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Upon reading this answer to this question, the last paragraph mentions the following. "Requiring the [almost complex] structure to be integrable corresponds to a certain PDE for this map." The map in question, I presume, can be taken to be some $f: M \to Gr(3, \mathbb{C}^\infty)$ for the almost complex $6$-dimensional manifold $M$, which corresponds to $TM$ by pullback of the tautological bundle.

I'm no expert in these topics so I am struggling to unpack this claim. What is this PDE and how do we obtain it? Furthermore, is there a reference to Yau's mentioned program or a survey of attempts/results in that direction?

EDIT

I've emailed Professor Yau and to my delight he responded. Here is what I've learned. The map $f$ above determines an almost complex structure $J$ on $TM$. Then, as Moishe mentioned in the comments, we have that the Newlander-Nirenberg theorem gives the integrable condition through the vanishing of the Nijenhuis tensor $N_J$, which takes as inputs two vector fields $X,Y$. In particular, we may write $$ N_J : \Gamma(TM) \times \Gamma(TM) \to \Gamma(TM) $$ which is the partial differential operator in question. Yau mentions that $N_J$ can be written in terms of the differential of $f$. This part eludes me.

So my updated question is, how can $N_J$ be written in terms of $df$?

EDIT 2

Perhaps the PDE in question relating to the Nijenhuis tensor is different than I explained, and it's "simply" a question of working locally. I'll try to sketch out the idea here. First, we have

$$ N_J (X,Y) = [X,Y] + J([JX,Y]+[X,JY]) - [JX,JY], $$

and writing it in local coordinates, this gives

$$ X = \sum_{i=1}^{n} X^i \partial_i, ~~~~ Y = \sum_{i=1}^{n} Y^i \partial_i ,$$

$$ N_J (X,Y) = \sum_{i=1}^{n} \sum_{j=1}^{n} \Big[ (X^j \partial_j Y^i - Y^j \partial_j X^i)\partial_i $$ $$ +J \Big( ((JX)^j \partial_j Y^i - Y^j \partial_j (JX)^i) \partial_i \Big) $$ $$ +J \Big( (X^j \partial_j (JY)^i - (JY)^j \partial_j X^i) \partial_i \Big) $$ $$ -\Big( (JX)^j \partial_j (JY)^i - (JY)^j \partial_j (JX)^i \Big)\partial_i \Big]. $$

The first and last term of the big equation are in terms of $\partial_i$ while the two middle terms are in terms of $J(\partial_i)$. Assuming that $\partial_i$ and $J(\partial_i)$ are linearly independent for all $i$, setting $N_J(X,Y) = 0$ gives us a system of two first order PDEs with $2n$ variables $X^i, Y^i,$ namely

$$ \sum_j (X^j \partial_j Y^i - Y^j \partial_j X^i) - ( (JX)^j \partial_j (JY)^i - (JY)^j \partial_j (JX)^i ) = 0$$ and $$ \sum_j ((JX)^j \partial_j Y^i - Y^j \partial_j (JX)^i) + (X^j \partial_j (JY)^i - (JY)^j \partial_j X^i) = 0. $$

We can (hopefully) solve this locally and then use a partition of unity argument to look for a global set of solutions.

The question still stands for how $N_J$ (or in particular the linear $J$) can be written in terms of $df$, but does the above make sense?

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    $\begingroup$ The PDE is the vanishing condition for the Nijenhuis tensor of the almost complex structure $J$. $\endgroup$ Commented Jul 9, 2022 at 5:58
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    $\begingroup$ @MoisheKohan How is that PDE related to the function $f$? $\endgroup$ Commented Jul 9, 2022 at 16:57
  • $\begingroup$ I do not know about your map, I was talking about a map $J: M\to C(M)$, which is a section of the bundle $E\to M$, whose fibers at $x\in M$ consist of almost complex structures on $T_xM$. From my reading the linked answer, it refers to the same map $J$. I cannot think of any other interpretation. $\endgroup$ Commented Jul 9, 2022 at 17:03
  • $\begingroup$ @MoisheKohan I see, perhaps that makes more sense. I'm curious why the linked answer mentions a map into a Grassmannian now $\endgroup$ Commented Jul 9, 2022 at 17:13
  • $\begingroup$ Then you should ping (or directly via email) Joel Fine and ask him. $\endgroup$ Commented Jul 9, 2022 at 17:15

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