# Examples of non-Kähler compact complex manifolds which satisfy the Dolbeault isomorphism

Let $X$ be a complex manifold and let $$f_X^n : \bigoplus_{p + q = n} H^{p, q}(X) \to H^n_{\mathrm{DR}}(X, \, \mathbb{C})$$ be the natural map from the Dolbeault cohomology to the de Rham cohomology with complex coefficients.

Question. Is there an example of a compact complex manifold $X$ which is not Kähler such that $f_X^n$ is an isomorphism for all $n$?

• arxiv.org/pdf/1302.0524.pdf
– user21574
Aug 18, 2016 at 13:28
• For curves the answer is clearly no. For surfaces, the answer is also no, because the classification implies that any compact complex surface with even first Betti number is necessarily a Kaehler manifold. Aug 18, 2016 at 13:35
• Hassan, do you have a page reference from Angella's thesis which answers my question?
– user94803
Aug 18, 2016 at 13:36

A compact complex manifold $X$ satisfies the Hodge decomposition $$H^k_{\mathrm{DR}}(X, \, \mathbb C) = \bigoplus_{p+q=k}H^{p, q}(X)$$ (possibly without Hodge symmetry) if its Frölicher spectral sequence $$E_1^{p,q} = H^{p,q}(X) \Rightarrow H^{p+q}(X)$$ degenerates at $E_1$. This happens for instance if $X$ satisfies the $\partial \bar{\partial}$-lemma.

All Moishezon manifolds satisfy it, because the $\partial \bar{\partial}$-lemma is true for projective manifolds, so it is true for any bimeromorphic modification of them. Furthermore, a Moishezon manifold admits a Kähler metric if and only if it is projective.

Thus, in order to obtain the example you are looking for, it suffices to take a non-projective Moishezon manifold (which only exists in dimension $\geq 3$).

For more details and references, you can look at

Dan Popovici, Deformation openness and closedness of various classes of compact complex manifolds; examples. Ann. Sc. Norm. Super. Pisa Cl. Sci. (5) 13 (2014), no. 2, 255--305.

I am slightly confused by Francesco's comment and answer, because the Frolicher spectral sequence does degenerate for compact complex surfaces. So why isn't a non-Kahler surface an example? Perhaps there is some subtlety in defining the groups $H^{p,q}(X)$ that has been swept under the rug. In any case, because the Frolicher spectral sequence degenerates at the $E_1$-page, Hodge decomposition (without Hodge symmetry!!) should hold, even for non-Kahler surfaces. An example would be the quotient of $\mathbb{C}^2\backslash 0$ by multiplication by $2$. This compact complex manifold is diffeomorphic to $S^1\times S^3$ so is non-Kahler (the first Betti number is odd). But it is the case that $$H^k(X,\mathbb{C})=\bigoplus_{p+q=k} H^{p,q}(X).$$ Also, there is a subtlety to phrasing your question, as in general, there is no natural map from Dolbeault to de Rham cohomology, only a spectral sequence from one to the other.

• Yes, this is right. Francesco was probably assuming the question was asking for more than it was. Aug 30, 2016 at 16:15