We know that there is a physical interpretation for symplectic manifolds (briefly, the fact that a sympletic form assigns to any Hamiltoninan a vector field which describes the motion of particles). My question is if there is a physical meaning to the fact that all symplectic forms locally look alike, i.e. they are standard locally,(for instance, for the cotangent bundle of a manifold, what does it mean that any non-standard symplectic form locally looks like the standard one- which basically imposes the Hamilton's equations on the hamiltonian H)?
[Math] physical intuition for Darboux’s theorem
sg.symplectic-geometrysymplectic-topology
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Let's start by answering the first question.
Let $M$ be any manifold. Consider a physical system consisting of a point-particle moving on $M$. What are the configurations of this physical system? The points of $M$. Hence $M$ is the configuration space.
Typically one takes $M$ to be riemannian and we may add a potential function on $M$ in order to define the dynamics. (More complicated dynamics are certainly possible -- this is just the simplest example.)
As an example, let's consider a point particle of mass $m$ moving in $\mathbb{R}^3$ under the influence of a central potential $$V= k/r,$$ where $r$ is the distance from the origin. The configuration space is $M = \mathbb{R}^3\setminus\lbrace 0\rbrace$.
Classical trajectories are curves $x(t)$ in $M$ which satisfy Newton's equation $$m \frac{d^2 x}{dt^2} = \frac{k}{|x|^2}.$$ To write this equation as a first order equation we introduce the velocity $v(t) = \frac{dx}{dt}$. Geometrically $v$ is a vector field (a section of the tangent bundle $TM$) and hence the classical trajectory $(x(t),v(t))$ defines a curve in $TM$ satisfying a first order ODE: $$\frac{d}{dt}(x(t),v(t)) = (v(t), \frac{k}{m|x(t)|^2})$$ This equation can be derived from a variational problem associated to a lagrangian function $L: TM \to \mathbb{R}$ given by $$L(x,v) = \frac12 m v^2 - \frac{k}{|x|}.$$ The fibre derivative of the lagrangian function defines a bundle morphism $TM \to T^*M$: $$(x,v) \mapsto (x,p)$$ where $$p(x,v) = \frac{\partial L}{\partial v}.$$
In this example, $p = mv$. The Legendre transform of the lagrangian function $L$ gives a hamiltonian function $H$ on $T^*M$, which in this example is the total energy of the system: $$H(x,p) = \frac{1}{2m}p^2 + \frac{k}{|x|}.$$
The equations of motion can be recovered as the flow along the hamiltonian vector field associated to $H$ via the standard Poisson brackets in $T^*M$:
$$ \frac{dx}{dt} = \lbrace x,H \rbrace \qquad\mathrm{and}\qquad \frac{dp}{dt} = \lbrace p,H \rbrace.$$
Being integral curves of a vector field, there is a unique classical trajectory through any given point in $T^*M$, hence $T^*M$ is a phase space for the system; that is, a space of states of the physical system. Of course $TM$ is also a space of states, but historically one calls $T^*M$ the phase space of the system with configuration space $M$. (I don't know the history well enough to know why. There are brackets in $TM$ as well and one could equally well work there.)
Not every space of states is a cotangent bundle, of course. One can obtain examples by hamiltonian reduction from cotangent bundles by symmetries which are induced from diffeomorphisms of the configuration space, for instance. Or you could consider systems whose physical trajectories satisfy an ODE of order higher than 2, in which case the cotangent bundle is not the space of states, since you need to know more than just the position and the velocity at a point in order to determine the physical trajectory.
It's late here, so I'll forego answering the bonus question for now.
Using the h-principle, Gromov showed that there is a symplectic form on $\mathbb{R}^6$ which admits $S^3$ as a Lagrangian submanifold. Using holomorphic curves, he showed that the standard symplectic form on $T^* \mathbb{R}^3$ does not admit any such Lagrangian. There is now a whole industry of building exotic symplectic forms on non-compact manifolds (see papers of Seidel-Smith, Mark McLean, ...).
Probably the only reasonable answer to characterising cotangent bundles uses the existence of a Lagrangian foliation by planes. If you have a foliation parametrised by a manifold which admits a Lagrangian section, then you have yourself an open subset of a cotangent bundle (this is just Weinstein's theorem). You can't drop the condition of the existence of a section precisely because you can add the pull back of a $2$-form on the base. If your symplectic form is "complete" then the existence of a Lagrangian section is a cohomological condition. Pick any section: If the pullback of $\omega$ doesn't vanish, then you don't have a cotangent bundle. If it vanishes in cohomology, you can use a primitive $1$-form to flow your section to a Lagrangian.
Added Remark:
I want to point out that the methods we have for producing different symplectic forms do not proceed by writing down different $2$-forms on the same space. Rather, you find some construction of symplectic manifolds (using some general notion of symplectic surgery) which produces a large class of symplectic manifolds, then you prove that some of these result in the same smooth manifold. The existence of a diffeomorphism is obtained abstractly, so I do not know of examples where we can write down a Hamiltonian whose dynamics for two different symplectic forms can be compared.
Best Answer
Another way to interpret this question is: Is there a 'heuristic' reason that all closed nondegenerate $2$-forms in $2n$ dimensions are locally equivalent? (It becomes more reasonable to ask this when you realize that nothing like this holds for, say, $3$-forms in general dimension, even for $3$-forms in dimension $5$.) There is a 'function count' heuristic reason for this local equivalence, which goes like this:
If $\omega$ is a closed $2$-form in dimension $n$, then, locally $\omega = \mathrm{d}\alpha$ for some $1$-form, by Poincaré's Lemma.
Now, in any local coordinates, $\alpha = a_i(x) \mathrm{d} x^i$, so $\alpha$ (and hence $\omega$) 'depends' on a choice of $n$ functions of $n$ variables, the $a_i$. However, local coordinate changes also depend on $n$ functions of $n$ variables, so you might expect that, by choosing coordinates cleverly, you could bring $\alpha$ into some standard form, just as you can put a (nonzero) vector field into a standard local form in so-called 'flow-box coordinates'.
It turns out that you can't quite do this for $1$-forms (another piece of evidence that you shouldn't mentally identify $1$-forms with vector fields as people often do when they first start thinking about differential forms); there is an algebraic invariant, the rank $\rho(p)$ of $\mathrm{d}\alpha$ as a tensor at each point $p$. However, if this rank is locally constant, then, yes, you can choose coordinates so as to reduce $\mathrm{d}\alpha$ to a standard normal form. That is the content of Darboux' Theorem.
To reduce $\alpha$ itself to standard form, you'd need a little more information. That is the content of the Pfaff Theorem. However, because $\alpha$ wasn't uniquely determined by $\omega$ (you could always add a term $\mathrm{d}f$ to $\alpha$), this extra information gets thrown away. (In fact, it is this 'gauge' ambiguity that shows that you should still have one arbitrary function left over in your choice of normalizing coordinates once you put $\omega$ in normal form. Hence the arbitrary function, i.e., the arbitrary choice of a Hamiltonian, in finding vector fields whose flows preserve $\omega$.)
If you try to do the same count for, say, closed $3$-forms in dimensions above $4$, you'll see immediately that there is no hope for such a simple normal form.