I am working on 2D Liouville field theory and trying to follow mostly Harold Erbin's note on 2d quantum gravity and Liouville theory.
I have a really simple question:
One consider the Euclidean Liouville action which is given by
$S_L = \frac{1}{4\pi}\int d^2\sigma\sqrt{h}\left(h^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi+QR\phi+4\pi\mu e^{2b\phi}\right)$
Now, in order to get the e.o.m as well as the stress-energy tensor, we have to vary the action, which yields
\begin{equation}\delta_hS_L = \frac{1}{4\pi}\int d^2\sigma\sqrt{h}\delta h^{\mu\nu}\left[-\frac{1}{2}h_{\mu\nu}\left(h^{\rho\sigma}\partial_{\rho}\phi\partial_{\sigma}\phi+QR\phi+4\pi\mu e^{2b\phi}\right)\\+\left(\partial_{\mu}\phi\partial_{\nu}\phi+QR_{\mu\nu}\phi+Q(h_{\mu\nu}\Delta\phi-\nabla_{\mu}\nabla_{\nu}\phi)\right)\right] ,\end{equation} while the variation w.r.t. $\phi$ gives, \begin{equation}\delta_{\phi}S_L = \frac{1}{4\pi}\int d^2\sigma\sqrt{h}\delta \phi\left(-2\Delta\phi+QR+8\pi\mu be^{2b\phi}\right)\end{equation}The equation of motion for $\phi$ are given in the usual way and one obtain \begin{equation}QR[h]-2\Delta\phi=-8\pi\mu b e^{2b\phi}\end{equation} If one consider the flat metric, then this reduces to \begin{equation}\partial_{\mu}\partial^{\mu}\phi=4\pi\mu b e^{2b\phi}\end{equation}
The stress energy tensor is computed in the usual way using \begin{equation}T_{\mu\nu} = -\frac{4\pi}{\sqrt{h}}\frac{\delta S}{\delta h^{\mu\nu}}\end{equation} In the notes, it is claimed that this gives \begin{equation}T_{\mu\nu}= -\left(\partial_{\mu}\phi\partial_{\nu}\phi-\frac{1}{2}h_{\mu\nu}h^{\rho\sigma}\partial_{\rho}\phi\partial_{\sigma}\phi\right)+Q(-h_{\mu\nu}\Delta\phi+\nabla_{\mu}\nabla_{\nu}\phi)+2\pi\mu be^{2b\phi}h_{\mu\nu}\end{equation}
$\textbf{Question 1:}$ Why does the contribution proportional to $R$ vanishes? I have the feeling that this is the stress energy tensor for flat space but not for curved space of do they vanish in every case?
Then, we go to complex plane (section 6.5.2). The metrix is given by $ds^2 = dzd\bar{z}$. This means that
\begin{equation}g_{zz}=g_{\bar{z}\bar{z}}=0, g_{z\bar{z}}=g_{\bar{z}z}=\frac{1}{2}\end{equation} Also, the complex coordinates derivative are easily found to be $\partial_{z}= \frac{1}{2}(\partial_0-i\partial_1)$.
$\textbf{Question 2:}$ I don't fully understand how to transform the e.o.m as well as the stress-energy tensor in those new coordinates. The result should be:
\begin{equation}\partial\bar{\partial}\phi = 4\pi \mu be^{eb\phi}\end{equation} \begin{equation}T(z) = T_{zz} = -(\partial\phi)^2+Q\partial^2\phi+2\pi\mu e^{eb\phi}\end{equation}
Can someone give me a hint please?