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Along Hamilton's equations, spatial derivatives of the Hamiltonian function cancel in its total time derivative. Thus $dH/dt={\partial }_{t}H$. An autonomous Hamiltonian function is conserved; explicit time dependence generally prevents conservation even though phase space is still preserved.

A moment map assigns to each element of a Lie algebra a Hamiltonian function for its fundamental vector field, and is equivariant for the coadjoint action. For an abelian Lie group, constant shifts are allowed; the numerical scale also depends on the chosen Lie algebra generators and the normalization of the symplectic form.

For the symplectic form $\omega =idz\wedge d\bar{z}/(1+|z{|}^2{)}^2$, the circle group rotates $z$ by $e^{i\theta }$. Its real generator is $K=iz{\partial }_z-i\bar{z}{\partial }_{\bar{z}}$. The contraction is ${\iota }_K\omega =-dH$, so $H$ is a Hamiltonian function in this sign convention. In the other chart $w=1/z$, $H=1/(1+|w{|}^2)$, proving smoothness at infinity. Its image is $[0,1]$ with the endpoints at the fixed poles. The symplectic area is $2\pi$; rescaling to the round area $4\pi$ would double this moment function.

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.

For the spatial supremum-norm convention, a normalized Hamiltonian function gives the path lengthOn an open manifold use compactly supported Hamiltonians; on a closed manifold use zero-mean Hamiltonians, which fixes their additive time-dependent constants. The Hofer metric from the spatial supremum norm isEquivalently one minimizes this length over Hamiltonian paths joining the two maps. Inversion and conjugation preserve the generator's supremum norm, so the distance is symmetric and bi-invariant; concatenation gives the triangle inequality. Its nondegeneracy is the geometric assertion proved in Question 3, not a consequence of these formal properties alone.

Another common convention for Hofer's metric uses the oscillation length ${\ell }_{\mathrm{osc}}(H)={\int }_0^1(\max H_t-\min H_t)dt$, giving a distance $d_H$. Normalized or compactly supported Hamiltonians have values straddling zero, soThus these conventions give equivalent distances, and agree on whether an energy vanishes. We keep the same chosen convention in every energy estimate.

The Hamiltonian displacement energy of a subset $A$ isThe infimum is $+\infty$ if there is no displacing map. In the oscillation convention replace ${\rho }_{\infty }$ by $d_H$.

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.

Set $\tau =\epsilon t$ and use the stated weighted spatial and parameter scales. Dividing the two equations by ${\epsilon }^3$ and ${\epsilon }^4$ respectively giveswhere prime now means differentiation with respect to $\tau$. The Hamiltonian function satisfiesThus $H=v^2/2+\alpha u-u^3/3$ is conserved for $\epsilon =0$.

For $\alpha >0$, let $s=\sqrt{\alpha }$. The conservative system has a center at $(-s,0)$ with $H_c=-2{\alpha }^{3/2}/3$ and a saddle at $(s,0)$ withClosed contours in the left potential well have $H_c<H<H_h$. At $H_h$ the left-hand separatrix is a homoclinic orbit, sinceIt leaves and returns to $(s,0)$, with its left turning point at $u=-2s$. The right separatrix branches are unbounded. The energy levels above the saddle value are open, as are the separate right-hand components of lower energy levels. For $\alpha \le 0$ there is no nondegenerate potential well and no such saddle loop.

When $a=0$, define the Hamiltonian functionThen $\dot{x}=-H_y$ and $\dot{y}=H_x$, so $\dot{H}=0$. At $(\sqrt{b},0)$ the Hessian of $H$ is $2\sqrt{b}I$, a strict nondegenerate minimum; at $(-\sqrt{b},0)$ it is $-2\sqrt{b}I$, a strict maximum. Small regular level curves around either point are closed curves, and the vector field is nonzero on them. Motion around each curve therefore returns after a finite period, proving the existence of periodic orbits and classifying both equilibria as center equilibria.

The separatrix level is $H=0$, consisting of $x=0$ and $x^2/3+y^2=b$. The two arcs of this ellipse run from the lower saddle equilibrium to the upper one, while the vertical segment runs from the upper saddle to the lower one. They enclose the two families of periodic orbits, as shown in the last panel of the phase portrait. Outside the loops, regular levels are unbounded.

