In $(\mathbb{R}^4,\omega_{std})$ is positive symplectic area enough to guarantee a pseudoholomorphic disc representative?

Question

I will present my question in the context that I encountered it, although I believe it probably applies in general context. Consider $\mathbb{R}^4 \cong \mathbb{C}^2$ with the standard symplectic form $\omega = \sum dp_k\wedge dq_k$. Consider a Lagrangian $L\cong S^1 \times S^1$ given by the product of two circles, $A$ and $B$, each lying in an $\mathbb{R}^2$ factor of $\mathbb{R}^4$. Consider a class $\alpha\in\pi_2(\mathbb{R}^4,L)$, such that $\omega(\alpha)>0$.

1) Does there exists an almost complex structure $J$ compatible with $\omega_{std}$ such that $\alpha$ is represented by a $J$-holomorphic disc $u$ with boundary mapped to $L$?

Note that for the integrable complex structure $J_{std}$ that comes from $\mathbb C$ this is true only for $\alpha$ of the form $kA+mB$, where $k,m\ge0$ (here I abused notation and referred to the discs bounded by the circles $A$ and $B$ using the same letters)

2) If the answer to 1 is positive, can $J$ be chosen to be regular? (in the sense that all the moduli spaces of curves are being cut out transversally).

Thank you! :)

Answer 1

1) Yes. The boundary of a holomorphic disc on $S^1 \times S^1$ is subcritical isotropic, and we can use Gromov's h-principle to move its boundary (by a Hamiltonian isotopy) to the sought curve on $a S^1 \times b S^1$, given that the latter curve bounds the same symplectic area. (I.e. we must first rescale the initial disc appropriately.) The new disc has boundary on $a S^1 \times b S^1$ and is pseudoholomorphic for the push-forward of the standard complex structure under the Hamiltonian diffeomorphism (this complex structure is also compatible with $\omega_0$!).

1b) Since there is an h-principle also for open Lagrangian submanifolds, under the additional condition that Maslov index of the two curves on the respective tori coincide, one can arrange so that even an open neighbourhood of the initial Lagrangian $S^1 \times S^1$ is mapped into target Lagrangian $aS^1 \times bS^1$ by the Hamiltonian diffeomorphism.

2) In some explicit cases it is obviously not possible to achieve regularity. For instance, the symplectic embedding $z \mapsto (a \overline{z},bz)$ from the unit disc to a disc of Maslov index zero with boundary on $aS^1 \times bS^1$, $b>a>0$, can be made pseudoholomorphic, but is never regular.

The refined construction in (1b) does however produce regular discs, given that the initial disc was regular.