Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 64 (Applications of Differential Geometry to Physics)

An orthonormal coframe writes a metric as $g={\eta }_{ab}{\theta }^a\otimes {\theta }^b$ with constant diagonal signature matrix $\eta$. In Cartan's first structure equation, solveThese are the torsion-free and metric-compatibility conditions and determine the Levi-Civita connection. Then Cartan's second structure equation givesContract $R^a{}_{bad}$ to obtain the Ricci tensor and then contract again for the scalar curvature. This fixes the curvature convention; reversing the definition of the Riemann curvature tensor reverses the resulting curvature signs.

On $z>0$, choose ${\theta }^0=dt/z$, ${\theta }^1=dx/z$, ${\theta }^2=dy/z$, ${\theta }^3=dz/z$, with $\eta =\mathrm{diag}(-1,1,1,1)$. For $a=0,1,2$,A torsion-free metric connection therefore hasFor example, ${\Omega }^a{}_3=-d{\theta }^a=-{\theta }^a\wedge {\theta }^3$, while for $a,b<3$,Metric antisymmetry supplies the remaining components, giving uniformlyThus the metric has constant sectional curvature $-1$. In four dimensions,It is an Einstein manifold, with Einstein constant $-3$ in the stated curvature convention.

These are Poincare coordinates on anti-de Sitter spacetime of unit radius. Constant curvature gives the maximal ten-dimensional local isometry algebra $\mathfrak{so}(3,2)$. This can also be checked directly: set $u=(t,x,y)$, ${\eta }_{ab}=\mathrm{diag}(-1,1,1)$ and $u_a={\eta }_{ab}{u}^b$. The ten independent Killing vector fields areTranslations and Lorentz transformations leave the numerator and $z$ unchanged. A dilation rescales numerator and denominator equally. For $K_a$, direct differentiation gives ${\mathcal{L}}_{K_a}{\eta }^{(4)}=4u_a{\eta }^{(4)}$ and $K_a(z)=2u_{a}z$, so the conformal factor cancels and ${\mathcal{L}}_{K_a}g=0$.

Hence the maximally extended Anti-de Sitter spacetime has connected isometry group locally $S{O}_0(3,2)$, with the appropriate covering group if one unwraps its time coordinate. The displayed coordinates cover only a patch. The ten local generators do not all give globally complete flows preserving that patch; for example special conformal flows can cross its horizon. The manifest complete patch symmetries include the boundary Poincare transformations and positive dilations. This distinguishes local maximal symmetry from the global isometries of a chosen coordinate domain.

Let $\mathcal{J}=\star J$ be the current $(d-1)$-form on an oriented $d$-dimensional spacetime. For a slab bounded by two spacelike hypersurfaces ${\Sigma }_1,{\Sigma }_2$ and a side boundary $B$, the Stokes theorem givesThus $Q(\Sigma )={\int }_{\Sigma }\mathcal{J}$ is conserved if the side flux vanishes, for example for a spatially compact current or appropriate falloff at infinity. With side flux it instead obeys the corresponding charge-balance equation. This is the differential-form version of current conservation.

For the Abelian Chern--Simons theory, take $A\mapsto A+d\chi$ with a smooth compactly supported gauge parameter. The change in the gauge-field term is a boundary term because $d\chi \wedge dA=d(\chi dA)$. For the source,Discarding the boundary terms, the action variation isGauge invariance for every such $\chi$ and $c\ne 0$ therefore gives $d\star J=0$. If $c=0$, the current decouples and gauge invariance places no condition on it. On a spacetime with boundary, boundary conditions or boundary degrees of freedom are also needed to handle the discarded terms; on a nontrivial compact gauge bundle, large-gauge invariance is an additional global issue.

For the field equation, useThe one-form $\delta A$ commutes with the two-form $\mathcal{J}$ under the wedge product, soup to the boundary term. ConsequentlyApplying $d$ and using $d^2=0$ again gives current conservation for $c\ne 0$. The factor two comes from varying both occurrences of $A$.

Orient a spatial region $D$ so its boundary is $\gamma$, and define $Q_D={\int }_D\mathcal{J}$. Applying Stokes theorem and the field equation yieldsFor a U(1) connection, in a convention with unit minimal charge, the gauge-invariant quantity is its holonomy $e^{i\Phi }$, the Wilson-loop or Aharonov-Bohm phase. A large gauge transformation can change the chosen representative of $\Phi$ by $2\pi n$. Therefore $\Phi$ is a phase angle modulo $2\pi$, rather than an absolute gauge-invariant real number. A particle of charge $q$ has phase $e^{iq\Phi }$ in the corresponding normalization. The flux-charge relation thus attaches a gauge phase to enclosed charge.

