Example of a symplectic but non-Hamiltonian vector field on $\mathbb{T}^{2n}$

I want to show that there exists a symplectic vector field on the $$2n$$ torus $$\mathbb{T}^{2n}$$, endowed with the unique symplectic form $$\omega$$ that pullsback to the canonical symplectic form $$\omega_0$$ on $$\mathbb{R}^{2n}$$ under the quotient map $$\pi:\mathbb{R}^{2n}\to\mathbb{T}^{2n}$$, which is not Hamiltonian. We identify $$T_x\mathbb{T}^{2n}\cong\mathbb{R}^{2n}$$ for all $$x\in\mathbb{R}^{2n}$$ and we define the vector field $$X\in\mathcal{X}(\mathbb{T}^{2n})$$ by $$X(x)=v$$ for some fixed $$0\neq v\in\mathbb{R}^{2n}$$. Then I want to show that $$d\iota_X\omega=0$$.

I consider the vector field $$\tilde{X}$$ on $$\mathbb{R}^{2n}$$ defined by $$\tilde{X}(x)=v$$. Then $$\tilde{X}$$ is symplectic and satisfies $$d\pi_x\tilde{X}(x)=X_{\pi(x)}$$, so $$d\iota_{\tilde{X}}\omega_0=0$$. But $$d\iota_{\tilde{X}}\omega_0=d\omega_0(\tilde{X},\cdot)=d(\pi^*\omega(\tilde{X},\cdot))=d\omega_{\pi(\cdot)}(d\pi\tilde X,d\pi\cdot)\underbrace{=}_{?}d\omega(X,\cdot)=d\iota_X\omega,$$ so $$X$$ is symplectic. I am not sure about the step with a question mark, as the $$d\pi$$ in the second argument disappears. Is there any justification for this?

Now, I want to show that $$X$$ is not Hamiltonian. As always, we assume it is, so there exists a smooth map $$H:\mathbb{T}^{2n}\to\mathbb{R}$$ such that $$\iota_X\omega=dH$$. But I don't see how to arrive at some contradiction now.

• $di_{\tilde{X}}\omega_0$ is a 1-form on $\mathbb{R}^{2n}$, while $di_X\omega$ is a 1-form on $\mathbb{T}^{2n}$, so you wouldn't expect them to be equal. Instead for $x\in\mathbb{R}^{2n}$, $(di_{\tilde X}\omega_0)_x = (di_X\omega)_{\pi(x)}\circ (d\pi)_x$ (which can also be written $di_{\tilde X}\omega_0 = \pi^*(di_X\omega)$). And this is what you have shown. For your second question, think about what happens if you integrate $di_X\omega$ around some closed non-simply connected loop in $\mathbb{T}^{2n}$ (e.g. take $v=e_1$, and integrate around the loop $t\mapsto te_2$). Mar 22 '20 at 11:57
• I understand your first point, but I don't quite get what you mean with integrating $d\iota_x\omega$, as I thought we just concluded that $d\iota_X\omega=0$. Don't you want to show that $\iota_X$ is a representative of a non-zero cohomology class in $H_{\text{dR}}^1(\mathbb{T}^{2n})$?
– user680806
Mar 22 '20 at 13:17
• Apologies, I meant integrate $i_X\omega$ (to show that it couldn't possibly equal $dH$ for some $H$). Mar 22 '20 at 15:50
• Isn't the answer given by Tsemo Aristide correct as well, which is very elegant?
– user680806
Mar 22 '20 at 19:51
• Yes, Tsemo's answer is correct, and I agree more elegant. I guess it's a matter of taste: for example, the answer I gave would still apply for the case $\mathbb{T}^p\times\mathbb{R}^{2n-p},\ p\ge 1$, whereas Tsemo's wouldn't (since the manifold is no longer compact). So it's perhaps more fundamental. Mar 23 '20 at 11:12

Hint: On a compact manifold, an Hamiltonian vector field has a fixed point. Let $$H$$ be the Hamiltonian, there exists $$x$$ such $$H(x)$$ is a maximum, and $$dH(x)=0=i_{X(x)}\omega_x$$ implies that $$X(x)=0$$.
On $$\mathbb{T}^n$$ take any vector field induced by $$\phi_t(x)=x+ta, a\in\mathbb{R}^{2n}-0$$, $$\phi_t(x)$$ is a symplectic vector of $$\mathbb{R}^{2n}$$ which induces an symplectic vector on $$\mathbb{T}^{2n}$$.
• Aren't you done after the first part: If the vector field $X$ defined by $X(x)=v$ is Hamiltonian, the fact that there exists $x\in\mathbb{T}^{2n}$ such that $dH(x)=0$ implipes that $X(x)=0$, so $v$ must be zero in order for $X$ to be Hamiltonian?
• Because of the non-degeneracy by $\omega_x$