Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 72 / 1

Use sequential semicolon indices: $X^a{}_{;kl}={\nabla }_l({\nabla }_k{X}^a)$, where the second covariant derivative also differentiates the lower index of ${\nabla }_k{X}^a$. For a torsion-free connection, direct expansion givesSubtract the expression with $k,l$ exchanged. Commuting partial derivatives cancel the first term; the first-derivative terms cancel in pairs; symmetry of the lower Christoffel symbol indices cancels both terms containing ${\Gamma }^p{}_{kl}$. What remains isThis is the Ricci identity with the curvature convention printed on the PDF's first page. In terms of derivative operators it is $[{\nabla }_l,{\nabla }_k]X^a=R^a{}_{bkl}{X}^b$; writing the operator order explicitly avoids reversing the sign.

The lower-index rule follows by applying the commutator to the scalar contraction $Y_a{X}^a$. Its second derivatives commute, so the product rule implies $Y_{a;kl}-Y_{a;lk}=-R^p{}_{akl}{Y}_p$. The derivative commutator is a derivation on tensor products, since the cross terms cancel. It therefore acts once on every index, with a positive sign on each upper index and a negative sign on each lower index. For an arbitrary tensor of valence $(r,s)$,Local sums of decomposable tensors establish the formula for every tensor field, rather than only for a single tensor product.

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.