Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 309 / 2

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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