Let
$$
(M,\omega)=(T^2,dx\wedge dy),
\qquad
L=\{y=0\}.
$$
For any sufficiently small nonzero $\delta$, translation
$$
\phi_t(x,y)=(x,y+t\delta)
$$
is a <symplectic isotopy> with $\phi_1(L)\cap L=\varnothing$. 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 $L'$ 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 area
$$
\int_A\omega,
$$
which is nonzero. This contradicts vanishing Hamiltonian flux, so $L$ is not Hamiltonian displaceable.

Solution

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

{title2=$\operatorname{Flux}$}

For a symplectic isotopy generated by $X_t$, its flux is
$$
\operatorname{Flux}(\phi_t)=\left[\int_0^1\iota_{X_t}\omega\,dt\right]\in H^1(M;\mathbb R).
$$
It vanishes for a <Hamiltonian isotopy>.

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 Incoming links: Flux homomorphism