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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?

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  • $\begingroup$ 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 Dec 19 '16 at 15:11
  • $\begingroup$ @Amdeberhan Thanks! Do we know more? For instance, is the universal cover smoothly bounded? Is it convex? $\endgroup$ – Jaikrishnan Dec 20 '16 at 4:49

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