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Let $D\subset\mathbb R^2\subset\mathbb R^n$ be a unit planar disc in $\mathbb R^n$. Let $S$ be an orientable two-dimensional surface in $\mathbb R^n$ such that $\partial S=\partial D$. Of course, we have $area(S)\ge area(D)$. Assume that $area(S)< area(D)+\delta$ where $\delta>0$ is small.

Then $S$ is close to $D$ in the following sense: there is a 3-dimensional surface $F$ filling the gap between $S$ and $D$ such that $volume(F)<\varepsilon(\delta)$ where $\varepsilon(\delta)\to 0$ as $\delta\to 0$ ($n$ is fixed). "Filling the gap" means that $\partial F=S-D$.

This fact immediately follows from the compactness theorem for flat norms. But this proof is indirect and does not answer the following questions (I am especially interested in the second one):

1) Are there explicit upper bounds for $\varepsilon(\delta)$? How do they depend on $\delta$ and $n$?

2) Can $\varepsilon(\delta)$ be independent of $n$? Or, equivalently, does the above fact hold true in the Hilbert space?

In the unlikely event that 2-dimensional surfaces are somehow special, what about the same questions about $m$-dimensional surfaces, for a fixed $m$?

Remarks: "Surfaces" here are Lipschitz surfaces or rectifiable currents or whatever you prefer to see in this context. Rather than talking about the filling surface $F$, one could equivalently say that the integral flat norm of $S-D$ is less than $\varepsilon(\delta)$.

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Something's wrong with this "gap filling" stuff. Even in the easiest cases were $S$ is just a disc that has a small "dent" (say the surface {$(x,y,z) | x^2+y^2\leq 1, z=c\cdot(1-x^2+y^2)$}) the boundary of $F$ is $S\cup D$ not $S-D$. –  Johannes Hahn Mar 11 '10 at 22:41
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No, $S-D$ is not $S\setminus D$. It is the formal sum of $S$ and $-D$ where $-D$ is $D$ taken with the opposite orientation. As a set, it equals $S\cup D$ for a generic surface. –  Sergei Ivanov Mar 11 '10 at 22:58
    
Okay, but there is still something wrong with this. If $S$ is not entirely above or below $D$ (more wave-like), then the gap between $S$ and $D$ is no manifold. –  Johannes Hahn Mar 12 '10 at 0:38
    
Yes a surface may be singular, it is just a map from a manifold. If n is large enough, you can approximate everything by immersions if you really need. On the constructive side, it is easier to work with singular chains rather than surfaces. –  Sergei Ivanov Mar 12 '10 at 5:04
    
Did you try mean-curvature-flow? If yes (and if it works) what does it give you when $n$ grows? –  Anton Petrunin Mar 12 '10 at 18:32
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1 Answer

up vote 4 down vote accepted

There is Almgren's isoperimetric inequality:

Let $\Sigma$ be a $k$-surface in $\mathbb R^n$. Assume $vol _k \Sigma \le vol_k S^k$. Then one can fill $\Sigma$ by a $(k+1)$-surface with volume $\le vol_{k+1} B^{k+1}$. (Here the "surfaces" might have singularities.)

I will use it to show that there is an estimate $\epsilon(\delta)$ which does not depend on $n$.

Take $r$-nbhd $Z_r$ of $D$. Note* that one can give an explicite estimate of $r$, independent of $n$ so that total area of $S$ outside of $Z_r$ is very small. Moving a bit $r$, one can make the length of intersection curve $\gamma=\partial Z_r\cap S$ sufficiently small. Use Almgren to fill $\gamma$ by a surface; it breaks $S$ into two pieces $S=S_1+S_2$;

  • the surface $S_1$ lies in $Z_r$ and $\partial S_1=\partial D$,
  • the surface $S_2$ has small area and $\partial S_2=0$.

Fill both $S_1$ and $S_2$ separately:

  • taking all segments from point on $S_1$ to its projection on $D$ gives a filling of $S_1-D$
  • fill $S_2$ using Almgren again.

(*)There is a map $\mathbb R^n\to D$ which decrease distances by some factor $k=k(r)<1$ outside of $Z_r$ and $k(r)$ can be found explicitly. So if an essential piece of $S$ is outside of $Z_r$ then the area of $S$ is essentially bigger that $area(D)$.

Say, take $f(x)=$ "sum of maximal and minimal distance to the points in $D$". This function is convex and it is constant on $D$. Take Sharafutdinov retruction for the level sets of this function.

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This r should go to zero along with $\delta$, shouldn't it? Then how do you prove that the area outside the r-neighborhood is small? –  Sergei Ivanov Mar 13 '10 at 21:00
    
@Sergei. I add something to the answer. –  Anton Petrunin Mar 13 '10 at 22:12
    
Great! And this works in far more general setting than in the question. –  Sergei Ivanov Mar 13 '10 at 23:08
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