# Quantum Gravity Hamiltonian in $AdS$

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Consider the global $AdS_{d+1}$ metric given by

$$ds^2 = \frac{1}{cos^2 \rho}[-dt^2 + d\rho^2 + sin^2 \rho d\Omega_{d-1}^2]$$

Now we follow the statements as made in Page 4 of . Here one now looks at quantum gravity in asymptotically $AdS$ spaces. For this case, the metric is expanded in the following manner:

$$g_{\mu \nu} = g^{AdS}_{\mu \nu} + h_{\mu \nu}$$

near the boundary, where one chooses the Fefferman-Graham gauge $h_{\rho \mu} = 0$. My problem is in the next statement where the author writes that the canonical Hamiltonian is given by:

$$H^{can} = \lim_{\rho \to \pi/2} (cos\rho)^{2-d}\int d^{d-1}\Omega \dfrac{h_{tt}}{16\pi G_N}$$

The author refers to  and  for making the above statement but I don't see where this statement is made in the above papers. Please shed some light on why this statement is correct.

asked Dec 28, 2018

## 1 Answer

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The answer to my question is as follows. The Hamiltonian given by
$$H^{can} = \lim_{\rho \to \pi/2} (cos\rho)^{2-d}\int d^{d-1}\Omega \dfrac{h_{tt}}{16\pi G_N}$$

is a surface term, which is characteristic of gravity. In general the Hamiltonian can have a volume term which has contribution from the Hamiltonian density and a surface term which has contribution from the surface integral. However if one does an ADM decomposition of the Einstein-Hilbert action, then one finds that the Hamiltonian density is a constraint which is set to zero. Therefore there is no volume contribution to the Hamiltonian.

So one is left with the surface term only. It was shown by [Regge and Teitelboim] that in order to properly implement the variational principle for the Hamiltonian so as to get the equations of motion for pure gravity, the surface term of the above form is necessary. This was shown not only for asymptotically $AdS$ spacetime for which we have the above Hamiltonian but also for any spacetime with a boundary in the linked paper.

answered Jan 13 by (85 points)

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