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Recently I face up with a problem, which I realized that have close connection with the following problem. $\{ f_{n} \}_{n=1}^{\infty}$ is analytic map from $C^{n}$ to $C^{n} $\ $U$ where U is open neighborhood of 0, whether f is a normal family.

I know when n=1, this is really Montel Normal family criterion, However I did know whether it is true for high dimension. also I heard that the for any two topological equivalent simple connected domain in$ C^{n}$ $(n\geq 2)$, the probability of holomorphic equivalent for this two domain is 0. I want to know what is the precise statement for this theorem.

Any advice and comments will be appreciated.

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up vote 2 down vote accepted

The key phrase to keep in mind is "taut manifold": a complex manifold X which is taut has the property that Hol(Y,X) is normal for every complex manifold Y. A fact that may be of interest to you is that a taut domain in $\mathbb{C}^n$ is necessarily pseudoconvex.

More information, including criteria of tautness and the relation with the notion of hyperbolicity can be found in the following books: by Marco Abate,

and by Shoshichi Kobayashi,

Hyperbolic Complex Spaces, Springer, 1998

As for lack of biholomorphic equivalence for domains in $\mathbb{C}^n$, I am not sure if your "probabilistic" statement can be put on rigorous footing. It is true that an euclidean ball and a polydisk in $\mathbb{C}^n$ are not biholomorphically equivalent when $n \geq 2$. The result goes back to Poincare and has several proofs. You can learn more from the text by S. Krantz in several comlex variables; there is also a good sketch of proof among the exercises in the text by Grauert and Fritzsche, "From holomorphic functions to complex manifolds."

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There are uncountably many biholomorphically inequivalent domains in C^n which are as close to the unit ball as you like – Steven Gubkin Oct 12 '12 at 22:54
Right, but "uncountably many" is not the same as "with probability one", even if this latter notion makes sense in this context (which I doubt). This discussion could be made much more precise and to the point if the two questions (about normal families and about biholomorphic non-equivalence) were asked separately. – Margaret Friedland Oct 13 '12 at 18:24
Thank you for your answer and statement. I know the proof of unit ball and polydisk are not holomorphically equivalent. – yaoxiao Oct 14 '12 at 13:49

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