Square roots of Hamiltonian diffeomorphisms (Q462168)

Let \((M, \omega)\) be a closed symplectic manifold, and let \(\text{Symp} (M, \omega)\) be the group of symplectic diffeomorphisms of \((M, \omega)\). Then the group \(\text{Ham}(M, \omega)\) of Hamiltonian diffeomorphisms on \((M, \omega)\) is a normal subgroup of \(\text{Symp}(M, \omega)\) and is an infinite-dimensional Lie group. It is well known by a result of Banyaga that \(\text{Ham} (M, \omega)\) is simple, i.e., contains no non-trivial normal subgroups.NEWLINENEWLINEThe paper under review is concerned with the structure of \(\text{Ham} (M, \omega)\). To be precise, the authors prove that there exists an arbitrarily \(C^\infty\)-small Hamiltonian diffeomorphism \(\phi\) which does not admit any square root, i.e., \(\phi\neq \psi^2\) for any \(\psi\in \text{Ham} (M, \omega)\). One crucial step of the paper is to construct a Hamiltonian diffeomorphism \(\phi\) which has exactly one \(2k\)-cycle. The main result then follows from Milnor's observation that an obstruction to the existence of a square root is an odd number of \(2k\)-cycles. It would be interesting to answer the question of whether an arbitrarily given \(C^\infty\)-small Hamiltonian diffeomorphism admits no square root. Finally, we remark that there are several misprints in the paper that can be easily noticed.