Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 53 / 1 / a / ii / Solution

With the $+---$ signature and $n^2=1$, direct expansion of the two spatial projection tensors givesThis restriction is negative definite on spatial vectors. The positive spatial induced metric is instead $h_{\alpha \beta }=n_{\alpha }{n}_{\beta }-g_{\alpha \beta }=-q_{\alpha \beta }$, whose pullback to a time slice is the $h_{ij}$ used in the line element. The spatial metric sign for a unit timelike normal is important here: the plus sign in the PDF's claimed equality is incompatible with its normal normalization: $g_{\alpha \beta }+n_{\alpha }{n}_{\beta }$ is not even transverse to $n^{\alpha }$.

The spatial covariant derivative of a spatial tensor projects every index, including its derivative index. Projection only on the derivative index is sufficient for a scalar but not for a general tensor. Using metric compatibility of the spacetime Levi-Civita connection, we obtainEach term contains a normal contracted with its spatial projection tensor. Consequently $\boxed{D_i{h}_{jk}=0}$, and the negative spatial restriction also satisfies $D_i{q}_{jk}=0$. This proves the requested metric compatibility of the spatial covariant derivative after correcting the source's metric sign. Contracting indices gives the particular expression written in the question.