section 7 8

7. The direct products of k vectors in Λk 9.14.15 This is a repeat of Section 5 where we translate to the dual space. Rather than repeat all the words, we just summarize the results. Our task is now to generalize the direct products from V*2 to V*k, where V*k ≡ V* x V* x .... x V* // k component spaces, each one is V* A generic element of Vk is λ1 λ2 ..... λk . all λi ϵ V* The basis elements of Vk are λi λ ..... λi . If n = dim(V*), there are a total number of such basis elements is nk. In the full set of such direct product basis elements, two or more of the λk might be the same. This will always be the case if k > n where n ≡ dim(V). A rank-k tensor T in V*k has this general expansion T = Σii....i Φii....i (λiλi .....λi) . Using the notion of a multiindex I (an ordinary multiindex), I ≡ {i1, i2, .....ik} is = 1,2....n and a shorthand notation for the basis vectors λI ≡ λiλi .....λi the general rank-k tensor in Vk can be expanded in the following compact notation T = ΣI ΦI λI. The direct product of k vectors is "k-multilinear" meaning it is linear in each of its k factors. For example λ1(λ2 +λ2')λ3.....λk = λ1λ2λ3 .....λk + λ1λ2'λ3 .....λk λ1(sλ2)λ3 ..... λk = s(λ1λ2λ3 .....λk) s = scalar Here we show linearity in the 2nd factor. All the other factors have similar equations. We impose this k-multilinearity by fiat, but it is the only reasonable and consistent way to handle the direct products. For k = 2, for example, one would have λ1(λ2 +λ2') = λ1λ2 + λ1λ2' λ1(sλ2) = s(λ1λ2) s = scalar consistent with earlier "rules". The above equations are meaningful for any integer k, regardless of the value n = dim(V*). Normally one does not add apples and oranges, so one does not add items of the form ab to those of the form abc. However, as one writer notes, fruit salad is good, and so we could define a very large vector space of the form V*(V) = V*0 + V*1 + V*2 + V*3 + .... Here V*0 = the space of scalars, V*1 = V* the space of vectors (that is, linear functionals on V), V*2 = V* x V* = the space of "bivectors" (bilinear functionals on V), and so on. The most general element of the space V*(V) would have the form T = s + ΣiΦi λi + Σij Φij λiλj + Σijk Φijk λiλjλk + ...... This large space V*(V) is in fact itself a vector space. It should be clear to the reader that it is closed under addition and has the right scalar rule. For example, writing if k1 and s are scalars, k1 + a + bc + fgh = sum of 4 elements of V*(V) = an element of V*(V) s(k1 + a + bc + fgh) = (sk1) + (sb) c + f(sg)h = element of V*(V) The space is also closed under the multiplication operation . For example (bc)(fgh) = bcfgh = ϵ V*5 = ϵ V*(V) . One could make the following definitions with regard to the space V*(V) k1 scalar 0-blade 0 a vector 1-blade 1 ab bivector 2-blade 2 abc trivector 3-blade 3 abcd quadvctor 4-blade 4 ..... arb. element of T multivector linear combination of any the above However, these terms are really reserved for objects within the wedge space we shall define below. Since V*(V) is closed under the operations + and , it is "an algebra" (the space Vk alone is not an algebra because it is not closed under ). The V*(V) algebra is different from that of the reals, due its definition as a sum of vector spaces. The elements of V*(V) have different "grades" as shown in the right column above, and is known therefore as a "graded algebra". Sometimes V*(V) is called "the tensor algebra" over V. The dimensionality of the space V*(V) is as follows, where n = dim(V*), dim(V*(V)) = 1 + n + n2 + n3 + ... = ∞ 8. The wedge products of k vectors in V*k This section is a mimic of Section 6 above. Turning now to wedge products, we want to define the wedge product of k vectors in V*, α1^ α2^ .....