Keep the smooth $\beta=\beta_H$ geometry fixed away from the horizon but identify Euclidean time with arbitrary period $\beta$. The horizon then has deficit angle $2\pi(1-\beta/\beta_H)$. Its delta-function curvature contributes
$$
\int_{\mathcal M}\sqrt g\,R\big|_{\rm tip}
=4\pi\left(1-\frac\beta{\beta_H}\right)A,
$$
and hence
$$
I_{\rm tip}=-\frac{A}{4G_N}
\left(1-\frac\beta{\beta_H}\right).
$$
Using $\log Z=-I_{\rm grav}$,
$$
\boxed{S_{\rm BH}
=\left.(1-\beta\partial_\beta)\log Z\right|_{\beta_H}
=\frac{A}{4G_N}
=\frac{2\pi^2r_H^3}{4G_N}.}
$$

All smooth bulk terms, including the cosmological-constant volume term, are proportional to the Euclidean time period and are annihilated by $1-\beta\partial_\beta$. The asymptotic <Gibbons–Hawking–York boundary term> and holographic counterterms are likewise smooth and linear in $\beta$. Only the curvature singularity at the fixed point of the Euclidean time circle survives.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 354 / 2 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-354/2/b/i/solution