Back to symplectic geometry. So far, everything I did in my last post only used the fact that the symplectic form was skew symmetric, not that it was closed. Indeed the “closed” property is rather mysterious, (as far as I’m concerned, although in the literature it is called “geometric”), since I don’t know of any really good geometric intuition for the action of exterior derivative on 2-forms. Still, it is a hugely important condition, and key to many of the special properties of symplectic geometry, notably for us, the Darboux theorem. Note that there is no real equivalent condition for Riemannian structures, and therefore it takes us in a whole new direction. I would love to have a better sense of how to “explain” why certain symplectic arguments don’t work in the Riemannian world (eg Darboux theorem), but I haven’t delved deeply enough into the proof to do this, since it’s not coherent to just say “the Riemannian form isn’t closed.”

For an application of closedness, recall the Lie derivative of a tensor with respect to a vector field : it’s just the derivative of along the flow lines of . In the case that is a differential form, one can prove formally, using induction and certain algebraic properties of , that

where of an -form is just contraction of with that form (ie evaluate that form on to get an form). Again, this formula means nothing to me other than some algebra (I wish I could change that!), but it’s called “Cartan’s (magic) formula” and it’s very nice. In particular, suppose is Hamiltonian, and we want to evaluate ; contraction of and gives by definition which is exact and therefore closed; what’s more, because is closed, the other term in Cartan’s formula is 0, and we see that —the symplectic form is invariant under all Hamiltonian flows, which is to say, the flow of is a one parameter group of symplectomorphisms (diffeomorphisms which preserve the symplectic form, the analogy of isometries in Riemannian geometry)! The equivalent concept for Riemannian geometry is a Killing vector field, and generically, these are very hard to find, which one can see because a generic Riemannian manifold has a finite group of isometries (right?). But in the symplectic case we’ve just produced lots of (one-parameter subgroups of!) symplectomorphisms, indeed the space is infinite dimensional, and in this way is much more akin to volume preserving smooth maps (of which it is a subset, since is a volume form). In particular, we’ve shown that for every smooth function we get a family of symplectomorphisms, tangent to . In general a vector field whose flow is a family of symplecticmorphisms is called symplectic—by Cartan’s formula, this is the case if contracted with , it yields a closed 1-form—and we see that Hamiltonian vector fields correspond exactly to the case that this closed form is exact.

Questions to do with Hamiltonian symplectomorphisms (those that come from Hamiltonian vector fields), and related subjects, are at the heart of symplectic geometry. For example, one can ask about their fixed points, which is the Arnold conjecture, and there is the related Weinstein conjecture, just proved by Taubes. For now I won’t go into these, and will let the comparison with Riemannian geometry be the main motivation.

Concerning that comparison, we remarked earlier that on a fixed vector space , both symmetric and skew-symmetric forms, when non-degenerate, can always be put into a standard form, in other words, their only invariant is the dimension of the underlying space. Therefore at any point in our manifold, we know what the form looks like at that point. However, in Riemannian geometry, if we could construct some neighborhood around a point for which, in local coordinates, the metric restricted to every point in that neighborhood was standard, then we would have shown that that neighborhood was isometric to Euclidean space, and therefore flat, for example. But not all Riemannian manifolds are locally flat! Thus even locally, Riemannian geometry is quite complex. The same thing, remarkably, is NOT true for symplectic manifolds. This is Darboux’s theorem.

The proof I will give starts with a lemma called Moser’s trick; there is also a more hands on and dirty, “down-to-earth geometric” proof, I believe. Anyway, suppose is compact, and we have a smooth family of symplectic (closed, non-degenerate) forms , (here is the space of differential 2-forms), with an “exact derivative,” meaning that if we look at for each (which we can define in the usual way, since is an -vector space), then the resulting 2-form can be written as for some smooth family of 1-forms. This is the assumption. Then Moser tells us that this path of 2-forms, which lives only in some associated space of forms, can actually be realized geometrically on via a family of diffeomorphisms ( such that ) (NOT symplectomorphisms, duh, since they change the symplectic form). The proof is quick: each is non degenerate, so it gives us a vector field for each , which is just the contraction (as an anecdote, note here that is actually , so we’re plugging the primitive into its own derivative! This always seemed a little incestual to me. On a more serious note, it comes up all the time: is there some more concise or intuitive way to describe what it’s measuring?). Since is compact, we have a flow of diffeomorphisms generated by . And this is the desired family of maps! Because

for all ! (the last equality follows from the linearity of , and the first is one of these fancy Lie-derivative identities (applied to the flow of a time-dependant vector field) which one simply proves directly). So it’s constant, and it starts at (because is the identity).

Note that the assumption that is exact implies that is constant. Indeed, this latter assumption is actually sufficient for the theorem. So this says that any two “isotopic forms” (cohomologous symplectic forms connected by a path of cohomologous symplectic forms) are actually “strongly isotopic,” meaning the path of forms can be realized by a path of diffeomorphisms, as above.

