$\mathbb{S}^1$-action over $T^2$ is symplectic but not Hamiltonian

differential-geometrygroup-actionssymplectic-geometry

I've read the following assertion in Ana Cannas' Lectures in Symplectic Manifolds:

"On the $2$-torus $(T^2,d\theta_1\wedge d\theta_2)$, the vector fields $X_1=\frac{\partial}{\partial\theta_1}$ and $X_2=\frac{\partial}{\partial\theta_2}$ are symplectic but not Hamiltonian."

Here's my attempt to prove it.

I'm used to writing $T^2=\mathbb{R}^2/\mathbb{Z}^2$, so I'll use $x,y$ for the coordinates and the symplectic form will be $dx\wedge dy$.

Clearly the flows for $X_1,X_2$ are $\phi_t^1(x,y)=(x+t,y)$ and $\phi_t^2(x,y)=(x,y+t)$ respectively, so $(\phi_t^1)^*(dx\wedge dy)=d(x+t)\wedge dy=dx\wedge dy$ and $(\phi_t^2)^*(dx\wedge dy)=dx\wedge dy$.

This means that both $X_1,X_2$ are symplectic. On the other hand, $i_{X_1}(dx\wedge dy)=dy$ and $i_{X_2}(dx\wedge dy)=dx$, so it looks like both are also Hamiltonian.

I guess what I'm doing wrong is that $dx,dy$ are perfectly ok in $\mathbb{R}^2$ but not in $T^2$ because of the quotient by $\mathbb{Z}^2$. But in that case, is the first argument with the pullbacks also wrong?

Best Answer

Let me give you a different proof:

Claim: every Hamiltonian field on a compact symplectic manifold $(M,\omega)$ vanishes at some point.

Proof: Assume that $X = X_f$ for some smooth $f\colon M \to \Bbb R$. By compactness of $M$, the function $f$ must have a maximum value at some point $p \in M$. Then ${\rm d}f_p = 0$, and so $X(p) = X_f(p) = 0$. $\square$

Consequence: $X_1$ and $X_2$ on $T^2$ are not Hamiltonian.

Proof: $X_1$ and $X_2$ never vanish, and $T^2$ is compact. $\square$

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