Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 354 / 2

Table of contents
- a 2
  - Solution a
- b 2
  - Solution b
- c 2
  - Solution c
- d 2
  - Solution d
- e 2
  - Solution e

a

Solution

Varying the gauge field and integrating by parts gives

\nabla_{a} F^{ab} = 0, \nabla_{[a} F_{b c]} = 0,

the second identity following from

F = d A

. Varying the inverse metric, using

δ F^{2} = 2 F_{a c} F_{b}^{c} δ g^{ab},

and discarding the Einstein--Hilbert boundary term gives

R_{ab} - \frac{1}{2} R g_{ab} + Λ g_{ab} = 2 (F_{a c} F_{b}^{c} - \frac{1}{4} g_{ab} F_{c d} F^{c d}) .

For AdS

_{4}

,

Λ = - 3/ L^{2}

.

Solved by gpt-5.6-sol high.

b

Solution

The horizon equation gives

2 M = \frac{r _{+}^{3}}{L ^{2}} + \frac{Q ^{2}}{r _{+}} .

With

Q = r_{+}^{2} μ / L

g (z) = 1 - (1 + μ^{2}) z^{3} + μ^{2} z^{4} .

The coordinate changes in the question give

d s^{2} = \frac{L ^{2}}{z ^{2}} [- g (z) d τ^{2} + \frac{d z ^{2}}{g ( z )} + d X^{2} + d Y^{2}],

A = μL (1 - z) d τ .

The horizon is at

z = 1

, and all nontrivial dimensionless dependence is through

μ

;

L

remains only as the overall curvature scale.

Solved by gpt-5.6-sol high.

c

Solution

Choose the Fefferman--Graham coordinates radial variable from

\frac{d Z}{Z} = \frac{d z}{z g ( z )} .

Near the boundary,

Z = z [1 + \frac{1 + μ ^{2}}{6} z^{3} + O (z^{4})], z = Z + O (Z^{4}) .

Consequently, through

O (Z^{2})

,

d s^{2} = \frac{L ^{2}}{Z ^{2}} [d Z^{2} - d τ^{2} + d X^{2} + d Y^{2} + O (Z^{3})],

A_{τ} = μL - μL Z + O (Z^{4}) .

By the holographic dictionary, the leading boundary value of

A_{τ}

is the source for charge density. Thus

μ

is the dimensionless holographic chemical potential; the coefficient of the normalizable linear term is proportional to the CFT charge density,

ρ = μL / (4 π G)

with the action normalization and outward-orientation convention used here.

Solved by gpt-5.6-sol high.

d

Solution

The transformation is a boundary Lorentz boost of rapidity

β

. It preserves

- d τ^{2} + d X^{2} = - d T^{2} + d W^{2}

and describes the charged thermal state in a frame moving with speed

∣ v ∣ = tanh β

. The gauge field becomes

A = μL (1 - z) (cosh β d T + sinh β d W) .

If the rest-frame contravariant current is

J^{μ} = (ρ, 0, 0)

, then in the new coordinates

J^{T} = ρ cosh β, J^{W} = - ρ sinh β, J^{Y} = 0,

where the sign of the spatial component follows from the stated passive coordinate transformation. Reversing the boost convention reverses that sign. For small

β

, the induced spatial current is

J^{W} ≃ - ρβ

.

Solved by gpt-5.6-sol high.

e

Solution

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. Therefore

σ_{DC} = \infty .

Equivalently, the real optical conductivity contains a delta function at zero frequency and its imaginary part has a

1/ ω

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.

Solved by gpt-5.6-sol high.

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