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.
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Find the 1-form $\omega(X,.)$, where $X$ is the vector field generating the action. Check that it's not exact (e.g. by computing an integral and not getting $0$). – user8268 Sep 01 '13 at 15:51
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A Hamiltonian circle action on a compact manifold always has fixed points at the critical set of a Hamiltonian potential. – Andrew D. Hwang Sep 01 '13 at 16:25
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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.
Henry T. Horton
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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.
Mücahit Meral
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In fact, any nontrivial circle group $\mathbb{S}^1$-action on torus $(\mathbb{T}^2,\omega)$ cannot be Hamiltonian.
Since if there exists a Hamiltonian action, we will have moment map: $$\mu:\mathbb{T}^2\longrightarrow\mathbb{R}\cong\mathrm{Lie}(\mathbb{S}^1)^*$$ since $\mathbb{T}^2$ is compact, $\mu$ must have (isolated) critical points, these critical points correspond to the zeros of fundamental vector fields $\underline{X}$, where $X\in\mathrm{Lie}(\mathbb{S}^1)$, and they are fixed points of $\mathbb{S}^1-$action.
Denoted by $(\mathbb{T}^2)^{\mathbb{S}^1}$ the set of fixed points of $\mathbb{S}^1-$action, if the action is Hamiltonian, then by previously statement, $(\mathbb{T}^2)^{\mathbb{S}^1}$ has positive Euler characteristic numbver, according to a theorem of Kobayshi (see DeVito's answer here): $$0<\chi\left((\mathbb{T}^2)^{\mathbb{S}^1}\right)=\chi(\mathbb{T}^2)$$ but the later equals to $0$, a contradiction.
Z. Liu
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