Vanishing of Nijenhuis tensor given complex linearity?

Question

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.

Answer 1

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 $$J[JX,Y]=-J[Y,JX]=-[JY,JX]=[JX,JY]\;.$$

Answer 2

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 $N_J(X, Y) = 0$.

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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.