The Riemann mapping theorem states that there is conformal bijection between the open disk and any simply connected open subset of the complex number plane. My question is, for which subsets of the complex number plane is there a conformal covering map from the disk to that space?

Example: In this image, a ring model of the hyperbolic plane is given. It is given by $w=e^{za}$, where $w$ is a point in the ring, and $z$ is a point in the band model of the hyperbolic plane, for appropriate $a$. It only works if the image in the hyperbolic plane has the appropriate symmetry, since some different points in the band model get mapped to the same point in the ring model. Essentially, it corresponds to a covering map from the hyperbolic plane to the ring. Since the hyperbolic plane is conformally equivalent to the open disk, that means there is also a conformal covering map from the disk to the ring.

So, what other spaces besides the ring work?


1 Answer 1


I'm assuming you are considering open subsets of a plane. Each connected open subset $U$ of the complex plane is a Riemann surface, and so is covered by its universal cover $\tilde U$ which is a simply connected Riemann surface.

The Uniformisation Theorem states that up to isomorphism, the only simply connected Riemann surfaces are $D$ (the unit disc), $\Bbb C$ and $\Bbb C_\infty$ (the Riemann sphere). Certainly $\tilde U$ won't be the Riemann sphere, so the answer to your problem is "yes unless $\tilde U$ is $\Bbb C$".

For which $U$ will $\tilde U$ be the complex plane. The only quotients of $\Bbb C$ by a discrete subgroup of automorphisms acting discontinuously are $\Bbb C$ itself, $\Bbb C^*$ and compact tori. So apart from $\Bbb C$, and complements of a point in $\Bbb C$ each connected open set in the complex plane is covered conformally by $D$.

  • $\begingroup$ So would a complement of two points work? $\endgroup$ Jan 18, 2018 at 5:47
  • $\begingroup$ Yes, indeed the classical modular function $\lambda$ is a covering map from the upper half plane (conformally equivalent to the disc) to $\Bbb C-\{0,1\}$. $\endgroup$ Jan 18, 2018 at 5:49
  • $\begingroup$ One interesting application of this is that any set which meets the conditions of your last sentence corresponds to a hyperbolic surface. Moreover, that set will be conformally equivalent to that surface. $\endgroup$ Jan 18, 2018 at 5:55
  • $\begingroup$ Speaking of which, every hyperbolic surface has a corresponding Fuchsian group, is there a simple way to figure out which fuschian group corresponds to disk covered region of the complex plane? $\endgroup$ Jan 18, 2018 at 5:57
  • $\begingroup$ What universally covers the punctured plane? If it's the plane, what does $0+0i$ get mapped to? $\endgroup$ Jan 18, 2018 at 22:27

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