I'm trying to understand the derivation of the Einstein field equations through the stationary action principle. I've found a pretty clear explanation on the wikipedia article for the Einstein-Hilbert action(https://en.wikipedia.org/wiki/Einstein%E2%80%93Hilbert_action), which uses calculations based on those found in "Spacetime and Geometry : An introduction to General Relativity" by Sean M. Carroll.
I'm having trouble, though, with the notation used in the article : is the $\delta$-operator used in the article the functional derivative operator?
For example, i am interpreting the variation $\delta S$ of the action $S(g) = \int [\sqrt{-g}R + \sqrt{-g}L_m]d^4x $ as $ \lim_{\epsilon \to 0} \frac{d(S_{\epsilon})}{d\epsilon} $, where $ \epsilon \mapsto \sigma^{\epsilon}$ is a curve in the space of the metrics on spacetime, based in $g$, and $ S_{\epsilon}$ is $S(\sigma^{\epsilon})$
Then, i have interpreted terms like $\frac{\delta \sqrt{-g}}{\delta g^{\mu \nu}}$ as defined through the relation $ \delta \sqrt{-g} = \int \frac{\delta\sqrt{-g}}{\delta g^{\mu \nu}}\delta g^{\mu \nu} $ ( I have found this definition on the wikipedia article for functional derivatives, https://en.wikipedia.org/wiki/Functional_derivative .)
But when we calculate $ \delta R = R_{\mu \nu} \delta g^{\mu \nu}$ and the article follows with $\frac{\delta R}{\delta g^{\mu \nu}} = R_{\mu \nu} $.
I am having problems understanding this step : why are we just "dividing" both sides of the equations for $\delta g^{\mu \nu}$ and forgetting about the integral? Am i wrong in understanding the meaning of the $\delta$ symbol? If so, where can i find a more mathematically sound explanation?
Thank you very much for your time.
The functional derivative $DF$ of a functional $F$ (usually non-linear) is a linear functional such that, $$ \langle DF[\phi], \eta \rangle = \left. \frac{d}{d\lambda} F[\phi+\lambda\eta] \right|_{\lambda=0} $$ for all $\phi$ and $\eta$ in the domain of $F$ (or a sub-domain when it comes to $\eta$).
Often there exists a function $f$ such that the left hand side, the application of $DF[\phi]$ on $\eta,$ can be written as an integral, $$ \langle DF[\phi], \eta \rangle = \int f(x) \, \eta(x) \, dx. $$
In this case we can identify $DF[\phi]$ with $f$ and it's common to use Leibniz notation, $$ DF[\phi](x) = f(x) = \frac{\delta F[\phi]}{\delta \phi(x)}. $$
So, $\delta R = R_{\mu \nu} \, \delta g^{\mu \nu}$ means that (with the dependency of $R$ on $g$ made explicit), $$ \left. \frac{d}{d\lambda} R[g+\lambda\eta] \right|_{\lambda=0} = \int R_{\mu\nu}[g](x) \, \eta^{\mu\nu}(x) \, dx $$ and $$ \frac{\delta R[g]}{\delta g^{\mu\nu}} = R_{\mu\nu}[g]. $$