In the 2-torus with symplectic form $\omega=d\theta\wedge d\varphi$, is the vector field $X=\partial_\theta$ Hamiltonian?

Question

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?)

Answer 1

Remark: $\theta$ and $\varphi$ are regarded as the angle coordinates of $S^1 \subset \mathbb C$.

First we should recall that the 1-form $d\varphi$ is not the exterior derivative of $\varphi$ on $X$. Namely $\varphi$ is not defined on the whole space $T$. But it is defined on arbitrary simply connected open subset on $T$ (after choosing a branch).

For any exact 1-form $df$ (on $T$) and any closed curve $\gamma:[0,1] \to T$, we always have $$\int_{[0,1]} \gamma^*df = \int_{[0,1]} d(f \circ \gamma) = f(\gamma(1))-f(\gamma(0)) = 0.$$

On the other hand, considering the closed curve $\gamma(t):=(0,e^{2\pi t})$, we have $$\int_{[0,1]} \gamma^* d\varphi = \int_0^1 2\pi dt = 2\pi.$$ Therefore $d\varphi$ is not exact.

Answer 2

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.

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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 $