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This subgroup uses Hamiltonians whose support in $[0,1]\times M$ is compact. Their flows are the identity outside a fixed compact set. On closed manifolds this support condition imposes no extra restriction. On an open symplectic manifold, it excludes nontrivial whole-fiber cotangent fiber translations, even when their generating one-form is exact.

The cotangent fiber translation $f(q,p)=(q,p+{\alpha }_q)$ is a smooth diffeomorphism with inverse $(q,p)\mapsto (q,p-{\alpha }_q)$, and $\pi \circ f=\pi$. For a tangent vector $v$ at $(q,p)$, the intrinsic definition of the canonical one-form on a cotangent bundle givesThus $f^{\ast }\theta =\theta +{\pi }^{\ast }\alpha$. Taking its exterior derivative with the fixed sign convention yieldsIf $\alpha$ is a closed differential one-form, the last term vanishes, proving that $f$ is a symplectomorphism. In fact the converse holds too: the pullback by the zero section shows that ${\pi }^{\ast }(d\alpha )=0$ implies $d\alpha =0$.

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.

Yes. Choose a global differential one-form $\beta$ with $d\beta =\sigma$, and use the cotangent fiber translationIt is a diffeomorphism with inverse $T_{-\beta }$, and $\pi \circ T_{\beta }=\pi$. Directly from the Liouville one-form,The sign convention $\omega =-d\alpha$ therefore givesThus $T_{\beta }:(T^{\ast }X,\omega )\to (T^{\ast }X,{\omega }_{\sigma })$ is the required symplectomorphism. Equivalently, $T_{-\beta }^{\ast }\omega ={\omega }_{\sigma }$ gives the inverse direction. This translation equivalence of exact twisted cotangent bundles is global even when $X$ is not compact; no completeness argument for a Moser theorem flow is needed.

For any positive-dimensional $X$, choose a smooth function $h$ with $dh$ not identically zero. A nonconstant smooth bump in a coordinate chart supplies one on an arbitrary smooth manifold. Translate by this exact differential form:The cotangent fiber translation identities giveThus a nontrivial exact fiber translation is a symplectomorphism that fails to preserve the Liouville one-form. For example, on $T^{\ast }\mathbb{R}$ one can take $T(q,p)=(q,p+1)$: it preserves $dq\wedge dp$ but changes $pdq$ to $(p+1)dq$.

If dimension zero is admitted, $T^{\ast }X=X$ and $\alpha =0$, so every diffeomorphism preserves $\alpha$ and this requested distinction is impossible. The example therefore applies in positive dimension, as required for a nonzero differential one-form.

The cotangent fiber translation $f$ is a diffeomorphism, with inverse $(x,\xi )\mapsto (x,\xi -{\beta }_x)$. Because $\pi \circ f=\pi$, the definition of the canonical one-form on a cotangent bundle givesTaking the exterior derivative yieldsIf $\beta$ is a closed differential form, this is $f^{\ast }\omega =\omega$, proving that the cotangent fiber translation is a symplectomorphism.

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 $\sigma =d\beta$, cotangent fiber translation by $+\beta$ gives a symplectomorphism from $(T^{\ast }X,-d\alpha )$ to $(T^{\ast }X,-d\alpha +{\pi }^{\ast }\sigma )$. Its pullback changes $\alpha$ by ${\pi }^{\ast }\beta$, so the extra two-form cancels with $-{\pi }^{\ast }d\beta$. The opposite translation gives the inverse direction, globally without compactness.