^ αk . We impose the requirement that this wedge product changes sign when any two vectors are swapped. This leads to the following candidate wedge product definition which we write in three equivalent forms : α1^ α2^ .....^ αk = Σii....i εii....i (αi αi ..... αi) = ΣP (-1)S(P) ( αP(1) αP(2) ..... αP(k)) = α1 α2 ..... αk + all signed permutations In the first line, ε is the usual permutation tensor (here with k indices) and the sum over each index ranges from 1 to k. If the first term in this sum is taken to be that term which has ir= r, one obtains the third line since ε123..k = 1. On the second line the sum is over all permutations P of the set of integers {1,2,3....k}, and S(P) is the number of pair swaps involved in the permutation P. For example P{1,2,3...k} = {2,1,3...k} has S(P) = 1. If the first term in this sum involves the identity permutation P = I, then again one obtains the third line. The third line is really just a vague symbolic way of writing out either of the first two lines, where we expose the first term. We shall verify the "changes sign" rule momentarily, thus vetting our candidate expressions above. Meanwhile, as encouragement, here is the first expansion above for k = 2 and k = 3: α1^ α2 = Σa,b =12 εab αa αb // 2! = 2 terms = α1 α2 - α2 α1 // agrees with *** α1 ^ α2 ^ α3 = Σa,b,c =13 εabc αa αb αc // 3! = 6 terms = α1α2α3 - α1α3α2 + α3α1α2 - α3α2α1 + α2α3α1 - α2α1α3 . The wedge product is k-multilinear. Since the wedge product is a linear combination of direct products, the fact that the operator is k-multilinear implies that the wedge ^ operator also has this property. Thus, for example, the rules given above for become α1^(α2 +α2')^α3^.....^αk = α1^α2^α3^ .....^αk + α1^α2'^α3^ .....^αk α1^(sα2)^α3^ .....^ αk = s(α1^α2^α3^ .....^αk) s = scalar and these rules apply independently to every vector in the wedge product. The sum contains k! terms, The number of permutations P of k objects is k!, so the second line says there are k! terms. In the first line, although there are kk terms in the sum, only those terms for which ε has all different indices contribute, so again there are k! terms. Verification of the "changes sign under pair swap" rule. Our candidate equivalent forms from above are α1^ α2^ .....^ αk = Σii....i εii....i (αi αi ..... αi) = ΣP (-1)S(P) ( αP(1) αP(2) ..... αP(k)) . For the first line, consider a swap of 1 and 2: α2^ α1^ .....^ αk = Σii....i εii....i (αi αi ..... αi) = Σii....i εii....i (αi αi ..... αi) // swap dummy indices i1↔i2 = Σii....i [-εii....i] (αi αi ..... αi) // swap ε indices i1↔i2 = - α1^ α2^ .....^ αk . For the second line we start off with α2^ α1^ .....^ αk = ΣP (-1)S(P) ( αP(2) αP(1) ..... αP(k)) Define the permutation P' by P'{1,2...k} = {2,1...k} which has S(P') = 1. Then the above sum my be written this way, since for example 2 = P'(1) and then P(2) = P(P'(1)) = PP'(1), = ΣP (-1)S(P) ( αPP'(1) αPP'(2) ..... αPP'(k)) . The swap count S(PP') = S(P) + S(P') = S(P) + 1 so that (-1)S(PP') = - (-1)S(P). Then = - ΣP (-1)S(PP') ( αPP'(1) αPP'(2) ..... αPP'(k)) . Finally we can use the "rearrangement theorem" of group theory applied to the permutation group. This theorem says that ΣP f(PQ) = ΣP f(P) where Q is any group element. The first sum is just a reordering of the second sum and so equals the second sum. Using Q = P' we continue to get, = - ΣP (-1)S(P) ( αP(1) αP(2) ..... αP(k)) = - α1^ α2^ .....