I am trying to derive equation
$$ G_{\mu\nu}=2\nabla_{\mu}\nabla_{\nu}\phi-2\nabla_{\mu}\phi\nabla_{\nu}\phi+3g_{\mu\nu}(\nabla\phi)^2-2g_{\mu\nu}\Box\phi-\lambda^2g_{\mu\nu}e^{2\phi} $$
from these notes.
The given metric is $$ ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}+\frac{1}{\lambda^2}e^{-2\phi}d\Omega^2 $$
I tried calculating the Christoffel symbols and derive the equation using $$ G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu} $$ but I'm not sure on how to handle the $d\Omega^2$ components of the metric.
Edit: The action for the given equations of motion is $$ \mathcal{S}=\frac{1}{2\pi}\int{d^2x\sqrt{-g} e^{-2\phi}\left(R+2\nabla_{\mu}\phi\cdot\nabla^{\mu}\phi+2\lambda^2e^{2\phi}\right)} $$
My Answer: Special thanks to ApolloRa I litterally used multiple answers from you so thanks!
Firstly, to obtain the Einstein tensor we have to vary the action and from the equations of motion solve for $R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}$. Now the action reads
$$ \mathcal{S}=\frac{1}{2\pi}\int{d^2x \sqrt{-g}e^{-2\phi}\left(R+2g^{\mu\nu}\nabla_{\mu}\phi\nabla_{\nu}\phi+2\lambda^2e^{2\phi}\right)} $$ the variation w.r.t to $g_{\mu\nu}$ is $$ \delta\mathcal{S}=\frac{1}{2\pi}\int{d^2x}\left(\delta\sqrt{-g}e^{-2\phi}\left(R+2g^{\mu\nu}\nabla_\mu\phi\nabla_\nu\phi+2\lambda^2e^{2\phi}\right)+\sqrt{-g}\left(\delta(e^{-2\phi}R)+2\delta(e^{-2\phi}g_{\mu\nu}\nabla_\mu\phi\nabla_\nu\phi\right)\right) $$ Now I show $\delta(e^{-2\phi}R)=\delta(e^{-2\phi}g^{\mu\nu}R_{\mu\nu})$ where $$ \delta(e^{-2\phi}g^{\mu\nu}R_{\mu\nu}) = e^{-2\phi}R_{\mu\nu}\delta g^{\mu\nu} + g^{\mu\nu}e^{-2\phi}\delta R_{\mu\nu} $$
Then we use the Palatini identity where we have $$ e^{-\phi}\delta R_{\mu\nu} g^{\mu\nu} = e^{-2\phi}(g_{\mu\nu}\Box - \nabla_\mu\nabla_\nu)\delta g^{\mu\nu} $$ Now for these 2 terms we use the Leibniz rule as many times as needed to get the $\delta g_{\mu\nu}$ outside the derivative.
When we do that we get $$ e^{-2\phi} g_{\mu\nu}\Box \delta g^{\mu\nu} = \text{total derivative} -2e^{-2\phi}g_{\mu\nu}\delta g^{\mu\nu}\Box \phi + 4e^{-2\phi}g_{\mu\nu}\delta g^{\mu\nu}(\nabla\phi)^2 $$ and
$$ e^{-2\phi}\nabla_\mu\nabla_\nu\delta g^{\mu\nu} = \text{total derivative} +\nabla_\mu\nabla_\nu(e^{-2\phi}) \delta g^{\mu\nu} $$
Now combining these together we get $$ \delta(e^{-2\phi}R) = \text{total derivatives} +e^{-2\phi}R_{\mu\nu}\delta g^{\mu\nu}-2e^{-2\phi}g_{\mu\nu}\Box\phi\delta g^{\mu\nu}+4e^{-2\phi}g_{\mu\nu}(\nabla\phi)^2\delta g^{\mu\nu}+2\nabla_\mu\nabla_\nu e^{-2\phi}\delta g^{\mu\nu}-4\nabla_\mu\phi\nabla_\nu\phi e^{-2\phi}\delta g^{\mu\nu} $$ Now the variation of the action becomes $$ \delta\mathcal{S} = \frac{1}{2\phi}\int{d^2x\left(-\frac{1}{2}\sqrt{-g}\delta g^{\mu\nu}g_{\mu\nu}e^{-2\phi}(R+2(\nabla\phi)^2+2\lambda^2e^{2\phi})+\sqrt{-g}\left(\text{total derivatives} + e^{-2\phi}R_{\mu\nu}\delta g^{\mu\nu}-2e^{-2\phi}g_{\mu\nu}\Box\phi\delta g^{\mu\nu}+4e^{-2\phi}g_{\mu\nu}(\nabla\phi)^2\delta g^{\mu\nu}+2\nabla_\mu\nabla_\nu\phi e^{-2\phi}\delta g^{ \mu\nu}-2\nabla_\mu\phi\nabla_\nu\phi e^{-2\phi}\delta g^{\mu\nu}\right)+\sqrt{-g} 2 e^{-2\phi}\nabla_\mu\phi\nabla_\nu\phi\delta g^{\mu\nu}\right)} $$ Imposing $\delta\mathcal{S}=0$ and that the total derivatives vanish we get $$ -\frac{1}{2}g_{\mu\nu}(R+2(\nabla\phi)^2+2\lambda^2e^{2\phi})+ R_{\mu\nu}-2g_{\mu\nu}\Box\phi+4g_{\mu\nu}(\nabla\phi)^2+2\nabla_\mu\nabla_\nu\phi - 4\nabla_\mu\phi\nabla_\nu\phi+2\nabla_\mu\phi\nabla_\nu\phi = 0 $$ Now solving for $R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=G_{\mu\nu}$ I get $$ G_{\mu\nu}=-2\nabla_\mu\nabla_\nu\phi +2\nabla_\mu\nabla\nu\phi-3g_{\mu\nu}(\nabla\phi)^2+2g_{\mu\nu}\Box\phi+\lambda^2g_{\mu\nu}e^{2\phi} $$ which is almost correct if you ignore the minus sign. I have been searching for the sign error for too long now. Where is the error?
PS:ApolloRa thank you again!
