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1

It must be something like $(-1)^{\text{parity of the permutation}}$, where the parity of the permutation can be defined (there are several possible definitions) as the parity of the number of transpositions into which the permutation is decomposed. One shows that the number of factors of any two such decompositions has the same parity. Another possible ...


2

The exterior algebra (generated by a vector space $V$) is an algebra. It is the quotient of the tensor algebra $TV$ by the relation $xy+yx=0$ for all $x,y\in V$. I think you are confusing what is the algebra and what is the vector space. $V$, in this context, does not form an algebra under the wedge product. However, the vector space $\Lambda(V)$ does. We ...


1

First of all, $V$ in $\wedge V$ is a vector space, not vector field (just to fix the terminology). As egreg pointed out, if $u,v\in V$, then $u\wedge v$ is not an element of $V$, but it is an element of $\wedge^2 V\subset \wedge V$. The algebra $\wedge V$ is really constructed in such a way that you will have a correctly defined operation $\wedge$ on it. In ...


0

In response to a comment you posed: $$x_{j}=\sum_{i=1}^{4}a_{ji}y_{i},\;\;\;j=1,...,6$$ Pulling back the given form I obtained $$B^{*}v=\sum_{k,i=1}^{4}(a_{1i}a_{4k}+a_{2i}a_{5k}+a_{3i}a_{6k})y_{i}\wedge y_{k}$$ To make this vanish we respect the antisymmetric nature of the wedge product. All we need is for $a_{1i}a_{4k}+a_{2i}a_{5k}+a_{3i}a_{6k}$ to be ...


0

Let $y_{k}=x_{k}=x_{k+3}$, $k=1,2,3$ $$B^{*}v=dy_{1}\wedge dy_{1}+dy_{2}\wedge dy_{2}+dy_{3}\wedge dy_{3}=0$$ Now, there are a few remarks to be made. The image of $\mathbb{R}^{3}$ under the map is not $\mathbb{R}^{6}$ but this should be expected. The best we could hope for by such a linear map is that the image is a 3-dimensional subset of ...


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Answer to 1: The kernel of $\wedge^n \phi$ is precisely the antisymmetric subspace generated by the elements you describe. In particular, we have $$ \ker \wedge^n\phi = A \wedge \left(\wedge^{n-1} E\right) $$ Answer to 2: Perhaps you should consider the map $$ \Phi = \phi \wedge \left(\wedge^{n-1} I\right) $$ where $I$ denotes the identity mapping over ...


0

The wedge product is define as: let $\omega\in \Omega^k(V)$ and $\tau\in \Omega^l(V)$; then $$\omega \wedge \tau=\frac{1}{k!l!}A(\omega\otimes\tau);$$ where $$A(\omega)(x_1,...,x_k)=\sum_{\sigma\in S_{k}}sgn(\sigma)\omega(x_{\sigma(1)},...x_{\sigma(k)})=\sum_{\sigma\in S_{k}}sgn(\sigma)\sigma\omega.$$ $$ $$ LEMMA: If $\omega\in\Omega^k(V)$, ...



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