I'm reading "Lectures on black holes and the $AdS_3/CFT_2$ correspondence" by Kraus.


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.

  • $\begingroup$ Was your question answered by Jon? If so, please consider marking it as correct. If you obtained more information from the authors, then consider adding detail to the question, or providing an answer of your own based on your correspondence. $\endgroup$ – qubyte Dec 26 '11 at 3:55

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.

  • $\begingroup$ What would you substitute for the stress-energy tensors $T_{ww}, T_{\bar w \bar w}$ and the currents $J_w, \tilde J_{\bar w}$? $\endgroup$ – Satoshi Nawata Nov 24 '11 at 15:16
  • $\begingroup$ Eqs.(6.8) in the Kraus' review. $\endgroup$ – Jon Nov 24 '11 at 19:46
  • $\begingroup$ I thought plugging 6.8 into 7.11 would give us 7.12. But it does not. How would you explain the relative sign between $A^2_w$ and $\widetilde A^2_w$? Why there are no term like $\tau A_wA_{\bar w}$ even though you have $\tau_2$ in 7.11? – $\endgroup$ – Satoshi Nawata Nov 25 '11 at 2:16
  • $\begingroup$ I have just expanded the computation in my previous answer. $\endgroup$ – Jon Nov 25 '11 at 9:20
  • $\begingroup$ Thank you very much. But could you explain more elaborately how you obtain from the last equation you wrote to 7.12 in the paper? I have obtained the last equation you wrote before I post this question on the website. I would like to know the gap between this equation and 7.12. I am sorry about my stupid question, but I still don't understand it. $\endgroup$ – Satoshi Nawata Nov 25 '11 at 18:03

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