Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 16 (Symplectic Topology)

Let $Q$ be the configuration smooth manifold, and let $L(t,q,v)$ be a smooth Lagrangian on its tangent bundle. For a path with fixed endpoints, its action isThe principle of stationary action requires the first variation to vanish for every fixed-endpoint variation. In a coordinate chart take $q_s=q+s\eta$, with $\eta (a)=\eta (b)=0$. Differentiation under the integral and integration by parts giveRepeated coordinate indices are summed. The endpoint term vanishes. Since compactly supported variations can be chosen independently in each coordinate, the fundamental lemma of the calculus of variations yields the Euler-Lagrange equationsConversely these equations make the displayed first variation zero for every fixed-endpoint variation. Variations localized in coordinate charts establish the same assertion for paths on $Q$. Thus the Euler-Lagrange equations are exactly the stationary-path condition, not necessarily a condition for an action minimum.

The passage to Hamiltonian mechanics uses the Legendre transform in mechanics. Assume a regular Lagrangian: the velocity Hessian matrix $({\partial }^{2}L/\partial v^i\partial v^j)$ is invertible. Thendefines a locally invertible map from the tangent bundle to the cotangent bundle. Write its local inverse as $v=v(t,q,p)$ and define the HamiltonianDifferentiating this expression, all terms containing $dv$ cancel because $p_i=L_{v^i}$. HenceThe Euler-Lagrange equations become Hamilton's equations:Conversely, a solution of Hamilton's equations satisfies $\dot{q}=v(t,q,p)$, hence $p=L_v(t,q,\dot{q})$, and $\dot{p}=L_q$ recovers the Euler-Lagrange equations. This proves local equivalence of the two descriptions.

On the cotangent bundle, choose the canonical symplectic form ${\omega }_{\mathrm{can}}={\sum }_{i}d{q}^i\wedge d{p}_i=-d\lambda$ and the convention ${\iota }_{X_H}{\omega }_{\mathrm{can}}=dH$. Then $X_H={\sum }_i(H_{p_i}{\partial }_{q^i}-H_{q^i}{\partial }_{p_i})$, so its integral curves are exactly the phase-space equations above. This sign convention is used throughout these solutions. For a hyperregular Lagrangian, the Legendre transform in mechanics is globally invertible and gives global equivalence; regularity alone only gives local equivalence. A singular velocity Hessian matrix may instead produce constraints, so the ordinary unconstrained argument does not apply to every Lagrangian.

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.

There are no symplectic structures on the Möbius strip. A symplectic form on a surface would be a nowhere-vanishing two-form, hence would provide an orientation. The Möbius strip is nonorientable: transport around its core reverses a transverse direction and therefore reverses any local orientation. This is incompatible with such a two-form. The argument applies whether the boundary is included or only the interior is considered.

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 Hamiltonian flow of $H$ solvesCompactness and the absence of a boundary make this smooth vector field complete, so the flow exists for all real $t$. The defining equation and Cartan's magic formula giveTherefore the pullback of a differential form differentiation rule yieldsso $({\varphi }_H^t{)}^{\ast }\omega =\omega$. Pullback commutes with the wedge product of differential forms, and consequentlyThus the flow preserves the symplectic volume, in fact the entire symplectic form.

Moser's trick turns variation of symplectic forms into an equation for a time-dependent vector field. Let ${\omega }_t$, $0\le t\le 1$, be a smooth path of symplectic forms on a compact manifold without boundary, with a constant de Rham cohomology class. Choose a smooth family of one-forms ${\alpha }_t$ such that ${\dot{\omega }}_t=d{\alpha }_t$. Such a smooth choice can be made using a fixed auxiliary metric; the essential requirement is this exactness throughout the path.

Nondegeneracy uniquely determines $X_t$ byLet $f_t$ be its flow, with $f_0=\mathrm{id}$. Compactness ensures existence over the whole parameter interval. By Cartan's magic formula,ThusThe path is made constant by a diffeomorphism moving with $X_t$. Every interpolating form must be nondegenerate; equal endpoint cohomology alone does not ensure that every linearly interpolated form is a symplectic form. On a noncompact manifold one instead needs completeness of this flow, or restricts to a sufficiently small neighborhood, as in the local argument below.

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 Poincare 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.