^ αk . Wedge product vanishes if any two vectors are the same. Given a sign change for any pair swap of vectors in the wedge product, we know that α1^ α2^ .....^ αk = 0 if any two (or more) vectors are the same. For example. a ≡ α2^ α1^ .....^ αk = - α1^ α2^ .....^ αk = -a if 1 = 2, so a = 0. We can then write αj ^ αj ^ .... ^ αj = εjj....j α1^ α2^ .....^ αk . If the subscripts jr are a permutation of {1,2...k}, the above expression generates the correct sign for the wedge product relative to α1^ α2^ .....^ αk . If one or more subscripts jr are the same, the ε forces the result to be zero. Our two expressions for the wedge product stated above then become αj ^ αj ^ .... ^ αj = εjj....j Σii....i εii....i (αi αi ..... αi) = εjj....j ΣP (-1)S(P) ( αP(1) αP(2) ..... αP(k)) . Wedge product vanishes if vectors are linearly dependent. We have just shown that the wedge product vanishes if any two vectors are the same. It is also true that the wedge product α1^ α2^ .....^ αk vanishes if the vectors αi are linearly dependent. Linear dependence means one can write at least one vector in the set as a linear combination of the others, so perhaps α2 = ( Σi≠2 aiαi). Then α1^ α2^ .....^ αk = α1^ ( Σi≠2 aiαi)^ .....^ αk = Σi≠2 ai (α1^ αi ^ .....^ αk) // since ^ is k-multilinear, see above But (α1^ αi ^ .....^ αk) = 0 for any i ≠2 since then two vectors are the same. QED. Wedge product vanishes if k > n . If dim(V*) = n, we know there can be at most n linearly independent vectors in V*. If k > n, any set of k vectors must be linearly dependent. Thus, the wedge product of any set of k vectors must vanish if k > n. Therefore for a given vectors space V* of dimension n, the only wedge products of interest are those for k = 1,2,3....n. Basis elements for Λk. Consider the following objects in Λk obtained by wedging together k basis elements of V*, where each λi is selected from the set of n available for V* which has dimension n, (λj ^ λj ^ .... ^ λj) . Of these putative nk objects, only n*(n-1)*...*(n-k+1) = n!/(n-k)! are non-zero because all the others have at least two vectors the same, so we can assume that all the subscripts jr are different. Now there exists a unique permutation P of the all-different subscripts {jr} such that { j1, j2....jk} = P{ i1, i2....ik} where i1 < i2 < ..... < ik If this permutation involves S pairwise swaps of indices, then we know that (λj ^ λj ^ .... ^ λj) = (-1)S (λi ^ λi ^ .... ^ λi) where i1 < i2 < ..... < ik because each pairwise swap of vectors in a wedge product creates a minus sign. Since there are k! possible permutations P, there are k! equations like the above which relate different objects to the same object (λi ^ λi ^ .... ^ λi) which has i1 < i2 < ..... < ik .Thus, if we want to count our number of independent basis elements of Λk, we have to divide our earlier count of n!/(n-k)! non-vanishing objects by k!. The conclusion is that there are (n,k) independent basis elements for Λk and they all have this form (λi ^ λi ^ .... ^ λi) where i1 < i2 < ..... < ik basis elements Example: for k = 3 and n = 6 we have λ1^λ3^λ5 = (-1)0 λ1^λ3^λ5 = +λ1^λ3^λ5 135 λ1^λ5^λ3 = (-1)1 λ1^λ3^λ5 = - λ1^λ3^λ5 153→135 λ3^λ1^λ5 = (-1)1 λ1^λ3^λ5 = - λ1^λ3^λ5 315 →135 λ3^λ5^λ1 = (-1)2 λ1^λ3^λ5 = +λ1^λ3^λ5 351→315→135 λ5^λ1^λ3 = (-1)2 λ1^λ3^λ5 = +λ1^λ3^λ5 513→153→135 λ5^λ3^λ1 = (-1)3 λ1^λ3^λ5 = - λ1^λ3^λ5 531→513→153→135 Each of this group of 3! objects is equal to + or - the same object. We now define Λk to be the space whose objects can be written in this form T = Σi<i<....