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A symplectic capacity assigns $c(X,\omega )\in [0,\infty ]$ to symplectic manifolds in a fixed dimension, is monotone under symplectic embeddings, satisfies $c(X,a\omega )=ac(X,\omega )$ for $a>0$, and is nontrivial in the sense thatThe second domain is the standard symplectic cylinder. A normalized symplectic capacity assigns $\pi$ to both unit domains, hence $\pi R^2$ to both radius-$R$ domains. The general axioms already give invariance under symplectomorphisms and quadratic scaling under coordinate dilation. For the rigidity argument, use the assumed existence of capacities in each dimension, including after adding a symplectic plane; normalization is not needed.

Let ${\varphi }_j$ be symplectomorphisms converging locally uniformly to a smooth diffeomorphism $\varphi$. Work in Darboux charts. For a sufficiently small solid ellipsoid $E$ centered at a chosen point and $0<a<1$, uniform convergence and degree of a continuous mapping give, for large $j$,For the second inclusion, ${\varphi }^{-1}{\varphi }_j$ is uniformly close to the identity on the larger ellipsoid. Its boundary stays outside $\overline{E}$ and is homotopic there to the identity, so it has degree one about every point of $E$ and must cover $E$. Monotonicity and scaling now give $a^{2}c(E)\le c(\varphi (E))\le a^{-2}c(E)$. Letting $a$ tend to one proves capacity preservation under uniform limits.

At the chosen point, write $T=D\varphi$. The rescaled maps $(\varphi (tz)-\varphi (0))/t$ preserve capacities of ellipsoids and converge uniformly on compact sets to $T$. Applying the same sandwich argument proves that $T$ preserves the capacities of all solid ellipsoids; so does $T^{-1}$.

We prove the required linear capacity rigidity explicitly. Write the standard symplectic form as $\Omega$. Suppose $\Omega (u,v)=1$ andComplete $u,v$ to a symplectic basis, and complete $T^{T}u/\lambda$, with the appropriately signed $T^{T}v/\lambda$, to another symplectic basis. If their basis matrices are $P,P^{\prime }$, then $A=(P^{\prime }{)}^{-1}{T}^{T}P$ has first two columns $\lambda e_1,\pm \lambda f_1$. Hence $A^T$ sends the unit ball into the symplectic cylinder of radius $\lambda$. This matrix differs from $T$ by symplectic linear maps, so it preserves capacities of ellipsoids. Its first coordinate pair is multiplied by $\lambda$, with a possible sign, and therefore $(A^T{)}^k$ sends the unit ball into the cylinder of radius ${\lambda }^k$. Monotonicity givesfor every $k$, contradicting finiteness of the right-hand unit-cylinder capacity as $k$ tends to infinity. For a normalized capacity, the first iterate already gives the contradiction.

A zero pairing is handled by a small perturbation of $u,v$. A pairing of absolute value greater than one gives the same contradiction for $T^{-1}$, after normalizing the transformed pair. Thusfor every pair. The squared bilinear forms agree, so their difference times their sum is the zero polynomial. The real polynomial ring is an integral domain; consequently one factor vanishes identically. Therefore $T$ is either symplectic or anti-symplectic.

To eliminate the negative sign, repeat the argument for ${\varphi }_j\times {\mathrm{id}}_{{\mathbb{R}}^2}$, which are symplectomorphisms for $\omega \oplus {\omega }_{{\mathbb{R}}^2}$. If ${\varphi }^{\ast }\omega =-\omega$ at the chosen point, the limit pulls this product form back to $-\omega \oplus {\omega }_{{\mathbb{R}}^2}$, which is neither the product form nor its negative. This contradicts the preceding derivative classification. Thus ${\varphi }^{\ast }\omega =\omega$ everywhere. We have proved the Eliashberg–Gromov rigidity theorem: the symplectomorphism group is closed in the $C^0$ topology on diffeomorphisms.

Finally put $E=W^{\perp }$, with perpendicularity taken for the standard Euclidean inner product. Its dimension is two, so being nonisotropic means that the restricted symplectic form is nondegenerate. For the standard compatible complex structure $J$, $W^{\omega }=JE$. If $Jv\in W\cap JE$, then $v\in E$ and $\omega (v,e)=0$ for every $e\in E$, forcing $v=0$. Thus $W$ is a symplectic subspace and admits a symplectic basis extended to the whole space.

