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.

  • $\begingroup$ $di_{\tilde{X}}\omega_0$ 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 $t\mapsto te_2$). $\endgroup$
    – user17945
    Mar 22 '20 at 11:57
  • $\begingroup$ 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 $H_{\text{dR}}^1(\mathbb{T}^{2n})$? $\endgroup$
    – user680806
    Mar 22 '20 at 13:17
  • $\begingroup$ Apologies, I meant integrate $i_X\omega$ (to show that it couldn't possibly equal $dH$ for some $H$). $\endgroup$
    – user17945
    Mar 22 '20 at 15:50
  • $\begingroup$ Isn't the answer given by Tsemo Aristide correct as well, which is very elegant? $\endgroup$
    – user680806
    Mar 22 '20 at 19:51
  • 1
    $\begingroup$ 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. $\endgroup$
    – user17945
    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 $\mathbb{T}^{2n}$.

  • $\begingroup$ 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? $\endgroup$
    – user680806
    Mar 22 '20 at 19:24
  • $\begingroup$ Because of the non-degeneracy by $\omega_x$ $\endgroup$
    – user680806
    Mar 22 '20 at 19:30

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