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Let $X$ be a (smooth, non-compact) complex space. By a compactification of $X$ we mean a compact complex space $\overline X$ which contains a dense open subset biholomorphic to $X$ (we shall identify this subset with $X$) and such that $\overline X\setminus X$ is a proper analytic subset (with no conditions on its codimension).

In particular, given two (different) compactifications of $X$, they always contain biholomorphic dense open subsets.

My question is: given two compactifications of $X$, are they necessarily bimeromorphic?

More precisely, does the biholomorphism between the two dense open subsets always extend to a global bimeromorphic map?

I guess the answer is no, but after a moment of reflection I cannot find any counterexample. Perhaps, it would suffice to look at compactifications of $\mathbb C^2$...

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This seems related: mathoverflow.net/questions/68421/… –  M P Jan 18 '12 at 16:36
    
I think that the tag ag.algebraic-geometry could be also appropriate here. I added it –  Francesco Polizzi Jan 18 '12 at 17:51
    
As you wish! The power of points... :) –  diverietti Jan 18 '12 at 18:13
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up vote 6 down vote accepted

As you guessed, the answer is no.

A counterexample in dimension $2$ can be found in Vo Van Tan's paper On the compactification of strongly pseudoconvex surfaces III: the Stein surface $\mathbb{C}^* \times \mathbb{C}^*$ admits two algebraic compactifications $M_1$ and $M_2$ which are not bimeromorphic.

In fact, $M_1=\mathbb{P}^1 \times \mathbb{P}^1$ is a rational surface whereas $M_2$ is a $\mathbb{P}^1$-bundle over an elliptic curve.

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Another example, different but diffeomorphic to Polizzi's, is given by
$E \times \mathbb C$, where $E$ is a fixed elliptic curve say $ \mathbb C^* / \lbrace z \mapsto 2z \rbrace$. It can be compactified as the projective surface $E \times \mathbb P^1$ or as the (non-Kähler) Hopf surface $\mathbb C^2-\{0\} / \lbrace (z,w) \mapsto (2z,2w) \rbrace$.

Notice that the two compactifications have fields of meromorphic functions of different transcendence degree over $\mathbb C$, i.e., they have different algebraic dimensions.

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