TITLE: $\Omega=x\;dy \wedge dz+y\;dz\wedge dx+z\;dx \wedge dy$ is never zero when restricted to $\mathbb{S^2}$
QUESTION [9 upvotes]: We define a 2-form $\Omega$ on $\mathbb{R^3}$ by $\Omega=x\;dy \wedge dz+y\;dz\wedge dx+z\;dx \wedge dy$. How can I show that $\Omega|_\mathbb{S^2}$ is nowhere zero? Before proving that how can I compute the restriction $\Omega|_\mathbb{S^2}$?

REPLY [14 votes]: Observe that wedging $\Omega$ with $\omega := x dx + y dy + z dz$ gives $(x^2 + y^2 + z^2) dx \wedge dy \wedge dz$, a positive multiple of the volume form.  Now $TS^2 = \ker \omega\vert_{S^2}$.  Now let $v, w \in T_uS^2$ and think of $u$ as the radial vector on $\mathbb R^3$.  Then these three vectors are linearly independent so that $\Omega \wedge \omega (u,v,w) \ne 0$.  But $\Omega \wedge \omega(u,v,w)$ is proportional to $\Omega(v,w)$ since $v,w \in \ker \omega$.