Vanishing of Nijenhuis tensor given complex linearity? I believe this is a very simple question but I do get stuck here. Given the assertion that Lie bracket is complex linear for $v\to[v,w]$ (i.e. commutes with almost complex structure $J$), how can I show that the Nijenhuis tensor $\mathcal{N}(X,Y)=[JX,JY]-J[JX,Y]-J[X,JY]-[X,Y]$ vanishes?
I did in the following way but did not see where it goes wrong:
$$
[X,JY]=-[JY,X]=-J[Y,X]=J[X,Y]=[JX,Y]
$$
where the first and third equalities use anticommutativity of brackets, second and fourth use complex linearity. 
I saw this Proof that the Nijenhuis tensor vanishes in a complex manifold before, but still do not quite understand the calculation. Thanks in advance.
 A: First of all, $N_J$ is $J$-antilinear in the first argument:
\begin{align*}
N_J(JX, Y) &= [J^2X, JY] - J[J^2X, Y] - J[JX, JY] - [JX, Y]\\
&= -[X, JY] + J[X, Y] - J[JX, JY] - [JX, Y]\\
&= -J[JX, JY] - [JX, Y] - [X, JY] + J[X, Y]\\
&= -J([JX, JY] - J[JX, Y] - J[X, JY] - [X, Y])\\
&= -JN_J(X, Y).
\end{align*}
Now if $N_J$ is assumed to be $J$-linear in the first argument, we have for any $X$ and $Y$
$$JN_J(X, Y) = N_J(JX, Y) = -JN_J(X, Y)$$
so $JN_J(X, Y) = 0$ and hence $N_J(X, Y) = 0$.

It is not necessary for the question, but note that $N_J$ is skew-symmetric:
\begin{align*}
N_J(Y, X) &= [JY, JX] - J[JY, X] - J[Y, JX] - [Y, X]\\
&= -[JX, JY] + J[X, JY] + J[JX, Y] + [X, Y]\\
&= -([JX, JY] - J[X, JY] - J[JX, Y] - [X, Y])\\
&= -N_J(X, Y).
\end{align*}
Therefore, $N_J$ is actually $J$-antilinear in both arguments:
$$N_J(X, JY) = -N_J(JY, X) = JN_J(Y, X) = -JN_Y(X, Y).$$
Likewise, if $N_J$ is assumed to be $J$-linear in one argument, it is automatically $J$-linear in the other argument as well.
A: From $[JX, Y]=[X, JY]$, by replacing $X$ with $JX$, you get
$$-[X,Y]=[-X, Y]=[JX, JY]$$
so
$$N(X,Y)=2[JX, JY]-2J[JX,Y]\;.$$
But
$$J[JX,Y]=-J[Y,JX]=-[JY,JX]=[JX,JY]\;.$$
