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Let $M$ be a compact complex connected [but not necessarily kähler] $n$-manifold, and suppose we have a holomorphic map $$(\mathbb{C}^*)^n \to M$$ such that the image is open. Is the image necessarily dense in $M$?

Motivation: My intuition (which comes from the algebraic world) says that the answer ought to be "yes." On the other hand, I know that many properties of smooth algebraic varieties do not hold for complex manifolds in general. Knowing whether this statement has a counterexample would improve my intuition about the complex world.

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2 Answers 2

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An example exists already for $M={\mathbb C}P^2$, furthermore, there exists an injective holomorphic map $f: {\mathbb C}^2\to {\mathbb C}^2\subset {\mathbb C}P^2$ whose image is open but not dense. Recall that a domain $\Omega$ in ${\mathbb C}^2$ is called a Fatou-Bieberbach (FB) domain if $\Omega\ne {\mathbb C}^2$ and there exists a biholomorphic map $f: {\mathbb C}^2\to \Omega$. First examples of FB domains were constructed by Fatou and Bieberbach and it is a bit of an industry to construct FB domains with interesting properties. For instance, B. Stensönes constructed in her paper ("Fatou-Bieberbach domains with smooth boundary", Annals of Math, vol. 145, 1997, 365-377) FB domains in ${\mathbb C}^2$ whose boundaries are smooth. If you look at Proposition 3.1 (part v) of her paper, it shows that the complement of her FB domain has nonempty interior. I am pretty sure that one can find earlier examples as well. Now, if you want the domain of $f$ to be $({\mathbb C}^\times)^2$, just restrict the above holomorphic map.

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  • $\begingroup$ Wow - this is surprising. I would have bet money, that this is impossible! $\endgroup$ Commented May 9, 2012 at 1:05
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    $\begingroup$ Surprising... That just shows how different algebraic geometry is from complex geometry. Is there an intuitive way to explain how to construct such a domain? $\endgroup$
    – temp
    Commented May 9, 2012 at 2:49
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    $\begingroup$ Nice answer, but Berit Stensones is a a woman (and was in 1997). $\endgroup$
    – Igor Rivin
    Commented May 9, 2012 at 7:13
  • $\begingroup$ Oops! Corrected now. $\endgroup$
    – Misha
    Commented May 9, 2012 at 12:17
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Fatou-Bieberbach domains were first constructed by Fatou in 1927; the construction was improved by Bieberbach in 1932. The idea is linearizing coordinates at attracting cycles.

Let $f:\bf C^2 \to C^2$ be an analytic automorphism, such that $(0,0)$ is an attracting fixed point, for instance the Hénon mapping $$ f: (x,y) \mapsto (x^2-y,x/2). $$ Denote by $L$ the derivative of $f$ at the origin.

There is then (usually, in particular in the case above) a unique analytic mapping $\phi:\bf C^2\to C^2$ with $\phi(0,0)=(0,0)$, $D\phi(0,0)=id$ and $L\circ \phi= \phi\circ f$.

In this case (and in many others), $\phi^{-1}$ is defined on the basin of attraction of the origin by the formula $$ \phi(x,y)= \lim_{m \to \infty} L^{-\circ m} \circ f^{\circ m} (x,y). $$

The map $\phi$ is injective, but its image is certainly not dense. The image is the basin of attraction of the origin. There are lots of points not attracted to the origin, for instance the set $$ U=\{(x,y) |\ |x|\ge 2, |y|\le x\}. $$ it is easy to see that $f(U)\subset U$, and that if you iterate $f$ in $U$ all orbits tend to infinity.

Showing that the limit above exists and defines an analytic map is not hard; I will write a proof if you are interested.

There is an enormous literature about linearization at periodic points.

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