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Define the characteristic vector (fields) of a form as $\iota_v\omega=0$. By the properties of the interior product if both $v_1$ and $v_2$ are c.v., then so are their linear combinations, $\text{span}(v_1,v_2)$. Furthermore, if the rank of the form $r$, the number of linearly independent characteristic vectors is $n-r$. One can describe a tangent plane to ...
Perhaps a more understandable way to see this is to parametrize $\Bbb R^3-\{0\}$ by spherical coordinates: $$(x,y,z) = f(r,\phi,\theta)= (r\sin\phi\cos\theta,r\sin\phi\sin\theta,r\cos\phi).$$ Then $f^*\beta = (r\cos\phi)\dfrac{r^3\sin\phi\, d\phi\wedge d\theta}{r^4} = \cos\phi\sin\phi\,d\phi\wedge d\theta$. So it's clear, then, that $f^*(d\beta) = ... 2 Note that, in each term, we've written for instance$df \wedge dy \wedge dz$. Because$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z} dz$, and$dy \wedge dy = 0, dz \wedge dz = 0$, we see that $$df \wedge dy \wedge dz = \frac{\partial f}{\partial x}dx \wedge dy \wedge dz.$$ So ultimately for this ... 2 Incomplete thoughts: Thinking of the elements as written in an$n \times m$grid, send it to the element$\alpha = \bigotimes_{j=1}^m (v_{1j} \wedge \cdots \wedge v_{nj}) \otimes \bigotimes_{i=1}^n (w_{i1} \wedge \cdots \wedge w_{im})$. (so, putting the$v$'s together in columns and the$w$'s in rows.) To see that this is well-defined, it suffices to ... 2 Now apply the magic formula again: $$L_X(\eta)=i_Xd\eta+di_X\eta$$ where$\eta=i_Yd\omega+di_Y\omega$and get $$L_X(L_Y\omega)=i_Xd(i_Yd\omega+di_Y\omega)+di_X(i_Yd\omega+di_Y\omega)$$ use that$d$is linear $$i_Xd(i_Yd\omega)+i_X(di_Y\omega)+di_X(i_Yd\omega)+di_X(di_Y\omega)$$ Subtract$L_Y L_X\omega$and compare with$\$ ...