Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 16 (Symplectic Geometry and Hamiltonian Systems)

Use the Hamiltonian vector field convention ${\iota }_{X_{H_t}}\omega =-d{H}_t$. A smooth time-dependent Hamiltonian function $H:[0,1]\times M\to \mathbb{R}$ with flow defined over the whole interval generates the Hamiltonian isotopy determined byThe group of Hamiltonian diffeomorphisms isThese maps are symplectomorphisms: by Cartan's magic formula, ${\mathcal{L}}_{X_{H_t}}\omega =d{\iota }_{X_{H_t}}\omega =0$. Thus $d({\varphi }_H^t{)}^{\ast }\omega /dt=0$.

The term “group” is justified explicitly. If ${\varphi }_t,{\psi }_t$ are generated by $H_t,K_t$, their product path is generated by $H_t+K_t\circ {\varphi }_t^{-1}$, since the pushforward of $X_{K_t}$ by a symplectomorphism is $X_{K_t\circ {\varphi }_t^{-1}}$. The inverse path is generated by $-H_t\circ {\varphi }_t$. The zero Hamiltonian gives the identity.

On a closed symplectic manifold, all these flows exist on the finite time interval. On an open manifold one often means the compactly supported Hamiltonian diffeomorphism group ${\mathrm{Ham}}_c$, using Hamiltonians with compact support in $[0,1]\times M$. We use that subgroup for the finite-energy definitions on Euclidean space below. The unrestricted complete-flow group is needed for the global cotangent translations in Question 4; those translations are generally not compactly supported.

Take $\varphi$ in the identity component ${\mathrm{Symp}}_0(M,\omega )$ and choose a smooth symplectic isotopy ${\varphi }_t$ from the identity to $\varphi$. Let its generating vector field be $X_t={\dot{\varphi }}_t\circ {\varphi }_t^{-1}$. Differentiating ${\varphi }_t^{\ast }\omega =\omega$ givesThus ${\eta }_t={\iota }_{X_t}\omega$ is a closed differential one-form. By the vanishing of the first de Rham cohomology, it is exact, so choose $H_t$ with ${\eta }_t=-d{H}_t$.

The primitives can be chosen smoothly in both $t$ and the point. On each connected component fix a base point $x_0$ and define $H_t(x)=-{\int }_{x_0}^x{\eta }_t$. Exactness makes this independent of the chosen path. In a coordinate neighbourhood of any point, use a fixed path to the neighbourhood's centre followed by a smoothly varying short path; this shows smooth dependence locally, and hence globally. One may subtract a time-dependent mean without changing the Hamiltonian vector field. Therefore the original symplectic isotopy is Hamiltonian, and $\varphi \in \mathrm{Ham}(M,\omega )$.

Here the usual description of ${\mathrm{Symp}}_0$ by smooth isotopies agrees with the topological identity component. To see the local path-connectedness behind that description, use a Weinstein neighborhood of the diagonal in $(M\times M,-\omega \oplus \omega )$. The graph of a symplectomorphism close to the identity corresponds to the graph of a small closed one-form on the diagonal. Multiplying that one-form by $t\in [0,1]$ gives Lagrangian graphs; for sufficiently small forms their first projections remain diffeomorphisms, so they give symplectomorphisms connecting the original map to the identity. These local paths, translated in the group and concatenated with smooth time reparametrizations, identify its connected and smooth-path components.

Conversely every Hamiltonian isotopy is symplectic by part (a), and connects its time-one map to the identity. Hence

For the spatial supremum-norm convention, a normalized Hamiltonian function gives the path lengthOn an open manifold use compactly supported Hamiltonians; on a closed manifold use zero-mean Hamiltonians, which fixes their additive time-dependent constants. The Hofer metric from the spatial supremum norm isEquivalently one minimizes this length over Hamiltonian paths joining the two maps. Inversion and conjugation preserve the generator's supremum norm, so the distance is symmetric and bi-invariant; concatenation gives the triangle inequality. Its nondegeneracy is the geometric assertion proved in Question 3, not a consequence of these formal properties alone.

Another common convention for Hofer's metric uses the oscillation length ${\ell }_{\mathrm{osc}}(H)={\int }_0^1(\max H_t-\min H_t)dt$, giving a distance $d_H$. Normalized or compactly supported Hamiltonians have values straddling zero, soThus these conventions give equivalent distances, and agree on whether an energy vanishes. We keep the same chosen convention in every energy estimate.

The Hamiltonian displacement energy of a subset $A$ isThe infimum is $+\infty$ if there is no displacing map. In the oscillation convention replace ${\rho }_{\infty }$ by $d_H$.

