Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 2 / 11F (Geometry)

Withthe coordinate tangent vectors satisfyThus the first fundamental form isGeometrically, $S$ is the half of a helicoid with positive radius, winding upwards indefinitely.

Consider the upper half of the catenoid, parametrized as the surface of revolutionIt has the same first fundamental form, so $(u,v)\mapsto Y(u,v)$ defines a local isometry. By the invariance of Gaussian curvature under local isometry, the curvatures agree; directly, the conformal metric above gives

For an oriented regular surface with unit normal $N$, the Gauss map sends each point $p$ to $N(p)\in S^2$. The outward normal to the catenoid isIts image is the open southern hemisphere, apart from its limiting equator and south pole. The Gauss map is one-to-one after taking $v$ modulo $2\pi$ and reverses orientation. Since the Jacobian of the Gauss map is the Gaussian curvature, its signed spherical area is the total Gaussian curvature:Equivalently, $KdA=-{\mathrm{sech}}^{2}ududv$, whose integral over $u>0$ and $0\le v<2\pi$ is $-2\pi$.

The helicoid has no angular identification: its height coordinate is $v$. ConsequentlyThe catenoid has total curvature $-2\pi$. A global Riemannian isometry preserves both the Gaussian curvature and the area element, hence preserves total curvature. ThereforeThe same obstruction is visible in the parameter map: increasing $v$ by $2\pi$ gives a different point of the helicoid but the same point of the catenoid.