Question about Nijenhuis tensor In Mcduff&Salamon's Book, they assume $\partial_X Y=\sum_j \xi_j \frac{\partial Y}{\partial x_j}$, when $X=\sum_j \xi_j \frac{\partial }{\partial x_j}$.
There is a conclusion I can not understand is that $\partial_X (JY)=(\partial_XJ)Y+J\partial Y$ (Leibniz's rule in some sense). But what is the definition of $(\partial_XJ)$ in $(\partial_XJ)Y$.
Another question is, after complicated calculations, we have
$N_J(X,Y)=(\partial_{JX}J)Y-(\partial_{JY} J)X+(\partial_X J)JY-(\partial_Y J)JX$.
They stated that $N_J(.,).$ is function linear, I do not know why since I do not know the definition of $(\partial_XJ)$ in $(\partial_XJ)Y$.
 A: In my version of the book (third edition), they make it clear that $\partial_XY$ is just compact notation for the local expression $\sum_j \xi_j \frac{\partial Y}{\partial x_j}$. They don't explicitly define $\partial_X J$, but I think you should infer that it's essentially defined by the expression
$$
\partial_X(JY) = (\partial_XJ)Y+J\partial_XY.
$$
In local coordinates, with $X=\sum_i\xi^i\frac{\partial}{\partial x^i}$, $Y=\sum_i\eta^i\frac{\partial}{\partial x^i}$, the $i$th compenent of the LHS is
$$
\sum_j\xi^j\frac{\partial (JY)^i}{\partial x^j} = \sum_{jk}\xi^j\frac{\partial (J^i_{\ k}\eta^k)}{\partial x^j} = \sum_{jk}\xi^j\left(\frac{\partial J^i_{\ k}}{\partial x^j}\eta^k+J^i_{\ k}\frac{\partial \eta^k}{\partial x^j}\right).
$$
For this to equal the $i$th compenent of the right hand side, it must be that as a linear operator in local coordinates
$$
(\partial_XJ)^i_{\ k} = \sum_j\xi^j\frac{\partial J^i_{\ k}}{\partial x^j}.
$$
This doesn't mean $\partial_XJ$ is a tensor, i.e. that it has an invariant geometric meaning - it's just an expression in local coordinates.
I have to say, though, this seems like a really unenlightening way to prove $N_J$ is a tensor (i.e. satisfies $N_J(fX, Y) = fN_J(X,Y)$, and similarly for the $Y$ slot). You can prove this much more directly by using the result that
$$
[fX,Y] = f[X,Y] - (Yf)X,
$$
the tensorial nature of $J$, and the fact that $J^2=-1$.
