$M$ is non-Kähler complex manifold. Assume that $\omega$ is $\partial$-exact and $\bar\partial$-exact $(p,q)$-form.



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> **Question**. Is $\omega$ also $\partial\bar\partial$-exact?
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Based on fabulous David Speyer's answer to [this MO question](https://mathoverflow.net/questions/59479/partial-bar-partial-lemma-for-contractible-domains) I suspect that the answer is elementary or very hard.



Are $\partial$-exact and $\bar\partial$-exact forms also $\partial\bar\partial$-exact?
I think that the answer is in general *no* and that a counterexample can be constructed as follows. I will refer to the paper by D. Angella, G. Dloussky, A. Tomassini 


[On Bott-Chern cohomology of compact complex surfaces](http://link.springer.com/article/10.1007/s10231-014-0458-7), *Annali di Matematica Pura ed Applicata* **195** (2016), pp 199–217.


Let us consider a non-Kähler surface $S$ of Inoue type. Then we have (see Table 1 p. 210 of the aforementioned paper) 
$$h^{1, \, 1}\_{\bar{\partial}}(S)=0, \quad \textrm{hence} \quad h^{1, \, 1}\_{\partial}(S)=0 \quad (\spadesuit) $$
where the second equality is obtained by conjugation and duality induced by the Hodge $\ast$-operator associated to a given Hermitian metric.


Moreover, we have $$h^{1, \,1}\_{\textrm{BC}}(S)=1 \quad (\clubsuit)$$
where $h^{\bullet, \, \bullet}\_{\textrm{BC}}$ denotes the dimension of Bott-Chern cohomology group $H^{\bullet, \, \bullet}\_{\textrm{BC}}$, namely $$H^{\bullet, \, \bullet}\_{\textrm{BC}}(S) = \frac{\ker \partial \cap \ker \bar{\partial}}{\mathrm{im}\, \partial \bar{\partial}} .$$


Now, $(\clubsuit)$ means that there exists a $(1, \, 1)$-form $\omega$ on $S$ which is both $\partial$-closed and $\bar{\partial}$-closed, but *not* $\partial \bar{\partial}$-exact. 


On the other hand, $(\spadesuit)$ implies that any $\partial$-closed (resp. $\bar{\partial}$-closed) $(1,\, 1)$-form on $S$ is actually $\partial$-exact (respectively, $\bar{\partial}$-exact).


Summing up, $\omega$ is a $(1, \, 1)$-form on $S$ which is both $\partial$-exact and $\bar{\partial}$-exact, but not $\partial \bar{\partial}$-exact.


