Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 53 / 1

Write $h_{ij}={}^{(3)}{g}_{ij}$ for the positive spatial metric tensor and $h^{ij}$ for its inverse. Expanding the 3+1 decomposition gives $g_{00}=N^2-h_{ij}{N}^i{N}^j$, $g_{0i}=-h_{ij}{N}^j$ and $g_{ij}=-h_{ij}$. The inverse isFor example, $g^{00}{g}_{00}+g^{0i}{g}_{i0}=1$, and $g^{00}{g}_{0j}+g^{0i}{g}_{ij}=0$. The spatial block similarly gives $g^{i0}{g}_{0j}+g^{ik}{g}_{kj}={\delta }^i{}_j$. These checks determine all blocks without treating the spatial block alone as the inverse of the four-metric.

Raising the normal covector with this inverse metric tensor yieldsTake $N>0$ so it is future-pointing. Its norm is $n_{\mu }{n}^{\mu }=1$. The spatial projection tensor obeys $P^{\mu }{}_{\nu }{n}^{\nu }=0$, and multiplication givesThus it is an idempotent linear projection onto vectors tangent to the spatial hypersurface. The negative spatial components of the spacetime metric tensor do not change this idempotence.

With the $+---$ signature and $n^2=1$, direct expansion of the two spatial projection tensors givesThis restriction is negative definite on spatial vectors. The positive spatial induced metric is instead $h_{\alpha \beta }=n_{\alpha }{n}_{\beta }-g_{\alpha \beta }=-q_{\alpha \beta }$, whose pullback to a time slice is the $h_{ij}$ used in the line element. The spatial metric sign for a unit timelike normal is important here: the plus sign in the PDF's claimed equality is incompatible with its normal normalization: $g_{\alpha \beta }+n_{\alpha }{n}_{\beta }$ is not even transverse to $n^{\alpha }$.

The spatial covariant derivative of a spatial tensor projects every index, including its derivative index. Projection only on the derivative index is sufficient for a scalar but not for a general tensor. Using metric compatibility of the spacetime Levi-Civita connection, we obtainEach term contains a normal contracted with its spatial projection tensor. Consequently $\boxed{D_i{h}_{jk}=0}$, and the negative spatial restriction also satisfies $D_i{q}_{jk}=0$. This proves the requested metric compatibility of the spatial covariant derivative after correcting the source's metric sign. Contracting indices gives the particular expression written in the question.

Differentiate the spatial projection tensor before making any contractions:For the spatial projector derivative identity, its first term vanishes after projection on $\mu$. The definition of the extrinsic curvature of a spatial hypersurface then gives the stronger tensor identityHere $\lambda$ remains a free index. For a normal to a genuine foliation, the extrinsic curvature is symmetric: projecting ${\nabla }_{[\nu }{n}_{\mu ]}$ gives zero because locally $n_{\mu }$ is a scalar multiple of a time gradient. This is hypersurface orthogonality implies symmetric extrinsic curvature.

Contract $\rho$ with $\lambda$ in the stronger identity to obtain exactly the contraction displayed in the PDF:The last equality follows from the transversality of the extrinsic curvature. Thus the literal printed identity is valid, although both its sides vanish; the uncontracted identity explains its geometric origin.

For a centered Gaussian random field, Wick's theorem says that every odd moment vanishes and every even moment is the sum over all pairings of products of two-point functions. For six fields there are $15$ pairings. In the connected three-point function at nonzero external momenta, each of the three external fields must pair with a distinct field at the cubic interaction. There are $3!=6$ such Wick contractions. Choosing the undifferentiated field gives three possibilities; interchanging the two differentiated fields gives a further factor of two. The remaining nine pairings involve an external-external pair and an internal pair and belong to tadpole/disconnected contributions. Define the background so the one-point function vanishes, or equivalently subtract these contributions.

It is useful to keep a coefficient $\mathcal{C}$ multiplying the cubic Hamiltonian: its literal printed value is $1$, whereas the standard dimensionally normalized curvature interaction has $\mathcal{C}=M_{\mathrm{Pl}}^2$. This distinction will matter for the final amplitude. Withand $dt=ad\tau$, $\dot{\zeta }={\zeta }^{\prime }/a$, the interaction entering the time integral isIn the interaction picture, the unequal-time vacuum contraction required by the in-in formalism isFor a differentiated internal field replace $u_k^{\ast }$ by $u_k^{\ast \prime }$. The equal-time power spectrum fixes its magnitude; the displayed free De Sitter curvature mode functions and vacuum choice fix its unequal-time phase.

