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Under the coordinate transformation $\bar x=x+\varepsilon$, the variation of the metric $g^{\mu\nu}$ is: $$ \delta g^{\mu\nu}(x)=\bar g^{\mu\nu}(x)-g^{\mu\nu}(x)=-\frac{\partial{ g^{\mu\nu}}}{\partial x^{\alpha}}\varepsilon^{\alpha}+ g^{\mu\beta}\frac{\partial \varepsilon^{\nu}}{\partial x^{\beta}}+g^{\alpha\nu}\frac{\partial \varepsilon^{\mu}}{\partial x^{\alpha}} $$ the right hand side is equal to $$- {g^{\mu\nu}}_{,\alpha}\varepsilon^{\alpha}+ {\varepsilon^{\mu,\nu}}+{\varepsilon^{\nu,\mu}}=\varepsilon^{\mu;\nu}+\varepsilon^{\nu;\mu}$$ I have problem with the proof of the last equality. $$ \varepsilon^{\mu;\nu}+\varepsilon^{\nu;\mu}=g^{\alpha\nu}{\varepsilon^{\mu}}_{;\alpha}+g^{\alpha\mu}{\varepsilon^{\nu}}_{;\alpha}= $$

$$ g^{\alpha\nu}({\varepsilon^{\mu}}_{,\alpha}+\Gamma_{\beta\alpha}^{\mu}\varepsilon^{\beta})+g^{\alpha\mu}({\varepsilon^{\nu}}_{,\alpha}+\Gamma_{\beta\alpha}^{\nu}\varepsilon^{\beta})= $$

$$ \varepsilon^{\mu,\nu}+g^{\alpha\nu}\frac{1}{2}g^{\mu\gamma}(g_{\gamma\beta,\alpha}+g_{\gamma\alpha,\beta}-g_{\beta\alpha,\gamma})\varepsilon^{\beta}+ \varepsilon^{\nu,\mu}+g^{\alpha\mu}\frac{1}{2}g^{\nu\gamma}(g_{\gamma\beta,\alpha}+g_{\gamma\alpha,\beta}-g_{\beta\alpha,\gamma})\varepsilon^{\beta}= $$ Considering the summation over the repeated indeces each of the three items in both brackets gives the same quantity coupling with the respective indeces as: A(B+C-D)E, ABE=ACE=ADE, then A(B+C-D)E=ACE. I chose ACE $$ \varepsilon^{\mu,\nu}+\varepsilon^{\nu,\mu}+g^{\alpha\mu}g^{\nu\gamma}g_{\gamma\alpha,\beta}\varepsilon^{\beta}={g^{\mu\nu}}_{,\beta}\varepsilon^{\beta}+{\varepsilon^{\mu}}^{,\nu}+{\varepsilon^{\nu}}^{,\mu} $$ I have the first term with plus sign, opposite to the original one. What I did wrong? What am I missing?

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    $\begingroup$ Please see Lie derivative - any standard treatment will show the partial derivatives can be everywhere be replaced by covariant derivatives because the (Levi-Civita) connections cancel $\endgroup$ – lux Jan 31 at 3:58
  • $\begingroup$ On my phone so I can’t write out equations, but (derivative of inverse metric is equal to minus the raised index version of the derivative of metric. $\endgroup$ – Prahar Mar 22 at 2:30
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I have done such calculation, I think it will be helpful:

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    $\begingroup$ Please use mathjax for typesetting. $\endgroup$ – Johan Liebert Jan 31 at 2:47
  • $\begingroup$ Thank you, Nikita. Could you do it with upper indices? $\endgroup$ – Constantin Jan 31 at 3:28
  • $\begingroup$ I can easily up indices because of covariant derivatives $\endgroup$ – Nikita Jan 31 at 4:20
  • $\begingroup$ Hello, Nikita. Why are you starting from the metric and not from the inverse one? How to derive for upper indices case not using the result of lower indices case? $\endgroup$ – Constantin Feb 14 at 15:29
  • $\begingroup$ I started with law indices due to canonical defenition of metric:$$ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu}$$ Of course, you can repeat all my calculations in case with upper indices. All calculations will be the same. If you wanna, I can do this calculation for you. $\endgroup$ – Nikita Feb 14 at 19:25

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