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Consider the problem of trying to identify which $n$-dimensional compact complex manifolds can be endowed with a Kähler metric.

$\underline{n = 1}:$ Any hermitian metric on a Riemann surface is a Kähler metric as the Kähler form is a two-form, and all two-forms are closed. (This is also true of non-compact Riemann surfaces).

$\underline{n = 2}:$ A necessary and sufficient condition for a complex surface to admit a Kähler metric is that its odd Betti numbers are even.

The condition that odd Betti numbers are even is necessary for a complex manifold to admit a Kähler metric, but not sufficient for $n \geq 3$. To see it is not sufficient, consider the example of Hironaka which gives a deformation of three-dimensional Kähler manifolds to a non-Kähler manifold. The Betti numbers are invariant under diffeomorphisms, so the central fibre of the deformation has odd Betti numbers even (as it is diffeomorphic to a Kähler manifold) but it is not itself Kähler.

In general, for $n \geq 3$, it is not easy to determine when a compact complex manifold can be equipped with a Kähler metric. In the simplest of these cases, $n = 3$, are we any closer to solving this problem?

What conditions (necessary or sufficient) are there for a three-dimensional compact complex manifold to admit a Kähler metric? How far are we from a single necessary and sufficient condition (like we have for $n = 1$ and $2$)?

Of course I'm interested in results which apply for all $n$, but I'm guessing that there are some results specifically for $n = 3$.

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3 Answers

up vote 8 down vote accepted

The main obstruction to existence of Kahler metric (in addition to Lefschetz SL(2)-action and Riemann-Hodge relations in cohomology) is homotopy formality: the cohomology ring of a Kahler manifold is related to its de Rham algebra by a chain of homomorphisms of differential graded algebras inducing isomorphisms on cohomology. This is proven by Deligne-Griffiths-Morgan-Sullivan in 1970-ies.

This is a very strong topological condition; for instance, no nilmanifold (except torus) is homotopy formal. There are symplectic nilmanifolds satisfying hard Lefschetz and the rest of Riemann-Hodge conditions for cohomology.

Another obstruction is existence of a positive, exact current. As shown by Peternell, all non-Kahler Moishezon manifolds admit a positive, exact (n-1,n-1)-current, hence they are not Kahler. However, Moishezon manifolds are homotopy formal ([DGMS]), and often satisfy the Riemann-Hodge. This argument is also used to prove that twistor spaces of compact Riemannian 4-manifolds are not Kahler, except CP^3 and flag space (Hitchin).

The sufficient condition in this direction is obtained by Harvey-Lawson: they proved that a manifold is Kahler if and only if it does not admit an exact (2n-2)-current with positive, non-zero (n-1, n-1)-part.

Finally, Izu Vaisman has shown that any compact locally conformally Kahler manifold (a manifold with Kahler metric taking values in a non-trivial 1-dimensional local system) is non-Kahler.

Also, a complex surface is Kahler if and only if its $b_1$ is even. This was known from Kodaira classification of surfaces, and the direct proof was obtained in late 1990-ies by Buchdahl and Lamari using the Harvey-Lawson criterion.

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(actually, symplectic structures on nilmanifolds never satisfy the Hard Lefschetz Condition, as in the proof of Theorem A by Benson and Gordon, ams.org/mathscinet-getitem?mr=976592. In fact, the map $[\omega^{n-1}]: H^1_{dR}(X;\mathbb{R})\to H^{2n-1}_{dR}(X;\mathbb{R})$ is not surjective for a $2n$-dimensional nilmanifold $X$ with a symplectic structure $\omega$) –  daniele Jul 15 '13 at 13:16
    
thanks for a correction; my memory fails me, mea culpa –  Misha Verbitsky Jul 15 '13 at 21:28
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However, there are examples of non-formal (hence, non-Kahler) symplectic manifolds satisfying hard Lefschetz; see arxiv.org/abs/math/0403067, "The Lefschetz property, formality and blowing up in symplectic geometry", by Gil R. Cavalcanti –  Misha Verbitsky Jul 15 '13 at 21:36
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A partial answer.

From the cohomological point of view, a compact Kähler manifold satisfies the "$dd^{\mathcal{J}}$-Lemma" with respect to any of its structures. In fact, such properties do not characterize Kählerness: for example, Fujiki class $\mathcal{C}$ manifolds and Moishezon manifolds satisfy $\partial\overline{\partial}$-Lemma, as Gunnar recalled.

