dual non-dual comparison

7. The Wedge Product of k vectors in the dual space : the vector spaces Λk and Λ(V) 1 7.1 Definition of the wedge product of k dual vectors 3 7.2 Properties of the wedge product of k dual vectors 3 7.3 The dual vector space Λk and its basis 4 7.4 Tensor Expansions for a dual tensor in Λk 5 7.5 The alt, Alt and Sym operators 6 7.6 Expansions for the wedge product of k dual vectors 6 7.7 Number of elements in Λk compared with V*k. 7 7.8 Multiindex notations 8 7.9 The Dual Exterior Algebra Λ(V) 8 7.10 The Wedge Product of two dual tensors 8 7.11 The Wedge Product of two or more dual tensors in Multiindex Notation 9 7. The Wedge Product of k vectors in the dual space : the vector spaces Λk and Λ(V) Before continuing, we pause to make a comparison of non-dual and dual equations for tensors of rank 1 and 2. The non-dual equations are on the left, the dual ones on the right. The first section deals with vectors, the second with rank 2 tensor products, and the third with rank 2 wedge products. In each case, we have tried to make equations look as parallel as possible. Sometimes sums are implied and sometimes they are written out. On the right, sometimes functionals are shown, and other times functions. Various equation numbers are quoted in italics. ________________________________________________________________________________ Rank 1: V Rank 1: V* Vectors {ei} = basis of V (4.1.1) → {λi} = basis of V* (4.2.1) a = Σiaiei (2.5.1) vector → α = Σiαiλi (2.11.8) vector b = Σibiei (2.5.1) vector → β = Σiβiλi (2.11.8) vector a ei = ai → α(ei) = αi (2.11.9) a v = Σiaivi α(v) = Σiαivi (ei)j = δij (2.6.4) → λi(ej) = δij (2.11.6) ei v = vi → λi(v) = vi ________________________________________________________________________________ Rank 2: V2 Rank 2: V*2 Tensor Product (eiej)ab = δia δjb → (λiλj)(ea,eb) = δai δbj (eiej)ab vawb = viwj → (λiλj)(v,w) = viwj (2.11.15) T ≡ Σij Tij eiej (4.1.8) → T = Σij Tij λiλj (2.11.17) T = ΣI TI eI → T = ΣI TIλI (2.11.15) T (eiej) = Tij → T(ei,ej) = Tij (2.11.19) T (vw) ≡ Σij Tijviwj → T(v,w) = Σij Tij viwj (2.11.18) (a b)ij = aibj (3.1.8) → (α β)(ei,ej) = αiβj = (α β)ij ab = Σijaibj eiej → α β = Σijαiβj λiλj (ab) (vw) = (av)(bw) → (α β)(v,w) = α(v)β(w) T = V0 V V2 V3 ... → L = V*0 V* V*2 V*3 ... tensor algebra dual tensor algebra ________________________________________________________________________________ Rank 2: L2 Rank 2: Λ2 Wedge Product ei^ej = [eiej - ejei]/2 → λi^λj = [λiλj - λjλi]/2 (ei^ej)ab = [δaiδbj - δajδbi]/2 → (λi^λj)(ea,eb) = [δiaδjb - δjaδib]/2 (ei^ej) (vw) = [viwj - vjwi]/2 → (λi^λj)(v,w) = [viwj - vjwi]/2 a ^ b ≡ (ab - ba)/2 (4.3.1) → α ^ β ≡ (α β - β α)/2 (4.4.1) (a ^ b)ij = (aibj- ajbi)/2 → (α ^ β)(ei,ej) = (αiβj - αjβi)/2 (a ^ b)(vw) = [(av)(bw)-(aw)(bv)]/2 → (α ^ β)(v,w) = [α(v)β(w) - α(w)β(v)]/2 