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Let $U\subset \mathbb R^n$ be an open.

Let $f: S^1 \times U \to \mathbb C$ be a smooth map (i.e. $C^\infty$) s.t. for every $x\in U$ the map $f_x:=f|_{S^1\times \{x\}}:S^1\to \mathbb C$ is a smooth embedding (i.e. with everywhere non-zero derivative) that goes around the origin with winding number $+1$.

For every $x\in U$ there's a unique Riemann map $\Phi_x$ : [unit disc] $\to$ [interior of $f_x(S^1)$] that sends $0\mapsto 0$ and whose derivative at zero is in $\mathbb R_{\ge 0}$.

Is $\Phi:=\coprod_{x\in U}\Phi_x:$ [unit disc] $\times U\to\mathbb C$ smooth?

I'm also interested in the same question where the open unit disc is replaced by the closed unit disc (i.e. smoothness all the way to the boundary).

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  • $\begingroup$ Smooth means $C^\infty$? $\endgroup$ Commented Apr 23, 2019 at 12:04
  • $\begingroup$ Yes, smooth means $C^\infty$. One can of course formulate variants of my question for any regularity class, but I'm currently most interested in the $C^\infty$ version. $\endgroup$ Commented Apr 23, 2019 at 13:36
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    $\begingroup$ Unless I am mistaken, the method based on Green function gives you smooth dependence on parameters. The method of solving Riemann mapping problem using Green functions is described in M.E. Taylor "Partial differential equations I", section 5.4 in first edition. I did't check that the Green function depends on domain smoothly, but it's a solution of Dirichlet problem ... $\endgroup$ Commented Apr 23, 2019 at 16:48
  • $\begingroup$ I should add, that this question is a very simple case of "gluing analytic disks" method. Usually one is interested in attaching analytic disks to a totally real (except possibly finitely many complex points) submanifold of $\mathbb{C}^n$. Recently (after Gromov), it's usually attaching pseudoholomorphic curves to lagrangian submanifold. In your case, it's $S^1\times U\subset \mathbb{C}^{n+1}$. $\endgroup$ Commented Apr 23, 2019 at 19:11
  • $\begingroup$ @OlegEroshkin: If you could explain in what sense my question is a special case of "gluing analytic disks", that would be very appreciated. (Maybe you could even put that as an answer, as opposed to a mere comment.) $\endgroup$ Commented Apr 24, 2019 at 9:45

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The conformal map $\phi$ between the unit disk $D$ and the simply connected domain $\Omega$ with a smooth boundary, subject to normalization $\phi(0)=a$ and $\phi'(0)>0$, depends smoothly on the domain $\Omega$. It's proved, for example, in Bell's book "The Cauchy Transform, Potential Theory and Conformal Mapping", see Theorem 28.1.

There it's proved by integral transform method. It can be proved using other approaches to Riemann Mapping problem. For example, it's easy to prove using method based on Green's function as described in section 5.4 in Taylor's "Partial Differential Equations 1. Basic Theory", 1st ed.

I don't know who was the first to establish that the conformal map depends smoothly on the domain. Quite possible it was Riemann himself, because to find how that map changes under small perturbation of the domain one need to solve Riemann-Hilbert problem.

Now, the relation to gluing analytic disks. That's definitely an overkill for the problem in hands, but it may be useful. Let's consider $\mathbb{C}^{n+1}$ with the standard symplectic form. Then the image $f(S^1\times U)\subset \mathbb{C}\times\mathbb{R}^n\subset\mathbb{C}^{n+1}$ is a Lagrangian submanifold. $\Phi$ is a family of analytic disks attached to that manifold. In this setting, there is no natural normalization, so it's not completely equivalent. Nevertheless, appropriate version of smoothness of the gluing map is classical (and easy). The difficult part (in general) is the existence of at least one disk and some version of compactness (existence of limits).

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    $\begingroup$ Regarding smooth dependence of Riemann map on the domain, the first to consider explicitly such a problem was Gaston Julia, Sur une équation aux dérivées fonctionelles liées à la représentation conforme, Annales scientifiques de l’É.N.S. 3e série, tome 39 (1922), p. 1-28, He used the method of Green function and earlier results of Hadamard. See numdam.org/article/ASENS_1922_3_39__1_0.pdf $\endgroup$ Commented Apr 24, 2019 at 20:41

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