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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.

The base ring must be specified in the algebraic definition. For a commutative $k$-algebra $R$, work in ${\mathrm{End}}_k(R)$ and let $m_f$ mean multiplication by $f\in R$. Set $D_k^{-1}(R)=0$ and define recursivelyThese are the operators of order at most $i$. The order of a nonzero differential operator relative to a base ring is the least $i$ for which it belongs to $D_k^i(R)$; the zero operator can be assigned order $-\infty$. An operator of order zero commutes with all multiplications and is itself multiplication by $P(1)$, so $D_k^0(R)=R$. The union $D_k(R)={\bigcup }_{i\ge 0}{D}_k^i(R)$ is the ring of finite-order operators, with composition as multiplication. The identityand induction on the sum of orders give $D_k^p{D}_k^q\subseteq D_k^{p+q}$. Sums have order at most the maximum of their orders, proving that this union is a ring.

For an arbitrary commutative ring, the absolute version uses $k=\mathbb{Z}$, so the ambient endomorphisms are additive. For the polynomial ring in this question, the intended Weyl-algebra identification is the relative one with $k=\mathbb{C}$. Absolute operators would also include nonzero derivations of the coefficient field $\mathbb{C}$: a derivation on a rational function field in a transcendence basis over $\mathbb{Q}$ extends to its algebraic closure in characteristic zero, and hence to $\mathbb{C}$. Extending it coefficientwise to $\mathbb{C}[x]$ gives an order-one additive operator which is not complex-linear. Such an operator cannot belong to the complex Weyl algebra. This explains why the linearity convention matters.

Now take $R=\mathbb{C}[x_1,\dots ,x_n]$. Coordinate multiplication has order zero, and each partial derivative ${\partial }_i$ has order one because $[{\partial }_i,m_f]=m_{{\partial }_{i}f}$. They satisfy the defining Weyl algebra relations, giving an algebra homomorphism $A_n\to D_{\mathbb{C}}(R)$. Reordering spans $A_n$ by $x^{\alpha }{\partial }^{\beta }$. If an ordered expression acted as zero, commutation with the coordinates according to a maximal derivative multiindex would give multiplication by $\beta !f_{\beta }(x)=0$; removing the top coefficients successively proves that every coefficient is zero. Thus the homomorphism is injective, as in the ordered monomial basis of a Weyl algebra proof.

For surjectivity, let $P$ have order at most $m$, and put ${\delta }_{i}P=[P,m_{x_i}]$. The operations ${\delta }_i$ commute, since the coordinate multiplications commute. A product of more than $m$ of them kills $P$. Define polynomial coefficientsCommuting $P$ through $m_{x^{\alpha }}$ repeatedly gives the finite identityIt follows by induction on $|\alpha |$ from $P{m}_{x_i}=m_{x_i}P+{\delta }_{i}P$ and Pascal's identity. Terms with $|\beta |>m$ vanish. Sincefor $\beta \le \alpha$, we obtain $P(x^{\alpha })={\sum }_{|\beta |\le m}{a}_{\beta }(x){\partial }^{\beta }(x^{\alpha })$. The monomials are a complex basis of $R$, so complex linearity yields equality of the operators on all of $R$. HenceFor a nonzero ordered expression ${\sum }_{\beta }{a}_{\beta }(x){\partial }^{\beta }$, its order is exactly ${\max }_{a_{\beta }\ne 0}|\beta |$. The upper bound follows from composition; the lower bound follows by taking that many coordinate commutators to extract a nonzero multiplication operator. This describes the order filtration of a Weyl algebra:

The associated graded ring is thereforeThe symbol of ${\partial }_i$ is ${\xi }_i$. In a product of ordered operators, moving derivatives past coefficients introduces terms of lower differential order. Thus top-order symbols multiply as ordinary polynomials, and the ordered basis proves there are no further relations. In particular $[D^p,D^q]\subseteq D^{p+q-1}$, with $D^{-1}=0$: the top-order symbols of the two products agree and cancel. Unlike the Bernstein filtration, the order filtration's degree-zero part is the entire polynomial ring and its pieces are infinite-dimensional over $\mathbb{C}$.

