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I am wondering if, for instance, 2-torus can be considered symplectic manifold. On one hand, it seems easy to construct a `symplectic 2-form', which is given by \begin{equation} \omega=d\theta\wedge d\varphi \end{equation} where $(\theta,\varphi)$ parametrizes the circles $\mathbb T=S^1\times S^1$. However, because it is a compact manifold, we also know that it does not have a global symplectic potential i.e. $\omega=d\phi$ for some 1-form $\phi$, otherwise it will contradict Stokes' theorem.

I originally suspected it may not admit symplectic structure because 'visually', I could draw a line from the centre (origin) of the torus such that it is tangential to the torus. A vector $p$ along this line would be 'tangential' to the tangent space at the point where the line touches the torus and hence it leads to degenerate 2-form. But I think there may be a hole in this argument.

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    $\begingroup$ remember that $\theta$ and $\phi$ are not global on the torus. They just exists locally. $\endgroup$ – L.F. Cavenaghi Jan 27 '17 at 19:12
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    $\begingroup$ Yes, the torus is symplectic, but like you said, the symplectic form is not exact. $\endgroup$ – Matthew Leingang Jan 27 '17 at 19:33
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    $\begingroup$ If $M$ is a surface, then a symplectic form on $M$ is the same thing as a volume form. Hence symplectic surfaces are exactly the orientable ones. $\endgroup$ – Pedro Jan 27 '17 at 22:04
  • $\begingroup$ If you look at the torus as $\mathbb{R}^2/\mathbb{Z}^2$ then you just need to find an equivariant 2-form on the plane $\endgroup$ – 54321user May 5 '17 at 6:12
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(From the comments) If $M$ is a surface, then a symplectic form on $M$ is the same thing as a volume form. Hence a surface admits a symplectic structure iff it's orientable.

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