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Let $C_\pm$ be the two circles obtained by intersecting the cylinder $x^2+y^2=R^2$ with the planes $z=\pm 1$, on which we mark four points $A_\pm:(R,0,\pm 1)$ and $B_\pm:(-R,0,\pm 1)$. Assume that $R$ is big enough such that minimal surface between $C_\pm$ is a catenoid.

I want to find a minimal surface of disk type $f:D \to \mathbb{R}^3$ such that $\partial D$ is mapped to the curve $\gamma$ as follows: $\gamma$ starts from $A_+$, then around $C_+$ by $+3\pi$ (clockwise) to $B_+$, then straight down to $B_-$, around $C_-$ by $-3\pi$ (anticlockwise) to $A_-$, finally straight back to $A_+$.

This curve is not a Jordan curve. I wonder if there are any results on "Plateau problem" with this kind of boundary.

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  • $\begingroup$ If it helps, remove the $z$ axis and pass to the universal cover. $\endgroup$ – Fan Zheng Feb 14 '16 at 17:01
  • $\begingroup$ @FanZheng: Sorry, I'm slow, could you expand? $\endgroup$ – Hao Chen Feb 14 '16 at 17:03
  • $\begingroup$ I'm just saying there is a covering map from $\mathbb R^3$ to $\mathbb R^3$ with the $z$ axis removed. Then you're solving the problem with a simple boundary, but not the standard metric. I'm not sure that's easier for you, though. $\endgroup$ – Fan Zheng Feb 14 '16 at 17:11
  • $\begingroup$ @FanZheng: It seems that I lose information dramatically by passing to the cover ... $\endgroup$ – Hao Chen Feb 14 '16 at 17:14
  • $\begingroup$ You also need to pull back the metric, so it's not the standard one. $\endgroup$ – Fan Zheng Feb 14 '16 at 17:19
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The paper below shows that a least area can be found spanning any rectifiable curve. Sometimes the disk is not immersed however, as with a disk bounding a figure 8 in the plane.

J. Hass, Singular curves and the Plateau problem, International J. of Math. 2, (1991) 1-16.

For your example, there is a more straightforward construction of a minimal disk with boundary $\gamma$ that gives a disk that is immersed. You assume that a catenoid spans $C_+$ and $C_-$. The union of this catenoid and the cylinder between $C_+$ and $C_-$ is a torus that bounds a solid torus. The boundary of this piecewise smooth solid torus has mean curvature that points into the solid torus (or at least never out of it) and the angles at the curves where the two surfaces intersect are less than $\pi$. This "mean convex" boundary condition suffices to allow the classic solution of Plateau's problem to be solved for simple curves, as shown in the Meeks Yau papers- see "Topology of Three Dimensional Manifolds and the Embedding Problems in Minimal Surface Theory" William H. Meeks and Shing-Tung Yau, Annals of Mathematics, Vol. 112, No. 3 (Nov., 1980), pp. 441-484.

Now your curve is not embedded on the boundary of the solid torus, but the curve lifts to an embedded curve in a double cover of the solid torus. Solve the Plateau problem there (using the induced metric, also mean convex) and project to get a minimal disk.

For large $R$ this will be the least area disk spanning $\gamma$. It stays close to the cylinder. For small $R$ I think there might be a smaller disk that cuts across the axis.

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  • $\begingroup$ Thank you Joel! This is exactly what I wanted. I have a question about your paper: Does the curve in the example admits any "area-conserving splitting"? As you said, the example can be argued in a more straightforward way, but I'm curious and want to understand the splitting. $\endgroup$ – Hao Chen Feb 16 '16 at 19:10
  • $\begingroup$ Let $I_C$ be the infimum of areas for disks spanning a curve $C$. The method shows that either there is an immersed disk of area equal to $I_C$ or there is a way of splitting at an intersection point that yields two curves $C', C''$ with $I_C = I_{C'} + I_{C''}$. $\endgroup$ – Joel Hass Feb 16 '16 at 19:54
  • $\begingroup$ Aha, I didn't understand the lemma correctly. Thanks. $\endgroup$ – Hao Chen Feb 16 '16 at 20:07

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