Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 354 (Gauge/Gravity Duality)

A Euclidean conformal transformation is a coordinate map whose Jacobian preserves angles:For ${\tilde{x}}^{\mu }=x^{\mu }+\delta {ϵ}^{\mu }+O({\delta }^2)$, its left side isTaking the trace identifies the infinitesimal scale change as $2\partial \cdot ϵ/d$. The traceless part must vanish, giving the conformal Killing equation

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Translation and rotation invariance make the two-point function depend only on $|x_{12}|$. Scale covariance then fixes its power, and inversion or special-conformal invariance requires equal dimensions for a nonzero scalar two-point function. For identical scalar primary operators,Only the normalization $C_2$ depends on the operator convention.

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The same symmetries fix the scalar three-point function toEach insertion acquires total scaling weight $2{\Delta }_i$ from the two distances containing it, while special conformal transformations leave no independent cross-ratio for three points.

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Under the AdS-CFT correspondence, each single-trace operator is represented by one bulk field. The leading two-point Witten diagram is a bulk propagator joining the two boundary insertions, equivalently two bulk-to-boundary legs contracted through the quadratic bulk action. The leading connected three-point diagram has one bulk cubic vertex integrated over AdS and three bulk-to-boundary propagators ending at $x_1,x_2,x_3$. The bulk kinetic normalization fixes $C_2$, while the cubic coupling fixes $C_{123}$.

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For three identical operators, conformal invariance givesWhen $|x|\ll |y|$, $|y-x|=|y|[1+O(|x|/|y|)]$, soInserting the operator product expansion, the identity term has zero expectation with $O(y)$, while the $O(0)$ term giveswhich has precisely the required position dependence.

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Comparing the coefficients in the short-distance limit givesIf the operator is normalized so that $C_2=1$, its OPE coefficient equals the three-point normalization.

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Varying the gauge field and integrating by parts givesthe second identity following from $F=dA$. Varying the inverse metric, usingand discarding the Einstein--Hilbert boundary term givesFor AdS${}_4$, $\Lambda =-3/L^2$.

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The horizon equation givesWith $Q=r_{+}^2\mu /L$, defineThe coordinate changes in the question giveThe horizon is at $z=1$, and all nontrivial dimensionless dependence is through $\mu$; $L$ remains only as the overall curvature scale.

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Choose the Fefferman--Graham coordinates radial variable fromNear the boundary,Consequently, through $O(Z^2)$,By the holographic dictionary, the leading boundary value of $A_{\tau }$ is the source for charge density. Thus $\mu$ is the dimensionless holographic chemical potential; the coefficient of the normalizable linear term is proportional to the CFT charge density, $\rho =\mu L/(4\pi G)$ with the action normalization and outward-orientation convention used here.

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The transformation is a boundary Lorentz boost of rapidity $\beta$. It preserves $-d{\tau }^2+d{X}^2=-d{T}^2+d{W}^2$ and describes the charged thermal state in a frame moving with speed $|v|=\tanh \beta$. The gauge field becomesIf the rest-frame contravariant current is $J^{\mu }=(\rho ,0,0)$, then in the new coordinateswhere the sign of the spatial component follows from the stated passive coordinate transformation. Reversing the boost convention reverses that sign. For small $\beta$, the induced spatial current is $J^W\simeq -\rho \beta$.

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The boost produces a persistent current without an applied electric field. More generally, at nonzero charge density a homogeneous electric field injects momentum, and exact translation invariance provides no mechanism to relax it. ThereforeEquivalently, the real optical conductivity contains a delta function at zero frequency and its imaginary part has a $1/\omega$ pole. This holographic D.C. conductivity is physically reasonable for the ideal translationally invariant model. A lattice, disorder, impurities, or another source of momentum relaxation is needed for finite D.C. resistivity.

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