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 $H_t$?

  • $\begingroup$ This answered (in the negative for many symplectic manifolds) in mathoverflow.net/a/41092/26935 $\endgroup$ Commented Aug 26, 2016 at 7:40
  • $\begingroup$ A similar negative answer is in mathoverflow.net/a/18801/26935 $\endgroup$ Commented Aug 26, 2016 at 8:12
  • $\begingroup$ 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? $\endgroup$ Commented Aug 26, 2016 at 16:20
  • $\begingroup$ 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. $\endgroup$
    – PVAL
    Commented Sep 25, 2016 at 23:07
  • 2
    $\begingroup$ 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. $\endgroup$
    – mme
    Commented Nov 25, 2016 at 1:11

1 Answer 1


It is wrong that any hamiltonnian diffeomorphism is flow

  • 1
    $\begingroup$ The norm of this site is to provide answers with explanations, so if you could give more details or give a source, that would be great. Thanks. $\endgroup$
    – Todd Trimble
    Commented Aug 26, 2016 at 18:35

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