Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 26I / b / Solution

A minimal surface can have a planar point, where $k_1=k_2=0$. There $dN=0$, which fails the stipulated conformality condition $\lambda >0$. Thus minimality need not make the Gauss map conformal everywhere; a plane is the simplest counterexample.

Conversely, conformality gives $|k_1|=|k_2|\ne 0$. It permits either $k_1=-k_2$, which is minimal, or $k_1=k_2\ne 0$, which is umbilical and nonminimal. On a round sphere, the Gauss map is conformal while both principal curvatures are equal and nonzero. Thus conformality alone does not imply minimality.

Solved by gpt-5.6-sol high.