In the resulting linear symplectic coordinates, $W$ is $\{q_1=p_1=0\}$. The projection of the bounded set $U$ to the $(q_1,p_1)$ plane is bounded, so $U+W$ lies in a symplectic cylinder of some finite radius $R$. As $U$ is nonempty and open, it contains a ball of some radius $r>0$, which is also contained in $U+W$. Monotonicity and scaling giveas in capacity of a bounded set thickened by a symplectic subspace. For a normalized symplectic capacity, the two bounds simplify to $\pi r^2$ and $\pi R^2$.

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 manifold is a smooth manifold with a smooth antisymmetric bivector $\pi =\frac{1}{2}{\pi }^{ij}{\partial }_i\wedge {\partial }_j$ whose function bracket $\{f,g\}={\pi }^{ij}({\partial }_{i}f)({\partial }_{j}g)$ satisfies the Jacobi identity. The bracket is automatically a derivation in each argument. A symplectic manifold has a nondegenerate differential two-form $\omega$ satisfying $d\omega =0$, that is, a closed differential form; nondegeneracy forces its dimension to be even. It determines a nondegenerate Poisson structure. We use matrix convention $\pi =-{\omega }^{-1}$, so that ${\iota }_{X_H}\omega =dH$ corresponds to $X_H(f)=\{f,H\}$.

Examples of genuinely nonsymplectic Poisson geometries include the zero bracket on ${\mathbb{R}}^3$ and the Lie-Poisson bracket $\{x_i,x_j\}={ϵ}_{ijk}{x}_k$ on ${\mathbb{R}}^3$. Their three-dimensional underlying manifold cannot carry a nondegenerate two-form. In the latter case the Poisson rank is two away from the origin and zero at the origin, with spheres and the origin as symplectic leaves. Even in even dimension a given Poisson structure need not come from a symplectic form: $\pi =x{\partial }_x\wedge {\partial }_y$ on ${\mathbb{R}}^2$ satisfies Jacobi but vanishes along $x=0$. The statement concerns this degenerate structure, not a claim that ${\mathbb{R}}^2$ admits no other symplectic structure.

For any candidate bivector, applying the Jacobi identity to coordinate functions gives the necessary conditionIt is sufficient as well: expanding $\{f,\{g,h\}\}$ and its cyclic companions cancels all terms with second derivatives by antisymmetry of $\pi$. What remains is $J^{ijk}({\partial }_{i}f)({\partial }_{j}g)({\partial }_{k}h)$. This is the coordinate Jacobi condition for a Poisson bivector.

For a symplectic form, differentiate $\omega \pi =-I$ to obtain ${\partial }_{\ell }\pi =\pi ({\partial }_{\ell }\omega )\pi$. Substituting this into the cyclic expression givesbecause $d\omega =0$. Thus the inverse symplectic structure always satisfies the Jacobi identity. Conversely, for a nondegenerate Poisson structure, multiplying by three inverse matrices proves closure of its associated two-form.

In momentum space, choose Hamiltonian $H(\mathbf{p})=E(\mathbf{p})$ and the constant Poisson bivectorIts Poisson bracket is $\{f,g\}=e\mathbf{B}\cdot (\nabla f\times \nabla g)$. The Hamiltonian vector field therefore givesexactly the prescribed motion. The sign here uses the signed charge $e$ as printed; an electron's physical charge may be negative. Since the magnetic field is uniform, every derivative of ${\pi }^{ij}$ vanishes, so every term in $J^{ijk}$ is zero. Jacobi holds independently of the dispersion function $E$. This is the momentum-space Poisson structure for a uniform magnetic field.

