So I am trying to learn about Hamiltonian systems from Peter Olver's book, applications of Lie groups to differential equations. Right now I am on the section talking about Poisson maps and there is this piece I do not quite understand
I understand the proposition and the proof but in the piece where we are after talking about the canonical example where they say rotations are Poisson maps (canonical maps since the manifold is symplectic) I am confused. I do not know too much about the rotation group S0(2) (i haven't seen representation theory only basic Lie groups and algebras), but the fact that H is a generator confuses me because $H$ maps from $R^2 \to R$ so I am not sure how it could be a generator for the rotations because from what I know about rotations in $R^2$ it following this formula: $(x,y) \to (x\cos(\theta) - y \sin(\theta), x \sin(\theta) +y \cos(\theta))$.
So then I was thinking to use the proposition above and calculate the flow of the vector field $p d_q - qd_p$ and got $\phi(t,(p,q)) = ( -qt +p, pt+q)$ but that doesn't give me anything (unless I made an error in computation). So I am kind of confused by this even though it seems relatively simple.
He means the Hamiltonian system generated by $H$, namely $$ \frac{dq}{dt} = \frac{\partial H}{\partial p} = p ,\qquad \frac{dp}{dt} = -\frac{\partial H}{\partial q} = -q . $$ In other words, it's the flow generated by the vector field $p \partial_q - q \partial_p$, just as you wrote. But I don't know how you tried to solve this, since the correct result is not at all what you wrote, but rather $$ \begin{pmatrix} q(t) \\ p(t) \end{pmatrix} = \begin{pmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{pmatrix} \begin{pmatrix} q_0 \\ p_0 \end{pmatrix} , $$ where $(q_0,p_0)=(q(0),p(0))$.
(Your solution, which can be rephrased as $(q(t),p(t))=(-q_0 t + p_0, p_0 t + q_0)$, satisfies $dq/dt = -q_0$, which is not equal to $p(t)$ as it should, and similarly $dp/dt$ isn't equal to $-q(t)$.)