Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 16 / 4

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.