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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 ...

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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 ...

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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 ...

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Let me reformulate the question. You are essentially asking if the isomorphism $\Lambda^2(V^*) \rightarrow (\Lambda^2 V)^*$ where $\varphi^* \wedge \psi^* \mapsto [v \wedge w \mapsto \varphi(v)\psi(w) - \psi(v)\varphi(w)]$ is "canonically" induced by the natural isomorphism $$V^* \otimes V^* \rightarrow (V \otimes V)^*$$ where \$\varphi \otimes \psi \mapsto ...

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