Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 24I / Solution

A geodesic is a curve with zero covariant acceleration in an affine parameter. For a unit-speed curve on an oriented surface with unit normal $N$, its geodesic curvature is $k_g=⟨{\gamma }^{\prime \prime },N\times {\gamma }^{\prime }⟩$: it measures the tangent-plane component of acceleration perpendicular to the tangent. A unit-speed curve is geodesic exactly when this component vanishes.

The exponential map ${\exp }_p(v)$ follows for unit time the geodesic starting at $p$ with initial velocity $v$, whenever it is defined. Since ${\exp }_p(tv)={\gamma }_v(t)$, its derivative at zero is $\boxed{d({\exp }_p{)}_0={\mathrm{id}}_{T_{p}S}}$. Choose an oriented orthonormal basis of $T_{p}S$ and let $v(\theta )$ be its unit circle. Geodesic polar coordinates are $F(r,\theta )={\exp }_p(rv(\theta ))$ in a normal neighborhood, with the usual angular singularity at $r=0$.

Put $T=F_r$ and $J=F_{\theta }$. Radial curves are unit-speed geodesics, so $E=⟨T,T⟩=1$ and ${\nabla }_{r}T=0$. The torsion-free connection gives ${\nabla }_{r}J={\nabla }_{\theta }T$, and thereforeSince $J(0,\theta )=0$, this proves $F=⟨T,J⟩=0$, the radial form of the Gauss lemma. Moreover $J(r,\theta )=r{v}^{\prime }(\theta )+O(r^2)$, with $|v^{\prime }|=1$, by the derivative of the exponential map. Hence $G=|J{|}^2=r^2+O(r^3)$, givingThe last derivative is the one-sided radial limit.

The global Gauss-Bonnet theorem for a compact oriented surface with piecewise smooth boundary isHere $K$ is Gaussian curvature, the boundary is positively oriented with the surface on its left, ${\epsilon }_j$ are signed exterior turning angles at corners, and $\chi$ is the Euler characteristic. Smooth geodesic boundary components contribute neither curvature nor corner terms.

A contractible simple closed geodesic on the cylinder would bound a disc, giving $\int KdA=2\pi$, impossible when $K<0$. Thus any such geodesic is an essential loop. If two distinct essential simple loops on an annulus intersect, the elementary bigon criterion gives two subarcs bounding an innermost disc: lifting the homotopic loops to the covering strip and choosing an innermost region supplies such a bigon. Distinct geodesics cannot have tangential intersections by uniqueness of the geodesic initial-value problem, so its two interior angles $\alpha ,\beta$ are positive. Gauss-Bonnet on this geodesic bigon gives $\int KdA=\alpha +\beta >0$, another contradiction. Therefore two distinct simple closed geodesics would be disjoint. They would then bound a compact annulus with geodesic boundary and Euler characteristic zero, forcing $\int KdA=0$, again impossible. There is at most one simple closed geodesic. Completeness of the noncompact surface is not needed for these compact-region arguments.