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A Darboux chart is a coordinate chart in which a symplectic form becomes the standard symplectic form.

The scalar circle group action on ${\mathbb{C}}^{n+1}$ with the standard symplectic form has moment map $\mu =-|z{|}^2/2$ in the convention ${\iota }_{X_H}\omega =dH$. Reduction of the radius-$R$ sphere gives Complex projective space and the form ${\omega }_R=(i{R}^2/2)\partial \bar{\partial }\log (1+|w{|}^2)$. Its area on a complex projective line is $\pi R^2$. Thus $R=\sqrt{2}$ yields the Fubini-Study form normalized to projective-line area $2\pi$, while $R=1$ yields half that form. The Hopf fibration identifies the quotient.

Choose coordinates $(q_1,\dots ,q_n,p_1,\dots ,p_n)$ with standard symplectic form ${\omega }_0={\sum }_{j}d{q}_j\wedge d{p}_j$, and regard the given sphere $S$ as lying in $\{p_n=0\}$. With ${\iota }_{X_H}{\omega }_0=-dH$, the linear Hamiltonian function $H_{\delta }=\delta q_n$ has Hamiltonian vector field $\delta {\partial }_{p_n}$ and translates in the missing coordinate.

To make this a permissible compactly supported map, take a smooth cutoff $\chi$ equal to one on an open neighbourhood of the fixed compact swept setChoose $\chi$ with compact support and let $R={\sup }_{\mathrm{supp}\chi }|q_n|<\infty$. For $0<\delta <1$ put $G_{\delta }=\delta q_n\chi$. On the whole swept region the cutoff is locally constant, so its derivatives vanish and $X_{G_{\delta }}=\delta {\partial }_{p_n}$. Thus the flow carries every $z\in S$ to $z+t\delta e_{p_n}$ for $0\le t\le 1$.

The time-one image lies in $\{p_n=\delta \}$ and is disjoint from the original sphere. Its Hofer metric from the spatial supremum norm obeysIn the oscillation convention the upper bound is $2R\delta$. Since $\delta$ can approach zero while the cutoff and $R$ stay fixed,This proves the zero displacement energy for a compact set in a hyperplane; it does not assert that any one nonidentity displacing map has zero norm.

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 $L$.

Let $\pi :T^{\ast }N\to N$ be the projection and $s_{\alpha }(q)=(q,{\alpha }_q)$ the section with image ${\Gamma }_{\alpha }$. For the canonical one-form on a cotangent bundle,Since $\pi \circ s_{\alpha }=\mathrm{id}$, evaluation on a tangent vector $v\in T_{q}N$ givesUse the same position-first standard symplectic form as before, $\omega =-d\theta$. Pullback commutes with the exterior derivative, soThe graph of a differential one-form is an embedded submanifold, because its projection is the inverse of the section on the graph. Its dimension is $\dim N$, half that of $T^{\ast }N$. Therefore the vanishing of the pulled-back symplectic form is exactly the Lagrangian submanifold condition, andThis proves that the graph of a closed one-form is Lagrangian using the intrinsic one-form rather than a coordinate guess.

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.

Use the standard symplectic form ${\omega }_0={\sum }_{j=1}^{n}d{q}_j\wedge d{p}_j$. A canonical transformation is a diffeomorphism $f$ preserving this symplectic form:Thus, writing $f(q,p)=(Q,P)$, the defining identity is ${\sum }_{j}d{Q}_j\wedge d{P}_j={\sum }_{j}d{q}_j\wedge d{p}_j$. For a local canonical transformation, require this on its open domain. In particular preservation of the symplectic form, rather than just of phase-space volume, is the condition in dimensions greater than two.

Throughout these solutions choose the Hamiltonian vector field convention ${\iota }_{X_F}\omega =dF$. It gives $\dot{q}=F_p$, $\dot{p}=-F_q$ and agrees with the positive coordinate Poisson bracket used below. The opposite convention for Hamiltonian vector fields changes signs in the Lie bracket of vector fields identity but none of the commuting or integrability conclusions.