a ^ b = - b ^ a (4.3.2) → α ^ β = - β ^ α (4.4.2) a ^ a = 0 (4.3.3) → α ^ α = 0 (4.4.3) T = Σij Tij ei ^ ej (4.3.5) → T = Σij Tij λi ^ λj (4.4.5) T = ΣI TI eI → T = ΣI TI λI T = Σi<j Aij (ei ^ ej) (4.3.10) → T = Σi<j Aij (λi ^ λj) (4.4.10) Aij ≡ (Tij - Tji) Aij ≡ (Tij- Tji) T = Σ'I AI eI → T = Σ'I AI λI T = (1/2) A → T = (1/2) A L = L0 L L2 L3 ... → Λ = Λ0 Λ Λ2 Λ3 ... exterior algebra dual exterior algebra a ^ b = Σi<j det (ei^ej) (4.3.12) → α ^ β = Σi<j det (λi ^ λj) (4.4.12) ________________________________________________________________________________ a ^ b ^ c = det(a,b,c) (e1^ e2^ e3) (4.3.15) → α ^ β ^ γ = det(α,β,γ) (λ1 ^ λ2 ^ λ3) (4.4.15) for V3 for V*3 ________________________________________________________________________________ Using the above table as a guide, we now convert Chapter 6 equations from non-dual to dual space. 7.1 Definition of the wedge product of k dual vectors The forms for the wedge product of k dual vectors from (6.1.2) and (6.1.3) with v → α are, αj^ αj^ .....^ αj = (1/k!) ΣP (-1)S(P) ( αP(j) αP(j) ..... αP(j)) (7.1.1) α1^ α2^ .....^ αk = (1/k!) Σii....i εii....i (αi αi ..... αi) ir = 1,2..k = (1/k!) ΣP (-1)S(P) ( αP(1) αP(2) ..... αP(k)) i = (1/k!) [ α1 α2 ..... αk + all signed permutations ] (7.1.2) 7.2 Properties of the wedge product of k dual vectors Most of Section 6.2 translates to the dual world in a completely trivial fashion, so we just quote the headings: 1. The sums in (7.1.1) and (7.1.2) have k! terms. (6.2.1) 2. The wedge product is k-multilinear. (6.2.2) α1^(s1α2 + s2α2')^α3^.....^αk = s1(α1^α2^α3^ .....^αk) + s2(α1^α2'^α3^ .....^αk) (7.2.1) (6.2.3) 3. The wedge product changes sign if any vector pair is swapped. (6.2.4) 4. Wedge product vanishes if any two vectors are the same. (6.2.5) 5. Wedge product vanishes if vectors are linearly dependent. (6.2.6) 6. Wedge product vanishes if k > n . (6.2.7) 7. Components. Here there is a change. For the non-dual space we had, vj^ vj^ .....^ vj = (1/k!) ΣP (-1)S(P) ( vP(j) vP(j) ..... vP(j)) (vj^ vj^ .....^ vj)ii...i = (1/k!) ΣP (-1)S(P) ( vP(j) vP(j) ..... vP(j))ii...i = (1/k!) ΣP (-1)S(P) (vP(j))i(vP(j))i ... (P(j))i (6.2.8) For the dual space, the corresponding equations for a wedge product of linear functionals are, αj^ αj^ .....^ αj = (1/k!) ΣP (-1)S(P) ( αP(j) αP(j) ..... αP(j)) (7.2.2) (αj^ αj^ .....^ αj)(vi, vi....vi) = (1/k!) ΣP (-1)S(P) ( αP(j) αP(j) ..... αP(j))(vi, vi....vi) = (1/k!) ΣP (-1)S(P) αP(j)(vi) αP(j)(vi)... αP(j)(vi) . (7.2.3) Because any α(v) is linear in v, α being a linear functional, we see that (7.2.3) represents a k-multilinear function. Since this sum has the form (A.4.4) where fP(j)i = αP(j)(vi), we know from (A.4.5) that (αj^ αj^ .....