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The 2-sphere, endowed with the round Riemann metric with constant curvature 1, is a symplectic manifolds.

My question is: Is the group of symplectic automorphisms of $S^2$ with respect to this symplectic structure connected?

My motivation to this question is as follows. If the answer to this question is yes, then any symplectic automorphism of $S^2$ can be connected by a path with the identity map. Then the study of the automorphism can be reduced to the studty of a family of smooth functions. And then we are possibly allowed to enter the area of complex analysis since smooth funtions on $S^2$ can be approximated by holomoprhic functions on some complex manifold containg $S^2$ as a totally real submanifold.

The above question can be asked for compact Hermitian symmetric spaces, or more generally for coadjoint orbits of semi-simple Lie groups. So could you recommand any references about the topological structure of symplectic automorphisms of these spaces?

Symplectic geometry is not my major. Thanks a lot for your help!

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    $\begingroup$ Close to a duplicate: mathoverflow.net/questions/9335/… $\endgroup$
    – Ryan Budney
    Commented Oct 23, 2014 at 17:23
  • $\begingroup$ @Entaou A symplectic form on a 2-dim manifold is the same as an area form, so the group of symplectomorphisms is the same as the group of area preserving diffeomorphisms (and preserving the orientation). In the case of the 2-sphere, the group $\mathrm{Diff}(S^{2}, \omega)$ is homotopy equivalent to $\mathrm{SO}(3)$ by Smale. In particular, it is connected. $\endgroup$
    – Oldřich Spáčil
    Commented Oct 25, 2014 at 21:10
  • $\begingroup$ @Entaou More generally, I understand that you are interested in the connected component of the identity in the group of symplectic diffeomorphisms. In the case of simply connected manifolds this connected component is exactly the group $\mathrm{Ham}(M, \omega)$ of Hamiltonian diffeomorphisms of $(M, \omega)$ -- these are the diffeomorphisms "generated by functions, indeed" (loosely speaking). And, as far as I can remember, all coadjoint orbits are simply connected. $\endgroup$
    – Oldřich Spáčil
    Commented Oct 25, 2014 at 21:17
  • $\begingroup$ @Oldřich Spáčil Thank you! I think that your exact meaning is that: Ham(M,ω) is generated by time-dependent functions. $\endgroup$
    – Entaou
    Commented Oct 27, 2014 at 13:21
  • $\begingroup$ @Entaou Yes, time-dependent functions, that's correct and that's why I wrote loosely speaking ;-) $\endgroup$
    – Oldřich Spáčil
    Commented Oct 27, 2014 at 17:13

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