Motivation for the Nijenhuis tensor

Question

I'm learning about complex and almost complex structures on smooth manifolds, in particular the Newlander-Nirenberg theorem. Recall that for an almost complex structure $J$ on a smooth manifold, the Nijenhuis tensor field of $J$ is defined by $$ N_J(X, Y) = [X, Y] + J[JX, Y] + J[X, JY] - [JX, JY], $$ and Newlander-Nirenberg tells us that non-vanishing of $N_J$ is the obstruction to an almost complex structure coming from a complex structure.

I'm wondering if anyone can shed some light on the motivation for this construction: is there any intuitive reason why we'd expect $N_J$ to capture this obstruction? Moreover, are there any good reasons or historical motivation as to why this expression was considered in the first place? In other words, how would one stumble upon this tensor naturally?

Answer 1

Let $J$ be an almost complex structure on the smooth $2n$-dimensional manifold $M$. We can extend $J$ complex linearly to the complexified tangent bundle $TM\otimes_{\mathbb{R}}\mathbb{C}$. Then $TM\otimes_{\mathbb{R}}\mathbb{C}$ decomposes as $TM\otimes_{\mathbb{R}}\mathbb{C} = T^{1,0}M\oplus T^{0,1}M$ where $T^{1,0}M$ and $T^{0,1}M$ are the $i$ and $-i$-eigenspaces of $J$ respectively. Explicitly

\begin{align*} T^{1,0}M &= \{v - iJv \mid v \in TM\}\\ T^{0,1}M &= \{v + iJv \mid v \in TM\}. \end{align*}

Moreover, given any $w \in TM\otimes_{\mathbb{R}}\mathbb{C}$, then $w = w^{1,0} + w^{0,1}$ where $w^{1,0} = \tfrac{1}{2}(w - iJw) \in T^{1,0}M$ and $w^{0,1} = \tfrac{1}{2}(w + iJw) \in T^{0,1}M$.

If $J$ is integrable, and $(U, (z^1, \dots, z^n))$ are holomorphic coordinates, then $\left\{\frac{\partial}{\partial z^1}, \dots, \frac{\partial}{\partial z^n}\right\}$ is a local basis of sections for $T^{1,0}M|_U$ and $\left\{\frac{\partial}{\partial \bar{z}^1}, \dots, \frac{\partial}{\partial \bar{z}^n}\right\}$ is a local basis of sections for $T^{0,1}M|_U$. Extending the Lie bracket complex bilinearly, it follows that sections of $T^{1,0}M$ are closed under Lie bracket, as are sections of $T^{0,1}M$, i.e. $[\Gamma(T^{1,0}M), \Gamma(T^{1,0}M)] \subseteq \Gamma(T^{1,0}M)$ and $[\Gamma(T^{0,1}M), \Gamma(T^{0,1}M)] \subseteq \Gamma(T^{0,1}M)$. Said another way, $\Gamma(T^{1,0}M)$ and $\Gamma(T^{0,1}M)$ are Lie subalgebras of $\Gamma(TM\otimes_{\mathbb{R}}\mathbb{C})$. This provides a necessary condition for integrability.

Returning to the general case, let $X-iJX, Y-iJY$ be local sections of $T^{1,0}M$. Their Lie bracket is

$$[X-iJX, Y - iJY] = [X, Y] -i[JX, Y] -i[X, JY] - [JX, JY],$$

so the $(0, 1)$-part satisfies

\begin{align*} &\ 2[X-iJX, Y - iJY]^{0,1}\\ =&\ ([X, Y] -i[JX, Y] -i[X, JY] - [JX, JY])\\ &\ + iJ([X, Y] -i[JX, Y] -i[X, JY] - [JX, JY])\\ =&\ ([X, Y] + J[JX, Y] + J[X, JY] - [JX, JY])\\ &\ + i(J[X, Y] -[JX, Y] - [X, JY] - J[JX, JY])\\ =&\ N_J(X, Y) + iJ([X, Y] + J[JX, Y] + J[X, JY] - [JX, JY])\\ =&\ N_J(X, Y) + iJN_J(X, Y). \end{align*}

Therefore $\Gamma(T^{1,0}M)$ is closed under Lie bracket if and only if $N_J = 0$. A similar computation shows that $\Gamma(T^{0,1}M)$ is closed under Lie bracket if and only if $N_J = 0$.

In conclusion, $N_J = 0$ is a necessary condition for integrability. The Newlander-Nirenberg Theorem shows that it is also sufficient.

Answer 2

A partial answer, regarding the history:

As far as I know, and this is also claimed in “Nijenhuis geometry” by Bolsinov, Konyaev & Matveev (arXiv:1903.04603 [math.DG]), the Nijenhuis tensor first appeared in Albert Nijenhuis's article “$X_{n-1}$-forming sets of eigenvectors” (Indag. Math. 1951, MR43540), in connection with a different problem, with no mention of complex structures: Suppose the $(1,1)$ tensor field $J$ is diagonalizable at every point. Then the Nijenhuis tensor $N_J$ is zero iff there is a local coordinate system in which $J$ is diagonal.