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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.

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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 $\frac{1}{|\kappa|}$.

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    @AymanHourieh Can you give an explanation of why $N_{uv}=N_{vu}$. Is it following from Schwarz's theorem? Thank you.2018-07-13