For $a>0$ the divergence of the original vector field is exactly $-a$. If a periodic orbit existed, uniqueness of trajectories would make it a simple closed curve, tangent everywhere to the vector field. The flux across its boundary would be zero, whereas the divergence theorem would give $-a$ times its positive enclosed area. This contradiction is the Bendixson-Dulac criterion. Hence $\boxed{\text{periodic orbits exist if and only if }a=0}$.

Set $s=|z{|}^2$. Contraction of the symplectic form with the rotation vector field givesWith the convention fixed in the unheaded solution, ${\iota }_K\omega =-dH$, so the Hamiltonian function isIts derivative is $dH=ds/(1+s{)}^2$. In the chart $w=1/z$, $H=1/(1+|w{|}^2)+C$, so it extends smoothly across infinity; the additive constant is the kernel freedom already identified. Thus $K=X_H$ globally, proving that the action is a Hamiltonian group action. With $C=0$, the moment map for rotation of the complex projective line takes values in $[0,1]$, reaching its endpoints at the two fixed poles.

One can check the normalization in polar coordinates: $\omega =2r(1+r^2{)}^{-2}dr\wedge d\varphi$ has total symplectic area $2\pi$, not $4\pi$, and $H=r^2/(1+r^2)$. Equivalently, in a polar angle with $r=\tan (\vartheta /2)$, $H=(1-\cos \vartheta )/2$. This explains the half-height factor that would be lost by replacing the printed form with the unscaled round-sphere area form. Under the opposite convention ${\iota }_{X_H}\omega =dH$, the Hamiltonian is the negative of this function, up to a constant.

A Hamiltonian group action is a smooth Lie group action of $G$ on $(M,\omega )$ by symplectomorphisms, together with an equivariant moment map $\mu :M\to {\mathfrak{g}}^{\ast }$. For $\xi \in \mathfrak{g}$, let ${\xi }_M(p)={\frac{d}{dt}|}_0\exp (t\xi )\cdot p$ be its fundamental vector field. In our sign convention the defining identities areHere the left coadjoint action means $({\mathrm{Ad}}_g^{\ast }\nu )(\xi )=\nu ({\mathrm{Ad}}_{g^{-1}}\xi )$. Each component ${\mu }^{\xi }$ is therefore a Hamiltonian function for the corresponding infinitesimal action. Equivariance is part of the definition; merely requiring each infinitesimal generator to be a Hamiltonian vector field is the weaker condition of a weakly Hamiltonian action. Reversing the defining sign of Hamiltonian vector fields reverses the moment map sign as well.

For an autonomous Hamiltonian function, the chain rule and Hamilton's equations giveThus the energy is constant along each characteristic:The time-independent hypothesis is implicit in the displayed expression in this subpart. For the time-dependent Hamiltonian function allowed earlier, the correct Hamiltonian energy balance is insteadFor example $H(t,x,v)=t+|v{|}^2/2$ has a unique global Hamiltonian flow, but its value along a curve increases at unit rate. Thus conservation cannot be claimed for arbitrary time-dependent $H$.

A Poisson bivector is a smooth antisymmetric contravariant two-tensorwhose bracket on smooth functions,satisfies the Jacobi identity. Bilinearity and antisymmetry are immediate, and the product rule makes the bracket a derivation in each argument. Together these properties define a Poisson manifold. Unlike a symplectic form, the Poisson bivector need not be nondegenerate.

Apply the Jacobi identity to the coordinate functions. Since $\{x^i,x^j\}={\omega }^{ij}$,Changing ${\omega }^{im}$ to $-{\omega }^{mi}$ and rearranging the three summands gives the printed coordinate Jacobi condition for a Poisson bivector:This is also sufficient: expanding the Jacobiator of three arbitrary smooth functions, all terms involving second derivatives cancel in pairs by antisymmetry. The remaining coefficient of ${\partial }_{i}F{\partial }_{j}G{\partial }_{k}H$ is the coordinate-function Jacobiator displayed above.

