wedge scraps

6.10 The Wedge Product of two or more general tensors Start with general tensor T of rank k (space Lk) and general tensor S of rank k' (space Lk'), T = Σi<i<....<i k! Aii....i (ei^ ei .....^ ei) . rank k, T ϵ Lk S = Σj<j<....<j k'! Bjj....j (ej^ ej .....^ ej) . rank k', S ϵ Lk' (6.10.1) Each tensor is presented in the ordered expansion of (6.4.1) where coefficients are A and B as shown. We wish to write T and S in terms of the symmetric expansion (6.4.3). In order to do that, we need to know the tensor components Tii...i and Sjj....j. We know from (6.5.2) that [Alt(T)]ii...i = Aii...i [Alt(S)]ii...i = Bii...i . (6.10.2) There are many T tensors for which Alt(T) = A. The obvious simplest choice is to select T and S as T = A S = B (6.10.3) so that the T and S tensors are already totally antisymmetric. The symmetric expansions are then T = Σii....i Aii....i (ei^ ei .....^ ei) . rank k, T ϵ Lk S = Σjj....j Bjj....j (ej^ ej .....^ ej) . rank k', S ϵ Lk' . (6.10.4) We then form the wedge product of these two tensors in a manner similar to the equations leading to (5.6.6), T^S = Σii....i Σjj....jAii....i Bjj....j(ei^ ei .....^ ei) ^(ej^ ej .....^ ej) = Σii....ijj....jAii....i Bjj....j(ei^ ei .....^ ei^ej^ ej .....^ ej) = Σii....iii....i[Aii....i Bii....i] (ei^ ei ......^ ei) = Σii....i[Aii....i Bii....i] (ei^ ei ......^ ei) = Σii....i[AB]ii....iii....i(ei^ ei ......^ ei) (6.10.5) where, similar to (5.6.7), we use in the last line our standard outer product notation, [AB]ii...iii...i = Aii...i Bii...i . (6.10.6) Eq. (6.10.5) shows that the product T^S is an element of Lk+k'with the following tensor components, (T^S)ii....i = [AB]ii...i. (6.10.7) The next step is to express (6.10.5) as an ordered sum rather than as a symmetric sum. Let us momentarily replace AB by the symbol T, so that (6.10.5) reads, T^S = Σii....iTii...i(ei^ ei ......^ ei) (6.10.8) According to (6.4.1), this symmetric sum can be replaced by the ordered sum (6.4.1), T^S = Σi<i<....<i (k+k')! Aii...i(ei^ ei ......^ ei) (6.10.9) where, according to (6.4.12) and then (6.5.2), Aii...i = (1/(k+k')!) ΣP (-1)S(P) TP(i)P(i)...P(i) = [Alt(T)]ii...i // A = Alt(T) (6.10.10) Replacing T = AB we get this final result for the wedge product of tensors T and S, T^S = Σi<i<....<i { (k+k')! [Alt(AB)]ii...i } (ei^ ei ......^ ei) (6.10.11) Note that, even though A and B are each totally antisymmetric with respect to their own indices, the Alt operator is still needed to provide antisymmetry between the "cross indices" of A and B. Example 1: k = 1 and k' = 1. Eq (6.10.11) reduces to T ^ S = A ^ B = Σi<i { (2)! [Alt(AB)]ii } (ei^ ei) . From (6.5.1), [Alt(AB)]ii = ΣP (-1)S(P) AP(i)BP(i) = (1/2) [ AiBi - BiAi ] so that A ^ B = Σi<i (2)!(1/2) [ AiBi - BiAi ] (ei^ ei) = Σi<i [ AiBi - BiAi ] (ei^ ei) in agreement with (4.3.12). ********************************** Multiplying two or more tensors in L(V) I think this topic will fit here, but I don't have it ready yet. See below and see separate doc(s) on how do do these products. I am staying in analogy with the end of Section 4.5. I am stopping here pre-Hawaii. Suppose I start with what I already know from the previous section, TS = Σii....i[Tii....i Sii....i] (ei ei ...... ei) (4.5.23) (TS)ii....i = Tii....i Sii....i (4.5.24) I could treat (TS) as itself a tensor like T1 ≡ (TS) which has some A1 as per (4.6.22), T1 = Σi<i<....<i (k+k')! A1ii...i (ei ^ ei ^ .... ^ ei) (4.6.22) and I would know then that A1ii...i = ΣP (-1)S(P) T1P(i)P(i)...P(i) or A1 = Alt(T1) = Alt(TS) which looks a little like Spivak. The wedge product of two tensors follows the path shown in (4.5.22) through (4.5.24) with → ^ : T = Σii....i Tii....i (ei^ ei .....^ ei) . rank k, T ϵ Lk S = Σjj....j Sjj....j (ej^ ej .....^ ej) . rank k', S ϵ Lk' (4.6.47) T^S = Σii....i[Tii....i Sii....i] (ei^ ei ......^ ei) (4.6.48) (T^S)symii....i = Tii....i Sii....i (4.5.24) Clearly the wedge product T^S is an element of Lk+k'with the following tensor components, (T^S)symii....i = Tii....i Sii....i = (TS)ii....i (4.6.49) So I have T^S = Σii....i(TS)ii....i (ei^ ei ......^ ei) Now we want to cast this into an ordered sum form T^S = Σi<i<....<i (k+k')! (TS)Aii....i (ei^ ei ......^ ei) I may know how to do this using (4.6.30), (TS)Aii....i = ΣP (-1)S(P) (TS)P(i)P(i)...P(i) or (TS)A = [Alt(TS)] Then the above reads T^S = Σi<i<....<i (k+k')! [Alt(TS)]ii....i (ei^ ei ......^ ei) Now we are getting still closer to Spivak. Now back up to this T = Σii....i Tii....i (ei^ ei^ .....^ ei) . rank k, T ϵ Lk I could write this in ordered form as T = Σi<i<....<i k! TAii....i (ei^ ei^ .....^ ei) . rank k, T ϵ Lk where I know that TAii....i = (1/k!) ΣP (-1)S(P) TP(i)P(i)...P(i) or TA = Alt(T) Similarly SA = Alt(S) and there you see those other factorials of interest. How about this Alt(TA) = k! Alt(Alt(T)) = k! Alt(T) Although the above product is valid, it is not really what we want because the expansions all contain huge numbers of redundant terms (since the basis elements have k! related forms). What we really want is for everything to have an ordered sum, not a symmetric sum. Carrying out the above products using the ordered expansions requires more work. We might start with T = Σi<i<....<i k! Aii....i (ei^ ei^ .....^ ei) . rank k, T ϵ Lk S = Σj<j<....<j k'! Bjj....j (ej^ ej .....^ ej) . rank k', S ϵ Lk' (4.6.51) The connection between the T,S coefficients in (4.5.22) and the A,B coefficients in (4.5.27) is given from (4.6.30) as Aii...i = (1/k!)ΣP (-1)S(P) TP(i)P(i)...P(i) or A = Alt(T) Bii...i = (1/k'!) ΣP (-1)S(P) SP(i)P(i)...P(i) or B = Alt(S) But clearly one cannot write Σi<i<....<i Σj<j<....<j = Σi<i<....<i< j<j<....<j // wrong so the product AB cannot be naively formed as it was above. The easy way is to go ahead and do the symmetric expansions above and then make use of (4.6.30) where (T^S)Aii....i = ΣP (-1)S(P) (T^S)P(i)P(i)...P(i) or (A^B)ii....i = [Alt(T^S)]ii....i or (A^B) = Alt(T^S) Then the product of the tensors in (4.5.22) is given by A^B = Σi<i<....<i [ (k+k')! Alt(T^S)ii....i ] (ej^ ej .....^ ej) Problem: User is given that A,B coefficients, not the T,S ones, so how is user supposed to do the calculation shown above? above. *************************************8 To make this match the factor on the first line, we have to slide the entire red group of basis vectors to the left. As we first slide ei to the left through all the black basis vectors, we pick up a sign (-1)k' because this involves k' swaps according to rule (6.2.6). We then slide ei through the black group so it comes to rest just to the right of T^S = Σii....i[Tii....i Sii....i] (ei^ ei ......^ ei) S^T = Σii....i[Sii....i Tii....i] (ei^ ei ......^ ei) (6.10.17) In the second expression, re-order the dummy summation indices in this elaborate manner: i → i // add k to this group of indices i → i ... i → i i → i // subtract k' from this group of indices i → i ... i → ik (6.10.18) This reordering has no effect on Σii....i and causes the square bracket in S^T to equal the square bracket in T^S. The question remains: what does this reordering do to the basis vector wedge product? Consider, (ei^ ei ......^ ei) = (ei^ ei .............^ ei^ ei .......^ ei) → (ei^ ei ......