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Independent functions $Q_1,\dots ,Q_n$ on a $2n$-dimensional symplectic manifold can locally form the position half of canonical coordinates exactly when $\{Q_i,Q_j\}=0$ throughout a neighbourhood. Their commuting Hamiltonian vector fields span the fibres of $Q$. Flow from a transverse section gives coordinates $(Q,t)$ with $\omega =\sum d{t}_i\wedge d{Q}_i+\beta (Q)$. Write $\beta =d\alpha$ locally and take $P_i=-t_i-{\alpha }_i(Q)$. Then $\omega =\sum d{Q}_i\wedge d{P}_i$.

For a hypersurface $N$ of a symplectic manifold, the restriction of the symplectic form has a one-dimensional kernel, its characteristic line field. On a regular value level $N=H^{-1}(c)$, this line is generated by the Hamiltonian vector field of $H$.

Suppose a symplectic Lie group action on a connected symplectic manifold has global component Hamiltonians and a common fixed point $p_0$. Choose their constants linearly so ${\mu }_{\xi }(p_0)=0$. The difference $c(\xi ,\eta )=\{{\mu }_{\xi },{\mu }_{\eta }\}-{\mu }_{[\xi ,\eta ]}$ has zero Hamiltonian vector field because fundamental fields and Hamiltonian fields have compatible Lie-bracket conventions; it is therefore constant. At $p_0$, all fundamental fields vanish, so both terms in $c$ vanish. Thus $c=0$ everywhere and the Poisson brackets of the components reproduce the group Lie algebra. For a connected group this also yields coadjoint equivariance. A linear unitary action on a vector space has the origin as such a fixed point.

The metric defines this globally smooth Hamiltonian on the cotangent bundle. With $\omega =d{x}^i\wedge d{p}_i$ and ${\iota }_{X_H}\omega =dH$, its Hamiltonian vector field projects to affinely parametrized geodesics. The momentum is $p_i=g_{ij}{\dot{x}}^j$, and its conserved value is half the squared speed.

If $H$ is $C^2$ and its spatial Hessian matrix is uniformly bounded on every finite time interval, its Hamiltonian vector field is globally Lipschitz continuous in phase space with at most linear growth. The Picard-Lindelöf theorem and Gronwall inequality give a unique Hamiltonian flow for all finite forward and backward times. Bounds on ${\nabla }_{z}H(t,0)$ follow from time continuity on compact intervals.

A Hamiltonian function is a smooth function whose differential determines a Hamiltonian vector field.

A Hamiltonian group action is a Lie group action on a symplectic manifold with an equivariant moment map $\mu :M\to {\mathfrak{g}}^{\ast }$. In the sign convention ${\iota }_{X_H}\omega =dH$, each component satisfies $d⟨\mu ,X⟩={\iota }_{X_M}\omega$, where $X_M$ is the fundamental vector field. The opposite Hamiltonian vector field convention reverses the sign.

On a symplectic manifold, use ${\iota }_{X_f}\omega =-df$ and $\{f,g\}=\omega (X_f,X_g)=X_f(g)$ for the Poisson bracket. Cartan's magic formula gives ${\mathcal{L}}_{X_f}\omega =0$. Taking the interior product of a differential form with $[X_f,X_g]$ then yields $-d\{f,g\}$, proving the homomorphism. Its kernel consists of locally constant functions, since $X_f=0$ precisely when $df=0$. On a connected manifold these are the constants. The quotient of smooth functions by this kernel is therefore the Lie algebra of Hamiltonian vector fields. A different contraction sign requires a compatible bracket sign.

Transport a phase-space density along Hamilton's equations. The Hamiltonian vector field is $b=({\nabla }_{v}H,-{\nabla }_{x}H)$ and the equation is ${\partial }_{t}f+b\cdot {\nabla }_{x,v}f=h$. Its zero divergence gives volume-preserving transport. For $h=0$, the density is constant along characteristics; a source is integrated by the Duhamel formula for Hamiltonian transport. This is distinct from the elliptic Liouville equation.

Compactly supported Hamiltonian vector fields can attain any prescribed tangent vector at a point of a symplectic manifold: prescribe the derivative of a local Hamiltonian and multiply by a bump function. Composing flows for a tangent basis gives an open orbit by the inverse function theorem. Every orbit is open, so on a connected manifold the compactly supported Hamiltonian transformations act transitively.

