# Kähler metric on a Zariski open subset of a non-Kähler manifold

I may be missing an obvious example, but here goes...

Let $X$ denote a complex manifold of dimension $n$. If $X$ is Kähler, then the induced metric on any complex submanifold is also Kähler. If instead $X$ is non-Kähler, we can at least say that any coordinate neighborhood $(U,\varphi)$ inherits a Kähler metric from the induced Euclidean metric on an open subset of $\mathbb{C}^n$. If $X$ is already equipped with a metric, note that this metric on $U$ may not be compatible with it. However, I am unsure of what can be said about the metric structure of $U$ if instead we allow something a little more exotic. For example, what if we take $U$ to be the complement of an analytic set? This is the situation in which I am most interested.

My question is:

Is there an example of a non-Kähler manifold, $X$, and a Zariski open subset $U\subset X$, such that $U$ admits a Kähler metric?

Added: Francesco has provided a class of examples in dimensions $\geq 3$ and BS has provided an example in dimension 2 as a comment to Benoît's answer. I am going to write up BS's example, with a few minor additions, as an answer. In case he decides to write his own answer, I'll delete mine and encourage everyone to vote his up.

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One can consider the following example.

A Moishezon manifold $M$ is a compact connected complex manifold such that the field of meromorphic functions on $M$ has transcendence degree equal to the complex dimension of $M$. Complex algebraic varieties have this property, but the converse is not true if the dimension is at least $3$.

In 1967, Boris Moishezon showed that a Moishezon manifold is a projective algebraic variety if and only if it admits a Kaehler metric. Now take a non-projective (hence non-Kaehler) Moishezon manifold $M$. One can prove that such a manifold admits a bimeromorphic modification $\pi \colon \widetilde{M} \to M$ such that $\widetilde{M}$ is projective.

This implies that there exist an analytic subset $S \subset M$ such that $U:=M - S$ is biholomorphic to an open subset of a projective variety (one can take as $S$ a $1$-codimensional analytic subset).

Hence $U$ admits a Kaehler metric.

For further details see the paper by Shanyu Ji Currents, metrics and Moishezon manifolds, Pacific J. Math. Volume 158, Number 2 (1993), 335-351.

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This is a great answer! Thanks. –  Kevin Jul 15 '12 at 0:47
Any chance you'd know where to look for an example in dimension 2? It seems the above construction breaks down there because all Moishezon surfaces are Kahler, by the Chow-Kodaira Theorem. –  Kevin Jul 15 '12 at 1:38
You are welcome. Right, this construction cannot be applied in dimension $2$; one needs another idea. I should think about it. –  Francesco Polizzi Jul 15 '12 at 8:05
More concretely, any smooth non-projective toric manifold is such an example. Use Chow's lemma you can find an open which is projective, so Kahler. –  temp Aug 26 '12 at 18:58

This example is due to BS -- see the edit to the original question.

For an example in dimension two, consider the Hopf surface $X = \left(\mathbb{C}^2\setminus \{0\}\right)/\mathbb{Z}$, where $\mathbb{Z}$ acts as $n\cdot(x,y) = (2^nx,2^ny)$. One can show that $X$ is diffeomorphic to $S^3\times S^1$. It follows that $X$ is non-Kähler because $H^2(X,\mathbb{Z}) = 0$. The $\mathbb{Z}$ action restricts to an action on $\mathbb{C}^*$, which is identified with $\mathbb{C}^*\times \{0\}$, and the smooth curve $E = \mathbb{C}^*/\mathbb{Z}$ of genus 1 includes into $X$ as a submanifold. $Y = X\setminus E$ is then biholomorphic to $E\times \mathbb{C}$ via the map $[(x,y)]\rightarrow \left([y],x/y\right)$. Thus $Y$ admits a Kähler structure. In fact, $Y$ is a quasiprojective variety, as it is isomorphic to an open subset of $E\times \mathbb{P}^1$.

More generally, the Hopf manifolds $X_n = \left(\mathbb{C}^n\setminus \{0\}\right)/\mathbb{Z},\ \ n\geq 2$, are all non-Kähler, as $X_n\simeq S^{2n-1}\times S^1$. The identification $\mathbb{C}^{n-1}\simeq \mathbb{C}^{n-1}\times \{0\}\subset \mathbb{C}^n$ gives rise to an embedding $X_{n-1}\subset X_n$ and, by an argument identical to the one above, we get an isomorphism $Y_n = X_n\setminus X_{n-1} \simeq E\times \mathbb{C}^{n-1}$. Again, $Y_n\subset E\times \mathbb{P}^{n-1}$ is quasiprojective. This gives an example in every dimension $n\geq 2$.

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This is only a slight nitpick: "... and the smooth curve $E$ of genus 1 includes canonically into $X$". In fact $X$ is the total space of a family of elliptic curves $E$ (all isomorphic between them) over the projective line, so the inclusion of $E$ into $X$ is not very canonical (unless we can agree on one point of $\mathbb P^1$---a homogeneous manifold---being more canonical than all the others!). –  Gunnar Þór Magnússon Jul 19 '12 at 4:46
Good point. I'll tone down the wording a bit. –  Kevin Jul 19 '12 at 5:38