^ ei^ ei ......... ^ ei) (6.10.19) To get this wedge product back to standard order, we first slide ei to the left k' places which generates a sign (-1)k'. We then slide ei through the same left group of basis vectors picking up another sign (-1)k' . By the time ei has been slid through this group, the total sign is [(-1)k']k = (-1)kk' . ********************************** Definition and usage of A and alt Above we have used the standard definitions of coefficients Aii...i and of the operator Alt(). Here we provide alternate coefficients Aii...i (italic A) and a lower-case operator alt() merely to simplify expressions which will come up below. This is just a matter of removing annoying factors k! We define Aii...i ≡ k! Aii...i (6.5.12) alt(T) ≡ k! Alt(T) (6.5.13) Using these new coefficients and new operator, we rewrite various equations above: T = Σ1≤i<i<....<i≤n Aii...i (ei ^ ei ^ .... ^ ei) (6.4.1) (6.5.14) T = Σ'I AI eI // ordered multiindex version of the above (6.5.15) [alt(T)]ii...i ≡ ΣP (-1)S(P) TP(i)P(i)...P(i) (6.5.1) (6.5.16) Aii...i = ΣP (-1)S(P) TP(i)P(i)...P(i) (6.4.12) (6.5.17) = [alt(T)]ii...i // i1 < i2 < ..... < ik A = alt(T) (6.5.2) (6.5.18) *******************************8 The multiindex notations have already been introduced earlier in several places. These notations take some getting used to, so here we shall start off by translating all the equations of Section 6.10 into multiindex notation as a sort of practical exercise. The equation numbers below are chosen to match those of Section 6.10. We use ΣI to indicate an ordinary multiindex sum, and Σ'I to indicate an ordered sum. Then we show how this powerful notation yields useful results quickly. One should pay careful attention to the definition of symbols I, I' and I below. T = Σ'I AI eI I = {i1, i2, .. ik} eI ≡ (ei^ ei .....^ ei) rank k S = Σ'J BJ eJ J = {j1, j2, .. jk'} eJ ≡ (ej^ ej .....^ ej) rank k' (6.11.1) AI = [alt(T)]I BJ = [alt(S)]J (6.11.2) T = ΣI TI eI // symmetric expansions S = ΣJ SJ eJ (6.11.3) T^S = (ΣITIeI) ^ (ΣJSJeJ) = ΣIΣJTISJ eI^eJ (a) = ΣI,JTISJ eI^eJ (b) = ΣITISI' eI (d) I' = {ik+1, ik+2, .. ik+k'}, I = {i1, i2, .. ik+k'} = ΣI(TS)I eI (e) eI ≡ (ei^ ei ......^ ei) (6.11.4) [TS]I = TISI' I' = {ik+1, ik+2, .. ik+k'} (6.11.5) [T^S]I = [TS]I (6.11.6) T^S = ΣI T I eI // symmetric expansion T ≡ TS (6.11.7) T^S = Σ'I AI eI // ordered expansion (6.11.8) AI = ΣP (-1)S(P)T P(I) = [alt(T )]I // A = alt(T ) (6.11.9) The final result is T ^ S = Σ'I [alt(TS)]I eI (6.11.10) where [alt(TS)]I = ΣP (-1)S(P) TP(I)SP(I') TP(I) ≡ TP(i)P(i)...P(i) SP(I') ≡ SP(i)P(i)...P(i) . (6.11.11) As our first application, we rederive the graded commutivity rule (6.10.19): T^S = ΣIΣJTISJ eI^eJ S^T = ΣJΣISJTI eJ^eI = ΣIΣJTISJ [ eJ^eI ] = ΣIΣJTISJ [ (-1)kk'eI^eJ ] = (-1)kk' ΣIΣJTISJeI^eJ = (-1)kk'T^S . (6.11.12) Next, consider the wedge product of three tensors where the third tensor R has rank k" T = ΣI TI eI rank k // symmetric expansions S = ΣJ SJ eJ rank k' R = ΣK RK eK rank k" T^S^R = (ΣITIeI) ^ (ΣJSJeJ) ^ (ΣKRKeK) = ΣI,J,K TISJRK (eI^eJ^eK) . (6.11.13) We pause at this point to state some commutivity rules by inspection of the above expression S^T^R = ΣI,J,K TISJRK (eJ^eI^eK) = (-1)kk' T^S^R T^R^S = ΣI,J,K TISJRK (eI^eK^eJ) = (-1)k'k" T^S^R R^S^T = ΣI,J,K TISJRK (eK^eJ^eI) = (-1)(k"k+k'k+k"k')T^S^R (6.11.14) The last case requires a small piece of work: (eK^eJ^eI) = (-1)kk'(eK^eI^eJ) = (-1)kk'(-1)kk"(eI^eK^eJ) = (-1)kk'(-1)kk"(-1)k'k"(eI^eJ^eK) = (-1)(k"k+k'k+k"k')(eI^eJ^eK) (6.11.15) Other permutations of T^S^R can be obtained in a similar fashion. Each required basis-vector-group swap generates a factor of the form (-1)κκ'. Resuming the above development, T^S^R = ΣI,J,K TISJRK (eI^eJ^eK) = ΣITISI'RI" eI I = {i1, i2, .. ik+k'+k"}, eI ≡ ei^ ei .....^ ei I = {i1, i2, .. ik}, I' = {ik+1, ik+2, .. ik+k'}, I" = {ik+k'+1, ik+k'+2, .. ik+k'+k"} = ΣI(TSR)I eI // outer product notation (6.11.10) and here are related equations, T^S^R = ΣI T I eI = ΣI [TSR] I eI // symmetric expansion T^S^R = Σ'I AI eI = Σ'I [alt(TSR)]I eI // ordered expansion, A = alt(T) [alt(TSR)]I = ΣP (-1)S(P) TP(I)SP(I')RP(I") where ΣP is over all permutations of {i1, i2, .. ik+k'+k"} . To go further, we really need a more systematic notation. Let Ti = tensor of rank ki Ii = multindex range of ir values for tensor Ti For the product of two tensors we write T1 = tensor of rank k1 I1 = {i1, i2.....ik} T2 = tensor of rank k2 I2 = {ik+1, ik+2.....ik+k} and then T1^T2 = Σ'I [alt(T1T2)]I eI eI ≡ ei ei ..... ei where [alt(T1T2)]I = ΣP (-1)S(P) T1P(I)T2P(I) T1P(I) ≡ T1P(i)P(i)...P(i) T2P(I) ≡ T2P(i)P(i)...P(i) . For the product of three 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 eI eI = ei ei ..... ei where [alt(T1T2T3)]I = ΣP (-1)S(P) T1P(I)T2P(I)T3P(I) T1P(I) ≡ T1P(i)P(i)...P(i) T2P(I) ≡ T2P(i)P(i)...P(i) T3P(I) ≡ T3P(i)P(i)...P(i) For the product of N tensors Ti of rank ki the pattern is the same and we get this ordered sum, T1^T2^...^TN = Σ'I [alt(T1T2...TN)]I eI eI = ei ei ..... ei where [alt(T1T2...TN)]I = ΣP (-1)S(P) T1P(I)T2P(I)...TNP(I) T1P(I) ≡ T1P(i)P(i)...P(i) T2P(I) ≡ T2P(i)P(i)...P(i) ..... TNP(I) ≡ TNP(i)P(i)...P(i) ok to here *********************************** The evaluations of these two functional objects are as follows (α1 α2 ..... αk)(v1, v2 ....vk) = α1(v1)α2(v2) ... αk(vk) (5.1.3) (λi λi ..... λi)(v1, v2 ....vk) = λi(v1)λi(v2) ... λi(vk) = (v1)i (v2)i... (vk)i (5.1.4) (α1 α2 ..... αk)(vj, vj ....vj) = α1(vj)α2(vj) ... αk(vj) (5.1.3) (α1 α2 ..... αk)(ej, ej ....ej) = α1(ej)α2(ej) ... αk(ej) = (α1)j(α2)j ... (αk)j (λi λi ..... λi)(vj, vj ....vj) = λi(vj)λi(vj) ... λi(vj) = (vj)i (vj)i ... (vj)i (5.1.4) (λi λi ..... λi)(ej, ej ....ej) = λi(ej)λi(ej) ... λi(ej) = (ej)i (ej)i ... (ej)i = δji δji ... δji (α1 α2 ..... αk)(vj, vj ....vj) = α1(vj)α2(vj) ... αk(vj) (6.1.3)' (5.1.4) (5.1.3) (λi λi ..... λi)(vj, vj ....vj) = λi(vj)λi(vj) ... λi(vj) = (vj)i (vj)i ... (vj)i (5.1.4) Evaluated at the basis vectors we get, (α1 α2 ..... αk)(e1, e2 ....ek) = α1(e1)α2(e2) ... αk(ek) (5.1.3) (λi λi ..... λi)(v1, v2 ....vk) = λi(v1)λi(v2) ... λi(vk) = (v1)i (v2)i... (vk)i (5.1.4) (α1 α2 ..... αk)(vj, vj ....vj) = α1(vj)α2(vj) ... αk(vj) = (v1)j (v2)j .... (vk)j (5.1.3) (λi λi ..... λi) = (ei)j (ei)j .... (ei)j = δij δij .... δij . (5.1.4) ************** We know this is true since T(V*) = Σk=0∞ V*k and we showed in (6.3.2) that each V*k is a vector space. For example, the "0" element in T(V*) is the direct sum of the "0" elements of all the V*k. To show that T(V*) is an algebra, we must show that it is closed under both addition and multiplication. It should be clear to the reader that T(V*) is closed under addition and has the right scalar rule. For example, if k1 and s are scalars, k1 + α + βκ + ρση = sum of 4 elements of T(V*) = an element of T(V*) s(k1 + α + βκ + ρση) = (sk1) + (sβ) κ + ρ(sσ)η = element of T(V*) . (6.4.5) This additive closure is of course necessary for T(V*) be a vector space. The space is also closed under the multiplication operation . For example (βκ)(ρση) = βκρση = ϵ V*5 = ϵ T(V*) . // (βκ) ϵ V*2 ,(ρση) ϵ V3 (6.4.6) Here we have used the associative property (2.8.22) applied to vectors. This closure claim is stated more generally below (5.6.7). ********** v1w1 + v2w2 = + = = ******* Written more concisely (but less clearly) Σi=1nvi(ei0) + Σi=1n'wi(0e'i) = Σi=1n(viei0) + Σi=1n'(0wie'i) // (B.5) = [Σi=1n viei][Σi=1n 0] + [Σi=1n'0] [Σi=1n'wje'j] // (B.1.b) = [Σi=1n viei]0 +0 [Σi=1n'wje'j] = [Σi=1n viei] [Σi=1n'wje'j] //(B.4) = v w ************* Brief Digression: The Direct Sum of Vector Spaces The direct sum of two or more vector spaces is a very simple concept. Consider this example, G ≡ V2V3. : V2V3 → V5 : (v,w) ↦ z or z = vw v = ϵ V2 w = ϵ V3 z = vw = = ϵ V5 . (5.4.1) Here we visualize the direct sum vector z as a tall column vector which is the union of the two smaller column vectors v and w. In the tall column vector, v and w each occupy a private region. With this construction, one sees that vw ≠ wv because the resulting vectors in V5 are different, but in the context of Chapter 1 isomorphism (it does not matter which vector is put first) one often sees stated that vw = wv, as appropriate for an addition operator. And since α = , one has α(vw) = (αv)(αw). Whereas dim(AB) = dimA * dimB, it is clear that dim(AB) = dimA + dimB. The operator is a bilinear operator, v(αw1+βw2) = α(vw1) + β(vw2) ?? This would imply that v(αw1) = α(vw1) which is wrong!!!! So is NOT a bilinear operator. That only applies to multiplicative operators! How show that V2V3 is a vector space? associative : ok (ab)c = a(bc) communivave: Finally, since the spaces V2 and V3 are vector spaces, so is V2V3: The zero is 00, addition is closed because each subspace is closed, and so on. The fact that the subspaces are vector spaces is induced into the direct sum space, αz1+βz2 = α (v1w1) + β (v2w2) = (αv1) (αw1) + (βv2) (βw2) = (αv1 + βv2) + (αw1 + βw2) ************************************* 11.18.15 ************************ Definition: The permutation tensor εiii...i where ir ϵ {1,2,3...k} is defined by these simple rules: ε123..k = 1 εiii...i changes sign if any two indices are swapped, ir↔is . Therefore, εiii...i = 0 if two or more indices are the same Definition: A function f(1,2..k) is totally antisymmetric if it changes sign when any two arguments are swapped. Fact: The function f(1,2,3...k) = εiii...i is totally antisymmetric This follows from the two definitions above. Fact: Any totally antisymmetric function f(1,2,3...k) must be a multiple of εiii...i. The object ε is usually called "the permutation tensor" even though it is not really a tensor in the strong sense involving transformation. There exists a related object, also written ε, which is the "Levi Civita tensor" and it really is a tensor, see ****. The permutation tensor has very simple properties. If any two indices are swapped, it changes sign, and ε12..k = 1. Thus, εii...i = 0 if any two or more indices are the same. This permutation tensor makes no sense unless the ir are restricted to the range {1,2..k}. The number of indices on ε must match the range of values the indices can take. **************************** 11/19/15 But we recognize the sum from (A.1.7) as being this g1,2,..k(1,2...k) = [Alt(f1,2,..k)](1,2...k) = (1/k!) det(Hab) (A.2.20) f1,2,..k(1,2...k) = h1(1)h2(2).....hk(k) . (A.2.20) Then we have g1,2,..k(1,2...k) = [Alt(f1,2,..k)](1,2...k) = (1/k!) ΣP (-1)S(P)h1(P(1))h2(P(2)).....hk(P(k)) (A.2.21) Define a matrix H as follows Hab ≡ ha(b) (A.2.22) The above is then g1,2,..k(1,2...k) = [Alt(f1,2,..k)](1,2...k) = (1/k!) ΣP (-1)S(P)H1P(1)H2P(2)....HkP(k) (A.2.23) The Determinant Theorem (A.1.18) says ΣP (-1)S(P) H1P(1)H2P(2) ...HkP(k) = ΣP (-1)S(P) HP(1)1HP(2)2 ...HP(k)k (A.2.24) Then we can write g1,2,..k(1,2...k) = [Alt(f1,2,..k)](1,2...k) = (1/k!) ΣP (-1)S(P)hP(1)(1)hP(2)(2).....hP(k)(k) (A.2.25) Therefore, we have two different but equivalent ways to write Alt(f1,2,..k) : g1,2,..k(1,2...k) = [Alt(f1,2,..k)](1,2...k) = (1/k!) ΣP (-1)S(P)h1(P(1))h2(P(2)).....hk(P(k)) (a) = (1/k!) ΣP (-1)S(P)hP(1)(1)hP(2)(2).....hP(k)(k) (b) where f1,2,..k(1,2...k) = h1(1)h2(2).....hk(k) (A.2.26) In (a) the arguments (1,2...k) are permuted, while in (b) the labels 1,2,..k are permuted. We can re-express the sums (a) and (b) in these alternate notations (see (A.6.1) regarding ε ), (a) = (1/k!) [ h1(1)h2(2)h3(3) + all signed permutations of the arguments ] (a) = (1/k!) [ h1(1)h2(2)h3(3) + all signed permutations of the subscripts ] (A.2.27) (a) = (1/k!) Σii...i εii...i h1(i1) h2(i2).....hk(i) ir = 1,2..k (b) = (1/k!) Σii...i εii...i hi(1) hi(2)....hi(k) ir = 1,2..k (A.2.28) (a) = (1/k!) det(H) ir = 1,2..k (b) = (1/k!) det(H) ir = 1,2..k (A.2.28) Here are two examples to encourage the reader: Example for k = 2: (A.2.29) (a) = (1/2!) ΣP (-1)S(P)h1(P(1))h2(P(2)) = (1/2) [ h1(1)h2(2) - h1(2)h2(1)] (b) = (1/2!) ΣP (-1)S(P)hP(1)(1)hP(2)(2) = (1/2) [ h1(1)h2(2) - h2(1)h1(2) ] = (a) Example for k = 3: Here red is used to show which pair will be swapped to get the next term. (A.2.30) (a) = (1/3!) ΣP (-1)S(P)h1(P(1))h2(P(2))h3(P(3)) = (1/6) * [ h1(1)h2(2)h3(3)-h1(2)h2(1)h3(3)+h1(2)h2(3)h3(1)-h1(3)h2(2)h3(1)+h1(3)h2(1)h3(2)-h1(1)h2(3)h3(2)] 1 2 3 4 5 6 (b) = (1/3!) ΣP (-1)S(P)hP(1)(1)hP(2)(2)hP(3)(3) = (1/6) * [ h1(1)h2(2)h3(3)- h2(1)h1(2)h3(3)+ h2(1)h3(2)h1(3)- h3(1)h2(2)h1(3)+ h3(1)h1(2)h2(3)- h1(1)h3(2)h3(3)] 1 2 5 4 3 6 We have numbered the terms 1,2...6 in each sum, and one sees how they are merely reordered 3↔5. ******** A Special Case for the form of function f Recall this generic notation theorem (A.2.26) , g1,2,..k(1,2...k) = [Alt(f1,2,..k)](1,2...k) = (1/k!) ΣP (-1)S(P)h1(P(1))h2(P(2)).....hk(P(k)) (a) = (1/k!) ΣP (-1)S(P)hP(1)(1)hP(2)(2).....hP(k)(k) (b) where f1,2,..k(1,2...k) = h1(1)h2(2).....hk(k) (A.2.26) We now let g1,2,..k(1,2...k) = Gjj...jii...i f1,2,..k(1,2...k) = fji fji.... fji fab = ha(b) = some rank-2 mixed tensor Then (A.2.26) becomes Gjj...jii...i = (1/k!) ΣP (-1)S(P) fji fji.... fji (a) = (1/k!) ΣP (-1)S(P) fji fji.... fji (b) (A.5.13) Whenever one encounters a sum of the form (a), it can be replaced by (b), and vice versa. Example: Let fab = δab = δa,b. Then δ jj...jii...i ≡ (1/k!) ΣP (-1)S(P) δji δji.... δji (a) = (1/k!) ΣP (-1)S(P) δji δji.... δji (b) (A.5.13) For example δ jjii = ************* A Special Case for the form of function f Recall now the generic Alt expansion, g(1,2...k) = [Alt(f)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) . (A.2.1) Suppose the function f has some second set of static labels which we also call 1,2...k. Then the above becomes g1,2,..k(1,2...k) = [Alt(f1,2,..k)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)f1,2,..k( P(1),P(2)...P(k) ) .(A.2.19) Suppose also that the function f can be factored in the following manner, f1,2,..k(1,2...k) = H11H22.....Hkk . (A.2.20) Then (A.2.19) says, using (A.1.7) for the determinant, g1,2,..k(1,2...k) = [Alt(f1,2,..k)](1,2...k) = (1/k!) ΣP (-1)S(P)H1P(1)H2P(2).....HkP(k) = (1/k!) det(H**) (A.2.21) Since det(H**) = det(HT**) we end up with (1/k!) ΣP (-1)S(P)H1P(1)H2P(2).....HkP(k) = (1/k!) ΣP (-1)S(P)HP(1)1HP(2)2.....HP(k)k or det(H**) = det(HT**) (A.2.22) If H is a rank-2 tensor, we can raise the second index to get, (1/k!) ΣP (-1)S(P)H1P(1)H2P(2).....HkP(k) = (1/k!) ΣP (-1)S(P)HP(1)1HP(2)2.....HP(k)k or det(H**) = det(HT**) (A.2.23) ********* Fact: ΣP (-1)S(P) fji fji.... fji (A.5.13) = ΣP (-1)S(P) fji fji.... fji Proof: The above equation says k! det(f**) = k! det(fT**) where fab is a kxk matrix. See (A.1.19). ************ (7.2.8) (7.2.8) Thinking of Mab = (va)b , (A.1.19) tells us the the P (1/k!) ΣP (-1)S(P) (vj)P(i)(vj)P(i) ... (vj)P(i) where we have used the tensor outer product form (2.8.18). Since this matches the form (A.4.4), Fii...ijj...j = ΣP (-1)S(P) fP(j)i fP(j)i ...fP(j)i , (A.4.4) one can rewrite (7.2.8) with the P operators moved to the ir indices, (vj^ vj^ .....