Is there an explicit example of non-Kahler manifold $M$ such that $M$ satisfies the dd^c lemma ?
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4$\begingroup$ Any Moishezon manifold would do. So perhaps asking for explicit examples of Moishezon manifolds would answer your question. $\endgroup$– José Figueroa-O'FarrillCommented Apr 21, 2011 at 9:34
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6$\begingroup$ complementing Jose's comment: ... or in fact anything bimeromorphic to a K\"ahler manifold, see Deligne, Griffiths, Morgan, Sullivan, Real homotopy type of K\"ahler manifolds. By the way, a explicit example of a Moishezon space is given in Appendix B of Hartshorne's Algebraic geometry. $\endgroup$– algoriCommented Apr 21, 2011 at 11:48
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1$\begingroup$ Let $X$ be a compact complex manifold. Then, for every $k\in\mathbb N$, the following equality holds $$∑_{p+q=k}(dim_{\mathbb C}H^{p,q}_{BC}(X)+dim_{\mathbb C}H^{p,q}_A(X))=2dim_{\mathbb C}H^k_{dR}(X,\mathbb C),$$ if and only if we have $\partial\bar\partial$-Lemma $\endgroup$– user21574Commented Jul 21, 2017 at 4:04
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1$\begingroup$ In previous comment, Bott-Chern and Aeppli cohomologies defined as $$H^{∙,∙}_{BC}(X):=\frac{ker∂∩ker\bar ∂}{im∂\bar ∂}$$ and $$H^{∙,∙}_A(X):=\frac{ker∂\bar ∂ }{im∂+im\bar ∂}$$. $H^k_{dR}(X,\mathbb C)$ is the de Rham cohomology. See Angella, DanieleTomassini, Adriano On the$ ∂\bar ∂$-lemma and Bott-Chern cohomology. Invent. Math. 192 (2013), no. 1, 71–81. $\endgroup$– user21574Commented Jul 21, 2017 at 4:13
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$\begingroup$ Moreover to give reference to first comment of @JoséFigueroa-O'Farrill: see (5.23) Corollary. of the paper of Griffiths-Deligne et al link.springer.com/article/10.1007/BF01389853 $\endgroup$– user21574Commented Nov 17, 2017 at 19:22
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3 Answers
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Here is an example of a Moishezon manifold which is easy to visualize. Take a high degree (e.g. a quintic) hypersurface $Z$ in $\mathbb{P}^{4}$ which has a single ordinary double point. Let $X$ be a small resolution of $Z$. Explicitly, a small analytic neighborhood of the singularity can be identified with the vertex of a cone over a two dimensional quadric and you just need to blow-up the Weil divisor which is the preimage of one ruling. The threefold $X$ is compact complex manifold and does not admit any Kaehler structure. The last statement follows for instance from a theorem of Smith-Thomas-Yau which states that a threefold with a single node will admit a symplectic small resolution only if the three sphere that vanishes at the node is homologous to zero. The high degree condition on $Z$ ensures that the vanishing cycle is not homologous to zero, hence the statement.
answered Apr 21, 2011 at 21:19
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$\begingroup$ Tony, this is a nice example. $\endgroup$ Commented Apr 22, 2011 at 12:24
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This does not answer your question, but it seems interesting to point out.
A non-example would be any complex threefold diffeomorphic to the six-sphere (if such a manifold exists).
To see this, first note that the $dd^c$-lemma holds for $(1,1)$-forms on a compact complex manifold iff $b_1=2h^{0,1}$ (see for example the paper of Paul Gauduchon "La 1-forme de torsion d'une variete hermitienne compacte", Math.Ann. 1984, page 504).
Now any threefold diffeomorphic to the six-sphere must have $h^{0,1}\geq 1$ by a result of Alfred Gray ("A property of a hypothetical complex structure on the six-sphere", Bollettino U.M.I. 1997, Theorem 5), and therefore the $dd^c$-lemma for $(1,1)$-forms fails since $b_1=0$.
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4$\begingroup$ No access to Gray's paper where I am, so no idea if he does the following, but here's a short proof that $h^{0,1}\geq 1$ for those who are curious. The argument holds whenever both integer cohomology in degree 2 and $b_n$ vanish on a complex $n$-fold. Since $H^2(Z)=0$, the exponential sequence shows $H^{0,1}$ surjects onto the group of line bundles. So it's enough to find a non-trivial holomorphic line bundle. Now the canonical bundle is nontrivial, for if not a holomorphic $n$-form $\Omega$ would have $\int \Omega \wedge \bar{\Omega} >0$ so $[\Omega]$ would be non-zero contradicting $b_n=0$. $\endgroup$ Commented Apr 22, 2011 at 3:26
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I think this is an example.
There is an integrable complex structure on X=S^3 xS^3. For topological reasons we have Td(X)[X]=1. Thus h^0,1 =0. It's an easy exercise that a complex manifold with h^0,1 =0 satisfies the d\bar{d} lemma.
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5$\begingroup$ The exercise is not easy for me. The $dd^c$ lemma has very strong consequences, such as degeneration of various spectral sequences, formallity etc. So I find the last statement surprising. $\endgroup$ Commented Apr 21, 2011 at 16:03
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$\begingroup$ I think that perhaps $H^{0,1}=0$ implies the d-dbar lemma for (1,1)-forms, but I can't see how it would help for higher degrees. $\endgroup$ Commented Apr 22, 2011 at 2:12
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1$\begingroup$ @Yang-Mills, from what you say it follows that $h^{0,1} = 0$ DOES imply the $dd^c$-lemma for $(1,1)$-forms. For if $h^{0,1}=0$ then it follows that $b_1=0$ hence certainly $b_1 = 2h^{0,1}$. But there is a much more prosaic way to see it. Let $a = db$ be an exact real (1,1) form. Then $\bar{\partial}b^{0,1} = 0$ so $b^{0,1} = \bar{\partial} f$. From here it follows that $a = 2 i \partial\bar{\partial} g$ where $g$ is the imaginary part of $f$. $\endgroup$ Commented Apr 22, 2011 at 2:26
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$\begingroup$ so it seems that $S^3\times S^3$ does not satisfy the $dd^c$-lemma after all, since $b_1\neq 2h^{0,1}$. Also, it has $h^{1,1}=1$, see the same paper of Hofer, so there are nontrivial $\overline{\partial}$-closed $(1,1)$-forms $\endgroup$ Commented Apr 22, 2011 at 2:28
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1$\begingroup$ It's possible that Craig intended the statement only for (1,1) forms as in the follow up discussion, it's hard to say. However, I am sure that the stronger statement is false. The reason I think so is that by Deligne-Griffiths-Morgan-Sullivan, a compact complex manifold satisfying the $dd^c$-lemma should be formal. Now, I think one can use a twistor construction (Taubes, JDG 1992) to build a compact complex manifold which is not formal but with $b_1=0$. $\endgroup$ Commented Apr 22, 2011 at 12:59