Is the intersection of such a triple of minimal surfaces in the 3-ball a single point?

Question

Let $S_1,S_2,S_3$ be three simple closed curves on the 2-sphere $\mathbb{S}^2$. (With no smoothness or rectifiability assumption)

For each $i$, let $M_i$ denote the minimal surface (i.e. disc) bounded by $S_i$, as provided by Douglas. Note that $M_i$ is contained in the interior of $\mathbb{S}^2$ in $\mathbb{R}^3$.

Suppose that the intersections of the $S_i$ follow the same pattern as the intersections of the equator of the earth with two distinct Meridian circles; that is, $S_i \cap S_j$ is a pair of points for every $i \neq j$, and $S_i$ separates the two points of $S_{i+1} \cap S_{i+2}$ for every $i$, where addition is modulo 3.

Question: Must $M_1 \cap M_2 \cap M_3$ be a single point?

Answer 1

This isn't an answer, but is too long for a comment.

Your question has a negative answer in general -- see below.

One initial comment: There is no reason in general for there to be a unique Douglas-Rado disk (i.e. an area minimizer in the class of disks) so you can't really speak of "the the minimal surface (i.e. disc) bounded by $S_i$, as provided by Douglas."

That being said, there are, in general, also many other minimal disks spanning a curve that are not Douglas-Rado solutions (i.e. are not area minimizing disks but just minimal).

If you consider solutions of the latter type, your question is definitely false.

As an example: Consider $S_1$ to be the curve given by starting with the three parallel circles $\mathbb{S}^2\cap \{x_3=0, \pm \epsilon\}$ (for $\epsilon>0$ small) which we denote by $C_0$ and $C_{\pm \epsilon}$. Form $S_1$ by cutting out small pieces from the three circles around the intersection with the $x_1=0$ plane and then gluing in two "necks" connecting the three circles. The one connecting $C_\epsilon$ and $C_0$ is on the $x_2>0$ side and the connecting $C_{0}$ and $C_{-\epsilon}$ is on the $x_2<0$ side. This gives the curve $S_1$

By the bridge principle for stable minimal surfaces as long as the necks are small enough there is a stable minimal disk, $M_1$, spanning $S_1$. It looks like three stacked disks joined by small bridges.

For $S_2$ and $S_3$ take two small rotations around the $x_3$-axis of the circle $\{x_1=0\}\cap \mathbb{S}^2$. If you choose the rotations properly (in particular small enough compared to the size of the necks used in constructing $S_1$) then the three curves satisfy your hypothesis. In this case there are unique minimal disks $M_2$ and $M_3$ spanning $S_2$ and $S_3$ which are flat. It is not hard to see that $M_1\cap M_2\cap M_3$ consists of three points.

This doesn't quite mean you can't solve your problem, but it does mean you have to use the area minimizing property. Indeed, in this picture there should be at least two area-minizing disks spanning $S_1$ and for both the triple intersection should be a disk

Edit:

In fact, as pointed out by @LeoMoos, this construction can be turned into a counterexample to the original question. The needed modification is as follows:

Pick $\epsilon\in (0,1)$ sufficiently large so any minimal surface spanning $C_0\cup C_{-\epsilon}\cup C_\epsilon$ has three components and hence is three disks. Such an $\epsilon$ can be seen to exist by using catenoid barriers. In particular, the area minimizer in the GMT sense (more formally the $\mathbb{Z}_2$-minimizing integral current) with this boundary is the union of the three flat disks.

It follows that in this case, if the necks are thin enough, then the Douglas-Rado solution with boudary $S_1$ looks like the three disks joined by thin bridges. To see this just send the neck size to zero in this case $S_1$ converges as a $\mathbb{Z}_2$ current to the three circles which means $M_1$ converges as a $\mathbb{Z}_2$ current to the three disks.

The rest of the argument is the same.