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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

With the conventionthe time-dependent Hamiltonian vector field is uniquely determined because $\omega$ is nondegenerate. If ${\varphi }_t$ is its local flow, Cartan's magic formula givesThus ${\varphi }_t^{\ast }\omega =\omega$ wherever the flow is defined, which is the Hamiltonian flow preserves the symplectic form property.

Write $v=(a,b)$ and ${\omega }_0=dx\wedge dy$. The linear Hamiltonianhas Hamiltonian vector field $X_{H_0}=v$. Choose a smooth cutoff $\rho$ that is one on $B(r+|v|)$ and zero outside $B(r+|v|+ϵ)$, and put $H=\rho H_0$. Every trajectory beginning in $B(r)$ and following $v$ remains in $B(r+|v|)$ for time $0\le t\le 1$, so the time-one map translates that ball by $v$. Outside $B(r+|v|+ϵ)$ its vector field vanishes, so the map is the identity. This is a compactly supported Hamiltonian translation and hence a compactly supported symplectomorphism.

For the connected-sum construction, choose a Darboux chart about the unique transverse intersection and straighten the two Lagrangian sheets to ${\mathbb{R}}^2$ and $i{\mathbb{R}}^2$ in ${\mathbb{C}}^2$. Remove small disks from the two sheets and join their boundary circles by the standard Lagrangian neckin ${\mathbb{R}}_x^2\oplus {\mathbb{R}}_y^2$, where $(a(t),b(t))$ follows a smooth arc from one positive coordinate ray to the other and agrees with those rays near its ends. Its pullback of ${\sum }_{j}d{x}_j\wedge d{y}_j$ vanishes because $u\cdot u^{\prime }=0$. Gluing this neck to the unchanged surfaces performs Lagrangian surgery. Topologically it is their connected sum, so it gives a Lagrangian connected sum

Finally, choose an immersed circle ${\gamma }_l:S^1\to {\mathbb{R}}^2$ with exactly $l$ transverse double points and no other multiple points; one may add $l$ small figure-eight kinks to an embedded circle. Let $c:S^1\to {\mathbb{R}}^2$ be an embedded circle and defineThis Product Lagrangian immersion satisfies ${\iota }^{\ast }{\omega }_0=0$. If $p$ is a double point of ${\gamma }_l$, its two local branches times $c(S^1)$ meet along $\{p\}\times c(S^1)$. Their tangent spaces intersect precisely in the tangent line to that circle, so the intersection is clean. The $l$ double points therefore give exactly $l$ disjoint clean self-intersection circles.

Near each clean circle, perturb one Lagrangian sheet by the graph of $\delta dh$ in its Weinstein neighborhood, where $h:S^1\to \mathbb{R}$ is a Morse function with one minimum and one maximum. The clean circle is replaced by two transverse double points. Resolve both by Lagrangian surgery. Each resolution attaches one one-handle and lowers the Euler characteristic by two, so resolving all $l$ clean circles changes the Euler characteristic of the original torus from zero toChoose the orientation-reversing neck at one double point; the resulting connected surface is nonorientable, while all the surgeries remove their double points and leave an embedding. Thus the Givental construction of nonorientable Lagrangian surfaces gives a closed connected nonorientable surface of Euler characteristic $-4l$ Lagrangian embedded in $({\mathbb{R}}^4,{\omega }_0)$.