Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 115 / 4

A horizontal lift of $\gamma$ from $p$ is a curve $\tilde{\gamma }$ in $E$ projecting to $\gamma$, starting at $p$, and tangent to the horizontal distribution of the connection. In a local frame write $\tilde{\gamma }(t)=(\gamma (t),v(t))$. Horizontality is the linear ordinary differential equationwhose initial-value theorem gives local existence and uniqueness; successive trivializations continue the lift.

A geodesic satisfies ${\nabla }_{\dot{\gamma }}\dot{\gamma }=0$, andfor the geodesic with initial velocity $v$. Since $d({\exp }_p{)}_0=1$, the inverse function theorem makes ${\exp }_p$ a diffeomorphism near zero; its inverse gives normal coordinates. A geodesic sphere is ${\exp }_p(\{v:|v|=r\})$ inside such a normal neighborhood.

The Gauss lemma statesFor the variation ${\gamma }_s(t)={\exp }_p(t(v+sw))$, let $J={\partial }_s{\gamma }_s{|}_{s=0}$. The coordinate vector fields commute, so metric compatibility and constant geodesic speed giveSince $J(0)=0$, evaluation at $t=1$ proves the formula. In particular radial and spherical directions are orthogonal.

Solved by gpt-5.6-sol high.

Choose a normal ball on which ${\exp }_p$ is a diffeomorphism, and take $\epsilon$ smaller than half its radius. For $|X|<\epsilon$, $t\mapsto {\exp }_p(tX)$ is a geodesic of length $|X|$. If a competing curve remains in the normal ball, write it in polar form. The Gauss lemma makes radial and angular velocities orthogonal, so its speed is at least the absolute radial speed and its length is at least $|X|$. A curve leaving the larger normal ball already accumulates more than $|X|$ in radial variation. Thus the radial geodesic minimizes length.

At $q$, both $X$ and the geodesic sphere $\Sigma$ are hypersurfaces. If their tangent hyperplanes were distinct, they would be transverse. The transverse intersection theorem would then make $X\cap \Sigma$ a submanifold of dimensionFor $m\ge 3$ this has positive dimension near $q$, contradicting that the intersection is the singleton $\{q\}$. Therefore $T_{q}X=T_q\Sigma$.

The conclusion fails in dimension two because a transverse intersection is zero-dimensional and may be isolated. In the Euclidean plane, let $\Sigma$ be the unit circle, $q=(1,0)$, and let $X=\{(x,0):1/2<x<3/2\}$. Then $X\cap \Sigma =\{q\}$, but $T_{q}X$ is horizontal whereas $T_q\Sigma$ is vertical.

Solved by gpt-5.6-sol high.