Activities in the Dual Direct Product Space v2

Activities in the Dual Tensor product Space PhL 9.9.15 this doc is closed, do not edit! 2. The Wedge Product 1 2.1. Two vector spaces V, W and their tensor product space VW 1 2.2. Two vector spaces V*, W* and their tensor product space V* x W* 4 2.3. The wedge product in the tensor product space V x V 7 2.4. The wedge product in the tensor product space V* x V* 11 2. The Wedge Product We now back up a bit and reconsider the space VW and its elements vw. The goal of the next two subsections is to establish the parallelism between the vector space VW and the "dual" vector space V*W*. Some repetition is used to reinforce earlier stated facts. Then the following two subsections introduce the wedge product in a similar parallel fashion. 2.1. Two vector spaces V, W and their tensor product space VW Consider two vector spaces V and W of dimension n and n'. Let {ei} = basis of V dim(V) = n v = Σi=1n vi ei = general vector in V {ei'} = basis of W dim(W) = n' w = Σj=1n'wj ej' = general vector in W {eie'j} = basis for the tensor product space VW dim(VW) = n*n' vw = a pure "vector" in the tensor product space VW The basis vectors of V and W. The basis vectors of V are denoted ei. Expanding a vector v on these basis elements, one writes v = Σi viei. The coefficients vi take values in the field K (normally reals). These expansion coefficients vi are the "components of vector v in the ei basis". Since ek is itself a vector in V, one can write ek = Σi (ek)iei . Notice that k is a label on ek, whereas the i on (ek)i is a component index. So the components of the vector ek in the ei basis are (ek)i. Given any two vectors v1 and v2 in V, consider the sum f(v1,v2) ≡ Σk=1n(v1)k(v2)k which clearly is an element of the field K. This function f: VxV → K is manifestly bilinear in its two vector arguments, see *** . This sum can of course be applied to any two basis vectors, so that fij ≡ f(ei,ej) = Σk=1n(ei)k(ej)k We now make two definitions: The {ei} are orthogonal if fij = giδi,j : Σk=1n(ei)k(ej)k = giδi,j The {ei} are orthonormal if fij = δi,j : Σk=1n(ei)k(ej)k = δi,j In our work below, we sometimes use a simple basis ej where (ej)k = δj,k. Each such basis vector has all components zero except for the component whose index matches the label of the basis vector, and that component has magnitude 1. If V = Rn one would think of the en as en = which are axis-aligned unit vectors. This simple basis is orthonormal according to the above definition, since Σk=1n(ei)k(ej)k = Σk=1nδi,kδj,k = δi,j . In the discussion below where ei appear, we do NOT assume the ei are orthogonal or orthonormal. We just assume they are linearly independent, as must be true for any basis. The exact same comments apply to the basis vectors e'i of W. Comment: Notice that in the above paragraphs there is no mention of metrics, metric tensors, metric spaces, norms, normed linear spaces, inner scalar or dot products, Hilbert spaces, Cartesian space, Euclidean space or any of the topics that arise when a vector space "grows up" and applies itself to some useful purpose. We keep the discussion at the level of the primordial vector space. Outer Product Revisited. The notion of an outer product was discussed in Section 1.3 above. We had for example (where ai and bj are the components of vectors a and b), (a b)ij = aibj // outer product of two vectors (here a,b are vectors) (A B)abcd = AabBcd // outer product of two rank-2 tensors (here a,b,c,d are indices) The outer product of two vectors a and b may be written in vector/matrix notation as follows, a b = abT = ( b1. b2....bn) = (a b)ij = (abT)ij and we have seen in ** how the outer product of AB can be written as a Kronecker product. The same vector/matrix notation used above can also be used to express the function f(a,b) appearing in ***, f(a,b) = aTb = ( a1. a2....an) = Σk=1n akbk = Σk=1n bkak = bTa = f(b,a) Tensor Product Revisited. By convention one represents an element of a tensor product space using the symbol. It is a certain kind of "product" between a vector in one vector space and a vector in another vector space. On can treat as an operator : VxW → (VW) in the sense that (v,w) = (v) (w) = (vw) = element of tensor product space (VW). Certain rules were declared in Section 1 which make the tensor product be a vector space, and which in an intuitive sense just seem "reasonable", (kv) w = v (kw) = k (vw) // k = scalar (ϵ K) v (w1+ w2) = vw1 + vw2 // left distributive property (v1 + v2) w = v1w + v2w // right distributive property In the last two equations, the inside + represents addition in either W or V, whereas the + on the right side represents addition in VW. These lines say that multiplication "distributes" over addition +. The scalar rule can be combined with the distributive rules to obtain this equivalent rules restatement: v (k1w1+ k2w2) = k1(vw) + k2(vw2) // k1,k2 = scalar (ϵ K) (k1v1 + k2v2) w = k1(v1w)+ k2(v2w) . // k1,k2 = scalar (ϵ K) The above rules in effect say that defines a "bilinear" operation -- it is linear separately in each of its operands. Notice that the following two rules are incorrect: v w = w v // wrong! (unless V = W and v = w) (kv) (kw) = k (vw) // wrong! (unless k = 1) As noted earlier last rules apply to the direct sum , not the tensor product . Using the correct "rules" above, one may write v w = ( Σiviei)( Σjwjej') = Σijviwj (eie'j) showing how this pure tensor product vector can be expressed in terms of the basis functions. A general "vector" (tensor) in W x V can be written as a linear combination of the basis vectors, T ≡ ΣijFij eie'j T ϵ V x W . Now if it happens that the two spaces being tensor-multiplied are the same, so W = V, we refer to the space V x W = V x V as V2, and the most general "vector" of this tensor product space may be then written as in terms of an arbitrary basis as, T ≡ ΣijFij eiej . T ϵ V x V = V2 Although we have said T is a "vector" in the abstract sense that a vector space (even a tensor product vector space) has "vectors" as elements, the usual terminology is to say that T is a "rank-2 tensor" in the space V2. The word "tensor" has a weak and a strong meaning as noted earlier. In the weak meaning, a rank-2 tensor is something that has components with two indices like Fij. In the strong meaning, a rank-2 tensor is a set of components Fij which transform in a certain manner with respect to some transformation. As noted in ***, the components Fij are linear combinations of the simple-basis components Tij and so both component sets transform the same way under a transformation. For example under rotation R the rotated components T'ij are related to the unrotated components by T'ij = Σab RiaRjbTab F'ij = Σab RiaRjbFab . For a general non-linear transformation, Ria is the "differential" of the transformation. It is shown in tensor doc Appendix Q that if the {ei} are the "tangent base vectors" of that transformation, then the Fij are the "contravariant components" of the tensor T, which in standard notation are usually written Tij . On the other hand, if the {ei} are taken to be the "reciprocal base vectors" {ei), then the coefficients Fij are the "covariant components" of the tensor T, written as Tij. Written in proper up/down tensor notation, one then has T ≡ ΣijTij eiej ei = tangent base vectors Tij = contravariant components T ≡ ΣijTij eiej ei = reciprocal base vectors Tij = covariant components The other two combinations result in coefficients which are "mixed" tensors having an upper contravariant index and a lower covariant index, T ≡ ΣijTij eiej T ≡ ΣijTij eiej The reciprocal base vectors ei are just the dual vectors qi discussed earlier and in the next section. Notice again that it is the object T which is the real "tensor", and it can be expanded in various ways on different choices of basis elements resulting in components of various natures. 2.2. Two vector spaces V*, W* and their tensor product space V*W* Let V* and W* be the "dual spaces" of V and W. In our treatment of the dual space side of things, we shall attempt to use Greek letters for all vectors and tensors encountered. Vectors in V or W or VW will continue to be represented by Latin letters. Assume that {λi} = basis of V* {λi'} = basis of W* Like all elements of V*, the object λi ϵ V* is a "linear functional over V". This is a linear function which, when evaluated at a point in V, produces a scalar: λi : V → K. As usual, we think of K as the reals R, but we try to stay more general by having K be an arbitrary field. In normal calculus, if f: V → R, one refers to f as a function, and f(v) as that function evaluated at some point in V, though loosely speaking f(v) is also called a function. To emphasize the distinction, we shall refer to f as a "functional" and f(x) as a "function". In particular, if the n vectors {ei} are a basis for V (basis vectors are linearly independent), one