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Let $D$ be a domain in $\mathbb{C}$ with $n$ boundary components. From the work of Koebe, we know that $D$ can be conformally mapped to a parallel slit domain of a specified angle of inclination (relative to the real axis).

Question: What boundary regularity does this map possess?

Further remarks: This map, in the same way that one proves the Riemann mapping theorem, is constructed variationally. Following Chapter VII of Nehari's book on the subject: Let $S_{\zeta}$ be the class of functions $f$ which are univalent in $D$ and have a simple pole of residue $1$ at $z = \zeta$ in $D$. Hence, near $z=\zeta$, we can write $$f(z) = \frac{1}{z-\zeta} + a_0 + a_1(z-\zeta) + \cdots.$$ If $\zeta = \infty$, we write $$f(z) = z + a_0 + \frac{a_1}{z} + \frac{a_2}{z^2} + \cdots.$$

Within the class $S_{\zeta}$, the conformal map is then obtained as the maximiser of $$\max_{f \in S_{\vartheta}} \text{Re} \{ e^{-2i\vartheta}a_1 \},$$ where $a_1$ is as above.

Results: Bell and Krantz (1987) proved that (univalent) conformal maps between bounded domains with smooth boundaries extend smoothly to the boundary. Does this imply that the map taking $D$ to the parallel slit domain is smooth up the boundary? I doubt this! Maybe, I'm not understanding the geometry of this map well enough.

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  • $\begingroup$ The theory is analogous to the theory of the boundary behaviour of conformal isomorphisms between the disc and connected domains. This is the subject of Pommerenke's book "Boundary behaviour of conformal maps", where you can find a wealth of information. $\endgroup$ – Lasse Rempe Jul 24 '20 at 12:53
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EDIT. I just realized that my previous answer was incorrect. The slits in the Kobe theorem that you consider are bounded, since the image under the mapping function with your normalization has $\infty$ inside the domain. Therefore the map is smooth if you assume that the boundary of $D$ is smooth.

The smoothness depends on the boundary of $D$. If $\partial D$ is smooth, then the map is smooth at all points of $\partial D$.

In general, smoothness of a conformal map from $D_1$ to $D_2$ at a point $z_0\in\partial D_1$ depends only on the behaviour of $\partial D_j$ in arbitrarily small neighborhoods of $z_0$ and $f(z_0)$. Assuming $\partial D_1$ is smooth, at the points which are mapped to the straight parts of the slits, the map is smooth, same at preimages of the tips of the slits (where the map is like $z^2$ in local coordinates).

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  • $\begingroup$ Is it clear that the tips of the slits are given locally by $z^2$? $\endgroup$ – Very Confused Jul 24 '20 at 21:16
  • $\begingroup$ Can one prevent the map from taking the value $\infty$? $\endgroup$ – Very Confused Jul 24 '20 at 21:45
  • $\begingroup$ I do not understand your last comment. What do you mean by "prevent"? Slit plane has infinity on the boundary. There are standard domains on which any n-connected region can be mapped. But you were asking about slit domain. $\endgroup$ – Alexandre Eremenko Jul 24 '20 at 22:59
  • $\begingroup$ On your first comment: yes this is clear. Compose your map with $\sqrt{z-a}$ where $a$ is the tip. You obtain a map between smooth domains. $\endgroup$ – Alexandre Eremenko Jul 24 '20 at 23:00
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    $\begingroup$ @AmorFrati: I confirm. $\endgroup$ – Alexandre Eremenko Jul 25 '20 at 21:45
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I started to write this as comments, but really it is too long for a comment. It is something of an elaboration on Alex's answer.

The theory is analogous to the theory of the boundary behaviour of conformal isomorphisms between the disc and connected domains. This is the subject of Pommerenke's book "Boundary behaviour of conformal maps", where you can find a wealth of information.

You probably want to map from the "simple" domain to the more complicated one; so you want to map TO the domain $D$. (If you are really interested in the boundary behaviour the map from $D$, then you should still be able to deduce anything you want to ask from the below.) Moreover, you probably want to use a circle domain (the complement of finitely many closed discs in the sphere) rather than a slit domain. (This is OK since you only have finitely many components.) The reason for this is two-fold: firstly, the boundary behaviour at the end-points of the slits will differ from that at the interior points, even if the geometry at the corresponding points of the boundary of $D$ is the same. This is because the two maps essentially differ by a square root (which takes a radial slit starting at 0 to a line). Secondly, with circles you do not have to worry about different branches coming from both sides of a slit.

So now the question becomes: Suppose that $\phi$ is a conformal isomorphism from $U$ to $D$, where $U$ is the complement of $n$ disjoint round discs, and $D$ is the complement of $n$ disjoint compact and connected sets. What can be said about the boundary behaviour of $\phi$?

The answer is that essentially anything that is true for $n=1$ (where $U$ is the unit disc) also holds for $n>1$. (More precisely, this is the case as long as the statements are invariant under maps that are conformal near the boundaries of $U$ and $D$.) You can find this information in Pommerenke's book mentioned above. In particular:

  • The map $\phi$ has a continuous extension to $\overline{U}$ if and only if every connected component of $\partial D$ is locally connected. (This is the Carathéodory-Torhost theorem.)
  • The extension is analytic (with non-zero derivative) if and only if $\partial D$ consists of analytic curves.
  • As Alex says, the map is $C^{\infty}$ if and only if the boundary is.
  • For lower degrees of smoothness, things are a little more subtle, but the answers exist there also. I don't have Pommerenke's book to hand at the moment, but the relevant term is "Dini smoothness".

If you are interested in the behaviour of a conformal map between two other $n$-connected domains, neither of which is bounded by disc, map them both to a circle domain and apply the theory above in order to understand what is going on. :)

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