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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.

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Just by information, it is a good exercise trying to prove the following statement: let $M$ be a manifold and $(T^*M,\omega_0)$ the cotangent bundle with the cannonical symplectic form. Then, if we take $\mu$ to be a $1$-form on $M$, $(x,\xi)\mapsto(x,\xi+\mu)$ is a symplectomorphism if and only if $\textrm{d}\mu=0$ –  matgaio May 10 '12 at 5:36

2 Answers 2

up vote 4 down vote accepted

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


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

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This is just a comment to matgaio answer - sorry, I cannot do comments. To me, the physical intuition behind it is the following. Thinking of the points in $S^1$ as the position of a particle, and the vectors of the cotangent bundle as momentum, then this flow describes a particle that stays put in time but is gaining a lot of velocity (more precisely, momentum), as time goes on. –  Espinho da Flor Apr 30 '12 at 20:39
I don't know much about the physics of the problem. I think this way because of some restrictions on the existence of moment maps for some symplectic actions of lie groups. Good to know there is a possible physical interpretation. –  matgaio Apr 30 '12 at 22:36
The statement "the gradient field of $H$ should be everywhere non-zero" is not obvious to me. The Hamiltonian vector field can be generated by a family of smooth maps $C\times \Bbb R \rightarrow \Bbb R$ (not just a single map) , and certainly each of these maps can have some vanishing gradient. Moreover if the Hamiltonian vector field being nonzero (for some time t) somehow implied that the gradient vector field of some function was nonzero at that point, we would have a 1 line proof of Arnold's conjecture. –  PVAL Aug 5 at 3:46

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

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.

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I don't know much about the physics of the problem. I think this way because of some restrictions on the existence of moment maps for some symplectic actions of lie groups. Good to know there is a possible physical interpretation. –  matgaio Apr 30 '12 at 22:36

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