Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 15 / 1

In cotangent bundle coordinates $(q^1,\dots ,q^n,p_1,\dots ,p_n)$, the Liouville one-form and the canonical symplectic form areThe twisted cotangent symplectic form is closed: pullback of a differential form commutes with the exterior derivative, soTo prove the form is nondegenerate, write a tangent vector in these coordinates as $(u,v)$ and test its pairing with an arbitrary vertical vector $(0,w)$. The base differential two-form has zero pairing with vertical vectors, henceA vector in the kernel of a linear map therefore has $u=0$. Pairing it next with every $(z,0)$ gives $-z^i{v}_i=0$, so $v=0$. Thus the kernel is zero at every point, independently of the size or rank of $\sigma$. Together with closedness, this proves ${\omega }_{\sigma }$ is a symplectic form. The argument is local only for convenience; the nondegenerate property is intrinsic and the coordinate charts cover the whole cotangent bundle.

The differential one-form $\theta$ defines a smooth section of a vector bundle, and its one-form graph is an embedded copy of $X$ because $\pi \circ \theta ={\mathrm{id}}_X$. The defining property of the Liouville one-form gives ${\theta }^{\ast }\alpha =\theta$. ConsequentlyThe one-form graph has dimension $n$, half the dimension of the cotangent bundle. It is therefore a Lagrangian submanifold precisely when this pullback of a differential form vanishes:This is the twisted Lagrangian graph criterion.

Now suppose the projection restricts to a diffeomorphism on a Lagrangian submanifold $L$. Its inverse followed by the inclusion defines a smooth section $s:X\to T^{\ast }X$, hence a global differential one-form $\theta$ with image $L$. The criterion forces $\sigma =d\theta$, so its class in de Rham cohomology is zero. Thus $[\sigma ]\ne 0$ rules out every such Lagrangian section. This cohomological obstruction to a Lagrangian section concerns graphs over the entire base; it does not claim that all Lagrangian submanifolds are absent.

Yes. Choose a global differential one-form $\beta$ with $d\beta =\sigma$, and use the cotangent fiber translationIt is a diffeomorphism with inverse $T_{-\beta }$, and $\pi \circ T_{\beta }=\pi$. Directly from the Liouville one-form,The sign convention $\omega =-d\alpha$ therefore givesThus $T_{\beta }:(T^{\ast }X,\omega )\to (T^{\ast }X,{\omega }_{\sigma })$ is the required symplectomorphism. Equivalently, $T_{-\beta }^{\ast }\omega ={\omega }_{\sigma }$ gives the inverse direction. This translation equivalence of exact twisted cotangent bundles is global even when $X$ is not compact; no completeness argument for a Moser theorem flow is needed.