^ vj)ii...i = (1/k!) ΣP (-1)S(P) (vj)P(i)(vj)P(i) ... (vj)P(i) . (7.2.9) According to (A.4.5), then, we have ************ This wedge product can be expanded using (7.1.2), (ej ^ ej ^ .... ^ ej) = (1/k!) ΣP (-1)S(P) ( eP(j) eP(j) ..... eP(j)) . (7.3.8) The components of the above equation are (ej ^ ej ^ .... ^ ej)ii...i = (1/k!) ΣP (-1)S(P)( eP(j) eP(j) ..... eP(j))ii...i = (1/k!) ΣP (-1)S(P) (eP(j))i ( eP(j))i...( eP(j))i = (1/k!) ΣP (-1)S(P) δP(j)i δP(j)i... δP(j)i . (7.3.9) = (1/k!) ΣP (-1)S(P) δjP(i) δjP(i)...δjP(i) // from (A.4.4) (7.3.10) and from (A.4.5) we conclude that, Fact: (ej^ ej^ .....^ ej)ii...i is totally antisymmetric in both the labels jr and the indices ir. (7.3.11) ***************** This is the Title PhL 1.11.15 Note that page numbering is turned on in this template and view is 125%, located in phil/roaming/microsoft/templates size about 219K. Consider G = Alt(F) or Gii...i = [Alt(F)]ii...i = (1/k!) ΣP (-1)S(P) Fii...i Suppose F carries some static labels jj...j. Then the above reads: Gjj...jii...i = (1/k!) ΣP (-1)S(P) Fjj...jii...i I want to show that Gjj...jii...i = (1/k!) ΣP (-1)S(P) Fjj...jii...i In my generic world g(1,2...k) = [Alt(f)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) (A.2.1) If f carries extra labels then g12..k(1,2...k) = [Alt(f12..k)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)f12..k( P(1),P(2)...P(k) ) ************** Argument why might be true: Recall now the generic Alt expansion. g(1,2...k) = [Alt(f)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) (A.2.1) Suppose the function f has some second set of static labels which we also call 1,2...k. Then the above becomes g1,2,..k(1,2...k) = [Alt(f1,2,..k)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)f1,2,..k( P(1),P(2)...P(k) ) The static labels are shown in red. We could alternately consider the (1,2..k) labels to be static, and the lower labels 12..k to be associated with a permutation sum. We then get instead g1,2,..k(1,2...k) = [Alt(f1,2,..k)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)fP(1),P(2),..P(3)(1,2...k) where again the static labels are shown in red. Comparing the two equations, we conclude that ΣP (-1)S(P)f1,2,..k( P(1),P(2)...P(k) ) = ΣP (-1)S(P)fP(1),P(2),..P(3)(1,2...k) Example with k = 2: ΣP (-1)S(P)f1,2( P(1),P(2) ) = f1,2(1,2) - f1,2(2,1) ΣP (-1)S(P)fP(1),P(2)(1,2) = f1,2(1,2) - f2,1(1,2) NOT TRUE!!!!!! So my argument is bogus. Who says the two separate Alt definitions would be the same! Try again and specialize to the case where f(1,2..k) = h(1)h(2)...h(k) = product of identical factors. g(1,2...k) = [Alt(f)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) (A.2.1) = (1/k!) ΣP (-1)S(P) h( P(1)) h( P(2)) .......h( P(k)) gabc...q(1,2...k) = (1/k!) ΣP (-1)S(P) ha( P(1)) hb( P(2)) .......hq( P(k)) This then allows my determinant argument! But try directly Assume that f12..k(1,2...k) = h1(1)h2(2).....hk(k) where the red labels are static. Then g12..k(1,2...k) = (1/k!) ΣP (-1)S(P)f12..k( P(1),P(2)...P(k) ) = (1/k!) ΣP (-1)S(P) h1(P(1))h2(P(2)).....hk(P(k)) Now define Hij ≡ hi(j) Then the above reads g12..k(1,2...k) = (1/k!) ΣP (-1)S(P)H1P(1) H2P(2) ..... *********** 7.4 Tensor Expansions for a tensor in Lk Lk can be regarded as the vector space whose "vectors" (rank-k tensors) can be written in this ordered-sum form, T = Σ1≤i<i<....<i≤n Aii...i (ei ^ ei ^ .... ^ ei) . (7.4.1) This is so because the set (ei ^ ei ^ .... ^ ei) with 1 ≤ i1 < i2 < ..... < ik ≤ n forms a complete basis for Lk, as discussed just above. Example: If n = 3 and k = 2, then T = Σ1≤i<i≤3 Aii (ei^ ei) = A12 (e1^e2) + A13 (e1^e3) + A23 (e2^e3) . (7.4.2) On the other hand, one can also expand this same tensor T in a (redundant) symmetric sum as, T = Σii...i Tii...i (ei ^ ei ^ .... ^ ei) ir = 1,2..n (7.4.3) as in (4.3.5) for k = 2. Here Tii...i are a set of coefficients. What then is the connection between the Aii...i and the Tii...i coefficients? Start with the symmetric form (7.4.3), T = Σii...i Tii...i (ei ^ ei ^ .... ^ ei) ir = 1,2..n = Σi≠i≠...≠i Tii...i (ei ^ ei ^ .... ^ ei) . // (7.2.5) (7.4.4) Partition the summation space as follows (1 ≤ ir ≤ n), Σi≠i≠...≠i = [ Σi<i<...<i + Σi<i<...<i + many similar reorderings ] . (7.4.5) The total sum can be written in this manner, using the permutation sum notation, Σi≠i≠...≠i = ΣP ΣP(i)<P(i)<...<P(i) (7.4.6) where P are the k! permutations of the k integers {1,2,...k}. Using the form (7.4.6), the sum (7.4.4) may be rewritten as, T = ΣP ΣP(i)<P(i)<...<P(i) Tii...i (ei ^ ei ^ .... ^ ei) . (7.4.8) In (A.7.1) it is shown that, ΣP [ΣP(i)<P(i)<...<P(i)] fii...i = Σi<i<...<i [ΣP fP(i)P(i)...P(i)] . (A.7.1) Within the ΣP permutation sum, the permutation operators P have moved from the summation indices to the summand indices. One then has from (7.4.8), T = Σi<i<...<i ΣP [ TP(i)P(i)...P(i) (eP(i) ^ eP(i) ^ .... ^e P(i)) ] . (7.4.9) But we know from (7.3.5) that ( eP(i) ^ eP(i) ^ .... ^e P(i)) = (-1)S(P) (ei ^ ei ^ .... ^ ei) (7.4.10) where S(P) is the number of swaps associated with permutation P. Thus. T = Σi<i<...<i [ΣP (-1)S(P) TP(i)P(i)...P(i)] (ei ^ ei ^ .... ^ ei) (7.4.11) which we can compare with the ordered sum (7.4.1), T = Σi<i<....<i Aii...i (ei ^ ei ^ .... ^ ei) . (7.4.1) Thus, since the basis is complete, the relation between the A and T coefficients is given by Aii...i = ΣP (-1)S(P) TP(i)P(i)...P(i) i1 < i2 < ..... < ik = [ Tii...i + all signed permutations ] // k! terms (7.4.12) The Aii...i appear in the expansion (7.4.1) only for index values 1 ≤ i1 < i2 < ..... < ik ≤ n, but we can interpret (7.4.12) as defining Aii...i for all index values. Since (7.4.12) has the general form shown in (A.3.1), we conclude that Fact: Aii...i is a totally antisymmetric tensor. (7.4.13) Comment: Tii...i and Aii...i are both rank-k tensors, see (5.5.3). Examples: (relating the A and T coefficients) Aab = Tab - Tba // as in (4.3.10) k = 2 Aabc = Tabc - Tacb + Tcab - Tcba + Tbca - Tbac k = 3 (7.4.14) Reconsider now (7.4.3) with ir → jr, T = Σjj...j Tjj...j (ej ^ ej ^ .... ^ ej) . jr = 1,2..n (7.4.4) We may compute the components of T by applying ii...i to both sides, Tii...i = Σjj...j Tjj...j (ej ^ ej ^ .... ^ ej)ii...i = (1/k!) ΣP (-1)S(P) Σjj...j Tjj...j δjP(i) δjP(i)...δjP(i) // (7.3.10) = (1/k!) ΣP (-1)S(P) TP(i)P(i)...P(i) = (1/k!) Aii...i . // (7.4.12) (7.4.15) which says T = (1/k!)A (7.4.15a) and from (7.4.13) it follows that T must be a totally antisymmetric rank-k tensor. This shows that the expansions (7.4.1) and (7.4.3) for T are capable only of describing a totally antisymmetric tensor T. Fact: The space Lk is the space of totally antisymmetric rank-k tensors T. (7.4.16) In contrast, the space Vk is the space of all rank-k tensors T, so Lk Vk. See Section 7.7 below. Recall that the pure vector wedge product v1^ v2^ .....