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For a smooth nonconstant holomorphic disk $h$ in ${\mathbb{C}}^n$ with boundary on a Lagrangian submanifold $L$, its symplectic area is ${\int }_{D^2}|{\partial }_{s}h{|}^{2}dsdt>0$. The Cauchy-Riemann equations give ${\partial }_{t}h=J{\partial }_{s}h$, so this is the pullback of the standard symplectic form. The Stokes theorem identifies the area with the period of a primitive along the boundary loop. Thus the Liouville class of a Lagrangian submanifold evaluates nontrivially on this loop and cannot vanish.

Fix a primitive $\lambda$ of the standard symplectic form on ${\mathbb{R}}^{2n}$. With the coordinates of Question 2 one convenient choice isFor a Lagrangian submanifold with inclusion $i:L\hookrightarrow {\mathbb{R}}^{2n}$, $d(i^{\ast }\lambda )=i^{\ast }{\omega }_0=0$. Its Liouville class of a Lagrangian submanifold isAny two primitives differ by a closed one-form on the contractible ambient space, hence by an exact form, so the class is independent of the primitive. The cotangent primitive convention $\theta ={\sum }_j{p}_{j}d{q}_j$, with ${\omega }_0=-d\theta$, gives the negative of this class; rationality, positive period size and nonvanishing do not depend on that sign.

The period homomorphism sends an integral one-cycle $c$ to ${\int }_c{i}^{\ast }\lambda$. Its image $\mathrm{Per}(L)\subset \mathbb{R}$ is an additive subgroup. A rational Lagrangian submanifold is one for which this subgroup is discrete. In the nonzero case it has a unique positive generator:Equivalently ${\ell }_L$ is a real multiple of an integral cohomology class. The invariant $\gamma (L)$ is its least positive Liouville period, also expressible as $\inf \{a>0:a\in \mathrm{Per}(L)\}$. If the period group is zero, that infimum is $+\infty$ by the empty-set convention; some treatments instead reserve “rational” for the nonzero discrete case. Because the ambient space is contractible, the Stokes theorem identifies these periods with signed symplectic areas of disks whose boundaries lie on $L$.

Write the standard complex coordinates as $z_j=q_j+i{p}_j$ and let $J$ denote multiplication by $i$. Use the standard symplectic form ${\omega }_0={\sum }_{j}d{q}_j\wedge d{p}_j$, for which ${\omega }_0(v,Jv)=|v{|}^2$. In real coordinates $s,t$ on the disk, holomorphicity and the Cauchy-Riemann equations implyTherefore the pulled-back two-form isIf this integral were zero, the continuous nonnegative integrand would vanish everywhere in the interior. The displayed Cauchy-Riemann relation would then make both first derivatives zero, forcing $h$ to be constant on the connected disk and, by continuity, on its boundary. For the nonconstant map under consideration the symplectic area is consequently strictly positive:Take the primitive $\lambda =\frac{1}{2}{\sum }_j(q_{j}d{p}_j-p_{j}d{q}_j)$ from Question 3. Let $c=h{|}_{\partial D^2}$, oriented as the boundary of the disk. Since $c$ lies in $L$, the Stokes theorem givesThe restriction $\lambda {|}_L$ is closed because $L$ is a Lagrangian submanifold. If its Liouville class of a Lagrangian submanifold were zero, it would be an exact differential form, and its integral along every closed curve in $L$ would vanish. The boundary loop above contradicts this. The Liouville class is therefore nonzero. This proves that a holomorphic disk detects a nonzero Liouville class directly by positivity and Stokes; no disk-existence theorem is being assumed beyond the disk given here.

Fix $p\in X$. A linear change of coordinates first identifies ${\omega }_p$ with the standard symplectic form ${\omega }_0$. After shrinking to a star-shaped neighborhood, everyis nondegenerate. The Poincaré lemma gives $\omega -{\omega }_0=d\sigma$, with $\sigma (p)=0$. Define $X_t$ by ${\iota }_{X_t}{\omega }_t=-\sigma$ and let ${\varphi }_t$ be its local flow. Cartan's magic formula givesThus ${\varphi }_1^{\ast }\omega ={\omega }_0$, proving the Darboux theorem.

For a smooth function $H$ on a closed symplectic manifold, its Hamiltonian vector field is defined by ${\iota }_{X_H}\omega =-dH$. The same formula gives ${\mathcal{L}}_{X_H}\omega =-d^{2}H=0$, so the Hamiltonian flow preserves the symplectic form. To move one point to another in a connected $X$, join them by a path, cover the path by finitely many Darboux charts, and in each chart use a cutoff linear Hamiltonian to perform a small translation. Composing these compactly supported Hamiltonian diffeomorphisms proves that symplectomorphisms act transitively on each connected component.

In ${\mathbb{R}}^2$, every embedded curve is a Lagrangian submanifold. Let $L_1,L_2$ be circles enclosing different Euclidean areas. A plane symplectomorphism preserves area and carries the bounded complementary component of one circle to that of its image, so no symplectomorphism maps $L_1$ to $L_2$.

The same phenomenon exists in every ${\mathbb{R}}^{2n}$. Withtake the product toriThe Liouville class of a Lagrangian submanifold has respective period vectors $\pi (1,\dots ,1)$ and $2\pi (1,\dots ,1)$ on these tori. Every symplectomorphism of ${\mathbb{R}}^{2n}$ preserves the Liouville class up to the induced integral change of basis on $H_1(T^n)$, because its pullback changes $\lambda$ only by an exact form. An integral automorphism sends a primitive vector to a primitive vector and therefore cannot send the first period vector to the second. Hence