Some time ago I needed to solve the following problem:
We have a torus with $\mathbb{T}^2$ with coordinates $(x \ mod \ 1, y \ mod \ 1)$ and symplectic form $\omega = dx \wedge dy$. Given function $F : \mathbb{T}^2 \to \mathbb{T}^2$ $F(x,y) = (x + \sin^2 (2\pi y), y)$. Prove that there is no such Hamiltonian flow $\Phi^t$ such that $\Phi^1 = F$.
Here is my solution:
Consider the function $H : \mathbb{T}^2 \to \mathbb{R}$. Symplectic gradient of $H$ induces the flow $\Phi^t$ i.e. $\frac{d}{dt} \Phi^t = sgrad H$, where $sgrad H$ is such vector field that $\omega (sgrad H, V) = dH(V)$.
It is easy to check that with standart symplectic form $dx \wedge dy$ $sgrad H = (\frac{\partial H}{\partial y}, -\frac{\partial H}{\partial x})$.
Now, consider some trajectory of $\Phi^t$: $\gamma(t) = (x(t), y(t)) = \Phi^t(x,y)$.
This trajectory conserves the Hamiltonian:
$H(x(t), y(t)) = const$.
Assume that $\Phi^1 = F$, then $(x(t+1), y(t+1)) = F(x(t), y(t)) = (x(t) + \sin^2(2\pi y(t)), y(t))$.
$H(x(t) + \sin^2(2\pi y(t)), y(t)) = const \Rightarrow \frac{d}{dt}H(x(t) + \sin^2(2\pi y(t)), y(t)) = 0$.
$\frac{d}{dt}H(x(t) + \sin^2(2\pi y(t)), y(t)) = dH(\dot{x}(t) + \dot{y}(t) 2\pi \sin(4 \pi y(t)), \dot{y}(t)) = dH(\dot{x}(t),\dot{y}(t)) + dH(\dot{y}(t) 2\pi \sin(4 \pi y(t)), 0) $.
The first term $dH(\dot{x}(t),\dot{y}(t))$ is zero, because it is just the derivative of $H(x(t), y(t))$.
Then we get an equation $dH(\dot{y}(t) 2\pi \sin(4 \pi y(t)), 0) = 0$, which is equivalent to $\frac{\partial H}{\partial x} \dot{y}(t) 2\pi \sin(4 \pi y(t)) = $, but $\dot{y} = -\frac{\partial H}{\partial x}$, because (x(t), y(t)) is the flow induced by symplectic gradient of $H$.
So it follows that $\frac{\partial H}{\partial x} = 0$ and $H(x,y) = H(y)$.
Then the flow $\Phi^t(x,y) = (x + t \frac{\partial H}{\partial y}, y)$. If $\Phi^1 = F$, then $\frac{\partial H}{\partial y} = \sin ^2 (2\pi y)$, but since H is a function on torus $H(1) - H(0) = \int_0^1 \sin ^2 (2\pi y) dy$ must be equal to $0$, but it is impossible because $\sin ^2 (2\pi y)$ is positive function.
But I think that there is a mistake in the solution:
When I considered the identity $\frac{d}{dt}H(x(t) + \sin^2(2\pi y(t)), y(t)) = dH(\dot{x}(t) + \dot{y}(t) 2\pi \sin(4 \pi y(t)), \dot{y}(t)) $, I said that $dH(\dot{x}(t), \dot{y}(t)) = 0$, because it is derivative of $H(x(t), y(t))$, but I think there is something wrong with this argument:
The differential is applied at point $(x(t) + \sin^2(2\pi y(t)), y(t))$ (which is just $(x(t+1), y(t+1))$, so for my solution to work I need that $d_{(x(t+1), y(t+1))}(\dot{x}(t), \dot{y}(t)) = 0$, but we know only that $d_{(x(t), y(t))}(\dot{x}(t), \dot{y}(t)) = 0$ (from the conserving of Hamiltonian along the trajectory).
So my solution does not work.
Is there a way to modify this solution so it becomes right? Also I would like to know, if there is another approach to solve this problem?