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

This group consists of time-one maps of Hamiltonian isotopies whose flows are defined throughout the specified time interval. With ${\iota }_{X_H}\omega =-dH$, composition of isotopies generated by $H_t,K_t$ has generator $H_t+K_t\circ {\varphi }_t^{-1}$, where ${\varphi }_t$ is the first isotopy. The inverse path has generator $-H_t\circ {\varphi }_t$. These formulas prove closure under composition and inversion. On a closed symplectic manifold, every smooth time-dependent Hamiltonian has a complete flow over a bounded time interval.

If ${\iota }_{X_{H_t}}\omega =-d{H}_t$ and ${\varphi }_t$ is its flow, then Cartan's magic formula givesThus every defined time slice of a Hamiltonian isotopy is a symplectomorphism.

The displayed distance uses the oscillation length of a compactly supported Hamiltonian isotopy. On a closed manifold one may normalize each Hamiltonian to have zero mean. Additive time-dependent constants do not change the flow or its oscillation. Pullback by a symplectomorphism preserves extrema, so this distance is bi-invariant. Nondegeneracy requires an additional geometric argument; it does not follow merely from the definition as an infimum of path lengths.

Use the Hamiltonian vector field convention ${\iota }_{X_{H_t}}\omega =-d{H}_t$. A smooth time-dependent Hamiltonian function $H:[0,1]\times M\to \mathbb{R}$ with flow defined over the whole interval generates the Hamiltonian isotopy determined byThe group of Hamiltonian diffeomorphisms isThese maps are symplectomorphisms: by Cartan's magic formula, ${\mathcal{L}}_{X_{H_t}}\omega =d{\iota }_{X_{H_t}}\omega =0$. Thus $d({\varphi }_H^t{)}^{\ast }\omega /dt=0$.

The term “group” is justified explicitly. If ${\varphi }_t,{\psi }_t$ are generated by $H_t,K_t$, their product path is generated by $H_t+K_t\circ {\varphi }_t^{-1}$, since the pushforward of $X_{K_t}$ by a symplectomorphism is $X_{K_t\circ {\varphi }_t^{-1}}$. The inverse path is generated by $-H_t\circ {\varphi }_t$. The zero Hamiltonian gives the identity.

On a closed symplectic manifold, all these flows exist on the finite time interval. On an open manifold one often means the compactly supported Hamiltonian diffeomorphism group ${\mathrm{Ham}}_c$, using Hamiltonians with compact support in $[0,1]\times M$. We use that subgroup for the finite-energy definitions on Euclidean space below. The unrestricted complete-flow group is needed for the global cotangent translations in Question 4; those translations are generally not compactly supported.

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

Write the exact differential form as $\alpha =du$ for a smooth real-valued function $u$ on $N$. On $T^{\ast }N$ take the Hamiltonian function $H=u\circ \pi$. With $\omega ={\sum }_{j}d{q}_j\wedge d{p}_j$ and ${\iota }_{X_H}\omega =-dH$, the Hamiltonian vector field isIts complete flow is explicit:The base point stays fixed, so this flow exists for all real $t$, even if $u$ is unbounded on a noncompact base. Its time-one map is exactly $f$, proving that the cotangent fiber translation is a Hamiltonian diffeomorphism in the unrestricted complete-flow group.

This construction also justifies the hint for this family, without treating exactness as a general substitute for a Hamiltonian isotopy. For any Hamiltonian isotopy ${\psi }_t$ on this cotangent bundle, Cartan's magic formula, $d\theta =-\omega$ and ${\iota }_{X_{H_t}}\omega =-d{H}_t$ giveIntegrating proves that ${\psi }_1^{\ast }\theta -\theta$ is exact. For the present translation this difference is ${\pi }^{\ast }\alpha$, which is exact if and only if $\alpha$ is exact: one direction is pullback of $du$, and the other is pullback by the zero section. Necessity therefore follows from the differential identity, while sufficiency follows from the complete flow just constructed.

Compact support is a separate issue. If $du$ is nonzero at some $q$, the map moves every point of the entire unbounded fiber over $q$, so it is not a member of the compactly supported Hamiltonian diffeomorphism group. For example $N=\mathbb{R}$, $u(q)=q$ gives $(q,p)\mapsto (q,p+1)$. Thus the claim about this global translation uses the full Hamiltonian group, not ${\mathrm{Ham}}_c$; its flow can be cut off to agree on a prescribed compact swept set, but such a cutoff does not give the same map on the entire cotangent bundle.

Differentiating the assumed primitive identity and using Cartan's magic formula givesApply the pullback of a differential form by ${\rho }_t^{-1}$ and rearrange:This smooth family of Hamiltonian functions proves that ${\rho }_t$ is a Hamiltonian isotopy in the convention of part (b).

To connect this with the preceding assertion that every ${\rho }_t^{\ast }\lambda -\lambda$ is an exact differential form, the smooth choice of primitives causes no additional obstruction. Fix $p_0\in M$ and normalize $F_t(p_0)=0$. Define $F_t(p)$ by the line integral of ${\rho }_t^{\ast }\lambda -\lambda$ from $p_0$ to $p$. The fundamental theorem for line integrals makes this independent of the path. Near any $p$, use one fixed path to a nearby chart center followed by coordinate line segments; this expression is smooth jointly in $t,p$. Thus these normalized primitives form the required smooth family.

Write the exact differential form as $\beta =dh$ and take the cotangent fiber translation family ${\rho }_t(x,\xi )=(x,\xi +td{h}_x)$. Its generating vector field is vertical and in local coordinates isWith $\omega ={\sum }_{j}d{x}_j\wedge d{\xi }_j$ and the Hamiltonian vector field convention from Question 3,Thus ${\rho }_t$ is a Hamiltonian isotopy and $f={\rho }_1$ is a Hamiltonian diffeomorphism. The local flow is explicitly defined for every real $t$, regardless of compactness. With the alternative convention ${\iota }_{X_H}\omega =-dH$, the same family is generated by $H=h\circ \pi$.

The definition in the question does not require compact support for the Hamiltonian function; the displayed Hamiltonian diffeomorphism uses that definition.

If $H{|}_L$ is constant, then $dH(v)=0$ for every $v\in TL$. By the definition of the Hamiltonian vector field,Since $L$ is Lagrangian, $(TL{)}^{\omega }=TL$, so $X_H$ is tangent to $L$. Its Hamiltonian flow preserves $L$. This is the Hamiltonian flow preserves a constant-level Lagrangian principle.

The cotangent lift is functorial:Since ${\varphi }_{t+s}={\varphi }_t\circ {\varphi }_s$, it follows thatThus $({\psi }_t)$ is a flow with infinitesimal vector field $V_{\#}$ as stated in the question.

Finally, a cotangent lift sends the conormal bundle $N^{\ast }Y$ to $N^{\ast }f(Y)$. Explicitly, if $\xi$ annihilates $T_{x}Y$, then $(d{f}_x^{-1}{)}^{\ast }\xi$ annihilates $T_{f(x)}f(Y)$. Since ${\varphi }_t(Y)=Y$, we haveso the conormal bundle is invariant.

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.