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Let $(M, \omega)$ be a $2n$-dimensional symplectic manifold and let $L_2(M,|\omega^n|)$ be the Hilbert space of complex-valued functions on $M$ that are square integrable with respect to the Liouville measure.

A classical (but still wonderful) remark due to Koopman is that given a measure preserving map $\phi: M \rightarrow M$, the operator $$ U_\phi : L_2(M,|\omega^n|) \longrightarrow L_2(M,|\omega^n|) $$ defined by $f \mapsto f \circ \phi$ is unitary. The operator $U_\phi$ is sometimes called the Koopman operator of the map $\phi$. Since symplectomorphisms are measure preserving, it seems natural to ask the following

Vague question. Is there anything at all particular about Koopman operators of symplectomorphisms?

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    $\begingroup$ Assuming that by "uniform closure", you mean "norm closure", then operators that you describe form a closed discrete set (any tow of them have distance 2). $\endgroup$ Commented Jun 27, 2014 at 12:39
  • $\begingroup$ Of course !! It was a dumb way to try to make precise a vague question. I'm editing the question to leave it vague. ;-) $\endgroup$ Commented Jun 27, 2014 at 12:48
  • $\begingroup$ What is the motivation for your question? $\endgroup$ Commented Jun 27, 2014 at 15:32
  • $\begingroup$ @AndréHenriques: I'm trying to see if there is any symplectic version of ergodic theory, one that uses that the maps are not just measure preserving. $\endgroup$ Commented Jun 27, 2014 at 17:13
  • $\begingroup$ Which results/concepts in ergodic theory do you expect to be affected by the fact that the transformation is a symplectomorphism? $\endgroup$ Commented Jun 27, 2014 at 19:05

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The Koopman operator is bracket preserving with respect to the canonical Poisson bracket on $C^{\infty}(M)$. Formally, this implies the dual operator (the Frobenius-Perron operator) is a Poisson automorphism with respect to the Lie-Poisson structure on $\bigwedge^{2n}(M)$.

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  • $\begingroup$ Thanks. This is the usual "functional" description of symplectic maps, but how to "marry" this with ergodic theory? Perhaps we have to change the space of square integrable functions for a Sobolev space and have the Koopman operator act on it instead. $\endgroup$ Commented Aug 5, 2014 at 10:12

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