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For a symplectic ball centered at the origin, its symplectic blowup replaces a smaller concentric ball by the disk bundle in ${\mathcal{O}}_{{\mathbb{CP}}^{n-1}}(-1)$. The zero section is the exceptional divisor $E\cong {\mathbb{CP}}^{n-1}$; the blowdown collapses $E$ to the original point and is a symplectomorphism away from $E$. The new symplectic form agrees with ${\omega }_0$ outside the surgery region, while its integral over a line in $E$ specifies the blowup size. In real dimension four, $E\cong {\mathbb{CP}}^1$ and $E^2=-1$.

A Lefschetz pencil on the compact oriented four-manifold $X$ is a mapwith finite base locus $B$, such that orientation-compatible complex coordinates give the local model $f(z_1,z_2)=[z_1:z_2]$ at a base point and $f(z_1,z_2)=z_1^2+z_2^2$ at each critical point. Blowing up all $b=|B|$ base points produces a Lefschetz fibration $\tilde{X}\to S^2$ whose regular fiber is the closed smooth fiber of the pencil.

If that fiber has genus $g$ and there are $m$ critical points, thenEvery blowup raises the Euler characteristic by one, so $\chi (\tilde{X})=\chi (X)+b$. Therefore the Euler characteristic of a Lefschetz pencil is

Now let $X={\mathbb{CP}}^2$ with its Fubini-Study form. A symplectic smooth fiber $F$ represents $dH\in H_2({\mathbb{CP}}^2;\mathbb{Z})$ for some positive integer $d$, because its symplectic area is positive. The Symplectic adjunction formula, using $K_{{\mathbb{CP}}^2}=-3H$, givesand henceThese values are $0,0,1,3,6,10,\dots$. They are all attained: two sufficiently general homogeneous polynomials of degree $d$ generate a pencil of complex plane curves with only the singularities allowed in a Lefschetz pencil, and its smooth fibers are symplectic. Thus the displayed list is exactly the list described by the Genus of a symplectic Lefschetz pencil on the complex projective plane.

The special orthogonal group in three dimensions isDifferentiating $A(t{)}^{T}A(t)=I$ at the identity shows that its Lie algebra is the space of skew-symmetric matrices. A convenient basis iswith $[J_a,J_b]={ϵ}_{abc}{J}_c$.

The infinitesimal action of $\exp (t{J}_a)$ on a point $x\in {\mathbb{R}}^3$ is $J_{a}x=e_a\times x$. It therefore generates the vector fieldFundamental vector fields for this left action form an antihomomorphism with the stated convention, and direct differentiation givesso their span is closed under the Lie bracket of vector fields.

The brackets $\{x,y\}=z$, $\{y,z\}=x$, and $\{z,x\}=y$ define the rotational Lie-Poisson structure on R3. With the convention that a Hamiltonian vector field acts by $X_H(f)=\{f,H\}$,Hence the required Hamiltonians are simplyThe quadratic functionsatisfies $\{F,x_a\}=0$ for all $a$, so it is a Casimir function of a Poisson manifold. Its nonzero regular level sets $F=R^2$ are spheres. The Poisson tensor has rank two there and is tangent to each level set, so it inverts to a symplectic form; each sphere is a symplectic leaf. Rotations preserve both $F$ and the alternating tensor ${ϵ}_{abc}$, hence preserve the restricted Poisson tensor and its inverse symplectic form. The $SO(3)$ action therefore restricts to a symplectic action on every sphere $S_R^2\subset {\mathbb{R}}^3$.

A transformation of phase space is canonical when it preserves the symplectic form, or equivalently all Poisson brackets. For one degree of freedom this isequivalently $\{Q,P{\}}_{q,p}=1$.

Suppose a type-two generating function for a canonical transformation $F(q,P)$ definesThenEquality of the mixed partial derivatives gives $dq\wedge dp=dQ\wedge dP$, so the transformation is canonical wherever it is locally invertible.

Forthe fundamental theorem of calculus and differentiation under the integral sign giveThusand in particular $P=(p^2+q^2)/2$.

