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Extend $J$ complex-linearly to the complexification of a real vector bundle $T_{\mathbb{C}}X=TX{\otimes }_{\mathbb{R}}\mathbb{C}$. Since $J^2=-\mathrm{id}$, its $+i$ and $-i$ eigenbundles have smooth projectionsEach has complex rank $n$: complex conjugation interchanges them, and together they have rank $2n$. These are the type decomposition of the complexified tangent bundle. For an arbitrary almost complex structure, they are smooth complex vector bundles; a holomorphic vector bundle structure requires integrability.

When $J$ comes from a holomorphic atlas, write a holomorphic coordinate as $z_j=x_j+i{y}_j$. Its Wirtinger derivatives areThe induced almost complex structure has $J{\partial }_{x_j}={\partial }_{y_j}$ and $J{\partial }_{y_j}=-{\partial }_{x_j}$. Hence the displayed vector fields are respectively $+i$ and $-i$ eigenvectors. They are linearly independent, and each collection has $n$ elements. Thus they give local frames for $T^{1,0}X$ and $T^{0,1}X$ respectively, with $T^{1,0}X$ the holomorphic tangent bundle. The repeated $T^{0,1}X$ in the first sentence of the printed item must be read as the two complementary eigenbundles.

Use the complex-bilinear extension of the Nijenhuis tensor expression to complex vector fields. If $U,V\in T^{1,0}X$, then $JU=iU$ and $JV=iV$, soFor $U,V\in T^{0,1}X$, the same substitution with $-i$ gives $N(U,V)=-8[U,V{]}^{1,0}$. For inputs of opposite types, the two leading Lie brackets cancel, as do the two terms involving $J$, so $N(U,V)=0$. Applying the type decomposition of the complexified tangent bundle to both inputs consequently givesThus $N=0$ precisely when the Lie bracket of any two sections of either eigenbundle remains in that same eigenbundle. In other words, $N=0$ if and only if both tangent-type distributions are involutive distributions. This proves the equivalence directly and does not assume that an arbitrary almost complex structure already has holomorphic coordinates.

We use complex-valued smooth differential forms. The sheaf of smooth differential forms is ${\mathcal{A}}^k(U)=\Gamma (U,{\bigwedge }^k(T^{\ast }X\otimes \mathbb{C}))$, with the usual restriction maps. Dualizing the type decomposition of the complexified tangent bundle and taking exterior powers decomposes this bundle into the summandsTheir smooth sections form the sheaf of differential forms of type (p, q) ${\mathcal{A}}^{p,q}$. Locally a section is a sum of $f_{I,J}d{z}_I\wedge d{\bar{z}}_J$ with $|I|=p$, $|J|=q$ and smooth coefficients. Holomorphic transition maps preserve types, so the local decompositions agree globally. HenceThe exterior derivative splits as $d=\partial +\bar{\partial }$, with bidegrees $(1,0)$ and $(0,1)$. Its square being zero gives ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$. The Dolbeault cohomology is thereforeForms in negative or out-of-range bidegrees are understood to be zero.

On a complex manifold, this sheaf consists of smooth differential forms of type (p, q). The type decomposition of the complexified tangent bundle gives ${\mathcal{A}}^k={⨁}_{p+q=k}{\mathcal{A}}^{p,q}$.