^ vk is k-multilinear in the vi, and so is the underlying tensor product v1 v2 ..... vk. For k = 1, (7.4.15) says Ti = Σj Tj(ej)i = ΣjTjδji = Ti (7.4.17) so for a vector there is no distinction between Tk and Tk . ********************* entire dumped section 7.5 7.5 The alt, Alt and Sym operators One can regard the sum shown in (7.4.12) as being an operation "alt" performed on the tensor T to produce another tensor A which is totally antisymmetric. If one defines [alt(X)]ii...i ≡ ΣP (-1)S(P) XP(i)P(i)...P(i) (7.5.1) then (7.4.12) becomes Aii...i = [alt(T)]ii...i or A = alt(T) . (7.5.2) It is useful to define another version of the alt operator which has a scaling factor 1/k! , [Alt(X)]ii...i ≡ (1/k!) ΣP (-1)S(P) XP(i)P(i)...P(i) (7.5.3) so that Alt(X) = (1/k!)alt(X) . (7.5.4) It is shown in (A.3.1,2) that alt(X) and therefore Alt(X) are totally antisymmetric tensors for any tensor X. If X is already totally antisymmetric, one finds that alt(X) = k! X // since all k! terms in (7.5.1) are the same X = totally antisymmetric (7.5.5) Alt(X) = X . X = totally antisymmetric (7.5.6) We showed in (7.4.15a) that T = (1/k!) A . (7.5.7) so (7.5.2) can also be written T = Alt(T) . // (1/k!) A = (1/k!) alt(T) (7.5.8) One can define a total symmetrizing operator Sym in a manner similar to (7.5.3) but where the (-1)S(P) is omitted, [Sym(X)]ii...i ≡ (1/k!) ΣP XP(i)P(i)...P(i) (7.5.9) and then the tensor Sym(X) is totally symmetric for any tensor X. If X is already totally symmetric, then Sym(X) = X X = totally symmetric (7.5.10) Fact: The operators Sym and Alt are both projection operators, meaning that Sym2 = Sym and Alt2 = Alt when applied to any tensor X. (7.5.11) Proof: Alt(Alt(X)) = Alt(A) // where A ≡ Alt(X) = totally antisymmetric = A // by (7.5.6) = Alt(X) // since A ≡ Alt(X) Sym(Sym(X)) = Sym(S) // where S ≡ Sym(X) = totally symmetric = S // by (7.5.10) = Sym(X) // since S ≡ Sym(X) Fact: The Sym and Alt projection operators are orthogonal : Sym(Alt(X)) = 0 Alt(Sym(X)) = 0 (7.5.12) These seemingly reasonable claims are proven in Appendix A.9. We can define a third projection operator this way, Else() ≡ 1 - Alt() - Sym() // projection operator Else(X) = X - Alt(X) - Sym(X) . // applied to X (7.5.13) One can then decompose an arbitrary tensor X into three pieces, X = Alt(X) + Sym(X) + Else(X) = A + S + E (7.5.14) Then Alt(X) = Alt(Alt(X)) + Alt(Sym(X)) + Alt(Else(X)) = Alt(X) + 0 + Alt(Else(X)) Alt(Else(X)) = 0 Alt(E) = 0 . (7.5.15) Sym(X) = Sym(Alt(X)) + Sym(Sym(X)) + Sym(Else(X)) = 0 + Sym(X) + Sym(Else(X)) Sym(Else(X)) = 0 Sym(E) = 0 . (7.5.16) This verifies that the "else" piece E of a tensor has neither a totally antisymmetric nor a totally symmetric component. Examples: For a rank-2 tensor Xab one finds Aab = (Xab - Xba) Sab = (Xab + Xba) Eab = Tab - Aab - Sab = 0 (7.5.17) so the leftover else piece Eab is null. On the other hand, for a rank-3 tensor Xabc, Aabc = (Xabc - Xacb + Xcab - Xcba + Xbca - Xbac) Sabc = (Xabc + Xacb + Xcab + Xcba + Xbca + Xbac) Eabc = Xabc - Aabc - Sabc = Xabc - (Xabc - Xacb + Xcab - Xcba + Xbca - Xbac) - (Xabc + Xacb + Xcab + Xcba + Xbca + Xbac) = Xabc - (Xabc + Xcab + Xbca ) = Xabc - (Xcab + Xbca ) (7.5.18) so in this case the leftover piece Eabc is not null. Notice that Xcab = Xabc if X is either totally symmetric or totally antisymmetric Xbca = Xabc if X is either totally symmetric or totally antisymmetric and for this reason (7.5.18) shows that Eabc = 0 if X is either totally symmetric or totally antisymmetric. Fact: (vj^ vj^ .....^ vj) = Alt(vj vj ..... vj) (7.5.19) Proof: Recall from definition (7.1.2) vj^ vj^ .....^ vj = (1/k!) ΣP (-1)S(P) ( vP(j) vP(j) ..... vP(j)) . (7.1.2) Therefore (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......(vP(j))i // outer product = (1/k!) ΣP (-1)S(P) (vj)P(i)(vj)P(i)......(vj)P(i) // (A.6.1) = (1/k!) ΣP (-1)S(P) (vj vj ..... vj)P(i)P(i) ...P(i) // outer product = Alt(vj vj ..... vj)ii....i // (7.5.3) def of Alt and therefore (vj^ vj^ .....^ vj) = Alt(vj vj ..... vj). QED Corollary: (ej^ ej^ .....^ ej) = Alt(ej ej ..... ej) (7.5.20) ********************************** For example, Aii...i = [Alt(v1v2...vk)]ii...i = [v1 ^ v2 ^ ... ^ vk]ii...i = det[ (vj)i] = det = det Aii...i = det = det . (7.5.8) As noted in (A.4.9), for k < n the second determinant above is a minor of matrix Q ≡ [v1,v2....vk] whose columns are the vectors vi. The minor is the full width of Q but only has the rows specified by i1...ik. When k = n, the minor is the full det(Q). See the k=2 example in Fig (4.3.13). We then have these three variations of the vector wedge product expansion: v1 ^ v2 ^ ... ^ vk = Σii...i (v1)i (v2)i ... (vk)i (ei ^ ei ^ .... ^ ei) (7.5.2) v1 ^ v2 ^ ... ^ vk = Σi<i<....<i [alt(v1v2...vk)]ii...i (ei ^ ei ^ .... ^ ei) (7.5.7) v1^ v2^ .....^ vk = Σi<i<....<i det (ei ^ ei ^ .... ^ ei). (7.5.9) ******** Example: Evaluate each of these three expansions for k = 2: T^ = Alt(T) = v1 ^ v2 = Alt(v1v2) = (v1v2 - v2v1)/2 v1 ^ v2 = A = Alt(v1v2) = v1 ^ v2 = det[ (vj)i] = = v1 ^ v2 = Σii(v1)i (v2)i(ei ^ ei) = [Σi(v1)iei] ^ [Σi(v2)iei] = v1 ^ v2 v1 ^ v2 = Σi<i [alt(v1v2]ii(ei ^ ei) = [(v1)i(v2)i - (v1)i(v1)i] (ei ^ ei) v1^ v2 = Σi<i det . (7.5.10) The first line is an identity while the last two lines agree with (4.3.12). ******************* 5.6 The Tensor Product of two or more tensors in T(V) The tensor algebra T(V) shown in (5.4.1) is closed under both + and . It seems evident how one would add two tensors of T(V) of the form (5.4.2), but how would one multiply two tensors? Consider two tensors of rank k and k' expanded as in (5.2.1), T = Σii....i Tii....i (ei ei ..... ei) . rank k, T ϵ Vk (5.6.1) S = Σjj....j Sjj....j (ej ej ..... ej) rank k', S ϵ Vk' . (5.6.2) Multiplying these together with one gets, using the rules (5.3.1), TS = [Σii....iTii....i(ei ei ..... ei)][Σjj....jSjj....j(ej ej ..... ej)] = Σii....i Σjj....jTii....i Sjj....j(ei ei ..... ei) (ej ej ..... ej) (5.6.3) = Σii....ijj....jTii....i Sjj....j(ei ei ..... ei ej ej ..... ej) (5.6.4) = Σii....iii....i[Tii....i Sii....i] (ei ei ...... ei) . (5.6.5) Notice that the (2.8.22) associativity of is used going from (5.6.3) to (5.6.4). In the last step (5.6.5), we renamed the dummy jr summation indices so that j1 = ik+1 , j2 = ik+2 and so on. Rewriting the last equation, TS = Σii....i[Tii....i Sii....i] (ei ei ...... ei) . (5.6.6) Clearly the product TS is an element of Vk+k'with the following tensor components, (TS)ii....i = Tii....i Sii....i . (5.6.7) Thus the tensor on the left is the outer product of the two tensors on the right, similar to (3.1.13). Both sides of this equation of course transform in the same manner in the sense of (2.1.6). We have just shown that if T ϵ Vk and S ϵ Vk', then TS ϵ Vk+k. Thus we have strengthened the claim made in (5.4.5) that T(V) is closed under the operation : the tensor product of an Vk tensor with an Vk' tensor lies in Vk+k' which is in T(V). It is easy then to show that this is true for the tensor product of any two multivectors as defined below (5.4.6). The tensor product of three or more tensors works in the same fashion. For example, if R has rank k" then we find that TSR ϵ Vk+k'+k" with the following outer product relation, (TSR)ii....i = Tii....i Sii....iRii....i . (5.6.8) Using the ordinary multiindices of (5.2.2-4), the above equations can be considerably compacted : T = ΣI TI eI S = ΣJ SJ eJ I = {i1, i2, .. ik} J = {j1, j2, .. jk'} (5.6.9) (5.6.1) (5.6.2) eI ≡ ei ei ..... ei eJ ≡ ej ej ..... ej TI = Tii....i SJ = Sjj....j TS = ΣI,J TISJ eIeJ = ΣI (TS)I eI (TS)I = TISI' (5.6.10) (5.6.3) (5.6.6) (5.6.7) I = {i1, i2, .. ik+k'} eI ≡ ei ei ..... ei I = {i1, i2, .. ik} I' = {ik+1, ik+2, .. ik+k'} TSR = ΣI,J,K TISJRK eIeJeK = ΣI (TSR)I eI (TSR)I = TISI'RI" (5.6.11) (5.6.8) I = {i1, i2, .. ik+k'+k"} eI ≡ ei ei ..... ei I = {i1, i2, .. ik} I' = {ik+1, ik+2, .. ik+k'} I" = {ik+k'+1, ik+k'+2, .. ik+k'+k"} In the more systematic notation discussed at the end of Section 6, the tensor product of N tensors Ti of rank ki is given by T1T2...TN = ΣI (T1IT2I .... TNI) eI = ΣI (T1T2....TN)I eI (5.6.12) Ti = tensor of rank ki Ii = multiindex range of ir values for tensor Ti I1 = {i1, i2.....ik}, I2 = {ik+1, ik+2.....ik+k}, etc. I = I1 I2 .... IN = {i1, i2......ik+k+...k} eI = ei ei ..... ei The