More precisely:

  • with respect to the complex structure, a compact Kähler manifold satisfies the classical $\partial\overline{\partial}$-Lemma, stating that every $\partial$-closed $\overline{\partial}$-closed $d$-exact form is also $\partial\overline{\partial}$-exact, namely, the natural map $H^{\bullet,\bullet}_{BC}(X)\to H^{\bullet}_{dR}(X;\mathbb{C})$ induced by the identity is injective (here, $H^{\bullet,\bullet}_{BC}(X):=\frac{\ker\partial\cap\ker\overline{\partial}}{\mathrm{imm}\partial\overline{\partial}}$ is the Bott-Chern cohomology of $X$, and its dual $H^{\bullet,\bullet}_{A}(X):=\frac{\ker\partial\overline{\partial}}{\mathrm{imm}\partial+\mathrm{imm}\overline{\partial}}$ is the Aeppli cohomology). As a consequence, one gets the Hodge decomposition. In particular, $\dim_{\mathbb{C}} H^{\bullet}_{dR}(X;{\mathbb{C}}) = \sum_{p+q=k} \dim_{\mathbb{C}} H^{p,q}_{\overline{\partial}}(X)$ and $\dim_{\mathbb{C}} H^{p,q}_{\overline{\partial}}(X)=\dim_{\mathbb{C}} H^{q,p}_{\overline{\partial}}(X)$ for every $k$, for every $p,q$. Note that such equalities do not characterize even the validity of the $\partial\overline{\partial}$-Lemma. Furthermore, one has that a compact complex manifold satisfies the $\partial\overline{\partial}$-Lemma if and only if $2\dim_{\mathbb{C}}H^k_{dR}(X;\mathbb{C}) = \sum_{p+q=k} \left(\dim_{\mathbb{C}} H^{p,q}_{BC}(X) + \dim_{\mathbb{C}} H^{p,q}_{A}(X)\right)$ for every $k$.

  • with respect to the symplectic structure, a compact Kähler manifold satisfies the Hard Lefschetz Condition, which is the symplectic counterpart of the $\partial\overline{\partial}$-Lemma. By definition, Hard Lefschetz Condition states that $[\omega^k]\colon H^{n-k}_{dR}(X;\mathbb{R})\to H^{n+k}_{dR}(X;\mathbb{R})$ is an isomorphism for every $k$, where $\omega$ is the Kähler form and $2n$ denotes the real dimension of $X$. By considering the symplectic differential $d^{\Lambda}:=\left[d, -\iota_{\omega^{-1}}\right]$ and defining the Tseng and Yau symplectic Bott-Chern cohomologies as in the complex case by using $d$ and $d^\Lambda$ instead of $\partial$ and $\overline{\partial}$, one has that the Hard Lefschetz Condition is equivalent to the $dd^\Lambda$-Lemma, which is equivalent to the equality $\dim_{\mathbb{R}} H^k_{BC}(X) + \dim_{\mathbb{R}} H^k_{A}(X) = 2\dim_{\mathbb{R}} H^k_{dR}(X;\mathbb{R})$ for every $k$.

  • Looking at a Kähler manifold as a generalized-complex manifold (in the sense of Hitchin, Gualtieri, Cavalcanti,) one gets further numerical properties. For example, one gets that the $dd^{\mathcal{J}}$-Lemma with respect to the generalized-complex structure of type $n$ is equivalent to the equalities $\dim_{\mathbb{C}}H_{dR}^{k}(X;\mathbb{C})= \sum_{p+q=k}\dim_{\mathbb{C}}H^{p,q}_{\overline{\partial}}(X)$ and $\sum_{p-q=k}\dim_{\mathbb{C}}H^{p,q}_{BC}(X) = \sum_{p-q=k}\dim_{\mathbb{C}}H^{p,q}_{\overline{\partial}}(X)$ for every $k$ (note, "$p-q=k$" instead of "$p+q=k$").

Obviously, as said, these cohomological properties are quite far from being sufficient.. Notwithstanding, Kähler manifolds are characterized inside some special classes of compact manifolds, such as, for example, nilmanifolds and solvmanifolds.

Note also a related question: Fundamental Groups of compact Complex manifolds?

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One necessary condition for the existence of a Kaehler structure on a manifold is Theorem 1.8 (the hard Lefschetz property) from here. According to the author (Claire Voisin) these are the only known obstructions.

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That is very probably not sufficient. Consider Hironaka's example $X$ of a non-Kahler deformation of projective manifolds. That manifold is Moishezon since it is a deformation limit of projective manifolds (by Popovici, I think). Then there exists a modification $\epsilon X \to X'$ where $X'$ is projective. It follows that the cohomology ring of $X$ admits a Hodge decomposition. I would think that you could then find some $(1,1)$-class on $X$ to play the role of a Kahler class in the hard Lefschetz theorem, the only difference being that it wouldn't be represented by a Kahler form. –  Gunnar Magnusson Jul 8 '13 at 15:59
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@Andrei : There are other obstructions known on the fundamental group see e.g. homepages.math.uic.edu/~jaca2009/notes/Arapura.pdf , plus, in the paper by Voisin you cite, her goal is to show that there are other obstructions, coming from the algebra structure of the cohomology. –  BS. Jul 9 '13 at 9:10
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