Why is the wedge product of vectors used in exterior algebra?

[JG89]

I've learned that the wedge product is a product operation on two alternating tensors that yields another alternating tensor, but sometimes while surfing the net I see people using the wedge product for two vectors. For example, on the wiki page titled "Exterior algebra" it says that "using the standard basis \{ e_1, e_2, e_3 \}, the wedge product of a pair of vectors u and v is ..." (the result is an alternating tensor, which seems correct)

I didn't write out the formula because it's irrelevant. My question is, why are they computing the wedge product of two vectors if the wedge product is defined on the set of alternating tensors?

 

[tiny-tim]

Hi JG89!

… why are they computing the wedge product of two vectors if the wedge product is defined on the set of alternating tensors?

Same as "why do we call vectors vectors when they're really first-order tensors" and "why do we call angular momentum a vector when it's really a pseuodvector".

Vectors are isomorphic to first-order alternating tensors …

so if it looks like a vector, quacks like a vector, and is isomorphic to a vector,

then let''s call it a vector! ​

 

[Hootenanny]

so if it looks like a vector, quacks like a vector, and is isomorphic to a vector,

then let''s call it a vector! ​

Brilliant line!

 

[JG89]

A first-order tensor over a vector space V would be a linear transformation from V to the set of real numbers. Suppose V = \mathbb{R}^n. I don't see how a vector in \mathbb{R}^n would be associated in any natural way with a linear transformation from \mathbb{R}^n to \mathbb{R}

 

[Hootenanny]

A first-order tensor over a vector space V would be a linear transformation from V to the set of real numbers. Suppose V = \mathbb{R}^n. I don't see how a vector in \mathbb{R}^n would be associated with a linear transformation from \mathbb{R}^n to \mathbb{R}

In actuality, a first order tensor can act as a map from the 0-order tensor space to a vector space OR as a map from a vector space to the 0-order tensor space. For example, let \boldsymbol{a}\in\mathbb{R}^n (i.e. a first order tensor). Then if b\in\mathbb{R} (i.e. 0-order tensor), then

\begin{aligned}\boldsymbol{a} & :\mathbb{R}\mapsto\mathbb{R}^n\\&:b\mapsto b\boldsymbol{a}\;.\end{aligned}

Equally, let \boldsymbol{a}\in\mathbb{R}^n (i.e. a first order tensor). Then if \mathbb{b}\in\mathbb{R}^n (i.e. 0-order tensor), then

\begin{aligned}\boldsymbol{a} & :\mathbb{R}^n\mapsto\mathbb{R}\\&:b\mapsto \boldsymbol{a}\cdot\boldsymbol{b}\;.\end{aligned}

 

[JG89]

So suppose u and v are two vectors in \mathbb{R}^n, then we can think of both u and v as the 1-tensors defined by u(x) = xu and v(x) = xv where x is a real number. So when we talk about the wedge product of the vectors u and v, then I can use my good old definition given in my book for the wedge product of the 1-tensors u and v, correct?