7
$\begingroup$

Hi,

Perhaps it's a stupid question, in that case i'll delete it. Let M be a compact orientable smooth (Kahler if changes things) manifold of dimension $dim_{\mathbb{R}}(M)=2n$ with $n\geq1$, let $\gamma$ be a simple smooth closed curve that lies in a (holomorphic) coordinate chart and that can be taken as small as necessary (one can choose $\gamma$ such $diam(\gamma)<\epsilon$ with $\epsilon>0$). If i want to solve Plateau problem for a $\gamma$ so small such that the solution is contained in the coordinate chart, do i get something that can be parametrized by closed disc?

Thank you in advance.

Edit:

I'll try to clarify my question, this is what i wanted to know. Let $U$ be a sufficiently small geodesically convex set of a manifold $M$ and $\gamma$ a smooth simple closed curve lying in $U$ (no other assumptions on $\gamma$).

1) Can $\gamma$ be the boundary of an embedded closed disk?
2) If $\gamma$ can be the boundary of a closed disk, then can it be the boundary of a minimal (as a surface, not only among the disks that it bounds) embedded closed disk?

I anticipate that i couldn't see works of Douglas so i don't know if the answer to my question is there.

My suspect was that the answer could be yes for dimension 2 (i think about jordan curve theorem), Professor Thusrston example of the knotted curve suggests me that in dimension 3 i need additional assumptions on the curve not only on the linking number. But what happens for dimension $n\geq4$?

$\endgroup$
4
  • $\begingroup$ I think this is a good question. I haven't worked out the details myself, but I am pretty sure that you can prove this on any Riemannian manifold using standard tools in differential geometry. I'm hoping that a real expert will provide a reference, more explicit details, or a counterexample. The idea is to use the inverse function theorem applied to the appropriate functional to demonstrate the existence of a minimal surface parameterized by the closed disk. Then use a comparison argument to show existence and minimality. $\endgroup$
    – Deane Yang
    Commented Jan 14, 2011 at 16:53
  • $\begingroup$ Just to echo what has been said, you should be more precise in your question. Namely, given your $\gamma$ there is a solution that is an (immersed) minimal disk spanning $\gamma$ namely a Douglas-Rado solution (see Bill's great answer for some details). This is because the $\gamma$ you describe is spanned by some disk and hence by direct methods in the calculus of variations is spanned by a minimal disk (of course easier said then proved). However, there may be many other minimal surfaces (or weaker concepts) spanning the curve so it is not correct to speak of THE solution. $\endgroup$
    – Rbega
    Commented Jan 15, 2011 at 18:01
  • $\begingroup$ I edited the question, i hope it is more clear now $\endgroup$
    – Italo
    Commented Jan 16, 2011 at 0:39
  • $\begingroup$ Bill Thurston's answer (which is quite good) shows that my comment is not right. At best the approach outlined by me might work if the initial curve is "sufficiently flat", whatever that means. $\endgroup$
    – Deane Yang
    Commented Jan 16, 2011 at 3:24

1 Answer 1

13
$\begingroup$

I've been waiting for someone with more expertise than me to answer, but since they haven't (so far) I'll say something.

There are different versions of Plateau-like problems; I'm not sure if there's a specific single one that's generally accepted as "the Plataeu problem". One can ask for a mapped-in disk with minimal area having the given boundary, a mapped-in surface of minimal area, a current of minimal area, an integral current of minimal area, an embedded "minimal surface" meaning that it's just a critical point for area among embedded (or mapped in if you prefer) surfaces, a minimal disk ... A lot is known about these different questions, and the answers aren't the same.

First: even for a curve in Euclidean space, there might not be an embedded minimal disk. The easiest examples are for a knotted curve in R^3. However, there are also unknotted curves in R^3 that do not bound a disk in their convex hull, so they do not bound any embedded minimal disk. The minimum area of a disk bounding an unknotted curve grows exponentially in an appropriate measure of the complexity of the curve; the genus of a minimum area surface also grows exponentially. (Fred Almgren and I once wrote a paper about this). The same examples can be transported to any higher dimension, e.g. taking the product with another manifold.

Second: Jesse Douglas showed how to find mapped-in minimal disks in great generality. This will work locally within any manifold of dimension. The basic technique is that for any parametrization of the curve by the boundary of a disk, first find an energy-minimizing map of the disk that extends this parametrization --- a harmonic map. Now consider the energy as a function of the parametrization (which is a kind of Teichmuller space). The critical points of the harmonic energy with respect to the Teichmuller space are minimal surfaces: the basic insight is that critical points of the harmonic maps within the Teichmuller space are when the harmonic map is conformal. When it is not conformal, you can change the conformal structure on the disk (which is equivalent to giving a reparametrization, by the uniformization theorem) and reduce energy.

Third: minimal surfaces of any type (inclding currents) whose boundary is inside a convex set are always contained in the convex set. Here convex means that geodesic arcs between nearby points on the boundary are contained within the set. Even weaker conditions are sufficient, but this is good enough for your purposes --- a metric ball of small radius in any Riemannian manifold is convex in this sense.

Fourth: there always exist minimizing objects of some sort, but they might not be things that you are happy with: one thing that happens is that $k$ times a curve might bound a surface of area less than $k$ times that for the original curve (in dimensions > 3). A limit of $1/k$ times the minimum area for $k$ times the curve might be a diffuse current, spread out in an entire region. I think this can happen even locally in a Kahler manifold, but I'm not sure. Even without using fractional weights, the minimizing object would in general be an integral current that is like a surface but with singularities. I don't know the classification for the 2-d case, but i'm sure the experts do.

On the other hand, if you take a nearly round circle in a small coordinate chart, there should be an embedded minimal disk --- basically because nearly flat minimal surfaces are stable, so changing the metric a little bit gives you a new minimal surface by implicit function theorem type arguments (in the space of surfaces, which in this case can be described as graphs of functions). In R^3, there's a theorem that any curve whose total curvature is less than 4$\pi$ bounds an embedded minimal disk; I suspect there are known estimates likke this in Riemannian manifolds of higher dimension, using curvature assumptions about some convex set contining the curve, but I don't actually know.

Is there an expert who can correct or extend what I've said?

$\endgroup$
2
  • $\begingroup$ Thank you very much for the detailed answer! I'll wait a little more before accepting it because i want to see if you or someone has anything else to say. $\endgroup$
    – Italo
    Commented Jan 16, 2011 at 0:45
  • 1
    $\begingroup$ Not an expert, but I'll add one bit to (part 4 of) this excellent answer: Given a boundary, you can always find an area-minimizing rectifiable current with the given boundary. Almgren's big regularity paper says that the interior of this minimizer is an embedded smooth manifold away from a Hausdorff codimension two singular set. In the case where the boundary is 1-dimensional, Sheldon Chang further showed that the singular points are isolated classical branch points. $\endgroup$
    – Dan Lee
    Commented Jan 24, 2011 at 22:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .