Let $M$ be a complex manifold, and $\omega$ be a $(p,q)$-form. Then $d\omega$ is an element of $\Omega^{p+1,q}(M)\oplus\Omega^{p,q+1}(M)$, so that $d = \partial + \overline{\partial}$, where $\partial$ and $\overline{\partial}$ are the Dolbeault operators.


Now let $M$ be *almost* complex. It is commonly stated that $d = \partial + \overline{\partial}$ only holds for complex manifolds, and not for almost complex manifolds. But why is this? After extending $d$ to also be complex linear, if $\omega = \sum\_i f(z)dz^i$ is a $(0,1)$-form, I'd say that we would have $ d\omega = \sum\_i df\wedge dz\_i 
= \sum\_{i,j} \frac{\partial f}{\partial z^j}dz^j\wedge dz^i + \frac{\partial f}{\partial \overline{z}^j}d\overline{z}^j\wedge dz^i,$
which clearly does not have a $(0,2)$-part. Why is this wrong?


On the other hand, let $X, Y$ be antiholomorphic tangent vectors, then $d\omega(X,Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y]) = -\omega([X,Y])$. Since $M$ is not nessecarily complex, $[X,Y]$ is not nessecarily also antiholomorphic, so that this term does not nessecarily vanish. But $d\omega$, being a 2-form, can only give a nonzero result if it has a $(0,2)$-part. So from this I can see that it has to have one, but I can't see why this contradicts the calculation of $d\omega$ above.



Exterior derivative on almost complex manifolds
In writing $\omega$ you used a symbol $dz$ which doesn't make sense unless there is a holomorphic coordinate. Your $dz$ should really be an element of a frame of (1,0) 1-forms, which need not be closed (as you have assumed).


