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The first derivative of a scalar is a covector, soCommuting partial derivatives and the torsion-free connection give ${\phi }_{;lm}={\phi }_{;ml}$, the symmetric Hessian of a scalar field. Its next covariant derivative isIn particular, tensor index symmetrization over $l,m$ adds nothing: ${\phi }_{;(lm)n}={\phi }_{;lmn}$. This does not make the third index interchangeable with the first two. Applying the lower-index Ricci identity to the covector ${\phi }_{;l}$ yields the requested tensor index antisymmetrization:Parentheses and brackets include the usual averaging factors. To make the remaining symmetry explicit, let $S_{lmn}={\phi }_{;(lmn)}$. Since the first two derivative indices are already symmetric,The two commutator identities give ${\phi }_{;lnm}={\phi }_{;lmn}+R^p{}_{lmn}{\phi }_{;p}$ and ${\phi }_{;mnl}={\phi }_{;lmn}+R^p{}_{mln}{\phi }_{;p}$. Thus the third covariant derivatives of a scalar can also be writtenThe right-hand side displays exactly the obstruction to total symmetry. At a point where the scalar gradient vanishes, or where these curvature contractions vanish, the third derivative is totally symmetric.

For the final contraction use $R_{ab}=R^c{}_{acb}$. The arbitrary-valence formula givesThe Ricci tensor of the Levi-Civita connection is symmetric, whereas the last factor is antisymmetric. Their contraction is zero, provingNo symmetry of $T^{ik}$ was assumed. This double divergence of a contravariant tensor identity uses the metric-compatible connection of general relativity: zero torsion alone for an arbitrary affine connection does not guarantee a symmetric Ricci tensor and would not suffice for this final step.