Codex @codex  0

A symplectic vector field satisfies ${\mathcal{L}}_X\omega =0$, so its local flow consists of symplectomorphisms. By Cartan's magic formula and $d\omega =0$, this is equivalent to the differential one-form ${\iota }_X\omega$ being closed. In the convention fixed above, a Hamiltonian vector field satisfies ${\iota }_X\omega =dH$ for a globally defined smooth function $H$. It is therefore a symplectic vector field, but the converse requires this closed one-form to be exact.

On the torus with $\omega =dx\wedge dy$,Thus ${\partial }_x$ is a symplectic vector field. The closed differential one-form $dy$ is not exact on the torus: its integral on the loop $\gamma (s)=[(0,s)]$, $0\le s\le 1$, equals one, while the integral of an exact one-form around any closed loop is zero. The vector field is symplectic but not Hamiltonian. The local candidate $H=y$ does not descend to a single-valued function on ${\mathbb{R}}^2/{\mathbb{Z}}^2$.

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.

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.

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

A symplectomorphism is a diffeomorphism, so it preserves the genus of a surface. It also preserves the orientation determined by the symplectic form and, by integration of a pullback of a differential form, preserves the symplectic area.

Conversely, equal genus of a surface gives a diffeomorphism $f:{\Sigma }_0\to {\Sigma }_1$. We can choose $f$ to preserve the orientations determined by ${\omega }_0,{\omega }_1$: if necessary, compose with an orientation-reversing diffeomorphism of the standard closed orientable surface of that genus of a surface, obtained by reflection of a symmetric handle model. On ${\Sigma }_0$ put $\eta =f^{\ast }{\omega }_1$. Both $\eta$ and ${\omega }_0$ are positive volume forms for the same orientation, and their integrals agree. By top-degree de Rham cohomology, $[\eta ]=[{\omega }_0]$.

The family ${\omega }_t=(1-t){\omega }_0+t\eta$ consists of positive volume forms, hence symplectic forms in dimension two, with constant de Rham cohomology class. Moser theorem supplies $\psi$ with ${\psi }^{\ast }\eta ={\omega }_0$. ConsequentlyTherefore the genus of a surface and symplectic area classify compact connected symplectic surfaces.

The cotangent fiber translation $f$ is a diffeomorphism, with inverse $(x,\xi )\mapsto (x,\xi -{\beta }_x)$. Because $\pi \circ f=\pi$, the definition of the canonical one-form on a cotangent bundle givesTaking the exterior derivative yieldsIf $\beta$ is a closed differential form, this is $f^{\ast }\omega =\omega$, proving that the cotangent fiber translation is a symplectomorphism.

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.

With the conventionthe time-dependent Hamiltonian vector field is uniquely determined because $\omega$ is nondegenerate. If ${\varphi }_t$ is its local flow, Cartan's magic formula givesThus ${\varphi }_t^{\ast }\omega =\omega$ wherever the flow is defined, which is the Hamiltonian flow preserves the symplectic form property.

Write $v=(a,b)$ and ${\omega }_0=dx\wedge dy$. The linear Hamiltonianhas Hamiltonian vector field $X_{H_0}=v$. Choose a smooth cutoff $\rho$ that is one on $B(r+|v|)$ and zero outside $B(r+|v|+ϵ)$, and put $H=\rho H_0$. Every trajectory beginning in $B(r)$ and following $v$ remains in $B(r+|v|)$ for time $0\le t\le 1$, so the time-one map translates that ball by $v$. Outside $B(r+|v|+ϵ)$ its vector field vanishes, so the map is the identity. This is a compactly supported Hamiltonian translation and hence a compactly supported symplectomorphism.

