Contracted Bianchi identity

The Ricci tensor and scalar satisfy ${\nabla }^{\alpha }{R}_{\alpha \beta }=\frac{1}{2}{\nabla }_{\beta }R$.

For the Levi-Civita connection, the second Bianchi identity gives ${\nabla }_i{R}^{ik}{}_{mn}={\nabla }_m{R}^k{}_n-{\nabla }_n{R}^k{}_m$. Applying another covariant derivative and the contracted Bianchi identity reduces its divergence to a commuting scalar Hessian plus curvature commutators on the Ricci tensor. The latter are the difference of symmetric tensors: a square of the Ricci tensor and $R^{ab}{R}_{anbm}$, which is symmetric in $m,n$ by curvature pair exchange and Ricci symmetry. Both differences vanish. This identity does not require vacuum field equations and is independent of the overall curvature sign convention.

The Einstein tensor isThe contracted Bianchi identity and metric compatibility give
${\nabla }^{\nu }{G}_{\mu \nu }=0$, ensuring compatibility of the Einstein equations with covariant stress-energy conservation.

For $d{s}^2=-d{t}^2+a(t{)}^{2}d{\mathbf{x}}^2$, the nonzero connection coefficients areWith $G_t^t=-3H^2$ and
$G_i^j=-(2\ddot{a}/a+H^2){\delta }_i^j$, direct substitution into the mixed-tensor covariant derivative gives zero.

The vacuum equation with positive cosmological constant gives
$H^2=\Lambda /3$. Its expanding flat-slicing solution is

In a spatially flat universe dominated by a positive constant energy density $ϵ$, the Friedmann equations give $H^2=8\pi Gϵ/(3c^2)$. On the expanding branch $a(t)=a_i{e}^{H(t-t_i)}$, so the scale factor doubles after $(\log 2)/H$. This concerns a linear length scale; the enclosed physical volume grows by a factor of eight over the same interval.