For $e\mathbf{B}\ne 0$ this bivector has rank two and $C(\mathbf{p})=\mathbf{B}\cdot \mathbf{p}$ is a Casimir function of a Poisson manifold: $\{C,f\}=0$ for every $f$. Its symplectic leaves are the planes $C=c$. For axes with $\mathbf{B}=B_0{\mathbf{e}}_3$, the leaf form is $\omega =(e{B}_0{)}^{-1}d{p}_1\wedge d{p}_2$ in our inverse-sign convention. Also $\dot{E}=\{E,E\}=0$, so trajectories on a Fermi surface lie on its intersections with these planes, typically curves. The energy surface is invariant for this Hamiltonian, but need not be a Poisson submanifold; the bracket has been defined on the whole momentum space. If $e\mathbf{B}=0$, the Poisson structure and motion vanish.

A Lagrangian submanifold of a $2n$-dimensional symplectic manifold $(M,\omega )$ is an $n$-dimensional submanifold $L$ such that the pullback of $\omega$ to $L$ is zero. Equivalently, each tangent space $T_{x}L$ is its own symplectic orthogonal. The dimension condition makes it maximally isotropic: in a nondegenerate alternating vector space no isotropic subspace has dimension greater than $n$.

In geometric quantization, a line bundle with a connection on a vector bundle whose curvature is proportional to $\omega /\hslash$ provides the initial quantum-state space. A real polarization foliates phase space locally by Lagrangian leaves; requiring sections to be covariantly constant along these leaves reduces their independent variables by half. The curvature restricts to zero on a Lagrangian leaf, allowing local flat sections. Globally its holonomy can impose the familiar Bohr-Sommerfeld quantization condition. Thus Lagrangian submanifolds describe choices of quantum configuration data and the support of semiclassical states, rather than requiring simultaneous sharp values of all position and momentum coordinates.

For the thermodynamic application, regard $(S,V,N)$ as base coordinates $q^i$ and $(T,-P,\mu )$ as fiber coordinates $p_i$ on $T^{\ast }{\mathbb{R}}^3$. The canonical one-form on a cotangent bundle and the position-first symplectic form areThe negative sign in the pressure coordinate is essential. A smooth potential $M(S,V,N)$ determines the embeddingwhere the last three coordinates are ordered as $(T,-P,\mu )$: explicitly $T=M_S$, $P=-M_V$, and $\mu =M_N$, so the second fiber coordinate is $-P=M_V$. Accordingly, in the actual $p_i$ coordinates the graph isProjection to the base is the identity, so this is an embedded three-dimensional submanifold, even if the Hessian of $M$ is degenerate. The first law of thermodynamics gives ${\iota }_M^{\ast }\theta =dM$. HenceIt has half the dimension of the ambient cotangent bundle and is therefore Lagrangian. This proves the thermodynamic potential as a Lagrangian graph construction for every smooth choice of $M$, locally on any domain on which the potential is defined. No invertibility, extensivity or stability assumption is needed for this geometric conclusion.

In components the same vanishing follows from equality of mixed derivatives. The thermodynamic conjugate variables satisfyThese cancel the three coefficients of the pulled-back two-form. They are the integrability relations behind the local thermodynamic potential, and demonstrate directly why the first-law graph is Lagrangian.

Hamilton's equations are ${\dot{q}}_j=\partial H/\partial p_j$ and ${\dot{p}}_j=-\partial H/\partial q_j$. The Arnold-Liouville theorem states that $n$ functionally independent first integrals in involution on a $2n$-dimensional symplectic manifold give local integrability; a compact connected regular common level is an $n$-torus. Near such a torus there are action-angle variables $(I,\vartheta )$ with $H=H(I)$, $\dot{I}=0$ and $\dot{\vartheta }=\partial H/\partial I$.

For the displayed diagonal oscillator Hamiltonian set $E_k=(p_k^2+{\omega }_k^2{q}_k^2)/2$. Each is constant, since ${\dot{q}}_k=p_k$ and ${\dot{p}}_k=-{\omega }_k^2{q}_k$. Different $E_k$ depend on disjoint canonical pairs, so their Poisson brackets vanish. Their differentials are independent wherever every $d{E}_k$ is nonzero, an open dense set; this proves complete integrability. For nonzero frequencies put ${\Omega }_k=|{\omega }_k|$. A positive-energy level in each pair is an ellipse with semiaxes $\sqrt{2E_k}$ and $\sqrt{2E_k}/{\Omega }_k$. Its enclosed area is $2\pi E_k/{\Omega }_k$, so the action variables areFor the usual positive oscillator frequencies this is $I_k=E_k/{\omega }_k$. If a printed “constant” frequency is allowed to be zero, that pair is a free particle: $p_k$ is a commuting integral, but its generic level is noncompact and has no closed-cycle oscillator action. Complete integrability persists, while the torus/action formula requires nonzero frequencies.

