Homework 3

#exercise-problems #math

Problem Statement

Show that if $γ (t) = (x (t), y (t), z (t))$ is a curve on $S^{2}$ then:

d N (γ^{'} (t)) = (- x^{'} (t), - y^{'} (t), - z^{'} (t))

where $N$ is the Gauss map of $S^{2}$

Let $X (u, v) = (\begin{matrix} s i n u c o s v \\ s i n u s i n v \\ c o s u \end{matrix})$ be a parameterization of $S^{2}$ , then $γ (t) = X (u (t), v (t))$ and the basis of the tangent vectors are:

\begin{aligned} X_{u} & = {(\begin{array}{c} s i n u c o s v \\ s i n u s i n v \\ c o s u \end{array})}_{u} = (\begin{array}{c} c o s u c o s v \\ c o s u s i n v \\ - s i n u \end{array}) \\ X_{v} & = {(\begin{array}{c} s i n u c o s v \\ s i n u s i n v \\ c o s u \end{array})}_{v} = (\begin{array}{c} - s i n u s i n v \\ s i n u c o s v \\ 0 \end{array}) \end{aligned}

The first fundamental form of $S^{1}$ :

\begin{aligned} E & = X_{u} \cdot X_{u} \\ = c o s^{2} u c o s^{2} v + c o s^{2} u s i n^{2} v + s i n^{2} u \\ = 1 \\ F & = X_{u} \cdot X_{v} \\ = - (c o s u c o s v) (s i n u s i n v) + (c o s u s i n v) (s i n u c o s v) \\ = 0 \\ G & = X_{v} \cdot X_{v} \\ = s i n^{2} u s i n^{2} v + s i n^{2} u c o s^{2} v \\ = s i n^{2} u \end{aligned}

Second fundamental form of $S^{1}$ :
We first compute the unit normal vector $n$ :

\begin{aligned} n & = \frac{X_{u} \times X_{v}}{| | X_{u} \times X_{v} | |} \\ = \frac{1}{s i n u} (\begin{array}{c} c o s v s i n^{2} u \\ s i n v s i n^{2} u \\ c o s u s i n u \end{array}) \\ = (\begin{array}{c} c o s v s i n u \\ s i n v s i n u \\ c o s u \end{array}) \end{aligned}

and then the second derivatives:

\begin{aligned} X_{u u} & = (\begin{array}{c} - s i n u c o s v \\ - s i n u s i n v \\ - c o s u \end{array}) \\ X_{u v} & = (\begin{array}{c} - c o s u s i n v \\ c o s u c o s v \\ 0 \end{array}) \\ X_{v v} & = (\begin{array}{c} - s i n u c o s v \\ - s i n u s i n v \\ 0 \end{array}) \end{aligned}

Thus:

\begin{aligned} L & = X_{u u} \cdot n \\ = - s i n^{2} u c o s^{2} v - s i n^{2} u s i n^{2} v - c o s^{2} v \\ = - 1 \\ M & = X_{u v} \cdot n \\ = - (c o s u s i n v) (c o s v s i n u) + (c o s u c o s v) (s i n v s i n u) \\ = 0 \\ N & = - s i n^{2} u c o s^{2 v} - s i n^{2} u s i n^{2} v \\ = - s i n^{2} u \end{aligned}

We know that $d N (w_{1}, w_{2}) = (\begin{matrix} a_{1}^{1} & a_{2}^{1} \\ a_{1}^{2} & a_{2}^{2} \end{matrix}) (\begin{matrix} w_{1} \\ w_{2} \end{matrix})$ , where

(\begin{matrix} a_{1}^{1} & a_{2}^{1} \\ a_{1}^{2} & a_{2}^{2} \end{matrix}) = {(\begin{matrix} E & F \\ F & G \end{matrix})}^{- 1} (\begin{matrix} L & M \\ M & N \end{matrix})

Thus:

\begin{aligned} d N (w_{1}, w_{2}) & = (\begin{array}{c} a_{1}^{1} & a_{2}^{1} \\ a_{1}^{2} & a_{2}^{2} \end{array}) (\begin{array}{c} w_{1} \\ w_{2} \end{array}) \\ = - (\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}) (\begin{array}{c} w_{1} \\ w_{2} \end{array}) \\ = (\begin{array}{c} - w_{1} \\ - w_{2} \end{array}) \end{aligned}

Thus for $γ^{'} (t) = (\begin{matrix} u^{'} (t) \\ v^{'} (t) \end{matrix})$ it is clear that

\begin{matrix} ◻ & d N (γ^{'} (t)) = (\begin{matrix} - u^{'} (t) \\ - v^{'} (t) \end{matrix}) = (- x^{'} (t), - y^{'} (t), - z^{'} (t)) \end{matrix}