The Marsden-Weinstein theorem says that for a Hamiltonian group action, if $c$ is a regular value fixed by the coadjoint action and $G$ acts freely and properly on ${\mu }^{-1}(c)$, thenis a symplectic manifold with a unique form characterized by ${\pi }^{\ast }{\omega }_c={\iota }^{\ast }\omega$, where $\iota$ includes the level set and $\pi$ is the quotient projection. Its dimension is $\dim M-2\dim G$.

Equip ${\mathbb{C}}^{n+1}$ with the standard symplectic formLet the circle group act by $e^{i\theta }\cdot z=e^{i\theta }z$. Its generator is $X={\sum }_j(-y_j{\partial }_{x_j}+x_j{\partial }_{y_j})$, andConsequently the moment map in our convention is $\mu (z)=-|z{|}^2/2$. It is invariant, hence equivariant because the circle group is abelian. The level ${\mu }^{-1}(-1)$ is the sphere of radius $\sqrt{2}$; it is regular, the circle group acts freely there, and properness follows from compactness of the group. The quotient is Complex projective space, by $z\mapsto [z]$, a scaled Hopf fibration. The Marsden-Weinstein theorem therefore produces a reduced symplectic form on ${\mathbb{CP}}^n$.

To identify it rather than only assert its existence, use the primitiveOn the affine chart choose the local section $s(w)=\sqrt{2}(1,w)/\sqrt{S}$ of the quotient. Direct substitution givesDifferentiating, using $\bar{\partial }\partial =-\partial \bar{\partial }$, givesThis is the Fubini-Study form from circle reduction. The unit sphere instead produces half this form; our radius $\sqrt{2}$ is exactly what gives the $2\pi$ line-area normalization used above.

No nonconstant J-holomorphic curve with this closed connected source exists. The standard symplectic form on ${\mathbb{R}}^{2n}$ is exact; for instanceThe Energy identity for a J-holomorphic curve and the Generalized Stokes theorem giveThe target metric from the compatible almost complex structure is positive definite, so the nonnegative continuous energy density must vanish everywhere. Thus $du=0$. Connectedness of $\Sigma$ makes $u$ constant. This argument uses exactness and compatibility, and works even when $J$ is nonintegrable; it does not require the ordinary holomorphic maximum principle on the target.

Use ${\mathbb{C}}^2$ with the standard symplectic form ${\omega }_0=d{x}_1\wedge d{y}_1+d{x}_2\wedge d{y}_2$ and the compatible almost complex structure given by multiplication by $i$. Consider the embedded submanifoldIn coordinates $(x_1,y_1,x_2,y_2)$, its tangent space is $W=\{(a,b,b,0):a,b\in \mathbb{R}\}$. Applying $J$ gives $J(a,b,b,0)=(-b,a,0,b)$. If this is again in $W$, its fourth coordinate forces $b=0$, and its second and third coordinates then force $a=0$. Thus $W\cap JW=0$ and ${\dim }_{\mathbb{R}}L=2$: $L$ is a totally real submanifold.

However, the pullback of a differential form under the parametrization isThus this totally real submanifold is not a Lagrangian submanifold.

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

Normalize the Fubini-Study form so that a complex projective line has symplectic area $\pi$. The displayed map takes the unit symplectic ball in ${\mathbb{C}}^n$ diffeomorphically onto the complement of the last-coordinate hyperplane. Its lift to the unit sphere has last coordinate positive real, so that coordinate contributes zero to the pulled-back standard symplectic form. The map is therefore a symplectomorphism.

An invertible real matrix preserving the standard symplectic form. With $J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$, it satisfies $A^{T}JA=J$, equivalently $AJ{A}^T=J$. The first identity describes preservation of the form; the second describes preservation of coordinate Poisson brackets.

Use the standard symplectic form ${\sum }_{j}d{q}_j\wedge d{p}_j$ and convention ${\iota }_{X_H}\omega =-dH$. The Hamiltonian $H=\delta q_n$ translates $p_n$ by $\delta$. A fixed compactly supported cutoff equal to one near the set swept for $|\delta |\le 1$ leaves that translation unchanged on the original compact set. Its supremum norm is at most $C|\delta |$, so the translation displaces the set at arbitrarily small cost. This proves zero Hamiltonian displacement energy for either Hofer normalization.