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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 ...


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Let $X=z(x,y,z)/r^2$. Then $\beta=i(X)\Omega$ where $\Omega=dx\wedge dy \wedge dz$ is the volume element of $\mathbb{R}^3$ and $i$ is the interior product. Then the definition of divergence operator says $$d(i(X)\Omega)=\operatorname{div}(X)\Omega$$ So to show $d\beta=0$ is equivalent to show $\operatorname{div}(X)=0$ and we have ...


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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) = ...


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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 ...


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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 ...


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If $\omega$ and $\tau$ are closed forms (which you have not used), then $$d(\omega \wedge \tau) = d\omega \wedge \tau + (-1)^{p-1}\omega \wedge d\tau = 0,$$ hence $\omega \wedge \tau$ is also closed, hence it corresponds to a class in cohomology. Let us see that this class depends only on the classes of $\omega$ and $\tau$. Let us suppose that $\omega = ...


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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 $$ ...


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$B \subset C$ doesn't imply $A/B \subset A/C$. It implies $A/B$ surjects to $A/C$. The argument in Dummit & Foote must not be finished. To show what you want (which presumably comes later in Dummit & Foote): The key is to use $(x+y) \otimes (x+y) = x \otimes y + y \otimes x + x\otimes x + y\otimes x$. This shows that $x\otimes y + y \otimes x$ is ...


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If $\alpha$ is a $1$-form then $i_X\alpha=\alpha(X).$ If $\alpha$ is a $2$-form then $i_X\alpha$ is the $1$-form $\alpha(X,\cdot).$ Thus, $$i_Yi_X\alpha=i_Y(i_X\alpha)=i_Y(\alpha(X,.))=\alpha(X,Y).$$ In the particulat case of $\alpha=d\omega$ you get the desired result. Note that $i_X(d\omega)$ is a $1$-form and thus ...


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I believe this theorem that you mentioned only works for 2-vectors. If $w$ is a 2-vector then $w$ has rank $p$, if $\wedge^pw\neq 0$ and $\wedge^{p+1}w= 0$. Actually, let us prove that $w=e_0\wedge(e_1\wedge e_2+e_3\wedge e_4)$ is not decomposable, if $e_0,e_1,e_2,e_3,e_4$ are linear independent vectors. First, $v_1,\ldots,v_n$ are linear independent ...



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