In three-dimensional autonomous mechanics, a regular common level set of $N$ independent isolating integrals has dimension $6-N$ in phase space. If an orbit is dense in a $d$-dimensional invariant torus and the integrals characterize its closure, $N=6-d$. A generic three-frequency regular orbit has three integrals; a spherical nonclosed rosette has energy and three components of specific angular momentum, with two-dimensional closure and four integrals. A periodic nonstationary orbit has one-dimensional closure and can be isolated by five independent orbit labels on the invariant family. The maximum is five wherever the Hamiltonian vector field is nonzero, since all integral differentials annihilate that field. A single resonant orbit need not yield an additional smooth single-valued integral on an open neighborhood; global superintegrability is stronger than orbital closure.

Start from a Casimir function $H_0$ of the first Poisson structure. Whenever possible, choose $H_n$ whose first Hamiltonian vector field is minus the second Hamiltonian vector field of $H_{n-1}$. A sequence extending indefinitely is the coefficient sequence of a formal Casimir function for the Poisson pencil. Its functions pairwise commute for both Poisson brackets. Existence at each stage is an additional solvability requirement; compatibility alone does not guarantee it.

For a weakly Hamiltonian left Lie group action on a connected symplectic manifold, choose ${\iota }_{X_f}\omega =df$. Fundamental vector fields and Hamiltonian vector fields are both antihomomorphisms, so the displayed difference is constant. The Jacobi identity for the Poisson bracket makes it a two-cocycle. Changing ${\mu }_X$ by $b(X)$ changes the difference by $-b([X,Y])$. Thus its class in Lie algebra cohomology obstructs choosing an equivariant moment map. For a connected group, vanishing of this class gives infinitesimal and group equivariance.

The next step of a Lenard-Magri recursion requires $-{\Pi }_{2}d{H}_{n-1}$ to be a Hamiltonian vector field for ${\Pi }_1$. It must be tangent to the symplectic leaves, and its contraction with the leaf symplectic form must have a global potential, with a compatible smooth extension between leaves. Even the first step may fail: on ${\mathbb{R}}^2$, take ${\Pi }_1=0$, ${\Pi }_2={\partial }_x\wedge {\partial }_y$ and $H_0=x$. Both Poisson structures are compatible, but ${\Pi }_{1}d{H}_1=-{\Pi }_{2}dx$ cannot hold, because its left side is zero and its right side is nonzero.

A symplectic manifold is a smooth even-dimensional manifold $M$ with a differential two-form $\omega$ that is closed, $d\omega =0$, and nondegenerate at every point. Nondegeneracy identifies vector fields with differential one-forms by the interior product $X\mapsto {\iota }_X\omega$. In this solution use the explicit Hamiltonian vector field conventionThis contraction sign differs from the negative-sign convention sometimes used for Hamiltonian vector fields; all subsequent signs are fixed by the displayed choice. In Darboux coordinates with $\omega ={\sum }_{a}d{x}_a\wedge d{y}_a$ it givesThe Poisson bracket is bilinear and antisymmetric, and the ordinary product rule gives $\{f,gh\}=\{f,g\}h+g\{f,h\}$. Thus it remains to check the Jacobi identity for the Poisson bracket.

The Cartan formula for the Lie derivative and closedness of the symplectic form give ${\mathcal{L}}_{X_f}\omega =d({\iota }_{X_f}\omega )+{\iota }_{X_f}d\omega =d^{2}f=0$. Henceand nondegeneracy givesApplying both sides to an arbitrary smooth function $h$ gives $\{f,\{g,h\}\}-\{g,\{f,h\}\}=\{\{f,g\},h\}$, which is exactly the Jacobi identity for the Poisson bracket. Consequently the smooth functions, with their pointwise product and this Poisson bracket, form a Poisson algebra. Constants have zero Hamiltonian vector field. If $f,g$ are components of moment maps, the same displayed relation relates the brackets of their Hamiltonian vector fields to the Poisson bracket of those components. In the opposite contraction convention the corresponding relation is the Hamiltonian Lie algebra homomorphism with the compatible bracket sign.

For the point vortices, remove the collision diagonals from $({\mathbb{R}}^2{)}^N$, since the logarithmic Hamiltonian function is singular there. Set $s_{ab}=(x_a-x_b{)}^2+(y_a-y_b{)}^2>0$ for $a\ne b$. Differentiating each pair contribution givesThe Hamiltonian vector field therefore gives the equations of motionEach contribution is perpendicular to the displacement from vortex $b$ to vortex $a$ and has magnitude $2/|{\mathbf{r}}_a-{\mathbf{r}}_b|$. This is the velocity induced by a point vortex of positive circulation, with the normalization $\Gamma /(2\pi )=2$. Each vortex is advected by the velocity of the others, without a self-induced singular velocity. Pairwise contributions cancel in the total velocity, so ${\sum }_a{x}_a$ and ${\sum }_a{y}_a$ are conserved. For two vortices the separation is constant and they rotate counterclockwise about their midpoint, as expected for identical positive point vortices.

