Homework 6

#exercise-problems #math

Problem Statement

Let $X (u, v)$ be an isothermal parameterization of a minimal surface $S$ (that is $F = 0$ and $E = G$ ) and let

\begin{matrix} (1) & φ = \frac{1}{2} (X_{u} - i X_{v}) \end{matrix}

Show that:

\begin{aligned} X_{u} \times X_{v} & = 4 I m (φ_{2} {\bar{φ}}_{3}, φ_{3} {\bar{φ}}_{1}, φ_{1} {\bar{φ}}_{2}) \\ = 2 I m (φ \times \bar{φ}) \end{aligned}

From (1) we can write $X_{u}$ and $X_{v}$ as:

\begin{aligned} X_{u} & = φ + \bar{φ} \\ X_{v} & = \frac{\bar{φ} - φ}{i} \end{aligned}

Therefore computing $X_{u} \times X_{v}$ we obtain that:

\begin{aligned} X_{u} \times X_{v} & = (φ + \bar{φ}) \times (\frac{\bar{φ} - φ}{i}) \\ = \frac{2}{i} (φ \times \bar{φ}) \end{aligned}

Hence if $φ = (φ_{1}, φ_{2}, φ_{3})$ then:

\begin{aligned} φ \times \bar{φ} & = (\begin{array}{c} φ_{2} {\bar{φ}}_{3} - φ_{3} {\bar{φ}}_{2} \\ φ_{3} {\bar{φ}}_{1} - φ_{1} {\bar{φ}}_{3} \\ φ_{1} {\bar{φ}}_{2} - φ_{2} {\bar{φ}}_{1} \end{array}) \\ = 2 i (\begin{array}{c} I m (φ_{2} {\bar{φ}}_{3}) \\ I m (φ_{3} {\bar{φ}}_{1}) \\ I m (φ_{1} {\bar{φ}}_{2}) \end{array}) \end{aligned}

That is $φ \times \bar{φ} = i I m (φ \times \bar{φ})$ , the real part is equal to $0$ since $\overset{―}{φ \times \bar{φ}} = - φ \times \bar{φ}$ .

Thus we have shown:

\begin{aligned} X_{u} \times X_{v} & = \frac{2}{i} \cdot 2 i (\begin{array}{c} I m (φ_{2} {\bar{φ}}_{3}) \\ I m (φ_{3} {\bar{φ}}_{1}) \\ I m (φ_{1} {\bar{φ}}_{2}) \end{array}) \\ = 4 I m (\begin{array}{c} φ_{2} {\bar{φ}}_{3} \\ φ_{3} {\bar{φ}}_{1} \\ φ_{1} {\bar{φ}}_{2} \end{array}) \\ = 2 I m (φ \times \bar{φ}) \end{aligned}