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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.

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.

A simple closed curve dividing a symplectic two-sphere into two regions of equal area cannot be displaced by a symplectic isotopy. Any disjoint image would lie in one complementary disc, yet it would still have to bound a disc of half the total area.