Performing the three momentum integrations imposes ${\mathbf{p}}_a=-{\mathbf{k}}_a$ for each assignment and leaves one overall momentum delta function. Put $U=u_{k_1}(0)u_{k_2}(0)u_{k_3}(0)$, which is real here, and defineThe upper early-time contour ${\mathcal{C}}_{+}$ runs from $-\infty (1-i\delta )$ to $0$, with $\delta >0$. The Hamiltonian's minus sign and the six connected Wick contractions then giveThis is equivalently $\mathrm{Re}[-4i\mathcal{C}{ϵ}^{2}U\sum I_i]$ with unconjugated De Sitter curvature mode functions and the conjugate lower contour ${\mathcal{C}}_{-}$. Before momentum integration, the same result consists of the three cyclic delta assignments, each with the extra factor of two for the identical differentiated legs.

The PDF's intermediate formula writes only three cyclic assignments without that factor of two. Taken literally it undercounts the connected contractions. Its unconjugated modes must also use the conjugate contour, rather than the upper contour of the original in-in expression. Both points are required for a consistent in-in bispectrum conjugation rule.

Let $A_k=H/(2\sqrt{ϵ}{M}_{\mathrm{Pl}}{k}^{3/2})$ and $K=k_1+k_2+k_3>0$. For the unconjugated integrand on the lower contour, the De Sitter curvature mode functions giveHere the two derivative factors supply ${\tau }^2$, cancelling $a^2=1/(H^2{\tau }^2)$. The vacuum prescription for inflationary in-in integrals can be implemented by a factor $e^{\delta \tau }$ on the real negative axis, with $\delta >0$, followed by $\delta \downarrow 0$. In this notationThereforeAn undamped boundary evaluation on the real axis at $-\infty$ is not valid. On the upper contour the complex-conjugate integrand instead yields the conjugate value. Conjugating the modes without conjugating the contour would produce exponential growth.

Using all six connected Wick contractions from the preceding part and $U=A_{k_1}{A}_{k_2}{A}_{k_3}$ gives the literal-Hamiltonian resultThe connected correlator is $(2\pi {)}^3{\delta }^{(3)}(\sum {\mathbf{k}}_i)B_{\mathcal{C}}$. The three assignments constitute the curvature bispectrum from a zeta zeta-prime-squared interaction. The numerator contains six powers of $H$ from the six modes, two of which are cancelled by $a^2$; the six powers of $M_{\mathrm{Pl}}$ remain unless the vertex supplies two.

Consequently the printed final expression, with $M_{\mathrm{Pl}}^{-4}$, is obtained for $\boxed{\mathcal{C}=M_{\mathrm{Pl}}^2}$. With the literal printed Hamiltonian $\mathcal{C}=1$, the answer instead has $\boxed{M_{\mathrm{Pl}}^{-6}}$. The Planck normalization of a cubic curvature interaction requires a factor $M_{\mathrm{Pl}}^2$ in the Hamiltonian if its final bispectrum is intended. In addition, integrating the intermediate expression literally with only three assignments gives half the fully contracted amplitude. These are normalization defects, rather than changes in the momentum shape. For dimensionless comoving curvature perturbations, the usual interaction normalization is also required by the mass dimension of the action.

For the standard normalization, $P_{\zeta }(k)=H^2/(4ϵM_{\mathrm{Pl}}^2{k}^3)$, so $B/P_{\zeta }^2$ is of order $ϵ$. This slow-roll suppression means that the primordial non-Gaussianity from this interaction alone is not expected to be detectably large. For example, in the squeezed bispectrum configuration $k_1\ll k_2\simeq k_3$, matching this contribution to the local convention gives $f_{\mathrm{NL}}=5ϵ/24$. This is a single-vertex contribution, not the complete single-field slow-roll inflation prediction; other vertices and field redefinitions contribute at the same slow-roll order. A claim of detectability for enhanced interactions requires a model beyond this approximation.