For a Lie algebra, the natural global space carrying the proposed linear bracket is its dual space ${\mathfrak{g}}^{\ast }$. If $v^i$ is the chosen basis, define its linear coordinate function by $x^i(\ell )=\ell (v^i)$. The Lie-Poisson bracket iswhere $d{F}_{\ell },d{G}_{\ell }$ are elements of $\mathfrak{g}=({\mathfrak{g}}^{\ast }{)}^{\ast }$. On a general manifold the same coordinate expression gives a local construction; a global one requires compatible transition rules. The use of ${\mathfrak{g}}^{\ast }$ supplies that compatibility intrinsically.

Here ${\omega }^{ij}=c_r^{ij}{x}^r$ and ${\partial }_m{\omega }^{jk}=c_m^{jk}$. The left side of the coordinate Jacobi condition for a Poisson bivector is thereforeThe final coefficient is the negative of the coefficient of $v^r$ in the Lie algebra identity $[v^i,[v^j,v^k]]+[v^j,[v^k,v^i]]+[v^k,[v^i,v^j]]=0$. Thus the Lie-Poisson bracket satisfies the Jacobi identity.

It remains to find the Lie algebra structure constants for the printed rotation fields. Distinguish their original spatial coordinates $(x,y,z)$ from the coordinates $(x^1,x^2,x^3)$ on the dual space. Use the conventional Lie bracket of vector fields. Their component vectors are $(0,-z,y)$, $(z,0,-x)$ and $(-y,x,0)$, respectively. For example,Similarly,The negative sign is essential: these are the fundamental fields of a left rotation action with the stated Lie bracket of vector fields, and consequently have the infinitesimal left-action sign convention discussed above. The first field here has component $y{\partial }_z$, as printed in the PDF.

The Lie-Poisson bracket on the dual space consequently hasFor the evolution convention $\dot{F}=\{F,H\}$, the Hamiltonian function has derivatives $(2A{x}^1,2B{x}^2,2C{x}^3)$. Substitution gives the quadratic rotational Lie-Poisson dynamicsThese are Euler-type Hamilton's equations on a noncanonical Poisson manifold. When $A=1/(2I_1)$, $B=1/(2I_2)$ and $C=1/(2I_3)$ with positive principal inertias, they are the Euler equations for a torque-free rigid body in body angular-momentum coordinates. As a check, $H$ is conserved by antisymmetry of the Poisson bracket, and $C_0=(x^1{)}^2+(x^2{)}^2+(x^3{)}^2$ is a Casimir function of a Poisson manifold. Direct differentiation of $C_0$ in the three equations cancels the terms $4[(C-B)+(A-C)+(B-A)]x^1{x}^2{x}^3$. The Hamiltonian flow therefore lies on both an energy level and a sphere, a symplectic leaf of this signed rotational bracket.

Differentiating the assumed primitive identity and using Cartan's magic formula givesApply the pullback of a differential form by ${\rho }_t^{-1}$ and rearrange:This smooth family of Hamiltonian functions proves that ${\rho }_t$ is a Hamiltonian isotopy in the convention of part (b).

To connect this with the preceding assertion that every ${\rho }_t^{\ast }\lambda -\lambda$ is an exact differential form, the smooth choice of primitives causes no additional obstruction. Fix $p_0\in M$ and normalize $F_t(p_0)=0$. Define $F_t(p)$ by the line integral of ${\rho }_t^{\ast }\lambda -\lambda$ from $p_0$ to $p$. The fundamental theorem for line integrals makes this independent of the path. Near any $p$, use one fixed path to a nearby chart center followed by coordinate line segments; this expression is smooth jointly in $t,p$. Thus these normalized primitives form the required smooth family.

Write the exact differential form as $\beta =dh$ and take the cotangent fiber translation family ${\rho }_t(x,\xi )=(x,\xi +td{h}_x)$. Its generating vector field is vertical and in local coordinates isWith $\omega ={\sum }_{j}d{x}_j\wedge d{\xi }_j$ and the Hamiltonian vector field convention from Question 3,Thus ${\rho }_t$ is a Hamiltonian isotopy and $f={\rho }_1$ is a Hamiltonian diffeomorphism. The local flow is explicitly defined for every real $t$, regardless of compactness. With the alternative convention ${\iota }_{X_H}\omega =-dH$, the same family is generated by $H=h\circ \pi$.

The definition in the question does not require compact support for the Hamiltonian function; the displayed Hamiltonian diffeomorphism uses that definition.