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Suppose we have a sympletic toric manifold $(M,\omega)$ of dimension $4$ and let $\triangle$ be its corresponding Delzant polytope. Suppose that this polytope is "nice" enough so that we are able to defined a map, using action-angle coordinates $\Psi:\triangle \times \mathbb{T}^2\rightarrow \triangle \times \mathbb{T}^2$ \begin{equation} \Psi(x_1,x_2,\theta_2,\theta_2)=(-x_1,-x_2,-\theta_1,-\theta_2). \end{equation} Examples of Delzant polytopes where we could define this map would be a square, octagon, etc.… We would just need to have enough symmetries. We are assuming that the polygon is centered at the origin.

Now this map will be a symplectomorphism, and what I have been wondering is if this could be a Hamiltonian symplectomorphism ? I tried some approaches to prove this but I got nowhere. Basically my idea was to use Banyaga's theorem, since we know that $H_1(M)=\{0\}$, we would just need to prove that $\Psi$ is in the connected component of the identity in $\operatorname{Symp}(M)$, however I was not able to construct such a family of maps.

Then, I remebered the Arnol'd conjecture, and when we can talk about Floer homology, this basically gives us a lower bound on the number of fixed points of an Hamiltonian symplectomorphism in terms of Betti numbers of $M$. So here $\Psi$ has $4$ fixed points, and so if the topology of $M$ is complicated enough we could come into some trouble, assuming that we were in a position to talk about Floer homology.

So right now I'm inclining to the fact that generally this won't be a Hamiltonian symplectomorphism because I'm not sure how $H_2(M)$ behaves, but I would like to hear some input on this. What do you think ?

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Sometimes it is, sometimes it isn't.

The spheres living over the edges generate the second homology, so you can read off the action on $H_2$ from that. For $S^2\times S^2$ (square) the action on homology is trivial (because opposite edges are homologous) and the symplectomorphism is indeed Hamiltonian. For a hexagon (3-point blow up of $P^2$) the action on homology is nontrivial.

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  • $\begingroup$ Thanks for the answer, it really helps to get a better picture of what is going on. I just have a few questions that come from my lack of knowledge I suppose. So if we have Hamiltonian diffeomorphism, this will induce the trivial map on homology ? So it kinda generalizes the fact that the hamiltonian diffeomorphism preserves the symplectic form ? Also the sphere over the edges generate the second homology but has it happens in the product of spheres some of them are homologous. And so does that generalize to other examples, say if the normal vectors of 2 facets $\endgroup$
    – Someone
    Jan 14, 2023 at 14:44
  • $\begingroup$ satisfy some condition then we know that the spheres generated by them are homologous ? Sorry for the questions I'm just trying to get a good idea of what is going. Furthermore if instead you know of a reference for these facts that would also be of great help. Thank you ! @JonnyEvans $\endgroup$
    – Someone
    Jan 14, 2023 at 14:45

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