Nijenhuis tensor

Remarks

The map $N_J$ is $C^\infty(M)$-bilinear and antisymmetric, so it is a tensor field of type $(1,2)$, a $2$-form with values in $TM$. It equals $\frac{1}{2}[J, J]$, where $[J, J]$ is the Frölicher–Nijenhuis bracket of $J$ with itself. For an almost complex structure, where $J^2 = -\id$, the formula reads $N_J(X, Y) = [JX, JY] - J[JX, Y] - J[X, JY] - [X, Y]$. Some authors use a different constant factor.

Example: on $\R^2$ let $J \partial_x = \partial_y$ and $J \partial_y = -\partial_x$. For $X, Y$ among $\partial_x, \partial_y$, all of $X, Y, JX, JY$ are constant combinations of $\partial_x$ and $\partial_y$, so all brackets vanish. By bilinearity over functions, $N_J = 0$. More generally, for an almost complex structure one has $N_J(JX, Y) = -J N_J(X, Y)$, so $N_J(JX, X) = 0$; on a $2$-manifold $X$ and $JX$ span each tangent space where $X \ne 0$, hence $N_J = 0$ on every surface.

Non-example: the expression $[JX, JY] - [X, Y]$ alone is not a tensor. Replacing $X$ by $fX$ for a function $f$ gives $f([JX, JY] - [X, Y]) - (JY)(f)\, JX + Y(f)\, X$, since $[fA, B] = f[A, B] - B(f) A$. The last two terms are not zero in general, so the expression is not $C^\infty(M)$-linear in $X$; the two middle terms of $N_J$ cancel them.