For the connected-sum construction, choose a Darboux chart about the unique transverse intersection and straighten the two Lagrangian sheets to ${\mathbb{R}}^2$ and $i{\mathbb{R}}^2$ in ${\mathbb{C}}^2$. Remove small disks from the two sheets and join their boundary circles by the standard Lagrangian neckin ${\mathbb{R}}_x^2\oplus {\mathbb{R}}_y^2$, where $(a(t),b(t))$ follows a smooth arc from one positive coordinate ray to the other and agrees with those rays near its ends. Its pullback of ${\sum }_{j}d{x}_j\wedge d{y}_j$ vanishes because $u\cdot u^{\prime }=0$. Gluing this neck to the unchanged surfaces performs Lagrangian surgery. Topologically it is their connected sum, so it gives a Lagrangian connected sum

Finally, choose an immersed circle ${\gamma }_l:S^1\to {\mathbb{R}}^2$ with exactly $l$ transverse double points and no other multiple points; one may add $l$ small figure-eight kinks to an embedded circle. Let $c:S^1\to {\mathbb{R}}^2$ be an embedded circle and defineThis Product Lagrangian immersion satisfies ${\iota }^{\ast }{\omega }_0=0$. If $p$ is a double point of ${\gamma }_l$, its two local branches times $c(S^1)$ meet along $\{p\}\times c(S^1)$. Their tangent spaces intersect precisely in the tangent line to that circle, so the intersection is clean. The $l$ double points therefore give exactly $l$ disjoint clean self-intersection circles.

Near each clean circle, perturb one Lagrangian sheet by the graph of $\delta dh$ in its Weinstein neighborhood, where $h:S^1\to \mathbb{R}$ is a Morse function with one minimum and one maximum. The clean circle is replaced by two transverse double points. Resolve both by Lagrangian surgery. Each resolution attaches one one-handle and lowers the Euler characteristic by two, so resolving all $l$ clean circles changes the Euler characteristic of the original torus from zero toChoose the orientation-reversing neck at one double point; the resulting connected surface is nonorientable, while all the surgeries remove their double points and leave an embedding. Thus the Givental construction of nonorientable Lagrangian surfaces gives a closed connected nonorientable surface of Euler characteristic $-4l$ Lagrangian embedded in $({\mathbb{R}}^4,{\omega }_0)$.

Normalize the Fubini-Study form so that a complex projective line has symplectic area $\pi$. The displayed map takes the unit symplectic ball in ${\mathbb{C}}^n$ diffeomorphically onto the complement of the last-coordinate hyperplane. Its lift to the unit sphere has last coordinate positive real, so that coordinate contributes zero to the pulled-back standard symplectic form. The map is therefore a symplectomorphism.

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.

A symplectic capacity is monotone under symplectic embeddings, scales by $c(X,a\omega )=ac(X,\omega )$ for $a>0$, and has positive unit-ball capacity and finite unit-cylinder capacity. It is invariant under symplectomorphisms. A normalized symplectic capacity additionally assigns $\pi$ to both unit domains.

If $x=\chi (x^{\prime })$ is an angular diffeomorphism and $\eta \cdot dx$ is a closed differential one-form, the map $y=\eta (\chi (x^{\prime }))+D\chi (x^{\prime }{)}^{-T}{y}^{\prime }$ is a symplectomorphism. It combines a cotangent lift of a diffeomorphism with translation by a closed differential one-form. In particular $\eta =du+a$ is allowed on a torus even when the constant one-form $a\cdot dx$ is not exact.

A symplectic isotopy is a smooth path ${\varphi }_t$ of symplectomorphisms beginning at the identity.

If $\sigma =d\beta$, cotangent fiber translation by $+\beta$ gives a symplectomorphism from $(T^{\ast }X,-d\alpha )$ to $(T^{\ast }X,-d\alpha +{\pi }^{\ast }\sigma )$. Its pullback changes $\alpha$ by ${\pi }^{\ast }\beta$, so the extra two-form cancels with $-{\pi }^{\ast }d\beta$. The opposite translation gives the inverse direction, globally without compactness.