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I recently learned the following formulation of the Darboux theorem in a class.

Theorem: Suppose $\omega_t$ is a smoothly varying family of symplectic forms on a closed manifold $M$ such that the cohomology class of $\omega_t$ is independent of $t$. Then there is a smoothly varying family of diffeomorphisms $F_t$ of $M$ such that $F_0$ is the identity and $F_t^* \omega_t = \omega_0$.

The classical formulation of the Darboux theorem is obtained as a corollary in the following way. The condition that a closed 2-form is nondegerate is an open condition, so given a symplectic form $\omega_0$ on $M$ there is a neighborhood in the space of representatives of the cohomology class of $\omega_0$ which consists only of nondegenerate forms. This neighborhood can be taken to be path connected, so the theorem guarantees that for any symplectic form $\omega_1$ which is sufficiently close to $\omega_0$ and which belongs the same cohomology class there is automatically a diffeomorphism which pulls back $\omega_1$ to $\omega_0$.

My question: is the assumption that $\omega_1$ is sufficiently close to $\omega_0$ really necessary? In the proof that I learned, we really do need a path $\omega_t$ of nondegenerate forms because the idea is to use the Poincare lemma to write $\omega_t = \omega_0 + d \beta_t$ and then obtain a time dependent vector field $X_t$ satisfying $\iota_{X_t}\omega_t = -\frac{d}{dt}\beta_t$. To say that $X_t$ exists we need $\omega_t$ to be nondegenerate for all time, and the diffeomorphisms $F_t$ are obtained from the flow of $X_t$.

I guess the real problem here is that I don't know all that many interesting examples of symplectic manifolds to begin with, and even on those examples that I know I can never produce more than one symplectic structure. Can anyone help?

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Could you make this question more precise? When you ask if the assumption is necessary, necessary for what? –  Michael Bächtold Oct 16 '10 at 12:18
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up vote 5 down vote accepted

(Sorry I don't have enough reputation to make this a comment.)

I don't fully understand your question, in particular since I don't know which formulations of Darboux's theorem you are concerned with. The version I would describe as the "classical" one is that each point of a symplectic manifold admits a neighbourhood diffeomorphic to a standard symplectic ball in $R^{2n}$. In the proof I know, you need to shrink the size of your neighbourhood twice: the first time, in order to have that $\omega_1$ and $\omega_0$ are connected through symplectic forms, and the second time, to guarantee that the flow $\mathbf{F}_t$ exists to time $t=1$.

Whatever ways you may have of sidestepping one or both of these, you definitely need to shrink the size the neighbourhood somehow. Indeed, the answer to the more global question (how big a neighbourhood can you embed?) is a very subtle one related to symplectic capacities and to symplectic packing problems. For instance, Gromov's non-squeezing theorem shows that in the case of the symplectic cylinder $D^2 \times \mathbb{R}^2$, you can't get a symplectic ball any bigger than the one of radius 1. You can, however, easily smoothly embed a ball of much bigger radius (even in a volume preserving way). When you pull back the symplectic form to this ball, you will not be able to deform it to the standard one, except in a neighbourhood of a given point.

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Based on your answer and Tim Perutz's link, I think my question was indeed really all about the non-squeezing theorem, which I was aware of but did not consider when I asked the question. Thanks for helping me to get oriented! –  Paul Siegel Oct 21 '10 at 12:53
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