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I Wedge product of basis vectors

  1. Nov 30, 2016 #1
    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
     
  2. jcsd
  3. Nov 30, 2016 #2

    fresh_42

    Staff: Mentor

    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}##
     
  4. Nov 30, 2016 #3
    Are those components or basis vectors? Can you explain that in English?
     
  5. Nov 30, 2016 #4

    fresh_42

    Staff: Mentor

    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
     
  6. Dec 3, 2016 #5

    Stephen Tashi

    User Avatar
    Science Advisor

    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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