 Solution

Write $h_{ij}={}^{(3)}{g}_{ij}$ for the induced metric and lower the shift vector with it, $N_i=h_{ij}{N}^j$. In the stated negative-shift convention the four-dimensional components and inverse components areMultiplying the given future unit normal by this metric tensor gives $n_{\mu }=(-N,0,0,0)$. In particular the tangential normal components vanish, but their covariant derivatives do not:The relevant Christoffel symbol isFor the second equality, the last parenthesis is $2h_{k\ell }{}^{(3)}{\Gamma }^{\ell }{}_{ij}$, so its contraction subtracts the two spatial connection terms in the symmetrized derivative of $N_i$. ThereforeThis is extrinsic curvature with a negative shift. Both the future-normal orientation and the definition $K=-\nabla n$ matter: replacing the threading by $d{x}^i+N^{i}dt$ reverses the shift terms. The spatial induced metric is generally a function of time as well as position; restricting it to time-independent data would simply make the displayed time derivative vanish.