# Question on Banyaga's Theorem

Let $(M^{2n}, \omega)$ be a compact symplectic manifold. Suppose $\phi$ is a Hamiltonian diffeomorphism of $M$. In other words there exists a one-parameter family of smooth functions $H_t : M \to \mathbb R, 0 \leq t \leq 1$ such that if we define the vector fields $X_{H_t}$ via $d H_t = X_{H_t} \lrcorner \omega$, and we define a one-parameter family of diffeomorphisms $\psi_t$ via

$\frac{\partial \psi}{\partial t} = X_{H_t}\\ \psi_0 = \mbox{Id}$,

then $\phi = \psi_1$. My understanding is that it is a theorem of Banyaga that there exists a single smooth function $f$ such that $\phi$ can be expressed as the time $1$ flow of the vector field $X_f$ defined as above via $d f = X_f \lrcorner \omega$.

My question is: is this correct, and if so can you give a precise reference? Moreover, is it possible to explicitly construct the requisite function $f$ from the one-parameter family $H_t$?

• This answered (in the negative for many symplectic manifolds) in mathoverflow.net/a/41092/26935 – Peter Michor Aug 26 '16 at 7:40
• A similar negative answer is in mathoverflow.net/a/18801/26935 – Peter Michor Aug 26 '16 at 8:12
• I will try to understand those answers. If I may, here is a link to some notes from Banyaga: personal.psu.edu/auw4/dakar.pdf He defines $\mathcal H$ on page 1 as the group of time 1 autonomous Hamiltonian diffeomorphisms, and $\mbox{Ham}$ as the group of all Hamiltonian diffeomorphisms on page 5. The Corollary on page 7 claims that, for a compact manifold, these two are the same. I don't think I am misunderstanding the statement, and Banyaga is apparently an expert in the field. What is the issue? – Thisquestionisreallyhard Aug 26 '16 at 16:20
• If I recall correctly, the set of autonomous Hamiltonians is closed under conjugation from any element of $\mbox{Ham}$. So by a theorem of Banyaga ($\mbox{Ham}$ is simple), any time there is a non-autonomous Hamiltonian, the set of autonomous ones fail to be a group. – PVAL Sep 25 '16 at 23:07
• He said the group generated by autonomous Hamiltonians. There's no reason to believe the product is autonomous. Then the interesting result is that you can express any Hamiltonian diffeomorphism as the composition of finitely many autonomous Hamiltonian diffeomorphisms. – Mike Miller Nov 25 '16 at 1:11