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I have a problem with derivation of the transformation law for Christoffel symbols: two different approaches give me two different results.

I assume that the equation for the covariant derivative of a vector shall be transformed as a tensor and transform it and those parts in it which I know. Here is what I get ($A^i_{\,;l}$ denoting covariant derivative): $$A^i_{\,;l} = \frac{\partial A^i}{\partial x^l}+\Gamma^i_{\,kl}A^k = \frac{\partial x^{'m}}{\partial x^l} \frac{\partial}{\partial x^{'m}} \left(A^{'n} \frac{\partial x^i}{\partial x^{'n}} \right) + \Gamma^i_{\,kl} A^{'n}\frac{\partial x^k}{\partial x^{'n}} =\\ =\frac{\partial x^{'m}}{\partial x^l} \frac{\partial A^{'n}}{\partial x^{'m}} \frac{\partial x^i}{\partial x^{'n}}+ \frac{\partial x^{'m}}{\partial x^l}A^{'n} \frac{\partial^2 x^i}{\partial x^{'n}\partial x^{'m}} + \Gamma^i_{\,kl} A^{'n}\frac{\partial x^k}{\partial x^{'n}} $$ On the other hand, $$ A^i_{\,;l} = A^{'m}_{\enspace;n}\frac{\partial x^i}{\partial x^{'m}}\frac{\partial x^{'n}}{\partial x^{l}} = \frac{\partial x^i}{\partial x^{'m}}\frac{\partial x^{'n}}{\partial x^{l}} \frac{\partial A^{'m}}{\partial x^{'n}} + \frac{\partial x^i}{\partial x^{'m}}\frac{\partial x^{'n}}{\partial x^{l}} \Gamma^{'m}_{\enspace nk} A^{'k}$$ From here I get, as far as $A^{'n}$ can be any vector, $$\frac{\partial x^{'m}}{\partial x^l}A^{'n} \frac{\partial^2 x^i}{\partial x^{'n}\partial x^{'m}} + \Gamma^i_{\,kl} A^{'n}\frac{\partial x^k}{\partial x^{'n}} = \frac{\partial x^i}{\partial x^{'m}}\frac{\partial x^{'n}}{\partial x^{l}} \Gamma^{'m}_{\enspace nk} A^{'k} = \frac{\partial x^i}{\partial x^{'m}}\frac{\partial x^{'k}}{\partial x^{l}} \Gamma^{'m}_{\enspace kn} A^{'n} \\ \Gamma^i_{\,kl} \frac{\partial x^k}{\partial x^{'n}} = \frac{\partial x^i}{\partial x^{'m}}\frac{\partial x^{'k}}{\partial x^{l}} \Gamma^{'m}_{\enspace kn} - \frac{\partial x^{'m}}{\partial x^l} \frac{\partial^2 x^i}{\partial x^{'n}\partial x^{'m}} \\ \Gamma^{i}_{\,kl} = \frac{\partial x^i}{\partial x^{'m}} \frac{\partial x^{'n}}{\partial x^{k}} \frac{\partial x^{'k}}{\partial x^{l}}\Gamma^{'m}_{\enspace,nk} - \frac{\partial x^i}{\partial x^{'m}}\frac{\partial^2 x^{'m}}{\partial x^{k}\partial x^{l}}$$ Meanwhile, if I express $A^{'m}_{\enspace;n}$ through $A^{i}_{\,;l}$ I get a different result: $$ A^{'m}_{\enspace;n} = \frac{\partial A^{'m}}{\partial x^n}+\Gamma^{'m}_{\enspace nk} A^{'k} = ... =\frac{\partial x^{r}}{\partial x^{'n}} \frac{\partial A^{i}}{\partial x^{r}} \frac{\partial x^{'m}}{\partial x^{i}}+ A^{i}\frac{\partial x^{r}}{\partial x^{'n}} \frac{\partial^2 x^{'m}}{\partial x^{i}\partial x^{r}} + \Gamma^{'m}_{\enspace nk}A^{i} \frac{\partial x^{'k}}{\partial x^{i}} \\ A^{'m}_{\enspace;n} = A^{i}_{\,;l}\frac{\partial x^{'m}}{\partial x^{i}}\frac{\partial x^{l}}{\partial x^{'n}} = \frac{\partial x^{'m}}{\partial x^{i}}\frac{\partial x^{l}}{\partial x^{'n}} \frac{\partial A^{i}}{\partial x^{l}} + \frac{\partial x^{'m}}{\partial x^{p}}\frac{\partial x^{l}}{\partial x^{'n}} \Gamma^{p}_{\enspace li} A^{i} \\ \Gamma^{i}_{\,kl} = \frac{\partial x^i}{\partial x^{'m}} \frac{\partial x^{'n}}{\partial x^{k}} \frac{\partial x^{'k}}{\partial x^{l}}\Gamma^{'m}_{\enspace,nk} + \frac{\partial x^i}{\partial x^{'m}}\frac{\partial^2 x^{'m}}{\partial x^{k}\partial x^{l}}$$

I know that the correct formula is the second one. Could you explain me what is wrong with the first approach?

I am sorry for a really silly question, as I guess.

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  • $\begingroup$ IN your first derivation, how did you get from line 2 to line 3? $\endgroup$ Commented Sep 8, 2016 at 23:47
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    $\begingroup$ I equal two equations that I have for covariant derivative. The first terms get terminated, as they are the same. Then I move the term with second-order derivative to the other side (here is where I get minus), rename indices so that I have A'k in every of the three terms, and just get rid of it. Finally, I multiply both sides by dx'n/dxu and come with transformation law. I will Tex it as soon as I will have access to my PC. $\endgroup$
    – user108687
    Commented Sep 9, 2016 at 5:23
  • $\begingroup$ @MarkFischler I updated the question $\endgroup$
    – user108687
    Commented Sep 9, 2016 at 17:22

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