archived sections 4_3 and 4_4 on 1_22_16

Surgically removed sections 4.3 and 4.4 on 1.22.16, archiving here PhL 1.22.16 These sections were replaced with those I have been holding in a separate doc which uses T^ to describe a wedge-space tensor. It took me a while to be happy with this notation, but now I am. 4.3 The wedge product of 2 vectors (a) Definition of the wedge product of 2 vectors and the space L2 Momentarily jumping ahead, consider this equation, v ^ w = (vw - wv)/2 . v ϵ V and w ϵ W If V and W are different vector spaces, this makes no 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 use 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" ("exterior product") of two vectors a,b ϵ V, a ^ b ≡ (ab - ba)/2 . dim(V) = n (4.3.1) 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. Some authors write L2 as V^V (projmec ref), but this notation seems uncommon. The above definition trivially implies that a ^ b = - b ^ a a, b ϵ V (4.3.2) and a ^ a = 0 a ϵ V . (4.3.3) In (1.1.5) we stated certain scalar and distributive properties of the operator. These properties are passed through to the wedge ^ operator by the above definition. For example, (ka) ^ b = [ (ka)b - b(ka)]/2 = k [ ab - ba ]/2 = k (a ^ b) k = scalar (a+c) ^ b = [(a+c)b - b(a+c)]/2 = [ab + cb - ba - bc]/2 = [ ab - ba ]/2 + [ cb - bc ]/2 = (a ^ b) + (c ^ b) distributive and similarly for a ^ (kb) and a ^ (b + c). To summarize, we have a set of rules as follows: (ka) ^ b = k (a ^ b) (a+c) ^ b = (a ^ b) + (c ^ b) k ϵ K a ^ (kb) = k (a ^ b) a ^ (b + c) = (a ^ b) + (a ^ c) a,b,c ϵ V (4.3.4) The operator ^ is then seen to be "bilinear" over elements of V: it is separately linear in each operand. 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 Tij ei ^ ej . Σij ≡ Σi=1n Σj=1n (4.3.5) where the Tij are the expansion coefficients. For example, if Tij = aibj this would be, T = Σij aibj ei ^ ej = (Σiaiei) ^ ( Σjbjej) = a ^ b (4.3.6) and then a ^ b is included in L2 for any vectors a and b in V. We can take the ab component of (4.3.5) as follows Tab = Σij Tij (ei ^ ej)ab = Σij Tij(eiej - ejei)ab/2 = Σij Tij[ (eiej)ab - (ejei)ab] / 2 = Σij Tij [ eiaejb - ejaeib] / 2 = Σij Tij [ δia δjb - δja δib] / 2 = (1/2)[ Tab -Tba] Tab = - Tba (4.3.7) This shows that the expansion (4.3.5) can only represent an antisymmetric rank-2 tensor T. One could rearrange the n2 basis vectors of VV into these two groups, (ei^ ej) = [eiej - ejei]/2 n(n-1)/2 independent elements in this set (4.3.8) (ei * ej) ≡ [eiej+ ejei]/2 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 just the first set of basis vectors. 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 obviously closed under addition of vectors since Σij Tij ei ^ ej + Σij T'ij ei ^ ej = Σij (Tij+T'ij) ei ^ ej . (4.3.9) 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 Tij (ei ^ ej) = Σi≠ j Tij (ei ^ ej) // (ei ^ ei) = 0 = Σi<j Tij (ei ^ ej) + Σi>j Tij (ei ^ ej) = Σi<j Tij (ei ^ ej) + Σj>i Tji (ej ^ ei) // i↔j in second sum = Σi<j Tij (ei ^ ej) - Σi<j Tji (ei ^ ej) // (ej ^ ei) = - (ei ^ ej) = Σi<j (Tij - Tji) (ei ^ ej) (4.3.7) = Σi<j Aij (ei ^ ej) Aij ≡ (Tij - Tji) = 2Tij Aij = - Aji . (4.3.10) 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 antisymmetric matrices A. Below we use this terminology, T = Σij