Form the Rees ring of a filtered algebraThe central element $h$ is a non-zero-divisor. This ring is generated by $h$, the $x_i$, and $y_i=h{\partial }_i$, withIts quotient modulo $h$ is the associated graded ring just computed. Explicitly the degree-$m$ component of $h\mathcal{A}$ is $D^{m-1}{h}^m$, so the degree-$m$ quotient is $D^m/D^{m-1}$. The quotient modulo $h-1$ recovers $A_n$.

Since $\mathcal{A}/h\mathcal{A}$ is commutative, any two lifts $F,G\in \mathcal{A}$ of symbols have $[F,G]\in h\mathcal{A}$. DefineDivision by $h$ is unique because $h$ is a non-zero-divisor. If $F$ is replaced by $F+hU$, the quotient changes by $[U,G]$, which is zero modulo $h$; changing $G$ works the same way. Thus the bracket is well defined. Antisymmetry and bilinearity follow from the commutator. The identity $[F,GH]=[F,G]H+G[F,H]$ proves the Leibniz rule. For the Jacobi identity, lift each inner bracket by $[G,H]/h$; the cyclic sum of the outer expressions is the commutator Jacobi identity divided by $h^2$, hence zero. These verifications prove that the graded polynomial ring is a Poisson algebra. This is the Poisson bracket from a central deformation parameter construction.

The generators give $\{{\xi }_i,x_j\}={\delta }_{ij}$ and all coordinate-coordinate and symbol-symbol brackets zero. The Leibniz rule therefore determines the answer on every polynomial:The sign here follows the chosen convention $[P,Q]=PQ-QP$: in particular $\{{\xi }_i,x_j\}={\delta }_{ij}$, rather than $\{x_i,{\xi }_j\}={\delta }_{ij}$. For homogeneous symbols the bracket has order degree $p+q-1$, and can equivalently be written $\{{\sigma }_p(P),{\sigma }_q(Q)\}={\sigma }_{p+q-1}([P,Q])$.

Write ${\xi }_P(p)={\frac{d}{dt}|}_0\exp (t\xi )\cdot p$ for the fundamental vector field of a left Lie group action, and use ${\iota }_{X_f}\omega =df$ throughout. Such fundamental fields satisfy $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$. A moment map has linear components ${\mu }_{\xi }=⟨\mu ,\xi ⟩$ satisfyingFor the action to be Hamiltonian, these components can be chosen equivariantly, or, infinitesimally,This identifies the span of the component Hamiltonians with the Lie algebra image, rather than identifying the entire infinite-dimensional Poisson algebra $C^{\infty }(P)$ with $\mathfrak{g}$. For an actual isomorphism with $\mathfrak{g}$, injectivity is also needed; a faithful infinitesimal action supplies it.

Preservation of $\omega$ alone gives $d{\iota }_{{\xi }_P}\omega =0$, not necessarily exactness. For example, translation generated by ${\partial }_q$ on the symplectic two-torus has contraction $dp$, whose period is nonzero. That action has no globally defined component Hamiltonian. Once the contractions are exact, choose Hamiltonians linearly in $\xi$. On each connected component of $P$, the discrepancyis constant: apply $[X_f,X_g]=-X_{\{f,g\}}$ and the fundamental-field bracket. The Jacobi identity makes $\kappa$ a Lie algebra two-cocycle. Replacing ${\mu }_{\xi }$ by ${\mu }_{\xi }+b(\xi )$ changes it to $\kappa (\xi ,\eta )-b([\xi ,\eta ])$. Thus the exactness condition and vanishing of this moment-map equivariance obstruction are precisely the required infinitesimal conditions.