rank of this product tensor is then K = Σi=1N ki and the tensor is an element of VK. Example 1: The tensor product of two rank-1 tensors. TS = Σii[Ti Si] (ei ei) = Σij TiSj (eiej) = Σij (TS)ij (eiej) (TS)ab = Σij TiSj (eiej)ab = Σij TiSj δiaδjb = TaSb (5.6.13) Example 2: The tensor product of two rank-2 tensors. TS = Σiiii Tii Sii (ei ei ei ei ) (TS )abcd = TabScd (5.6.14) In both examples the evaluation of components produces the expected outer product forms. Special cases of the tensor product TS. Assume T and S have rank k and k'. If S = κ' ϵ K = a scalar, then rank(S) = k' = 0 and (5.6.4) reads, TS = Σii....ijj....jTii....i Sjj....j(ei ei ..... ei ej ej ..... ej) = Σii....iTii....i (κ') (ei ei ..... ei) = κ'T and ST = Σjj....jii....i Sjj....jTii....i (ej ej ..... ej ei ei ..... ei) = Σii....i (κ') Tii....i ( ei ei ..... ei) = κ'T so we find that TS = ST = κ'T . If T = κ and S = κ', the result above would be TS = κκ' and ST = κ'κ and so TS = ST = κκ'. Thus, TS = κS = ST = Sκ = κS if T = κ ϵ V0 TS = Tκ' = ST = κ'T = κ'T if S = κ' ϵ V0 TS = κκ' = ST = κ'κ = κκ' if T,S = κ,κ' ϵ V0 (5.6.15) *********** Consider these Eq. (7.4.7) ordered expansions of general tensors T^ and S^ of rank k and k', T^ = Σi<i<....<i Aii....i (ei^ ei .....^ ei) rank k, T^ ϵ Lk (7.9.1) Σ'IAIe^I S^ = Σj<j<....<j Bjj....j (ej^ ej .....^ ej) rank k', S^ ϵ Lk' (7.9.2) Σ'JBJe^J where from (7.4.16), Aii...i = k! [Alt(T)]ii...i or A = k! Alt(T) Bjj....j = k! [Alt(S)]jj....j or B = k! Alt(S) . (7.9.3) **************** The next step is to express (7.9.4) as an ordered sum rather than a symmetric sum. Momentarily replace TS by the symbol T,' so that (7.9.4) reads, T^^ S^ = Σii....iT ' ii...i(ei^ ei ......^ ei) . (7.9.7) This symmetric sum can be replaced by the ordered sum (7.4.7) T^^ S^ = Σi<i<....<i A' ii...i(ei^ ei ......^ ei) (7.9.8) where, according to (7.4.16), A' ii...i= k! [Alt(T ' )]ii...i . // A ' = alt(T ') (7.9.9) Restoring T ' = TS gives this ordered expansion for the wedge product of tensors T^ and S^, T^^ S^ = Σi<i<....<i [k!Alt(TS)]ii...i (ei^ ei ......^ ei) . with (7.9.10) [k!Alt(T^^ S^)]ii...i = ΣP (-1)S(P) Tii...i Sii...i = ΣP (-1)S(P) [TS]ii...i Here ΣP is over all permutations of (the subscripts of) {i1, i2, ....ik+k'}. Example 1: k = 1 and k' = 1 so that T and S are just vectors (n ≥2). The above equations reduce to T^^ S^ = Σi<i [2!Alt(T^^ S^)]ii (ei^ ei) . [2!Alt(TS)]ii = ΣP (-1)S(P) TiSi = TiSi - SiTi (7.9.11) so that T^^ S^ = Σi<i [TiSi - SiTi](ei^ ei) = Σi<j [TiSj - SiTj] (ei^ ej) (7.9.12) in agreement with (4.3.12). Example 2: k = 2 and k' = 1 so that T is a rank-2 tensor and S is still a vector (n ≥ 3) T^^ S^ = Σi<i<i3 [alt(TS)]iii (ei^ ei^ ei) . (7.9.13) From (7.9.10), [3!Alt(TS)]iii = ΣP (-1)S(P) TiiSi) (7.9.14) [3!Alt(TS)]abc = TabSc - TacSb + TcaSb - TcbSa + TbcSa - TbaSc // see (7.4.14) so that T^^ S^ = Σa<b<c {TabSc - TacSb + TcaSb - TcbSa + TbcSa - TbaSc} (ea^ eb^ ec) (7.9.15) where the summation range is 1 ≤ a < b < c ≤ n. In the case dim(V) = n = 3, there is only one term which has a = 1, b = 2 and c = 3, T^^ S^ = {T12S3 - T13S2 + T31S2 - T32S1 + T23S1 - T21S3} (e1^ e2^ e3) // n = 3 **************** (c) Multiindex Notation Rehash of Section 7.10 For "practice" and for use in the next section, we give an abbreviated copy, paste and edit rehash of Section 7.10 above using the multiindex notation. Whenever there is a confusion, one must write things out in detail. Equation numbers from above are shown in italics. __________________________________________________________________________________ Consider these Eq. (7.4.7) ordered expansions of general L(V) tensors T^ and S^ of rank k and k', T^ = Σ'I AI e^I I = {i1, i2, .. ik} e^I ≡ (ei^ ei .....^ ei) rank k S^ = Σ'J BJ e^J J = {j1, j2, .. jk'} e^J ≡ (ej^ ej .....^ ej) rank k' (7.9.1) where from (7.5.2), AI = [alt(T)]I BJ = [alt(S)]J . (7.9.2) The corresponding symmetric expansions (7.4.4) of T^ and S^ are given by, T^ = ΣI TI e^I // symmetric expansions rank k, T^ ϵ Lk S^ = ΣJ SJ e^J . rank k', S^ ϵ Lk' (7.9.3) We form the wedge product of these two tensors in a manner similar to the equations leading to (5.6.6), T^^S^ = [ ΣITI e^I] ^ [ ΣJ SJ e^J] (a) = ΣI ΣJTI SJ (e^I^e^J) (b) = ΣI,JTISJ (e^I^e^J) I = {i1, i2, .. ik} (c) = ΣI,I'TI SI' (e^I^e^I') I ≡ {i1...ik+k'}, I' ≡ {ik+1...ik+k'} so I I' = I (d) = ΣI TI SI' e^I e^I ≡ (ei^ ei ......^ ei) (e) = ΣI (TS)I e^I // symmetric expansion of T^S (7.9.4) where, as in (5.6.7), we use in the last line the standard outer product notation, (TS)I = TI SI' . (7.9.5) Eq. (7.9.4) shows that the product T^^S^ is an element of Lk+k', T^^S^ = ΣI (T^S)I e^I with (T^S)I = (TS)I . (7.9.6) The next step is to express (7.9.4) as an ordered sum rather than a symmetric sum. Momentarily replace TS by the symbol T,' so that (7.9.4) reads, T^^S^= ΣIT 'I e^I . (7.9.7) This symmetric sum can be replaced by the ordered sum (7.4.1) T^^S^ = Σ'I A' I e^I (7.9.8) where, according to (7.4.12) and then (7.5.2), A' I = [k!Alt(T ' )]I . // A ' = k! Alt(T ') (7.9.9) Restoring T ' = TS gives this ordered expansion for the wedge product of tensors T^ and S^, T^^S^ = Σ'I [k!Alt(TS)]I e^I. (7.9.10) [k!Alt(TS)]I = ΣP (-1)S(P) TP(I)S P(I') where ΣP is over all (k+k')! permutations of (the subscripts of) {i1, i2, ....ik+k'}. Commutivity Rule for the Wedge Product of two tensors T^ and S^ Recall the expansion of T^S from (7.9.4)(b), T^^S^ = ΣI,J TISJ (e^I ^ e^J) (7.9.17) Swapping T↔S , k↔k' and I ↔ J gives the following form for the wedge product S^^T^ , S^^T^ = ΣJ,I SJTI (e^J ^ e^I) = ΣI,J TISJ (e^J ^ e^I) . (7.9.18) Equations (7.9.17) and (7.9.18) are identical except for the last factor involving the basis vectors. (e^J ^ e^I) = [(-1)k']k (e^I ^ e^J) // slide e^J left through e^I , as in (7.9.19) (7.9.19) S^^T^ = (-1)kk'T^^S^ ranks of the two tensors are k and k' . (7.9.20) ________________________________________________________________________________ ************* 7.10 The Wedge Product of two tensors (a) Wedge Product of two tensors T and S Consider these Eq. (7.4.1) ordered expansions of general tensors T and S of rank k and k', T = Σi<i<....<i Aii....i (ei^ ei .....^ ei) rank k, T ϵ Lk S = Σj<j<....<j Bjj....j (ej^ ej .....^ ej) rank k', S ϵ Lk' (7.10.1) where from (7.5.2), Aii...i = [alt(T)]ii...i or A = alt(T) Bjj....j = [alt(S)]jj....j or B = alt(S) . (7.10.2) The corresponding symmetric expansions (7.4.3) of T and S are given by, T = Σii....i Tii....i (ei^ ei .....^ ei) . rank k, T ϵ Lk S = Σjj....j Sjj....j (ej^ ej .....^ ej) . rank k', S ϵ Lk' . (7.10.3) We form the wedge product of these two tensors in a manner similar to the equations leading to (5.6.6) : T^S = [ Σii....iTii....i (ei^ ei .....^ ei)]^[ Σjj....j Sjj....j (ej^ ej .....^ ej)] (a) = Σii....i Σjj....jTii....i Sjj....j(ei^ ei .....^ ei) ^ (ej^ ej .....^ ej) (b) = Σii....ijj....jTii....i Sjj....j(ei^ ei .....^ ei^ ej^ ej .....^ ej) (c) = Σii....iii....i[Tii....i Sii....i] (ei^ ei ......^ ei) (d) = Σii....i[Tii....i Sii....i] (ei^ ei ......^ ei) (e) = Σii....i[TS]ii....iii....i(ei^ ei ......^ ei) (7.10.4) where, as in (5.6.7), we use in the last line the standard outer product notation, [TS]ii...iii...i = Tii...i Sii...i . (7.10.5) Notice that the (7.9.4) associativity of ^ is used going from (a) to (b). Eq. (7.10.4) shows that the product T^S is an element of Lk+k' with the following tensor components, T^S = Σii....i (T^S)ii....i(ei^ ei ......