Non-vanishing volume form on $S^2$ I have been given the form $\mu=x\,dy\wedge dz+y\,dz\wedge dx+z\,dx\wedge dy\in\Omega^2(S^2)$. I am asked to prove that this form is never vanishing. That is, $\nexists p\in S^2$ such that $\mu_p=0$. Intuitively, I would say that th eonly point where $\mu$ vanishes is $(0,0,0)$ which is not in $S^2$, so $\mu$ is nowhere 0. However, I know there are expressions of the form $xdx+ydy+zdz$ which are 0 on $S^2$. How could I prove that $\mu$ is nowhere vanishing?
 A: HINT: What is the unit normal vector to the sphere at $p=(x,y,z)$? So, if $v,w$ give a basis for $T_pS^2$, what is $\mu(p)(v,w)$?
A: Let $P=(x_0,y_0.z_0)$ be a point on the sphere. Then we can regard $\omega
=x_0\,dy\wedge dz+y_0\,dz\wedge dy+z_0\,dz\wedge dx$ as a bilinear map
on the tangent space $T_P(S^2)$ at $P$. Let $u_1=x_1\partial_x+y_1\partial_y
+z_1\partial_z$ and $u_2=x_2\partial_x+y_2\partial_y+z_2\partial_z$ be vectors
in $T_P(S^2)$. Then $\omega$ takes the pair $(u_1,u_2)$ of tangent vectors to
$\det A$ where
$$A=\pmatrix{x_0&y_0&z_0\\x_1&y_1&z_1\\x_2&y_2&z_2}.$$
As $x_0^2+y_0^2+z_0^2=1$ we can choose an orthogonal matrix $A$ with first
row $(x_0,y_0,z_0)$. Then the remaining rows define tangent vectors
$u_1$ and $u_2$, and $\omega(u_1,u_2)=\det A=\pm1\ne0$. So $\omega\ne0$.
A: Hint Use that $$\mu = \iota_E (dx \wedge dy \wedge dz) , \qquad E := x \partial_x + y \partial y + z \partial_z ,$$ and that the restriction $E\vert_{S^2}$ is the unit normal vector field along $S^2$.
Alternatively, if you know a little about group actions, show that $\mu$ is preserved by the usual action of $SO(3)$ on $S^2$. Since that action is transitive, it suffices to check that $\mu_p \neq 0$ for any single $p \in S^2$.