For $L=\frac{1}{2}{\sum }_a(x_a^2+y_a^2)$, compute its Poisson bracket with $H$ directly:The last sum vanishes by pairing $(a,b)$ with $(b,a)$, since their numerators are opposite and denominators equal. Thus $L$ Poisson commutes with $H$ and is conserved. Equivalently, the Hamiltonian function depends only on pairwise distances and is invariant under simultaneous rotations.

The standard positive generator of the simultaneous circle group $SO(2)$ action isIts interior product with the symplectic form isUse the moment map convention $d⟨\mu ,\xi ⟩=-{\iota }_{Y_{\xi }}\omega$ for the usual positive rotation generator. Identifying $\mathfrak{so}(2{)}^{\ast }$ with $\mathbb{R}$, this proves $L$ is the moment map for the standard rotation action. It is invariant under the action, hence equivariant because circle group $SO(2)$ is abelian. Additive constants are possible; the natural normalization is $L=0$ at the origin of the ambient space. On the collision-free space this polynomial is its restriction. With the solution's positive contraction convention, the rotation generator is $Y=-X_L$. Confusing these two signs would reverse the final augmented Hamiltonian.

Let ${\mathcal{R}}_{\phi }$ denote simultaneous counterclockwise rotation through $\phi$, and introduce the co-rotating coordinates by $\mathbf{r}(t)={\mathcal{R}}_{\Omega t}{\mathbf{r}}^{\prime }(t)$. Both the symplectic form and $H$ are rotation-invariant, so pulling back the inertial Hamiltonian vector field leaves $X_H$ unchanged. Differentiating the rotation gives the co-rotating equationThe co-rotating positions are constant precisely when this Hamiltonian vector field vanishes. By nondegeneracy of the symplectic form, that is equivalent toConversely a collision-free critical point of this augmented Hamiltonian function produces a rigidly rotating solution. This is a rotating point-vortex relative equilibrium.

As a sign and scaling check, scaling every position by $s>0$ changes $H$ by $-N(N-1)\log s$. Therefore ${\sum }_a{\mathbf{r}}_a\cdot {\nabla }_{a}H=-N(N-1)$. At a rotating point-vortex relative equilibrium, ${\nabla }_{a}H+\Omega {\mathbf{r}}_a=0$, so $2\Omega L=N(N-1)$. For $N\ge 2$ the angular velocity is positive in this circulation convention. An equilateral three-vortex configuration of circumradius $\rho$ has $L=3{\rho }^2/2$ and $\Omega =2/{\rho }^2$, agreeing directly with the equations of motion.

Use the Hamiltonian vector field convention ${\iota }_{X_{H_t}}\omega =-d{H}_t$. A smooth time-dependent Hamiltonian function $H:[0,1]\times M\to \mathbb{R}$ with flow defined over the whole interval generates the Hamiltonian isotopy determined byThe group of Hamiltonian diffeomorphisms isThese maps are symplectomorphisms: by Cartan's magic formula, ${\mathcal{L}}_{X_{H_t}}\omega =d{\iota }_{X_{H_t}}\omega =0$. Thus $d({\varphi }_H^t{)}^{\ast }\omega /dt=0$.

The term “group” is justified explicitly. If ${\varphi }_t,{\psi }_t$ are generated by $H_t,K_t$, their product path is generated by $H_t+K_t\circ {\varphi }_t^{-1}$, since the pushforward of $X_{K_t}$ by a symplectomorphism is $X_{K_t\circ {\varphi }_t^{-1}}$. The inverse path is generated by $-H_t\circ {\varphi }_t$. The zero Hamiltonian gives the identity.

On a closed symplectic manifold, all these flows exist on the finite time interval. On an open manifold one often means the compactly supported Hamiltonian diffeomorphism group ${\mathrm{Ham}}_c$, using Hamiltonians with compact support in $[0,1]\times M$. We use that subgroup for the finite-energy definitions on Euclidean space below. The unrestricted complete-flow group is needed for the global cotangent translations in Question 4; those translations are generally not compactly supported.

Take $\varphi$ in the identity component ${\mathrm{Symp}}_0(M,\omega )$ and choose a smooth symplectic isotopy ${\varphi }_t$ from the identity to $\varphi$. Let its generating vector field be $X_t={\dot{\varphi }}_t\circ {\varphi }_t^{-1}$. Differentiating ${\varphi }_t^{\ast }\omega =\omega$ givesThus ${\eta }_t={\iota }_{X_t}\omega$ is a closed differential one-form. By the vanishing of the first de Rham cohomology, it is exact, so choose $H_t$ with ${\eta }_t=-d{H}_t$.

