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I have been doing some calculation on variation of Ricci's tensor with respect to the metric, that, according with S. Carroll (An Introduction to General Relativity: Spacetime and Geometry, equation 4.62) is \begin{align}\delta R_{\mu \nu}= \nabla_{\rho}(\delta \Gamma_{\mu \nu}^{\rho}) - \nabla_{\nu}(\delta \Gamma_{\lambda \mu}^{\lambda}) \tag{4.62} \end{align} and I got a explicit equation in terms of the metric, which is \begin{equation} g^{\mu \nu} \delta R_{\mu \nu} = -\nabla_{\mu}\nabla_{\nu}\delta g^{\mu\nu} + g_{\mu \nu}\square\delta g^{\mu \nu} \tag{1}\end{equation} that agrees with the result obtained by Prahar here Variation of modified Einstein Hilbert Action. The thing is that, as far as I am concerned, \begin{align} g^{\mu \nu}\delta R_{\mu \nu}= g^{\mu \nu}\frac{\delta R_{\mu \nu}}{\delta g^{\alpha \beta}}\tag{2} \end{align} because it is the variation respect to the metric. So, if we write explicitly the variation $\delta$ as $\frac{\delta}{\delta g^{\alpha \beta}}$ in (1) we would get\begin{align}g^{\mu \nu} \frac{\delta}{\delta g^{\alpha \beta}} R_{\mu \nu} = -\nabla_{\mu}\nabla_{\nu}\frac{\delta}{\delta g^{\alpha \beta}} g^{\mu\nu} + g_{\mu \nu}\square\frac{\delta}{\delta g^{\alpha \beta}} g^{\mu \nu}\end{align} But \begin{align} \frac{\delta g^{\mu \nu}}{\delta g^{\alpha \beta}} = \frac{1}{2} \left(\delta_{\alpha}^{\mu}\delta_{\beta}^{\nu} + \delta_{\alpha}^{\nu}\delta_{\beta}^{\mu} \right) \tag{3} \end{align} and \begin{align}\nabla_{\mu}\delta_{\beta}^{\alpha}=0 \tag{4} \end{align} Therefore, plugging (3) and (4) into (1), we get \begin{align} \frac{\delta R_{\mu \nu}}{\delta g^{\alpha \beta}} = 0. \end{align} Is it true? if not, why does this fall? I think this time notation is messing around with me.

Thanks.

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This confusion can be settled with a much simpler example. Let $$ F(x) = \frac{d}{dx}f(x)\,, $$ Now $$ \delta F(x) = \frac{d}{dx}\delta f(x)\,. $$ With your reasoning one would say $$ \frac{\delta F}{\delta f} = \frac{d}{dx} \frac{\delta f}{\delta f} = \frac{d}{dx} 1 = 0\,, $$ which is nonsence. But actually $$ \frac{\delta f(x)}{\delta f(y)} = \delta(x-y)\,, $$ so $$ \frac{\delta F(x)}{\delta f(y)} = \frac{d}{dx} \delta(x-y)\,. $$ The derivative of the Dirac distribution is also a distribution. The same goes with your case, you have to put a Dirac delta in $$ \frac{\delta g^{\mu\nu}(x)}{\delta g^{\alpha\beta}(x')} = \frac12\left(\delta^{\mu}_\alpha\delta^\nu_\beta + \delta^{\nu}_\alpha\delta^\mu_\beta\right) \,\delta^d(x-x')\,. $$

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  • $\begingroup$ Thank you so much! This resolves my doubt. $\endgroup$ Mar 24, 2020 at 21:55
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    $\begingroup$ the correct way to say "thank you" is to upvote and/or accept the answer $\endgroup$
    – Yukterez
    Mar 30, 2020 at 6:53

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