – Noumeno Jul 11 at 14:30
  • $\begingroup$ Because you can't just multiply by ${\partial x^\lambda}/{\partial x^{\lambda'}}$. You would have three $\lambda'$ indices. $\endgroup$ – mike stone Jul 11 at 15:49
  • $\begingroup$ And regarding the plus sign in my book, as the photo demonstrates? $\endgroup$ – Noumeno Jul 11 at 16:30
  • $\begingroup$ All I can say is that my copy has a minus sign in eq 3.10. As far as I can tell mine is a first printing copyright 2004. The error must have crept in in a later print run. I'll seeif I can upload an image to may answer! $\endgroup$ – mike stone Jul 11 at 17:05

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