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)?


1 Answer 1


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.

  • $\begingroup$ 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

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