TITLE: Exterior derivative on almost complex manifolds
QUESTION [6 upvotes]: 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.

REPLY [9 votes]: Just to follow up on Eric's correct answer: when you have an almost complex structure $J$, you can decompose $1$-forms into type $(1,0)$ and $(0,1)$. Locally, you can find a local basis $e^1, \ldots, e^n$ of $(1,0)$-forms, but these are not of the form $dz^1, \ldots, dz^n$. Indeed, as Eric mentioned, we do not have local holomorphic coordinates. Then $\bar e^1, \ldots, \bar e^n$ are a local basis of $(0,1)$ forms. Now if we compute $de^i$, it is a $2$-form, so it can be written in the form
\begin{equation*}
de^i = a^i_{jk} e^j \wedge e^k + b^i_{jk} e^j \wedge \bar e^k + c^i_{jk} \bar e^j \wedge \bar e^k.
\end{equation*}
The almost complex structure $J$ is integrable if and only if all the $c^i_{jk}$'s are zero.