Let $X$ be an almost complex manifold. A well-known result says that $\bar{\partial}^{2}=0$ implies the integrability of the almost complex structure. My question is what about $\partial\bar{\partial}+\bar{\partial}\partial=0$? Does it imply the integrability?
I have proved that $\partial\bar{\partial} f+\bar{\partial}\partial f=0$ always holds for any function $f$, but does it hold for any form?
The existence of an almost complex structure $J$ implies that one can decompose the space of $1$-forms into forms of type $(1,0)$ and of type $(0,1)$. If $\varepsilon^k$ defines a local frame for $\Lambda^{1,0}$ (the bundle of $(1,0)$-forms), then $$d\varepsilon^k = A_{ji}^k \varepsilon^j \wedge \varepsilon^i + B_{j\overline{i}}^k \varepsilon^j \wedge \overline{\varepsilon}^i + C_{\overline{j} \overline{i}}^k \overline{\varepsilon}^j \wedge \overline{\varepsilon}^i.$$ The almost complex structure is integrable if and only if $C_{\overline{j} \overline{i}}^k =0$ for all $i,j,k$. This is equivalent to $\bar{\mu} = \mu =0$, where the exterior derivative decomposes as $d = \mu + \overline{\mu} + \partial + \bar{\partial}$.
Hence, the question is whether $\partial \bar{\partial} + \bar{\partial} \partial$ implies that $\mu = 0$. I do not think this is true, but I'll give it some more thought.
In the meantime, this article is a nice read: https://www.sciencedirect.com/science/article/pii/S0001870821004096?ref=cra_js_challenge&fr=RR-1