Question about the Newlander-Nirenberg theorem for almost complex manifolds

Question

While studying the definition of almost complex structure, I have thought about almost complex structures which are not complex structures.

To ask a more complete question, I write the following commutative diagram which exists for almost complex manifold $(M,J)$: $\require{AMScd}$ \begin{CD} T_pM @>{\varphi}>> T_pM \otimes \mathbb{C} \simeq T_{p}^{1,0}M \oplus T_{p}^{0,1}M \\ @VVV \swarrow \pi_{1} \\ T_{p}^{1,0}M \end{CD}

for all $p \in M$, where $T_{p}^{1,0} = \operatorname{Span}\left\{ \partial/\partial z_i \right\}$, and $T_{p}^{0,1} = \operatorname{Span}\left\{ \partial/\partial \overline{z}_i \right\}$.

To begin with, Wikipedia referred to the following fact: "The converse question, whether the almost complex structure implies the existence of a complex structure is much less trivial, and not true in general". (The following example is a counterexample, but now I am not interested in such a counterexample.)

Anyway, according to the lecturer, generally, if $X,Y$ is contained in $TM$, then the Lie bracket $[X, Y]$ is also in $TM$. In case of almost complex manifold, however, the Lie bracket $[X,Y]$ is not necessarily an element of $T^{1,0}M$ if $X,Y \in T^{1,0}M$. Thus, the lecturer said the almost complex structure $(M,J)$ is called a complex structure if the Lie bracket is also in $T^{1,0}M$. Remarkably, we have the Newlander-Nirenberg theorem.

Newlander-Nirenberg Theorem: Let $(M,J)$ be an almost complex manifold. The following are equivalent:

$1.$ for all $X,Y \in T^{1,0}M$, $[X,Y] \in T^{1,0}M$,

$2.$ for all $X,Y \in T^{0,1}M$, $[X,Y] \in T^{0,1}M$,

$3.$ $d=\partial+ \overline{\partial}$ (so-called $J$ is integrable),

$4.$ $\overline{\partial}^2 = 0$,

$5.$ $N \equiv 0$ where $N(X,Y):=[JX,JY]-J[X,Y]-J[JX,Y]-J[X,JY]$ (the tensor $N$ is called the Nijenhuis tensor).

The most curious point is the statement of $3$ (or $4$). Shouldn't it always hold for any $(M,J)$, even if it is not a complex manifold? For example, according to the Newlander-Nirenberg theorem, if a manifold is an almost complex manifold but not a complex manifold, then $\overline{\partial}^2 \neq 0$. However, is it possible for $\overline{\partial}^2 \neq 0$ to hold? Does the non-equality violate the Dolbeault cohomology?

Note: The Newlander-Nirenberg Theorem I described is dependent on the lecturer (on Youtube). Thus, I do not believe thoroughly that the statement is the right one. Actually, I corrected some statements via google search. If the answerer decides that statement is not right, please rectify the statement.

Answer 1

Your definition of $T^{1,0}_p$ and $T^{0,1}_p$ only makes sense on a complex manifold: on an almost complex manifold, you do not have a notion of holomorphic coordinates $z^1, \dots, z^n$. In general, you can define these spaces as $i$ and $-i$-eigenspaces of $J$ respectively (where $J$ acts on $TM\otimes_{\mathbb{R}}\mathbb{C}$ by extending complex-linearly).

A smooth $(p, q)$-form $\alpha$ can locally be written as a sum of terms of the form $f\, \eta^{i_1}\wedge\dots\wedge\eta^{i_p}\wedge\bar{\eta}^{j_1}\wedge\dots\wedge\bar{\eta}^{j_q}$ where $f$ is a smooth function, $\eta^i$ are smooth $(1, 0)$-forms (sections of $(T^{1,0}M)^*$), and $\bar{\eta}^j$ are smooth $(0,1)$-forms (sections of $(T^{0,1}M)^*$); if $J$ is integrable, then there are holomorphic coordinates $z^1, \dots, z^n$ and we can take $\eta^i = dz^i$ and $\bar{\eta}^j = d\bar{z}^j$. Since $d$ satisfies the Leibniz rule (i.e. $d(\beta\wedge\gamma) = d\beta\wedge\gamma + (-1)^{|\beta|}\beta\wedge d\gamma$) and

