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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?

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up vote 24 down vote accepted

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.

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What do you mean by "various tops" in the above? – David Carchedi Jun 18 '10 at 21:06
Top = spinning top. Generalizations of the spinning tops of Euler, Lagrange, Kovalevskaya correspoding to the case $\mathfrak{g}=so_3\ltimes \mathbb{R}^3.$ See Michel Audin's book, for example. – Victor Protsak Jun 19 '10 at 1:52
Thanks for this nice answer. It covers everything I asked for :) – Jan Weidner Jun 23 '10 at 11:26

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

  1. Parameters and stability. Sometimes it is easier to analyze stability of the dynamics on a Poisson manifold than on its symplectic leaves.

  2. 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.

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An interesting article, although of course his view on the present question is constrained by his topic. One small remark: in 2, I think you mean $so(3)^*$ (or $so(3)$ -- they are isomorphic), not the group SO(3). – Victor Protsak Jun 18 '10 at 5:45
Victor, I think that that you and David are saying the same thing. The configuration space of the rigid rotator is SO(3). The hamiltonian dynamics are taking place in the cotangent bundle, which is, of course, $\mathrm{SO}(3) \times \mathfrak{so}(3)^*$. – José Figueroa-O'Farrill Jun 18 '10 at 13:06
José, I don't think so, because the cotangent bundle is a symplectic manifold, so it wouldn't justify Poisson point of view. I am a bit unclear on point 2: "coadjoint orbit" implies that the Poisson manifold is $\mathfrak{g}^*.$ Of course, $\mathfrak{g}$ need not be $so(3),$ it may be, for example, its semidirect product with $\mathfrak{R}^3.$ – Victor Protsak Jun 19 '10 at 2:00
Of course I wasn't accurate and the Poisson manifold underlying the rigid rotator's dynamics is the dual of the Lie algebra of SO(3). But what I really meant is that the Euler angle description of the rigid body is considered to be more natural than working on the coadjoint orbit (in this case the unit sphere S^2 in so(3)*). – David Bar Moshe Jun 19 '10 at 5:08
Thanks, this article looks interesting! – Jan Weidner Jun 23 '10 at 11:29

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.

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"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." This is not quite right. Surely it is $T^*Q$ which admits a canonical symplectic structure. To define we start by defining the canonical 1-form $\theta$ on $T^*Q$. A point $\alpha \in T^*Q$ can be thought of as 1-form (also denoted) $\alpha$ on $Q$ at the point $\pi(\alpha)$, with $\pi:T^*Q \to Q$ the canonical projection. Finally, $\theta_\alpha = \pi^*\alpha$. – José Figueroa-O'Farrill Jun 17 '10 at 23:20
I realised that I ran out of room before defining the symplectic structure $\omega$ on $T^*Q$: $\omega = -d\theta$ (in my conventions). – José Figueroa-O'Farrill Jun 18 '10 at 0:25
Ah. Yes, I stand corrected. I didn't want to get technical with this point (as you have in your comment), as I assumed the questioner was familiar with the symplectic structure, from the way he phrased the question. But I see my attempt at hand-waving failed, so thank you for your correction. – Aasmund Ervik Jun 18 '10 at 9:36

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