 Solution

Double contraction of the projected Gauss equation, with the timelike-normal sign and curvature convention in the question, givesHere $q^{\mu \nu }=g^{\mu \nu }+n^{\mu }{n}^{\nu }$. Expanding the two projectors givesThe four-normal contraction vanishes by Riemann antisymmetry. Because $g_{\mu \nu }{n}^{\mu }{n}^{\nu }=-1$, the right side is $2G_{\mu \nu }^{(4)}{n}^{\mu }{n}^{\nu }$. Rearranging proves the Hamiltonian constraint identity:For flat FRW, $R^{(3)}=0$, $K^2=9H^2$ and $K_{\mu \nu }{K}^{\mu \nu }=3H^2$, so $G_{nn}=3H^2$. Einstein's equation is therefore the first Friedmann equation, $3H^2=8\pi G\rho$, or $3H^2=8\pi G\rho +\Lambda$ if a cosmological constant is placed separately in Einstein's equation.

The scalar $R^{(3)}$ measures intrinsic spatial curvature, $K^2$ measures squared volume expansion, and the norm of $K_{ij}$ measures directional expansion including shear. More generally $K_{ij}=K{q}_{ij}/3+s_{ij}$ with trace-free $s_{ij}$ gives $K_{ij}{K}^{ij}=K^2/3+s_{ij}{s}^{ij}$. The shear vanishes in FRW, leaving the positive $3H^2$ contribution. This is an instantaneous constraint on initial data, not the separate acceleration equation.