Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 311 / 1

Because $U$ is a null tangent, metric compatibility givesBecause $U$ is tangent to an affinely parametrized null geodesic congruence,by the geodesic equation. Thus $B$ has no component along $U$ in either index.

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With $U^2=N^2=0$ and $U\cdot N=-1$, define the screen-space projectorDirect contraction gives $P^a{}_b{U}^b=P^a{}_b{N}^b=0$, while a vector orthogonal to both $U$ and $N$ is unchanged. Moreover $P^a{}_c{P}^c{}_b=P^a{}_b$ and $P^a{}_a=2$, so its image is precisely the two-dimensional transverse space.

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Decompose the optical tensor on the two-dimensional screen space asThe null expansion, null shear, and null twist are respectivelyThe shear is symmetric and trace-free, while the twist is antisymmetric.

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Differentiate $B_{ab}={\nabla }_b{U}_a$ along an affinely parametrized generator. Commuting covariant derivatives and using ${\nabla }_{U}U=0$ gives the optical evolution equationTaking its screen trace and usingproduces the Null Raychaudhuri equationThe generators lie in a null hypersurface, so they are hypersurface orthogonal. The Frobenius theorem therefore gives ${\hat{\omega }}_{ab}=0$, leaving

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Contracting the Einstein field equations with the null tangent eliminates the trace term and givesby the null energy condition. The screen metric is positive definite, so ${\hat{\sigma }}_{ab}{\hat{\sigma }}^{ab}\ge 0$. The Null Raychaudhuri equation consequently impliesWhile $\theta <0$ this is equivalent toIf $heta({\lambda }_0)={\theta }_0<0$, integration givesThe right-hand side reaches zero after affine distance $2/|{\theta }_0|$. A finite negative expansion cannot pass through this value, so $heta\to -\infty$ no later than that point. This is the null focusing theorem.

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Choose a coordinate sphere at late conformal time ${\eta }_0$ and comoving radius $r_0$. Its areal radius is $R=a(\eta )r$, so its area is $A=4\pi a^2{r}^2$. Future-directed radial null generators have $dr/d\eta =\pm 1$. Up to a common positive normalization of their affine tangents, their expansions therefore have the signs ofThe outgoing sign is positive because $a^{\prime }>0$. The ingoing sign is also positive wheneverAn arbitrarily large sphere exists because the spatial topology is ${\mathbb{R}}^3$. Both future null expansions are positive on such a sphere, so it is an anti-trapped surface.

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Apply the time-reversed Penrose singularity theorem. The two past-directed null congruences orthogonal to the compact anti-trapped surface have negative expansion. The null energy condition, through the Einstein field equations, supplies the null convergence condition, and the null focusing theorem forces each generator to acquire a conjugate point within finite affine length if it can be extended that far.

If every past-directed null generator were complete, the boundary of the causal past of the surface would therefore be generated only for a bounded affine interval. Compactness of the initial surface and continuous dependence of geodesics on initial data would make that achronal boundary compact. A globally hyperbolic spacetime provides a Cauchy hypersurface and a timelike flow projecting the boundary onto it. The standard Penrose argument then makes its image both open and closed, forcing the connected Cauchy hypersurface to be compact. This contradicts its stipulated topology ${\mathbb{R}}^3$.

Hence at least one past-directed null generator ends after finite affine parameter: the universe is null-geodesically incomplete to the past. Global hyperbolicity controls the causal boundary, the noncompact spatial topology supplies the contradiction, and the energy condition supplies focusing.

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