Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 309 (General Relativity)

A Killing vector field $K$ obeys ${\mathcal{L}}_{K}g=0$, or equivalently ${\nabla }_{(a}{K}_{b)}=0$. Along an affinely parametrized geodesic with tangent $U$,The second term vanishes by the geodesic equation, while the first contracts the symmetric tensor $U^a{U}^b$ with the antisymmetric part selected by the Killing equation. Thus $g(K,U)$ is a geodesic conserved quantity from a Killing vector. In Minkowski spacetime, translational Killing fields give conserved energy-momentum and rotational or boost Killing fields give the corresponding angular-momentum and boost charges; the same construction works for any spacetime isometry.

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Write ${\mathcal{L}}_{X}Y=[X,Y]$. Expanding the definition on $fZ$ gives terms proportional to derivatives of $f$ with coefficientAll remaining terms carry an overall factor $f$, soThe same argument gives linearity in $Y$, confirming that the Lie derivative of an affine connection is tensorial.

Using the torsion-free identities $[X,Y]={\nabla }_{X}Y-{\nabla }_{Y}X$ and $[X,Z]={\nabla }_{X}Z-{\nabla }_{Z}X$, expandThis is the required formula.

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The flow of a Killing vector field consists of local isometries. Isometries preserve both the metric and its unique torsion-free metric-compatible Levi-Civita connection, hence ${\mathcal{L}}_K\nabla =0$. Set $X=K$ and $Y=Z=U$ in part (b). Since ${\nabla }_{U}U=0$,and antisymmetry of the first two curvature arguments givesThis is the geodesic deviation equation with connecting field $K$: applying the isometry to the original geodesic produces a neighboring geodesic, and curvature determines their relative acceleration.

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An infinitesimal coordinate change generated by ${\xi }_a$ acts on the trace-reversed metric perturbation asIts divergence changes bySolving $\square {\xi }_b={\partial }^a{\bar{h}}_{ab}$ therefore imposes Lorenz gauge in linearized gravity, ${\partial }^a{\bar{h}}_{ab}=0$. Transformations satisfying $\square {\xi }_b=0$ remain and form the residual gauge symmetry of linearized gravity.

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For ${\bar{h}}_{ab}=A_{ab}{e}^{ik\cdot x}$, the vacuum wave equation and Lorenz condition requireA residual plane-wave gauge parameter ${\xi }_a=B_a{e}^{ik\cdot x}$ obeys $\square {\xi }_a=0$ because $k$ is null and changes the polarization byIn four dimensions its trace changes by $\delta A^a{}_a=2ik\cdot B$. Choosing the longitudinal component of $B$ appropriately sets $A^a{}_a=0$. The remaining residual transformations have $k\cdot B=0$.

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For $v,w\in k^{\perp }$, define $\mathcal{A}([v],[w])=A_{ab}{v}^a{w}^b$. Transversality gives $A(k,w)=0$, so changing either representative by a multiple of $k$ does not change the value; hence this is a symmetric bilinear form on $V=k^{\perp }/⟨k⟩$. After imposing $A^a{}_a=0$, the remaining gauge parameters obey $k\cdot B=0$, and the formula in part (b) gives $\delta A(v,w)=0$ for $v,w\in k^{\perp }$. Thus the form is gauge invariant.

Choose a null vector $\ell$ with $k\cdot \ell =-1$ and orthonormal spacelike vectors $e_1,e_2$ representing a basis of $V$. Since $A(k,\ell )=0$,The spacetime trace condition therefore says precisely that the induced bilinear form on the two-dimensional quotient $V$ is trace-free.

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After the trace gauge, $h_{ab}={\bar{h}}_{ab}=A_{ab}{e}^{ik\cdot x}$. The linearized Riemann curvature operator is built from terms containing two factors of $k$ and one factor of $A$:Contracting with $k^{[c}{v}^{d]}$ gives zero by $k^2=0$ and $k^a{A}_{ab}=0$, so $\mathcal{R}(k\wedge v)=0$. If ${\omega }_{cd}{k}^d=0$, every term in $R_{ab}^{(1)}{}^{cd}{\omega }_{cd}$ also contains such a contraction and vanishes.

To count the kernel, use the null basis $k,\ell ,e_1,e_2$. The forms $k\wedge v$ span the three-dimensional spaceThe condition ${\omega }_{ab}{k}^b=0$ defines the three-dimensional spaceTheir intersection has dimension two, so their sum is a four-dimensional subspace of $\ker \mathcal{R}$. Since $\dim {\Lambda }^2=6$, the rank-nullity theorem gives

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An orthonormal frame in spacetime is a local basis $\{e_a\}$ satisfying $g(e_a,e_b)={\eta }_{ab}$. The metric determines it only up to a point-dependent Lorentz transformation, with orientation and time orientation optionally restricting the component of the Lorentz group. The orthonormal coframe in spacetime is defined byand the metric is

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The connection 1-forms are defined by $\nabla e_b={\omega }^a{}_b\otimes e_a$. Metric compatibility gives ${\omega }_{ab}=-{\omega }_{ba}$, while vanishing torsion gives Cartan's first structure equationThe curvature 2-forms arewhich displays their relation to the Riemann curvature tensor.

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ChooseCartan's first structure equation gives the independent nonzero connection 1-formswith the remaining forms obtained by metric antisymmetry. Cartan's second structure equation then givesThese signs follow the curvature convention stated in Question 1.

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In the orthonormal frame choose the null vector $k=e_0+e_2$, which has a nonzero angular component. The curvature forms giveFor a null vector, the trace term in the Einstein field equations drops out, so the null energy condition implies $R_{ab}{k}^a{k}^b\ge 0$. Therefore

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A diffeomorphism $\varphi :M\to M$ is a smooth bijection with smooth inverse. It acts covariantly on a differential form by the pullback of a differential form:For vector fields $X_0,\dots ,X_p$,The coordinate definition of the exterior derivative, or its characterization as the unique natural graded derivation extending the differential of functions, gives

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The Hodge star operator is defined byfor forms of the same degree. In four dimensions, under $\hat{g}=e^{2\psi }g$, the inner product on $p$-forms scales by $e^{-2p\psi }$ and the volume form scales by $e^{4\psi }$. HenceFor $p=2$ the exponent vanishes, so the Hodge star on two-forms is conformally invariant.

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Transform both fields geometrically, $g\mapsto {\varphi }^{\ast }g$ and $A\mapsto {\varphi }^{\ast }A$. Naturality givesandEach Lagrangian four-form therefore transforms by pullback. Integration of a top-degree form is unchanged under an orientation-preserving diffeomorphism, so both the Einstein-Hilbert and Maxwell terms, and hence the full action, are invariant.

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For the infinitesimal diffeomorphism generated by $X^{\nu }$,Substitute these into the given first variation and integrate by parts. Symmetry of $E^{\mu \nu }$ givesBoundary terms vanish by assumption. Since $X^{\nu }$ is arbitrary and diffeomorphism invariance says $\delta S=0$, the coefficient must vanish:When the Maxwell equations $E^{\mu }=0$ hold, their divergence also vanishes and the identity reduces to ${\nabla }_{\mu }{E}^{\mu }{}_{\nu }=0$, the off-shell origin of stress-energy conservation.

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