<i Tii...i (λi ^ λi ^ .... ^ λi) because the set (λi ^ λi ^ .... ^ λi) with i1 < i2 < ..... < ik forms a complete basis for Λk. The coefficients of the (n,k) basis elements of the sum are the Tii...i where i1 < i2 < ..... < ik. On the other hand, we know we can also expand this same Q as T = Σii...i Φii...i (λi ^ λi ^ .... ^ λi) What then is the connection between the Tii...i and the Φii...i coefficients? Let's start with this last form: T = Σii...i Φii...i (λi ^ λi ^ .... ^ λi) = Σi≠i≠...≠i Φii...i (λi ^ λi ^ .... ^ λi) We then partition the summation space as follow: Σi≠i≠...≠i = [ Σi<i<...<i + Σi<i<...<i + many similar reorderings ] The total sum can be written in this manner, using the permutation sum notation, Σi≠i≠...≠i = ΣP ΣP(i)<P(i)<...<P(i) where P are the k! permutations of the k integers {1,2,...k}. Thus, we can write T = ΣP ΣP(i)<P(i)<...<P(i) Φii...i (λi ^ λi ^ .... ^ λi) . Just below we shall prove the following Lemma, which in the meantime we hope seems at least plausible to the reader, ΣP [ΣP(i)<P(i)<...<P(i)] fii...i = Σi<i<...<i [ΣP fP(i)P(i)...P(i)] . Within the ΣP permutation sum, the index permutation operators have moved from the summation indices to the summand indices. Accepting this Lemma, we then have T = Σi<i<...<i ΣP [ ΦP(i)P(i)...P(i) (λP(i) ^ λP(i) ^ .... ^λ P(i)) ] . But we know that ( λP(i) ^ λP(i) ^ .... ^λ P(i)) = (-1)S(P) (λi ^ λi ^ .... ^ λi) where S(P) is the number of swaps associated with permutation P. Thus. T = Σi<i<...<i [ΣP (-1)P FP(i)P(i)...P(i)] (λi ^ λi ^ .... ^ λi) which we can compare with our other sum T = Σi<i<....<i Tii...i (λi ^ λi ^ .... ^ λi) . Thus, the relation between the T and Φ coefficients is given by Tii...i = ΣP (-1)P ΦP(i)P(i)...P(i) i1 < i2 < ..... < ik = Φii...i - Φii...i + other signed permutations = a total of k! terms Examples: Tab = Φab - Φba k = 2 a < b Tabc = Φabc - Φacb + Φcab - Φcba + Φbca - Φbac k = 3 a < b < c Exercise: Show that an arbitrary wedge product of k vectors lies in the space Lk Consider the expansion T = Σii...i Φii...i (λi ^ λi ^ .... ^ λi) where Φii...i = αi βi ... Then get = Σii...i αi βi ...(λi ^ λi ^ .... ^ λi) = α ^ β ^ ... ^ q To obtain this same result using the other expansion, one sets Tii...i = ΣP (-1)P αP(i)βP(i)...qP(i) = αiβi...qi - αibi...qi + other signed terms Number of elements in Λk compared with V*k. From above we have found that dim(V*k) = nk // number of basis elements of Vk dim(Λk) = // number of basis elements of Λk If we set the number of reals to some large integer N, then ratio = = = = / nk . For a given n, this is a strongly decreasing function of k. For example, for n = 10 we can plot the log of the ratio for k = 0 to 10, Multi-index notations. This is done in two different ways. First, using the "redundant" expansion, T = Σii...i Φii...i (ei ^ ei ^ .... ^ ei) T = ΣI ΦI eI where eI ≡ ei ^ ei ^ .... ^ ei and I ≡ 1 ≤ i1, i2,.... ik ≤ n = ordinary multiindex n = dim(V) The more significant notation involves the other expansion which has only one term for each linearly independent basis element T = Σi<i<....<i Tii...i (ei ^ ei ^ .... ^ ei) T = ΣI TI eI where eI ≡ ei ^ ei ^ .... ^ ei and I ≡ 1 ≤ i1 < i2 <....