TITLE: Square root for Hamiltonian diffeomorphisms
QUESTION [7 upvotes]: Let $\psi_t: X\to X$, $t \in [0,1]$, be a path Hamiltonian diffeomorphism on a symplectic manifold $X$, given by functions $H_t$. If $H_t \equiv H$ is independent of $t$ then 
$$ \psi_1 = \psi_{\frac{1}{2}}^2 $$
and therefore the Hamiltonian diffeomorphism $\psi_1$ has a Hamiltonian square root.
Is the same thing true for any arbitrary Hamiltonian $\psi_1$, i.e. is there another Hamiltonian $\phi$ such that $\phi^2 = \psi_1$ ?

REPLY [2 votes]: In a short paper posted last week, Peter Albers and Urs Frauenfelder prove that if $(M,\omega)$ is any closed symplectic manifold, then in any $\mathcal{C}^\infty$-neighborhood of the identity in $\text{Ham}(M,\omega)$ there is a Hamiltonian diffeomorphism that does not have a square root in $\text{Ham}(M,\omega)$ (the square root is not required to lie in this neighborhood).
The key is an observation of Milnor's that for any $k > 0$, an obstruction to a self-diffeomorphism of a manifold having a square root is that it has an odd number of $2k$-cycles.