For the unit-frequency simple harmonic motion, the transformed Hamiltonian is simply $H=P$. Hamilton's equations becomeThereforeand transformation back gives the familiar solution

Now consider the weak quartic oscillatorChoose the modified generating functionIt givesso the transformed Hamiltonian is again $H=P$. Hence $P=E$ and $Q=t-t_0$. Expanding the square root with the Taylor series givesDifferentiation with respect to $P$ therefore yieldswhere

Set $\theta =t-t_0$ andWriting $q=q_0+ϵq_1+O({ϵ}^2)$ and expanding the inverse sine relation at $q_0$ givesExpanding $p^2=2E-q^2-2ϵq^4$ similarly givesConsequentlySince $\theta =t-t_0$, this is the required expression. The formula is understood away from turning points where $p_0=0$ and the ratio $q/p$ itself is singular.

Let $J_0$ be the standard complex structure on ${\mathbb{R}}^{2n}$, let ${\omega }_0$ be the standard symplectic form, and let $g_0$ be the Euclidean inner product. They satisfyThe two-out-of-three property for unitary structures says that a real linear map preserving any two of these structures preserves the third. In terms of the corresponding matrix groups,

For example, if $A$ preserves $J_0$ and ${\omega }_0$, thenso $A$ preserves $g_0$. If it preserves $J_0$ and $g_0$, the second displayed identity shows that it preserves ${\omega }_0$. Finally, $J_0$ is uniquely determined by $g_0(J_{0}u,v)={\omega }_0(u,v)$, so preservation of $g_0$ and ${\omega }_0$ implies $A{J}_0=J_{0}A$. The common intersection is therefore the unitary group.

Any symplectic form on $S^2$ orients its tangent bundle. Choose a compatible complex structure and inner product; this reduces the structure group to $SO(2)$. Since every line in an oriented symplectic plane is Lagrangian,is an oriented circle bundle.

If a transition function of $T{S}^2$ rotates vectors through an angle $\alpha$, its action on unoriented lines rotates the coordinate $e^{2i\theta }$ through $2\alpha$. The Euler class of the Lagrangian-line bundle of an oriented plane bundle therefore givesBy the Poincaré-Hopf theorem,for the chosen orientation, up to changing both signs. Hence the Euler number of $\mathcal{L}Gr(T{S}^2)$ is $\pm 4$, which is nonzero. A smoothly trivial oriented circle bundle has zero Euler class, so $\mathcal{L}Gr(T{S}^2)\to S^2$ cannot be smoothly trivial for any choice of symplectic form.

Take the toruswith the translation-invariant symplectic formThe coordinate vector fields give a global symplectic frame of $TM$, so $TM$ is symplectically trivial. Passing fiberwise to the Lagrangian Grassmannian bundle giveswhich is a smooth trivialization over the compact symplectic manifold $T^{2n}$.

Moser's trick states that if $M$ is compact and ${\omega }_t$, $0\le t\le 1$, is a smooth family of symplectic forms whose de Rham cohomology class is independent of $t$, then there is an isotopy ${\varphi }_t$ withSince $[{\dot{\omega }}_t]=0$, choose a smooth family of one-forms ${\sigma }_t$ with ${\dot{\omega }}_t=d{\sigma }_t$. Nondegeneracy of ${\omega }_t$ uniquely determines a vector field $X_t$ byCompactness makes its flow ${\varphi }_t$ exist for the whole interval. Cartan's magic formula and $d{\omega }_t=0$ givewhich proves the theorem.

Smooth degree-$d$ hypersurfaces form the complement of the discriminant in the projective space of degree-$d$ homogeneous polynomials. This complement is path connected, so $X$ and $X^{\prime }$ lie in a smooth one-parameter family. The Ehresmann fibration theorem identifies the fibers smoothly. Under such an identification, the restrictions of the Fubini-Study form form a family ${\omega }_t$ whose cohomology class is the fixed restricted hyperplane class. Moser's trick therefore gives the symplectic equivalence of smooth projective hypersurfaces.

