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Suppose I have a one degree of freedom analytic Hamiltonian $H(q,p)$ defined on a semi-infinite cylinder, i.e. $(q,p) \in \mathbb{T} \times \mathbb{R}^{+}$, such that all level sets $H(q,p)=c$ are closed curves. I also add the condition that there are no equilibria. I think that then the time-1 map of the flow will be twist (with integrability following from the integrability of $H(q,p)$), but how do I prove the twist property?

Moreover, could one relax the assumptions on the Hamiltonian (i.e. defined on a cylinder, closed orbits, no equilibria) and still retain the twist property?

I have a one degree of freedom analytic Hamiltonian $H(q,p)$ defined on a semi-infinite cylinder, i.e. $(q,p) \in \mathbb{T} \times \mathbb{R}^{+}$, such that all level sets $H(q,p)=c$ are closed curves. I also impose the condition that there are no equilibria. 

I think that then the time-1 map of the flow will be twist (with integrability following from the integrability of $H(q,p)$), but how do I rigorously prove the twist property?

Moreover, could one relax the assumptions on the Hamiltonian (i.e. defined on a cylinder, and all orbits are closed, no equilibria) and still retain the twist property?

The related scientific literature I've seen either deals with 1.5 degree of freedom Hamiltonian, or Poincare maps of 2 d.o.f. systems, not with the type of Hamiltonians I wrote above...

Suppose I have a one degree of freedom analytic Hamiltonian $H(q,p)$ defined on a semi-infinite cylinder, i.e. $(q,p) \in \mathbb{T} \times \mathbb{R}^{+}$, such that all level sets $H(q,p)=c$ are closed curves. I think that then the time-1 map of the flow will be twist (with integrability following from the integrability of $H(q,p)$), but how do I prove the twist property?

Moreover, could one relax the assumptions on the Hamiltonian (i.e. defined on a cylinder, and all orbits are periodic) and still retain the twist property?

Suppose I have a one degree of freedom analytic Hamiltonian $H(q,p)$ defined on a semi-infinite cylinder, i.e. $(q,p) \in \mathbb{T} \times \mathbb{R}^{+}$, such that all level sets $H(q,p)=c$ are closed curves. I also add the condition that there are no equilibria. I think that then the time-1 map of the flow will be twist (with integrability following from the integrability of $H(q,p)$), but how do I prove the twist property?

Moreover, could one relax the assumptions on the Hamiltonian (i.e. defined on a cylinder, closed orbits, no equilibria) and still retain the twist property?

Suppose I have a one degree of freedom analytic Hamiltonian $H(q,p)$ defined on a semi-infinite cylinder, i.e. $(q,p) \in \mathbb{T} \times \mathbb{R}^{+}$, such that all level sets $H(q,p)=c$ are closed curves. I think that then the time-1 map of the flow will be twist (with integrability following from the integrability of $H(q,p)$), but how do I prove the twist property?

Moreover, could one relax the assumptions on the Hamiltonian (i.e. defined on a cylinder, and all orbits are periodic) and still retain the twist property?

Suppose I have a one degree of freedom analytic Hamiltonian $H(q,p)$ defined on a semi-infinite cylinder, i.e. $(q,p) \in \mathbb{T} \times \mathbb{R}^{+}$, such that all level sets $H(q,p)=c$ are closed curves. I think that then the time-1 map of the flow will be twist (with integrability following from the integrability of $H(q,p)$), but how do I prove the twist property?

Moreover, could one relax the assumptions on the Hamiltonian (i.e. defined on a cylinder, and all orbits are periodic) and still retain the twist property?