4
$\begingroup$

From an earlier question, the universal cover of a Kodaira fibered surface is a bounded domain in $\mathbb{C}^2$. It is also not the polydisk or the ball. Can we say more about the structure of the universal cover? In particular, I am interested in whether the domain has simple boundary or not.

We say that a domain $D \subset \mathbb{C}^n$ has simple boundary if whenever we have a holomorphic mapping $\phi : \mathbb{D} \to \overline{D}$, either $\phi(\mathbb{D}) \subset D$ or $\phi$ is constant. Strongly pseudoconvex domains and domain with real-analytic boundary are examples of domains with simple boundary. Product domains do not have simple boundary. Is it a possible for the universal cover to be a product domain in $\mathbb{C}^2$ or is it always a domain with simple boundary?

Improve this question
$\endgroup$
2
  • $\begingroup$ These universal covers are bounded, contractible, non-symmetric domains with discrete biholomorphism groups. That is one more way of justifying that Kodairas are not covered by the 2-ball. A failure of the Riemann Mapping Theorem in higher dimensions. $\endgroup$
    – T. Amdeberhan
    Commented Dec 19, 2016 at 15:11
  • $\begingroup$ @Amdeberhan Thanks! Do we know more? For instance, is the universal cover smoothly bounded? Is it convex? $\endgroup$
    – Jaikrishnan
    Commented Dec 20, 2016 at 4:49

0

Reset to default

You must log in to answer this question.