A Hamiltonian system on a $2d$-dimensional symplectic manifold is Liouville integrable when it has $d$ first integrals, including the Hamiltonian, whose differentials are independent on a dense regular region and which pairwise have zero Poisson bracket. The Arnold-Liouville theorem states that a compact connected regular common level set is a $d$-torus; near it there are action-angle variables $(I,\theta )$ in which the Hamiltonian depends only on $I$ and the motion is $\dot{I}=0$, $\dot{\theta }={\nabla }_{I}H$. Without compactness, the torus conclusion needs modification.

Here $q_1$ is a cyclic coordinate, so $p_1=\ell$ is conserved and $\{H,p_1\}=0$. The Hamilton equations areThe differentials $dH,d{p}_1$ are independent away from the circular-orbit critical set, establishing integrability. Put $r=q_2$; on a fixed angular-momentum level,For a bounded noncollision orbit require $\ell \ne 0$ and $-k^2/(2{\ell }^2)<E<0$. Its two turning radii areThey are positive and $\alpha <\beta$, and $p_2^2=(-2E)(r-\alpha )(\beta -r)/r^2$. In terms of $u=1/r$ and the angular coordinate, eliminating time gives $u^{\prime \prime }+u=k/{\ell }^2$. ThusAfter an angular change of $2\pi$ the phase-space state returns to itself, and the elapsed time is finite because $r$ is bounded away from zero and infinity. This proves periodicity; a torus alone would only guarantee quasiperiodic motion. At $E=-k^2/(2{\ell }^2)$ the orbit is circular and remains periodic. For $E\ge 0$ the allowed radial motion is unbounded. The collision case $\ell =0$ reaches the excluded boundary $r=0$ and is not a periodic orbit in the given phase space.

With the usual action convention $(2\pi {)}^{-1}\oint pdq$, the angular cycle gives $I_1=\ell$, and the two radial legs giveFor completeness, substitute $r=(\alpha +\beta )/2+(\beta -\alpha )\cos \theta /2$ to evaluate the integral as $\pi [(\alpha +\beta )/2-\sqrt{\alpha \beta }]$. Therefore $I_2=k/\sqrt{-2E}-|\ell |$ and $H=-k^2/[2(I_2+|I_1|{)}^2]$. The two frequencies have equal magnitude, consistent with the closed Kepler orbits. The circular limit has $I_2=0$ and $\alpha =\beta$.

The Arnold-Liouville theorem concerns a connected compact component $L$ of a regular common level of $n$ commuting independent smooth functions $F=(F_1,\dots ,F_n)$ on a $2n$-dimensional symplectic manifold. It states that $L$ is a Lagrangian torus, and that a neighbourhood of $L$ has action-angle variablesin which every $F_i$ depends only on $I$. If $H=F_1$ or, more generally, $H=h(F)$, its Hamiltonian flow satisfiesAngles have period $2\pi$ here. The theorem is local near this regular compact component; it does not assert globally defined action-angle variables across singular levels or over an entire base with monodromy.

First, $L$ has dimension of a manifold $n$ by the submersion theorem. The Hamiltonian vector fields $X_{F_i}$ are independent, tangent to $L$, and commute. They span $TL$. Moreoverso $L$ is a Lagrangian submanifold. Compactness makes these restricted vector fields complete. Their joint flow is an ${\mathbb{R}}^n$-action on $L$. Its orbits are open because the fields span $TL$, so connectedness gives one orbit. The stabilizer $\Lambda$ of a point is discrete by the inverse function theorem, andCompactness forces $\Lambda$ to be a full-rank Euclidean lattice: otherwise an unbounded linear coordinate transverse to its span would descend to the quotient. Thus $L$ is an $n$-torus.