Tij (ei ^ ej) = the "symmetric expansion" of T T = Σi<j Aij (ei ^ ej) = the "ordered expansion" of T Meanwhile, the most general element of V2 can be written T = Σij Tij (ei ej) . (4.1.9) Now each matrix Tij 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 - ) . (4.3.11) The conclusion is that L2 contains less than half the number of elements in V2. This ratio is of course the same as the (4.3.8) count ratio of L2 basis vectors to V2 basis vectors: [n(n-1)/2] / [n2] = (n-1)/2n. (c) Wedge products and determinants: the geometry connection From (4.3.6) and (4.3.10) with Tij = aibj we get, a ^ b = Σij aibj (ei ^ ej) = Σi<j (aibj- ajbi) (ei ^ ej) = Σi<j det (ei^ej) Aij = (aibj- ajbi) = det . (4.3.12) The determinants which appear here are 2x2 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. Below that matrix is shown on the left, and some of the 2x2 minors are shown in gray on the right: (4.3.13) If V = R2 (so n=2) there is only one term in the sum (4.3.12), the one with i=1 and j=2, so a ^ b = det e1^e2 = det(a,b) e1^e2 = [ a1b2 - a2b1] e1^e2 . (4.3.14) If one draws a parallelogram (2-piped) in the x-y plane with edges a and b, one knows that the area of that 2-piped is |a x b| which is then |a1b2 - a2b1| = |det(a,b)|. There is then some connection between the wedge product of two vectors in R2 and the geometry of R2. Later in (7.5.6) we will 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) (4.3.15) and here det(a,b,c) is the volume of the 3-piped spanned by the vectors a,b,c, so again there is a geometry connection. 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 + [a3b1 - a1b3] e3^e1 + [a2b3 - a3b2] e2^e3 . (4.3.16) The coefficients are those which appear in the normal "cross product" of two contravariant vectors, a x b = [a1b2 - a2b1] e3 + [a3b1 - a1b3] 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 (4.3.17) in which case one can say a ^ b = [a1b2 - a2b1] e1^e2 + [a3b1 - a1b3] e3^e1 + [a2b3 - a3b2] e2^e3 ↔ a x b = [a1b2 - a2b1] e3 + [a3b1 - a1b3] e2 + [a2b3 - a3b2] e1 (4.3.18) 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. With V = R4 for example, using the result (4.3.12) stated above, a ^ b = det e1^e2 + dete1^e3 + dete1^e4 + dete2^e3 +dete2^e4 + dete3^e4 . (4.3.19) 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...). (d) Components For the tensor product of two basis vectors we have, (eie'j)rs = (ei)r(e'j)s = δirδjs // VW (eiej)rs = (ei)r(ej)s = δirδjs // VV = V2 The components of (ei^ ej) are given by. (ei^ ej)rs = (1/2)[eiej - ejei]rs = (1/2)[(eiej)rs - (ejei)rs] = (1/2)[(ei)r(ej)s - (ej)r(ei)s] = (1/2) (δir δjs - δis δjr) (ei^ ej)rs = - (ei^ ej)sr = - (ej^ ei)rs . // two forms of antisymmetry (4.3.20) We now examine the pure wedge product a ^ b using the both the symmetric expansion (4.3.5) and the ordered expansion (4.3.10). Using the symmetric double sum expansion form (4.3.5) with Tij = aibj one has from (4.3.6) and (4.3.20), (a ^ b)rs = Σij aibj (ei ^ ej)rs = Σij aibj [δirδjs - δjrδis]/2 = (arbs - asbr)/2 . (4.3.24) Using the ordered double sum expansion (4.3.10) with Tij = aibj, we find instead (a ^ b)rs = Σi<j (aibj- ajbi) (ei ^ ej)rs = Σi<j det (ei ^ ej)rs = (1/2) Σ1≤i<j≤n det [δirδjs - δjrδis] . (4.3.25) For r = s, one clearly has (a ^ b)rs = 0. If r < s, then