$\Omega=x\;dy \wedge dz+y\;dz\wedge dx+z\;dx \wedge dy$ is never zero when restricted to $\mathbb{S^2}$ 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 $\Omega|_\mathbb{S^2}$?
 A: 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 $v,w \in \ker \omega$.
A: Use the spherical coordinates: 
$$(x,y,z)=(r\sin\phi\cos\theta,r\sin\phi\sin\theta,r\cos\phi),0<\theta<2\pi,0<\phi<\pi.$$
Restricted to $\mathbb{S}^2$, we have $r=1$, i.e. 
$$(x,y,z)=(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi),$$
which implies that
$$dx|_{\mathbb{S}^2}=-\sin\phi\sin\theta\,d\theta+\cos\phi\cos\theta\,d\phi,$$
$$dy|_{\mathbb{S}^2}=\sin\phi\cos\theta\,d\theta+\cos\phi\sin\theta\,d\phi,$$
$$dz|_{\mathbb{S}^2}=-\sin\phi\,d\phi.$$
This gives
$$dx\wedge dy\big|_{\mathbb{S}^2}=-\sin\phi\cos\phi\,d\theta\wedge d\phi,$$
$$dy\wedge dz\big|_{\mathbb{S}^2}=-\sin^2\phi\cos\theta\,d\theta\wedge d\phi,$$
$$dz\wedge dx\big|_{\mathbb{S}^2}=-\sin^2\phi\sin\theta\,d\theta\wedge d\phi.$$
Combining all these, we have
$$\Omega\big|_{\mathbb{S}^2}=(xdy \wedge dz+ydz\wedge dx+zdx \wedge dy)\big|_{\mathbb{S}^2}$$
$$=-\sin^3\phi\cos^2\theta\,d\theta\wedge d\phi-\sin^3\phi\sin^2\theta\,d\theta\wedge d\phi-\sin\phi\cos^2\phi\,d\theta\wedge d\phi$$
$$=-\sin\phi\,d\theta\wedge d\phi.$$
Since  $0<\phi<\pi$, we have $\Omega\big|_{\mathbb{S}^2}\neq 0$.
A: if $p=(x,y,z)$ is on the sphere, then any tangent to the sphere at that point is normal to the direction vector of $p$ -- eg the vector $(y, -x,0)^T$ (you can identify the tangent vectors to the sphere with their image under the tangent map, formally you need to pull back the form to the sphere). 
If you evaluate your form on pairs of vectors of that type (not equal to each other, because of antisymmetry) you can check whether it's nonzero by simple calculation (I did not).
(The restriction to the sphere is obtained kind automatically if you only insert vectors tangent to the Sphere).