^ αj)(vi, vi....vi) (7.2.4) is totally antisymmetric in both the jr and ir indices. Thus (7.2.4) is in fact an antisymmetric k-multilinear function. Evaluation at basis vectors gives, (αj^ αj^ .....^ αj)(ei, ei....ei) = = (1/k!) ΣP (-1)S(P) αP(j)(ei) αP(j)(ei)... αP(j)(ei) = (1/k!) ΣP (-1)S(P) [αP(j)]i[αP(j)]i ... [αP(j)]i (7.2.5) 7.3 The dual vector space Λk and its basis The basis vectors of Λk are these rank-k linear functionals, (λj ^ λj ^ .... ^ λj) (7.3.1) and this basis does in fact span a vector space by the argument of Section 6.3. Again there is a unique permutation such that (λj ^ λj ^ .... ^ λj) = (-1)S (λi ^ λi ^ .... ^ λi) where i1 < i2 < ..... < ik (7.3.2) and there are then independent basis vectors for Λk, counting as in Section 6.3. From (7.2.2), (λj ^ λj ^ .... ^ λj) = (1/k!) ΣP (-1)S(P) ( λP(j) λP(j) ..... λP(j)) . (7.3.3) Evaluation of the functional (7.3.3) at (vi, vi.....vi) gives, using (7.2.3), (λj ^ λj ^ .... ^ λj)(vi, vi.....vi) = (1/k!) ΣP (-1)S(P) ( λP(j) λP(j) ..... λP(j))(vi, vi.....vi) = (1/k!) ΣP (-1)S(P)(vi)P(j)(vi)P(j)...(vi)P(j) = (1/k!) ΣP (-1)S(P) (vi)P(j)(vi)P(j)...(vi)P(j) (7.3.4) As with (7.2.3) this object is totally antisymmetric in the jr and ir and is an antisymmetric k-multilinear basis function. The k-multilinearity is obvious from the form (7.3.4). Since (ei)P(j) = δiP(j), evaluation of (7.3.4) at (ei, ei.....ei) gives, (λj ^ λj ^ .... ^ λj)(ei, ei.....ei) = (1/k!) ΣP (-1)S(P) δiP(j)δiP(j)...δiP(j) (7.3.5) From (A.6.11) this can also be written (λj ^ λj ^ .... ^ λj)(ei, ei.....ei) = (1/k!) ΣP (-1)S(P) δP(j)i δP(j)i... δP(j)i (7.3.6) 7.4 Tensor Expansions for a dual tensor in Λk The dual version of the ordered expansion (6.4.1) is this, T = Σ1≤i<i<....<i≤n Aii...i (λi ^ λi ^ .... ^ λi) , (6.4.1) (7.4.1) T = Σ'I AI λI // multiindex while the symmetric expansion is (the second T is italic), T = Σii...i Tii...i(λi ^ λi ^ .... ^ λi) . (6.4.3) (7.4.2) T = ΣI TI λI // multiindex The connection between tensors A and T is Aii...i = ΣP (-1)S(P) TP(i)P(i)...P(i) (6.4.12) (7.4.3) which says A = alt(T ) = totally antisymmetric (6.5.2) (7.4.4) Result (6.4.14) becomes T = (1/k!) A . (6.4.14) (7.4.5) Evaluation of the symmetric expansion (7.4.2) using (7.3.4) gives T(vj, vj.....vj) = Σii...i Tii...i(λi ^ λi ^ .... ^ λi)(vj, vj.....vj) = (1/k!) Σii...i Tii...i ΣP (-1)S(P)(vi)P(j)(vi)P(j)...(vi)P(j) (7.4.6) We noted below (7.3.4) that (λi ^ λi ^ .... ^ λi)(vj, vj.....vj) is an antisymmetric k-multilinear function. Since (7.4.6) is a linear combination of such functions, we know that this most general function T(vj, vj.....vj) must also be an antisymmetric k-multilinear function. Therefore: Fact: The space Λk is equivalent to the space of all totally antisymmetric k-multilinear functions. (7.4.7) Strictly speaking, Λk is a space of linear functionals, but there is a 1-to-1 mapping between these functionals and their