only the δirδjs term can contribute to the ordered sum, since this will make i < j , otherwise only the second term contributes. Then using θ(Boolean) = 1 if true else 0, we can evaluate as follows, 2(a ^ b)rs = det θ(r<s) - det θ(s<r) = det θ(r<s) + det θ(s<r) // swap rows 2nd term = det [ θ(r<s) + θ(s<r)] = det = arbs - asbr . (4.3.26) Combining the results for r=s and r≠s we get (a ^ b)rs = (arbs - asbr)/2 // Ars = (Trs- Tsr)/2 (4.3.27) in agreement with (4.3.24) which used the symmetric sum. We now repeat this comparison for general elements of L2. Using the symmetric double sum (4.3.5), Trs = Σij Tij (ei ^ ej)rs = Σij Tij (δir δjs - δis δjr)/2 = (Trs- Tsr)/2 = Ars/2 (4.3.28) Using the ordered double sum (4.3.10), Trs = Σi<j Aij (ei ^ ej)rs = Σi<j Aij [δirδjs - δjrδis]/2 . (4.3.29) For r = s one has [..] = 0 so Trs = 0. Otherwise, 2 Trs = Ars θ(r<s) - Asrθ(s<r) = Ars θ(r<s) + Arsθ(s<r) = Ars[ θ(r<s) + θ(s<r)] = Ars (4.3.30) with the conclusion that Trs = Ars/2 for all r,s ϵ (1,n) (4.3.31) which agrees with (4.3.30) using the symmetric expansion. (e) Dot Products Since a^b is an element of V2 as well as of L2, we may use the V2 dot product to write (a^b) (cd) = {(ab-ba)/2} (cd) = (1/2) [ (ab)(cd) - (ba)(cd)] = [(ac)(bd) - (bc)(ad)]/2 (4.3.32) with this special case (ei^ej) (cd) = [(eic)(ejd) - (ejc)(eid)]/2 = [cidj - cjdi]/2 . (4.3.33) The dot product of two-vector wedge products is the same as (4.3.32), (a^b) (c^d) = {(ab-ba)/2} {(cd-dc)/2} = = (1/4) [ (ab)(cd) - (ba)(cd) - (ab)(dc) + (ba)(dc) ] = (1/4) [ (ac)(bd) - (bc)(ad) - (ad)(bc) + (bd)(ac) ] = [ (ac)(bd) - (bc)(ad) ]/2 = (a^b) (cd) = (ab) (c^d) (4.3.34) so then the special case is the same as (4.3.33), (ei^ej) (c^d) = [cidj - cjdi]/2 . (4.3.35) 4.4 The wedge product of 2 dual vectors Section 4.3 considered the wedge product of two vectors in V2. Here we consider the wedge product of two vectors in the dual space V*2. We mimic the approach of Section 4.3, omitting some details, and we match equation numbers even though this leaves some "holes" in the sequence. (a) Definition of the wedge product and the space Λ2 We start off by defining the following wedge product of two vectors (linear functionals) α and β of V*, α ^ β ≡ (α β - β α)/2 . (4.4.1) 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 which some authors call V* ^ V*. The proof that Λ2 is a subspace and not just a subset of V*2 is the same as in the Section 4.3 (a). The above definition trivially implies that α ^ β = - β ^ α α, β ϵ V* (4.4.2) and α ^ α = 0 α ϵ V* . (4.4.3) The "rules" for the ^ operator in Λ2 V*2 are found just as they were for L2 V2, namely : (kα) ^ β = k (α ^ β) (α+γ) ^ β = (α ^ β) + (γ ^ β) α ^ (kβ) = k (α ^ β) α ^ (β + γ) = (α ^ β) + (α ^ γ) (4.4.4) where α,β,γ are vectors in V* and k is a scalar in K. 