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Is there any obstruction that prevents a Hamiltonian diffeomorphism on some symplectic manifold to be realized as the time-one map of the Hamiltonian flow of an autonomous Hamiltonian?

In the same spirit, is there any obstruction that prevents a Hamiltonian diffeomorphism on $T^*M$ (with the canonical symplectic structure) to be realized as the time-one map of the Hamiltonian flow of a Tonelli Hamiltonian (i.e. a Hamiltonian which is fiberwise convex and superlinear)?

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I'm confused, what definition of Hamiltonian diffeomorphism are you using? I've seen them defined to be the time-one map of a Hamiltonian flow in Banyaga's on-line lecture notes. –  Ryan Budney Oct 4 '10 at 23:12
    
@Ryan: could you please give a link for the notes? I can't find anything like that on the Banyaga's home page. Thanks in advance! –  mathphysicist Oct 4 '10 at 23:38
    
Oh, it's probably because they're not on his web-page, sorry if I was confusing: math.psu.edu/wade/dakar.pdf –  Ryan Budney Oct 4 '10 at 23:41
    
@Ryan: Thanks again! –  mathphysicist Oct 4 '10 at 23:58
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Banyaga's work has some relevance here. He showed that for closed symplectic manifolds, the Hamiltonian group is simple. Yet the subgroup generated by the autonomous Hamiltonians is normal. Hence every Hamiltonian diffeo is a composite of autonomous Hamiltonian diffeos. –  Tim Perutz Oct 5 '10 at 16:11

2 Answers 2

A time-one map of a flow is homotopic (even isotopic) to the identity, so that provides a lot of homotopy and homology obstructions. In the two dimensional case, i.e., an oriented surface with a designated area element, you can get lots of explicit examples this way.

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However, hamiltonian diffeomorphisms are also always isotopic to the identity, aren't they? –  rpotrie Oct 5 '10 at 13:00
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YES ! Thank you for pointing that out. I was in fact unaware of the technical meaning of "Hamiltonian diffeomorphism" and just assumed that the OP was using it as a synonym for symplectomorphism. For others who may not have known the definition of Hamiltonian diffeomorphism, see the paper by Banyaga at www.math.psu.edu/wade/dakar.pdf –  Dick Palais Oct 5 '10 at 13:54

Can't you get such an obstruction by looking at periodic points? It's not difficult to show that generic Hamiltonian diffeomorphisms have only isolated fixed points, and indeed only isolated periodic points. On the other hand, if $\phi$ is a Hamiltonian diffeomorphism generated by an autonomous Hamiltonian $H$ (so $\phi^{\circ k}$ is generated by $kH$), then fixed points of $\phi^{\circ k}$ come in two types: either critical points of $H$ (which are also fixed points of $\phi$), or else points lying on nontrivial $k$-periodic orbits of the Hamiltonian vector field, and these latter fixed points are non-isolated (since every other point on the orbit is fixed as well).

Hence, given $\phi$, you can tell that $\phi$ isn't generated by an autonomous Hamiltonian if you find an isolated periodic point of $\phi$ whose minimal period is larger than $1$.

This seems fairly widely applicable--on many manifolds it's known that every Hamiltonian diffeomorphism has infinitely many periodic points (see for instance arXiv:1006.0372 and arXiv:0912.2064), and as noted above generic Hamiltonian diffeomorphisms have all their periodic points isolated.

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