Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 26I (Differential Geometry)

The Gauss map sends $p$ to the chosen unit normal $N(p)$. Differentiating $⟨N,N⟩=1$ shows that $d{N}_p$ maps $T_{p}S$ into itself. The shape operator $S_p=-d{N}_p$ is self-adjoint because its associated second fundamental form is symmetric. Its eigenvalues $k_1,k_2$ are the principal curvatures. A point is umbilical when $k_1=k_2$, and the surface is minimal when its mean curvature $H=(k_1+k_2)/2$ vanishes everywhere.

In an orthonormal principal basis, $d{N}_p$ has diagonal matrix $\mathrm{diag}(-k_1,-k_2)$. It is conformal exactly whenwhich is equivalent to $k_1^2=k_2^2\ne 0$. At a nonumbilical point this says $k_1=-k_2$, exactly the condition $H=0$. Hence, when there are no umbilical points, the surface is minimal if and only if its Gauss map is conformal.

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.

No. If the image lies in a great circle, there is a fixed unit vector $a$ with $⟨N,a⟩=0$, so $a$ is tangent to $S$ everywhere. Since the ambient derivative of this constant vector is zero, the shape operator satisfies $S(a)=0$. One principal curvature is therefore zero. Minimality makes their sum zero, so both vanish and $S=0$. The Gauss map is then locally constant, and connectedness makes it constant on all of $S$, contradicting its image being a great circle.