# What is the meaning of symplectic structure? [closed]

Answers can come in mathematical, physical, and philosophical flavors.

Edit: There seems to be a consensus that this question is not formulated well. I must respectfully disagree. My interest in the question is immaterial to the question itself. It is manifestly not a "what is" question. I see no reason to write more than is necessary for the formulation of the question, and I invite nothing more than a sentence giving me something to think about or an idea of where to look.

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## closed as not a real question by S. Carnahan♦, David Speyer, Qiaochu Yuan, Pete L. Clark, Greg KuperbergDec 12 '09 at 4:42

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

This will get much better answers if you include more detail (maybe something about why you're interested in the question, how much symplectic geometry you know, etc.) –  Ben Webster Dec 11 '09 at 20:23
What Ben said. As stated, it's an invitation for 100 times more work than it took to ask the question. –  Greg Kuperberg Dec 11 '09 at 20:56
@AK: Your assertion that "What is the meaning of symplectic structure?" is "manifestly not a 'what is' question" is manifestly confusing to me, as well as, I suspect, to many other readers. If you really just want a single sentence: a symplectic structure on a manifold M is given by a closed, nondegenerate 2-form \omega on M. See e.g. en.wikipedia.org/wiki/Symplectic_manifold This may not be what you wanted, but you gave us no clues. If the wikipedia reference does not satisfy you, you might try again with a new question. –  Pete L. Clark Dec 12 '09 at 3:54
-1 for claiming that 'It is manifestly not a "what is" question.'. Further, your interest in the question is not immaterial to the question itself! –  Scott Morrison Dec 12 '09 at 7:15
I think this question is perfectly valid, if not perfectly worded; at least I'm very interested in how different people motivate studying symplectic topology and would like to know any possible answers different from mine. However, I think it should be a community wiki question, as there might not be a unique correct answer. –  Ilya Grigoriev Dec 12 '09 at 23:31

Here's how I understand it. Classical mechanics is done on a phase space M. If we are trying to describe a mechanical system with n particles, the phase space will be 6*n*-dimensional: 3*n* dimensions to describe the coordinates of particles, and 3*n* dimensions to describe the momenta. The most important property of all this is that given a Hamiltonian function $H:M\to \mathbb R$, and a point of the phase space (the initial condition of the system), we get a differential equation that will predict the future behavior of the system. In other words, the function $H$ gives you a flow $\gamma:M\times \mathbb R \to M$ that maps a point p and a time t to a point $\gamma_t (p)$ which is the state of the system if it started at p after time t passes.

Now, if we do classical mechanics, we are very interested in changes of coordinates of our phase space. In other words, we want to describe M in a coordinate-free fashion (so that, for a particular problem, we can pick whatever coordinates are most convenient at the moment). Now, if $M$ is an abstract manifold, and $H:M\to\mathbb R$ is a function, you cannot write down the differential equation you want. In a sense, the problem is that you don't know which directions are "coordinates" and which are "momenta", and the distinction matters.

However, if you have a symplectic form $\omega$ on M, then every Hamiltonian will indeed give you the differential equation and a flow $M\times\mathbb R \to M$, and moreover the symplectic form is precisely the necessary and sufficient additional structure.

I must say, that although this may be related to why this subject was invented a hundred years ago, this seems to have little to do to why people are studying it now. It seems that the main reason for current work is, first of all, that new tools appeared which can solve problems in this subject that couldn't be solved before, and secondly that there is a very non-intuitive, but very powerful, connection that allows people to understand 3- and 4-manifolds using symplectic tools.

EDIT: I just realized that I'm not quite happy with the above. The issue is that the phase spaces that come up in classical mechanics are always a very specific kind of symplectic manifold: namely, the contangent bundlde of some base space. In fact, in physics it is usually very clear which directions are "coordinates of particles" and which are "momenta": the base space is precisely the space of "coordinates", and momenta naturally correspond to covectors. Moreover, the changes of coordinates we'd be interested in are always just changes of coordinates of the base space (which of course induce a change of coordinates of the cotangent bundle).

So, a symplectic manifold is an attempts to generalize the above to spaces more general than the cotangent bundle, or to changes of coordinates that mess up the cotangent-bundle structure. I have no idea how to motivate this.

UPDATE: I just sat in a talk by Sam Lisi where he gave one good reason to study symplectic manifolds other than the cotangent bundle. Namely, suppose you are studying the physical system of just two particles on a plane. Then, their positions can be described as a point in $P = \mathbb R^2 \times \mathbb R^2$, and the phase space is the cotangent bundle $M=T^* P$.

Notice, however, that this problem has a lot of symmetry. We can translate both points, rotate them, look at them from a moving frame, etc., all without changing the problem. So, it is natural to want to study not the space M itself, but to quotient it out by the action of some Lie group G.

Apparently, $M/G$ (or something closely related; see Ben's comment below) will still be a symplectic manifold, but it is not usually the cotangent bundle of anything. The difference is significant: in particular, the canonical one-form on $M=T^*P$ will not necessarily descend to $M/G$.

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Thanks for the answer, Ilya. I'll look further into your "necessary and sufficient" comment. What is a community wiki question? –  Andre Dec 12 '09 at 21:53
If you want to know anything in a less vague sense, you should read the first chapter of McDuff & Salamon. Really, there's not much philosophy here, all that's left is to learn Hamilton-Jacobi equations and how the Hamiltonian flow works. For what "community wiki" is read the FAQ - it's basically for questions that don't have a canonical answer. Finally, I realized I'm not quite happy with my answer, so I'll edit it. –  Ilya Grigoriev Dec 12 '09 at 23:19
It is not true that in Physics one is intersted only in canonical transformations which are induced from diffeomorphisms of the configuration space. cf. Hamilton-Jacobi theory. –  José Figueroa-O'Farrill Dec 12 '09 at 23:43
$M/G$ isn't symplectic. By Noether's theorem, the action of G corresponds to a conservation law (action by translations=conservation of linear momentum, action by rotations=conservation of angular momentum), and the phase space of things with a particular value of the conserved quantity mod G is symplectic. The key words here are "symplectic reduction." –  Ben Webster Jan 21 '10 at 0:32