show/hide this revision's text 3 This post is superseded by my newer post.

A (local) leaf $A_x \owns x$ parametrizes a family $B_y, y\in A_x$of the (local) leaves it intersects, and the foliation $A$ near $A_x$ induces a map of $A_x\to {\cal J}$ where ${\cal J}$ is the space of
complex structures on the (local) leaf $B_x$. (we just pullback restrictions of $J$ to $B_y$ by a flow in $A$)

But ${\cal J}$ itselfhas a complex structure, (as in the theory of moduli spaces).Claim: If your original $J$ is integrable then $M: A_x\to {\cal J}$ is holomorphic.

This should be straightforward to check I think.

ie this is a necessary condition.It's clearly a nontrivial condition, as we can choose $M$ quite freely.

Furthermore, if this is true for an open set of leaves $A_z, z\in B_x$,-- as well as the same property with $A,B$ switched,then it should imply integrability of $J$ on the associated nbhd,

ie this is sufficient.

One suggestion to verify this without a formal Nijenhuis tensor calculationwould be to use $M$ to reconstruct a foliation in a chart of $C^N, N=2n$,which realizes such an $M$, then to observe that it is diffeo to your double foliationwith $J$. For $n=1$ this is already an interesting geometric problem, (maybe an exercise) I don't know now how realistic it is for bigger $n$.

The general idea here is that besides the leafwise holomorphic structures, oneneeds some holomorphic dependence of the leaves on a transverse variable,and that this post is sufficient for full holomorphicitysuperseded by my newer answer.
show/hide this revision's text 2 added to necessary condition: same property with $A,B$ switched,

A (local) leaf $A_x \owns x$ parametrizes a family $B_y, y\in A_x$ of the (local) leaves it intersects, and the foliation $A$ near $A_x$ induces a map of $A_x\to {\cal J}$ where ${\cal J}$ is the space of
complex structures on the (local) leaf $B_x$. (we just pullback restrictions of $J$ to $B_y$ by a flow in $A$)

But ${\cal J}$ itself has a complex structure, (as in the theory of moduli spaces). Claim: If your original $J$ is integrable then $M: A_x\to {\cal J}$ is holomorphic. This should be straightforward to check I think.

ie this is a necessary condition. It's clearly a nontrivial condition, as we can choose $M$ quite freely.

Furthermore, if this is true for an open set of leaves $A_z, z\in B_x$, -- as well as the same property with $A,B$ switched, then it should imply integrability of $J$ on the associated nbhd,

ie this is sufficient.
(In particular it implies the same property with $A,B$ switched.)

One suggestion to verify this without a formal Nijenhuis tensor calculation would be to use $M$ to reconstruct a foliation in a chart of $C^N, N=2n$, which realizes such an $M$, then to observe that it is diffeo to your double foliation with $J$. For $n=1$ this is already an interesting geometric problem, (maybe an exercise) I don't know now how realistic it is for bigger $n$.

The general idea here is that besides the leafwise holomorphic structures, one needs some holomorphic dependence of the leaves on a transverse variable, and that this is sufficient for full holomorphicity.
show/hide this revision's text 1

A (local) leaf $A_x \owns x$ parametrizes a family $B_y, y\in A_x$ of the (local) leaves it intersects, and the foliation $A$ near $A_x$ induces a map of $A_x\to {\cal J}$ where ${\cal J}$ is the space of
complex structures on the (local) leaf $B_x$. (we just pullback restrictions of $J$ to $B_y$ by a flow in $A$)

But ${\cal J}$ itself has a complex structure, (as in the theory of moduli spaces). Claim: If your original $J$ is integrable then $M: A_x\to {\cal J}$ is holomorphic. This should be straightforward to check I think.

ie this is a necessary condition. It's clearly a nontrivial condition, as we can choose $M$ quite freely.

Furthermore, if this is true for an open set of leaves $A_z, z\in B_x$, then it should imply integrability on the associated nbhd,

ie this is sufficient.
(In particular it implies the same property with $A,B$ switched.)

One suggestion to verify this without a formal Nijenhuis tensor calculation would be to use $M$ to reconstruct a foliation in a chart of $C^N, N=2n$, which realizes such an $M$, then to observe that it is diffeo to your double foliation with $J$. For $n=1$ this is already an interesting geometric problem, (maybe an exercise) I don't know now how realistic it is for bigger $n$.

The general idea here is that besides the leafwise holomorphic structures, one needs some holomorphic dependence of the leaves on a transverse variable, and that this is sufficient for full holomorphicity.