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 Tij λi ^ λj . // Tij is in italics, T on the left is not (4.4.5) For example, if Tij = αiβj this would be T = Σij αiβj (λi ^ λj) = (Σiαiλi) ^ ( Σjβjλj) = α ^ β (4.4.6) and then α ^ β is included in Λ2 for any vectors α and β in V. (b) How big is the space Λ2 compared to the space V*2? Just as in Section 4.3 (b), we can show that T = Σij Tij (λi ^ λj) = Σi<j Aij (λi ^ λj) Aij ≡ (Tij- Tji) = 2Tij Aij = - Aji (4.4.10) where Aij is an antisymmetric n x n matrix. Using the same argument presented there, we find = = (1/2) = (1/2) (1 - ) . (4.4.11) (c) Wedge products and determinants From (4.4.6) and (4.4.10) with Tij = aibj we get, α ^ β = Σij αiβj (λi ^ λj) = Σi<j (αiβj- αjβi) (λi ^ λj) = Σi<j det (λi ^ λj) Aij = (αiβj- αjβi) = det . (4.4.12) If V* = R2 (so n=2) there is only one term in the sum (4.4.12), the one with i=1 and j=2, so α ^ β = det λ1 ^ λ2 = det(α,β) λ1 ^ λ2 = [ α1β2 - α2β1] λ1 ^ λ2 . (4.4.14) Later we will show that for V = R3 the triple wedge product of three vectors is given by, α ^ β ^ γ = det(α,β,γ) (λ1 ^ λ2 ^ λ3) (4.4.15) It does not seem useful to discuss "geometry" in the space of functionals. (d) Components (functions) For the tensor product of two evaluated basis functionals we have, (λiλ'j)(vr,vs) = λi(vr)λ'j(vs) = (vr)i(vs)j // V*W*; r and s are vector labels (λiλj)(vr,vs) = λi(vr)λj(vs) = (vr)i(vs)j // V*V* For (λi^λj), (λi^ λj)(vr,vs) = [(λiλj)(vr,vs) - (λjλi)(vr,vs)]/2 = [λi(vr)λj(vs) - λj(vr)λi(vs)]/2 = (1/2) [ (vr)i(vs)j - (vr)j(vs)i] // Λ2 = V* ^ V* (λi^ λj)(vr,vs) = - (λi^ λj)(vs,vr) = - (λj^ λi)(vr,vs) // two forms of antisymmetry (4.4.20) We now examine the pure wedge product α ^ β using the both the symmetric expansion (4.4.5) and the ordered expansion (4.4.10). Using the symmetric double sum expansion form (4.4.5) with Tij = αiβj one has, (α ^ β)(vr,vs) = Σij αiβj (λi ^ λj)(vr,vs) = Σij αiβj [ λi(vr)λj(vs) - λj(vr)λi(vs)]/2 = Σij αiβj[(vr)i(vs)j - (vr)j(vs)i]/2 = [ α(vr)β(vs) - α(vs)β(vr)]/2 . Evaluated at vr = er and vs = es this becomes, using (4.2.6), (α ^ β)(er,es) = Σij αiβj [δirδjs - δjrδis]/2 = (αrβs - αsβr)/2 . (4.4.24) Using the ordered double sum expansion (4.4.10) with Tij = αiβj, we find instead (α ^ β)(vr,vs) = Σi<j (αiβj- αjβi) (λi ^ λj)(vr,vs) = Σi<j det (λi ^ λj)(vr,vs) = Σi<j det [ λi(vr)λj(vs) - λj(vr)λi(vs)]/2 = Σi<j det (1/2) det . (4.4.25) Evaluation at vr = er and vs = es gives (α ^ β)(er,es) = Σi<j (1/2) det [δirδjs - δjrδis] (4.4.26) Repeating the argument (4.3.27) this becomes (α ^ β)(er,es) = (αrβs - αsβr)/2 // = [Alt(αβ)]rs (4.4.28) in agreement with (4.4.24) which used the symmetric sum. We now repeat this comparison for general elements of Λ2. Using the symmetric double sum (4.4.5), T(vr,vs) = Σij Tij (λi ^ λj)(vr,vs) = Σij Tij [ λi(vr)λj(vs) - λj(vr)λi(vs)]/2 (4.4.29) one finds that T(er,es) = Σij Tij (1/2) [δirδjs - δjrδis] = (Trs - Tsr)/2 = Ars // see (4.4.10) (4.4.30) Using the ordered double sum (4.4.10), T(vr,vs) = Σi<j Aij (λi ^ λj)rs = Σi<j Aij [δirδjs - δjrδis]/2 (4.4.31) For r = s one has [..] = 0 so T(er,es) = 0. Otherwise, 2T(vr,vs) = Ars θ(r<s) - Asrθ(s<r) = Ars θ(r<s) + Arsθ(s<r) = Ars[ θ(r<s) + θ(s<r)] = Ars (4.4.32) with the conclusion that T(vr,vs) = Ars/2 for all r,s ϵ (1,n) (4.4.33) Looking at (4.4.20), (4.4.23) and (4.4.29), one sees that (λi^ λj)(vr,vs), (α ^ β)(vr,vs) and T(vr,vs) are all antisymmetric bilinear functions of the two arguments vr, vs ϵ V. We conclude that : Fact: The vector space Λ2(V) is equivalent to the vector space of antisymmetric bilinear functions on V. (4.4.34) This may be compared to our earlier statement for the larger space V*2 = V*V* , Fact: The vector space V*2 is equivalent to the vector space of bilinear functions on V. (4.2.15)