Next choose a small ball of regular values near $F(L)$ and a neighbourhood $N$ of this component which is a product family of compact tori $L_c$. This local trivialization follows directly by choosing transverse vector fields $Z_j$ with $d{F}_i(Z_j)={\delta }_{ij}$ and lifting short radial paths in the base; compactness of $L$ gives a uniform neighbourhood where their flows exist. Hence $N$ deformation retracts onto $L$. The commuting joint flows on nearby fibres have smoothly varying full-rank Hamiltonian period lattices. A basis $t^{(j)}(c)\in {\mathbb{R}}^n$ of their periods can be chosen smoothly on this small ball: continue the return equations from a basis at the central fibre, using their nonsingular vertical flow derivatives and the implicit function theorem. No global choice of lattice basis is needed.

Because $\omega {|}_L=0$ and $N$ retracts onto $L$, its closed symplectic form is exact on $N$. Choose a one-form $\lambda$ with $d\lambda =-\omega$. Let ${\gamma }_j(c)$ be the smoothly continued cycles represented by the period basis and define the action integralsFor a transverse variation $v$ of the fibre, differentiating a cycle integral and using Cartan's magic formula eliminates the integral of the exact term. With the cycle parametrized by the joint-flow time $t^{(j)}$, this givesThusThe period matrix is nonsingular, so $I$ gives coordinates on the base. The corresponding Hamiltonian vector fields satisfyTheir time-$2\pi$ flows return along the basis cycles. They commute because each $I_j$ is a function of the commuting $F_i$. Consequently they give a free torus action on $N$.

Choose a section over the action-coordinate ball and use this torus action to define angles $\theta$. Since $X_{I_j}={\partial }_{{\theta }_j}$, the symplectic form takes the formThe closed base two-form $\beta$ is exact on the ball: write $\beta =d\alpha$, $\alpha ={\sum }_j{a}_j(I)d{I}_j$, by the Poincaré lemma. Changing the angular origins by ${\theta }_j^{\prime }={\theta }_j+a_j(I)$ removes this term, since ${\sum }_{j}d{\theta }_j^{\prime }\wedge d{I}_j=\omega$. This completes the construction of action-angle variables. The functions $F_i$ are constant on the fibres, so are functions of $I$ alone, and Hamilton's equations give the claimed straight-line motion. This proves both the topological and symplectic parts of the Arnold-Liouville theorem.

On a $2n$-dimensional symplectic manifold, a completely integrable Hamiltonian system has $n$ first integrals in involution $F_1=H,F_2,\dots ,F_n$ whose differentials are independent almost everywhere, hence on a dense open regular set:An equivalent formulation allows $H$ to be a function of the $n$ commuting independent integrals. The independence requirement is on the regular set, permitting exceptional critical points and singular fibres.

For example, on ${\mathbb{R}}^{2n}$ take $H={\sum }_{j=1}^n{E}_j$ with $E_j=(p_j^2+{\alpha }_j^2{q}_j^2)/2$ and ${\alpha }_j>0$. The disjoint coordinate pairs make all Poisson brackets $\{E_i,E_j\}$ zero, and the differentials are independent whenever no pair $(q_j,p_j)$ vanishes. Replacing $E_1$ by $H$ is an invertible linear map of this list. Thus $F_1=H,F_j=E_j$ for $j\ge 2$ give the required example. Irrational frequency ratios affect orbit density, but not complete integrability.

A symplectic manifold is a smooth manifold $M$ equipped with a differential two-form $\omega$ such thatThe first condition says it is a closed differential form; the second says it is pointwise nondegenerate. A nondegenerate skew-symmetric matrix has even size, so the dimension is $2n$. Equivalently, ${\omega }^n$ is a nowhere-zero top-degree differential form. It supplies an orientation and the volume form ${\omega }^n/n!$. These are the defining conditions on the symplectic form.

Odd-dimensional spheres cannot carry a symplectic form, by even-dimensionality. For $S^{2m}$ with $m\ge 2$, the second de Rham cohomology group vanishes. Any hypothetical symplectic form would therefore be $\omega =d\alpha$. The symplectic cohomology obstruction givesby the Generalized Stokes theorem. But the symplectic orientation makes ${\omega }^m$ a positive volume form, whose integral on this nonempty compact manifold is positive. This is a contradiction.