Choose coordinates $(q_1,\dots ,q_n,p_1,\dots ,p_n)$ with standard symplectic form ${\omega }_0={\sum }_{j}d{q}_j\wedge d{p}_j$, and regard the given sphere $S$ as lying in $\{p_n=0\}$. With ${\iota }_{X_H}{\omega }_0=-dH$, the linear Hamiltonian function $H_{\delta }=\delta q_n$ has Hamiltonian vector field $\delta {\partial }_{p_n}$ and translates in the missing coordinate.

To make this a permissible compactly supported map, take a smooth cutoff $\chi$ equal to one on an open neighbourhood of the fixed compact swept setChoose $\chi$ with compact support and let $R={\sup }_{\mathrm{supp}\chi }|q_n|<\infty$. For $0<\delta <1$ put $G_{\delta }=\delta q_n\chi$. On the whole swept region the cutoff is locally constant, so its derivatives vanish and $X_{G_{\delta }}=\delta {\partial }_{p_n}$. Thus the flow carries every $z\in S$ to $z+t\delta e_{p_n}$ for $0\le t\le 1$.

The time-one image lies in $\{p_n=\delta \}$ and is disjoint from the original sphere. Its Hofer metric from the spatial supremum norm obeysIn the oscillation convention the upper bound is $2R\delta$. Since $\delta$ can approach zero while the cutoff and $R$ stay fixed,This proves the zero displacement energy for a compact set in a hyperplane; it does not assert that any one nonidentity displacing map has zero norm.

Fix a primitive $\lambda$ of the standard symplectic form on ${\mathbb{R}}^{2n}$. With the coordinates of Question 2 one convenient choice isFor a Lagrangian submanifold with inclusion $i:L\hookrightarrow {\mathbb{R}}^{2n}$, $d(i^{\ast }\lambda )=i^{\ast }{\omega }_0=0$. Its Liouville class of a Lagrangian submanifold isAny two primitives differ by a closed one-form on the contractible ambient space, hence by an exact form, so the class is independent of the primitive. The cotangent primitive convention $\theta ={\sum }_j{p}_{j}d{q}_j$, with ${\omega }_0=-d\theta$, gives the negative of this class; rationality, positive period size and nonvanishing do not depend on that sign.

The period homomorphism sends an integral one-cycle $c$ to ${\int }_c{i}^{\ast }\lambda$. Its image $\mathrm{Per}(L)\subset \mathbb{R}$ is an additive subgroup. A rational Lagrangian submanifold is one for which this subgroup is discrete. In the nonzero case it has a unique positive generator:Equivalently ${\ell }_L$ is a real multiple of an integral cohomology class. The invariant $\gamma (L)$ is its least positive Liouville period, also expressible as $\inf \{a>0:a\in \mathrm{Per}(L)\}$. If the period group is zero, that infimum is $+\infty$ by the empty-set convention; some treatments instead reserve “rational” for the nonzero discrete case. Because the ambient space is contractible, the Stokes theorem identifies these periods with signed symplectic areas of disks whose boundaries lie on $L$.

Let $\varphi \in {\mathrm{Ham}}_c({\mathbb{R}}^{2n})$ be nonidentity and choose $x$ with $\varphi (x)\ne x$. Choose disjoint open neighbourhoods $U,V$ of $x,\varphi (x)$, respectively. Continuity gives a small Euclidean ball $B=B(x,r_0)$ with $B\subset U$ and $\varphi (B)\subset V$. Therefore $\varphi (B)\cap B=\varnothing$.

Inside this ball construct an explicit closed rational Lagrangian submanifold. For $0<a<r_0/\sqrt{n}$, take the product Lagrangian torusIts distance from $x$ is $\sqrt{n}a<r_0$. Each coordinate circle is one-dimensional, so the pullback of $d{q}_j\wedge d{p}_j$ to it is zero; summing proves that this $n$-dimensional product is a Lagrangian submanifold. It is compact without boundary.

On the $j$th fundamental circle, $q_j=a\cos t$, $p_j=a\sin t$, and the primitive in part (a) restricts to $\frac{1}{2}{a}^{2}dt$. Its integral is $\pi a^2$. Translation of the torus adds only an exact constant-coefficient one-form to the primitive, so it does not change the periods. Since $H_1(T^n;\mathbb{Z})$ is generated by these circles,The given displacement-energy bound applies to this explicitly constructed torus. Because $L\subset B$, the map $\varphi$ displaces $L$, and henceThis is Hofer nondegeneracy from rational Lagrangian displacement. For two different maps $\varphi ,\psi$, apply it to ${\varphi }^{-1}\psi \ne \mathrm{id}$ and use the distance definition. Therefore ${\rho }_{\infty }(\varphi ,\psi )>0$. The equivalent oscillation Hofer metric is nondegenerate as well.

Only the stated energy inequality was used; its conclusion was not used to presume positive energy of an arbitrary open set. The rational torus and its positive period have been constructed inside the particular displaced ball.

