TITLE: In the 2-torus with symplectic form $\omega=d\theta\wedge d\varphi$, is the vector field $X=\partial_\theta$ Hamiltonian?
QUESTION [7 upvotes]: Consider the 2-torus $T=S^1\times S^1$ with symplectic form $\omega=d\theta\wedge d\varphi$ and the vector field $X=\partial_\theta$. I wonder if $X$ is hamiltonian. In other words, is $\iota_X\omega$ exact on $T$?
By definition, we locally get $$ \iota_X\omega=\omega(X,-)=\omega(\partial_\theta,-)=d\varphi $$
How to show that this local relation fails globally on $T$? (and that $\iota_X\omega$ is not globally exact?)

REPLY [4 votes]: Preliminaries
The mapping $ \theta \times \varphi: \mathbb{T} \setminus \{ (1,1) \} \to (0,2 \pi) \times (0,2 \pi) \subseteq \mathbb{R}^{2} $ defined by
$$
\forall (a,b) \in (0,2 \pi) \times (0,2 \pi): \quad (\theta \times \varphi) \left( e^{2 \pi ia},e^{2 \pi ib} \right) \stackrel{\text{def}}{=} (a,b)
$$
is a homeomorphism. Hence, $ \mathbb{T} \setminus \{ (1,1) \} $ is seen to be a smooth $ 2 $-dimensional manifold.

Main Discussion
The symplectic form $ \omega = d{\theta} \wedge d{\varphi} $ is, technically speaking, only defined on $ \mathbb{T} \setminus \{ (1,1) \} $; we cannot immediately say that it is extendable to a symplectic form defined on all of $ \mathbb{T} $. The same must also be said of the vector field $ X = \partial_{\theta} $, which is only defined on $ \mathbb{T} \setminus \{ (1,1) \} $.

Theorem 1 $ X $ is a Hamiltonian vector field on $ \mathbb{T} \setminus \{ (1,1) \} $.

Proof:
\begin{align}
   X(\bullet)
&= {\partial_{\theta}}(\bullet) \\
&= (\partial_{\theta} \wedge \partial_{\varphi})(\bullet ~ \otimes d{\varphi}) \\
&= \Pi(\bullet ~ \otimes d{\varphi}) \\
&= {X_{\varphi}}(\bullet),
\end{align}
where $ \Pi $ is the bi-vector field that corresponds to $ \omega $. This immediately shows that $ X $ is the Hamiltonian vector field on $ \mathbb{T} \setminus \{ (1,1) \} $ corresponding to $ \varphi $. $ \quad \spadesuit $
As shown by H. Shindoh, $ d{\varphi} $ cannot be extended to a vector field on all of $ \mathbb{T} $, otherwise we would obtain an exact form on $ \mathbb{T} $, which leads to a contradiction if we integrate a carefully chosen closed contour in $ \mathbb{T} $.
This leads us to the grand finalé.

Theorem 2 Although $ X $ is a Hamiltonian vector field on $ \mathbb{T} \setminus \{ (1,1) \} $, it cannot be extended to a Hamiltonian vector field $ X_{f} $ on $ \mathbb{T} $.

Proof: Assume the contrary. Then we would have
$$
d{f|_{\mathbb{T} \setminus \{ (1,1) \}}} = d{\varphi},
$$
which leads to the same contradiction as shown in H. Shindoh’s solution. $ \quad \spadesuit $