Why circle action is not hamiltonian

Question

We have given an action $$S^1\times T^2\to T^2$$ $$(t,(\theta_1,\theta_2))\to (\theta_1+t,\theta_2)$$ Why this action can't be Hamiltonian? May i get some hint, comment suggestion. Thanks.

Answer 1

Write $S^1 = \mathrm{U}(1)$, and $\mathfrak{u}(1) = \mathrm{Lie}(\mathrm{U}(1)) = i\Bbb R$. Then given $\mathrm{U}(1)$ action on $T^2$ induces an infinitesimal action $$\rho: \mathfrak{u}(1) \times T^2 \longrightarrow T^2,$$ $$(\xi, (\theta_1, \theta_2)) \mapsto \left.\frac{d}{dt}\right|_{t = 0} \exp(t\xi).\!(\theta_1, \theta_2).$$ Part of the definition of a Hamiltonian action tells us that for each $\xi \in \mathfrak{u}(1)$, there exists a smooth function $$H_\xi: T^2 \longrightarrow \Bbb R$$ such that $$\iota_{\rho(\xi)}\omega = dH_\xi.$$ Since $\omega$ is nondegenerate, $\iota_{\rho(\xi)} \omega = 0$ if and only if $\rho(\xi) = 0$. For our given action, $\rho(\xi) \neq 0$ for $\xi \neq 0$. On the other hand, if some satisfactory $H_\xi$ existed, then since it is a smooth function on a compact manifold, it must have critical points, i.e. $dH_\xi$ will be zero at some point of $T^2$. Combining this with the above observations, we see that it is impossible to find suitable $H_\xi$ satisfying the definition of a Hamiltonian action for the given action.

Answer 2

I don't understand that $i_xw$ is closed and not exact. It is clear $i_xw=d\theta_2$ for $X=\frac{\partial}{\partial \theta_1}$. I don't know why $d\theta_2$ is closed.