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The Riemann mapping theorem says that if you have a simple closed curve in $\mathbb{C}$, then there is an essentially unique way to map a holomorphic disc to the interior. Is there any reasonable statement to make about holomorphic discs bounding a simple closed curve in $\mathbb{C}^2$?

(At first I thought you should just project to some appropriate choice of coordinate axes and reduce to the one dimensional case, but it is now not at all clear to me why some such choice should exist.)

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I think I remember a standard result that $\{ |z_1| < 1, |z_2| < 1 \}$ and $\{ |z_1|^2 + |z_2|^2 < 1 \}$ are conformally inequivalent domains in $\mathbb{C}^2$, and that in general Several Complex Variables behaves totally differently to one complex variable. I can't remember which book, but any good book on multivariable complex analysis should discuss this. It stated that the Riemann Mapping Theorem (along with almost all other single variable theorems) has basically no good multivariable analogue. –  Zen Harper Jan 26 '11 at 18:34
    
...of course my comment doesn't necessarily apply to your question, since you seem to be asking about domains with restricted kinds of boundary; but trying to think about boundaries of 4 (real) dimensional regions makes my head hurt! –  Zen Harper Jan 26 '11 at 18:53
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3 Answers 3

Here are two counterexamples:

(1) the set defined by $xy=1$, $|x|=1$,

(2) the set defined by $y^2=x^3-x$, $|x|=2$.

In both cases a compact Riemann surface bounded by the given curve would have to lie in the solution set of the given polynomial equation. In (1) this is impossible. In (2) it's possible but not with a disk.

EDIT Or an ordinary round circle in $\mathbb R^2\subset \mathbb C^2$. This is the same as (1) up to a complex-linear change of coordinates.

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If you want to generalise the Reimannian mapping theorem to higher dimensions the boundary condition should be different. Namely, it is more reasonable to ask that the boundary of the disk is mapped to a submanifold $M$ of $\mathbb C^n$ or real dimension $n$ such that $TM\oplus J(TM)$ span $\mathbb C^n$ at each point of $M$ (such submanifolds are called totally real). In other words, you can not force the boundary of a holomorphic disk in $\mathbb C^2$ to be an arbitrary curve. The best that you can do is to chose a two-dimensional surface in $\mathbb C^2$ and then the disk will chose a curve on the surface that can be its boundary. Under this condition you can expect to get a finite dimension set of solutions. This is what Gromov is doing in his seminal paper

Pseudo-holomorphic curves in symplectic manifolds

http://www.ihes.fr/~gromov/PDF/9[45].pdf

Example. Let us consider the torus $\mathbb T^2$ in $\mathbb C^2$ given by $|z|=|w|$. Then obviously the line $z=w$ intersects $\mathbb T^2$ in a circle and we get a holomorphic disk with boundary on $\mathbb T^2$. Let us prove that the curves on $\mathbb T^2$ isotopic this circle and bounding a holomorphic disk in $\mathbb C^2$ form a family of real dimension $3$. Indded let $D^2$ be a holomorphic disk in $\mathbb C^2$ with boundary on $\mathbb T^2$ isotopic to the above circle. Consider the projections of $D^2$ to the lines $z=0$ and $w=0$. Since these projections are holomorphic it is not hard to see that they are one to one to maps to disks $(|w|\le 1, z=0)$ and $(|z|\le 1, w=0)$. So $D^2$ is a graph of a holomorphic one to one map from one unit disk to the other. These maps (as we know well) form $SL(2,\mathbb R)$.

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For some curves this may not be possible. In fact, this is the original starting point of CR geometry. See this article for a hypersurface case.

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