evaluated functions. There is another 1-to-1 mapping between the functional T and the fully antisymmetric tensor Aii...i of (7.4.1). The objects T are rank-k tensors in the space Λk and are isomorphic to the antisymmetric rank-k tensors Aii...i in Lk. 7.5 The alt, Alt and Sym operators The definitions of these operators are exactly as in Section 6.5 and we have already stated the key result, A = alt(T ) = totally antisymmetric (6.5.2) T = Alt(T ) = totally antisymmetric (6.5.8) (7.5.1) This shows that the expansions (7.4.1) and (7.4.2) are capable only of representing a totally antisymmetric tensor T. In more detail, Aii...i = ΣP (-1)S(P)TP(i)P(i)...P(i) = [alt(T )]ii...i Tii...i = (1/k!) ΣP (-1)S(P)TP(i)P(i)...P(i) = [Alt(T )]ii...i (7.5.2) 7.6 Expansions for the wedge product of k dual vectors Taking T ii...i = (α1)i (α2)i ... (αk)i (6.6.1) (7.6.1) the symmetric expansion (7.4.2) becomes α1 ^ α2 ^ ... ^ αk = Σii...i(α1)i (α2)i ... (αk)i (λi ^ λi ^ .... ^ λi) (6.6.2) (7.6.2) The coefficient set Aii...i is given by A = alt(α1α2...αk) (6.6.4) (7.6.3) so the ordered expansion (7.4.1) can be written, α1 ^ α2 ^ ... ^ αk = Σi<i<....<i [alt(α1α2...αk)]ii...i (λi ^ λi ^ ... ^ λi) (6.6.5) (7.6.4) = Σi<i<....<i det (λi ^ λi ^ .... ^ λi) (6.6.8) (7.6.5) The last several equations can be written in partial multiindex notation as (Z = 1,2...k) α1 ^ α2 ^ ... ^ αk = ΣI(α1)i (α2)i ... (αk)iλI = ΣI (αZ)I λI (7.6.2a) α1 ^ α2 ^ ... ^ αk = Σ'I [alt(α1α2...αk)]I λI (7.6.4a) α1 ^ α2 ^ ... ^ αk = Σ'I det [(αZ)I] λI (7.6.5a) 7.7 Number of elements in Λk compared with V*k. dim(V*k) = nk // number of basis elements of V*k (5.1.3) (7.7.1) dim(Λk) = // number of basis elements of Λk (6.3.6) (7.7.2) If the number of elements of field K is N, then ratio = = = = / nk . (6.7.1) (7.7.3) 7.8 Multiindex notations These have already been displayed above, but here in more detail, T = Σii...i Tii...i (λi ^ λi ^ ... ^ λi) T = ΣI TI λI where λI ≡ λi ^ λi ^ ... ^ λi (6.8.1) and I ≡ {i1, i2,.... ik} with 1 ≤ ir ≤ n = ordinary multiindex, n = dim(V*) (7.8.1) T = Σi<i<....<i Aii...i (λi ^ λi ^ ... ^ λi) T = Σ'I AI λI where λI ≡ λi ^ λi ^ ... ^ λi (6.8.2) and I ≡ {i1, i2,.... ik} with 1 ≤ i1< i2<....< ik ≤ n = ordered multiindex, n = dim(V) (7.8.2) 7.9 The Dual Exterior Algebra Λ(V) The dual exterior algebra is written as Λ(V) = Λ0 Λ1 Λ2 Λ3 + .... // Λ(V) = Σk=0∞ Λk (6.9.1) (7.9.1) The table of objects is the same as (6.9.6) where we replace the Latin letters by Greek ones. For example, one would refer to α^β as a dual 2-blade, and linear combinations of such as a dual bivector. Finally, dim[Λ(V)] = 1 + n + + + ... + = Σk=0n = 2n = a finite number (6.9.7) (7.9.2) 7.10 The Wedge Product of two dual tensors This discussion is the same as in Section 6.10 where we take (ei^ ei .....