Differential Geometry and Its Application - Exercise 2.3.9

Problem:

For the cone $ x(u, v) = (v \cos u, v \sin u, v) $, computes the Gauss map and its derivative. Estimate the amount of area the image of the Gauss map takes up on the sphere.

Solution:

We compute the Gauss map step-by-step, again, we starts with the tangent vectors:

$ x_u = (-v\sin u, v \cos u, 0) $
$ x_v = (\cos u, \sin u, 1) $.

Then we compute the normal vector

$ x_u \times x_v = \left|\begin{array}{ccc}i & j & k \\ -v\sin u & v \cos u & 0 \\ \cos u & \sin u & 1\end{array}\right| = (v\cos u, v\sin u, -v)$

The make intuitive sense, the normal vector point radially outward and downwards (for $ v > 0 $)
Also note that when $ v = 0 $, the tangent vector is $ 0 $ and therefore the surface is not regular there, we will ignore that single point.

The unit normal vector is

$ \frac{1}{\sqrt 2}(\cos u, \sin u, -1) $.

So this is the Gauss map.

To compute the 'derivative', again, it is helpful to think of the Gauss map as a function from $ (x, y, z) $ to the unit sphere, so we can compute the $ 3 \times 3 $ Jacobian matrix.

$ \frac{\partial G_x}{\partial x} = \frac{\partial G_x}{\partial u}\frac{\partial u}{\partial x} = \frac{\partial G_x}{\partial u}\frac{1}{\frac{\partial x}{\partial u}} = \frac{-1}{\sqrt 2}\sin u \frac{1}{-v \sin u} = \frac{1}{\sqrt 2 v} $
$ \frac{\partial G_x}{\partial y} = \frac{\partial G_x}{\partial u}\frac{\partial u}{\partial y} = \frac{\partial G_x}{\partial u}\frac{1}{\frac{\partial y}{\partial u}} = \frac{-1}{\sqrt 2}\sin u \frac{1}{v \cos u} = \frac{-\tan u}{\sqrt 2 v} $
$ \frac{\partial G_x}{\partial z} = \frac{\partial G_x}{\partial u}\frac{\partial u}{\partial z} = \frac{\partial G_x}{\partial u}\frac{1}{\frac{\partial z}{\partial u}} = \infty $

Something seems wrong with the last line. We know that the derivative of the Gauss map is the Weingarten map, which should exist. Why am I getting an infinite there?

Alternatively, we can compute the Weingarten map using a shortcut. We know $ x_u[f] = f_u $, so we know:

$ S(x_u) = -\nabla_{x_u} U = -\frac{1}{\sqrt 2}(\frac{\partial \cos u}{\partial u}, \frac{\partial \sin u}{\partial u}, \frac{\partial (-1)}{\partial u}) = (-\sin u, \cos u, 0) = -\frac{1}{\sqrt 2 v} x_u $.

$ S(x_v) = -\nabla_{x_v} U = -\frac{1}{\sqrt 2}(\frac{\partial \cos u}{\partial v}, \frac{\partial \sin u}{\partial v}, \frac{\partial (-1)}{\partial v}) = (0, 0, 0) $.

As per the image of the Gauss map, it is a circle at $ z = -\frac{1}{\sqrt 2} $. It does not take up area of the surface of the sphere.