# Constant vector field on the torus $\mathbb{T}^{2n}$ is symplectic

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 $$\mathcal{L}_X\omega=0$$?

• Consider a lift $\tilde{X}$ of $X$ and compute $\mathcal{L}_{\tilde{X}}\omega_0$ Mar 24 '20 at 13:34
• 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 $\mathcal{L}_{\tilde{X}}\omega_0=\frac{d}{dt}\bigg|_{t=0}((\phi_{\tilde{X}}^t)^*\omega_0)=(D\omega_0)\circ d\phi_{\tilde{X}}^t+\omega_0(\partial_t d\phi_{\tilde{X}}^t,\phi_{\tilde{X}}^t)+\omega_0(d \phi_{\tilde{X}}^t,\partial_t d \phi_{\tilde{X}}^t)=0+0+0$
– user680806
Mar 24 '20 at 13:37
• I suppose so, yes Mar 24 '20 at 13:41
• why does $\mathcal{L}_{\overline{X}} \omega_0$ being zero imply that$\mathcal{L}_X \omega$ is also zero? Mar 24 '20 at 17:11

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: $$(\mathcal{L}_X \omega)_x(A,B) = \frac{\mathrm{d}}{\mathrm{d}t}\bigg|_{t=0} \omega_{x}(A,B) = 0$$