Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 15 (Symplectic Geometry)

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.

Put ${\xi }_t=\ker {\alpha }_t$, the contact distribution. If $\dim M=2r+1$, the contact form condition is ${\alpha }_t\wedge (d{\alpha }_t{)}^r\ne 0$. It implies that $d{\alpha }_t{|}_{{\xi }_t}$ is a symplectic form on each contact hyperplane: inserting a vector transverse to ${\xi }_t$ in that volume form leaves the nonzero top exterior power $(d{\alpha }_t{|}_{{\xi }_t}{)}^r$. Hence there is a unique smooth time-dependent vector field $Y_t\in {\xi }_t$ satisfyingIts smoothness follows by inverting the smoothly varying nondegenerate matrix of this contact distribution symplectic form. This is the horizontal generator of contact stability.

Let $R_t$ be the Reeb vector field of ${\alpha }_t$. The one-form ${\dot{\alpha }}_t+{\iota }_{Y_t}d{\alpha }_t$ vanishes on ${\xi }_t$, so it is $h_t{\alpha }_t$. Evaluating on $R_t$, using ${\alpha }_t(R_t)=1$ and ${\iota }_{R_t}d{\alpha }_t=0$, identifies the coefficient:By Cartan's magic formula, the Lie derivative of a differential form isbecause $Y_t$ lies in the contact hyperplane.

Let ${\rho }_t$ solve ${\partial }_t{\rho }_t=Y_t\circ {\rho }_t$, with ${\rho }_0={\mathrm{id}}_M$. Since $M$ is compact without boundary and $Y_t$ is smooth on the closed time interval, its solutions cannot escape and exist for the entire interval. Reversing the time-dependent equation supplies a smooth inverse for each ${\rho }_t$. Thus these flow maps give a smooth isotopy. Differentiating the pullback of a differential form along this isotopy yieldsFor each base point this is a scalar linear equation for a covector, initially ${\alpha }_0$. Its solution isBy finite-time flow completeness on a compact manifold, this construction gives the whole time interval. The contact conformal factor along an isotopy is smooth in $(t,x)$ and nowhere zero, proving Gray stability theorem. The precise domains here are ${\rho }_t:M\to M$ and $\rho :M\times [0,1]\to M$; the extra time factor attached to the already indexed ${\rho }_t$ in the source is a notational slip. The proof produces the required family on its whole specified interval. If a parameter domain of all $\mathbb{R}$ is intended, extend the smooth field slightly beyond $[0,1]$, multiply it by a time cutoff, and use compactness to obtain the extended isotopy $\rho :M\times \mathbb{R}\to M$. Its restriction to $[0,1]$ is unchanged.

We prove the Weinstein neighborhood theorem with the canonical sign convention ${\omega }_0=-d\alpha$. The essential first step is to match the two symplectic forms as bilinear forms on the entire tangent space along $X$, not only after pulling them back to $TX$.

Here is the needed symplectic splitting along a Lagrangian submanifold. Write $E=TM{|}_X$ and $L=TX$. Choose a smooth vector bundle complement $N$ of $L$, for example using a Riemannian metric. Since $L$ is a Lagrangian subspace in every fiber, the pairing mapis an isomorphism: its kernel is $N\cap L^{\omega }=N\cap L=0$, and both bundles have rank $n$. Let $s:L^{\ast }\to N$ be its inverse and put $b(\eta ,\zeta )=\omega (s\eta ,s\zeta )$. Define $a:L^{\ast }\to L$ byand set $\tilde{s}=s+a$. Nondegeneracy of the dual pairing gives a unique smooth $a$. Since $b$ is alternating,Thus $K=\tilde{s}(L^{\ast })$ is a smooth Lagrangian complement to $L$. The mapis a vector bundle isomorphism, is the identity on $TX$, and satisfiesThe last equality uses the canonical splitting of the tangent bundle of $T^{\ast }X$ along its zero section.

Choose a Riemannian metric near $X$ for which $K$ is orthogonal to $TX$. The tubular neighborhood construction using its Riemannian exponential map gives a diffeomorphismafter shrinking around the zero section. It restricts to the given inclusion of $X$ and has differential $A$ along it. Consequently $\beta ={\psi }^{\ast }\omega$ and ${\omega }_0$ agree pointwise as ambient bilinear forms at every point of the zero section.

We use the following precise relative Moser theorem. If two closed symplectic forms ${\gamma }_0,{\gamma }_1$ near a compact embedded submanifold $S$ agree as ambient bilinear forms along $S$, then, on smaller neighborhoods, there is a diffeomorphism $\kappa$ fixing $S$ pointwise with ${\kappa }^{\ast }{\gamma }_1={\gamma }_0$. One sufficient version allows a symplectic path ${\gamma }_t$ with ${\partial }_t{\gamma }_t=d{\eta }_t$ and ${\eta }_t=0$ as ambient covectors along $S$: solve ${\iota }_{Z_t}{\gamma }_t=-{\eta }_t$ and use its flow. The vector field vanishes on $S$, so the flow fixes $S$, and compactness permits a common smaller neighborhood for the full time interval.

For completeness, all hypotheses can be checked directly here. Put $\delta =\beta -{\omega }_0$ and ${\gamma }_t={\omega }_0+t\delta$. At the zero section, every ${\gamma }_t$ equals ${\omega }_0$. The nondegenerate property is open, and compactness of $X\times [0,1]$ allows a single smaller neighborhood on which all ${\gamma }_t$ are symplectic forms. Choose it fiberwise star-shaped. Write $H_s(q,p)=(q,sp)$ and let $E$ be the vertical radial vector field. The radial homotopy primitive near a zero sectionis smooth, is zero as an ambient covector along the zero section, and satisfiesThe radial homotopy operator gives this identity, using $d\delta =0$ and $H_0^{\ast }\delta ={\pi }^{\ast }{i}_0^{\ast }\delta =0$. Smoothness at $s=0$ follows because contraction with $E$ supplies a factor of $s$ after evaluation at $(q,sp)$; the vanishing of $\delta$ along $X$ supplies additional vanishing. Therefore solve ${\iota }_{Z_t}{\gamma }_t=-\eta$. By Cartan's magic formula, its flow maps ${\kappa }_t$ satisfyAfter shrinking the starting neighborhood, these flows exist for $0\le t\le 1$ and fix $X$. This is a relative local construction, not an assertion of completeness throughout the noncompact cotangent bundle.

Finally set $\phi =\psi \circ {\kappa }_1$, let $U_0$ be its sufficiently small domain, and let $U=\phi (U_0)$. ThenThus the map is a symplectomorphism of neighborhoods and agrees with the specified inclusion at every point of $X$.

For a covector $p\in T_x^{\ast }X$, define the cotangent lift of a diffeomorphism byThe inverse derivative here maps $T_{f(x)}X$ to $T_{x}X$. This is a smooth diffeomorphism, its inverse is $(f^{-1}{)}_{\#}$, and it satisfies $\pi \circ f_{\#}=f\circ \pi$.

For $v\in T_{(x,p)}{T}^{\ast }X$, the pullback of a differential form and the defining formula for the Liouville one-form giveHence $f_{\#}^{\ast }\alpha =\alpha$, and applying the exterior derivative also proves $f_{\#}^{\ast }\omega =\omega$. The cotangent lift of a diffeomorphism thus preserves both canonical forms, with no extra choice of metric or connection.

First observe that the Liouville one-form is zero as an ambient covector exactly on the zero section. Indeed, ${\alpha }_{(x,p)}=p\circ d\pi$, and $d\pi$ is surjective. Since $dg$ is invertible, $g^{\ast }\alpha =\alpha$ therefore implies that $g$ maps the zero section onto itself. It induces a diffeomorphism $f:X\to X$ defined by $g(x,0)=(f(x),0)$.

The map $g$ is a symplectomorphism, since it preserves $-d\alpha$. Let $E={\sum }_i{p}_i{\partial }_{p_i}$ be the vertical Liouville vector field. With the chosen sign convention,Preservation of both $\omega$ and $\alpha$ forces $g_{\ast }E=E$ because $\omega$ is nondegenerate. Its complete cotangent fiber-dilation flow isUniqueness of integral curves gives $g\circ D_s=D_s\circ g$. If $g(x,p)=(y,P)$, let $s\to -\infty$. Continuity and this commutation yieldThus $y=f(x)$ for every $p$: the whole cotangent fiber over $x$ maps into the fiber over $f(x)$. This step proves that $g$ covers $f$; it was not assumed.

Now write $g(x,p)=(f(x),P(x,p))$. For any $u\in T_{x}X$, choose a tangent vector to $T^{\ast }X$ projecting to $u$. Evaluating $g^{\ast }\alpha =\alpha$ on it givesSince $d{f}_x$ is invertible, $P(x,p)=p\circ (d{f}_x{)}^{-1}$. Thereforeand $f$ is unique. This rigidity of cotangent Liouville-form preservation uses smoothness at the zero section; it follows from the canonical form and its dilation dynamics on the entire cotangent bundle.

For any positive-dimensional $X$, choose a smooth function $h$ with $dh$ not identically zero. A nonconstant smooth bump in a coordinate chart supplies one on an arbitrary smooth manifold. Translate by this exact differential form:The cotangent fiber translation identities giveThus a nontrivial exact fiber translation is a symplectomorphism that fails to preserve the Liouville one-form. For example, on $T^{\ast }\mathbb{R}$ one can take $T(q,p)=(q,p+1)$: it preserves $dq\wedge dp$ but changes $pdq$ to $(p+1)dq$.

If dimension zero is admitted, $T^{\ast }X=X$ and $\alpha =0$, so every diffeomorphism preserves $\alpha$ and this requested distinction is impossible. The example therefore applies in positive dimension, as required for a nonzero differential one-form.