It remains to construct the finite subgroup for one convenient hypersurface. On the Fermat hypersurfacethe group $({\mu }_d{)}^{n+1}$ acts by diagonal coordinate multiplication. It preserves both $X_F$ and the Fubini-Study form. The kernel of its projective action is the diagonal subgroup ${\mu }_d$, so the effective Fermat hypersurface diagonal symmetry group isConjugating this action by a symplectomorphism $X_F\to X$ gives the required subgroup of $\mathrm{Symp}(X,{\omega }_{FS}{|}_X)$.

A smooth antisymmetric contravariant two-tensor defines a biderivation $\{F,G\}=\omega (dF,dG)$. It is a Poisson bivector when this bracket satisfies the Jacobi identity. Its rank may vary; where it is nondegenerate it inverts to a symplectic form. The coordinate Jacobi condition for a Poisson bivector makes the integrability constraint explicit.

Two closed symplectic forms agreeing as ambient bilinear forms along a compact embedded submanifold are equivalent near it by a diffeomorphism fixing it pointwise. Shrink until their affine interpolation is nondegenerate. A relative homotopy primitive $\eta$ for their difference vanishes as an ambient covector on the submanifold. Solving ${\iota }_{Z_t}{\omega }_t=-\eta$ gives a vector field vanishing there; its local flow fixes the submanifold and makes ${\omega }_t$ constant under pullback.

The symplectic area of a compact symplectic surface $(\Sigma ,\omega )$ is ${\int }_{\Sigma }\omega$, using the orientation determined by the symplectic form.

For real Klein-Gordon field solutions with suitable support or boundary conditions, $\Omega ({\varphi }_1,{\varphi }_2)={\int }_{\Sigma }({\varphi }_1{n}^a{\nabla }_a{\varphi }_2-{\varphi }_2{n}^a{\nabla }_a{\varphi }_1)d\Sigma$ is a conserved symplectic form. The field equation makes its current divergence-free. Its complexification gives $(u,v{)}_{KG}=i\Omega (u^{\ast },v)$, the Klein-Gordon inner product. A compatible complex structure on the Klein-Gordon solution space is additional data needed for a particle representation.

A symplectic leaf is a maximal connected immersed submanifold on which the Poisson bivector has constant full rank and therefore inverts to a symplectic form.

A symplectic form on a $2n$-dimensional smooth manifold determines the orientation in which the nowhere-zero volume form ${\omega }^n/n!$ is positive. Consequently every symplectic manifold is orientable. In dimension two this forbids a symplectic form on the Möbius strip.

For a Hamiltonian group action, a regular value level preserved by a free proper group action has a smooth quotient. The restricted symplectic form descends to a symplectic form on that quotient when the kernel is exactly the orbit directions; this is symplectic reduction.

A vector field on a symplectic manifold is symplectic when its flow preserves the symplectic form. Infinitesimally this is ${\mathcal{L}}_X\omega =0$. By Cartan's magic formula it is equivalent to $d({\iota }_X\omega )=0$. A Hamiltonian vector field has an exact contraction with the form, so is symplectic. On the standard two-dimensional torus, ${\partial }_x$ contracts to the closed but nonexact form $dy$, providing a symplectic non-Hamiltonian example.

The symplectic form on a $2n$-dimensional symplectic manifold determines the positive volume form ${\omega }^n/n!$ and its total integral, the symplectic volume, using the induced orientation.

The top-degree differential form ${\omega }^n/n!$ on a $2n$-dimensional symplectic manifold. Every canonical transformation preserves this form. For $n>1$, preservation of this volume alone does not imply preservation of the symplectic form.

With $\omega =-d\alpha$ and ${\omega }_{\sigma }=\omega +{\pi }^{\ast }\sigma$, the one-form graph of $\theta$ is a Lagrangian submanifold exactly when $d\theta =\sigma$. Indeed its pulled-back symplectic form is $\sigma -d\theta$ and its dimension is half that of the total space. This extends graph of a closed one-form is Lagrangian.