Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 25I (Differential Geometry)

For a compact oriented regular surface $S\subset {\mathbb{R}}^3$ without boundary, the Gauss-Bonnet theorem statesHere $K=k_1{k}_2$ is the Gaussian curvature, the product of the two principal curvatures; $dA$ is the area element induced by the first fundamental form; and $\chi (S)$ is the Euler characteristic. If $S$ is connected and has genus $g$, the classification theorem for surfaces gives $\chi (S)=2-2g$, so

Since $K\ge 0$, Gauss-Bonnet gives $4\pi (1-g)\ge 0$, hence $g\le 1$. If $g=1$, the integral of the continuous nonnegative function $K$ is zero, so $K$ vanishes everywhere.

That is impossible for a compact regular surface in ${\mathbb{R}}^3$. Choose $p\in S$ maximizing the Euclidean distance from the origin, after translating the origin off the surface if necessary. The sphere centered at the origin through $p$ is a supporting sphere. The supporting-sphere curvature bound shows that both principal curvatures at $p$ have the same sign and nonzero magnitude, so $K(p)>0$. Therefore $g\ne 1$, and $g=0$. The classification theorem for surfaces now shows that $S$ is diffeomorphic to the sphere.

Take the ring torus with minor radius $1$ and major radius $R_n=n+2$. With angular coordinate $\theta$ around the generating circle, its Gaussian curvature isand henceThus the displayed $\mathrm{limsup}$ condition holds, while every $S_n$ has genus one and is not diffeomorphic to the sphere.

False with the $\mathrm{limsup}$ printed in the paper. Let the even-indexed surfaces be one fixed unit sphere and the odd-indexed surfaces one fixed ring torus. Thenbut infinitely many surfaces are tori. The two fixed surfaces lie in one ball and have uniformly bounded area. Their injectivity radii have a positive common lower bound, and compactness gives a common bound on $|\nabla K|$, hence on every radial derivative $|{\partial }_{r}K|$. Thus this alternating sequence satisfies all four additional conditions as well.

The intended statement becomes true if $\mathrm{limsup}$ is replaced byHere is the proof. Put ${\delta }_n=\max (0,-{\inf }_{S_n}{K}_n)$, so ${\delta }_n\to 0$. Choose $p_n$ maximizing distance from the centre of the fixed containing ball. The supporting-sphere curvature bound givesWrite $K_0=K_n(p_n)$ and $j(r,\theta )=\sqrt{G(r,\theta )}$ in geodesic polar coordinates. Choose a fixed sufficiently small $c>0$ and set ${\rho }_n=c{K}_0^{-1/2}$. Conditions (3) and (4) allow $c$ to be chosen so that ${\rho }_n\le {ϵ}_0$ andThe Jacobi equation $j_{rr}=-K_{n}j$, with $j(0)=0$ and $j_r(0)=1$, and the Sturm comparison theorem then give $j(r,\theta )\ge c_{1}r$ on this ball for a constant $c_1>0$ independent of $n$. ConsequentlySince $K_n\ge -{\delta }_n$ elsewhere and $\mathrm{Area}(S_n)\le M$,for all sufficiently large $n$. Gauss-Bonnet forces $\chi (S_n)>0$, and the classification theorem for surfaces makes each such $S_n$ a sphere.

For the tori in part (c), the distance from the origin is as large as $R_n+1\to \infty$, so condition (1) fails. Their surface areas areso condition (2) fails as well.