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For a symplectic isotopy generated by $X_t$, its flux isIt vanishes for a Hamiltonian isotopy.

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.