Let \textbf{x}(u,v) be a regular parametrized surface and let z=u+iv denote the corresponding complex coordinate. Since u=\frac{z+\bar{z}}{2} and v=\frac{-i(z-\bar{z})}{2}, we may write \textbf{x}(z,\bar{z})=\left(x^1(z,\bar{z}),x^2(z,\bar{z}),x^3(z,\bar{z})\right).

We define the complex function \phi as follows,

\phi(z):=\left(\frac{1}{2}\left(\frac{\partial x^1}{\partial u}-i\frac{\partial x^1}{\partial v}\right),\frac{1}{2}\left(\frac{\partial x^2}{\partial u}-i\frac{\partial x^2}{\partial v}\right),\frac{1}{2}\left(\frac{\partial x^3}{\partial u}-i\frac{\partial x^3}{\partial v}\right)\right)

Theorem. Let M be a surface with patch \textbf{x}(u,v) and let \phi=\frac{\partial\textbf{x}}{\partial z}. Then \textbf{x}(u,v) is isothermal if, and only if, (\phi^1)^2+(\phi^2)^2+(\phi^3)^2=0.

Proof. Suppose that \textbf{x}(u,v) is isothermal. Note that (\phi^i)^2=\frac{1}{4}[(x^i_u)^2-(x^i_v)^2-2ix^i_ux^i_v]. Therefore,

= (\phi)^2

= \frac{1}{4}\left[\sum_{i=1}^{3}(x^i_u)^2-\sum_{i=1}^{3}(x^i_v)^2-2i\sum_{i=1}^{3} x^i_ux^i_v\right]

= \frac{1}{4}\left(|\textbf{x}_u|^2-|\textbf{x}_v|^2-2i\langle\textbf{x}_u,\textbf{x}_v\rangle\right)

= \frac{1}{4}\left(E-G-2iF\right)

= 0

Conversely, suppose that (\phi)^2=0. Then \frac{1}{4}\left(E-G-2iF\right)=0 and by properties of complex numbers this equations only holds if E-G=0 and F=0 which shows that \textbf{x}(u,v) is isothermal. \Box

Theorem. Suppose that M is a surface with patch \textbf{x}(u,v). Let \phi=\frac{\partial\textbf{x}}{\partial z} and suppose that (\phi)^2=0 (i.e. \textbf{x} is isothermal). Then M is minimal if, and only if, each \phi^i is analytic.

Proof. Suppose that M is minimal, then \textbf{x}(u,v) is harmonic; that is, \Delta\textbf{x}=0. Thus, \frac{\partial\phi}{\partial\bar{z}}=\frac{\partial}{\partial\bar{z}}\left(\frac{\partial\textbf{x}}{\partial z}\right)=\frac{1}{4}\Delta\textbf{x}=0. Therefore, each \phi^i=\frac{\partial\textbf{x}}{\partial z} is analytic. Conversely, the same calculation shows that if each \phi^i is analytic, then each x^i is harmonic, therefore, M is minimal. \Box

Corollary. x^i(z,\bar{z})=c_i+2\text{Re}\int\phi^i\;dz.

Proof. Since z=u+iv, we may write dz=du+idv. Then

\phi^idz=\frac{1}{2}\left[(x^i_u-ix^i_v)(du+idv)\right]=\frac{1}{2}\left[x^i_udu+x^i_vdv+i(x^i_udv-x^i_vdu)\right]

\bar{\phi}^idz=\frac{1}{2}\left[(x^i_u+ix^i_v)(du-idv)\right]=\frac{1}{2}\left[x^i_udu+x^i_vdv-i(x^i_udv-x^i_vdu)\right].

We then have that dx^i=\frac{\partial x^i}{\partial z}dz+\frac{\partial x^i}{\partial\bar{z}}d\bar{z}=\phi^idz+\bar{\phi}^id\bar{z}=2\text{Re}\phi dz and we can now integrate to get x^i. \Box

The problem of constructing minimal surfaces reduces to finding a tripple of analytic functions \phi=(\phi^1,\phi^2,\phi^3) with (\phi)^2=0. A nice we of constructing such a \phi is to take an analytic function f and a meromorphic function g with fg^2 analytic. Now, if we let f=\phi^1-i\phi^2 and g=\frac{\phi^3}{(\phi^1-i\phi^2)} then we have,

\phi^1=\frac{1}{2}f(1-g^2) \quad \phi^2=\frac{i}{2}f(1+g^2) \quad \phi^3=fg.

Note that f is analytic, g is meromorphic, and fg^2 is analytic since fg^2=-(\phi^1+i\phi^2). Furthermore, it is easily verified that (\phi)^2=0. Therefore, \phi determines a minimal surface.

Theorem. (Weierstrass-Enneper Representation I) If f is analytic on a domain D, g is meromorphic on D, and fg^2 is analytic on D then a minimal surface is defined by \textbf{x}(z,\bar{z})=\left(x^1(z,\bar{z}),x^2(z,\bar{z}),x^3(z,\bar{z})\right), where

\displaystyle x^1(z,\bar{z})=\text{Re}\int f(1-g^2)\;dz

\displaystyle x^2(z,\bar{z})=\text{Re}\int if(1+g^2)\;dz

\displaystyle x^3(z,\bar{z})=\text{Re}2\int fg\;dz.

Suppose that g is analytic and has an inverse g^{-1} in a domain D which is analytic as well. Then we can consider g as a new complex variable \tau=g with d\tau=g'dz. Define F(\tau)=\frac{f}{g'} and obtain F(\tau)d\tau=fdz. Therefore, if we replace g by \tau and fdz by F(\tau)d\tau we get the following.

Theorem. (Weierstrass-Enneper Representation II) For any analytic function F(\tau), a minimal surface is defined by \textbf{x}(z,\bar{z})=\left(x^1(z,\bar{z}),x^2(z,\bar{z}),x^3(z,\bar{z})\right), where

\displaystyle x^1(z,\bar{z})=\text{Re}\int(1-\tau^2)F(\tau)\;d\tau

\displaystyle x^2(z,\bar{z})=\text{Re}\int i(1+g^2)F(\tau)\;d\tau

\displaystyle x^3(z,\bar{z})=\text{Re}2\int\tau F(\tau)\;d\tau.

Note the corresponding \phi=\left(\frac{1}{2}(1-\tau^2)F(\tau),\;\frac{i}{2}(1+\tau^2)F(\tau),\;\tau F(\tau)\right).

This representaion tells us that any analytic function F(\tau) defines a minimal surface.

We can now use the Weierstrass-Enneper representation to produce minimal surfaces. For example, if (f, g) = (1, z) then we obtain a parametrization for Enneper’s surface. In fact, if (f,g) = (1,z^n) then an nth order Enneper’s surface is obtained.

First order Enneper surface

The Weierstrass-Enneper representation leads to infinite families of minimal surfaces and has proved fundamental in relating the study of minimal surfaces to the theory of complex analysis.

Advertisements