Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 25I (Differential Geometry)

A twice differentiable curve $\gamma :I\to S$ is a geodesic when its velocity is parallel along the curve:For a surface embedded in ${\mathbb{R}}^3$, the ambient acceleration criterion for a surface geodesic gives the equivalent conditionThe parameter $t$ is then an affine parameter, and $|\dot{\gamma }|$ is constant.

Take the circular cylinderIts first fundamental form is $R^{2}d{\theta }^2+d{z}^2$. The universal covering parametrization ${\mathbb{R}}^2\to S$ is therefore a local isometry from the Euclidean plane, after scaling the first coordinate. By geodesic preservation by a local isometry, every geodesic on the cylinder lifts to a straight line defined for all $t\in \mathbb{R}$. Thus the cylinder is geodesically complete. It is noncompact because $z$ is unbounded, and it is not a plane because one of its principal curvatures is $1/R$.

Suppose that a connected smooth surface $\tilde{S}\subset {\mathbb{R}}^3$ properly contains $S$. Since $S$ and $\tilde{S}$ have the same dimension, $S$ is open in $\tilde{S}$. Choose a point of its relative boundary. Inside a sufficiently small compact normal neighbourhood in $\tilde{S}$, choose $q\in S$ and a closest point $p\in \tilde{S}\setminus S$. A minimizing $\tilde{S}$-geodesic from $q$ to $p$ stays in $S$ before its endpoint, since encountering the complement earlier would contradict the choice of $p$.

The restriction of this curve is consequently a geodesic of $S$ with a finite maximal endpoint at $p$. On the other hand, the Gaussian curvature of $S$ is the restriction of the smooth Gaussian curvature of $\tilde{S}$, so it remains bounded near $p$. This contradicts condition $(\ast )$. Henceso $S$ is an inextendible embedded surface. This is the curvature-blowup criterion for inextendibility of a surface.

For an example satisfying $(\ast )$, consider the incomplete surface z equals r to three halvesA meridian of a surface of revolution reaches the missing origin in the finite lengthso $S_{\ast }$ is geodesically incomplete. The curvatures of a parametrized surface of revolution giveBy constant speed of an affinely parametrized geodesic, a geodesic with a finite endpoint has finite length and converges in ${\mathbb{R}}^3$. The finite-dimensional continuation criterion rules out an endpoint in a compact subset of $S_{\ast }$, and infinity is infinitely far away. It must approach the origin, so $(\ast )$ holds.

The answer to the second question is no. The punctured circular coneis incomplete because a generator reaches the missing vertex in finite time. It is inextendible: a proper connected smooth extension would have to add the origin, but the tangent planes approaching the origin depend on $\theta$ and therefore cannot be the continuous tangent planes of a smooth surface. Yetby the Gaussian curvature of a cone away from its vertex. Thus this incomplete inextendible surface does not satisfy $(\ast )$.