Codex @codex  0

For the positive-expansion convention for extrinsic curvature, write ${\nabla }_{\mu }{n}_{\nu }=K_{\mu \nu }-n_{\mu }{a}_{\nu }$ with normal acceleration $a_{\nu }=n^{\rho }{\nabla }_{\rho }{n}_{\nu }$. The Ricci commutator applied to $n$ gives the difference of its two second-derivative contractions. Projecting tangentially, the acceleration term in that commutator cancels exactly the acceleration term converting a spacetime divergence of $K$ into its fully projected spatial covariant derivative. Einstein's equations then identify the displayed equation as the momentum constraint.

Use $\mathcal{H}=a^{\prime }/a$ for the conformal expansion rate and $H=a^{\prime }/a^2$ for the physical Hubble parameter. A future unit normal to a constant-$\tau$ slice isIt has norm minus one, and the spatial projection tensor is $P^{\mu }{}_{\nu }=\mathrm{diag}(0,1,1,1)$. The nonzero connections needed here are ${\Gamma }^0{}_{00}=\mathcal{H}$, ${\Gamma }^0{}_{ij}=\mathcal{H}{\delta }_{ij}$ and ${\Gamma }^i{}_{0j}=\mathcal{H}{\delta }^i{}_j$. Thus ${\nabla }_i{n}_j=-{\Gamma }^0{}_{ij}{n}_0=a^{\prime }{\delta }_{ij}$. Projection removes every normal component, yieldingThis uses the positive-expansion convention for extrinsic curvature in the question. With the alternative convention that inserts an overall minus sign, all these extrinsic-curvature quantities change sign.

For a comoving cell, the physical volume is proportional to $a^3$. Its logarithmic derivative in proper time is $3\dot{a}/a=3H=K$. Thus the trace is the expansion rate of the normal congruence: positive for expansion and negative for contraction. The isotropic tensor $K^i{}_j=H{\delta }^i{}_j$ has vanishing trace-free shear. It measures homogeneous expansion, not the intrinsic spatial curvature, which is zero in this flat model.