Classical mechanics motivation for poisson manifolds? - MathOverflow

Classical mechanics motivation for poisson manifolds?

2010-06-17T16:34:22Z

Suppose I want to understand classical mechanics.

Why should I be interested in arbitrary poisson manifolds and not just in symplectic ones?

What are examples of systems best described by non symplectic poisson manifolds?

Answer by Aasmund Ervik for Classical mechanics motivation for poisson manifolds?

2010-06-17T17:38:48Z

Well, first of all, you would have to dive pretty far into classical mechanics in order to require either of these two tools. A typical introduction to the subject, based e.g. on the book of Goldstein, Poole and Safko, would not need such advanced mathematics.

That being said, non-symplectic Poisson manifolds do occur in classical mechanics, when studying the time evolution of systems, most commonly in the transition from classical mechanics to quantum mechanics.

A fundamental tool in quantum mechanics is the commutator between two operators, written e.g. [x,p]. The classical analogue of this is the Poisson bracket between two functions of the coordinates q_i, p_i and t, e.g. { f(q,p,t) , g(q,p,t) }.

Now for the geometry: suppose that your coordinates q_i lie in a Riemannian manifold, Q. Then Q together with the two-form w, which is the exterior derivative of the so-called canonical one-form on the phase space, usually form a symplectic manifold.

Now, a theorem (which I can't remember the name of) states that if (Q, w) is a symplectic manifold, then ( Q , {-,-} ) forms a Poisson manifold. This, again, turns out to be the requirement for constructing a Lie algebra homomorphism from the set of bounded sequences on Q to the set of vector fields on Q, and now we are very close to quantum mechanics.

For a specific system: consider standard 3D Euclidean space. This is odd dimension, so not symplectic. Construct the Poisson bracket such: {x,y} = z , {z,x} = y, {y,z} = x, where x,y,z, are the coordinate functions, i.e. this is the cross product in awkward notation. Now let F be in C^inf(Q), consider {F,-} acting as a derivation on the set of all polynomials. Then, by Weierstrass theorem, you can approximate a definition of {F,G} for all F, G in C^inf(Q).

So we have just constructed a very simple system (a point in Euclidean space) and represented it with a Poisson manifold. Note, however, that it has the set of all spheres centered at the origin as a foliation, and this is a foliation consisting of symplectic manifolds, which turns out to be a general feature.

Answer by Victor Protsak for Classical mechanics motivation for poisson manifolds?

2010-06-17T19:58:37Z

For many reasons and purposes, it is the Poisson bracket, not the symplectic form, that plays a primary role.

Equations of motion and, more generally, the evolution of observables have an easy form: $$ \frac{\partial f}{\partial t}=\{H,f\}.$$
Conserved quantities form a Poisson subalgebra: $$\{H,F\}=\{H,G\}=0 \implies \{H,\{F,G\}\}=0. $$
Since symmetries (Hamiltonian group actions) are Poisson by nature, the moment map is defined in the Poisson setting: $$ M\ni P\mapsto (X \mapsto H_X(P)).$$ Unlike in the symplectic case, both steps in the Hamiltonian reduction, restriction to the level set and factorization by the action of the stabilizer, are naturally carried out in the Poisson category, even for singular reduction.
Quasiclassical approximation in quantum mechanics (and conversely, quantization of classical mechanical systems) is expressed via the Poisson bracket: $$[\hat{F},\hat{G}]=ih\widehat{\{F,G\}} +O(h^2). $$

On the other hand, many natural systems have a degenerate Poisson bracket and/or are infinite-dimensional.

The phase space of various tops is the dual space $\mathfrak{g}^*$ of a Lie algebra. This is a universal example of a linear Poisson structure. Symplectic leaves are the coadjoint orbits and the Poisson center is given by the (classical) "Casimir functions".
Classical integrable systems such as KdV admit a bi-Hamiltonian structure (i.e. a pair of compatible Poisson brackets). This has no analogues in symplectic theory.
Some of these structures are obtained by a reduction from a linear Poisson structure on a suitable infinite-dimensional Lie algebra (loop algebra, algebra of matrix differential or pseudo-differential operators, etc).

Finally, a related practical consideration: even if you are interested in studying a symplectic manifold, frequently this is best accomplished by embedding it as a symplectic leaf into a simpler Poisson manifold, whether or not it has direct physical meaning. In particular, this applies to flag manifolds of semisimple Lie groups, which are topologically complicated objects, but can be identified with coadjoint orbits, thus embedded into a vector space with a linear Poisson structure.

Answer by David Bar Moshe for Classical mechanics motivation for poisson manifolds?

2010-06-18T05:02:37Z

The following article by J. Butterfield suggests 3 reasons motivating Poisson manifolds (over symplectic manifolds)(page 81)

Parameters and stability. Sometimes it is easier to analyze stability of the dynamics on a Poisson manifold than on its symplectic leaves.
Naturality: The rigid rotator problem is more natural to analyze on SO(3) than on its coadjoint orbit.

3.Reduction: When the configuration space is a Lie group, it is natural to use T*G/G as the reduced phase space. This is a Poisson manifold isomorphic to the dual of the Lie algebra and having a Lie Poisson structure.