 Solution

Use the compatible metrics $g_{\Sigma }(\cdot ,\cdot )={\omega }_{\Sigma }(\cdot ,j\cdot )$ and $g_X(\cdot ,\cdot )={\omega }_X(\cdot ,J\cdot )$. The energy of a smooth map isIt is a J-holomorphic curve when $du\circ j=J\circ du$. Splitting $du$ into its complex-linear and complex-antilinear parts gives the Energy identity for a J-holomorphic curveThe first term depends only on the homology class. It follows that a J-holomorphic map minimizes energy among all maps in its homology class.

One Monotonicity theorem for a J-holomorphic curve says that a nonconstant J-holomorphic curve through the center of a sufficiently small radius-$\rho$ ball, with boundary outside that ball, has area at least $c{\rho }^2$; in the standard complex ball one may take the sharp value $\pi {\rho }^2$. The Gromov non-squeezing theorem says

Write ${\mathbb{R}}^4=T^{\ast }{\mathbb{R}}^2$ with coordinates $(q,p)$ and form $d{q}_1\wedge d{p}_1+d{q}_2\wedge d{p}_2$. For $c>0$, the graphis a Lagrangian subspace. Points of $L_c\cap B(R)$ have $|q|\le R/\sqrt{1+c^2}$. Choose $c$ so large thatThen no two points of the $\epsilon$-neighborhood of $L_c\cap B(R)$ differ by a nonzero vector $(m,0)$ with $m\in {\mathbb{Z}}^2$, so quotienting $q$ modulo ${\mathbb{Z}}^2$ is injective there.

Relative to the Lagrangian splitting $L_c\oplus J{L}_c$, the map $(v,w)\mapsto (\lambda v,{\lambda }^{-1}w)$ is symplectic. It sends $B^4(r)$ into the indicated long thin neighborhood wheneverThese inequalities are compatible when $R>r^2/\epsilon$. Taking such an $R$ and then the quotient constructs the arbitrarily large symplectic balls in a cotangent cylinder: