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Take $\varphi$ in the identity component ${\mathrm{Symp}}_0(M,\omega )$ and choose a smooth symplectic isotopy ${\varphi }_t$ from the identity to $\varphi$. Let its generating vector field be $X_t={\dot{\varphi }}_t\circ {\varphi }_t^{-1}$. Differentiating ${\varphi }_t^{\ast }\omega =\omega$ givesThus ${\eta }_t={\iota }_{X_t}\omega$ is a closed differential one-form. By the vanishing of the first de Rham cohomology, it is exact, so choose $H_t$ with ${\eta }_t=-d{H}_t$.

The primitives can be chosen smoothly in both $t$ and the point. On each connected component fix a base point $x_0$ and define $H_t(x)=-{\int }_{x_0}^x{\eta }_t$. Exactness makes this independent of the chosen path. In a coordinate neighbourhood of any point, use a fixed path to the neighbourhood's centre followed by a smoothly varying short path; this shows smooth dependence locally, and hence globally. One may subtract a time-dependent mean without changing the Hamiltonian vector field. Therefore the original symplectic isotopy is Hamiltonian, and $\varphi \in \mathrm{Ham}(M,\omega )$.

Here the usual description of ${\mathrm{Symp}}_0$ by smooth isotopies agrees with the topological identity component. To see the local path-connectedness behind that description, use a Weinstein neighborhood of the diagonal in $(M\times M,-\omega \oplus \omega )$. The graph of a symplectomorphism close to the identity corresponds to the graph of a small closed one-form on the diagonal. Multiplying that one-form by $t\in [0,1]$ gives Lagrangian graphs; for sufficiently small forms their first projections remain diffeomorphisms, so they give symplectomorphisms connecting the original map to the identity. These local paths, translated in the group and concatenated with smooth time reparametrizations, identify its connected and smooth-path components.

Conversely every Hamiltonian isotopy is symplectic by part (a), and connects its time-one map to the identity. Hence

Take the usual convention that the smooth isotopy starts at ${\rho }_0=\mathrm{id}$. The exterior derivative commutes with a pullback of a differential form, soConsequentlyThe right-hand condition says precisely that ${\rho }_t$ is a symplectic isotopy.

Here an exact symplectic manifold is necessarily noncompact in positive dimension: if $\dim M=2n$ and $\omega =-d\lambda$, then ${\omega }^n=-d(\lambda \wedge {\omega }^{n-1})$, contradicting the positive symplectic volume and the Generalized Stokes theorem on a compact smooth manifold without boundary. Thus the paper's introductory word “closed” must here be read as “without boundary”, as its parenthesis suggests, rather than as including compactness.

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.