Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 59 / 1

The normalized timelike normal has $n_{\alpha }{n}^{\alpha }=-1$. Multiplying the linear projections givesAlso $P^{\mu }{}_{\alpha }{n}_{\mu }=n_{\alpha }+(n_{\mu }{n}^{\mu })n_{\alpha }=0$. ThusThe operator fixes every vector tangent to the spacelike hypersurface and annihilates its normal. The repeated-index typo in the TeX is resolved by the actual PDF's matrix-composition indices.

Use metric compatibility and differentiate the projector:Projecting the derivative and lower tensor index kills the first term, by part (a). The remaining term is the extrinsic curvature:The second projection may be inserted in the definition of $K$, since differentiating $n^{\beta }{n}_{\beta }=-1$ gives $n^{\beta }{\nabla }_{\alpha }{n}_{\beta }=0$.

For a tangential covector, the projected derivative is $D_{\nu }{W}_{\gamma }=P^{\beta }{}_{\nu }{P}^{\delta }{}_{\gamma }{\nabla }_{\beta }{W}_{\delta }$. Its second derivative isExpanding by the product rule, the derivative of the first inner projector gives $K_{\mu \nu }{n}^{\eta }{P}^{\kappa }{}_{\gamma }{\nabla }_{\eta }{W}_{\kappa }$. This contribution cancels on antisymmetrizing $\mu ,\nu$, because $K$ is symmetric.

The derivative of the second inner projector gives $K_{\mu \gamma }{n}^{\kappa }{P}^{\eta }{}_{\nu }{\nabla }_{\eta }{W}_{\kappa }$. Differentiating tangentiality, $n^{\kappa }{W}_{\kappa }=0$, rewrites it asThe remaining term is the projected four-dimensional curvature commutator. Using the curvature convention of the question,and replacing the tangential $W_{\xi }$ by $W_{\lambda }{P}^{\lambda }{}_{\xi }$, we obtainSince this holds for every tangential covector, the Gauss equation follows:The signs use a timelike unit normal and the specified commutator convention; reversing a curvature convention without changing the commutator would change the comparison.