Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 4 / 24H (Differential Geometry)

An isothermal parametrization is an immersion with first fundamental form $Ed{u}^2+2Fdudv+Gd{v}^2={\lambda }^2(d{u}^2+d{v}^2)$, where $\lambda >0$. Thus $E=G={\lambda }^2$ and $F=0$.

Write $W={\varphi }_{uu}+{\varphi }_{vv}$. Differentiating the metric identities givesFor example ${\varphi }_{vv}\cdot {\varphi }_u=({\varphi }_v\cdot {\varphi }_u{)}_v-{\varphi }_v\cdot {\varphi }_{uv}=-G_u/2$; the other identities are analogous. Hence $W$ is normal to the surface. For a unit normal $N$, its normal component is $e+g$. The supplied mean curvature formula reduces to $H=(e+g)/(2{\lambda }^2)$. ThereforeA surface is a minimal surface when its mean curvature vector vanishes. Since ${\lambda }^2>0$, the preceding identity proves minimality is equivalent to $\Delta \varphi =0$ in isothermal coordinates.

The null condition follows by expansion:It vanishes exactly when $E=G$ and $F=0$; the immersion hypothesis ensures their common value is positive. Thus it characterizes isothermal coordinates for an immersion.

With complex coordinate $\zeta =u+iv$, mixed derivatives commute andand similarly for $y,z$. By the Cauchy-Riemann equations, each ${\phi }_j$ is holomorphic exactly when the corresponding coordinate is harmonic. Combining with the preceding part gives an isothermal immersion is minimal exactly when all three ${\phi }_j$ are holomorphic. Without immersion the null condition could also describe a degenerate map, so the nondegeneracy hypothesis matters.

Let $\zeta =u+iv$. Direct differentiation givesThese are holomorphic polynomials and their squared sum isMoreover direct calculation of the metric gives $E=G=(1+u^2+v^2{)}^2>0$ and $F=0$, so the map is genuinely an immersion everywhere. The preceding characterizations prove that it is an isothermal parametrization of a minimal surface, the Enneper surface.