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If $\omega$ is a $1$-form on a symplectic manifold, will it be closed? It seems to be trivial that if $\sigma$ is symplectic structure on a manifold $M$, then the induced map $$\sigma^\vee: TM\to T^*M$$ is an isomorphism. Hence there exists some vector field $X$ such that $$\omega(Y)= \sigma(X,Y)$$ As $d\sigma=0$ implies $d\omega=0$. I don't see a mistake here. Is is true?

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You can surely give an example of a non-closed $1$-form on $\mathbb R^2$... –  Mariano Suárez-Alvarez Sep 5 '12 at 17:35

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up vote 4 down vote accepted

The mistake is that $d\sigma = 0$ does not imply $d\omega = 0$. By Cartan's formula, we have that \begin{align*} d\omega = & d(\iota_X \sigma) & & \\ = & \iota_x d\sigma - \mathcal{L}_X \sigma & & \text{(Cartan's formula)} \\ = & - \mathcal{L}_X \sigma, & & \text{($\sigma$ is closed)} \end{align*} where $\iota_X$ is the interior product with $X$, so that $(\iota_X \sigma)(Y) = \sigma(X, Y)$ for all vector fields $Y$, and $\mathcal{L}_X$ is the Lie derivative with respect to $X$, which by definition is $$\mathcal{L}_X \sigma = \left.\frac{d}{dt} \phi^\ast_t \sigma \right|_{t = 0},$$ where $\phi_t$ is the flow generated by the vector field $X$.

So if every $\omega$ were closed, then we would have that every vector field $X$ on $M$ is such that $\mathcal{L}_X \sigma = 0$, i.e. every vector field on $M$ is symplectic. But there is a one-to-one correspondence between flows and vector fields, and the flow of a symplectic vector field is a one-parameter group of symplectomorphisms. This of course is an incredibly strong requirement that is not often satisfied by any manifold $M$ (if ever).

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Moreover, the standard symplectic form on $\mathbb{R}^{2n}$ is $d\lambda$, where $\lambda = \sum_{i=1}^n x_i dy_i$, so $d\lambda \neq 0$. –  t.b. Sep 5 '12 at 17:34
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In response to your (if ever), the answer is never (unless $M$ is 0 dimenisonal). For, if $M$ is symplectic and dimension $\neq 0$, then its dimension at least 2. Then, in a small neighborhood of $M$ with coordinates $(x_1,..., x_n, y_1,...,y_n)$, one has the form $\omega = x_1 dy_1$. This obviously doesn't have $0$ differential. To get a form on the whole manifold, cut it off with a bump function. –  Jason DeVito Sep 5 '12 at 17:37
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No manifolds has all its $1$-forms closed. Closedness is a local property, so it is enough to give an example in $\mathbb R^n$ where it is easy. –  Mariano Suárez-Alvarez Sep 5 '12 at 17:37

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