^ ei) → (λi ^ λi ^ ... ^ λi) Aii....i → Aii....i Bii....i → Bii....i Tii....i → Tii....i Sii....i → Sii....i (T^S)ii....i → [TS]ii...i (7.10.1) and so on. Nothing new happens. The resulting symmetric wedge product of a rank k dual tensor T with a rank k' dual tensor S is given by, T ^ S = Σii....i[Tii....i Sii....i] (λi ^ λi ^ ... ^ λi) (6.10.4d) (7.10.2) = ΣI TISI' λI while the ordered expansion is T ^ S = Σi<i<....<i [alt(T S)]ii...i (λi ^ λi ^ ... ^ λi) . with (6.10.10) (7.10.3) [alt(T S)]ii...i = ΣP (-1)S(P) TP(i)P(i)...P(i)SP(i)P(i)..P(i) This wedge product is seen to be an element of Λk+k'. The graded commutativity rule is unchanged, S^T = (-1)kk'T^S k = rank(S) k' = rank(T) (6.10.19) (7.10.4) ok to here 7.11 The Wedge Product of two or more dual tensors in Multiindex Notation We translate only the last results of Section 6.11 in systematic notation: Ti = dual tensor of rank ki Ii = multindex range of ir values for tensor Ti For the product of two dual tensors we write T1 = dual tensor of rank k1 I1 = {i1, i2.....ik} T2 = dual tensor of rank k2 I2 = {ik+1, ik+2.....ik+k} and then T1^T2 = Σ'I [alt(T1T2)]I λI λI ≡ λi λi ..... λi where [alt(T1T2)]I = ΣP (-1)S(P) (T1)P(I)(T2)P(I) (T1)P(I) ≡ (T1)P(i)P(i)...P(i) (T2)P(I) ≡ (T2)P(i)P(i)...P(i) . For the product of three dual tensors we write T1 = tensor of rank k1 I1 = {i1, i2.....ik} T2 = tensor of rank k2 I2 = {ik+1, ik+2.....ik+k} T3 = tensor of rank k3 I3 = {ik+k+1, ik+k+2.....ik+k+k} and then T1^T2^T3 = Σ'I [alt(T1T2T3)]I λI λI ≡ λi λi ..... λi where [alt(T1T2T3)]I = ΣP (-1)S(P) (T1)P(I)(T2)P(I)(T3)P(I) (T1)P(I) ≡ (T1)P(i)P(i)...P(i) (T2)P(I) ≡ (T2)P(i)P(i)...P(i) (T3)P(I) ≡ (T3)P(i)P(i)...P(i) For the product of N dual tensors Tiof rank ki the pattern is the same and we get this ordered sum, T1^T2^...^TN = Σ'I [alt(T1T2...TN)]I λI λI ≡ λi λi ..... λi where [alt(T1T2...TN)]I = ΣP (-1)S(P) (T1)P(I)(T2)P(I)...(TN)P(I) (T1)P(I) ≡ (T1)P(i)P(i)...P(i) (T2)P(I) ≡ (T2)P(i)P(i)...P(i) ..... (TN)P(I) ≡ (TN)P(i)P(i)...P(i) ************** ADDER: I am now back here at 7:30 PM 11/6/15. *************** Start with the wedge product of two non-dual tensors T^S = Σii....i [Tii....i Sii....i] (ei^ ei ......^ ei) and translate this to ( ir = 1,2...n ) T^S = Σii....i Tii....iSii....i (λi ^ λi ^ .... ^ λi) . Now evaluate this equation at (v1,v2....vk, vk+1....vk+k') to get (T^S)(v1,v2....vk, vk+1....vk+k') = Σii....iTii....iSii....i (λi ^ λi ^ .... ^ λi)(v1,v2....vk, vk+1....vk+k') Use (7.3.4) to replace the second line, (T^S)(v1,v2....vk, vk+1....vk+k') = Σii....iTii....iSii....i (1/(k+k')!) ΣP (-1)S(P) (v1)P(i)(v2)P(i) ... (vk+k')P(i) = (1/(k+k')!) ΣP (-1)S(P) Σii....iTii....iSii....i (v1)P(i)(v2)P(i) ... (vk+k')P(i) . Using Corollary 1 (A.7.1) we are allowed to lower to P operators to get = (1/(k+k')!) ΣP (-1)S(P) Σii....iTii....iSii....i (vP(1))i(vP(2))i ... (vP(k+k'))i which we reorder to get = (1/(k+k')!) ΣP (-1)S(P) [ Σii...i Tii...i (vP(1))i(vP(2))i ... (vP(k))i] [ Σii...i Sii...i(vP(k+1))i(vP(k+2))i ... (vP(k+k'))i] . Meanwhile, back at the ranch, T(v1,v2....vk) = Σii...i Tii...i(λi ^ λi ^ .... ^ λi)(v1,v2....vk) = [Σii...i Tii...i(v1)i(v2)i ... (vk)i] and so T(vP(1),vP(2)....vP(k)) = [Σii...i Tii...i(vP(1))i(vP(2))i ... (vP(k))i] . For S we start with S(v1,v2....vk') = Σii...i Sii...i(v1)i(v2)i ... (vk')i but then we rename both the arguments vi→ vP(k+i) and the summation indices ii→ ik+i to get, S(vP(k+1),vP(k+2)....vP(k+k')) = [Σii...i Sii...i(vP(k+1))i(vP(k+2))i ... (vP(k+k'))i] Installing ** and ** into ** we then find (T^S)(v1,v2....vk+k') = (1/(k+k')!) ΣP (-1)S(P)T(vP(1),vP(2)....vP(k))S(vP(k+1),vP(k+2)....vP(k+k')) This is then the instruction for computing the wedge product of k-tensor T and a k'-tensor S, where we use the term tensor to apply to a function like T(v1,v2....vk) which is regarded as an element of Λk in the sense of ***. The resulting function (T^S)(v1,v2...vk+k') is then a (k+k')-tensor in Λk+k'. Taking (v1,v2....vk+k') → (vi,vi....vi) (since the argument vector labels are arbitrary), we get (T^S) (vi,vi...vi) = (1/(k+k')!) ΣP (-1)S(P)T(vP(i),vP(i)....vP(i))S(vP(i),vP(i)....vP(i)) Recall now our definition (6.5.3) of the Alt operator [Alt(X)]ii...i ≡ (1/k!) ΣP (-1)S(P) XP(i)P(i)...P(i) (6.5.3) If we lower indices and replace k→ k+k', this becomes [Alt(X)]ii...i ≡ (1/(k+k')!) ΣP (-1)S(P) XP(i)P(i)...P(i) Therefore one can write (T^S) (vi,vi...vi) = [Alt(X)]ii...i where Xii...i = T(vi, vi....vi) S(vi, vi....vi) In dense multiindex we might write the above as XI = T(vI)S(vI') and then (T^S)(vI) = [Alt(T(vI)S(vI'))]J = (1/(k+k')!) ΣP (-1)S(P)T(vP(I))S(vP(I')) Maybe compare this with T ^ S = ΣI TISI' λI But looking earlier I see (λi ^ λi ^ .... ^ λi) = (1/k!) ΣP (-1)S(P) ( λP(i) λP(i) ..... λP(i)) (7.3.3) [Alt(X)]ii...i ≡ (1/k!) ΣP (-1)S(P) XP(i)P(i)...P(i) so λI = [Alt(X)]I where Xii...i = λi λi ..... λi λI = [Alt(λλ...λ)]I Xii...i = (λ λ ..... λ)ii...i X = (λ λ ..... λ) Then you could say Example : Try an example where k = 1 and k' = 1, So (T^S)(v1,v2) = Σii=1n TiSi (1/2!) ΣP(-1)S(P) (v1)P(i)(v2)P(i) = Σii=1n TiSi (1/2) [ (v1)i(v2)i - (v1)i(v2)i ] = (1/2){ Σii=1n TiSi (v1)i(v2)i - Σii=1n TiSi (v1)i(v2)i } = (1/2) { [Σi=1nTi (v1)i] [Σi=1nSi (v1)i] - [][] } = (1/2) { T(v1) S(v2) - T(v2) S(v1) } as expected (finally) So at least it gives what I expect for this example. Where is the thread for the general case? Example : Try an example where k = 2 and k' = 1, So (T^S)(v1,v2,v3) = Σiii =1nTiiSi (1/3!) ΣP (-1)S(P) (v1)P(i)(v2)P(i)(v3)P(i) = (1/3!) Σiii =1nTiiSi [(v1)i(v2)i(v3)i - (v1)i(v2)i(v3)i + 4 more terms ] = (1/3!)[ T(v1,v2) S(v3) - T(v2,v1) S(v3) + etc ]