Past exam of the general relativity II course of the University of Oxford / 2017 / 1 / b

In the order $(u,v,x,y)$ the metric tensor and inverse metric areOnly $g_{uu}$ has nonzero partial derivatives. Substitution into the formula for the Levi-Civita connection,therefore givesEvery other Christoffel symbol vanishes. For example $g^{ud}$ selects $d=v$, and the metric tensor is independent of $v$ with constant $g_{av}$; hence ${\Gamma }^u{}_{bc}=0$.

Use the paper's Riemann curvature tensor conventionFor the independent components,The quadratic Christoffel symbol terms vanish here: any potentially contributing intermediate $u$ index gives a factor ${\Gamma }^u{}_{ab}=0$, and the remaining possible products have a lower $v$ index and vanish. Since the symmetric matrix $M$ gives $F_{,ij}=2M_{ij}$, lowering the first index givesAll remaining nonzero components follow from the Riemann curvature tensor symmetriesThere are no other independent components: terms involving a $v$ index vanish, while nonzero terms require one $u$ in each antisymmetric pair.

The Ricci tensor is consequentlyThe Ricci scalar vanishes even when the trace is nonzero, since $g^{uu}=0$. With zero cosmological constant, the vacuum Einstein field equations reduce to $R_{ab}=0$. Therefore the necessary and sufficient condition isThis is the Ricci tensor of a Brinkmann plane wave: a nonzero, trace-free profile can have nonzero Riemann curvature tensor despite vanishing Ricci tensor.