Classification of Willmore Surfaces with Vanishing Gaussian Curvature

Abstract

We classify simply-connected, orientable, complete Willmore surfaces with vanishing Gaussian curvature. We also study the Willmore cones in ${\mathbb {R}}^{3}$ and give a classification. As an application, we show that for a complete Willmore embedding $f:{\mathbb {R}}^{2} \rightarrow {\mathbb {R}}^{3}$, if its corresponding Gaussian curvature is nonnegative and the image of its Gauss map lies in $\big \{\theta \in {\mathbb {S}}^{2}: d_{{\mathbb {S}}^{2}}(\theta ,\theta _{0}) \le \alpha <\frac{\pi }{2} \big \}$, $f({\mathbb {R}}^{2})$ is a plane.

Appendix

[8, Lemma 2] plays an essential role in [8], we provide a proof of a 2-dimensional case here for the convenience of the reader.

Let $\Omega \subset {\mathbb {R}}^{2}$ be a domain. Let $f:\Omega \rightarrow {\mathbb {R}}$ be a $C^{2}$ function. For convenience, set $p=\nabla f=(p_{1}(x_{1},x_{2}),p_{2}(x_{1},x_{2}) )$. Clearly, p is a gradient map defined as in [8].

Let J(x) denote the Jacobian matrix $\displaystyle \Big ( \frac{ \partial p_{i} }{ \partial x_{j}} \Big )_{i,j}^{}$, r(x) denote the rank of J(x) and $r^{*}(x)$ denote the largest integer s with the property that every neighbourhood of x contains a point $x^{*}$ with $r(x^{*})=s$. By a standard calculation, we know that the Gaussian curvature of the graph induced by f vanishes if and only if $r \le 1$. If $r^{*}(x)=0$, there exists a neighbourhood U of x such that $\{(x,f(x)): x \in U\}$ is contained in a plane.

Lemma 4.3

Assume $\displaystyle \det J(x)$ is identically zero and at a point $x^{0}=(x_{1}^{0},x_{2}^{0})$, $r(x^{0})=1$. Set $S=\{ x \in \Omega : r(x)=1 \}$. Then p(x) is constant on $l(x^{0}) \cap S$, where $l(x^{0}) \subset {\mathbb {R}}^{2}$ is a unique straight line passing $x^{0}$. Furthermore, $r(x)=1$ on $l(x^{0}) \cap S$.

Proof

Step 1 We show that in a small neighbourhood U of $x^{0}$, $\{x \in U: p(x)=p(x^{0}) \}$ is the intersection of a straight line passing $x^{0}$ and U.

Since $\det J(x^{0})=0, r(x^{0})=1$, we may assume $\displaystyle \frac{\partial p_{1}}{\partial x_{1}}|_{x^{0}} \ne 0$ up to a linear transformation of the coordinates. Consider the map $q(x_{1},x_{2})=(q_{1}(x_{1},x_{2}), q_{2}(x_{1},x_{2}) )=( p_{1}(x_{1},x_{2}), x_{2} )$. The corresponding Jacobian matrix is $\begin{pmatrix} \frac{\partial p_{1}}{\partial x_{1}} &{} 0 \\ \frac{\partial p_{1}}{\partial x_{2}} &{} 1 \end{pmatrix}$, then we can introduce $(q_{1},q_{2})$ as new coordinates in q(U) up to contracting U. Then

$$\begin{aligned} x_{2}(q_{1},q_{2})=q_{2}; \quad p(q_{1},q_{2})=(q_{1}, p_{2}(q_{1},q_{2}) ). \end{aligned}$$

By chain rules, we obtain

$$\begin{aligned} \begin{pmatrix} 1 &{} 0 \\ \frac{\partial p_{2}}{\partial q_{1}} &{} \frac{\partial p_{2}}{\partial q_{2}} \end{pmatrix} = \begin{pmatrix} \frac{\partial p_{1}}{\partial q_{1}} &{} \frac{\partial p_{1}}{\partial q_{2}} \\ \frac{\partial p_{2}}{\partial q_{1}} &{} \frac{\partial p_{2}}{\partial q_{2}} \end{pmatrix} = \begin{pmatrix} \frac{\partial p_{1}}{\partial x_{1}} &{} \frac{\partial p_{1}}{\partial x_{2}} \\ \frac{\partial p_{2}}{\partial x_{1}} &{} \frac{\partial p_{2}}{\partial x_{2}} \end{pmatrix} \begin{pmatrix} \frac{\partial x_{1}}{\partial q_{1}} &{} \frac{\partial x_{1}}{\partial q_{2}} \\ 0&{} 1 \end{pmatrix}, \end{aligned}$$