The oriented area form on $S^2$ is nondegenerate and automatically closed, since a two-dimensional manifold has no nonzero three-forms. Thus among positive-dimensional spheres, exactly $S^2$ admits a symplectic structure:If zero-dimensional symplectic manifolds are admitted, $S^0$ also qualifies: its zero two-form is nondegenerate on the zero tangent spaces. This is a convention-dependent additional case, not another positive-dimensional example.

The Symplectic Darboux theorem says that every point of a $2n$-dimensional symplectic manifold has local coordinates $(q^1,\dots ,q^n,p_1,\dots ,p_n)$ in whichFirst choose linear coordinates at the point so the form there is standard. The required symplectic basis can be constructed inductively: choose $e,f$ with $\omega (e,f)=1$, split off their span, and repeat on its nondegenerate symplectic orthogonal complement. Extend these coordinates to a chart centered at zero.

Let ${\omega }_0$ be the constant standard form in this chart and $\eta =\omega -{\omega }_0$. The closed form $\eta$ vanishes at zero. On a small star-shaped ball the radial Poincaré lemma supplies a primitiveSince ${\eta }_0=0$, this primitive is $O(|x{|}^2)$. The interpolating forms ${\omega }_t={\omega }_0+t\eta$ are nondegenerate on a common smaller ball, because they all agree with ${\omega }_0$ at zero and $t$ ranges over a compact interval.

Apply the local version of Moser's trick: solve ${\iota }_{X_t}{\omega }_t=-\alpha$. The vector fields are $O(|x{|}^2)$ and fix zero. On a sufficiently small ball their flows exist for $0\le t\le 1$ and remain inside the coordinate chart; the quadratic bound makes their displacement smaller than the available margin. The same pullback calculation gives $f_1^{\ast }\omega ={\omega }_0$. Thus $f_1$ is a local symplectomorphism from the standard ball into $M$, and its inverse supplies the desired Darboux chart. There are no local symplectic invariants beyond dimension.

A Lagrangian submanifold of a $2n$-dimensional symplectic manifold is an embedded $n$-dimensional submanifold on which the symplectic form restricts to zero. The dimension requirement distinguishes it from a lower-dimensional isotropic submanifold.

A differential one-form $\sigma$ on $L$ gives an embedded section of its cotangent bundle, since $\pi \circ \sigma =\mathrm{id}$. Directly from the canonical one-form on a cotangent bundle,Its graph already has half the ambient dimension. ThereforeThis proves the graph of a closed one-form is Lagrangian criterion, in both directions.

Use the Weinstein neighborhood theorem: a neighborhood of a compact Lagrangian submanifold is symplectomorphic to a neighborhood of the zero section of its cotangent bundle, with the identification equal to the identity on that submanifold. Take the canonical sign $-d\lambda$ in this identification. We also use C1 openness of diffeomorphisms: on a compact manifold, all smooth self-maps sufficiently close in the $C^1$ topology to a fixed diffeomorphism are themselves diffeomorphisms.

In $M\times M$ put $\Omega =-{\mathrm{pr}}_1^{\ast }\omega +{\mathrm{pr}}_2^{\ast }\omega$. The diagonal $\Delta$ is Lagrangian. For a symplectomorphism $\varphi$, its graph ${\Gamma }_{\varphi }$ is also Lagrangian, because its pullback of $\Omega$ is $-\omega +{\varphi }^{\ast }\omega =0$.

If $\varphi$ is sufficiently $C^1$-close to the identity, ${\Gamma }_{\varphi }$ lies in the fixed Weinstein neighborhood of $\Delta$. Its image in $T^{\ast }\Delta$ is transverse to the cotangent fibers and is a section: the projection of that image to $\Delta \cong M$ is $C^1$-close to the identity, hence is a diffeomorphism on compact $M$. Reparametrizing by this projection identifies the image with $\mathrm{graph}(\sigma )$ for a small differential one-form $\sigma$.

The preceding graph of a closed one-form is Lagrangian criterion gives $d\sigma =0$. Since $H_{\mathrm{dR}}^1(M)=0$, the de Rham cohomology definition gives $\sigma =df$. Intersections with the zero section are precisely the critical points of $f$; under the neighborhood identification these are the intersections ${\Gamma }_{\varphi }\cap \Delta$, hence the fixed points of $\varphi$.

