Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 146 / 2

The Weinstein neighborhood theorem states that if $L$ is a compact Lagrangian submanifold of $(M,\omega )$, then neighborhoods of $L$ in $M$ and of the zero section in its cotangent bundle $T^{\ast }L$ are symplectomorphic. The symplectomorphism restricts to the identity on $L$, and the canonical form on $T^{\ast }L$ is taken with the sign matching the chosen convention.

Take $L=S^2$. For any Lagrangian embedding of $S^2$ into a symplectic four-manifold, the symplectic form identifies the normal bundle with $T^{\ast }{S}^2$. The self-intersection formula and the Euler characteristic giveup to the harmless orientation sign. If a smooth isotopy displaced $L$, its final image would represent the same homology class but have intersection number zero with $L$, a contradiction. This is the smooth non-displaceability from self-intersection.

For a compact ambient example, equip $S^2\times S^2$ with $\omega \oplus (-\omega )$. Its diagonalis Lagrangian because the two summands cancel on $T\Delta$, and the preceding argument shows that it is not smoothly displaceable.

Let $M=S^2$ with an area form and let $L$ be an equator dividing the sphere into two open hemispheres of equal area. Every curve in a symplectic surface is Lagrangian. A small normal push moves $L$ to a nearby latitude, so it is displaceable by a smooth isotopy.

Suppose a symplectic isotopy had final image $L^{\prime }$ disjoint from $L$. The curve $L^{\prime }$ must lie in one hemisphere. Of the two discs bounded by $L^{\prime }$, the one contained in that hemisphere has area strictly below half the total area, and the other has area strictly above half. On the other hand, a symplectomorphism maps the original two hemispheres to the two discs bounded by $L^{\prime }$ and preserves their areas, so both would have half the total area. This contradiction is the symplectic non-displaceability of an area bisector.

LetFor any sufficiently small nonzero $\delta$, translationis a symplectic isotopy with ${\varphi }_1(L)\cap L=\varnothing$. Choosing $|\delta |$ arbitrarily small makes the isotopy arbitrarily small in every $C^k$ norm.

This isotopy has nonzero flux homomorphism, represented by $\delta dx$. In contrast, every Hamiltonian isotopy has zero flux. If a Hamiltonian image $L^{\prime }$ were disjoint from $L$, the two homologous essential circles would bound an annulus $A$, and evaluation of the flux on $[L]$ would equal the signed symplectic areawhich is nonzero. This contradicts vanishing Hamiltonian flux, so $L$ is not Hamiltonian displaceable.

Again take a symplectic two-sphere, but let $L$ be a small latitude bounding a cap of area strictly below half the total area. A rotation carrying that cap to a disjoint cap carries $L$ to a disjoint Lagrangian circle. Rotations of the symplectic sphere are flows of Hamiltonian vector fields; for rotation about an axis, a height function is a Hamiltonian function. Thus this rotation is a Hamiltonian isotopy, and $L$ is Hamiltonian displaceable.