Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 25H (Differential Geometry)

WriteThe regularity of the profile makes $q>0$. The coordinate tangent vectors areso the first fundamental form of a surface of revolution has coefficientsChoose the unit normalTaking scalar products with the second derivatives of $\varphi$ gives the second fundamental form coefficientsThus the coordinate directions are principal directions, with principal curvaturesThe curvatures of a parametrized surface of revolution are thereforeReversing the orientation of a surface reverses the sign of $H$ but leaves $K$ unchanged.

A regular surface is a minimal surface when its mean curvature vanishes identically, equivalently when its two principal curvatures sum to zero everywhere.

The catenoidis an example. In part (a), take $f(v)=\cosh v$ and $g(v)=v$, so $q=\cosh v$. The formula there givesA plane and a circular cylinder have zero Gaussian curvature, as does a flat cone away from its vertex. Because a local isometry preserves Gaussian curvature, the catenoid is minimal and is not locally isometric to any of those surfaces; this is the catenoid is a minimal surface.

No. Fix $0<A<1$ and, near $s=0$, letThen $f^{\prime 2}+g^{\prime 2}=1$, so this is a regular unit-speed profile with $f>0$ on a sufficiently small interval. Rotate it around the $z$-axis. For an arc-length surface of revolution, $K=-f^{\prime \prime }/f$, and thereforeAt $s=0$, the formulas from part (a) give the principal curvaturesThey are unequal because $A\ne 1$, so the point is not an umbilical point. Every point of a sphere is umbilical. Consequently no neighbourhood of this point can lie on a sphere, giving a non-spherical surface of constant Gaussian curvature one.