early section 4

Wedge in the V*2 world do not edit, already incorporated. 4. The wedge product in the direct product space V* x V* We mimic the approach of Section 3 for the direct product space V x V, but now we have V* x V*. So, we start off by defining the following "wedge product" of two vectors α and β of V* α ^ β ≡ α β - β α . Notice therefore that α ^ β is an element of V* x V* = V*2, since it is a linear combination of elements of V* x V*. It is "antisymmetrized" under α ↔ β. Since not all elements of V* x V* can be written this way, the set of elements α ^ β exist in a subset of V* x V* which we shall call Λ2 ("wedge 2"), so Λ2 V2. The above definition implies that α ^ β = - β ^ α α, β ϵ V* and α ^ α = 0 α ϵ V* The "rules" for the ^ operator in V*2 are found as they were for V2, to wit : (kα) ^ β = k (α ^ β) (α+γ) ^ β = (α ^ β) + (γ ^ β) α ^ (kβ) = k (α ^ β) α ^ (β + γ) = (α ^ β) + (α ^ γ) where α,β,γ are vectors in V* and k is a scalar. All the above equations are valid for the special case where α,β,γ are basis vectors of V* : α = λi β = λj γ = λk. To more precisely define the space Λ2, we claim that the most general element of the space Λ2 can be written this way (that is, Λ2 is the space spanned by the λi ^ λj basis vectors) Φ = Σij Φij (λi ^ λj) . For example, if Φij = αiβj this would be Φ = Σij αiβj (λi ^ λj) = (Σiαiλi) ^ ( Σjβjλj) = α ^ β and then α ^ β is included in Λ2 for any vectors a and b in V. Just as in Section 3, we can show that Φ = Σij Φij (λi ^ λj) = Σi<j Aij (λi ^ λj) Aij ≡ (Φij- Φji) Aij = - Aji where Aij is an antisymmetric n x n matrix. We then reach the same conclusion which we restate in the current context, = = (1/2) = (1/2) (1 - ) Recall from Section 2 in the V* x W* discussion that. λiλ'j: V* x W* → R (λiλ'j)(v,w) = λi(v)λj'(w) . Now with W* = V* we have instead λi λj: V*2 → R (λiλj)(v1,v2) = λi(v1)λj(v2). and then λi^ λj: Λ2 → R (λi^ λj) = (λiλj) - (λjλi) // special case of ** above If we evaluate the V*2 functional (λi^ λj) at the V2 location (v1,v2) we find, (λi^ λj)(v1,v2) = (λiλj)(v1,v2) - (λjλi)(v1,v2) = λi(v1)λj(v2) - λj(v1)λi(v2) = det Notice that : (1) the basis function (λi^ λj)(v1,v2) is linear on both v1 and v2, so it is a "bilinear" function. This follows from the fact that the λk(v) functions are linear as was shown in ***. (2) the basis function (λi^ λj)(v1,v2) is antisymmetric under v1↔ v2, (λi^ λj)(v2,v1) = - (λi^ λj)(v1,v2) This is obvious from the determinant form since we switch two rows. As in Section 3, the wedge product of two vectors in V* can be expressed in terms of certain determinants which are minors of a "tall matrix" whose columns are α and β, α ^ β = Σij αiβj (λi ^ λj) = Σi<j (αiβj- ajβi) (λi ^ λj) = Σi<j det (λi ^ λj) . Because we are now in the dual space of linear functionals, we claim no particular geometric significance of the above wedge product when n = 2. One could of course blithely set V*2 = R2 and repeat the previous geometric discussion, but usually one does not discuss "geometry" in the dual space context. Instead, one notes that, if the linear functional α ^ β is evaluated at (v1,v2) in V2, one gets a statement about functions, (α ^ β)(v1,v2) = Σij αiβj (λi ^ λj)(v1,v2) = Σi<j αiβj (λi ^ λj)(v1,v2) = Σi<j (αiβj- ajβi) (λi ^ λj)(v1,v2) = Σi<j det (λi ^ λj)(v1,v2) = Σi<j det [ λi(v1)λj(v2) - λj(v1)λi(v2) ] = Σi<j det det The function (α ^ β)(v1,v2) is bilinear and antisymmetric because (λi ^ λj)(v1,v2) is bilinear and antisymmetric. As noted above, the most general functional in Λ2 may be written Φ = Σi<j Aij (λi ^ λj) Aij = - Aji so the most general function is then Φ(v1,v2) = Σij Φij (λi ^ λj)(v1,v2) = Σij Φij [ λi(v1)λj(v2) - λj(v1)λi(v2) ] Since this function is manifestly bilinear and antisymmetric, we conclude with this claim Fact: The space Λ2(V) of the wedge products of two linear functionals of V* over V is equivalent to the space of antisymmetric bilinear functions on V2.