Umbilic points on a connected smooth surface problem Let $S\subset \mathbb{R}^3$ be a connected smooth surface. Suppose that every point of $S$ is an umbilic point. Prove that $S$ is a subset of either a plane or a sphere in $\mathbb{R}^3$.
Here's a HW problem. I wonder how to prove it.
 A: Since this is homework and you haven't showed us your work, I'll give you the start and let you take it from there.
Let $\sigma(u, v)$ be a regular surface patch and $\mathbf{N}$ be the normal vector.
Since the surface is umbilic at every point, we have:
\begin{align*}
\mathcal{W}(\sigma_{u}) &= -\mathbf{N}_u = \kappa \ \sigma_{u} \\
\mathcal{W}(\sigma_{v}) &= -\mathbf{N}_v = \kappa \ \sigma_{v}
\end{align*}
Where $\mathcal{W}$ is the shape operator (Weingarten map), and $\kappa$ is the principal curvature.
By differentiating the first equation with respect to $v$ and the second with respect to $u$, we have:
\begin{align*}
- \mathbf{N}_{uv} &= \kappa_{v} \ \sigma_{u} + \kappa \ \sigma_{uv} \\
&= \kappa_{u} \ \sigma_{v} + \kappa \ \sigma_{uv}
\end{align*}
From that we get:
$$
\kappa_{v} \ \sigma_{u} = \kappa_{u} \ \sigma_{v}
$$
Since $\sigma_{u}$ and $\sigma_{v}$ are linearly independent, $\kappa_{u}$ and $\kappa_{v}$ must be 0. Hence, $\kappa$ is constant everywhere.
Can you take it from there? Use the equations in the beginning of my answer to find a relationship containing $\sigma$ and $\mathbf{N}$.

Rest of the proof (to reference in another question):
We have:
\begin{align*}
-\mathbf{N}_u &= \kappa \ \sigma_{u} \\
-\mathbf{N}_v &= \kappa \ \sigma_{v}
\end{align*}
$\kappa$ is constant. Hence:
$$
-\mathbf{N} = \kappa \ \sigma + v
$$
For a constant vector $v$.
Either $\kappa = 0$, so $\mathbf{N}$ is constant and the surface is (part of) a plane. Or $\kappa \ne 0$ and:
$$
\|\sigma + \frac{1}{\kappa} v \| = \|-\frac{1}{\kappa} N\| = \frac{1}{|\kappa|}
$$
And $\sigma$ is (part of) a sphere with radius $\frac{1}{|\kappa|}$.