Finally we use Moser’s trick to prove Darboux’s theorem. Let , here is not necessarily compact. Choose a basis in in which is standard (just at . Now we can extend the coordinates in to a local coordinate patch around , and we have two forms: the form , the one we started with, and the form defined to be the standard form at every point (note that by construction, these two forms agree at ). Now for our family of forms interpolating, we just take the line . We want to make sure these forms are non-degenerate (they are obviously closed), but because non-degeneracy is an open condition in the space of 2-forms, we can take our coordinate chart small enough so that we’re ok. Now note that the -derivative of is just independently of ; this form is closed, so locally it is exact (we’re doing everything locally) so let be an anti-derivative. Fixing up by a constant, we can assume it’s 0 at . Now we define to be the vector field which is contracted with . By restricting ourselves (again!) to a small set inside of our chart, we can define a global flow for (so this is slightly different than Moser’s trick, which was global but for which we used compactness of ). So indeed, applying the reasoning from Moser’s trick, interpolates between and , in particular, , which is the desired result. What’s more, because we fixed things up so that , and therefore its dual vector field , was zero at , the form is actually constant at as we apply the diffeomorphisms. Q.E.D.

Next time: ? Almost complex structures, contact geometry, J-holomorphic curves, examples, knot theory, something completely unrelated, nothing?

Here’s the best motivation I know for wanting your form to be closed. A nondegenerate 2-form on M has an inverse, a nondegenerate 2-vector-field. If the form is (skew)symmetric, so is its inverse. In local coordinates, a 2-form is a matrix, and its inverse is the inverse matrix, interpreted as a 2-vector (a section of the bundle that is the tensor-square of the tangent bundle).

Anyhoo, recall that any vector field is secretly a derivation C(M) \to C(M), where C(M) is the algebra of smooth functions on M. (A _derivation_ is an \R-linear map that satisfies the Leibniz rule.) Well, any 2-vector field is a biderivation, a map C(M)\otimes C(M) \to C(M) which is a derivation in each coordinate. Well, if you started with a nondegenerate antisymemtric 2-form, then you get an antisymmetric map. Which is almost a Lie bracket, and would be if it satisfied Jacobi.

And it turns out that Jacobi is equivalent the original form being closed.

By:

Theoon September 3, 2009at 4:04 PM

It is possible that the geometric structure implied by an arbitrary exterior 2-form is something worth studying. As far as I know, nobody has made a compelling case yet, but that doesn’t mean there isn’t one.

To me, the Darboux theorem motivates everything. It says that any symplectic manifold is locally equivalent to a cotangent bundle, so the local dynamics of a symplectic manifold are identical to that of a cotangent bundle. So locally, Hamiltonian mechanics works just like it does in classical physics. This means that any interesting phenomena is always due to the global structure of the symplectic manifold. No local curvature or connection to complicate life. No local differential geometric invariants to compute. So it is not so surprising that symplectic geometry has deeper interactions with topology than Riemannian geometry.

By:

Deane Yangon September 3, 2009at 9:33 PM

Regarding Lagrangian submanifolds:

Any symplectic manifold is locally equivalent to a tubular neighborhood of the zero section of a cotangent bundle. A cotangent bundle has both a natural zero section and a natural foliation by its vertical fibers. So a natural question is: which submanifolds of a symplectic manifold, via the Darboux theorem, can be either the zero section or a vertical fiber of a locally equivalent cotangent bundle?

The answer is: Lagrangian submanifolds in both cases. In particular, if you start with two transversal Lagrangian submanifolds intersecting in a point, then there is a local equivalence of the symplectic manifold with a cotangent bundle such that one Lagrangian submanifold is the zero section and the other is the vertical fiber passing through that point.

By:

Deane Yangon September 3, 2009at 9:39 PM

One more comment about Lagrangian submanifolds:

It is very instructive to understand how everything works on a cotangent bundle over, say, a manifold M.

The graph of a section of the cotangent bundle is an n-dimensional submanifold. Such a graph is a Lagrangian submanifold if and only if the section is the differential of a real-valued function on the manifold M.

Symplectic geometry can be viewed as originating from an effort to understand a scalar function by studying its differential instead. Of course, this motivation is totally gone when studying compact symplectic manifolds. But for me it makes Lagrangian submanifolds seem much less mysterious.

By:

Deane Yangon September 3, 2009at 9:47 PM

There is a natural corresponding condition for a Riemannian manifold — having curvature zero. This is a local condition about equality of derivatives which implies that your manifold is locally R^n with the standard Riemannian structure. The difference is that there are very few curvature zero Riemannian manifolds, while there are lots of symplectic examples.

By:

David Speyeron September 4, 2009at 5:44 PM

Regarding the remark by Speyer:

It’s not just that there are more examples of “flat symplectic manifolds”; *every* symplectic manifold is locally “flat”. This fact makes symplectic manifolds more analogous to arbitrary smooth manifolds than to Riemannian manifolds.

In particular, a flat Riemannian manifold has only a finite dimensional group of local isometries, while a symplectic manifold has an infinite dimensional group of local “isometries” (i.e. symplectomorphisms).

How symplectic manifolds are studied is more analogous to how (the topology of) smooth manifolds are studied than to how Riemannian manifolds are studied. The local flabbiness of a symplectic or smoth structure makes it very difficult to work with directly. Riemannian geometry is used as a tool (by imposing some local rigidity) to study either topological or symplectic manifolds, but should not be viewed as an analogy of either smooth or symplectic structures.

By:

Deane Yangon September 6, 2009at 6:38 PM