For a semisimple Lie algebra, both hold. First Cartan's magic formula givesSemisimplicity implies $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, so every generator is a sum of commutators and has a globally exact contraction. This proves Hamiltonian existence for a semisimple symplectic action.

Next use the nondegenerate Killing form to define $D$ by $B(D\xi ,\eta )=\kappa (\xi ,\eta )$. Antisymmetry makes $D$ skew for $B$. The cocycle identity and invariance of $B$ imply $D[\xi ,\eta ]=[D\xi ,\eta ]+[\xi ,D\eta ]$, so $D$ is a derivation. Choose $Z$ byThen, using cyclicity of trace and $[D,\mathrm{ad}\xi ]=\mathrm{ad}(D\xi )$,Hence $D=\mathrm{ad}Z$ and $\kappa (\xi ,\eta )=B(Z,[\xi ,\eta ])$. Taking $b(\xi )=B(Z,\xi )$ removes the cocycle and proves semisimple moment-map equivariance without merely invoking a cohomology theorem. For a connected group, integrate the infinitesimal equivariance along its one-parameter subgroups to obtain group equivariance. A disconnected group can require an additional discrete-equivariance check.

Use $\omega ={\sum }_{i}d{q}_i\wedge d{p}_i$, ${\iota }_{X_f}\omega =df$ andThen $X_f(g)=\{g,f\}$ and $[X_f,X_g]=-X_{\{f,g\}}$. For a left group action, let ${\xi }_P(p)={\frac{d}{dt}|}_0\exp (t\xi )\cdot p$; its fundamental fields obey $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$.

The components of a moment map satisfyThus ${\xi }_P=X_{{\mu }_{\xi }}$. A Hamiltonian moment map additionally has coadjoint equivariance,Here the right side evaluates on $\xi$ as $⟨\mu (p),{\mathrm{Ad}}_{g^{-1}}\xi ⟩$. With that convention its infinitesimal form is $\{{\mu }_{\xi },{\mu }_{\eta }\}={\mu }_{[\xi ,\eta ]}$. Some definitions reserve “moment map” for this equivariant version; choosing Hamiltonians for every generator first gives a weakly Hamiltonian action.

To derive the obstruction, compare the two antihomomorphism identities:On connected $P$, nondegeneracy of $\omega$ makesa constant alternating bilinear form. The two Jacobi identities implyReplacing ${\mu }_{\xi }$ by ${\mu }_{\xi }+b(\xi )$ changes $\kappa$ to $\kappa -b([\xi ,\eta ])$. Thus vanishing of this class in Lie algebra cohomology is the precise moment-map equivariance obstruction. In particular a connected group with $H^2(\mathfrak{g},\mathbb{R})=0$ permits constants to be chosen so that the component Poisson algebra is the Lie algebra. A locally effective action makes this realization faithful; otherwise it realizes the quotient by the infinitesimal kernel. The linear span of the components has the Lie brackets; the entire algebra of polynomial products of them is, of course, larger.

A useful sufficient condition is semisimplicity, which can be proved here without assuming the desired equivariance. Let $B$ be the nondegenerate Killing form and define $D$ by $B(D\xi ,\eta )=\kappa (\xi ,\eta )$. Invariance of $B$ and the cocycle identity giveso $D$ is a derivation. Choose $Z$ uniquely by $B(Z,\xi )=\mathrm{tr}(D\mathrm{ad}\xi )$. Since $[D,\mathrm{ad}\xi ]=\mathrm{ad}(D\xi )$, cyclicity of trace givesHence $D=\mathrm{ad}Z$ and $\kappa (\xi ,\eta )=B(Z,[\xi ,\eta ])$. The shift $b(\xi )=B(Z,\xi )$ removes the obstruction. This proves semisimple moment-map equivariance; for connected $G$ infinitesimal equivariance integrates along its one-parameter subgroups to group equivariance. It does not claim that arbitrary symplectic actions of arbitrary groups have moment maps.