On a nonempty compact manifold without boundary, $f$ has a maximum and a minimum. If it is nonconstant, these occur at distinct critical points. If it is constant, $df=0$ everywhere, so the whole graph is the diagonal and every point is fixed. For positive-dimensional $M$, there are at least two distinct fixed points. This is the nearby exact Lagrangian intersection lemma applied to the diagonal. No connectedness assumption is needed.

The usual positive-dimensional convention is necessary for the assertion: if zero-dimensional symplectic manifolds are allowed, a single-point $M$ has $H^1=0$ and only one fixed point. That is a literal exception to the printed statement.

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

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

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

Liouville integrability on a $2n$-dimensional symplectic manifold means that there are $n$ functionally independent first integrals, including $H$, whose pairwise Poisson brackets vanish. Near a compact connected regular common level set, the Arnold-Liouville theorem gives action-angle variables $(J_i,{\theta }_i)$ with $J_i=(2\pi {)}^{-1}\oint pdq$, ${\theta }_i$ periodic modulo $2\pi$, $H=H(J)$, ${\dot{J}}_i=0$ and ${\dot{\theta }}_i=\partial H/\partial J_i$.

For a frozen mass, the turning point is $q_{\ast }=(2E{)}^{1/(2k)}$. The action variable isAdiabatic invariance of the action holds to leading order when the mass varies slowly compared with an orbital period, away from degenerating orbits. Keeping $J$ fixed givesFor $k=1$ this agrees with $E=\omega J$ and $\omega \propto m^{-1/2}$.

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.

Prequantization realizes a symplectic manifold using a Hermitian line bundle with connection whose curvature is $i\omega /\hslash$ in the convention ${\iota }_{X_f}\omega =df$. Its existence requires integral symplectic periods. On sections, the Kostant-Souriau prequantum operator represents every classical observable and its Poisson bracket, but the resulting states depend on the full phase space. A polarization in geometric quantization removes this excess of variables.

A prequantum line bundle is a Hermitian line bundle over a symplectic manifold with unitary connection on a vector bundle of the displayed curvature. With $\omega =-d\theta$, a local connection is $\nabla =d-i\theta /\hslash$. Reversing the Hamiltonian-vector-field convention or passing to the dual line bundle reverses the corresponding curvature convention. Integrality of $\omega /(2\pi \hslash )$ on closed two-cycles is the topological existence condition.

A symplectic basis organizes a nondegenerate alternating bilinear form into independent pairs. Let $(V,\omega )$ be a finite-dimensional vector space over a field $K$, with $\omega$ a nondegenerate bilinear form satisfying $\omega (v,v)=0$ for every vector $v$. Its dimension is even, say $2n$, and a symplectic basis is an ordered basis $\mathcal{B}=(e_1,\dots ,e_n,f_1,\dots ,f_n)$ such thatHere ${\delta }_{ij}$ is the Kronecker delta. Each $(e_i,f_i)$ is a hyperbolic pair; different pairs have zero mutual pairing. These conventions apply in every characteristic. In characteristic two, $-1=1$, but the condition $\omega (v,v)=0$ is still essential: a symmetric bilinear form need not be an alternating bilinear form. The existence theorem and the characteristic caveat are given in Bilinear Forms by Keith Conrad (Keith Conrad, Bilinear Forms, §1, Theorem 1.6; §5, Theorems 5.1 and 5.4 and Definition 5.3).

The benefit is a common coordinate language for every symplectic vector space of the same finite dimension over the same field. The span of all $e_i$, and separately of all $f_i$, is a Lagrangian subspace; each paired span is a symplectic subspace. In Hamiltonian mechanics, the corresponding pairing of position and momentum coordinates describes phase space and canonical transformations. On a symplectic manifold, a symplectic basis describes one tangent space; Darboux coordinates additionally describe a whole local neighborhood.^[ref]

For a closed $2n$-dimensional symplectic manifold, a symplectic blowup of size $\epsilon =\pi R^2$ removes the symplectic volume of the radius-$R$ ball. Its exceptional divisor has lower dimension and contributes no ambient volume. The displayed volume loss shows that different size choices can give forms that are not related by a symplectomorphism.