= Solution
With the $+---$ signature and $n^2=1$, direct expansion of the two <spatial projection tensors> gives
$$
q_{\alpha\beta}=P^\mu{}_\alpha P^\nu{}_\beta g_{\mu\nu}
=g_{\alpha\beta}-n_\alpha n_\beta.
$$
This 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 obtain
$$
D_\alpha h_{\beta\gamma}=P^\rho{}_\alpha P^\sigma{}_\beta P^\lambda{}_\gamma\nabla_\rho h_{\sigma\lambda}
=P^\rho{}_\alpha P^\sigma{}_\beta P^\lambda{}_\gamma
\left[(\nabla_\rho n_\sigma)n_\lambda+n_\sigma\nabla_\rho n_\lambda\right]=0.
$$
Each term contains a normal contracted with its <spatial projection tensor>. Consequently $\boxed{D_i h_{jk}=0}$, and the negative spatial restriction also satisfies $D_iq_{jk}=0$. 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.
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