Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 2 / 24H (Differential Geometry)

A minimal surface has zero mean curvature $H=(k_1+k_2)/2$; equivalently its area is stationary under compactly supported normal variations. Its principal curvatures satisfy $k_2=-k_1$, so its Gaussian curvature isEquality occurs precisely where its second fundamental form vanishes.

Let $A=-dN$ be the self-adjoint shape operator. In an orthonormal principal basis, $⟨dN(v),dN(w)⟩=⟨Av,Aw⟩$, with diagonal coefficients $k_1^2,k_2^2$. The Gauss map is conformal with nonzero factor exactly when $k_1^2=k_2^2=\lambda >0$.

If the surface is minimal and has no umbilics, $k_2=-k_1$ and $k_1\ne 0$, so this equality holds. Conversely equality of the squares gives either $k_1=k_2$ or $k_1=-k_2$. The first possibility is excluded by the no-umbilic hypothesis. Hence $H=0$, proving minimality is equivalent to conformality of the Gauss map under the stated hypothesis. The factor is automatically positive because it equates positive definite inner products. The hypothesis is essential: a sphere has conformal Gauss map and nonzero mean curvature.

At a non-planar point of a minimal surface, the shape operator is not zero. Minimality gives nonzero opposite principal curvatures, so $K<0$ nearby and $dN$ is invertible. The inverse function theorem makes the Gauss map a local diffeomorphism onto a patch of the sphere; part (ii) makes it conformal.

Choose stereographic coordinates $(u,v)$ on that sphere patch. Their metric is $4(d{u}^2+d{v}^2)/(1+u^2+v^2{)}^2$. Pulling them back by $N$, the conformal identity $N^{\ast }{g}_{S^2}=\lambda g_S$ givesThe positive coefficient shows that these are local isothermal coordinates. No general isothermal-coordinate existence theorem is needed: the nondegenerate conformal Gauss map supplies them explicitly.