The background spatial metric is

g_{ij} = a^{2} δ_{ij}

. For the large spatial diffeomorphism

ϵ^{0} = 0, ϵ^{i} = ω^{i}_{j} x^{j}, ω_{ij} = ω_{ji}, ω_{i}^{i} = 0,

(1)

homogeneity and isotropy imply that the purely spatial background Christoffel symbols vanish. Hence

\nabla_{i} ϵ_{j} = a^{2} ω_{ji}, Δ γ_{ij} = - 2 ω_{ij} .

(2)

The shift is constant, transverse and traceless, and is therefore the zero-momentum adiabatic tensor mode.

A scalar transforms by its Lie derivative:

Δ φ (x) = - ω_{ij} x^{j} \partial_{i} φ (x) .

(3)

Fourier transformation and integration by parts in momentum space give

Δ φ (k) = ω_{ij} k_{i} \frac{\partial}{\partial k _{j}} φ (k) .

(4)

The term proportional to

ω_{i}^{i}

vanishes because the deformation is traceless.

ii

 0  0

Solution

 0  0

The equal-time canonical commutation relation for the transverse-traceless graviton and its canonical momentum is the transverse-traceless projector. At zero momentum the supplied polarization completeness relation gives

i [Q_{S}, γ_{mn}] = - 2 ω_{ij} \sum_{s} 2 ϵ_{ij}^{s} (0) ϵ_{mn}^{s} (- 0) = - 2 ω_{mn},

(1)

where symmetry and tracelessness of

ω_{ij}

remove the trace term. Thus the charge generates precisely the transformation in part i:

i [Q_{S}, γ_{mn}] = Δ γ_{mn} .

(2)

This is the soft, field-independent part of the Noether charge associated with the large diffeomorphism.

iii

 0  0

Solution

 0  0

Let

O = φ (k_{1}) φ (k_{2})

. Since

Q_{S} = - 2 ω_{ij} Π_{ij} (0)

i ⟨[Q_{S}, O]⟩ = 4 ω_{ij} Im ⟨ Π_{ij} (0) O ⟩ .

(1)

For a soft mode

γ^{s} = f_{q} a_{s} + f_{q}^{*} a_{s}^{†}

and

Π^{s} = g_{q} a_{s} + g_{q}^{*} a_{s}^{†}

, the vacuum obeys

⟨ 0∣ Π^{s} (q) = \frac{g _{q}}{f _{q}} ⟨ 0∣ γ^{s} (q) .

(2)

The mode-function Wronskian normalization

f_{q} g_{q}^{*} - f_{q}^{*} g_{q} = i

(3)

implies

Im \frac{g _{q}}{f _{q}} = - \frac{1}{2∣ f _{q} ∣ ^{2}} = - \frac{1}{2 P _{γ} ( q )},

(4)

where

P_{γ} (q)

is the graviton power spectrum per polarization. The equal-time bispectrum is real in this parity-even configuration, and therefore

i ⟨[Q_{S}, O] ⟩^{'} = - 2 ω_{ij} q \to 0 lim \frac{⟨ γ _{ij} ( q ) φ ( k _{1} ) φ ( k _{2} ) ⟩ ^{'}}{P _{γ} ( q )} .

(5)

This turns the charge insertion into the soft-graviton insertion used in the Soft graviton theorem.

iv

 0  0

Solution

 0  0

Applying the scalar transformation from part i to both fields gives

⟨ Δ O ⟩ = ω_{ij} \sum_{a = 1}^{2} k_{ai} \frac{\partial}{\partial k _{aj}} [(2 π)^{3} δ^{(3)} (k_{1} + k_{2}) P_{φ} (k_{1})] .

(1)

The derivative of the Dirac delta distribution is proportional to

δ_{ij}

and vanishes after contraction with traceless

ω_{ij}

. Removing that delta function leaves

⟨ Δ O ⟩^{'} = ω_{ij} k_{i} \frac{\partial}{\partial k _{j}} P_{φ} (k) .

(2)

Equating the two sides of the Ward-Takahashi identity and resolving the soft graviton into a polarization

s

yields the cosmological soft-graviton consistency relation

q \to 0 lim \frac{⟨ γ ^{s} ( q ) φ ( k ) φ ( - k - q ) ⟩ ^{'}}{P _{γ} ( q )} = - \frac{1}{2} ϵ_{ij}^{s} (q) k_{i} \frac{\partial}{\partial k _{j}} P_{φ} (k) .

(3)

For an isotropic power spectrum this is equivalently

- ϵ_{ij}^{s} k_{i} k_{j} \frac{\partial P _{φ}}{\partial k ^{2}} .

(4)

The long-wavelength adiabatic tensor mode acts on the short two-point function as an anisotropic rescaling of its momentum.

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Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 312 / 2

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