Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 354 / 2 / b

Let $R=R_{\mathrm{AdS}}$ and $r_c=\pi /2-ϵ$. For ${\mathrm{AdS}}_3$, $\mathcal{R}=-6/R^2$ and $\Lambda =-1/R^2$, whileThe bulk term isWith the inward normal $n^r=-\cos r/R$, the induced metric has $\sqrt{-h}=R^2\sin r/{\cos }^{2}r$ andThe Gibbons–Hawking–York boundary term is consequentlyThus, with the orientation and Lorentzian signs displayed in the question,

The divergence is local in the induced boundary metric and is cancelled by the leading holographic renormalization countertermOn the regulated cylinder this iswhich cancels the ${ϵ}^{-2}$ divergence of $I_{\mathrm{GR}}$. An overall sign changes if one uses the oppositely signed Euclidean generating functional; the local counterterm changes with it.

The stated finite valueis the Casimir energy of a two-dimensional conformal field theory on the unit spatial circle. The plane-to-cylinder conformal map shifts the Hamiltonian by $-c/12$, so the global ${\mathrm{AdS}}_3$ vacuum is dual to the CFT cylinder vacuum rather than to a state of zero cylinder energy. The sign assigned directly to the Lorentzian on-shell action depends on whether it is identified with the action density or with the vacuum-energy generating functional; the universal boundary datum is the vacuum energy $E_0=-c/12$ supplied in the question.