can find another set of n vectors {qi} in V such that, qiTej = δi,j. or ( (ej)1, (ej)2...) = or Σi=1n (qi)n(ej)n = δi,j A method for finding these unique qi from the ei is given in *** . The action of the linear functional λi can then be taken as λi(v) = qiTv which is manifestly linear since λi(αv) = αλi(v) and λi(v + v') = λ(v) + λ(v') α = scalar in field K Here v and v' are vectors in V. As noted earlier, the definition of linearity is often combined into a single statement, λi(αv + βv') = αλ(v) + βλ(v'), α,β = scalars in field K The vectors qi are sometimes called "covectors" or "reciprocal vectors" or "dual vectors" which form a "dual basis". From ** one then has λi(ej) = qiTej = δi,j . General linear functionals in V* and W* can be written as linear combinations of the basis functionals, α = Σiαiλi = general vector in V* α(v) = Σiαiλi(v) α: V → K β = Σjβjλj' = general vector in W* β(v) = Σiβiλi'(v) β: W → K where on the right we show the corresponding functions α(v) and β(v). There is a dual tensor product space V*W* which has the basis λiλ'j . We want this object λiλ'j to be a functional over the space V*W* such that λiλ'j: V*xW* → K . The natural way to accomplish this desire is to write (λiλ'j)(v,w) = λi(v)λj'(w) = scalar * scalar = scalar ϵ K This function is manifestly "bilinear" in that it is separately linear in each of the arguments v and w. For example, (λiλ'j)(v1+v2,w) = λi(v1+v2)λj'(w) = [λ(v1) + λ(v2)]λj'(w) = λ(v1)λj'(w) + λ(v2)λj'(w) = (λiλ'j)(v1,w) + (λiλ'j)(v2,w) . The "rules" for the operator in the space V*W* are the same as those for in the space VW, (kα) β = α (kβ) = k (α β) k = scalar α (β1+ β2) = α β1 + α β2 // distributive property (α1 + α2) β = α1β + α2β . // same idea as above A general element of the dual tensor product space V*W* can be written T = Σij Tij λiλ'j where the Tij are coefficients in the field K. Evaluating at a point (v,w) in VxW one gets, T(v,w) = Σij Tij (λiλ'j)(v,w) = Σij Tij λi(v) λj'(w) ϵ K so one may regard T : VxW → K. More specifically, consider α ϵ V*and β ϵ W* as shown above. Then α β = (Σiαiλi)(Σjβjλ'j) = Σij αiβj λiλ'j ϵ V*W* (α β)(v,w) = Σij αiβj (λiλ'j)(v,w) = Σij αiβj λi(v)λj'(w) ϵ K which then is just a particular example of the more general T(v,w) above. Another way to write the above line is (α β)(v,w) = Σij αiβj λi(v)λj'(w) = (Σiαi λi(v))(Σjβj λ'j(v)) = α(v) β(v) where now the bilinear function (α β)(v,w) is equal to the product of the two functions α(v) and β(w). If it happens that W = V, then W* = V* and we write V*W* = V*V* = V*2. Then the most general element T of V*2 can be expressed as T = Σij Tij (λiλj) T(v1,v2) = Σij Tij (λiλj)(v1,v2) = ΣijTij λi(v1)λj(v2) . In analogy with Section 2.1 above, one says that the linear functional T in the dual space V*2 is a "tensor" of rank 2, or order 2, or it is a 2-tensor. Whereas T earlier was a rank-2 tensor in V2, this object T is a rank-2 tensor in the dual space V*2. In the abstract, it is T which is the real tensor and it is shown expanded on a particular basis {λi} which in turn is associated with a particular basis {ei} of V as discussed above. What are the "components" of tensor T ? In a sense they are the coefficients Tij of the expansion shown above onto the basis (λiλj), in analogy with the non-dual space. But in the dual space this is not the sense of "component" we are so interested in. In the dual world, the function vector arguments play the role that vector component indices play in the outer products of the non-dual world. For example, compare these equations: (a b)ij = aibj // VW (α β)(v,w) = α(v)β(w) // V*W* T = Σij Fij eie'j // VW Tab = Σij Fij (eie'j)ab = Σij Fij (ei)a(e'j)b T = Σij Tij λiλ'j // V*W* T(v,w) = Σij Tij λiλ'j(v,w) = Σij Tij λi(v)λ'j(w) . So it is really the T(v,w) which are the "tensor components" of the tensor T, whereas the Tij are the coefficients in the above expansion, In the non-dual space the "tensor components" were Tij . One can then regard T(v,w) as the tensor of interest. For example, Spivak on page 75 refers (in effect) to our function T(v1,v2) as a being a 2-tensor. The space of bilinear functionals on V2 = VV (which includes any T above) is just V*2 = V*V*. As just noted, one can regard V*2 also as the space of all bilinear functions T(v1,v2) . Fact: The vector space V*2 of the tensor product of two linear functionals of V* over V is equivalent to the vector space of bilinear functions on V2. Comment: It is possible to emphasize the parallelism between the dual-world and the non-dual world by considering, in analogy with the functions λi(v), alternate functions vi(λ). Either world is then the dual of the other world. This is the approach taken on page 2 of Benn and Tucker. 