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  Interesting Hamiltonian System

+ 5 like - 0 dislike
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The definition of a Hamiltonian system I am working with is a triple $(X,\omega, H)$ where $(X,\omega)$ is a symplectic manifold and $H\in C^\infty(X)$ is the Hamiltonian function.

I am wondering if someone can give me an interesting, or useful, example of a Hamiltonian system for which $X$ is not the cotangent bundle of a manifold.

This post imported from StackExchange Physics at 2014-08-07 08:04 (UCT), posted by SE-user JonHerman
asked Aug 6, 2014 in Theoretical Physics by JonHerman (25 points) [ no revision ]
Same question on Mathoverflow: mathoverflow.net/q/147395/13917 Related (since the two-torus cannot be a cotangent bundle): physics.stackexchange.com/q/126676/2451 and physics.stackexchange.com/q/32095/2451

This post imported from StackExchange Physics at 2014-08-07 08:04 (UCT), posted by SE-user Qmechanic
Also possibly of interest on MO: mathoverflow.net/questions/35900/…

This post imported from StackExchange Physics at 2014-08-07 08:04 (UCT), posted by SE-user Chris White

1 Answer

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In many cases of interest, $X$ is a coadjoint orbit of a Lie group $G$, and $H$ an element in the corresponding Lie-Poisson algebra of the Lie algebra of $G$.  

These spaces describe in particular lots of exactly solvable problems - here $H$ is a sum of elements of the Lie algebra multiplied with Casimirs, plus a Casimir. Most nice exactly solvable problem can be cast in this form. See http://www.physicsoverflow.org/21556/coadjoint-orbits-in-physics?

For more on coadjoint orbits and their role in classical mechanics see 

J.E. Marsden and T.S. Ratiu,
Introduction to mechanics and symmetry,
Springer, New York 1994.

http://libgen.org/search.php?req=Marsden+symmetry

(A short notice is also in http://en.wikipedia.org/wiki/Coadjoint_representation )

answered Aug 7, 2014 by Arnold Neumaier (15,787 points) [ revision history ]
edited Aug 7, 2014 by Arnold Neumaier

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