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Let $\mathbb{T}^{2n}=\mathbb{R}^{2n}/\mathbb{Z^{2n}}$ be the $2n$-torus, which we equip with the unique symplectic form $\omega$ that pulls back to the standard symplectic form on $\mathbb{R}^{2n}$ under the natural projection $\pi:\mathbb{R}^{2n}\to\mathbb{R}^{2n}/\mathbb{Z^{2n}}$. We identify the tangent space $T_x\mathbb{T}^{2n}\cong\mathbb{R}^{2n}$ for all $x\in\mathbb{T}^{2n}$. Fix some $v\in\mathbb{R}^{2n}$ and define the vector field $X\in\mathcal{X}(\mathbb{T}^{2n})$ by $X(x)=v$. Then this is supposed to be an example of a vector field which is symplectic but not Hamiltonian for $v\neq 0$. I know how to show that it is not Hamiltonian. To show that it is symplectic, we have to show that $d\iota_X\omega=d(\omega(X,\cdot))=0$. By Cartan's magic formula and the closedness of $\omega$, this is equivalent to showing that $$ \mathcal{L}_X\omega=\frac{d}{dt}\bigg|_{t=0}((\phi_X^t)^* \omega)=0 $$ Thus, we have to compute the flow $\phi_X^t$. Note that $\frac{d}{dt}\phi_X^t(y)=X_{\phi_X^t(y)}=v$ for all $y$. Thus, do we have that $\phi_X^t(y)=y+vt$, where now we view $v\in\mathbb{T}^{2n}$? And do we have that $\mathcal{L}_X\omega=0$?

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    $\begingroup$ Consider a lift $\tilde{X}$ of $X$ and compute $\mathcal{L}_{\tilde{X}}\omega_0$ $\endgroup$ Mar 24 '20 at 13:34
  • $\begingroup$ Okay so $\tilde{X}$ is a lift, i.e. $\tilde{X}(x)=(d\pi)_x^{-1}X(\pi(x))=v$ as $\pi$ is locally trivial. Then $\phi_{\tilde{X}}^t(y)=y+vt$, and $(\phi_{\tilde{X}}^t)^*\omega_0=\omega_0\circ d\phi_{\tilde{X}}^t,$ so $\mathcal{L}_{\tilde{X}}\omega_0=\frac{d}{dt}\bigg|_{t=0}((\phi_{\tilde{X}}^t)^*\omega_0)=(D\omega_0)\circ d\phi_{\tilde{X}}^t+\omega_0(\partial_t d\phi_{\tilde{X}}^t,\phi_{\tilde{X}}^t)+\omega_0(d \phi_{\tilde{X}}^t,\partial_t d \phi_{\tilde{X}}^t)=0+0+0$ $\endgroup$
    – user680806
    Mar 24 '20 at 13:37
  • $\begingroup$ I suppose so, yes $\endgroup$ Mar 24 '20 at 13:41
  • $\begingroup$ why does $\mathcal{L}_{\overline{X}} \omega_0 $ being zero imply that$\mathcal{L}_X \omega$ is also zero? $\endgroup$
    – samlanader
    Mar 24 '20 at 17:11
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Yes, that's right, you do have $\phi_{X}^t(y) = y + v t$ where we're using the additive group structure of the torus: this group structure is what allows us to identify $T_x \mathbb{T}^{2n}$ with $\mathbb{R}^{2n}$ at every point $x \in X$. Then given $A,B \in T_x \mathbb{T}^{2n}$ we can calculate the Lie derivative from the definition: $$(\mathcal{L}_X \omega)_x(A,B) = \frac{\mathrm{d}}{\mathrm{d}t}\bigg|_{t=0} ((\phi_{X}^t)^{\ast}\omega)_x(A,B) = \frac{\mathrm{d}}{\mathrm{d}t}\bigg|_{t=0} \omega_{x+ tv}( D_x \phi_{X}^{t} (A), D_x \phi_{X}^{t}(B))$$ Under the identifications of the tangent spaces $T_x \mathbb{T}^{2n}$ and $T_{x+tv} \mathbb{T}^{2n}$ with $\mathbb{R}^{2n}$, the derivative $D_x \phi_{X}^{t}$ is simply the identity map (by definition) and $\omega_{x+tv}$ is $\omega_x$. Therefore when we differentiate a constant, we simply get zero: $$(\mathcal{L}_X \omega)_x(A,B) = \frac{\mathrm{d}}{\mathrm{d}t}\bigg|_{t=0} \omega_{x}(A,B) = 0$$

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