2.3. The wedge product in the tensor product space V x V (a) Definition of the wedge product and the space L2 Momentarily jumping ahead, we shall find that the wedge product of v ϵ V and w ϵ W is going to be v ^ w = vw - wv . If V and W are different vector spaces, this does not make any sense since the second term wv implies that w lies in the left space V and v lies in the right space W. So in our discussion of wedge products, we require that W = V. This being the case, instead of using letters v and w as representative vectors, we shall a and b. Then ei are the basis vectors for both component spaces in the tensor product space VV. So, we start off by defining the following "wedge product" of two vectors a and b of V, a ^ b ≡ ab - ba . dim(V) = n Notice therefore that a ^ b is an element of VV = V2, since it is a linear combination of elements of VV. It is "antisymmetrized" under a ↔ b. Since not all elements of VV can be written this way, the set of elements a ^ b exist in a subset of VV which we shall call L2, so L2 V2. The above definition implies that a ^ b = - b ^ a a,b ϵ V so a ^ a = 0 a ϵ V . In Section *.* we stated certain scalar and distributive properties of the operator. These properties are transferred onto the wedge ^ operator by the above definition. For example, (ka) ^ b = (ka)b - b(ka) = k [ ab - ba ] = k (a ^ b) k = scalar (a+c) ^ b = (a+c)b - b (a+c) = ab + cb - ba - bc = [ ab - ba ] + [ cb - bc ] = (a ^ b) + (c ^ b) distributive and similarly for a ^ (kb) and a ^ (b + c). To summarize: (ka) ^ b = k (a ^ b) (a+c) ^ b = (a ^ b) + (c ^ b) a ^ (kb) = k (a ^ b) a ^ (b + c) = (a ^ b) + (a ^ c) The operator ^ is then seen to be "bilinear" over elements of V: it is separately linear in each operand. All the above equations are valid for the special case where a,b,c are basis vectors of V, a = ei b = ej c = ek. To more precisely define the space L2, we claim that the most general element of the space L2 can be written this way, T = Σij Fij ei ^ ej . For example, if Fij = aibj this would be T = Σij aibj ei ^ ej = (Σiaiei) ^ ( Σjbjej) = a ^ b and then a ^ b is included in L2 for any vectors a and b in V. One could rearrange the n2 basis vectors of VV into these two groups: (ei^ ej) = [eiej - ejei] n(n-1)/2 independent elements in this set (ei * ej) ≡ [eiej+ ejei] n(n)/2 independent elements in this set for a total of n(n-1)/2+ n(n)/2 = n2 basis vectors. One would say then that L2 is spanned by the first set of basis vectors and none of the second set. It was noted above that L2 is a subset of V2. A stronger statement is that L2 is a subspace of V2. First of all, L2 is closed under addition of vectors since (a ^ b) + (a' ^ b') = Σij aibj (ei ^ ej) + Σij a'ib'j (ei ^ ej) = Σij [aibj+ a'ib'j] (ei ^ ej) = Σij Cij (ei ^ ej) ϵ L2 And if (a ^ b) is an element of L2 then so is k(a ^ b) = (kα) ^ b ϵ L2 . Finally, since a ^ a = 0, L2 includes the 0 element. So L2 then is a vector space which is a subspace of V2. (b) How big is the space L2 compared to the space V2? Consider this most general element of L2: T = Σij Fij (ei ^ ej) = Σi≠ j Fij (ei ^ ej) // (ei ^ ei) = 0 = Σi<j Fij (ei ^ ej) + Σi>j Fij (ei ^ ej) = Σi<j Fij (ei ^ ej) + Σj>i Fji (ej ^ ei) // i↔j in second sum = Σi<j Fij (ei ^ ej) - Σi<j Fji (ei ^ ej) // (ej ^ ei) = - (ei ^ ej) = Σi<j (Fij- Fji) (ei ^ ej) = Σi<j Aij (ei ^ ej) Aij ≡ (Fij- Fji) Aij = - Aji Thus, the number of elements in L2 is equal to the number of antisymmetric n x n matrices A one can construct which contain elements of field K. An n x n antisymmetric matrix has only n(n-1)/2 places to insert independent values since the diagonal is all zeros and one triangle is the negative of the other. If the scalar space K contains N elements ( N = ∞ for the reals), one could then construct exactly Nn(n-1)/2 matrices A. Meanwhile, the most general element of V2 can be written T = Σij Fij (ei ej) Now each matrix Fij defines an element of V2. Using the same counting method as above, the total number of elements of V2 is Nn2. We conclude that = = (1/2) = (1/2) (1 - ) The conclusion is that L2 contains less than half the number of elements in V2. (c) Wedge products and determinants: the geometry connection. For