Example of symplectic and hamiltonian diffeomorphism on $S^2$ and $T^2$

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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 non-trivial explicit examples of symplectic and Hamiltonian diffeomorphisms on compact surface? (at least examples for $S^2$ and $T^2$ ) NB : By surface I mean 2-dimensional manifold. Thanks a lot

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Sorry, what is a symplectic diffeomorphism on a manifold on which you haven't specified a symplectic structure? Do you mean a symplectic diffeomorphism of the cotangent bundle? – Qiaochu Yuan Jul 29 2011 at 15:45

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The standard area forms on the torus and $S^2$ are symplectic structures. I assume those would be good for explicit examples. – Elizabeth S. Q. Goodman Jul 30 2011 at 8:24
In fact as long as you endow a symplectic structure, it doesn't matter which one it is because they are all isotopic up to a scaling. For example if you give two symplectic structures $\omega_i$, $i=0,1$ which represent the same cohomology class, which can always be done by scaling, then you just apply standard Moser's trick on $\omega_t=(1-t)\omega_0+t\omega_1$. You can see everything gets through because of the dimension reason. – Weiwei Jul 30 2011 at 17:36
Thanks all for your interest to my question. First of all, sorry for not being precise. So could use on $S^2$ $\sin\varphi d\theta\wedge d\varphi)$ in spherical coordinate or $d\theta\wedge d z$ in cylindrical coordinate. Feel free to consider the one that can be helpful. Besides, for $T^2$, $d\theta\wedge d\varphi$ where $\theta, \varphi \in S^1$ are general angular coordinates. – Tatou Papora Aug 1 2011 at 15:43

3 Answers

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 non-displaceable: 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 quasi-states and quasi-measures, 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.)

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On an orientable 2-manifold (such as $S^2$ or $T^2$), the symplectic 2-form $\omega$ 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.

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The difference between hamiltonian diffeomorphism and symplectomorphism that is isotopic to the identity is quite small and describe by the flux homomorphism (The breakthrough by Banyaga). So any symplectic isotopy (path to identity in the group of symplectiomorphisms) with zero flux is ended by an Hamiltonian diffeo. Precisely speaking, the isotopy can be made homotopic to hamiltonian isotopy having the same ends. – Tatou Papora Aug 1 2011 at 16:01

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 Cotton-Clay helpful, for example http://arxiv.org/abs/0807.2488. But of course, if you just want some hands-on non-trivial symplectomorphisms, I think Dehn twists would be interesting enough.

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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/…. 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. – Tatou Papora Aug 1 2011 at 15:52
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. – Weiwei Aug 1 2011 at 18:51
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