Let $R=R_{\rm AdS}$ and $r_c=\pi/2-\epsilon$. For $\operatorname{AdS}_3$, $\mathcal R=-6/R^2$ and $\Lambda=-1/R^2$, while
$$
\sqrt{-g}=R^3\frac{\sin r}{\cos^3r}.
$$
The bulk term is
$$
I_{\rm bulk}=-\frac{R\Delta t}{4G}(\sec^2r_c-1).
$$
With the inward normal $n^r=-\cos r/R$, the induced metric has $\sqrt{-h}=R^2\sin r/\cos^2r$ and
$$
K=-\frac1R\left(\frac{\cos^2r}{\sin r}+2\sin r\right).
$$
The <Gibbons–Hawking–York boundary term> is consequently
$$
I_{\rm GHY}=\frac{R\Delta t}{4G}(1+2\tan^2r_c).
$$
Thus, with the orientation and Lorentzian signs displayed in the question,
$$
\boxed{I_{\rm GR}=\frac{R\Delta t}{4G}\sec^2r_c
=\frac{R\Delta t}{4G}\csc^2\epsilon.}
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-354/2/b/i/solution