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Aug 1, 2011 at 18:51 comment added Weiwei Sorry about the confusion. The point I mentioned this homotopy equivalence is to demonstrate that there aren't any specifically interesting Hamiltonians on $S^2$, and all hamiltonian are connected to one another. So unless you have further restrictions, e.g. requiring the symplectomorphism fixing a couple points etc., rigid motions are basically what you need to consider. But maybe you have more specific goals in your mind.
Aug 1, 2011 at 15:52 comment added Tatou Papora Sure. More than being homotopy equivalent it deformation retracts to $SO(3)$. One can find the proof in Mu-Tao's paper mrlonline.org/mrl/2001-008-005/2001-008-005-007.pdf. But the interst here is to handle an explicit diffeomorphism like one can write for example $f:\mathbb R\longrightarrow \mathbb R, x\mapsto \frac{x}{x^2+4}\sin(x)$ and whatever.
Jul 29, 2011 at 15:26 history answered Weiwei CC BY-SA 3.0