Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 354 / 2

The Ryu–Takayanagi formula gives the leading entropy of a static boundary region:On the $t=0$ slice of Poincaré ${\mathrm{AdS}}_4$,The minimal surface anchored on a boundary circle of radius $R_1$ is the hemisphereCutting it off at $z=ϵ$, its area isThereforeThe universal requirement was the perimeter-law divergence $\pi R_1/(2Gϵ)$; the finite constant depends on the stated pure ${\mathrm{AdS}}_4$ vacuum geometry and equals $-\pi /(2G)$ here. Since $G\sim N^{-3/2}$, both terms are of order $N^{3/2}$.

Let ${\gamma }_1$ and ${\gamma }_2$ be the individual Ryu–Takayanagi surfaces for $D_1$ and $D_2$. Their disconnected union ${\gamma }_1\cup {\gamma }_2$ is homologous to $D_1\cup D_2$ and is therefore an admissible competitor in the minimization that defines ${\gamma }_{12}$. Minimality givesDividing by $4G$ provesand hence the leading holographic mutual information obeysThis is the geometric realization of the nonnegativity of quantum mutual information.

There are two competing extremal-surface topologies. The disconnected candidate ${\gamma }_1\cup {\gamma }_2$ has area independent of the separation $L$ and gives $I_c=0$. A connected surface joining the two boundary circles can have smaller area when the gap is small; as $L\to 0$, short-distance entanglement across the nearby boundaries makes $I_c$ positive and divergent.

Increasing $L$ makes the connected candidate less favorable. At a critical separation $L=C(R_1,R_2)$ its area equals the disconnected area, and beyond that point the disconnected candidate is minimal. This entanglement-wedge phase transition gives

Write $S_{\mathrm{conn}}(L)$ for the connected candidate andfor the $L$-independent disconnected candidate. The physical entropy is the lower envelopeNear $L=C$, the connected branch rises to meet the horizontal disconnected branch. The entropy is continuous but generically has a kink there: its derivative jumps from the positive slope of $S_{\mathrm{conn}}$ to zero. Correspondingly, $I_c=S_{\mathrm{disc}}-S(D_1\cup D_2)$ falls to zero and remains zero.

A pair of boundary circles has one dimensionless invariant under the conformal field theory group. If their radii are $R_1,R_2$ and their center separation isa convenient invariant is the inversive distanceThe choice between connected and disconnected bulk surfaces can depend only on $\chi$. Let the transition occur at the single theory-independent numerical value $\chi =X$, with $X>1$. ThensoAll dependence on the two radii is fixed by conformal symmetry; only the pure number $X$ requires the explicit minimal-surface calculation.