# Gravity Hamiltonian in $AdS$

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 this paper. 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 this and this 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.

• Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files. – Qmechanic Dec 24 '18 at 14:12

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