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I am deriving the Einstein equation using the Einstein-Hilbert action: enter image description here

It is obvious that the variation in the Riemann Tensor is calculated from a variational product rule. What is not obvious to me is why variations obey this rule, and I'll like an explanation.

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    $\begingroup$ Elementary answer: Variations are derivatives of 1-parameter families of functions, thus they follow the product rule. Functional answer: Variations are exterior derivatives on the covariant phase space, and as such they follow the (anti-)product rule. Differential geometric answer: Variations are Lie derivatives on a jet bundle, and as such they follow the product rule. $\endgroup$ Nov 4, 2019 at 20:22

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The easiest way to go about this is to think of variation in terms of the background field method. Here, you write the metric as some background plus perturbation:

\begin{equation} g_{\mu \nu} = \bar{g}_{\mu \nu} + h_{\mu \nu} \end{equation}

and then you expand every quantity that depends on $g_{\mu \nu}$ as a power series in $h$:

$$ \Gamma = \bar{\Gamma} + \Gamma^{(1)} + \Gamma^{(2)} \cdots \\ R = \bar{R} + R^{(1)} + R^{(2)} \cdots $$

where the superscript $^{(n)}$ labels the $\mathcal{O} (h^n)$ perturbation. You can then easily see that

$$ \Gamma \Gamma = (\bar{\Gamma} + \Gamma^{(1)} + \Gamma^{(2)} \cdots)(\bar{\Gamma} + \Gamma^{(1)} + \Gamma^{(2)} \cdots) $$

Extracting the $\mathcal{O}(h)$ variation, like in your problem, we automatically get both the contributions:

$$ (\Gamma \Gamma)^{(1)} = \bar{\Gamma} \Gamma^{(1)} + \Gamma^{(1)} \bar{\Gamma} $$

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It's a fundamental property of infinitesimal variations $\delta$ that they are linear derivations, i.e. they obey Leibniz product rule. Short of a rigorous definition of infinitesimals, the proof is basically to consider variations as 1-parameter families of functions, and then differentiate wrt. the parameter, cf. above comment by Bence Racsko.

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