The metric variation requires specifying the independent source. The pure Chern-Simons term $A\wedge dA$ has no metric dependence, hence contributes zero stress-energy tensor. If the one-forms named in the question have fixed covariant components $A_{\mu },J_{\mu }$, the Hodge star in the source does depend on the metric:Using $\delta \sqrt{-g}=-\sqrt{-g}{g}_{\mu \nu }\delta g^{\mu \nu }/2$ givesUnder that literal fixed-one-form convention,There is another common convention in the topological source theory: hold the conserved two-form $\mathcal{J}=\star J$, equivalently the vector current density, fixed as the metric varies. Then both $A\wedge dA$ and $\mathcal{J}\wedge A$ are metric-independent, and $T_{\mu \nu }=0$ for this action. These are different variations, not contradictory calculations. The Chern-Simons source stress convention explains why a source prescription is necessary; a dynamical matter source would contribute its own action and stress as well.

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.

A principal bundle $\pi :P\to B$ with structure group $G$ has a smooth free right $G$-action, each fibre is a single orbit, and it is locally equivariantly isomorphic to $U\times G\to U$. A representation $\rho :G\to GL(V)$ defines the associated vector bundleLocal sections of $P$ identify $E$ with $U\times V$, and changes of section act by $\rho$ on the fibre coordinates.

A principal bundle is trivial if and only if it has a global smooth section. Given $s:B\to P$, the map $(b,g)\mapsto s(b)g$ is an equivariant isomorphism $B\times G\to P$: freeness and transitivity give its unique inverse on each fibre, and local trivializations make that inverse smooth. Conversely a product bundle has section $b\mapsto (b,e)$. This proves principal bundle trivialization by a global section. A contractible paracompact base is a sufficient condition for triviality of such a bundle; local sections alone are insufficient on a general base.

A manifold is parallelizable when its tangent bundle is trivial, equivalently when it possesses a global smooth frame. The circle $S^1$ is an example, with its nowhere-vanishing unit tangent field. The sphere $S^3$ is another: regard it as the unit quaternions. At a unit quaternion $q$, the three tangent vectors $qi,qj,qk$ are independent and smoothly depend on $q$. Quaternion multiplication preserves the norm, so these form an orthonormal frame. This gives the quaternionic left-invariant frame on the three-sphere.

For any Lie group $G$ of dimension $n$, choose a basis $E_1,\dots ,E_n$ of $T_{e}G$. Left translations defineThese fields are smooth; at every $g$, the differential of the diffeomorphism $L_g$ is invertible, so they are a basis of $T_{g}G$. Hence every Lie group manifold is parallelizable, as in parallelization of a Lie group by left translations.

For a real semisimple Lie group, let $B(X,Y)=\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)$ be the Killing form. It is nondegenerate and invariant under the adjoint action. Left translating $B$ gives a bi-invariant pseudo-Riemannian metric. For left-invariant fields, the Koszul formula reduces toThus ${\nabla }_{X}Y=[X,Y]/2$. The curvature convention of Question 1 givesTo obtain the Ricci tensor, trace the linear map $X\mapsto R(X,Y)Z$. Since $[[X,Y],Z]=\mathrm{ad}Z\mathrm{ad}YX$,This proves the Killing-form Einstein metric construction for every real semisimple group. If $G$ is compact semisimple, $-B$ is positive definite and its Ricci tensor is $-\frac{1}{4}B=\frac{1}{4}g$, giving a Riemannian Einstein metric. For a noncompact semisimple group the general Killing-form construction is indefinite. Thus the unrestricted assertion is understood in the pseudo-Riemannian sense appropriate here; positive definiteness is not obtained by simply taking $-B$ in the noncompact case.

The two unheaded requests are addressed here; the individual group-cover calculations are in the scoped parts below. A quaternion $q=a+bi+cj+dk$ has norm $|q{|}^2=a^2+b^2+c^2+d^2$. Thus the unit quaternions form exactly the sphere $S^3\subset {\mathbb{R}}^4$, with quaternion multiplication giving its smooth group structure. The explicit identification with the special unitary group $SU(2)$ isThe matrices representing $i,j,k$ satisfy the quaternion multiplication relations. The displayed matrix obeys $U^{†}U=|q{|}^{2}I$ and $\det U=|q{|}^2$; conversely every $SU(2)$ matrix has that form with $|q|=1$. Hence $S^3$ is the manifold of unit quaternions, and its group is $SU(2)$.

For the exterior-square double cover of SO(3,3), fix a basis $e_1,\dots ,e_4$ and volume element $v=e_1\wedge e_2\wedge e_3\wedge e_4$. Define the symmetric bilinear form $Q$ on the six-dimensional exterior square byIt is symmetric because two-forms commute under the wedge product. PutThen $Q(a_i,a_j)=Q(b_i,b_j)=0$ and $Q(a_i,b_j)={\delta }_{ij}$. The Gram matrix is $\left(\begin{array}{cc}0& I_3\\ I_3& 0\end{array}\right)$: the vectors $(a_i+b_i)/\sqrt{2}$ have norm $+1$, while $(a_i-b_i)/\sqrt{2}$ have norm $-1$. Therefore the quadratic form $\omega \mapsto \omega \wedge \omega$ has signature $(3,3)$.

For $A\in SL(4,\mathbb{R})$,Thus $\rho (A)={\Lambda }^{2}A$ preserves $Q$. The special linear group is connected: polar decomposition of an invertible real matrix writes $A=OP$ with $O\in SO(4)$ and $P$ positive definite of determinant one, and both factors can be joined to the identity. The image consequently lies in $S{O}_0(3,3)$.

If ${\Lambda }^{2}A=I$, then $Au\wedge Aw=u\wedge w$ for every pair, so $A$ preserves every two-plane. A line is an intersection of two such planes, hence every line is preserved. A linear map preserving every line is scalar: apply it to basis vectors and then to their pairwise sums. Thus $A=\lambda I$, and ${\Lambda }^{2}A=I$ forces ${\lambda }^2=1$. The kernel is exactly $\{\pm I\}$. It is discrete, so the derivative of $\rho$ is injective. Both Lie algebras have dimension $15$, making the image open; an open subgroup of a connected group is the whole group. HenceThe identity-component restriction is essential. The same principle of an explicit form-preserving representation, kernel calculation and dimension argument supplies the five remaining covers.

Identify ${\mathbb{R}}^4$ with the quaternions, and use the identification of unit quaternions with $SU(2)$ proved in the root Solution. The actionpreserves the Euclidean norm by multiplicativity of quaternion norms. It is a homomorphism $SU(2)\times SU(2)\to SO(4)$: its domain is connected, so the orthogonal image has determinant $+1$.

An element in the kernel fixes $v=1$, giving $q_L=q_R=q$. Fixing every other $v$ means $q$ commutes with all quaternions, so $q$ is real; since it has unit norm, $q=\pm 1$. Hence the kernel is the diagonal subgroup $\{(1,1),(-1,-1)\}$.

The differential acts as $v\mapsto av-vb$ for imaginary quaternions $a,b$. If it vanishes, $v=1$ gives $a=b$, and commutation with all $v$ makes $a$ real and imaginary, hence zero. The differential is injective, and both Lie algebras have dimension six. The image is therefore an open subgroup of connected $SO(4)$, hence all of $SO(4)$. Thuswhere ${\mathbb{Z}}_2$ acts diagonally, not separately on the two factors.

Identify a Minkowski vector $(t,x,y,z)$ with the Hermitian matrix representation of Minkowski four-vectorsFor $A\in SL(2,\mathbb{C})$, the real-linear action $X\mapsto AX{A}^{†}$ preserves Hermiticity and the determinant, so it preserves the Lorentz quadratic form. It also preserves the cone of positive-definite Hermitian matrices, hence the future timelike cone. The group $SL(2,\mathbb{C})$ is connected, by polar decomposition of an invertible complex matrix into $SU(2)$ and positive determinant-one Hermitian matrices. Thus its image lies in the proper orthochronous Lorentz group.

If $AX{A}^{†}=X$ for every Hermitian $X$, taking $X=I$ makes $A$ unitary; the remaining equations then say it commutes with every Hermitian matrix. Their real span is all Hermitian matrices and their complex span is all complex matrices, so $A$ is scalar. The determinant condition leaves precisely $A=\pm I$. A discrete kernel gives an injective differential, and the real dimensions of both groups are six. The image is therefore open and equals the connected Lorentz group:The quotient does not include the disconnected time-reversing component of $SO(3,1)$.

The space of real $2\times 2$ matrices has a determinant form of signature $(2,2)$: writeThe action $(A,B):X\mapsto AX{B}^{-1}$ of $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ preserves this determinant. Each factor is connected by polar decomposition of an invertible real matrix: its orthogonal factor lies in connected $SO(2)$ and its positive factor is an exponential of a symmetric traceless matrix. Thus the image is contained in $S{O}_0(2,2)$.

If the action is trivial, $X=I$ gives $A=B$, and fixing every $X$ makes $A$ a scalar matrix. Determinant one gives $A=B=\pm I$. The kernel is therefore the diagonal ${\mathbb{Z}}_2$. Its discreteness makes the differential injective; domain and target both have dimension six. An open subgroup of connected $S{O}_0(2,2)$ is the whole group, soAgain the kernel is diagonal, and the target is the identity component.

