Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 140 (Symplectic Topology)

Use polar coordinates $z_j=r_j{e}^{i{\theta }_j}$ on each complex coordinate plane. On $S^3$, where $r_1^2+r_2^2=1$,Direct substitution shows that ${\alpha }_s\wedge d{\alpha }_s$ is nowhere zero for $s>0$, so ${\alpha }_s$ is a contact form.

The vector fieldsatisfies ${\alpha }_s(R_s)=r_1^2+r_2^2=1$ and ${\iota }_{R_s}d{\alpha }_s=0$ on tangent vectors to the sphere. It is therefore the Reeb vector field. Its flow isIf both coordinates are nonzero, an orbit closes only if the two angular frequencies have rational ratio, equivalently if $s\in \mathbb{Q}$. For irrational $s$, the only closed orbits are $\{z_2=0\}$ and $\{z_1=0\}$, the two coordinate circles. This is the irrational contact ellipsoid flow on the three-sphere.

The Weinstein neighborhood theorem identifies a neighborhood of $X$ in $M$ with a neighborhood of the zero section in $T^{\ast }X$. If $j$ is sufficiently $C^1$-close to the inclusion, projection of its image to $X$ is a diffeomorphism. After reparametrization, $j(X)$ is therefore the graph of a small one-form $\beta$.

The graph of a closed one-form is Lagrangian criterion says that $j(X)$ is Lagrangian exactly when $d\beta =0$. Since $H^1(X;\mathbb{R})=0$, write $\beta =df$. Intersections of $j(X)$ with $X$ are the critical points of $f$. A smooth function on a compact manifold has a maximum and a minimum; if they coincide as points because $f$ is constant, every point is an intersection. Thus there are at least two intersection points, proving the nearby exact Lagrangian intersection lemma.

The cohomology hypothesis is necessary. Take the zero section in $T^{\ast }{S}^1$ and the graph of the arbitrarily small nowhere-zero closed one-form $\epsilon ,d\theta$. Both are Lagrangian and disjoint.

Define the cotangent lift of a diffeomorphismIf $\alpha$ is the canonical one-form and $W\in T_{(x,\xi )}{T}^{\ast }X$, thenHence $f_{\#}^{\ast }\alpha =\alpha$.

The area form $\sigma$ on $S^2$ is closed. For any closed two-form $B$ on a base, the twisted cotangent symplectic form $\omega +{\pi }^{\ast }B$ is closed and nondegenerate: pairing a putative kernel vector first with vertical vectors kills its horizontal component, and then pairing with horizontal vectors kills its vertical component. Thus both ${\omega }_{+}$ and ${\omega }_{-}$ are symplectic.

The cotangent bundle deformation-retracts onto its zero section. Their cohomology classes satisfyThey are therefore not strongly isotopic.

Let $f:S^2\to S^2$ be an orientation-reversing isometry. Then $f^{\ast }\sigma =-\sigma$, while its cotangent lift preserves the canonical form and satisfies $\pi \circ f_{\#}=f\circ \pi$. Consequentlyso the two forms are symplectomorphic.

If $H{|}_L$ is constant, then $dH(v)=0$ for every $v\in TL$. By the definition of the Hamiltonian vector field,Since $L$ is Lagrangian, $(TL{)}^{\omega }=TL$, so $X_H$ is tangent to $L$. Its Hamiltonian flow preserves $L$. This is the Hamiltonian flow preserves a constant-level Lagrangian principle.

The cotangent lift is functorial:Since ${\varphi }_{t+s}={\varphi }_t\circ {\varphi }_s$, it follows thatThus $({\psi }_t)$ is a flow with infinitesimal vector field $V_{\#}$ as stated in the question.

Finally, a cotangent lift sends the conormal bundle $N^{\ast }Y$ to $N^{\ast }f(Y)$. Explicitly, if $\xi$ annihilates $T_{x}Y$, then $(d{f}_x^{-1}{)}^{\ast }\xi$ annihilates $T_{f(x)}f(Y)$. Since ${\varphi }_t(Y)=Y$, we haveso the conormal bundle is invariant.