< ik ≤ n = increasing multiindex n = dim(V) The full wedge tensor algebra. We now repeat the step taken above for direct products of vectors for wedge products of vectors, cribbing the text and editing in changes. Define a very large vector space of the form, Λ(V) = Λ0 + Λ1 + Λ2 + Λ3 + .... Here Λ0 = the space of scalars, Λ1 = V* the space of vectors, Λ2 = V* x V* = the space of (wedge) "bivectors", and so on. The most general element of the space Λ(V) would have the form Λ = s + ΣiΦi ei + Σij Φij ei^ej + Σijk Φijk ei^ej^ek + ...... This large space Λ(V) is in fact itself a vector space. It should be clear to the reader that it is closed under addition and has the right scalar rule. For example, writing if k1 and s are scalars, k1 + α + β^c + δ^γ^h = sum of 4 elements of Λ(V) = an element of Λ(V) s(k1 + α + β^c + δ^γ^h) = (sk1) + (sα) ^δ + f^(sγ)^∞ = element of Λ(V) The space is also closed under the multiplication operation ^. For example (β^δ)^(φ^γ^κ) = β^δ^φ^γ^κ = ϵ Λ5 = ϵ Λ(V) . One then makes the following definitions with regard to the space L: k1 scalar 0-blade 0 α vector 1-blade 1 α^β bivector 2-blade 2 k = 2 α^β^κ trivector 3-blade 3 k = 3 α^β^κ^δ quadvector 4-blade 4 k = 4 ..... α^β^κ^δ^.... k-vector k-blade 4 k = k .... α^β^κ^δ^.... n-vector n-blade n k = n arbitrary element of Λ(V) multivector linear combination of any of the above Since Λ(V) is closed under the operations + and ^, it is "an algebra" (the space Λk alone is not an algebra because it is not closed under ^). The Λ(V) algebra is different from that of the reals, due its definition as a sum of vector spaces. The elements of Λ(V) have different "grades" as shown in the right column above, and is known therefore as a "graded algebra". Sometimes Λ(V) is called "the dual wedge tensor algebra" over V. Unlike in the direct product world, in the wedge product world, the above list is finite for a given n = dim(V*). For k = n, where n = dim(V*), there is exactly one linearly independent basis vector which is the wedge product of all the basis vectors of v. For k > n, all wedge products vanish since the vectors in the wedge product are linearly dependent. The dimensionality of the space L is as follows, dim(L(V)) = 1 + n + + + ... + = Σk=0n = 2n = a finite number Lemma: (as promised above). Show that : ΣP [ΣP(i)<P(i)<...<P(i)] fii...i = Σi<i<...<i [ΣP fP(i)P(i)...P(i)] . Rather than present a formal proof, we look at the two simplest cases. First, consider Q ≡ Σi≠j fij = [Σi<j + Σi>j] fij = Σi<jfij + Σj<ifij . Swapping the dummy summation indices in the second term gives Q = Σi<jfij + Σi<jfji = Σi<j [ fij + fji] . This can be written as Q = Σi<j [ΣP fP(i)P(j)] where ΣP is over all permutations of two objects, the first being the identity permutation. Consider next the triple sum case where there are six sums. For each of the last five sums, we rename the summation indices to cause that sum to have the same form as the first sum. Q ≡ Σi≠j≠k fijk = (Σi<j<k + Σi<k<j + Σj<i<k + Σj<k<i + Σk<i<j + Σk<j<i) fijk = Σi<j<k [ΣP fP(i)P(j)P(k)] where ΣP is now a sum over the permutations of three objects, such as P{1,2,3} = {2,1,3}. In the first line above, the sum of sums can itself be written as (Σi<j<k + Σi<k<j + Σj<i<k + Σj<k<i + Σk<i<j + Σk<j<i) = ΣP [ΣP(i)<P(j)<P(k)] So for the triple sum our Lemma can be stated as ΣP [ΣP(i)<P(j)<P(k)] fijk = Σi<j<k [ΣP fP(i)P(j)P(k)] The argument for a k-fold sum proceeds in the same manner, and we end up with ΣP [ΣP(i)<P(i)<...<P(i)] fii...i = Σi<i<...<i [ΣP fP(i)P(i)...P(i)] .