 Solution

Write $h_{ij}$ for the spatial metric tensor and lower the shift vector with $h_{ij}$. For the negative-shift convention, the four-dimensional metric and its inverse have componentsThe normal has covariant components $n_{\mu }=(-N,0,0,0)$, so $n_{\mu }{n}^{\mu }=-1$ and it annihilates every tangent vector to the spatial slice. Since $n_i=0$, its covariant derivative gives $n_{i;j}=N{\Gamma }^0{}_{ij}$ and therefore $K_{ij}=-N{\Gamma }^0{}_{ij}$.

Compute the required Christoffel symbol using the symmetric connection formula. Its time-index term and spatial-index term combine to giveThe spatial expression in parentheses is ${\dot{h}}_{ij}+D_i{N}_j+D_j{N}_i$, where $D$ is the covariant derivative of $h$. Thus the extrinsic curvature with a negative shift isThis derivation fixes both signs from the actual 3+1 decomposition of spacetime, rather than transferring a formula using the opposite shift vector convention. The connection formula used here has $g_{\lambda \kappa ,\nu }$ as its second differentiated term; the printed repeated-index version is a typographical error.