8
$\begingroup$

Let $D$ be a properly embedded free boundary disk in the closed unit ball $\mathbb{B}^3$ of $\mathbb{R}^3$. This means that $D$ is a smooth disk contained in this ball, $D \cap \partial \mathbb{B}^3 = \partial D$ and this intersection is orthogonal. By orthogonality here I mean this: if $N$ is a unit normal along $D$ (Gauss map), then $\langle N(x), x \rangle = 0$ for all $x \in \partial D$.

Assume that $D$ is strictly convex, that is to say, the principal curvatures are positive at each point of $D$ with respect to the fixed unit normal $N: D \to \mathbb{S}^2$. Does it follow that $N$ is a diffeomorphism onto its image? Equivalently, is $N$ injective?

The motivation is the following: if $S$ is a closed and connected surface in $\mathbb{R}^3$ which is also convex, then $N : S \to \mathbb{S}^2$ is a local diffeomorphism, hence a covering map. Since $\mathbb{S}^2$ is simply connected, this implies that $N$ is a global diffeomorphism. What happens when the surface has boundary?

$\endgroup$
12
  • $\begingroup$ I am not sure I get the image of what the set-up looks like. Is $D$ that you are describing the same as an embedded (or immersed?) copy of the closed unit disk? If so, what is the image to have in mind for how the boundary of $D$ intersects the boundary of the ball? I am having trouble in particular understanding what you mean by $D$ "meets the boundary $\partial \mathbb B^3$ orthogonally." $\endgroup$
    – Alex Ortiz
    Jul 10 '20 at 22:45
  • $\begingroup$ If $D$ intersects the boundary, but $D$ is also contained in the ball, I don't understand how the intersection can be orthogonal. This would imply to me that the disk has to leave the ball at least a little bit. $\endgroup$
    – Alex Ortiz
    Jul 10 '20 at 22:47
  • $\begingroup$ I think it is clear now $\endgroup$ Jul 10 '20 at 22:51
  • 1
    $\begingroup$ Unless I'm missing something, this question can be simplified to asking if Gauss map is injective on $D$. $\endgroup$
    – Kajelad
    Jul 11 '20 at 0:53
  • 1
    $\begingroup$ It seems like you could construct an inverse with the function which takes a $u\in\mathbb{S}^2$ to the point $x\in D$ which maximizes $u\cdot x$. $\endgroup$
    – Kajelad
    Jul 11 '20 at 1:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.