Showing that some symplectomorphism isn't Hamiltonian

Question

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.

Answer 1

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.

Answer 2

short answer: let $d\theta$ be the "angle" 1-form of $S^1$ (it is closed but not exact). The symplectic form you are considering is $\omega = d\theta \wedge dp$. If $\partial / \partial p$ was Hamiltonian, there would be a function $H$ satisfying: $$ dH = \omega(\partial / \partial p) = d\theta$$
which is imposible since $d\theta$ is not exact. Hence the field in not Hamiltonian. qed

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What I find interesting about this question is the following. I think of it in physical terms. Take the following with a graion of salt, if you wish. Imagine the following, you go out an measure two numbers, the position $x$ and momentum $p$ (to simplify, think of it as velocity) of a object constrained to move in a straight line. In my view, symplectic geometry asks you to do the following:

Step 1) Think of these two measuraments as two completely independent things. Assume that there are no interaction between these two numbers.

that is a strange take on the issue. We intuitively imagine that 3km/h in the "x-axis direction" will not let you move in the y-axis direction. What to do?

Step 2) We impose a relationship between x and p, not a the measurament (at the manifold level), but we re-state that there is a relationship between these two things at the tangent space of the $(x,p)$ plane. That is the symplectic form.

It is, in my view, a sort of 90 degrees rotation. It "rotates" the vector $\partial / \partial x$ to "$\partial / \partial p$" (more precisely, to $dp$). So, with the symplectic form, we can remember the relationship between $x$ and $p$.

Now that we assume that position and velocity are on a equal footing, what can go weird? Perhaps the simplest dynamics of all is the one given by the hamiltonian

$$ H = p^2/2m$$

using the symplectic form, this is the dynamics of the vector field $p \partial / \partial x$. The solution is a bunch of horizontal lines of the $(x,p)$ plane. So far, ok, nothing weird. Velocity is constant whereas the position is changing - it is just something moving with constant velocity. But now, symplectic geometry seems to have no preference for position over velocity; so, is there a hamiltonian system where position is fixed but velocity changes? Clearly yes: $$ H = x^2/2$$ will do the trick. To me, it is quite unphysical that symplectic geometry would allow such dynamics to exist. I asked several physics friends if they know of something in these lines taking place in physics (perhaps a it could be used as a calculation trick). Nobody knew. All I can think is that in symplectic geometry, if you see a parked car, don´t touch it! for it could be sitting there collecting a lot of linear momentum waiting for you to touch it and discharge on you a huge impact.