Differential Geometry and Its Application - Exercise 2.2.14

Problem:

Let $ M $ be the cylinder $ x^2 + y^2 = R^2 $ parametrized by $ x(u, v) = (R\cos u, R\sin u, v) $. Show that the shape operator on $ M $ is described on a basis by $ S(x_u) = -\frac{1}{R}x_u $ and $ S(x_v) = 0 $. Therefore, in the u-direction the cylinder resembles a sphere and in the v-direction a plane. Of course, intuitively, this is exactly right. Why?

Solution:

Let's compute the shape operator step by step, we start with the tangent vectors:

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

Then we compute the normal vector

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

Intuitively that make sense, the normal vector points radially outwards.

The unit normal vector is therefore $ (\cos u, \sin u, 0) $.

Next, we compute the directional derivatives:

$ v[\cos u] = \nabla \cos u \cdot v $.
$ v[\sin u] = \nabla \sin u \cdot v $.
$ v[\sin u] = \nabla 0 \cdot v = 0$.

At this point, it is useful to remind ourselves that when we talk about the gradient operator $ \nabla $, we are thinking the unit normal vector as a function of $ (x, y, z) $. Now $ \cos u = \frac{x}{R} $ and $ \sin u = \frac{y}{R} $, so the gradients can be computed easily as:

$ v[\cos u] = \nabla \frac{x}{R} \cdot v = (\frac{1}{R}, 0, 0) \cdot v = v_x $
$ v[\sin u] = \nabla \frac{y}{R} \cdot v = (0, \frac{1}{R}, 0) \cdot v = v_y $

Therefore the Weingarten map is

$ S_p(v) = -(\frac{1}{R}v_x, \frac{1}{R}v_y, 0) = \left(\begin{array}{ccc}\frac{-1}{R} & 0 & 0 \\ 0 & \frac{-1}{R} & 0 \\ 0 & 0 & 0\end{array}\right)\left(\begin{array}{c}v_x \\ v_y \\ v_z \end{array}\right) $.

This is representing the map as a 3 dimensional transform, but we also know the Weingarten map is a transform that takes tangent vector to tangent vectors, so we compute (simply by substituting) and get

$ S_p(x_u) = -\frac{1}{R}x_u $.
$ S_p(x_v) = 0 $.