An irrational flow on a symplectic torus is symplectic but not Hamiltonian

Small open rectangles of side lengths less than one in ${\mathbb{R}}^2$ project injectively to quotient charts of $T^2$; their transition maps are integer translations. They therefore define a smooth atlas on the topological torus of The two-dimensional torus $T^2=(\mathbb{R}\mathrm{/}\mathbb{Z}{)}^2$. Integer translations preserve $dx,dy$, so these forms descend, and $dx\wedge dy$ is closed and nondegenerate in every chart. The translation maps ${\phi }_t[(x,y)]=[(x+at,y+bt)]$ are well-defined and smooth for all real $t$, satisfy ${\phi }_{s+t}={\phi }_s{\phi }_t$ and ${\phi }_0=\mathrm{i}\mathrm{d}$, and have derivative in $t$ equal to $X$. Their coordinate differentials are the identity, so ${\phi }_t^{\ast }\omega =\omega$. The field $X=a{\mathrm{\partial }}_x+b{\mathrm{\partial }}_y$ is symplectic: ${\iota }_X\omega =ady-bdx$ is closed because its coefficients are constants, its complete flow preserves $\omega$ as just computed.

Its contraction is not exact: integrating $ady-bdx$ around the two generating loops gives the periods $a$ and $-b$, and at least one of them is nonzero because $(a,b)\mathrm{\ne }(0,0)$. By [F1] this one-form is not exact, so no Hamiltonian function exists.

Equivalently, the flow of $X$ is the action $t\cdot [(x,y)]=[(x+at,y+bt)]$ of $\mathbb{R}$ on $T^2$, which is symplectic by step 1.1; its fundamental field is $-X$ by [F3], and the moment equation for $\xi =1$ would require a function with $d{\mu }^1={\iota }_X\omega$, again impossible by step 2.1.

The irrational constant flow is therefore a symplectic action that is not Hamiltonian, refuting the statement.