Homework 7

#exercise-problems #math

Problem statement

Verify the Gauss Bonett Theorem holds for the sphere and the torus.

Sphere

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 the unit sphere $S^{1}$ embedded in $R^{2}$ .
The first fundamental form of $S^{1}$ :

\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}

Thus:

\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}

Gaussian Curvature of $S^{1}$ :

K = \frac{L N - M^{2}}{E G - F^{2}} = \frac{s i n^{2} u}{s i n^{2} u} = 1

The boundary $δ S^{1} = \emptyset$ (that is it does not have a boundary) thus:

\int_{S^{1}} K d A + \int_{δ S^{1}} k_{g} = \int_{S^{1}} 1 d A

Thus:

\begin{aligned} \int_{S^{1}} 1 d A & = \int_{0}^{2 π} \int_{0}^{π} s i n u d u d v \\ = 4 π \end{aligned}

Thus indeed $\int_{S^{1}} K d A + \int_{δ S^{1}} k_{g} = 2 π χ$ holds for the unit sphere since $χ (S^{1}) = 2$ .

Torus

Let $X (u, v) = (\begin{matrix} (R + r c o s v) c o s u \\ (R + r c o s v) s i n u \\ r s i n v \end{matrix})$ be a parameterization of the torus $T^{2}$ .

First fundamental form of $T^{2}$ :

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

Thus:

\begin{aligned} E & = (R + r c o s v)^{2} \\ F & = 0 \\ G & = r^{2} \end{aligned}

Second fundamental form of $T^{2}$ :

The unit normal vector is:

n = (\begin{matrix} c o s u c o s v \\ s i n u c o s v \\ s i n v \end{matrix})

and the second derivatives are:

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

Thus:

\begin{aligned} L & = - (R + r c o s v) c o s v \\ M & = 0 \\ N & = - r \end{aligned}

Gaussian curvature of $T^{2}$ :

K = \frac{c o s v}{r (R + r c o s v)}

The boundary $δ T^{2} = \emptyset$ thus:

\int_{T^{2}} K d A + \int_{δ T^{2}} k_{g} = \int_{T^{2}} \frac{c o s v}{r (R + r c o s v)} d A

Computing the right hand side:

\begin{aligned} \int_{T^{2}} \frac{c o s v}{r (R + r) c o s v} d A & = \int_{0}^{2 π} \int_{0}^{2 π} \frac{c o s v}{r (R + r c o s v)} r (R + r c o s v) d u d v \\ = \int_{0}^{2 π} \int_{0}^{2 π} c o s v d u d v \\ = 0. \end{aligned}

Thus indeed then since $χ (T^{2}) = 0$ , the theorem holds for the Torus.