2
$\begingroup$

This problem has to do with bulk reconstruction in AdS/CFT. It is given that the bulk-to-boundary propagator can be obtained from the bulk-to-bulk propagator by the following relation (c.f. https://ocw.mit.edu/courses/physics/8-821-string-theory-fall-2008/lecture-notes/ in lecture number 16)

$K(z,x;z^{'},x^{'}) \equiv \lim_{z \rightarrow 0} \frac{1}{\sqrt{\gamma}} \vec{n}. \partial G$

where $K$ is the bulk to boundary propagator, $G$ is the bulk to bulk propagartor (i.e the Green's function satisfying the following relation $(\Box - m^{2})G = \frac{1}{\sqrt{g}}\delta^{d+1}(x - x^{'})$ and $\gamma$ is the boundary metric (in AdS, given in Poincare coordinates).

Now, in the following paper in Appendix A, Equation number (69) has the following definition for the same (https://arxiv.org/abs/hep-th/0606141)

$\Phi(x^{'}) = \int d\tau d\Omega \frac{R^{D-2}}{\cos^{D-2}(\rho)}(\phi\partial_{\rho}G - G\partial_{\rho}\phi)|_{\rho \rightarrow \frac{\pi}{2}}$

which then defines $K$ via the definition

$\Phi(x^{'}) = \int d\tau d\Omega K(x|x^{'})\Phi_{0}(x^{'})$

where $\Phi_{0}$ is the boundary value of the field $\Phi$, which satisfies the equation $(\Box - m^{2})\Phi = 0$ everywhere, and $G$ is again the Green's function.

Therefore, we have two definitions of the bulk-to-boundary propagator, via the Green's function (One given in the MIT OCW lectures and the other in that hep-th arXiv paper). Clearly both must be the same. My problem is to derive one from the other. Or, at the very least, to find some logical relation between the two,

The first definition is in Poincare coordinates, where the AdS metric is as follows

$ds^{2} =\frac{dz^{2} + dx^{2}}{z^{2}}$

with $x \equiv (t,x_{i})$

And the second definition is in the Global coordinates, and so the metric there is

$ds^{2} = \frac{R^{2}}{\cos^{2}(\rho)}(-d\tau^{2} + d\rho^{2} + \sin^{2}(\rho)d\Omega^{2}_{d-1})$

Both results are in $d+1$ dimensional space-time. For ease of computation, $d$ is taken to be odd. Also, I've been told that part of the hint lies in the usage of Green's Second Identity. Couldn't do much with it myself.

Thank you

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.