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We say that a symplectic manifold $(M,\omega)$ is exact if $\omega=d\lambda$ for some $1$-form $\lambda$.

Consider a smooth manifold $S^1$. By standard construction, we know that a cotangent bundle $T^*S^1$ is a going to be a symplectic manifold with a symplectic form $\omega=dz\wedge d\theta$ where $\theta$ is an angle function and $z$ is the height function. Since $T^*S^1$ is trivializable, I think about $T^*S^1$ as a cylinder $$T^*S^1\cong M=\{(x,y,z)\in\mathbb{R}^3|x^2+y^2=1\}.$$

So, can I write that $\omega=d\lambda$ globally where $\lambda=zd\theta$ even if $\theta$ is a local parameter? In other words, is a cylinder an exact symplectic manifold?

eightc
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    Yes, actually ${\mathrm d} \theta$ is not a local form despite $\theta$ is a local coordinate. It is globally well-defined. – ChoMedit Feb 01 '22 at 11:23
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    The cotangent bundle is exact: the symplectic form constructed is $\omega = d\alpha$, where $\alpha$ is the tautological one form – Arctic Char Feb 01 '22 at 11:27
  • To elaborate on ChoMedit's comment: two different local angle functions $\theta_1$ and $\theta_2$ on $S^1$ are related via $\theta_2=\theta_1 + 2k\pi$ for some integer $k$ on the intersection of their domains, so ${\rm d}\theta_2={\rm d}\theta_1$ hold there. This means that to evaluate $({\rm d}\theta)(v)$ on some tangent vector to $S^1$, even though ${\rm d}\theta$ is not the exterior derivative of a globally defined function on $S^1$, one chooses an angle function $\theta_1$ defined near the base point of $v$ and computes $({\rm d}\theta_1)(v)$; ${\rm d}\theta_1$ is an actual differential. – Ivo Terek Feb 02 '22 at 21:23

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