the fixed points of symplectic diffeomorphism

Question

Let $(M,\omega)$ be a closed symplectic manifold with $\pi_2(M)=0$. Let {$f_t$}, $f_0=id$,$f_1=f\ne id$ be a Hamiltonian path on M generated by a Hamiltonian function F. Then how to prove that f has a pair of fixed points x and y s.t. their orbit {$f_tx$} and {$f_ty$} are contractible? The hint is to use Floer homology, but I cannot prove it.

Answer 1

The fact that there are two fixed points of the function $f$ and that the path $f_t(x), f_t(y)$ is contractible is equivalent to the fact that there exists two critical points of the standard actional functional

$\mathcal{A} : \mathcal{P}_0(M) \rightarrow \mathbb{R}$

where $\mathcal{P}_0(M)$ is the space of contractible loops in $M$. If one could get an isomorphism between coholomogy groups and floer (co-)homology groups then we will be done. Since we know that any symplectic manifold has sum of betti numbers atleast two and thus sum of dimension of floer homology groups will be atleast two. Thus we will have atleast two critical points of the actional functional. Now when then hamiltonian $F$ is small this result is true. It has something to do with thin-thick trajectories. I don't know in what generality the results holds. The condition on $\pi_2$ makes sure that the floer homology is defined.