[Math] Examples of compact complex non-Kähler manifolds which satisfy $h^{p,q} = h^{q,p}$

complex-geometryexampleskahler-manifolds

The existence of a Kähler metric on a compact complex manifold $X$ imposes restrictions on it's Dolbeault cohomology; namely, $h^{p,q}(X) = h^{q,p}(X)$ for every $p$ and $q$. I am looking for some explicit examples of compact complex manifolds which satisfy these restrictions, but are not Kähler (i.e. do not admit a Kähler metric).

In dimension 1, every compact complex manifold is Kähler ($d\omega$ is automatically zero).

In dimension 2, a compact complex surface $X$ is Kähler if and only if the first Betti number $b_1(X)$ is even. Therefore, if $h^{1,0}(X) = h^{0,1}(X)$, then $b_1(X) = h^{1,0}(X) + h^{0,1}(X) = 2h^{1,0}(X)$ so by the aforementioned result, $X$ is Kähler.

It follows that any examples must be of dimension at least three.

The only thing that has come to mind so far is Hironaka's example which is a deformation of Kähler manifolds such that the limiting fibre is not Kähler. As diffeomorphism type is preserved under deformations, I know that the limiting fibre has the same Betti numbers as a Kähler manifold (i.e. the odd Betti numbers are even), but I don't know if its Hodge numbers satisfy the desired symmetry.

Added Later:

  1. After consulting the book Cohomological Aspects in Complex Non-Kähler Geometry by Angella, I realised that the central fibre of Hironaka's deformation, which I will now call $X_0$, is indeed an example. We just need to show that $h^{p,q}(X_0) = h^{q,p}(X_0)$.

    First note that $X_0$ is a Moishezon manifold and is therefore a proper modification of a projective manifold. As projective manifolds satisfy the $\partial\bar{\partial}$-lemma, so do Moishezon manifolds; see Theorem 5.22 of Real Homotopy Theory of Kähler Manifolds by Deligne, Griffiths, Morgan, and Sullivan. In the same paper, they show that if a compact complex manifold satisfies the $\partial\bar{\partial}$-lemma, the Hodge-Frölicher spectral sequence degenerates at the first step, see Theorem 5.21. This in turn equivalent to the equality (for each $k$) $$b_k(X_0) = \displaystyle\sum_{p+q = k}h^{p,q}(X_0),$$ so $$b_k(X_a) = \displaystyle\sum_{p+q = k}h^{p,q}(X_a)$$ for all $a$ in the base of the Hironaka deformation. The map $a \mapsto h^{p,q}(X_a)$ is upper semicontinuous in general (see Corollary 9.19 of Voisin's Hodge Theory and Complex Algebraic Geometry I for example), but the map $a \mapsto b_k(X_a)$ is constant by Ehresmann's result. Due to the above equality, we see that $a \mapsto h^{p,q}(X_a)$ is actually constant for all $p$ and $q$. Therefore, $$h^{p,q}(X_0) = h^{p,q}(X_a) = h^{q,p}(X_a) = h^{q,p}(X_0)$$ where $a \neq 0$ and we use the fact that $X_a$ is projective and hence Kähler.

    The above result can be summarised as follows:

    Let $\mathcal{X} \to B$ be a deformation of compact complex manifolds with $0 \in B$. If $a \mapsto h^{p,q}(X_a)$ is constant on $B\setminus\{0\}$, and the Hodge-Frölicher spectral sequence of $X_0$ degenerates at the first step, then $a \mapsto h^{p,q}(X_a)$ is constant on $B$.

    In particular, given a deformation $\mathcal{X} \to B$ such that $h^{p,q}(X_a) = h^{q,p}(X_a)$ for all $a \in B\setminus\{0\}$, and the Hodge-Frölicher spectral sequence of $X_0$ degenerates at the first step, then $h^{p,q}(X_0) = h^{q,p}(X_0)$. If $X_0$ is not Kähler (as in Hironaka's example), $X_0$ is then an example of the type I was searching for.

  2. If $h^{p,q} = h^{q,p}$ for all $p$ and $q$, then $b_k$ is even for $k$ odd. So one may instead try to solve the easier problem of finding a non-Kähler manifold $X$ such that $b_k(X)$ is even for $k$ odd.

    As above, in the one or two dimensional case, if $b_k$ is even for $k$ odd, then $X$ is Kähler. In dimension three however, there is a simple example given by $S^3\times S^3$. A reference for the fact that $S^3\times S^3$ can be equipped with a complex structure is Complex Structures on $S^3\times S^3$ by Fujiki.

    The Kunneth formula gives $b_0 = 1$, $b_1 = 0$, $b_2 = 0$, $b_3 = 2$, $b_4 = 0$, $b_5 = 0$, and $b_6 = 1$. All odd Betti numbers are even, but $S^3\times S^3$ is not Kähler as $b_2 = 0$. In fact, all even Betti numbers of a Kähler manifold are positive ($\omega^k$ is not exact), so one might try to find an example of a non-Kähler manifold with $b_{2k} > 0$ and $b_{2k+1}$ even. By the discussion in point one above, an example of such a manifold is the central fibre of the Hironaka deformation.

Best Answer

Every compact complex manifold satisfying the $\partial\overline{\partial}$-Lemma has such a property. Particular examples are given by - as you already said - Hironaka example (and, more in general, Moishezon manifolds and manifolds in class C of Fujiki), or some deformations of twistor spaces (see LeBrun, Poon, Twistors, Kähler manifolds, and bimeromorphic geometry. II, J. Amer. Math. Soc. 5 (1992), no. 2, 317–325).

On the other side: in Ceballos, Otal, Ugarte, Villacampa, arXiv:1111.5873, Proposition 4.3, you find a concrete example of a compact complex manifold with the symmetry of Hodge diamond you require. This example does not satisfy $\partial\overline{\partial}$-Lemma.