Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 309 / 1

Fix the conventionEquivalently,This defines the Riemann curvature tensor because it is $C^{\infty }(M)$-linear in $U,V,X$: derivatives of a multiplying function cancel between the three terms. Thus its value at a point depends only on the three tangent vectors there, rather than their extensions.

Apply the definition to the coordinate basis, for which $[{\partial }_{\mu },{\partial }_{\nu }]=0$ and ${\nabla }_{\mu }{\partial }_{\tau }={\Gamma }^{\rho }{}_{\tau \mu }{\partial }_{\rho }$. Comparing coefficients givesequivalent to the paper's ordering after commuting scalar factors. Although the individual Christoffel symbols are not tensors, the preceding intrinsic definition proves that this complete combination is tensorial.

At an arbitrary point choose normal coordinates, so the Christoffel symbols vanish there. Torsion freedom and commutation of partial derivatives then give the algebraic first Bianchi identityDifferentiate the coordinate curvature expression and cyclically antisymmetrize. Third derivatives cancel, giving the second Bianchi identityBoth statements are tensorial and hence hold in every coordinate system. Contracting the differential identity, using the curvature symmetries and metric compatibility, yieldsthe contracted Bianchi identity.

Contract the isotropic-curvature formula on $a,c$. In dimension $n$ it givesSubstitution into the contracted Bianchi identity givesSince $n>2$, ${\nabla }_{b}K=0$, so $K$ is constant on each connected component. Thus this is constant sectional curvature, withThis argument is Schur theorem in pseudo-Riemannian geometry.

The vector $T^a$ is the tangent to a reference member of a one-parameter family of affinely parametrized geodesics. The Jacobi field $Y^a$ is the infinitesimal connecting vector from that geodesic to a neighboring one at equal parameter. The geodesic deviation equation states that curvature determines their relative acceleration.

Parallel propagation means ${\nabla }_T{e}_{\alpha }=0$, where $T=\dot{\lambda }$. By metric compatibility,The inner products therefore retain their initial values ${\eta }_{\alpha \beta }$, so the parallel-propagated orthonormal frame remains orthonormal.

In the propagated frame $T=e_0$, and the constant sectional curvature formula reduces geodesic deviation toThe stated temporal initial data give $y^0(t)=0$. Writing $\omega =\sqrt{-K}$ for $K<0$ and $\kappa =\sqrt{K}$ for $K>0$, the spatial displacement isThese formulas are valid to first order in the initial separation and relative velocity.