Differential Geometry and Its Application - Exercise 3.1.9

Problem:

Solution:

Let's compute the shape operator for the cylinder. First, we parametrize the cylinder as follow:

$ (R \cos u, R \sin u, v) $.

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

The normal vector is $ x_u \times x_v = (R \cos u, R \sin u, 0) $

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

$ S_{p}(x_u) = -\nabla_{x_u}U = (x_u[\cos u], x_u[\sin u], x_u[0]) = (\frac{\partial \cos u}{\partial u}, \frac{\partial \sin u}{\partial u}, \frac{\partial 0 u}{\partial u}) = (-\sin u, \cos u, 0) $

$ S_{p}(x_v) = -\nabla_{x_v}U = (x_v[\cos u], x_v[\sin u], x_v[0]) = (\frac{\partial \cos u}{\partial v}, \frac{\partial \sin u}{\partial v}, \frac{\partial 0 v}{\partial v}) = (0, 0, 0) $

Now, if we write a vector on the tangent plane using $ x_u $ and $ x_v $ as basis, then we know the shape operator can be written as a matrix $ \left(\begin{array}{cc}A & B\\C & D\end{array}\right) $

$ S_{p}(x_u) = \frac{1}{R}x_u \implies \left(\begin{array}{cc}A & B\\C & D\end{array}\right)\left(\begin{array}{c}1 \\ 0\end{array}\right) = \left(\begin{array}{c}\frac{1}{R} \\ 0\end{array}\right) $

$ S_{p}(x_v) = 0 \implies \left(\begin{array}{cc}A & B\\C & D\end{array}\right)\left(\begin{array}{c}0 \\ 1\end{array}\right) = \left(\begin{array}{c}0 \\ 0\end{array}\right) $

So the overall shape operator matrix is simply $ \left(\begin{array}{cc}\frac{1}{R} & 0\\0 & 0\end{array}\right) $, and therefore its eigenvalues are $ \frac{1}{R} $ and $ 0 $, so the mean curvature is $ \frac{1}{2R} $ and the Gaussian curvature is $ 0 $.

That is why the surface is flat, but not minimal.