Does any minimal surface which is regular has a conformal parameterization?

Question

Since given surface is regular and minimal, I have $$H=\frac{Eg-2Ff+Ge}{2(EG-F^2)}=0$$ and $$X_u\times X_v\ne0$$ Can I derive $E=G,\ F=0$ from these conditions?

Answer 1

It is not true that you can always derive $E=G$, $F=0$ on an arbitrary parametrization (even on a minimal surface). However, it is true that there is always a isothermal parametrization (that is, those satisfies your condition) for any surface (minimality not needed). The proof is a bit more difficult in general, but for minimal surface, it's easier.

The proof goes roughly as follows: by a rotation and translation if necessary, assume that the minimal surface is of the form

$$(x, y) \mapsto (x, y, u(x, y)),$$

where $u: U\to \mathbb R$ is a function on $U\subset \mathbb R^2$ and $U$ is simply connected. Then the minimal surface equation $H = 0$ will imply

$$\begin{cases} \frac{\partial}{\partial x} \left( \frac{1+|q|^2}{W}\right) = \frac{\partial}{\partial y} \left( \frac{p\cdot q}{W}\right) \\ \frac{\partial}{\partial x} \left( \frac{p\cdot q}{W} \right) = \frac{\partial}{\partial y} \left( \frac{1+|p|^2}{W} \right), \end{cases}$$

where $p = u_x$, $q = u_y$ and $W = \sqrt{1 + |p|^2 + |q|^2}$.

As $U$ is simply connected, there are functions $A, B$ so that

$$\frac{\partial}{\partial x} A = \frac{1+|q|^2}{W}, \frac{\partial}{\partial y} A = \frac{p\cdot q}{W},$$

$$\frac{\partial}{\partial x} B = \frac{p\cdot q}{W}, \frac{\partial}{\partial y} B = \frac{1+|p|^2}{W}.$$

One can then use this $A$, $B$ to define a new parametrization of the minimal surface, and under this new parametrization, we have $E = G$, $F=0$.

For the details you can check lemma 4.4 in the book "A survey of minimal surfaces" by R.Osserman, or see here.