Derivation of Eq. 7.12 in the review paper of Kraus I'm reading "Lectures on black holes and the $AdS_3/CFT_2$ correspondence" by Kraus.
http://arxiv.org/abs/hep-th/0609074
I don't know how one can obtain Eq.7.12. My stupid question is how to obtain this equation. After this equation, it is stated that "one has to take care to consider only variations consistent with the equations of motion and the assumed boundary conditions". What are the variations consistent with the equations of motion and the assumed boundary conditions? What Krasu says is as follows.
To compute the bulk functional integral, we need to evaluate the bulk action for the solutions 
which contribute, including boundary counterterms if necessary. For an on-shell solution around the $AdS_{3}$ vacuum, one can evaluate 
the action at the $AdS_3$ vacuum by using 
the variation of the action with respect to the boundary metric $g^{(0)}$ and the 
gauge fields $A^{(0)}, \tilde A^{(0)}$ 
\begin{equation}
\delta S=  \int   d^2x \sqrt{g}\,\left[
\frac12 T^{ij} \delta g_{ij}+\frac{i}{2\pi} J^{i} \delta  A_{i}
+\frac{i}{2\pi}\tilde J^{i} \delta \tilde A_{i} \right]~.
\end{equation}
where the superscript $(0)$ is omitted for brevity.
Reexpressing this in complex coordinates of the boundary metric, we obtain:
\begin{equation}
\delta S=4\pi i \left(T_{ww}\delta \tau+T_{\bar w\bar w}\delta \bar\tau
+\frac{\tau_2}{\pi} J_{w}\delta A_{\bar w}+\frac{\tau_2}{\pi} \tilde J_{\bar w}\delta \tilde A_{w}\right)~.
\end{equation}
One can integrate the above equation to get: 
\begin{eqnarray}
\mathcal{S}(\tau) & = & - 2 \pi i \tau \big(L_{0} - \frac{c}{24} \big) + 2 \pi i \bar\tau \big(\tilde L_{0} - \frac{\tilde c}{24} \big) \cr
&& \; - \frac{i\pi}{2} k \big( \tau A_{w}^{2} + \bar \tau A_{\bar w}^{2} + 2 \bar \tau A_{w} A_{\bar w} \big) 
+ \frac{i\pi}{2} \tilde k \big( \tau \tilde A_{w}^{2} + \bar\tau \tilde A_{\bar w}^{2} + 2  \tau \tilde A_{w} \tilde A_{\bar w} \big)  \, . 
\end{eqnarray}
I would like to know the derivation of the last equation.
 A: The answer is quite simple. You should use eqs.(6.8) in the paper. You put them into the last term of eq.(7.11) then, a straightforward integration, I mean something like $\delta\tau\rightarrow\tau$, $\delta A_{\bar w}\rightarrow A_{\bar w}$ and so on, should do the job.
So, let us consider (note that in your post there is a wrong sign)
$$
\delta S=(2\pi)^2 i \left(-T_{ww}\delta \tau+T_{\bar w\bar w}\delta \bar\tau
+\frac{\tau_2}{\pi} J_{w}^I\delta A_{I\bar w}+\frac{\tau_2}{\pi} \tilde J_{\bar w}^I\delta \tilde A_{Iw}\right)_{constant}~.
$$
(here "constant" means that only the zero mode is retained) and the corresponding eqs.(6.8) in Kraus' review
$$\eqalign{ T_{ww}&=-{k \over 8\pi} +{1 \over 8\pi}
A_w^2+{1 \over 8\pi} {\tilde A}_w^2~,  \cr T_{{\bar w}{\bar w}}&= -{{\tilde k} \over 8\pi}
+{1 \over 8\pi}A_{{\bar w}}^2+{1 \over 8\pi}{\tilde A}_{{\bar w}}^2~, \cr J^I_w &=
{i\over 2} k^{IJ} A_{Jw}~, \cr {\tilde J}^I_{{\bar w}}& = {i\over 2} {\tilde k}^{IJ}
{\tilde A}_{J{\bar w}}~.}$$
By substitution one has
$$ \delta S = (2\pi)^2  i\left[-\left(-{k \over 8\pi} +{1 \over 8\pi}
A_w^2+{1 \over 8\pi} {\tilde A}_w^2~\right)\delta\tau+\left(-{{\tilde k} \over 8\pi}
+{1 \over 8\pi}A_{{\bar w}}^2+{1 \over 8\pi}{\tilde A}_{{\bar w}}^2~\right)\delta{\bar\tau}\right.$$
$$\left.+\frac{i\tau_2}{2\pi}k^{IJ} A_{Jw}\delta A_{I\bar w}+\frac{i\tau_2}{2\pi}{\tilde k}^{IJ}{\tilde A}_{J{\bar w}}\delta \tilde A_{Iw}\right]_{constant}.$$
From this you get immediately the result when you note that the variation with respect to the gauge field just cancels out the $\tau_1$ contribution, having ${\bar\tau}-\tau$ that comes from the squared terms, and is recovered upon integration.
