3
$\begingroup$

There is a well-known "chaotic hypothesis" dating from 1995 or so in statistical physics that suggests that classical statistical-physical systems should be "effectively" Anosov. I won't get into the details of this, but I want to use it (and the observation that despite the Heisenberg group nilmanifold example, from the POV of statistical physics most "useful" Anosov systems are based on toral automorphisms or perturbations thereof) to motivate the following question:

Given a generic "nice" Hamiltonian $H$, at any point it should admit a quadratic approximation and the corresponding toral automorphism (which exists and is unique). Does anyone think that it is plausible to, or better still know of attempts (especially successful ones) to, work on the chaotic hypothesis by piecing these local approximations together (I'm naively thinking of taking connected sums of tori and patching together integral curves, which seems a little silly, but still...)?

$\endgroup$

1 Answer 1

1
$\begingroup$

Take the standard map with its break down of invariant circles and all, Cantorii and the like. How would you start to approximate by Anosov?

I am thinking of Aubry/Mather/Bangert results on 2 degree of freedom Hamiltonian systems, or, area preserving maps of an annulus to itself. There are various theorems to the effect that these systems are a kind of unentanglable mess of integrable and ``chaotic''.

On the flip side, there are results of Gole/Boyland asserting that IF a natural mechanical system: kinetic + potential, admit a hyperbolic metric: so the system is on $T^*Q$, and $Q$ admits a hyperbolic metric, then the system is ``semi-conjugate'' to the corresponding Anosov system.

$\endgroup$
1
  • $\begingroup$ After thinking about this for a while I realized that there are topological obstructions to these sorts of approximations. For example even with the quadratic Hamiltonian corresponding to the cat map the set $[0, .5)$ is a repeller. This is probably due to the same sort of thing you're mentioning. $\endgroup$ Commented Jan 8, 2010 at 22:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .