 Solution

Write $h_{ij}={}^{(3)}{g}_{ij}$ for the positive spatial metric tensor and $h^{ij}$ for its inverse. Expanding the 3+1 decomposition gives $g_{00}=N^2-h_{ij}{N}^i{N}^j$, $g_{0i}=-h_{ij}{N}^j$ and $g_{ij}=-h_{ij}$. The inverse isFor example, $g^{00}{g}_{00}+g^{0i}{g}_{i0}=1$, and $g^{00}{g}_{0j}+g^{0i}{g}_{ij}=0$. The spatial block similarly gives $g^{i0}{g}_{0j}+g^{ik}{g}_{kj}={\delta }^i{}_j$. These checks determine all blocks without treating the spatial block alone as the inverse of the four-metric.

Raising the normal covector with this inverse metric tensor yieldsTake $N>0$ so it is future-pointing. Its norm is $n_{\mu }{n}^{\mu }=1$. The spatial projection tensor obeys $P^{\mu }{}_{\nu }{n}^{\nu }=0$, and multiplication givesThus it is an idempotent linear projection onto vectors tangent to the spatial hypersurface. The negative spatial components of the spacetime metric tensor do not change this idempotence.