 Solution

For a scalar, ${\nabla }_a\varphi ={\partial }_a\varphi$, while its second covariant derivative is ${\nabla }_a{\nabla }_b\varphi ={\partial }_a{\partial }_b\varphi -{\Gamma }_{ab}^c{\partial }_c\varphi$. Partial derivatives commute and the Levi-Civita connection is torsion-free, ${\Gamma }_{ab}^c={\Gamma }_{ba}^c$, proving the asserted symmetry.

Write the Einstein equation without cosmological constant as $G_{ab}=\kappa T_{ab}$, $\kappa =8\pi G/c^4$. Tracing the given Ricci equation gives $R=g^{ab}{\partial }_a\varphi {\partial }_b\varphi$. HenceIn units $\kappa =1$ the prefactor is absent. Metric compatibility, the product rule and the just-proved symmetric Hessian of a scalar field giveThe contracted Bianchi identity makes the left side zero. Wherever the gradient covector is nonzero, at least one component is nonzero, forcing ${\nabla }_a{\nabla }^a\varphi =0$, even when that gradient is null.

Finally ${\Gamma }_{ab}^a={\partial }_b\log \sqrt{-g}$, obtained by differentiating the metric determinant. Thus for any vector $V$, ${\nabla }_a{V}^a=(\sqrt{-g}{)}^{-1}{\partial }_a(\sqrt{-g}{V}^a)$. Applying it to $V^a=g^{ab}{\partial }_b\varphi$ proves