Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 59 / 1 / a

Use the Levi-Civita connection interpretation of the curvature convention. Its Riemann curvature tensor satisfiesThe first antisymmetry comes from the derivative commutator; the second comes from compatibility with the metric tensor. The cyclic identity is the first Bianchi identity, and together these imply exchange symmetry of the two antisymmetric index pairs. Contracting gives the symmetric Ricci tensor, $\boxed{R_{ab}=R_{ba}}$. The differential Bianchi identity is ${\nabla }_{[a}{R}_{bc]de}=0$.

There is a hypothesis qualification. A metric connection need not be a torsion-free connection. For a general such connection, the tensorial derivative commutator contains $-T^c{}_{ab}{\nabla }_c$ as well as curvature; pair exchange, the stated cyclic identity and Ricci symmetry need not hold. The printed pure-curvature commutator, and the scalar identity requested later, use the torsion-free interpretation. Under that intended interpretation all the boxed symmetries follow. They are not assertions about arbitrary metric-compatible connections with torsion tensor.

For example, with a constant Minkowski metric a connection having only ${\Gamma }^1{}_{02}=1$ and ${\Gamma }^2{}_{01}=-1$ nonzero is metric compatible, since its connection matrix in direction zero is a spatial rotation. It has $T^1{}_{02}=1$. For the scalar $f=x^1$ the tensorial Hessian commutator is $[{\nabla }_0,{\nabla }_2]f=-1$, rather than zero. This explicitly demonstrates the missing zero-torsion condition used in the subsequent scalar proof.