Does $F_{A}^{0,2}=0$ for a connection $A$ on $TM$ almost complex give a complex structure?

Question

First, some motivation. Let $X$ be a complex manifold, and $A$ a Hermitian connection on some complex vector bundle $E$ over $X$. It is known that the existence of $A$ such that the $(0,2)$-part of the curvature form $F_A$ vanishes implies the existence of a holomorphic structure on $E$.

Now let $M$ be an almost complex manifold, and $A$ a Hermitian connection on the tangent bundle $TM$.

Does $F_{A}^{0,2}=0$ imply the existence of a complex structure on $M$?

After discussing this with my professor, he believed that this might not be the case, but I've yet to find a counterexample. The only place one would know to look so far is on an almost complex $4$-manifold that does not admit a complex structure. Otherwise, if the result turns out to be true, is there a standard reference on this matter?

Answer 1

Here is an example to think about: Let $S^6 = \mathrm{G}_2/\mathrm{SU}(3)$ be the $6$-sphere endowed with its $\mathrm{G}_2$-invariant almost Hermitian structure. There is a $\mathrm{G}_2$-invariant (special) Hermitian connection $A$ on the tangent bundle of $S^6$ whose curvature is of type $(1,1)$, but the only $\mathrm{G}_2$-invariant almost-complex structures on $S^6$ (and there are exactlty two) are not integrable.

The point is that the (special) Hermitian connection $A$ has torsion of type $(0,2)$, and that is what prevents the almost-complex structure from being integrable.

Of course, there might still exist some complex structure on $S^6$ (whether this is so is not currently known), but the existence of an Hermitian connection on its tangent bundle with curvature of type $(1,1)$ has proved of no help, so far, in finding a complex structure on $S^6$.

Here is an actual counterexample: S. T. Yau has shown that there are compact parallelizable 4-manifolds $M^4$ that support no complex structure. See S. T. Yau, Parallelizable manifolds without complex structure, Topology 15 (1976), 51–53. Since the tangent bundle is trivial, there is an almost-complex structure $J$ on $M$ for which the tangent bundle of $M$ is trivial as a complex vector bundle. Hence, the tangent bundle of $M$ supports a flat $J$-complex connection, i.e., a connection $A$ for which the complex structure is parallel and $F_A = 0$. A fortiori, we have $F^{0,2}_A = 0$, but $M$ does not admit a complex structure. Of course, the torsion of $A$ is non-zero and, in fact, must have a nontrivial $(0,2)$-piece.