Let $X$ be some almost complex manifold,with tangent bundle equipped with some complex structure $I$.
Then we know the exterior derivative decomposed into four part $d= \mu + \bar{\mu} + \partial + \bar{\partial}$ with $\mu $ component in $\mathcal{A}^{(p+2,q-1)}$ ,$\bar{\mu}$ component in $\mathcal{A}^{(p-1,q+2)}$.Where $\mathcal{A}^{p,q}$ means the differential form in type $(p,q)$.
I can't figure out the difference between complex manifold and almost complex manifold,we know on complex manifold $d = \partial +\bar{\partial}$.The proof is as follows,since:
$$d(f)=\sum_{i} \frac{\partial f}{\partial x_{i}} d x_{i}+\sum_{i} \frac{\partial f}{\partial y_{i}} d y_{i}=\sum_{i} \frac{\partial f}{\partial z_{i}} d z_{i}+\sum_{i} \frac{\partial f}{\partial \bar{z}_{i}} d \bar{z}_{i}$$
we have $$d(fdz_{I}\wedge d\bar{z}_J) = \sum_k \partial_{z_k}{f}dz_k\wedge dz_{I}\wedge d\bar{z}_J +\sum_l\partial_{\bar{z}_l}{f}d\bar{z}_l\wedge dz_{I}\wedge d\bar{z}_J$$
Hence $d = \partial + \bar{\partial}$.
The question is why almost complex manifold has $\mu$ component?It looks the proof is almost the same?
Best Answer
Recall that when you have an almost complex structure $J$, you can decompose any $1$-forms into type $(1,0)$ and type $(0,1)$ forms. Locally, this gives a basis $e^1,\dots,e^n$ of $\mathcal{A}^{(1,0)}(U)$ but they are not of the form $\mathrm{d}z^1,\dots,\mathrm{d}z^n$ since there is no guarantee their dual $e_i\in \mathfrak{X}^{(1,0)}(U)$ are involutive to let you integrate for the coordinates $z^1,\dots,z^n$. Indeed, if you decompose the $2$-form $\mathrm{d}e^i$ into its $(2,0)$-, $(1,1)$- and $(0,2)$-components $$ \mathrm{d}e^i=a^i_{jk} e^j\wedge e^k+b^i_{j\bar{k}} e^j\wedge\overline{e^k}+c^i_{\bar{j}\bar{k}} \overline{e^j}\wedge\overline{e^k} $$ By Frobenius's theorem, $J$ is integrable iff $c^i_{\bar{j}\bar{k}}=0$ for all $i,j,k$, or equivalently $\bar{\mu}=0$ (so $\mu=0$ too).