The second Bianchi identity and the vanishing divergence of the Einstein tensor

(Reduction to a point-normal chart.) Fix $p$ and take the point-normal chart of [L4]. By [L3], $R^a{}_{bcd;e}(p)={\mathrm{\partial }}_e{R}^a{}_{bcd}(p)$, because every term of the derivative formula carrying a Christoffel symbol vanishes at $p$. By [L2], $R^a{}_{bcd}(p)={\mathrm{\partial }}_c{\mathrm{\Gamma }}^a{}_{db}(p)-{\mathrm{\partial }}_d{\mathrm{\Gamma }}^a{}_{cb}(p)$, since the quadratic terms carry a factor $\mathrm{\Gamma }(p)=0$.

(Differentiate on the chart, then evaluate.) The full coordinate identity [L2] holds as an equality of smooth functions throughout the chart. Apply ${\mathrm{\partial }}_e$ to that identity before evaluating at $p$. Each derivative of a quadratic term is a sum of terms ${\mathrm{\partial }}_e({\mathrm{\Gamma }}^a{}_{cf}{\mathrm{\Gamma }}^f{}_{db})=({\mathrm{\partial }}_e{\mathrm{\Gamma }}^a{}_{cf}){\mathrm{\Gamma }}^f{}_{db}+{\mathrm{\Gamma }}^a{}_{cf}({\mathrm{\partial }}_e{\mathrm{\Gamma }}^f{}_{db})$, and similarly for the second product. Every such term contains an undifferentiated Christoffel symbol, which is zero at $p$ by [L4]. Since [L3] also gives $R^a{}_{bcd;e}(p)={\mathrm{\partial }}_e{R}^a{}_{bcd}(p)$, this yields $R^a{}_{bcd;e}(p)={\mathrm{\partial }}_e{\mathrm{\partial }}_c{\mathrm{\Gamma }}^a{}_{db}(p)-{\mathrm{\partial }}_e{\mathrm{\partial }}_d{\mathrm{\Gamma }}^a{}_{cb}(p)$, and cyclically $R^a{}_{bde;c}(p)={\mathrm{\partial }}_c{\mathrm{\partial }}_d{\mathrm{\Gamma }}^a{}_{eb}(p)-{\mathrm{\partial }}_c{\mathrm{\partial }}_e{\mathrm{\Gamma }}^a{}_{db}(p)$ and $R^a{}_{bec;d}(p)={\mathrm{\partial }}_d{\mathrm{\partial }}_e{\mathrm{\Gamma }}^a{}_{cb}(p)-{\mathrm{\partial }}_d{\mathrm{\partial }}_c{\mathrm{\Gamma }}^a{}_{eb}(p)$.

(Cyclic cancellation.) Adding the three displays of step 2.1, the six second partial derivatives of Christoffel symbols cancel in pairs by [L5]: ${\mathrm{\partial }}_e{\mathrm{\partial }}_c{\mathrm{\Gamma }}^a{}_{db}$ cancels $-{\mathrm{\partial }}_c{\mathrm{\partial }}_e{\mathrm{\Gamma }}^a{}_{db}$, $-{\mathrm{\partial }}_e{\mathrm{\partial }}_d{\mathrm{\Gamma }}^a{}_{cb}$ cancels $+{\mathrm{\partial }}_d{\mathrm{\partial }}_e{\mathrm{\Gamma }}^a{}_{cb}$, and $+{\mathrm{\partial }}_c{\mathrm{\partial }}_d{\mathrm{\Gamma }}^a{}_{eb}$ cancels $-{\mathrm{\partial }}_d{\mathrm{\partial }}_c{\mathrm{\Gamma }}^a{}_{eb}$. Hence $R^a{}_{bcd;e}+R^a{}_{bde;c}+R^a{}_{bec;d}=0$ at $p$; by [L6] this tensor identity holds everywhere. Lowering the first index using [L3] ($g_{af;e}=0$) gives $R_{abcd;e}+R_{abde;c}+R_{abec;d}=0$, and the antisymmetrized form is the same display rewritten. This proves (a).

(First contraction.) Contract the identity of step 3.1 with $g^{ac}$, using [L3] to move the metric inside the derivatives: $g^{ac}{R}_{abcd;e}={\mathrm{\nabla }}_e(g^{ac}{R}_{abcd})={\mathrm{\nabla }}_e{\mathrm{Ric}}_{bd}$; $g^{ac}{R}_{abde;c}={\mathrm{\nabla }}^a{R}_{abde}$; and $g^{ac}{R}_{abec;d}={\mathrm{\nabla }}_d(g^{ac}{R}_{abec})$. For the last, $R_{abec}=-R_{abce}$ by [L7], so $g^{ac}{R}_{abec}=-g^{ac}{R}_{abce}=-{\mathrm{Ric}}_{be}$ by [L1] and the Einstein summation convention. Hence ${\mathrm{\nabla }}_e{\mathrm{Ric}}_{bd}+{\mathrm{\nabla }}^a{R}_{abde}-{\mathrm{\nabla }}_d{\mathrm{Ric}}_{be}=0$.

(Second contraction and the Einstein divergence.) Contract the display of step 4.1 with $g^{bd}$, again moving the metric inside derivatives with [L3]. The first term gives ${\mathrm{\nabla }}_e(g^{bd}{\mathrm{Ric}}_{bd})={\mathrm{\nabla }}_{e}S$. For the second, renaming the summed pair $(b,d)$ as $(d,b)$ and using the skew-symmetry $R_{adbe}=-R_{dabe}$ of [L7], $g^{bd}{R}_{abde}=g^{db}{R}_{adbe}=-g^{bd}{R}_{dabe}=-{\mathrm{Ric}}_{ae}$, because the defining contraction of [L1] reads ${\mathrm{Ric}}_{ae}=g^{bd}{R}_{dabe}$ with its dummy pair renamed. Hence ${\mathrm{\nabla }}^a(g^{bd}{R}_{abde})=-{\mathrm{\nabla }}^a{\mathrm{Ric}}_{ae}$. The third term gives $-g^{bd}{\mathrm{\nabla }}_d{\mathrm{Ric}}_{be}=-{\mathrm{\nabla }}^b{\mathrm{Ric}}_{be}=-{\mathrm{\nabla }}^a{\mathrm{Ric}}_{ae}$ after renaming. Collecting, ${\mathrm{\nabla }}_{e}S-2{\mathrm{\nabla }}^a{\mathrm{Ric}}_{ae}=0$, that is ${\mathrm{\nabla }}^a{\mathrm{Ric}}_{ab}=\frac{1}{2}{\mathrm{\nabla }}_{b}S$ after renaming $e\to b$.

(Vanishing divergence.) By [L1], $G_{ae}={\mathrm{Ric}}_{ae}-\frac{1}{2}S{g}_{ae}$, so by [L3] and step 5.1, ${\mathrm{\nabla }}^a{G}_{ae}={\mathrm{\nabla }}^a{\mathrm{Ric}}_{ae}-\frac{1}{2}{g}^{ab}{g}_{ae}{\mathrm{\nabla }}_{b}S=\frac{1}{2}{\mathrm{\nabla }}_{e}S-\frac{1}{2}{\mathrm{\nabla }}_{e}S=0$. Since $e$ is arbitrary this is ${\mathrm{\nabla }}^a{G}_{ab}=0$ for every $b$, and the proof is complete.