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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)?

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3 Answers 3

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.

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This seems like a perfectly fine answer, but the question asked for physical intuition based on the application to Hamiltonian systems in which the phase space represents the motion of particles. I don't see any physics in this answer. –  Ben Crowell Feb 15 at 23:04
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@BenCrowell: Well, I used the word 'gauge', isn't that physics? :) Seriously, if one wanted, one could describe my answer in 'physics language' as saying that the symplectic form $\omega$ is just the 'field' of a 'potential' $\alpha$, and because the degree of freedom of the potential is the same as the degree of freedom of choice of coordinates, 'essentially all' potentials 'should' be (locally) equivalent up to change of coordinates. In this view, the 'flexibility' of symplectic geometry is just a manifestation of gauge equivalence of potentials '$\alpha$'. –  Robert Bryant Feb 16 at 10:03

Obviously, this is a question that could be interpreted in different ways. For me, Darboux's theorem is the symplectic analogue of the theorem that a flat Riemannian manifold (i.e. one where Riemann's curvature tensor vanishes) is locally the same as $\mathbb{R}^n$. For Darboux, the analogue of the Riemann curvature tensor is $d\omega$, the differential of the symplectic form.

So one variation on your question is "why should the symplectic form on phase space be closed?" which has a very clear answer:

If $d\omega\neq 0$, then the form $\omega$ is not invariant under time translation!

That is definitely not very physically realistic.

The way to compute this is by taking the Lie derivative of $\omega$ by a Hamiltonian vector field $X_f$, using Cartan's magic formula $\mathcal{L}=d\iota+\iota d$. One finds that $$\mathcal{L}_{X_f}\omega=d(\omega(X_f,-))+d\omega(X_f,-,-)=ddf+d\omega(X_f,-,-))=d\omega(X_f,-,-),$$ so closedness is essentially equivalent to invariance of $\omega$ under time translation.

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In Riemannian or semi-Riemannian geometry, the reason that our spaces don't all look alike in a neighborhood of a point is that parallelism fails, and the extent to which parallelism fails can be described as a curvature, which is invariant. For example, if two geodesics both start from a point and later on intersect again at some other point, this is a sign of curvature. It's invariant because intersection is invariant. It doesn't matter what coordinates you pick -- an intersection is an intersection. In general relativity, geodesics are the trajectories of test particles, and they can intersect in ways that imply curvature.

In the phase space for a Hamiltonian system, the integral curves never intersect, which is physically because preparing a physical system in a definite state is supposed to determine its future time evolution according to the laws of physics. (As in general relativity, the curves can be the trajectories of particles through phase space, but a point in phase space doesn't just tell you the particle's position, it tells you its entire physical state.) Since the integral curves can't intersect, if someone draws a set of integral curves for you, you can always bend and stretch the picture so that the integral curves look like parallel lines -- in some finite neighborhood.

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Thank you, this looks more like what I was looking for. Another picture came to my mind to compare the Riemannian vs symplectic manifolds: if you consider $T^*M$ (phase space) as a vector bundle over M (the physical space), then intersecting trajectories on M lift to integral curves on $T^*M$ which necessarily do not intersect (by what you said).Somehow roughly maybe we can say that Riemannian manifolds lift to symplectic manifolds in such a way that the effect of curvature vanishes on them (?). –  7779052 Feb 16 at 0:53
    
but what happens at the critical points of Hamiltonian, where the flow lines intersect? –  7779052 Feb 16 at 7:48
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For any vector field (not necessarily a Hamiltonian one with respect to some symplectic structure) on a manifold (e.g., the state space), distinct integral curves won't intersect; curvature has no effect on this, so your non-intersection property couldn't tell you anything about possible curvature of the symplectic structure. In Riemannian geometry, position doesn't tell you everything about the geodesic; once you take velocity into account as well, distinct geodesics don't intersect, even in the presence of curvature. Also, if the curvature is negative, geodesics can intersect only once. –  Robert Bryant Feb 16 at 9:53

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