2
$\begingroup$

Let $\gamma: S^1 \to \partial B \subset \mathbf{R}^3$ be a smooth, simple closed curve in the boundary of the unit ball. Suppose that $\gamma$ intersects every horizontal plane $\Pi_t = \{ z = t\}$ at most twice: \begin{equation} \# \gamma(S^1) \cap \Pi_t \leq 2 \quad \text{for all $t$.} \end{equation}

Does $\gamma$ bound a unique minimal disk (resp. minimal surface)?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

The answer is no.

Take two parallel circles of unit radius in $z=\pm \epsilon$ with $\epsilon$ small. Tilt the two circles very slightly toward one another. This satisfies your hypotheses. There are clearly at least three minimal surfaces with this boundary the tilted flat disks and a stable and unstable annulus (obtained by perturbing the appropriate pieces of a catenoid).

If you want non-uniqueness in the class of disks then form a small bridge between the two closest points on the circle. This boundary also satisfies your hypotheses. The bridge principle implies there is a stable minimal surfaces with this boundary that should look like the two flat disks joined by a small bridge. However, it is pretty clear the area minimizing disk is the "ribbon" solution going around the outside. Hence, you have two distinct minimal disks.

Improve this answer
$\endgroup$
3
  • $\begingroup$ That's neat, thanks! $\endgroup$
    – Leo Moos
    Jan 29 at 17:20
  • $\begingroup$ I thought I had seen a criterion like this in the literature, but I must have gotten it mixed up with another 'Rado-like' argument. $\endgroup$
    – Leo Moos
    Jan 29 at 17:31
  • 2
    $\begingroup$ If the boundary is a graph then the Rado argument would imply the surface is also a graph which gives uniqueness in a lot of situations. $\endgroup$
    – RBega2
    Jan 29 at 17:42

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.