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An almost complex structure is a smooth real vector bundle endomorphism $J:TM\to TM$ satisfying $J^2=-I$. In particular, $M$ has even real dimension $2n$. Extend $J$ complex-linearly to $T_{\mathbb{C}}M=TM{\otimes }_{\mathbb{R}}\mathbb{C}$. Its minimal polynomial divides $(u-i)(u+i)$, whose roots are distinct, so the type decomposition of the complexified tangent bundle isThe smooth projections are $(I-iJ)/2$ and $(I+iJ)/2$. Complex conjugation exchanges the two rank-$n$ eigenbundles. A $(1,0)$-covector annihilates $T^{0,1}M$ and a $(0,1)$-covector annihilates $T^{1,0}M$. Thus the differential forms of type (p, q) are the smooth sections ofHere the tensor product is identified with its image under the wedge product of differential forms. For a real tangent vector $v$, the vectors $v-iJv$ and $v+iJv$ exhibit the two types directly.

We now construct the almost complex structure from a decomposable volume form. Locally write $\Omega ={\theta }_1\wedge \cdots \wedge {\theta }_n$. The nonvanishing of $\Omega \wedge \overline{\Omega }$ says that the $2n$ covectors ${\theta }_1,\dots ,{\theta }_n,{\overline{\theta }}_1,\dots ,{\overline{\theta }}_n$ form a complex coframe. In particular, the span of the ${\theta }_j$ has the intrinsic descriptionIndeed, expansion in that coframe shows that every coefficient of a conjugate factor must vanish if $\alpha \wedge {\Omega }_x=0$. Therefore this span is independent of the chosen local factorization. The local coframes show that $W$ is a smooth rank-$n$ vector subbundle, andDefine $J_{\Omega }^{\ast }$ to be multiplication by $i$ on $W$ and by $-i$ on $\overline{W}$. It commutes with complex conjugation, so is the complexification of a real smooth bundle endomorphism, whose dual defines $J_{\Omega }$ on $TM$. It has square $-I$, and its $(1,0)$-covectors are exactly $W$. Conversely, any such almost complex structure must have this same action on the entire coframe. Thus $J_{\Omega }$ exists globally and is unique, and every factor in every allowed decomposition is a $(1,0)$-form.

Finally suppose $d\Omega =0$. Since ${\theta }_j\wedge \Omega =0$, the exterior derivative product rule givesDecompose $d{\theta }_j$ into types $(2,0)$, $(1,1)$ and $(0,2)$ using $J_{\Omega }$. The first two wedge products with the $(n,0)$-form $\Omega$ vanish by dimension. On $(0,2)$-forms, wedging with $\Omega$ is injective: the monomials $\Omega \wedge {\overline{\theta }}_a\wedge {\overline{\theta }}_b$ are linearly independent. For $n=1$ the $(0,2)$ space is zero, so the statement still holds. Hence every $d{\theta }_j$ has no $(0,2)$ component. For an arbitrary $(1,0)$-form $\alpha ={\sum }_j{a}_j{\theta }_j$, the extra terms $d{a}_j\wedge {\theta }_j$ likewise have only types $(2,0)$ and $(1,1)$. The bracket and differential-form criteria for integrability proved below now giveLocal decomposability has been used essentially; the nonvanishing top-degree product alone does not provide the required coframe.

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.

An almost complex structure $J$ gives the $+i$ and $-i$ eigenbundles of its complex-linear extension. Their projections are $(I-iJ)/2$ and $(I+iJ)/2$. For an integrable structure the first is the holomorphic tangent bundle.