Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 37D (General Relativity)

The antipodal action fixes the origin. Away from the origin it acts freely, so the quotient is locally Euclidean there. A small punctured neighborhood of the image of the origin, however, has linkthe real projective plane, whereas the link of a point in a three-dimensional smooth manifold is $S^2$. Equivalently, the neighborhood is a cone on ${\mathbb{RP}}^2$ and cannot be a three-ball. Thus

Removing the closed unit ball leaveswhich is an open subset of ${\mathbb{R}}^3$. Every point therefore has an ordinary Euclidean coordinate neighborhood, soIt has no boundary; the unit sphere was removed together with the ball.

The map takes one covector argument and returns two vector slots, so $S$ has two contravariant and one covariant index. Its tensor type isor type $(2,1)$, with components $S^{ab}{}_c$.

The covariant derivative adds one covariant index, so $\nabla S$ has type $(2,2)$. The outer product of two tensors adds their index counts. Hence

LetThe Euler-Lagrange equation givesSince ${\dot{x}}^{\mu }=Ld{x}^{\mu }/ds$ and $d/d\lambda =Ld/ds$, this becomesExpanding the derivative, multiplying by the inverse metric, and using the Christoffel symbolgives the geodesic equation