\begin{align*} df &\in \mathcal{E}_{\mathbb{C}}^1(M) = \mathcal{E}^{1,0}(M)\oplus\mathcal{E}^{0,1}(M)\\ d\eta^i &\in \mathcal{E}_{\mathbb{C}}^2(M) = \mathcal{E}^{2,0}(M)\oplus\mathcal{E}^{1,1}(M)\oplus\mathcal{E}^{0,2}(M)\\ d\bar{\eta}^j &\in \mathcal{E}_{\mathbb{C}}^2(M) = \mathcal{E}^{2,0}(M)\oplus\mathcal{E}^{1,1}(M)\oplus\mathcal{E}^{0,2}(M), \end{align*}

it follows that $d\alpha \in \mathcal{E}^{p+2,q-1}(M)\oplus\mathcal{E}^{p+1,q}(M)\oplus\mathcal{E}^{p,q+1}(M)\oplus\mathcal{E}^{p-1,q+2}(M)$. We define

\begin{align*} \mu &: \mathcal{E}^{p,q}(M) \to \mathcal{E}^{p+2,q-1}(M)\\ \partial &: \mathcal{E}^{p,q}(M) \to \mathcal{E}^{p+1,q}(M)\\ \bar{\partial} &: \mathcal{E}^{p,q}(M) \to \mathcal{E}^{p,q+1}(M)\\ \bar{\mu} &: \mathcal{E}^{p,q}(M) \to \mathcal{E}^{p-1,q+2}(M) \end{align*}

to be the projections of $d\alpha$ onto the four summands, so $d = \mu + \partial + \bar{\partial} + \bar{\mu}$. Note that $\mu$ is zero if and only if $d\eta^i$ has no $(0, 2)$-component, and $\bar{\mu}$ is zero if and only if $d\bar{\eta}^j$ has no $(2, 0)$-component; this is the case if $J$ is integrable, $\eta^i = dz^i$, and $\bar{\eta}^j = d\bar{z}^j$, in which case $d = \partial + \bar{\partial}$. So if $J$ is integrable, the condition $d^2 = 0$ is equivalent to

\begin{align*} \partial^2 &= 0\\ \partial\bar{\partial} + \bar{\partial}\partial &= 0\\ \bar{\partial}^2 &= 0. \end{align*}

On the other hand, if $J$ is non-integrable, the condition $d^2 = 0$ is equivalent to

\begin{align*} \mu^2 &= 0\\ \partial\mu + \mu\partial &= 0\\ \partial^2 + \bar{\partial}\mu + \mu\bar{\partial} &= 0\\ \partial\bar{\partial} + \bar{\partial}\partial + \mu\bar{\mu} + \bar{\mu}\mu &= 0\\ \bar{\partial}^2 + \partial\bar{\mu} + \bar{\mu}\partial &= 0\\ \bar{\partial}\bar{\mu} + \bar{\mu}\bar{\partial} &= 0\\ \bar{\mu}^2 &= 0. \end{align*}

In particular, we do not have $\partial^2 = 0$ or $\bar{\partial}^2 = 0$. There is no contradiction with the link you provide because, as pointed out there, they define Dolbeault cohomology for complex manifolds (which satisfy $\bar{\partial}^2 = 0$), not almost complex manifolds.

Finally, let me assure you that the five conditions you list are indeed equivalent. For example, see this answer for a proof of the fact that $1$ (or $2$) is equivalent to $5$. As shown above, $3$ implies $4$. Some careful computations show that $1$ implies $3$ and $4$ implies $2$, which completes the proof. However, that is not the statement of the Newlander-Nirenberg Theorem which instead states that $N \equiv 0$ if and only if there are local coordinates $(z^1, \dots, z^n)$ such that $J\frac{\partial}{\partial z^i} = i\frac{\partial}{\partial z^i}$ and $J\frac{\partial}{\partial \bar{z}^j} = -i\frac{\partial}{\partial \bar{z}^j}$; these coordinates are the holomorphic coordinates of a complex structure. Unlike the equivalence of the five statements in your post, the proof of the Newlander-Nirenberg Theorem is not just an application of the definitions, it is much more difficult.