^ ei) (T^S)ii....i = [TS]ii...i. (7.10.6) Thus we have strengthened the claim made in (7.9.10) that L(V) is closed under the operation ^ : the wedge product of an Lk tensor with an Lk' tensor lies in Lk+k' which is in L(V). It is easy then to show that this is true for the wedge product of any two multivectors as defined below (7.9.11). The next step is to express (7.10.4) as an ordered sum rather than a symmetric sum. Momentarily replace TS by the symbol T,' so that (7.10.4) reads, T^S = Σii....iT ' ii...i(ei^ ei ......^ ei) . (7.10.7) This symmetric sum can be replaced by the ordered sum (7.4.1) T^S = Σi<i<....<i A' ii...i(ei^ ei ......^ ei) (7.10.8) where, according to (7.4.12) and then (7.5.2), A' ii...i = ΣP (-1)S(P) T ' P(i)P(i)...P(i) = [alt(T ' )]ii...i . // A ' = alt(T ') (7.10.9) Restoring T ' = TS gives this ordered expansion for the wedge product of tensors T and S, T ^ S = Σi<i<....<i [alt(TS)]ii...i (ei^ ei ......^ ei) . with (7.10.10) [alt(TS)]ii...i = ΣP (-1)S(P) TP(i)P(i)...P(i)S P(i)P(i)..P(i) where ΣP is over all permutations of (the subscripts of) {i1, i2, ....ik+k'}. Example 1: k = 1 and k' = 1 so that T and S are just vectors (n ≥2). The above equations reduce to T ^ S = Σi<i [alt(TS)]ii (ei^ ei) . [alt(TS)]ii = ΣP (-1)S(P) TP(i)SP(i) = TiSi - SiTi (7.10.11) so that T ^ S = Σi<i [TiSi - SiTi](ei^ ei) = Σi<j [TiSj - SiTj] (ei^ ej) // remove italics due to (7.4.16) (7.10.12) in agreement with (4.3.12). Example 2: k = 2 and k' = 1 so that T is a rank-2 tensor and S is still a vector (n ≥ 3) T ^ S = Σi<i<i3 [alt(TS)]iii (ei^ ei^ ei) . (7.10.13) From (7.10.10), [alt(TS)]iii = ΣP (-1)S(P) TP(i)P(i)SP(i) (7.10.14) [alt(TS)]abc = TabSc - TacSb + TcaSb - TcbSa + TbcSa - TbaSc // see (7.4.14) so that T ^ S = Σa<b<c {TabSc - TacSb + TcaSb - TcbSa + TbcSa - TbaSc} (ea^ eb^ ec) (7.10.15) where the summation range is 1 ≤ a < b < c ≤ n. In the case dim(V) = n = 3, there is only one term which has a = 1, b = 2 and c = 3, T ^ S = {T12S3 - T13S2 + T31S2 - T32S1 + T23S1 - T21S3} (e1^ e2^ e3) // n = 3 Special cases of the wedge product T ^ S. Assume T and S have rank k and k'. If S = κ' ϵ K = a scalar, then rank(S) = k' = 0 and (7.10.4b) reads, T^S = Σii....ijj....jTii....i Sjj....j (ei^ ei .....^ ei^ ej^ ej .....^ ej) = Σii....iTii....i (κ') (ei^ ei .....^ ei) = κ'T and S^T = Σjj....jii....i Sjj....j Tii....i ( ej^ ej .....^ ej^ ei^ ei .....^ ei) = Σii....i (κ') Tii....i ( ei^ ei .....^ ei) = κ'T so we find that T^S = S^T = κ'T . If T = κ and S = κ', the result above would be T^S = κκ' and S^T = κ'κ and so T^S = S^T = κκ'. Thus, T^S = κ^S = S^T = S^κ = κS if T = κ ϵ V0 T^S = T^κ' = S^T = κ'^T = κ'T if S = κ' ϵ V0 T^S = κ^κ' = S^T = κ'^κ = κκ' if T,S = κ,κ' ϵ V0 (7.10.16) These special case results are seen to be the same as those for TS shown in (5.6.15). (b) Commutivity Rule for the Wedge Product of two tensors T and S Recall the expansion of T^S from (7.10.4)(b), T^S = Σii....ijj....jTii....i Sjj....j(ei^ ei .....^ ei^ej^ ej .....^ ej) (7.10.17) Swapping T↔S , T↔S, k↔k' and i ↔ j gives the following form for the wedge product S^T , S^T = Σjj....jii....iSjj....j Tii....i (ej^ ej .....^ ej^ei^ ei .....^ ei) = Σii....ijj....j Tii....i Sjj....j(ej^ ej .....^ ej^ei^ ei .....^ ei) (7.10.18) [ Note: In multiindex these equations are: T^S = ΣI,JTISJ eI^eJ and S^T = ΣI,JTISJ eJ^eI .] Equations (7.10.17) and (7.10.18) are identical except for the last factor involving the basis vectors. Consider the basis vector factor appearing in (7.10.18), (ej^ ej .....^ ej^ ei^ ei .....^ ei) . // eJ^eI To make this match the basis factor in (7.10.17), we have to slide all the red basis vectors to the left through all the black basis vectors. Each time a red passes through a black, we pick up a minus sign due to the rule (7.2.4). Thus, (ej^ ej .....^ ej^ ei^ ei .....^ ei) = (-1)k' ei ^ (ej^ ej .....^ ej^ ei .....^ ei) = (-1)k' (-1)k' ei ^ ei ^ (ej^ ej .....^ ej .....^ ei) = etc. = = [(-1)k']k ( ei^ ei .....^ ei ^ ej^ ej .....^ ej) . // (-1)kk' eI^eJ (7.10.19) Inserting (7.10.19) into (7.10.18) and comparing with (7.10.17) we arrive at this well-known result, sometimes called "graded commutivity" since the nature of the commutivity depends on the grades (ranks) of the two tensors, S^T = (-1)kk'T^S ranks of the two tensors are k and k' . (7.10.20) The wedge product of two tensors commutes if kk' is even, and anticommutes if kk' is odd. Example: If k = k' = 1, (-1)kk' = -1 and we recover the simple rule for vectors S^T = - T^S // S and T are rank-1 tensors (vectors) (7.10.21) as first stated in (4.3.2). One must keep in mind that the result S^T = - T^S is not valid for arbitrary tensors S and T. Examples: If k = 0 so T = κ, rule (7.10.20) says S^T = T^S, consistent with (7.10.16) line 1. If k=k'=0 so T = κ and S = κ', rule (7.10.20) says S^T = T^S, consistent with (7.10.16) line 3. (7.10.22) (c) Multiindex Notation Rehash of Section 7.10 For "practice" and for use in the next section, we give an abbreviated copy, paste and edit rehash of Section 7.10 above using the multiindex notation. Whenever there is a confusion, one must write things out in detail. Equation numbers from above are shown in italics. __________________________________________________________________________________ Consider these Eq. (7.4.1) ordered expansions of general tensors T and S of rank k and k', T = Σ'I AI eI I = {i1, i2, .. ik} eI ≡ (ei^ ei .....^ ei) rank k S = Σ'J BJ eJ J = {j1, j2, .. jk'} eJ ≡ (ej^ ej .....^ ej) rank k' (7.10.1) where from (7.5.2), AI = [alt(T)]I BJ = [alt(S)]J . (7.10.2) The corresponding symmetric expansions (7.4.3) of T and S are given by, T = ΣI TI eI // symmetric expansions rank k, T ϵ Lk S = ΣJ SJ eJ . rank k', S ϵ Lk' (7.10.3) We form the wedge product of these two tensors in a manner similar to the equations leading to (5.6.6), T^S = [ ΣITI eI] ^ [ ΣJ SJ eJ] (a) = ΣI ΣJTI SJ (eI^eJ) (b) = ΣI,JTISJ (eI^eJ) I = {i1, i2, .. ik} (c) = ΣI,I'[TI SI'] (eI^eI') I ≡ {i1...ik+k'}, I' ≡ {ik+1...ik+k'} so I I' = I (d) = ΣI [TI SI'] eI eI ≡ (ei^ ei ......^ ei) (e) = ΣI (TS)I eI // symmetric expansion of T^S (7.10.4) where, as in (5.6.7), we use in the last line the standard outer product notation, (TS)I = TI SI' . (7.10.5) Eq. (7.10.4) shows that the product T^S is an element of Lk+k', T^S = ΣI (T^S)I eI with (T^S)I = (TS)I . (7.10.6) The next step is to express (7.10.4) as an ordered sum rather than a symmetric sum. Momentarily replace TS by the symbol T,' so that (7.10.4) reads, T^S = ΣIT 'I eI . (7.10.7) This symmetric sum can be replaced by the ordered sum (7.4.1) T^S = Σ'I A' I eI (7.10.8) where, according to (7.4.12) and then (7.5.2), A' I = ΣP (-1)S(P) T ' P(I) = [alt(T ' )]I . // A ' = alt(T ') (7.10.9) Restoring T ' = TS gives this ordered expansion for the wedge product of tensors T and S, T ^ S = Σ'I [alt(TS)]I eI. (7.10.10) [alt(TS)]I = ΣP (-1)S(P) TP(I)S P(I') where ΣP is over all (k+k')! permutations of (the subscripts of) {i1, i2, ....ik+k'}. Commutivity Rule for the Wedge Product of two tensors T and S Recall the expansion of T^S from (7.10.4)(b), T^S = ΣI,J TISJ (eI ^ eJ) (7.10.17) Swapping T↔S , T↔S, k↔k' and I ↔ J gives the following form for the wedge product S^T , S^T = ΣJ,I SJTI (eJ ^ eI) = ΣI,J TISJ (eJ ^ eI) . (7.10.18) Equations (7.10.17) and (7.10.18) are identical except for the last factor involving the basis vectors. (eJ ^ eI) = [(-1)k']k (eI ^ eJ) // slide eJ left through eI (7.10.19) S^T = (-1)kk'T^S ranks of the two tensors are k and k' . (7.10.20) ________________________________________________________________________________ 7.11 The Wedge Product of N tensors in L(V) First, we mimic the multiindex development just above to obtain the wedge product of three tensors: T = ΣI TI eI rank k I = {i1, i2, .. ik} S = ΣJ SJ eJ rank k' I' ≡ {ik+1...ik+k'} R = ΣK RK eK rank k" I" = {ik+k'+1, ik+k'+2, .. ik+k'+k"} (7.11.1) T^S^R = (ΣITIeI) ^ (ΣJSJeJ) ^ (ΣKRKeK) = ΣI,J,K TISJRK (eI^eJ^eK) . // symmetric expansion = ΣI,I',I" TISI'RI" (eI^eI'^eI") = ΣI(TSR)I eI (TSR)I = TISI'RI" (7.11.2) = Σ'I [alt(TSR)]IeI // ordered expansion, eI ≡ (ei^ ei ......