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

Question. Is $\omega$ also $\partial\bar\partial$-exact?

Based on fabulous David Speyer's answer to this MO question I suspect that the answer is elementary or very hard.


1 Answer 1


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, 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.

  • $\begingroup$ Beautiful reasoning. Actually Angella's work led to this question. I read one of his paper and I found Bott-Chern/Aeppli cohomology interesting. $\endgroup$ Feb 7, 2017 at 16:33

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