Holomorphic Frobenius Theorem I'm trying to understand a proof of the Holomorphic Frobenius Theorem using the smooth version as seen in Voisin's Complex Geometry book: (pg 51)
http://www.amazon.com/Hodge-Theory-Complex-Algebraic-Geometry/dp/0521718015
She starts with a holomorphic distribution $E$ of dimension $k$ on a complex manifold which is closed under bracket. So, $[E,E]\subseteq E$. 
Then, to reduce to the real case, we take the real part of the distribution to get another distribution $\Re(E)$ of dimension $2k$ in the real tangent bundle.
What I can't understand is why this real distribution, $\Re(E)$, also satisfies the bracket condition. The books says this follows since $E$ is holomorphic and satisfies the bracket condition. My guess is that a local frame for it is given by the real and imaginary parts of a local frame $E$ but i'm not sure how to proceed from there.
Any help is much appreciated.
 A: I didn't check your link, but Voisin is probably restricting the standard complex-linear isomorphism
$$
X' \leftrightarrow X := \tfrac{1}{2}(X' \otimes 1 - JX' \otimes i) = \tfrac{1}{2}(X' - iJX')
\tag{1}
$$
between the real tangent bundle $(TM, J)$ and the holomorphic tangent bundle $(T^{1,0}M, i)$, a.k.a., the $i$-eigenspace of $J$ acting on the complexified bundle $TM \otimes \mathbf{C}$.
Because $J$ is integrable,
\begin{align*}
[X, Y]
  &= \tfrac{1}{4}[X' - iJX', Y' - iJY'] \\
  &= \tfrac{1}{4}\bigl([X', Y'] - i[JX', Y'] - i[X', JY'] - [JX', JY']\bigr) \\
  &= \tfrac{1}{2}\bigl([X', Y'] - iJ[X', Y']\bigr) \\
  &=: Z,
\end{align*}
the holomorphic vector field corresponding via (1) to $Z' = [X', Y']$. In words, the isomorphism (1) respects the complex bracket. If $E \subseteq T^{1,0}M$ is a holomorphic distribution closed under the bracket, the corresponding real distribution $E' \subset TM$ is also closed under the bracket.
It may also be helpful to read about the Nijenhuis tensor.
