Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 52 / 4 / b / i / Solution

Use the connection convention ${\nabla }_{\alpha }{Y}^{\rho }={\partial }_{\alpha }{Y}^{\rho }+{\Gamma }^{\rho }{}_{\mu \alpha }{Y}^{\mu }$, with the derivative index last. This is consistent with the printed curvature formula and the final formula in part (iv). The paper's $T_{\mu \nu }{}^{\rho }={\Gamma }^{\rho }{}_{\mu \nu }-{\Gamma }^{\rho }{}_{\nu \mu }$ is then the negative of the geometric torsion tensor defined by $\mathcal{T}(X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]$. Keep the paper's component convention throughout this question.

Set $A_{\mu \nu \rho }=g_{\rho \lambda }({\Gamma }^{\lambda }{}_{\mu \nu }-S^{\lambda }{}_{\mu \nu })$. The difference of affine connections is a tensor. Since the Levi-Civita connection $S$ is symmetric and metric compatible, expanding the nonmetricity tensor givesHere $T_{\mu \nu \rho }=g_{\rho \lambda }{T}_{\mu \nu }{}^{\lambda }$ is antisymmetric in its first two slots, while $N_{\mu \nu \rho }$ is symmetric in its first two slots. Cyclically permuting these identities and eliminating the other components givesRaise the last slot to obtain the connection decompositionwhereThe PDF is missing the factor $1/2$ in its definition of $K$. With the printed $K$, the decomposition would instead contain $-K/2$. In these formulas $T^{\rho }{}_{\mu \nu }$ raises the first slot of the fully lowered tensor; it must not be confused with $T_{\mu \nu }{}^{\rho }$, which raises the last slot.