Recovering topological invariants of symplectic manifold from the group of Hamiltonian diffeomorphisms?

Question

There is a famous result of Banyaga stating that if two closed symplectic manifolds $(M_1, \omega_1)$ and $(M_2, \omega_2)$ have isomorphic groups of Hamiltonian diffeomorphisms $\mathrm{Ham}(M_1, \omega_1)\simeq \mathrm{Ham}(M_2, \omega_2)$ then there exists a diffeomorphism $f:M_1\rightarrow M_2$ such that $f^*\omega_2=\lambda \omega_1$ for some $\lambda\in\mathbb{R}^{*}$. In other words, a symplectic structure on a closed manifold is determined (up to rescaling) by the group of Hamiltonian diffeomorphisms.

The result of Banyaga tells us that, in principle, it should be possible to understand the topology of $M$ only from $\mathrm{Ham}(M, \omega)$. However, as far as I understand, it's pretty hard to actually give an algorithm for doing that.

So my question is: are there any explicit procedures for recovery of algebro-topological invariants of $M$ from $\mathrm{Ham}(M, \omega)$? I am particularly interested in recovery of the fundamental group $\pi_1(M)$.

A weaker question would be: is there any explicit condition on $\mathrm{Ham}(M, \omega)$ which guarantees that $M$ is simply-connected?

I am aware of one comparatively weak result in this direction. Banyaga has constructed a map (called flux map) $f:\pi_1(\mathrm{Symp}_0(M, \omega))\rightarrow H^1(M, \mathbb{R})$. He has also shown that there is an isomorphism $$ \mathrm{Symp}_0(M, \omega)/\mathrm{Ham}(M, \omega) \simeq H^1(M, \mathbb{R})/\mathrm{ker}\:f. $$ Therefore, if we know both $\mathrm{Symp}^0(M, \omega)$ and $\mathrm{Ham}(M, \omega)$ we can at least recover a quotient of $H^1(M, \mathbb{R})$ (so we have a lower bound for its rank, for example).

P.S.: A clear exposition of some results on the group of Hamiltonian diffeomorphisms can be found in Polterovich's book "The geometry of the group of symplectic diffeomorphisms" (chapter 14 is particularly relevant to the question).

Answer 1

Let $M$ be compact and connected. For every closed $1$-form $\alpha$ on $M$ consider the Roger $2$-cocyle $\Psi_\alpha$ on the Lie algebra $\operatorname{ham}(M, \omega)$ defined by $$ \Psi_\alpha(X_f, X_g) = \int_M f \, \alpha(X_g) \, \frac{\omega^n}{n!}. $$ It was stated by Roger and proven by Janssens & Vizman that this map yields a bijection between $\operatorname{H}^1_{dR}(M)$ and the Lie algebra cohomology $\operatorname{H}^2(\operatorname{ham}(M, \omega), \mathbb R)$. This shows that you don't need to know the full group of Hamiltonian diffeomorphism but that you can actually extract the first de Rham cohomology of $M$ from the Lie algebra $\operatorname{ham}(M, \omega)$. In applications, it is however often more feasible to calculate the de Rham cohomology of $M$ than the Lie algebra cohomology of Hamiltonian vector fields.

A similar result holds for non-compact $M$, where $\operatorname{H}^2(\operatorname{ham}(M, \omega), \mathbb R)$ is isomorphic to the product of $\operatorname{H}^1_{dR, c}(M)$ and a $1$-dimensional subspace spanned by the Kostant-Souriau cocycle.