Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 312 / 1

Using the normalization of the timelike unit normal, the spatial projection tensor satisfiesLikewise $P^{\mu }{}_{\alpha }{n}^{\alpha }=0$ and $P^{\mu }{}_{\alpha }{P}^{\alpha }{}_{\nu }=P^{\mu }{}_{\nu }$. The spacetime form of the induced metric is thereforeIts restriction to the slice is the given three-metric. Metric compatibility of the four-dimensional Levi-Civita connection givesApply the three projectors defining the spatial covariant derivative. In the first summand $P^{\beta }{}_j{n}_{\beta }=0$, and in the second $P^{\alpha }{}_i{n}_{\alpha }=0$. Both summands vanish, soThis proves the metric compatibility of the spatial covariant derivative without treating the spatial projection tensor itself as covariantly constant.

Differentiate the normalization of the timelike unit normal with the metric-compatible covariant derivative:Thus the derivative of the normal is already orthogonal to the normal in its second index. Expanding the second spatial projection tensor in the definition of the extrinsic curvature of a spatial hypersurface givesTo prove symmetry using Frobenius theorem, set $F_{\alpha \beta }={\nabla }_{\alpha }{n}_{\beta }-{\nabla }_{\beta }{n}_{\alpha }$. Hypersurface orthogonality is equivalent toContract this with $n^{\alpha }{P}^{\beta }{}_{\mu }{P}^{\lambda }{}_{\nu }$. The last two terms vanish because a projected normal vanishes, while $n^{\alpha }{n}_{\alpha }=-1$ in the first term. HenceThe antisymmetric part of the doubly projected covariant derivative of $n$ is therefore zero. This proves that hypersurface orthogonality implies symmetric extrinsic curvature:No geodesic assumption for the normal congruence is needed; its normal acceleration may be nonzero.

Write $h_{ij}={}^{(3)}{g}_{ij}$, and retain only first-order scalar cosmological perturbations. The inverse induced metric and inverse lapse function areThe flat background spatial connection vanishes, and $N_i=-a^2{B}_{,i}$ is already first order. ConsequentlyThe comma on $B_{,i}$ matters: this is the gradient of the scalar shift vector potential, not an unrelated vector field. Substituting the time derivative of $h_{ij}$ into the negative extrinsic curvature convention givesRaising the first index with the perturbed inverse induced metric is essential. Its $\Phi$ and $E$ terms cancel the corresponding terms multiplying the background Hubble parameter, leavingFor the scalar shear potential $\chi =a^2(\dot{E}-B)/\bar{N}$, raised spatial derivatives mean ${\partial }^i=a^{-2}{\delta }^{ij}{\partial }_j$ to the required order. Thus the final term is $-{\partial }^i{\partial }_j\chi$. Taking the trace identifies the scalar expansion perturbation:Separating the trace and trace-free parts now yieldsAll signs follow from the source's $d{x}^i-N^{i}dt$ shift vector convention and negative extrinsic curvature definition.

The result above gives $K=-3H+\kappa$. Since the shear is trace-free and already first order, it cannot contribute linearly to the quadratic tensor contraction $K_{ij}{K}^{ij}=\mathrm{tr}(K^2)$. Thereforeup to second-order scalar cosmological perturbations. Combining these with the supplied spatial Ricci scalar gives the Hamiltonian constraintIts background part is the flat Friedmann equation $6H^2=16\pi G\bar{\rho }$. Subtracting that equation proves the linearized cosmological Hamiltonian constraint:Here $\rho$ is the energy density measured by the hypersurface normal, as required by the normal-normal projection of the Einstein field equations.