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With the stated curvature convention, the Ricci tensor is $R_{bd}=R^a{}_{bad}$ and the scalar curvature is $R=g^{bd}{R}_{bd}$. Contract the differential Bianchi identity once to give the divergence of the Riemann tensor, then again with the inverse metric. The curvature antisymmetries yield ${\nabla }^a{R}_{ab}-{\nabla }_{b}R+{\nabla }^a{R}_{ab}=0$, so ${\nabla }^a{R}_{ab}=\frac{1}{2}{\nabla }_{b}R$.

The Einstein field equations are $G_{ab}+\Lambda g_{ab}=(8\pi G/c^4)T_{ab}$, with $G_{ab}=R_{ab}-\frac{1}{2}R{g}_{ab}$. The contracted Bianchi identity makes the left side divergence-free, enforcing the local stress-energy conservation law ${\nabla }^a{T}_{ab}=0$ and ensuring consistency of the matter equations with the gravitational constraints.

For a scalar, second covariant derivatives commute. Commuting a further derivative past the resulting covector introduces Ricci curvature, giving ${\nabla }^2{\nabla }_a\varphi ={\nabla }_a{\nabla }^2\varphi +R_{ab}{\nabla }^b\varphi$.

Now $R_{ab}={\nabla }_a{\nabla }_b\varphi$ implies $R={\nabla }^2\varphi$. Taking its divergence and using that commutation formula gives $\frac{1}{2}{\nabla }_{b}R={\nabla }_{b}R+R_{ba}{\nabla }^a\varphi$, hence ${\nabla }_{b}R=-2R_{ba}{\nabla }^a\varphi$. Meanwhile ${\nabla }_b({\nabla }_a\varphi {\nabla }^a\varphi )=2R_{ba}{\nabla }^a\varphi$. Adding provesIn Riemannian signature this is the steady gradient Ricci soliton scalar identity, with soliton potential $f=-\varphi$.