A complex top form determine complex structure Let $M$ is a $2n$-dimensional real manifold and $\Omega\in \Lambda^{n}(T^{*}M\otimes \mathbb{C})$ be a complex top form. How does one determine a complex structure on the tangent space $TM$?
 A: Not every complex $n$-form on $M$ defines an almost complex structure. If $\Omega$ is such a form, it has to satisfy two additional conditions:


*

*$\Omega$ is decomposable, meaning that in a neighborhood of each point it can be written in the form $\Omega = \zeta^1 \wedge\dots \wedge \zeta^n$ for complex-valued $1$-forms $\zeta^1,\dots,\zeta^n$; and

*$\Omega\wedge \overline\Omega$ is nowhere zero.


Assuming these two conditions are satisfied, we can define complex subbundles $T'M,\ T''M \subseteq TM\otimes\mathbb C$ by 
\begin{align*}
T'M &= \lbrace v\in TM\otimes \mathbb C: v\ \lrcorner\  \overline\Omega  =0\rbrace,\\
T''M &= \overline{T'M} = \lbrace v\in TM\otimes \mathbb C: v\ \lrcorner\ \Omega  =0\rbrace.
\end{align*}
I'll show below that that $TM\otimes \mathbb C = T'M \oplus T''M$, so we can 
define the almost complex structure $J\colon TM\otimes C\to TM\otimes C$ by 
$$J(v \oplus w) = iv - i\overline w$$ 
for $v\in T'M$, $w\in T''M$. It is immediate from this definition that $J$
takes real vectors to real vectors and $J^2 = - \operatorname{Id}$, so it is an almost complex structure.
To see that this makes sense, we need to check that 
${TM \otimes \mathbb C = T'M\oplus T''M}$. 
Let $p\in M$ be arbitrary and choose complex $1$-forms $\zeta^1,\dots,\zeta^n$ such that $\Omega = \zeta^1\wedge\dots\wedge \zeta^n$ in a neighborhood of $p$. The fact that $\Omega \wedge \overline\Omega$ is nowhere zero guarantees that $(\zeta^1,\dots,\zeta^n,\overline {\zeta^1},\dots,\overline {\zeta^n})$ forms a local coframe for $T^*M\otimes \mathbb C$. Let $(Z_1,\dots,Z_n,\overline Z_1,\dots \overline Z_n)$ be the dual frame for $TM\otimes\mathbb C$. Then it is easy to check that $T'M = \operatorname{span}(Z_1,\dots,Z_n)$ and $T''M = \operatorname{span}(\overline Z_1,\dots \overline Z_n)$, from which the claim follows.
Conversely, if $M$ is endowed with an almost complex structure, we can just take $\Omega$ (locally, at least) to be any nonvanishing $(n,0)$-form. 
Of course, this only defines an almost complex structure. If you like this approach, I invite you to check that the almost complex structure is integrable if and only if $\Omega$ satisfies the following condition:
$$
\mathscr L_V \Omega \text{ is a function times $\Omega$ whenever $V$ is a smooth section of $T''M$}.
$$
In particular, this is the case if $\Omega$ is closed.
