# Symplectic 2-Torus

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?)

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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.

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I slightly changed the 2nd sentence to make my explanation more suitable. –  H. Shindoh Feb 10 '13 at 22:47

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. $\quad \spadesuit$

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