Let $\pi :T^{\ast }L\to L$ be the cotangent bundle projection. The canonical one-form on a cotangent bundle is intrinsically defined byNo metric or coordinate choice is needed. In local coordinates $\lambda ={\sum }_i{p}_{i}d{q}^i$. ChooseIt is closed because $d^2=0$, and its coordinate matrix is $\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$, which is invertible. The intrinsic definition of $\lambda$ makes the forms agree under all cotangent coordinate changes. This is the canonical symplectic form; choosing $d\lambda$ instead is the opposite common sign convention.

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.

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.

Use the normalization in which a projective line has area $2\pi$. On the affine chart of Complex projective space where $Z_0\ne 0$, set $w_j=Z_j/Z_0$ and $S=1+{\sum }_j|w_j{|}^2$. Define the Fubini-Study form byOn another chart the corresponding potential differs by $\log |h{|}^2$ for a nowhere-zero holomorphic function $h$, whose $\partial \bar{\partial }$ is zero. Thus these local differential forms glue to a global form. It is real and closed. Its Hermitian coefficient matrix is positive definite, since for $v\ne 0$ the Cauchy-Schwarz inequality givesHence it is a Kähler form and in particular a symplectic form.

On a complex projective line with $w=x+iy$ this is $2(1+|w{|}^2{)}^{-2}dx\wedge dy$, whose total area is $2\pi$. Thus $[{\omega }_{\mathrm{FS}}/(2\pi )]$ is the positive generator, equivalently $c_1(\mathcal{O}(1))$. Another common normalization uses half this form and gives line area $\pi$; the scale must be carried consistently into symplectic reduction.

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.

An almost complex structure is a smooth bundle endomorphism $J:TM\to TM$ with $J^2=-I$. It is an $\omega$-compatible almost complex structure whenThese conditions make $g_J(v,w)=\omega (v,Jw)$ a Riemannian metric. In particular it is symmetric: invariance and $J^2=-I$ give $\omega (Jv,w)=-\omega (v,Jw)$, and skew-symmetry then gives $g_J(w,v)=g_J(v,w)$. Positivity is the second condition. Moreover $J$ is an isometry for $g_J$. Compatibility does not require that the almost complex structure be integrable.

For a Riemann surface $(\Sigma ,j)$ and an almost complex manifold $(M,J)$, a J-holomorphic curve is a smooth map $u:\Sigma \to M$ satisfyingThus its differential is complex-linear at every point. In oriented local coordinates $s,t$ with $j{\partial }_s={\partial }_t$, this is $u_t=J{u}_s$, equivalently $u_s+J{u}_t=0$. No integrability of the target almost complex structure is required. Constant maps satisfy the definition; some usages reserve the word curve for nonconstant maps, so that restriction should be stated separately when intended.

Choose a Riemannian metric $h$ on $\Sigma$ compatible with its complex structure $j$, and use $g_J=\omega (\cdot ,J\cdot )$ on the target. The Dirichlet energy of a map isIt is independent of the particular conformal representative $h$: rescaling $h$ multiplies the squared differential norm by the inverse factor and the area element by the same factor. For a J-holomorphic curve, the Energy identity for a J-holomorphic curve isTo prove it, take an oriented $h$-orthonormal frame $e_1,e_2=j{e}_1$ and put $a=du(e_1)$, $b=du(e_2)$. The J-holomorphic curve equation says $b=Ja$. The pointwise energy density is therefore $\frac{1}{2}(|a{|}_{g_J}^2+|Ja{|}_{g_J}^2)=|a{|}_{g_J}^2$, while the pulled-back area density is $\omega (a,Ja)=|a{|}_{g_J}^2$. Integrating proves the identity.

More generally, the same frame givesWith the full tensor norm of ${\bar{\partial }}_{J}u$, its two frame components are $(a+Jb)/2$ and $(b-Ja)/2$, so their squared norms sum to $\frac{1}{2}|b-Ja{|}^2$. Hence the full identity isThis also fixes the normalization of the error term. The energy is nonnegative and vanishes exactly when $du=0$. The formulas apply whenever the relevant integrals are defined, in particular on compact source surfaces.

No nonconstant J-holomorphic curve with this closed connected source exists. The standard symplectic form on ${\mathbb{R}}^{2n}$ is exact; for instanceThe Energy identity for a J-holomorphic curve and the Generalized Stokes theorem giveThe target metric from the compatible almost complex structure is positive definite, so the nonnegative continuous energy density must vanish everywhere. Thus $du=0$. Connectedness of $\Sigma$ makes $u$ constant. This argument uses exactness and compatibility, and works even when $J$ is nonintegrable; it does not require the ordinary holomorphic maximum principle on the target.