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On wikipedia, the motivational example for introducting the wedge product $\wedge$ uses $\textbf{v}= a\textbf{e}_1+b\textbf{e}_2$ and $\textbf{w}=c\textbf{e}_1 + d\textbf{e}_2$ in $\mathbb{R}^2$, where $\lbrace \textbf{e}_1, \textbf{e}_2\rbrace$ is the usual cartesian basis. From here, they derive that $\textbf{v} \wedge \textbf{w} = (ad-bc) \textbf{e}_1 \wedge \textbf{e}_2$, where $ad-bc$ is the area of the parallelogram spanned by $\textbf{v}$ and $\textbf{w}$.

However, the general defintion of the wedge product of multilinear forms is

$v \wedge w (x_1, \dots, x_{k+n} ) := \frac{1}{k!} \frac{1}{n!} \sum_{\pi \in \text{perm} (k+n)} \text{sgn} (\pi) (v \otimes w) (x_{\pi (1)} , \dots x_{\pi (k+n)} ) $,

where $v$ is $k$-form ($v \in \Lambda^k (M)$) and $w$ is an $n$-form ($w \in \Lambda^n (M)$. It follows then that $v \wedge w = v \otimes w - w \otimes v$, if both $v$ and $w$ are 1-forms. How does this relate itself to the motivational example?

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  • $\begingroup$ If you think of the wedge product as simply an operation that satisfies a bunch of properties (in analogy, consider the inner product), then both the first and second definitions satisfy those properties and thus are wedge products. The second is just more general in that it applies in more situations than just $\Bbb R^n$. $\endgroup$
    – user137731
    Nov 3, 2016 at 14:33
  • $\begingroup$ Hey, thanks for your answer. What are the general properties that a wedge product must satisfy? I ask because I thought the second definition was THE general definition. Should it not be possible to derive the first definition from the second one or are those two "different" wedge products? $\endgroup$
    – L. Henry
    Nov 3, 2016 at 20:24
  • $\begingroup$ You can specify axioms just like you can for a vector space (/ any other algebraic structure). $\endgroup$
    – user137731
    Nov 8, 2016 at 3:34

2 Answers 2

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"different" wedge products?

It is the same operation, in the first example you apply it to two vectors, in the second -- to two 1-forms. But in the either case if you can write $w$ and $v$ in the basis of the corresponding space as $v= ae_1+be_2$, $w=ce_1 + de_2$ and define $$e_1\wedge e_2=e_1\otimes e_2-e_2\otimes e_1$$ then $$v\wedge w = v \otimes w - w \otimes v=\ldots= (ad-bc)e_1\wedge e_2$$ where I skipped some simple manipulations of opening brackets and cancelling symmetric tensor product.

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You can generalize the area of (oriented) parallelogram to determinant of a matrix - doesn't the $ad-bc$ ring a bell? If $k+n=dim(M)$ you would get something very similar.

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