For the Kepler calculation, put $k=Mm$ and work on $r=|\mathbf{r}|>0$, since the Hamiltonian is singular at the origin. The Hamilton equations areThe angular momentum therefore obeysFor the Runge-Lenz vector,The vector triple product identity gives $\mathbf{r}\times \mathbf{L}=\mathbf{r}(\mathbf{r}\cdot \mathbf{p})-r^2\mathbf{p}$. Thereforeand the terms cancel. Since $\dot{f}=\{f,H\}$, this proves

The conserved quantities consequently generate transformations preserving energy. Their brackets give the Kepler dynamical symmetry algebra. On $H<0$, define $\mathbf{A}=\mathbf{K}/\sqrt{-2H}$ as a function on that whole open region. Because $H$ Poisson commutes with both vectors, its brackets areWith ${\mathbf{J}}_{\pm }=(\mathbf{L}\pm \mathbf{A})/2$ one obtains two commuting $\mathfrak{so}(3)$ algebras:Thus the negative-energy algebra is $\mathfrak{so}(4)$. On $H>0$, using $\mathbf{A}=\mathbf{K}/\sqrt{2H}$ instead changes the last bracket to $-{ϵ}_{ijk}{L}_k$, yielding $\mathfrak{so}(3,1)$, with $\mathbf{L}$ rotations and $\mathbf{A}$ boosts. At zero energy the brackets of conserved functions on the characteristic orbit quotient give $\mathfrak{e}(3)=\mathfrak{so}(3)⋉{\mathbb{R}}^3$, with commuting $\mathbf{K}$.

There are two global qualifications. First, an energy hypersurface carries a presymplectic form: its restricted two-form has characteristic direction $X_H$. At $H=0$, the literal vector fields of $\mathbf{K}$ need only commute modulo $X_H$, since differentiating $\{K_i,K_j\}=-2H{ϵ}_{ijk}{L}_k$ still produces an $L_k{X}_H$ term. On nonzero-energy regions the energy-dependent normalization above gives an exact Hamiltonian Lie algebra; one must differentiate that normalization before restricting to a level. Second, completeness is required for a global group action. The collision-excluded Kepler phase space need not have complete hidden-symmetry flows, so the brackets establish local actions (or the appropriate simply connected covers), not an unconditional global action on every unregularized trajectory. Collision regularization supplies the familiar global bound-motion symmetry; this distinction is developed in math.berkeley.edu/~alanw/277papers00/tang.pdf . It does not change the three energy-dependent Lie algebras just derived.

Geometric quantization starts from a classical symplectic manifold $(P,\omega )$ and seeks a Hilbert space and quantum observables that reflect its geometry. It has two distinct stages: prequantization, which represents the full Poisson algebra on sections of a line bundle, and a choice of polarization in geometric quantization, which reduces the excessive phase-space dependence of those sections.

Use the conventions ${\iota }_{X_f}\omega =df$ and $\{f,g\}=\omega (X_f,X_g)$. A prequantum line bundle is a Hermitian line bundle $L\to P$ with unitary connection of curvatureExistence requiresfor every closed integral two-cycle $\Sigma$, equivalently integrality of the symplectic cohomology class. Necessity follows from the integral curvature periods of a unitary line bundle; conversely an integral class defines such a bundle, and one adjusts its connection to give the specified curvature. This is a restriction on the classical system, not a choice of local coordinates. For an exact cotangent symplectic form $\omega =-d\theta$, with $\theta =p_{j}d{q}_j$, the bundle can be trivial and a connection is $\nabla =d-i\theta /\hslash$. On a sphere with symplectic area $4\pi s$, the condition becomes $2s/\hslash \in \mathbb{Z}$. The sign of the curvature convention changes if one changes the Hamiltonian-vector-field convention or uses the dual bundle; the integrality condition is unaffected.

