Help in understanding the derivation of Einstein equations

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SUMMARY

The discussion focuses on the derivation of the variation of the Riemann tensor and the evaluation of a specific integral in General Relativity (GR). The variation of the Riemann tensor is expressed as $$\delta R^{\rho}_{\phantom{\rho}\sigma\mu\nu}=\nabla_{\mu}\left(\delta\Gamma^{\rho}_{\phantom{\rho}\nu\sigma}\right)-\nabla_{\nu}\left(\delta\Gamma^{\rho}_{\phantom{\rho}\mu\sigma}\right)$$, with the condition $$\Gamma^{\lambda}_{\phantom{\lambda}\nu\mu}\delta\Gamma^{\rho}_{\phantom{\rho}\lambda\sigma}-\Gamma^{\lambda}_{\phantom{\lambda}\mu\nu}\delta\Gamma^{\rho}_{\phantom{\rho}\lambda\sigma}=0$$ being crucial. The integral $$\int\nabla_{\rho}A^{\rho}\sqrt{-g}\,\mathrm{d}^4x=0$$ is confirmed to be zero due to the application of the divergence theorem, which states that total derivatives vanish when integrated over the entire spacetime.

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Homework Statement
I am working through the derivation of the Einstein field equations by varying the Einstein-Hilbert action. I need some help in understanding certain steps
Relevant Equations
Given below
There are two parts to my question.

The first is concerns the variation of the Reimann tensor. I am trying to show

$$\delta R^{\rho}_{\phantom{\rho}\sigma\mu\nu}=\nabla_{\mu}\left(\delta\Gamma^{\rho}_{\phantom{\rho}\nu\sigma}\right)-\nabla_{\nu}\left(\delta\Gamma^{\rho}_{\phantom{\rho}\mu\sigma}\right)$$

In order to show the above, it is necessary that ##\Gamma^{\lambda}_{\phantom{\lambda}\nu\mu}\delta\Gamma^{\rho}_{\phantom{\rho}\lambda\sigma}-\Gamma^{\lambda}_{\phantom{\lambda}\mu\nu}\delta\Gamma^{\rho}_{\phantom{\rho}\lambda\sigma}=0##. Why is this true?

The second part concerns the term ##\int\nabla_{\rho}A^{\rho}\sqrt{-g}\,\mathrm{d}^4x=0## where ##A^{\rho}=g^{\sigma\nu}\delta\Gamma^{\rho}_{\phantom{\rho}\nu\sigma}-g^{\sigma\rho}\delta\Gamma^{\mu}_{\phantom{\mu}\mu\sigma}##. Why is the integral zero?
 
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Thomas1 said:
In order to show the above, it is necessary that ##\Gamma^{\lambda}_{\phantom{\lambda}\nu\mu}\delta\Gamma^{\rho}_{\phantom{\rho}\lambda\sigma}-\Gamma^{\lambda}_{\phantom{\lambda}\mu\nu}\delta\Gamma^{\rho}_{\phantom{\rho}\lambda\sigma}=0##. Why is this true?
It's been a long time since I took GR, so I may be misremembering. But aren't Christoffel symbols symmetric in the bottom two indices?
 
Thomas1 said:
The second part concerns the term ##\int\nabla_{\rho}A^{\rho}\sqrt{-g}\,\mathrm{d}^4x=0## where ##A^{\rho}=g^{\sigma\nu}\delta\Gamma^{\rho}_{\phantom{\rho}\nu\sigma}-g^{\sigma\rho}\delta\Gamma^{\mu}_{\phantom{\mu}\mu\sigma}##. Why is the integral zero?
It's a total derivative so vanishes by the divergence theorem upon integration over the whole spacetime. First show that\begin{align*}
\int d^4 x \sqrt{-g} \nabla_{\rho} A^{\rho} = \int d^4 x \partial_{\rho} (\sqrt{-g} A^{\rho})
\end{align*}Then use the divergence theorem of ordinary calculus\begin{align*}
\int d^4 x \partial_{\rho} (\sqrt{-g} A^{\rho}) = \oint d^3 x \sqrt{-\gamma} n_{\rho} A^{\rho}
\end{align*}where the integral on the rhs is taken over the boundary 3-surface of induced metric ##\gamma_{ab}## and normal ##n_{\rho}##
 
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