Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 140 / 1 / Solution

Use polar coordinates $z_j=r_j{e}^{i{\theta }_j}$ on each complex coordinate plane. On $S^3$, where $r_1^2+r_2^2=1$,Direct substitution shows that ${\alpha }_s\wedge d{\alpha }_s$ is nowhere zero for $s>0$, so ${\alpha }_s$ is a contact form.

The vector fieldsatisfies ${\alpha }_s(R_s)=r_1^2+r_2^2=1$ and ${\iota }_{R_s}d{\alpha }_s=0$ on tangent vectors to the sphere. It is therefore the Reeb vector field. Its flow isIf both coordinates are nonzero, an orbit closes only if the two angular frequencies have rational ratio, equivalently if $s\in \mathbb{Q}$. For irrational $s$, the only closed orbits are $\{z_2=0\}$ and $\{z_1=0\}$, the two coordinate circles. This is the irrational contact ellipsoid flow on the three-sphere.