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Comments to the question (v2): One can define left and right exterior derivatives $$ d_L(\omega\wedge\eta)~=~(d_L\omega)\wedge\eta + (-1)^{|\omega|}\omega\wedge d_L\eta \tag{L}, $$ $$ d_R(\omega\wedge\eta)~=~(-1)^{|\eta|} (d_R\omega)\wedge\eta + \omega\wedge d_R\eta \tag{R}, $$ $$ d_R\omega~=~(-1)^{|\omega|}d_L \omega, \tag{C}$$ where $\omega, ...


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For any two vector spaces $V$ and $W$ over the same field, there is a canonical isomorphism $V^*\otimes W^* \cong (V\otimes W)^*$, given by $(a\otimes b)(v\otimes w) = a(v) b(w)$, extended by linearity. This is just a special case, where $W = V$ and we anti-symmetrise.


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It is precisely an infinitesimal change. But an infinitesimal difference in what? A differential 1 form measures the flux across an infinitesimal line situated at some point. By extension a differerential 0 form evaluated at a point in space measures the flux across an infinitesimal point (an infinitiseimAl point is the same as a finite point is the same ...


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So let's assume that $V$ has a non-degenerate bilinear form $\langle\cdot,\cdot\rangle$ with a basis $e_1,\dots,e_n$ such that $\langle e_i,e_j\rangle = \delta_{ij}$, the Kronecker delta. Let $*$ denote the Hodge star operator. Note that we have the formula $$ \langle x,y\rangle = *((*x)\wedge y) .$$ Let's identify any operator on $V$ with its matrix ...


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As by the question linked by Michael, it is generally proved in a course about differential forms that $\alpha\wedge\beta = (-1)^{pq} \beta\wedge\alpha$ where $p$ and $q$ are the degrees of $\alpha$, $\beta$. Using this the result is immediate. But you could try and prove it, at least in your specific case. Hint: it should not be difficult to prove it for ...


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There isn't much you need to do. The wedge product satisfies the relationship: $\alpha \wedge \beta = (-1)^{pq} \beta \wedge \alpha$ if $\alpha, \beta \in \Lambda^p, \Lambda^q$ respectively. In your case $$\omega \wedge \omega = (-1)^{(2q+1)(2q+1)} \omega \wedge \omega = -\omega \wedge \omega$$ That can only happen if $\omega \wedge \omega = 0$.


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1) The trick is to find the space $W\subset V$ of vectors $v=ae_1+be_2+ce_3+de_4\in V$ killing $\omega=e_1 \wedge e_2 + e_3 \wedge e_4 \in \bigwedge^2 V$ i.e. such that $v\wedge \omega=0$. The criterion for $\omega$ to be decomposable is that $W$ have dimension $2$, the grading in which we take the wedge product $\bigwedge ^2V$ (the theory predicts that we ...


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Given a $p$-form $\theta \in \bigwedge^p E^*$, we can define an alternating multilinear map $h \colon E^{n-p} \to \bigwedge^n E$ by $$h(u_1, \ldots, u_{n-p}) = \theta \wedge \tilde{u}_1 \wedge \ldots \wedge \tilde{u}_{n-p}.$$ Let $b \colon \mathbb{R} \to \bigwedge^n E$ be the linear map $$b(t) = t \omega.$$ Because $\bigwedge^n E$ is one-dimensional and ...



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