Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 309 / 2

Fix the conventionAlthough a single covariant derivative of a vector is tensorial, the second derivative contains connection-dependent terms. Their antisymmetric difference cancels every second derivative of a coordinate change. More intrinsically, the mapis $C^{\infty }(M)$-linear in each of $U,V,X$, so it defines the Riemann curvature tensor.

Apply the definition to a coordinate-basis vector, use $[{\partial }_c,{\partial }_d]=0$, and collect the coefficient of $X^b$. This givesup to the overall sign fixed in part i. The right side therefore transforms as a tensor even though its individual Christoffel-symbol terms do not. This identifies the displayed coordinate expression $K^a{}_{bcd}$ with $R^a{}_{bcd}$ after matching the paper's index and sign conventions.

At an arbitrary point choose normal coordinates, so ${\Gamma }^a{}_{bc}=0$ there. Torsion freedom and commuting partial derivatives immediately give the algebraic first Bianchi identityDifferentiating the coordinate curvature formula at that point and cyclically antisymmetrizing gives the differential identityBoth equations are tensorial, so validity in normal coordinates at every point proves them in every coordinate system.

Contract the differential identity on its first and third curvature indices and use the algebraic symmetries of the Riemann tensor. One obtains the contracted Bianchi identity

The left side of the stated identity must inherit $R_{abcd}=-R_{abdc}$. Exchange $c$ and $d$, reduce all curvature products using pair antisymmetry and the first Bianchi identity, and compare with the negative of the original expression. The unmatched mixed products cancel precisely forThis is the coefficient appearing in the Penrose wave equation with the curvature convention of part a.

The vacuum Einstein field equations imply $R_{ab}=0$ and hence $R=0$. Every term in the supplied general identity involving the Ricci tensor or its covariant derivatives therefore vanishes. Substituting $\alpha =2$ leavesThis nonlinear, gauge-independent curvature equation is the Penrose wave equation.

At $ϵ=0$ the metric is the constant Minkowski metric, whose Christoffel symbols and curvature vanish, soInsert $R=ϵR^{(1)}+O({ϵ}^2)$ into the Penrose wave equation. Every curvature-square term is $O({ϵ}^2)$, while the covariant wave operator reduces at first order to the flat d'Alembert operator. ThusThe linearized Riemann curvature operator is unchanged by $h_{\mu \nu }\mapsto h_{\mu \nu }+{\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }$, because the resulting third derivatives cancel pairwise. The equation therefore requires no gauge choice for $h_{\mu \nu }$.