By the assumption, we have $\displaystyle \frac{\partial p_{2}}{ \partial q_{2}}=0$, which implies $p_{2}$ depends only on $q_{1}$ and is independent of $q_{2}$. Also by $\displaystyle \frac{ \partial p_{1}}{ \partial x_{2} }=\frac{ \partial p_{2}}{ \partial x_{1} },$ we obtain

$$\begin{aligned} \frac{\partial x_{1}}{\partial q_{2}}+\frac{\partial p_{2}}{\partial q_{1}}=\frac{\partial p_{1}}{\partial q_{2}}\frac{\partial x_{1}}{\partial q_{1}}=0, \end{aligned}$$

then we obtain for some function $\beta (q_{1})$,

$$\begin{aligned} \beta (q_{1})=x_{1}+\frac{\partial p_{2}}{\partial q_{1}}q_{2}=x_{1}-\alpha (q_{1})x_{2}, \end{aligned}$$

(4.5)

where we set $\displaystyle \alpha =-\frac{\partial p_{2}}{\partial q_{1}}$.

Now for any $x=(x_{1},x_{2})$ such that $p(x)=p(x^{0})$, $q_{1}=q_{1}^{0}$ for some $q_{1}^{0}$, then locally, $\displaystyle p(x)=p(x^{0})$ is equivalent to $\displaystyle \beta (q_{1}^{0})=x_{1}+\alpha (q_{1}^{0}) x_{2}$, which implies in U, $\{x \in U: p(x)=p(x^{0}) \}$ is the intersection of a straight line passing $x^{0}$ and U.

Step 2 Let $l(x^{0})=\{ (x_{1},x_{2}) \in \Omega : \beta (q_{1}^{0})=x_{1}+\alpha (q_{1}^{0}) x_{2} \}$, we will show that p is constant on $l(x^{0}) \cap S$. By (4.5),

$$\begin{aligned} 1=\frac{ \partial p_{1}}{\partial q_{1}}&=\frac{ \partial p_{1}}{\partial x_{1}}\frac{ \partial x_{1}}{\partial q_{1}}+\frac{ \partial p_{1}}{\partial x_{2}}\frac{ \partial x_{2}}{\partial q_{1}} \nonumber \\&=\frac{ \partial p_{1}}{\partial x_{1}} \frac{ \partial ( \alpha (q_{1}) x_{2}+\beta (q_{1}) )}{ \partial q_{1} } \nonumber \\&=\frac{ \partial p_{1}}{\partial x_{1}} \Big (x_{2} \frac{\partial \alpha }{\partial q_{1} }+ \frac{\partial \beta }{\partial q_{1} } \Big ). \end{aligned}$$

(4.6)

We claim that $p(x)=p(x^{0})$ and $\displaystyle \frac{\partial p_{1}}{\partial x_{1}} \ne 0$ on $l(x^{0}) \cap S$. Assuming the contrary, there exists a curve $\gamma :[0,1] \rightarrow l(x^{0}) \cap S$ with $\gamma (0)=x^{0}, \gamma (1) \in \partial S$. Set

$$\begin{aligned} t_{0}=\sup \{ s: p(\gamma (t))=p(x^{0}), \frac{ \partial p_{1}}{\partial x_{1}} (\gamma (t))\ne 0, t \le s \} <1. \end{aligned}$$

Clearly, $p(\gamma (t_{0}))=p(x^{0})$. Furthermore, (4.6) holds in a small neighbourhood of each point $\gamma (s)$ for $s<t_{0}$. Along $\gamma |_{[0,s]}$, $\displaystyle \frac{\partial \alpha }{\partial q_{1} }, \frac{\partial \beta }{\partial q_{1} }$ is independent of s, which implies $\displaystyle \frac{\partial p_{1}}{\partial x_{1}}(\gamma (s))$ is bounded away from zero as $s \rightarrow t_{0}$. Applying the argument in Step 1 at the point $p( \gamma (t_{0}))$, we will obtain a contradiction to the maximality of $t_{0}$. We complete the proof. $\square$

The following corollary follows from Lemma 4.3 and an approximation argument immediately.

Corollary 4.4

If $r^{*}(x^{0})=1$ at a point $x^{0} \in \Omega$, then p(x) is constant on the intersection of a straight line passing $x^{0}$ and $\Omega$.