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

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.