I Wedge product of basis vectors

AI Thread Summary
The discussion explores the relationships and properties of the wedge product of basis vectors, highlighting its bilinearity, associativity, and the condition that the wedge of a vector with itself equals zero. It emphasizes that the wedge product is a tensor product that requires a different mathematical structure than the original vector space. The conversation contrasts the wedge product with the cross product, noting that the latter results in an element within the same vector space. The participants seek clarity on whether the wedge product results in components or basis vectors. Overall, the wedge product's unique characteristics set it apart from other vector operations.
Kevin McHugh
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Is there a set of relationships for the wedge product of basis vectors as there are for the dot product and the cross product?

i.e. e1*e1 = 1
e1*e2 = 0

e1 x e2 = e3
 
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It's bilinear, associative and ##v \wedge v=0## (if char ##\mathbb{F} \neq 2##). In general it's ##a \wedge b = (-1)^{nm} b \wedge a## for ##a \in \Lambda^n(V)=
\underbrace{V \wedge V \wedge \ldots \wedge V}_{n\ times}## and ##b \in \Lambda^m(V)= \underbrace{V \wedge V \wedge \ldots \wedge V}_{m\ times}##
 
Are those components or basis vectors? Can you explain that in English?
 
It is meant as for arbitrary vectors. So especially for basis vectors, too, if you like. In coordinates, it would be some equivalence classes of tensor products, but I don't know how to press this equivalence relation into coordinates. It means basically that the wedge-product or better exterior product is a tensor product, but some tensors are considered to be equal, because the relations below have to be met.

If we have a vector space ##V## over a field ##\mathbb{F}## in which ##1+1 \neq 0## holds, then for ## a,b,c \in V## and ##\lambda \in \mathbb{F}## the following is true:
  1. ##(a+b) \wedge c = a\wedge c + b \wedge c##
  2. ##\lambda (a \wedge b) = (\lambda a) \wedge b = a \wedge (\lambda b)##
  3. ##a \wedge (b \wedge c) = (a \wedge b) \wedge c##
  4. ##a \wedge a = 0##
  5. ##a_1 \wedge \ldots \wedge a_n \wedge b_1 \wedge \ldots \wedge b_m = (-1)^{nm} b_1 \wedge \ldots \wedge b_m \wedge a_1 \wedge \ldots \wedge a_n##
The first two are called linearity, which together with the fifth becomes multi-linearity (linear in all "factors"), the third one is associativity, the fourth is a special case of the fifth together with the fact, that ##1+1 \neq 0##, which is said as the characteristic of ##\mathbb{F}## is not two, and the fifth alone can be called graduated commutativity, i.e. it determines what happens, if we change the order of "factors". I don't know how to put the formulas in other English words as their names are.

Perhaps you want to read the Wikipedia entry on it: https://en.wikipedia.org/wiki/Exterior_algebra
 
Kevin McHugh said:
Are those components or basis vectors?

The first question is whether ##a \wedge b## is something (e.g. a vector or a scalar) in the same vector space that contains ##a## and ##b##.

I'd say no. For ##a## and ##b## in a a vector space ##V##, in order to define ##a \wedge b##, you must define a different mathematical structure than ##V## itself. ( As an analogy, we can use two real numbers x1, x2 to define an interval [x1,x2], but "an interval" is a different thing than a single real number. )

By contrast, the cross product operation (in 3 dimensions) ##a \times b## does produce a result that is also a element of the the same vector space ##V## that contains ##a## and ##b##.
 
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