The imaginary quaternions form Euclidean ${\mathbb{R}}^3$. Conjugation by a unit quaternion acts byIt preserves the imaginary subspace and its norm, and continuity from $q=1$ gives determinant $+1$. The kernel consists of quaternions commuting with all imaginary quaternions, hence is $\{\pm 1\}$.

For a unit imaginary quaternion $n$, take $q=\cos (\theta /2)+n\sin (\theta /2)$. If $v$ is perpendicular to $n$, quaternion multiplication giveswhile the component parallel to $n$ is unchanged. This produces the rotation through angle $\theta$ about axis $n$, and every element of $SO(3)$ has such an axis-angle description. The map is surjective, yieldingThe half-angle also explains why $q$ and $-q$ give the same rotation.

Use the real three-dimensional space of traceless matricesThus $-\det X$ is a quadratic form of signature $(2,1)$. The adjoint action $X\mapsto AX{A}^{-1}$ of $SL(2,\mathbb{R})$ preserves trace and determinant. Connectedness puts its image in $S{O}_0(2,1)$.

A kernel element commutes with every traceless matrix and with the identity, hence with every real matrix; it is therefore scalar. Determinant one leaves $\pm I$. The derivative is injective and both groups have dimension three, so the image is open and equals the identity component. Consequently

The Marsden-Weinstein theorem needs a Hamiltonian action, not merely an action by symplectomorphisms. Assume an equivariant moment map $\mu :P\to {\mathfrak{g}}^{\ast }$ exists, choose a coadjoint-fixed regular value $a$ (zero is the usual choice), and assume the action on $C={\mu }^{-1}(a)$ is free and proper. Regularity makes $C$ a submanifold of codimension $g=\dim G$. Equivariance and invariance of $a$ make $C$ invariant under $G$, and freeness and properness give a smooth quotient $P^{\prime }=C/G$. Thus

Let $\iota :C\hookrightarrow P$ and $\pi :C\to P^{\prime }$. For $v\in T_{p}C$ and $\xi \in \mathfrak{g}$,Moreover the conormal of $C$ is spanned by the independent $d{\mu }_{\xi }$, so nondegeneracy of $\omega$ gives $(T_{p}C{)}^{\omega }=\{{\xi }_P(p)\}$. The coadjoint-fixed value makes these orbit directions tangent to $C$. Hence the kernel of ${\iota }^{\ast }\omega$ is exactly the orbit tangent space. The restricted form is invariant and horizontal, so descends uniquely to a two-form ${\omega }^{\prime }$ withIt is closed because $\omega$ is closed. If a quotient tangent vector annihilates ${\omega }^{\prime }$, any lift lies in the kernel just identified and is vertical; the quotient vector is therefore zero. This proves nondegeneracy and completes symplectic reduction.

If the action is not free, singular strata or orbifold phenomena may occur. At a general noncentral regular value one quotients by its coadjoint stabilizer $G_a$, not by all of $G$; the regular dimension is $2n-g-\dim G_a$. These are genuine hypotheses behind the stated dimension formula.

Both proposed examples can be given explicitly. For symplectic reduction of an isotropic oscillator, take two modes withIts Hamiltonian flow is $z_j\mapsto e^{-it}{z}_j$, a circle action with moment map $H$. For $E>0$, $H^{-1}(E)$ is the sphere $S^3$ of radius $\sqrt{2E}$, and the action is free. The Hopf fibration identifies its quotient with ${\mathbb{CP}}^1\cong S^2$. On the chart $z_1\ne 0$, use $w=z_2/z_1$ and choose a section with $z_1=\sqrt{2E/(1+|w{|}^2)}$ real positive. Pulling back $\omega$ to this section givesWriting $w=x+iy$ gives $2Edx\wedge dy/(1+x^2+y^2{)}^2$, whose integral is $2\pi E$. This is the Fubini-Study form scaled by $E$ in the convention of area $2\pi$ at unit scale. The reduced dimension is $4-2=2$. At $E=0$, the circle fixes the origin and the smooth regular-level hypotheses fail.

For a unit-mass free particle in the plane, use the rotation circle action with angular-momentum moment map $\ell =x{p}_y-y{p}_x$. Choose ${\ell }_0\ne 0$ so the entire level avoids $r=0$ and the action is free. Polar canonical coordinates haveFixing $\ell ={\ell }_0$ and quotienting the rotation angle leavesThis planar rotational symplectic reduction replaces a planar free trajectory by radial motion with the centrifugal potential. A nonzero moment value is allowed because the circle is abelian and every value is coadjoint-fixed.

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.