Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 309 / 1

Use signature $(-,+,+,+)$, so the future-directed unit normal obeys $n^{\mu }{n}_{\mu }=-1$. The spatial projection tensor and the induced metric areIndeed $P^{\mu }{}_{\alpha }{n}^{\alpha }=0$ and $P^{\mu }{}_{\rho }{P}^{\rho }{}_{\alpha }=P^{\mu }{}_{\alpha }$: this is an idempotent linear map onto the tangent space of the spatial hypersurface. For a covariant tensor both slots must be projected:Contracting either slot with $n$ supplies the vanishing factor $P^{\mu }{}_{\alpha }{n}^{\alpha }$. Thus both normal contractions vanish, without requiring symmetry of $T$.

Take the orientation induced by the future-directed unit normal and positive lapse function $\alpha$. In coordinates adapted to the 3+1 decomposition of spacetime, any component ${ϵ}_{\ell ijk}$ with all four indices spatial vanishes: only three distinct spatial indices are available. The shift vector therefore contributes nothing, andHere $[ijk]$ is the Levi-Civita symbol, and the last equality uses the metric determinant in a 3+1 decomposition, $|g|={\alpha }^2\gamma$. The right side is precisely the spatial metric volume tensor, with component $\sqrt{\gamma }$ at $ijk=123$, the appropriate permutation sign, and zero for repeated indices. HenceMore invariantly, the spatial volume form is the restriction to the hypersurface of the contraction of the spacetime volume form with $n$.

For the Vacuum Einstein equations with zero cosmological constant, $R_{\alpha \beta }=0$, and its tensor contraction gives $R=0$. Every Ricci tensor and scalar curvature term in the Weyl tensor consequently disappears:Thus the Weyl tensor contains all vacuum curvature. Here vacuum means Ricci-flat; a nonzero cosmological constant would retain a constant-curvature contribution to the Riemann curvature tensor.

The electric part of the Weyl tensor has one normal in each antisymmetric pair of the Weyl tensor. ThereforeFor the magnetic part of the Weyl tensor, contracting the first slot gives $n^{\alpha }{n}^{\lambda }{ϵ}_{\alpha \lambda \mu \nu }=0$ by antisymmetry of the metric volume tensor; contracting the second gives $n^{\beta }{n}^{\rho }{C}^{\mu \nu }{}_{\beta \rho }=0$ by antisymmetry of the last curvature pair. Consequently both tensors are entirely spatial:These spatial projection tensor identities actually hold for any Weyl tensor; the vacuum assumption is needed subsequently to replace it by the Riemann curvature tensor.

All free indices below are spatial, and $K={\gamma }^{\mu \nu }{K}_{\mu \nu }$. Contract the vacuum Ricci tensor using $g^{\mu \nu }={\gamma }^{\mu \nu }-n^{\mu }{n}^{\nu }$:Reversing both antisymmetric curvature pairs identifies the last term as $-E_{\alpha \beta }$. The spatial Gauss–Codazzi equations for a spatial hypersurface then giveFor the magnetic part of the Weyl tensor, moving the normal to the first slot of the metric volume tensor introduces a minus sign:The remaining curvature indices are spatial, so the normal projection in the Gauss–Codazzi equations for a spatial hypersurface givesThe interchange $\mu \leftrightarrow \nu$ makes the two spatial covariant derivative terms equal. Thus the constants, with the paper's orientation and extrinsic-curvature convention, areIn particular, the sign of the magnetic expression must include the minus from moving $n$ past the first volume-tensor index. These formulas reconstruct the spatial electric part of the Weyl tensor and magnetic part of the Weyl tensor from the induced metric and the extrinsic curvature of a spatial hypersurface.