Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 25I (Differential Geometry)

For $v$ in a sufficiently small neighbourhood of $0\in T_{p}S$, let ${\gamma }_v$ be the geodesic satisfying ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$. The exponential map at $p$ isChoose an oriented orthonormal basis of $T_{p}S$, write $e(\theta )=(\cos \theta ,\sin \theta )$, and define geodesic polar coordinates by

The Gauss lemma says that radial and angular coordinate curves are orthogonal. Since $r\mapsto \varphi (r,\theta )$ is a unit-speed geodesic,Moreover, torsion-freeness of the surface connection gives $D_r{\varphi }_{\theta }=D_{\theta }{\varphi }_r$, and thereforeAt $r=0$, ${\varphi }_{\theta }=0$, so $⟨{\varphi }_r,{\varphi }_{\theta }⟩=0$ for every sufficiently small $r$. Thus the first fundamental form isand $\sqrt{G}/r\to 1$ as $r\downarrow 0$ because $d({\exp }_p{)}_0$ is the identity.

An orientation of $S$ chooses a unit normal at every point. The Gauss map isFor the unit tangent $T=\dot{\alpha }$, the surface covariant derivative is the tangential projectionThe signed geodesic curvature isBecause $|T|=1$, $D_{s}T$ is perpendicular to $T$ and henceConsequently ${\kappa }_g=0$ means $D_{s}T=0$: the tangent is parallel along $\alpha$, so $\alpha$ is a geodesic.

Let $M$ be a compact oriented surface with piecewise smooth, positively oriented boundary. If ${\theta }_j$ are the signed exterior turning angles at its corners, the global Gauss-Bonnet theorem with boundary isFor a smooth boundary the corner sum is absent, and for a closed surface both boundary terms are absent.

Suppose that two closed geodesics ${\gamma }_1$ and ${\gamma }_2$ were disjoint. Under the usual convention that a closed geodesic curve is simple, the Jordan curve theorem on the sphere shows that they bound an annulus $A\subset S$. Its Euler characteristic is zero. Give $A$ the induced orientation and apply the Gauss-Bonnet theorem:Both boundary components are geodesics, so their geodesic curvature vanishes. It follows thatThis is impossible because $K>0$ everywhere and $A$ has positive area. Hence the two closed geodesics must intersect, as summarized by the intersection of closed geodesics on a positively curved sphere.