Showing that some symplectomorphism isn't Hamiltonian I have the next symplectomorphism $(x,\xi)\mapsto (x,\xi+1)$ of $T^* S^1$, and I am asked if it's Hamiltonian symplectomorphism, i believe that it's not, though I am not sure how to show it. I know that it's Hamiltonian when there's a hamiltonian isotopy $\phi_t$ s.t $\phi_0=Id \ \phi_1=\psi$ where $\psi$ is the above symplectomorphism, and its vector field associated with it is Hamiltonian. But I don't see how to relate it to the question above. I was given a hint to calculate the Jacobian of this transformation, but don't see relevancy here. Any tips? Thanks, depressed MP.

This is not a Hamiltonian symplectomorphism.

First, once $S^1$ is a Lie group, its _cotangent_ bundle is a product and, in fact, can be thought as a cylinder. So the transformation in question is a translation of this cylinder $C$. If we wad a Hamiltonian

$$H:C\rightarrow\mathbb{R}$$

then its gradient will be orthogonal to the symplectic gradient. The symplectic gradient must be the field wich will give rise to the isotopy. But the gradient field of $H$ should be everywhere-non zero and tangent to $S^1$, which is impossible once $S^1$ is a compact manifold and any differentiable real function on compact manifolds must have critical points. !Hamiltonian versus Gradient flows