What is a symplectic form intuitively?

Hi, to completely describe a classical mechanical system, you need to do three things:

-Specify a manifold $X$, the phase space. Intuitively this is the space of all possible states of your system.

-Specify a hamilton function $H:X\rightarrow \mathbb{R}$, intuitivly it assigns to each state its energy.

-Specify a symplectic form $\omega$ on $X$. What is $\omega$ intuitively? What kind of information about physics does it capture?

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Maybe this article here is also very useful: research.microsoft.com/en-us/um/people/cohn/Thoughts/… –  Jan Weidner Jul 23 '10 at 11:53

To elaborate on a comment of Steve Huntsman: the symplectic form turns a form $d H$ into a flow $X_H$ with a number of properties, but other types of forms can do a similar job. Indeed, there are a number of situations in physics where the relevant $\omega$ is not symplectic, for example for the following reasons:

• $\omega$ might be degenerate in the sense that $i_X \omega = 0$ for certain $X \ne 0$. This occurs for instance when you pull back $\omega$ to a constraint surface in phase space. Or you might be working on the Lagrangian side, taking the pull-back of $\omega$ along a non-invertible Lagrangian.

• In non-holonomic mechanics, $\omega$ is sometimes not closed, with the derivative $d \omega$ being related to the non-integrability of the constraint distribution.

The point is that such forms all lead to valid generalizations of Hamilton's equations, so using a symplectic form to write down Hamilton's equations is to a large extent motivated by the fact that it "just works". The physical properties offered by using a symplectic form instead of an arbitrary two-form are the following:

• Non-degeneracy: the evolution vector field $X_H$ is determined uniquely by the Hamiltonian $H$. By contrast, if you have gauge freedom, there will typically be constraints in the phase space, hence a degenerate symplectic form (see above), leading to a non-unique evolution (which is what gauge freedom is --- several mathematically distinct evolutions being physically the same).

• Closedness: the system preserves the symplectic form $$L_{X_H} \omega = d i_{X_H} \omega + i_{X_H} d \omega = 0$$ if $\omega$ is closed. In the classical literature, this gives rise to a series of conservation laws called the "Poincare invariants". Again, non-holonomic systems typically don't exhibit this property, leading to all sort of weirdness.

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Thank you! This is very interesting :) –  Jan Weidner Mar 31 '10 at 13:13

$\omega$ lets you turn $H: X \rightarrow \mathbb{R}$ into a vector field and then a flow by establishing $H \rightarrow X_H$ via $dH(Y) = \omega(X_H,Y)$. The almost complex structure provided by $\omega$ always locally looks the same (viz. $\mathbb{R}^{2n}$ with $(x,p)$ coordinates) by the Darboux theorem and canonical transformations are just symplectomorphisms.

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Thanks, so you say the symplectic form captures precisely the data needed to translate an energy function into its infinitesimal flow? –  Jan Weidner Mar 31 '10 at 11:43
Almost any 2-form would turn a function into a flow. But a symplectic form does it in a way that is compatible with Hamilton's equations. –  Steve Huntsman Mar 31 '10 at 11:52
Actually there are lots of almost complex structures that are tame with respect to $\omega$. –  Charlie Frohman Mar 31 '10 at 15:17
If this were all it is, then you should never use symplectic forms, just Poisson forms. Unfortunately, Poisson geometry is not as well understood. There are symplectic leaves, but if the Poisson structure can jump in rank, I think that there is no known "Darboux theorem" giving the Poisson structure in some sort of normal form. –  Theo Johnson-Freyd Mar 31 '10 at 16:29

My intuition for the symplectic form in mechanics is that it tells you which coordinates are conjugate. By Darboux's Theorem, you can always write it as $\sum dx_i\wedge dp_i$, and being able to match a "position" coordinate with a "momentum" coordinate is essential to being able to do classical mechanics and to have equations of motion.

More concretely and rigorously, Steve's answer says essentially the same thing, about turning the Hamiltonian into a vector field so that there will be a flow.

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Okay thanks, let me try to rephrase/expand your answer a bit to see whether I understand it correctly. Say our system discribes the movement of a particle in some space. Then the symplectic form tells us, which direction in the phase space is the "position direction" and which direction is the "moment direction"? –  Jan Weidner Mar 31 '10 at 11:47
Not quite. Though given a direction that we want to think of as position, it tells us what the momentum direction is. What it gives us are pairs: given one coordinate, there's a second so that the pair are canonically conjugate, which means that, with respect to each other, they will act like position and momentum, and that they'll ignore the "other direction" (that is, things will Poisson commute) –  Charles Siegel Mar 31 '10 at 12:18
This is very interesting to me. So like Theo Johnson-Freyd said, in an arbitrary symplectic manifold we don't know which coordinate is momentum and which is position, the symplectic form lets us just find the other, when we already know one. –  Jan Weidner Mar 31 '10 at 17:07
Yeah, we really, honestly and truly do not know which coordinates are position and which are momentum, all we can get is pairs of coordinates, one of which is each. –  Charles Siegel Mar 31 '10 at 17:14

Incidentally, I more or less disagree that symplectic geometry captures what I would consider "classical mechanics". The reason is that in all the examples that I think deserve to be called "classical mechanics", I actually have a configuration space $N$, and your symplectic manifold is $X = {\rm T}^*N$ the cotangent bundle. Then, of course, the symplectic form is precisely (part of) the cotangent structure.

This is not to say that symplectic geometry isn't interesting — it's led to great mathematics, and certainly captures some of "classical mechanics". From the physics perspective, what I think makes it most interesting is that it shows that there are strange symmetries between mechanical systems, when you have a symplectomorphism ${\rm T}^*N \to {\rm T}^*N'$ that does not arise from a diffeomorphism $N \to N'$.

But physics is not invariant under all symplectomorphisms. Otherwise, how would I know which coordinates are "position" and which are "momentum"? And I do believe that I know this, although maybe I'm wrong. You and I should get together and compare if our Darboux coordinates differ only by a map $N \to N'$, or by some more interesting symplectomorphism.

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Thanks, thats an interesting viewpoint. What, if you start with a cotangent bundle and do hamiltonian reduction? You end up with a symplectic manifold, that is not a cotangent space anymore, though I would still call it mechanics. –  Jan Weidner Mar 31 '10 at 17:19
Hmm. Why should there be a difference between position and momentum to be preserved by phisical symmetries? –  Mariano Suárez-Alvarez Mar 31 '10 at 17:50
@MSA: I just mean that the actual real physical world is (pretty close to) a cotangent bundle, with a specified base. So if I think about it just as a symplectic space, I lose some information. –  Theo Johnson-Freyd Mar 31 '10 at 19:10
To say that all canonical transformations are symplectomorphisms is not to say that all symplectomorphisms are canonical transformations. –  Steve Huntsman Mar 31 '10 at 19:34
I think I'm with Mariano in this one. One of the lessons learnt from the dualities paradigm of the last 15 years or so is that the distinction between "position" and "momenta" (and I use quotes because the statement is much more general) is largely a matter of interpretation. The same <em>physics</em> can be described using different choices of dynamical variables. –  José Figueroa-O'Farrill Apr 1 '10 at 7:32