The primitives can be chosen smoothly in both $t$ and the point. On each connected component fix a base point $x_0$ and define $H_t(x)=-{\int }_{x_0}^x{\eta }_t$. Exactness makes this independent of the chosen path. In a coordinate neighbourhood of any point, use a fixed path to the neighbourhood's centre followed by a smoothly varying short path; this shows smooth dependence locally, and hence globally. One may subtract a time-dependent mean without changing the Hamiltonian vector field. Therefore the original symplectic isotopy is Hamiltonian, and $\varphi \in \mathrm{Ham}(M,\omega )$.

Here the usual description of ${\mathrm{Symp}}_0$ by smooth isotopies agrees with the topological identity component. To see the local path-connectedness behind that description, use a Weinstein neighborhood of the diagonal in $(M\times M,-\omega \oplus \omega )$. The graph of a symplectomorphism close to the identity corresponds to the graph of a small closed one-form on the diagonal. Multiplying that one-form by $t\in [0,1]$ gives Lagrangian graphs; for sufficiently small forms their first projections remain diffeomorphisms, so they give symplectomorphisms connecting the original map to the identity. These local paths, translated in the group and concatenated with smooth time reparametrizations, identify its connected and smooth-path components.

Conversely every Hamiltonian isotopy is symplectic by part (a), and connects its time-one map to the identity. Hence

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.

Write the exact differential form as $\alpha =du$ for a smooth real-valued function $u$ on $N$. On $T^{\ast }N$ take the Hamiltonian function $H=u\circ \pi$. With $\omega ={\sum }_{j}d{q}_j\wedge d{p}_j$ and ${\iota }_{X_H}\omega =-dH$, the Hamiltonian vector field isIts complete flow is explicit:The base point stays fixed, so this flow exists for all real $t$, even if $u$ is unbounded on a noncompact base. Its time-one map is exactly $f$, proving that the cotangent fiber translation is a Hamiltonian diffeomorphism in the unrestricted complete-flow group.

This construction also justifies the hint for this family, without treating exactness as a general substitute for a Hamiltonian isotopy. For any Hamiltonian isotopy ${\psi }_t$ on this cotangent bundle, Cartan's magic formula, $d\theta =-\omega$ and ${\iota }_{X_{H_t}}\omega =-d{H}_t$ giveIntegrating proves that ${\psi }_1^{\ast }\theta -\theta$ is exact. For the present translation this difference is ${\pi }^{\ast }\alpha$, which is exact if and only if $\alpha$ is exact: one direction is pullback of $du$, and the other is pullback by the zero section. Necessity therefore follows from the differential identity, while sufficiency follows from the complete flow just constructed.

Compact support is a separate issue. If $du$ is nonzero at some $q$, the map moves every point of the entire unbounded fiber over $q$, so it is not a member of the compactly supported Hamiltonian diffeomorphism group. For example $N=\mathbb{R}$, $u(q)=q$ gives $(q,p)\mapsto (q,p+1)$. Thus the claim about this global translation uses the full Hamiltonian group, not ${\mathrm{Ham}}_c$; its flow can be cut off to agree on a prescribed compact swept set, but such a cutoff does not give the same map on the entire cotangent bundle.

A Poisson manifold has a skew bilinear bracket on smooth functions which is a derivation in each argument and satisfies the Jacobi identity. Locally a Poisson bivector $\Pi$ represents it byThe derivation property is automatic for this expression. On applying the Jacobi identity to coordinate functions, and then using the derivation property for arbitrary functions, the necessary and sufficient condition isEquivalently its Schouten-Nijenhuis bracket $[\Pi ,\Pi ]$ vanishes.

A symplectic manifold has a closed nondegenerate two-form $\omega$. Write its coefficient matrix as ${\Omega }_{ij}$ and set ${\Pi }^{ij}=({\Omega }^{-1}{)}^{ij}$. It is skew and defines the bracket above. Differentiating $\Pi \Omega =I$ gives ${\partial }_{\ell }\Pi =-\Pi ({\partial }_{\ell }\Omega )\Pi$. Substitution into the Jacobi condition giveswhere the last equality is $d\omega =0$. Thus every symplectic manifold is Poisson. In this convention the Hamiltonian vector field satisfies ${\iota }_{X_f}\omega =-df$, and $\{f,g\}=X_{g}f$; this will also be the convention below.

The converse fails because a Poisson bivector need not be invertible. On ${\mathbb{R}}^2$, take $\Pi =0$. Its bracket is identically zero and satisfies every Poisson axiom, but it cannot be the inverse of any nondegenerate two-form. This is an even-dimensional counterexample, so odd dimension is not the only obstruction.