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Fix a reference path ${\gamma }_{\ast }$ from $L_0$ to $L_1$. By path connectedness, a path $\gamma$ can be joined to it by a smooth homotopy strip $u:[0,1{]}_s\times [0,1{]}_t\to M$ withSmooth homotopies may be used by the relative smoothing theorem; piecewise smooth homotopies give the same integrals. Orient the square by $ds\wedge dt$ and define the Lagrangian path-area functional byHere $\mathcal{A}$ denotes the functional called $A$ in the question. The minus sign is the convention for which downward gradient flow is $J$-holomorphic on the standard oriented strip.

For relative symplectic-area independence, glue two choices, with one orientation reversed, along their common reference and final paths. The resulting oriented surface is a relative $2$-cycle $Z$ in $(M,L_0\cup L_1)$. Since $H_2(M,L_0\cup L_1)=0$, it bounds a relative $3$-chain: as an absolute chain, $Z=\partial B+C$ for a chain $C$ supported in $L_0\cup L_1$. The symplectic form is closed, and it restricts to zero on each Lagrangian submanifold. Using piecewise smooth relative chains, Stokes theorem givesEquivalently, integration of the closed symplectic form defines the zero pairing on the zero relative homology group. The two strips therefore define the same value of $\mathcal{A}$.

Choosing a different reference path changes every value by the negative area of one fixed connecting strip, independent of $\gamma$. Changing the initially assigned reference value also adds a constant. Path connectedness ensures that this is one global constant, rather than an independent constant on each component.

For the first variation, let $\xi (t)$ be a variation vector field along $\gamma$, with $\xi (0)\in T_{\gamma (0)}{L}_0$ and $\xi (1)\in T_{\gamma (1)}{L}_1$. Differentiate the defining strip integral. Stokes theorem, or the variation formula for the integral of a closed differential form, reduces it to the final edge; the side edges give zero because their variations and tangent vectors lie in the Lagrangian submanifolds. With the chosen sign,If this vanishes for every admissible $\xi$, it in particular vanishes for every field supported in $(0,1)$. Nondegeneracy of the symplectic form, together with the fundamental lemma of the calculus of variations, forces $\dot{\gamma }(t)=0$ in the interior and hence everywhere by smoothness. Conversely, a constant path makes the displayed integral zero. The constant must lie in both endpoint submanifolds. ThusNo transverse-intersection assumption is required; the critical set can be empty or have positive dimension.

A compatible almost complex structure is a smooth bundle map $J:TM\to TM$ such thatThese conditions makea Riemannian metric. Its symmetry follows from the $J$-invariance and skew symmetry of $\omega$, and its positivity is the final compatibility condition. It also satisfies $g(Jv,w)=\omega (v,w)$.

Use the formal $L^2$ metric on the path space,The variation formula becomesThus a formal downward flow $s\mapsto u(s,\cdot )$ satisfiesPut the standard complex structure $j$ on the strip, with $j{\partial }_s={\partial }_t$. The J-holomorphic curve equation $du\circ j=J\circ du$ says $u_t=J{u}_s$, which is equivalent to the displayed flow equation because $J^2=-I$. The converse is identical: a smooth J-holomorphic curve on the strip with the specified Lagrangian boundary conditions gives a formal downward trajectory of $\mathcal{A}$.

As a sign check, along such a strip,The drop in $\mathcal{A}$ therefore equals the positive symplectic area swept out by the strip. The word formal matters: the $L^2$ calculation gives the interior equation and its Lagrangian boundary conditions; it does not assert that arbitrary smooth initial paths produce a well-posed ordinary flow on the smooth path space. In particular, the expression $J\dot{\gamma }$ need not satisfy the endpoint tangent restrictions for an arbitrary initial path. The claimed correspondence concerns smooth solutions of the strip equation.