Hello everybody. For a purpose of consolidation of some result I am trying to set down, I need to construct an example to sustain the theory and I am looking for symplectic and Hamiltonian diffeomorphisms. So does someone can help me writing some nontrivial explicit examples of symplectic and Hamiltonian diffeomorphisms on compact surface? (at least examples for $S^2$ and $T^2$ ) NB : By surface I mean 2dimensional manifold. Thanks a lot

For $S^2$ all symplectic diffeomorphisms are hamiltonian, and the symplectomorphism group is homotopy equivalent to $SO(3)$. For other surfaces, I find the papers by Andrew CottonClay helpful, for example http://arxiv.org/abs/0807.2488. But of course, if you just want some handson nontrivial symplectomorphisms, I think Dehn twists would be interesting enough. 


On an orientable 2manifold (such as $S^2$ or $T^2$), the symplectic 2form $\omega$ can be given by the signed area. In this case, symplectic diffeomorphisms are just those which preserve area and orientation. Now, let's set aside the difference between diffeomorphisms and flows of vector fields (which is a complicated enough issue by itself), and just focus on the difference between symplectic and Hamiltonian vector fields. A vector field $X$ is Hamiltonian if $ i_X \omega = dH $ for some Hamiltonian function $H$ (where $i_X$ is the contraction by $X$ and $d$ is the exterior derivative). On the other hand, $X$ is symplectic if the Lie derivative $L_X \omega$ vanishes. Applying Cartan's "magic formula" $L_X \omega = di_X \omega + i_X d \omega = d i_X \omega$ (since $\omega$ is closed), this means that $X$ is symplectic when $d i_X \omega = 0$. In summary, $X$ is Hamiltonian when $i_X \omega$ is exact, and symplectic when $i_X \omega$ is closed. The difference between these is given by the 1st de Rham cohomology of the manifold in question. So, as Weiwei said, "symplectic" and "Hamiltonian" are identical on $S^2$, since $S^2$ is simply connected. On the other hand, they're not the same on $T^2$, since the torus has nontrivial 1st cohomology. Putting coordinates $(\theta,\phi)$ on $T^2$, the vector fields $\partial/\partial \theta$ and $\partial/\partial \phi$ are both symplectic but not Hamiltonian. 


Ari's answer is good because you can see how even flowing along a symplectic vector field is not enough, but you could add a nice touch to the picture. Because a Hamiltonian diffeomorphism is exact, a simple closed curve $\gamma$ on a torus is nondisplaceable: i.e., under a Hamiltonian flow $\phi^t$ the image $\phi^t(\gamma)$ will intersect $\gamma$, because together the two curves will bound a region of zero signed area. (If you're curious, the concept for this comes from symplectic quasistates and quasimeasures, which tell you when subsets can be displaced by Hamiltonian flows. I haven't learned much about them.) On the other hand, again as mentioned, rotation along one of the two angles of the torus is a perfectly good symplectic action that can never be Hamiltonian because it displaces simple closed curves. (I was saddened to hear that because this action is not Hamiltonian, the torus is not a toric manifold.) 