the case discussed above where Fij = aibj we found a ^ b = Σij aibj (ei ^ ej) = Σi<j (aibj- ajbi) (ei ^ ej) = Σi<j det (ei^ej) . The determinants which appear here are minors of a matrix having n rows and 2 columns. The two columns are the vectors a and b, each of which has n components. That matrix is shown on the left, and some of the minors are shown in gray on the right: If V = R2 (so n=2), the above matrix is square and there is then only one minor which is the determinant of the entire matrix, a ^ b = det e1^e2 = det(a,b) e1^e2 = [ a1b2 - a2b1] e1^e2 It is easy to show (draw a picture) that det(a,b) is the area of the parallelogram (2-piped) spanned by vectors a and b. There is then some connection between the wedge product of two vectors in R2 and the geometry of R2. Later we shall show that for V = R3 the triple wedge product of three vectors is given by, a ^ b ^ c = det(a,b,c) e1^e2^e3 and here det(a,b,c) is the volume of the 3-piped spanned by the vectors a,b,c. However, for R3 the wedge product of two vectors is more complicated. Using the above expression, we find a ^ b = det e1^e2 + e1^e3 + e2^e3 = [a1b2 - a2b1] e1^e2 + [a1b3 - a3b1] e1^e3 + [a2b3 - a3b2] e2^e3 . The coefficients are (apart from a sign) those which appear in the normal "cross product" of two vectors, a x b = [a1b2 - a2b1] e3 - [a1b3 - a3b1] e2 + [a2b3 - a3b2] e1 . We do not wish, however, to identify for example e1^e2 with e3. After all, e3 is a basis vector in V, whereas e1^e2 is a vector in the tensor product space VV. One can, on the other hand, define a correspondence of sorts where one says (each line in cyclic order, and ↔ means "corresponds to") e1^e2 ↔ e3 e2^e3 ↔ e1 e3^e1 ↔ e2 // = - e1^e3 in which case one can say a ^ b = [a1b2 - a2b1] e1^e2 + [a1b3 - a3b1] e1^e3 + [a2b3 - a3b2] e2^e3 ↔ a x b = [a1b2 - a2b1] e3 + [a1b3 - a3b1] e2 + [a2b3 - a3b2] e1 so there is then a correspondence between the wedge product and the cross product in R3. This correspondence was described by Scottish mathematician William Hodge (1903-1975) around 1941 and the relationship ↔ is formalized by the "Hodge dual operator * " where for example *(e1^e2) = e3 in R3. For Rn with n > 3 there is no cross product of two vectors, but there is a wedge product. For V = R4 for example, using the result stated above, a ^ b = det e1^e2 + dete1^e3 + dete1^e4 + dete2^e3 +dete2^e4 + dete3^e4 . There are enthusiastic workers (e.g. Denker) who recommend deep-sixing the cross product altogether and replacing it with the wedge product for the study of topics like angular momentum (Ref...). 2.4. The wedge product in the tensor product space V* x V* We mimic the approach of Section 2.3 for the tensor product space VV, but now we have V*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*V*. It is "antisymmetrized" under α ↔ β. Since not all elements of V*V* = V*2 can be written this way, the set of elements α ^ β exist in a subspace of V*2 which we shall call Λ2, so Λ2 V*2. The proof that Λ2 is a subspace and not just a subset of V*2 is the same as in the previous section. The above definition implies that α ^ β = - β ^ α α, β ϵ V* so 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 in K. 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) T = Σij Φij (λi ^ λj) . For example, if Φij = αiβj this would be T = 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 T = Σij Φij (λi ^ λj) = Σi<j Tij (λi ^ λj) Tij ≡ (Φij- Φji) Tij = - Tji where Tij is an antisymmetric n x n matrix. Using the same argument of Section 2.3, we find = = (1/2) = (1/2) (1 - ) Recall from Section ** in the V*W* discussion that. (λiλ'j)(v,w) = λi(v)λj'(w) // λiλ'j: VxW → K Now with W* = V* we have instead (λiλj)(v1,v2) = λi(v2)λj(v2) // λiλj: VxV → K 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 T = Σi<j Tij (λi ^ λj) Aij = - Aji so the most general function is then T(v1,v2) = Σi<j Tij (λi ^ λj)(v1,v2) = Σi<j Tij [ λi(v1)λj(v2) - λj(v1)λi(v2) ] Since this function is manifestly bilinear and antisymmetric, we conclude with this claim: Fact: The vector space Λ2(V) of the wedge products of two linear functionals of V* over V is equivalent to the vector space of antisymmetric bilinear functions on V2. This may be compared to our earlier statement: Fact: The vector space V*2 of the tensor products of two linear functionals of V* over V is equivalent to the vector space of bilinear functions on V2. ok to here ***************************************************