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.