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For $k < 2$ and $k > \dim V - 2$ it's always true that a $k$-form $\omega$ both is decomposable and satisfies $\omega \wedge \omega = 0$. For $k = 2$ (and provided the field underlying $V$ does not have characteristic $2$), the condition $\omega \wedge \omega = 0$ is both necessary and sufficient for decomposability. (Proving this is a nice exercise.) ...
If $$\omega = \sum_{i=1}^n\frac{(-1)^{i-1}x_i}{\|x\|^n}\,dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n,$$then: $$d\omega = \sum_{i=1}^n\sum_{j=1}^n\frac{\partial}{\partial x_j}\left(\frac{(-1)^{i-1}x_i}{\|x\|^n}\right) dx_j \wedge dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n.$$Now, the only surviving term is when $j = ... 1 By definition, the wedge product of$1$-forms$\alpha, \beta$is $$\alpha \wedge \beta = \alpha \otimes \beta - \beta \otimes \alpha .$$ When$\beta = \alpha$, this is zero. The wedge product of$2$-forms has a different formula, so this does not apply. Indeed, for any finite-dimensional vector space$\Bbb V$of dimension$\geq 4$, fix a cobasis$(e^a)$of ... 1 I don't know what you mean by the regular representation of a Lie algebra, but the connection is the following. It's a bit confusing because there are a bunch of$3$-dimensional vector spaces that all get identified. Let$V$be a real inner product space. The inner product induces a canonical isomorphism$V \cong V^{\ast}$giving an isomorphism ... 1 Three vectors$x,y,z$in a plane are necessarily linearly dependent. Without loss of generality, we may assume$z$is a linear combination of$x$and$y$, i.e.$z=\lambda x+\mu y\$. Then $$(x\wedge y)z+(y\wedge z)x+(z\wedge x)y=\lambda(x\wedge y)x+\mu(x\wedge y)y+\lambda (y\wedge x)x+\mu(y\wedge y)x+\lambda(x\wedge x)y+\mu(y\wedge x)y.$$ Using skew-symmetry ...