Let $\pi :T^{\ast }N\to N$ be the projection and $s_{\alpha }(q)=(q,{\alpha }_q)$ the section with image ${\Gamma }_{\alpha }$. For the canonical one-form on a cotangent bundle,Since $\pi \circ s_{\alpha }=\mathrm{id}$, evaluation on a tangent vector $v\in T_{q}N$ givesUse the same position-first standard symplectic form as before, $\omega =-d\theta$. Pullback commutes with the exterior derivative, soThe graph of a differential one-form is an embedded submanifold, because its projection is the inverse of the section on the graph. Its dimension is $\dim N$, half that of $T^{\ast }N$. Therefore the vanishing of the pulled-back symplectic form is exactly the Lagrangian submanifold condition, andThis proves that the graph of a closed one-form is Lagrangian using the intrinsic one-form rather than a coordinate guess.

The cotangent fiber translation $f(q,p)=(q,p+{\alpha }_q)$ is a smooth diffeomorphism with inverse $(q,p)\mapsto (q,p-{\alpha }_q)$, and $\pi \circ f=\pi$. For a tangent vector $v$ at $(q,p)$, the intrinsic definition of the canonical one-form on a cotangent bundle givesThus $f^{\ast }\theta =\theta +{\pi }^{\ast }\alpha$. Taking its exterior derivative with the fixed sign convention yieldsIf $\alpha$ is a closed differential one-form, the last term vanishes, proving that $f$ is a symplectomorphism. In fact the converse holds too: the pullback by the zero section shows that ${\pi }^{\ast }(d\alpha )=0$ implies $d\alpha =0$.

Write the exact differential form as $\alpha =du$ for a smooth real-valued function $u$ on $N$. On $T^{\ast }N$ take the Hamiltonian function $H=u\circ \pi$. With $\omega ={\sum }_{j}d{q}_j\wedge d{p}_j$ and ${\iota }_{X_H}\omega =-dH$, the Hamiltonian vector field isIts complete flow is explicit:The base point stays fixed, so this flow exists for all real $t$, even if $u$ is unbounded on a noncompact base. Its time-one map is exactly $f$, proving that the cotangent fiber translation is a Hamiltonian diffeomorphism in the unrestricted complete-flow group.

This construction also justifies the hint for this family, without treating exactness as a general substitute for a Hamiltonian isotopy. For any Hamiltonian isotopy ${\psi }_t$ on this cotangent bundle, Cartan's magic formula, $d\theta =-\omega$ and ${\iota }_{X_{H_t}}\omega =-d{H}_t$ giveIntegrating proves that ${\psi }_1^{\ast }\theta -\theta$ is exact. For the present translation this difference is ${\pi }^{\ast }\alpha$, which is exact if and only if $\alpha$ is exact: one direction is pullback of $du$, and the other is pullback by the zero section. Necessity therefore follows from the differential identity, while sufficiency follows from the complete flow just constructed.

Compact support is a separate issue. If $du$ is nonzero at some $q$, the map moves every point of the entire unbounded fiber over $q$, so it is not a member of the compactly supported Hamiltonian diffeomorphism group. For example $N=\mathbb{R}$, $u(q)=q$ gives $(q,p)\mapsto (q,p+1)$. Thus the claim about this global translation uses the full Hamiltonian group, not ${\mathrm{Ham}}_c$; its flow can be cut off to agree on a prescribed compact swept set, but such a cutoff does not give the same map on the entire cotangent bundle.

Write the standard complex coordinates as $z_j=q_j+i{p}_j$ and let $J$ denote multiplication by $i$. Use the standard symplectic form ${\omega }_0={\sum }_{j}d{q}_j\wedge d{p}_j$, for which ${\omega }_0(v,Jv)=|v{|}^2$. In real coordinates $s,t$ on the disk, holomorphicity and the Cauchy-Riemann equations implyTherefore the pulled-back two-form isIf this integral were zero, the continuous nonnegative integrand would vanish everywhere in the interior. The displayed Cauchy-Riemann relation would then make both first derivatives zero, forcing $h$ to be constant on the connected disk and, by continuity, on its boundary. For the nonconstant map under consideration the symplectic area is consequently strictly positive:Take the primitive $\lambda =\frac{1}{2}{\sum }_j(q_{j}d{p}_j-p_{j}d{q}_j)$ from Question 3. Let $c=h{|}_{\partial D^2}$, oriented as the boundary of the disk. Since $c$ lies in $L$, the Stokes theorem givesThe restriction $\lambda {|}_L$ is closed because $L$ is a Lagrangian submanifold. If its Liouville class of a Lagrangian submanifold were zero, it would be an exact differential form, and its integral along every closed curve in $L$ would vanish. The boundary loop above contradicts this. The Liouville class is therefore nonzero. This proves that a holomorphic disk detects a nonzero Liouville class directly by positivity and Stokes; no disk-existence theorem is being assumed beyond the disk given here.