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Let $M$ be a complex manifold of dimension $N\ge2$ such that

$\qquad$(1) $M$ is diffeomorphic to $R^{2N}$,

$\qquad$(2) There is a compact set $K\subseteq M$ such that $M\setminus K$ is biholomorphic to $C^N\setminus \bar B_1$.

Must $M$ be biholomorphic to $C^N$?

I don't know if the problem is open, or easy. I'm interested in related problems and references.

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up vote 4 down vote accepted

Final edit

The answer to your question is positive for $n>1$, and this follow just from the fact that a holomorphic function defined on the complement to a pseudoconvex domain can be always extended to the domain for $n>1$. For $n=1$ the statement not true (as Pietro Majer says correctly says it). There is a reference now given in a comment by Margaret Friedland that justifies this answer.

In the case that you consider there is a holomorphic map from $M\setminus K$ to $\mathbb C^n\setminus \bar B_1$, i.e. you have $n$ holomorphic functions on $M\setminus K$. Each of these functions can be extended to $M$ provided $n>1$ since $K$ is pseudoconvex in $M$. So you get a proper holomorphic map from $M$ to $\mathbb C^n$. Moreover this map is birational (or of degree one in other words), so $M$ is a contractible topological space only if the map is an isomorphism (otherwise there will be some exceptional divisors on $M$ that will be contracted to points by the map and so the topology of $M$ will be non-trivial).

Here is the mathscient citation for the reference given by Margaret Friedland

"The authors prove the following: If $M$ is a finite complex manifold with connected boundary $bM$ such that the Levi form has one positive eigenvalue everywhere on $bM$, then every function on $bM$ which satisfies the “tangential Cauchy-Riemann equations” on $bM$ has a holomorphic extension to the whole of $M$."

Note that the boundary of a ball in $\mathbb C^n$ has positive Levi form (for $n>1$) and the "tangential Cauchy-Riemann" equation is automatically satisfied provided the function is defined and holomorphic in a neighbourhood of $bM$. Clearly we can assume the later in our case.

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In fact you don't need pseudoconvexity in order to extend across $K$. It is a compact set, you just use Hartog's extension theorem! – diverietti Nov 4 '12 at 13:09
How is it possible to use Hartogs' extension theorem when it is not yet known whether the domain $M$ is biholomorphic to $C^N$? Or is there a version of Hartogs for $f:M\setminus K\to C$ where $M$ is any complex manifold? I have a similar worry for the argument involving the pseudoconvex domain, because we only know about the local geometry of $\partial K$ but nothing yet about $K$. But here I don't know the subject as well. – Talio Nov 4 '12 at 16:37
The reference for (even more general) extension result is: Kohn, J. J.; Rossi, Hugo On the extension of holomorphic functions from the boundary of a complex manifold. Ann. of Math. (2) 81 1965 451–472. – Margaret Friedland Nov 5 '12 at 1:15
More specifically, if $M$ is a finite complex manifold with connected boundary $bM$ such that the Levi form has one positive eigenvalue everywhere on $bM$, then every function on $bM$ satisfying the tangential Cauchy-Riemann equations on $bM$ has a holomorphic extension to the whole of $M$. ``A finite complex manifold" means a pair of complex manifolds $(M,M')$, $M$ a submanifold of $M′$, with certain conditions imposed on the boundary $bM$ of $M$. – Margaret Friedland Nov 5 '12 at 1:19
There is a survey of results of this kind: Laurent-Thiébaut, Christine: Phénomène de Hartogs-Bochner dans les variétés CR. (French) Topics in complex analysis (Warsaw, 1992), 233–247, Banach Center Publ., 31, Polish Acad. Sci., Warsaw, 1995. Generally, they involve assumptions either on the boundary of the domain or the function itself. Hartogs proved a version for functions that are $\mathcal{C}1$ on the boundary; Bochner considered $L2$ functions. Because of these early results, this extension property is sometimes called the Hartogs-Bochner phenomenon. – Margaret Friedland Nov 6 '12 at 1:35

Consider, for $N=1$, the case where $M$ is the open unit disk $B_1$: diffeomeorphic to $\mathbb{R}^2$, not biholomorphic to $\mathbb{C}$. The compact $K:=\{0\}$ is actually such that $M\setminus K$ is biholomorphic to $\mathbb{C}\setminus \bar{B_1}$ via $z\mapsto 1/z$, disproving the conjecture.

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You are completely correct, and I overlooked this case. I've corrected the question in the easiest way to get at what I am really trying to figure out. – Talio Nov 4 '12 at 23:26

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