The Kostant-Souriau prequantum operator for a real observable isIts derivative part describes the classical Hamiltonian flow and the multiplication term corrects the bracket. Indeed,Since $X_f(g)=-\{f,g\}$, expanding the operator commutator givesWith the Hermitian bundle metric and Liouville measure, these operators are formally symmetric for real $f$ because Hamiltonian flows preserve symplectic volume. On noncompact spaces, domains and self-adjointness still require attention; the algebraic identity initially holds on an appropriate common smooth domain.

Prequantization alone is too large. Sections depend on $2n$ phase-space variables, and the canonical commutation relations are represented reducibly. A polarization is an involutive maximal isotropic distribution $\mathcal{F}\subset T_{\mathbb{C}}P$ of complex rank $n$. Polarized sections satisfyThe curvature vanishes on pairs of polarization vectors because $\omega {|}_{\mathcal{F}}=0$, and involutivity makes the equations locally compatible. A real polarization is tangent to a Lagrangian foliation. A complex polarization in a Kähler manifold chooses one complex tangent type. With the curvature sign used here, choose the holomorphic tangent distribution, giving antiholomorphic wavefunctions. The conjugate line-bundle and sign convention gives the more usual holomorphic description.

For $T^{\ast }Q$, the vertical real polarization is spanned locally by ${\partial }_{p_j}$. Since the canonical one-form has no vertical component, its polarized sections are functions of $q$ alone. In the above trivialization,One does not simply integrate these states over every momentum to retain the prequantum norm: they are constant along noncompact polarization leaves and that integral would diverge. The polarized inner product instead uses the configuration-space measure or an intrinsic half-density description. This recovers the Schrödinger realization.

Only observables whose Hamiltonian flows preserve $\mathcal{F}$ act directly on the polarized space. For example, the free-particle flow generated by $p^2/2$ sends vertical tangent vectors into directions with a configuration-space component, so its raw prequantum operator does not preserve the vertical polarization. To obtain the usual kinetic operator one needs an additional procedure, such as transporting the polarization and pairing back, rather than pretending every classical observable restricts automatically. The Blattner-Kostant-Sternberg pairing and related projection methods address this problem. The impossibility of quantizing all classical observables with every desired algebraic property is another reason to specify the admissible observable algebra.

A metaplectic correction tensors $L$ with a square root of the polarization's canonical determinant bundle when such a square root exists. These half-forms make the pairing more intrinsic and supply corrections to operator transport. They account for familiar Maslov and zero-point shifts; their existence is an extra topological condition. For a flat phase plane, take $z=q+ip$ and replace $\theta =pdq$ by the gauge-equivalent ${\theta }_s=(pdq-qdp)/2$. ThenThe antiholomorphic function therefore has Gaussian norm, the complex-conjugate Bargmann-Fock space. This explicit calculation fixes which polarization matches our curvature sign. Choosing the opposite tangent type with the same sign would give growing Gaussian sections, not the desired Hilbert space. On compact positive Kähler examples, the corresponding holomorphic description (or its conjugate in these conventions) gives finite-dimensional state spaces.

Finding a suitable polarization is therefore a substantial part of the construction. A globally smooth real Lagrangian foliation may fail to exist; for example a regular real polarization on $S^2$ would require a line field that its tangent topology forbids. Singular foliations and noncompact leaves complicate both states and inner products. Although the restricted prequantum connection is locally flat on a Lagrangian leaf, a global nonzero parallel section requires trivial holonomy. In an integrable system this selects the Bohr-Sommerfeld quantization leaves; including the half-form correction gives the familiar action conditionin the usual Maslov-index convention. Complex polarizations require compatible global complex geometry and positivity. Different choices can lead to quite different realizations, and a natural unitary comparison is not automatic. Thus integral curvature provides prequantization; polarization and its global compatibility determine the actual quantum state space.

A Poisson algebra is a commutative associative algebra with a Lie bracket that is a derivation in each argument. Smooth functions on a symplectic manifold are its standard example. The linear span of moment-map components may form a finite-dimensional Lie subalgebra, while closing those components under multiplication normally produces a much larger Poisson algebra.