Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 3 / 24H / iii / Solution

At $p\in D$, each of the given ambient rotations fixes $p$, rotates its tangent plane and preserves the local surface. Naturality of the shape operator under that rotation makes $S_p$ commute with every planar rotation. A symmetric real two-by-two operator commuting with all rotations is scalar, so every point of $D$ is an umbilical point. The trace-free part of the shape operator is continuous. Density therefore makes it zero everywhere on $S$.

Locally write $S=\kappa I$. The Codazzi equation gives $(X\kappa )Y=(Y\kappa )X$ for tangent fields $X,Y$; choosing independent fields proves $d\kappa =0$. Thus the surface is locally a plane if $\kappa =0$, or has constant spherical centersince its differential is $I-S/\kappa =0$. A change of normal reverses both $N$ and $\kappa$, leaving this center unchanged, so these local descriptions patch across the connected surface. In the flat case the whole surface lies in one plane, where a regular surface without boundary is open; a nonempty such set cannot also be compact and closed. Hence the flat case is excluded. In the other case $S$ lies on one sphere of radius $1/|\kappa |$. Its inclusion in that sphere is locally open, since both have dimension two, and its compactness makes the image closed. Connectedness of the sphere forces the image to be the entire sphere. Therefore $S$ is a sphere.