^ ei) where [alt(TSR)]I = ΣP (-1)S(P) TP(I)SP(I')RP(I") (7.11.3) Sample reordering rule: S^T^R = ΣI,J,K TISJRK (eJ^eI^eK) = ΣI,J,K TISJRK [ (-1)kk'(eI^eJ^eK)] = (-1)kk' T^S^R (7.11.4) Systematic Tensor Products To develop a more systematic approach, consider the first three tensors in a product sequence, 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} . (7.11.5) Define the following "cumulative ranks", κ1 = k1 // cumulative ranks, as in κ2 = k1+ k2 κ3 = k1+ k2 + k3 ... κN = k1 + k2 + ... + kN = Σi=1N ki . (7.11.6) Then rewrite (7.11.5), T1 = tensor of rank k1 I1 = {i1, i2.....iκ} T2 = tensor of rank k2 I2 = {iκ+1, iκ+2.....iκ} T3 = tensor of rank k3 I3 = {iκ+1, iκ+2.....iκ} ... TN = tensor of rank kN IN = {iκ+1,iκ+2.....iκ} . (7.11.7) Define, T1P(I) ≡ T1P(i)T1P(i)...T1P(i) T2P(I) ≡ T2P(i)T2P(i) ....T2P(i) T3P(I) ≡ T3P(i)T3P(i) ....T3P(i) ... TNP(I) ≡ TNP(i)TNP(i) ....TNP(i) . (7.11.8) We can now write out the product of any number of tensors. In each case we show the symmetric expansion first, then the ordered expansion. Wedge Product of 2 Tensors (7.11.9) T1^T2 = ΣII T1IT2I (eI^eI) = ΣI (T1T2)IeI T1^T2 = Σ'I [alt(T1T2)]I eI eI ≡ ei^ ei .....^ ei where [alt(T1T2)]I = ΣP (-1)S(P) T1P(I)T2P(I) Wedge Product of 3 Tensors (7.11.10) T1^T2^T3 = ΣIII T1IT2IT3I (eI^eI^eI) = ΣI (T1T2T3)I eI T1^T2^T3 = Σ'I [alt(T1T2T3)]I eI eI ≡ ei^ ei .....^ ei where [alt(T1T2T3)]I = ΣP (-1)S(P) T1P(I)T2P(I)T3P(I) Wedge Product of N Tensors (7.11.11) T1^T2^...^TN = ΣII...I T1IT2I....TNI (eI^eI ... ^eI) = ΣI (T1T2 ....TN)I eI T1^T2^...^TN = Σ'I [alt(T1T2...TN)]I eI eI ≡ ei^ ei .....^ ei where [alt(T1T2...TN)]I = ΣP (-1)S(P) T1P(I)T2P(I)...TNP(I) . Sign Rule for swapping two tensors Swapping two tensors in a tensor product results in either + or - the same tensor, as shown for example in (7.11.4) Consider an example where we have a wedge product of 9 tensors. The eI basis function groups are eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI which goes with T1 ^ T2 ^ T3 ^ T4 ^ T5 ^ T6 ^ T7 ^ T8 ^ T9 . The sign caused by swapping T3 ↔ T7 will be the same as the sign swapping eI ↔eI in the basis function. We do it one step at a time, first sliding the group eI to the left eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI = eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI (-1)kk = eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI (-1)(k+k)k = eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI (-1)(k+k+k)k = eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI (-1)(k+k+k+k)k Now with this as a starting point, we slide eI to the right, one group at a time, eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI = eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI (-1)kk = eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI (-1)k(k+k) = eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI (-1)k(k+k+k) and now we have successfully swapped eI ↔ eI so alsoT3 ↔ T7. The total sign is sign = (-1)m where m = (k6+ k5+ k4+ k3)k7 + (k4+k5+k6)k3 = (k4+k5+k6)(k3+k7)+ k3k7 . (7.11.12) Based on this result, we claim that : Fact: In a product of tensors T1^T2^T3.... of rank k1, k2, k3 ... , if two tensors are swapped Tr ↔ Ts (with r < s), the resulting tensor incurs the following sign relative to the starting tensor, sign = (-1)m where m = (kr+1+kr+2 ...+ks-1)(kr+ks) + krks . (7.11.13) If the sum of the ranks of the two swapped tensor is even, in effect m = krks . Example 1: T1 ^ T2 ^ T3 = (-1)m T2 ^ T1 ^ T3 r = 1 s = 2 m = (0)(k1+k2) + k1k2 = k1k2 (-1)m = (-1)kk (7.11.16) Example 2: T1 ^ T2 ^ T3 = (-1)m T3 ^ T2 ^ T1 r = 1 s = 3 m = (k2)(k1+k3) + k1k3 = k1k2 + k1k3 + k2k3 (-1)m = (-1)kk+kk+kk (7.11.17) Example 3: Suppose all the tensors are vectors with rank = 1. Then the sum of the ranks of any two tensors is 2 which is even, so swapping two of these tensors produces a minus sign phase = (-1)m where m ≈ krks = 1*1 = 1 in agreement with (7.2.4). (7.11.18) ************ Application to Tensors In the realm of tensors, we apply the above theorems identifying, f[z] = f[1,2....k] = Tii...i = components of a rank-k tensor T (C.1.7) F[Z] = F [k+1,k+2....k+k'] = Sii...i = components of a rank-k' tensor S (C.3.7) (C.5.2) The first theorem becomes: 1. ΣP (-1)σS(P) f[P(z)] F[P(Z)] = ΣP (-1)σS(P) f^[P(z)] F[P(Z)] (C.1.10) or ΣP (-1)σS(P) TP(i)P(i)...P(i) SP(i)P(i)....P(i) = ΣP (-1)σS(P) T^P(i)P(i)...P(i) SP(i)P(i)....P(i) or ΣP (-1)σS(P) (TS)P(i)P(i)...P(i) // see (5.6.7), outer product idea = ΣP (-1)σS(P))(T^S)P(i)P(i)...P(i)) (C.5.3) Using Alt and Sym from Chapter 7, [Alt(X)]ii...i ≡ (1/k!) ΣP (-1)S(P) XP(i)P(i)...P(i) (7.5.3) [Sym(X)]ii...i ≡ (1/k!) ΣP XP(i)P(i)...P(i), (7.5.9) (C.5.4) we can state the above theorem as σ = 1: Alt[(TS)]ii...i = Alt[(T^S)]ii...i or Alt[(TS)] = Alt[(T^S)] (C.5.5) σ = 0: Sym[(TS)]ii...i = Sym[(TsS)]ii...i or Sym[(TS)] = Sym[(TsS)] . (C.5.6) Here we slightly alter our notation, so T^ applies only for σ = 1 and is a totally antisymmetrized tensor, whereas we use T^ → Ts for σ = 0 to indicate a totally symmetric tensor. The three theorems are all similar and we can then summarize everything in two lines: Alt[(TS)] = Alt[(T^S)] = Alt[(TS^)] = Alt[(T^S^)] Sym[(TS)] = Sym[(TsS)] = Sym[(TSs)] = Sym[(TsSs)] . (C.5.7) Intuitively these equations are easily interpreted: If one is going to totally (anti) symmetrize a tensor product, the act of pre-(anti)-symmetrizing one or more tensors of the product makes no difference. This is what we expect, but now we have a formal proof for the case of two tensors. ************* C.6 Applications of the Theorems to Tensor Functions This section is very similar to Section C.5 above. Details are not repeated so the equation number sequence has holes in it. In the realm of tensor functions, we apply the theorems (C.5.1) this time identifying, f[z] = f[1,2....k] = T(vi,vi, .... vi) = components of a rank-k tensor function T (C.1.7) F[Z] = F [k+1,k+2....k+k'] = S(vi,vi, .... vi) = components of a rank-k' tensor T (C.3.7) (C.6.2) The first theorem becomes: 1. ΣP (-1)σS(P) f[P(z)] F[P(Z)] = ΣP (-1)σS(P) f^[P(z)] F[P(Z)] (C.1.10) or ΣP (-1)σS(P) T(vi,vi, .... vi) S(vi,vi, .... vi) = ΣP (-1)σS(P) T^(vi,vi, .... vi) S(vi,vi, .... vi) or ΣP (-1)σS(P) (TS)(v1,v2...vk+k') // see (6.7.1) = ΣP (-1)σS(P))(T^S)(v1,v2...vk+k') (C.6.3) Using Alt and Sym from Chapter 8, [Alt(X)](v1,v2....vk) ≡ (1/k!) ΣP (-1)S(P)X(vP(1),vP(2)....vP(k)) (8.5.11) [Sym(X)](v1,v2....vk) ≡ (1/k!) ΣP X(vP(1),vP(2)....vP(k)) (8.5.11) (C.6.4) we can state the above theorem as σ = 1: Alt[(TS)](v1,v2...vk+k') = Alt[(T^S)](v1,v2...vk+k') or Alt[(TS)] = Alt[(T^S)] (C.6.5) σ = 0: Sym[(TS)](v1,v2...vk+k') = Sym[(TsS)](v1,v2...vk+k') or Sym[(TS)] = Sym[(TsS)] (C.6.6) Our three theorems *** are all similar and we can then summarize everything in two lines: Alt[(TS)] = Alt[(T^S)] = Alt[(TS^)] = Alt[(T^S^)] Sym[(TS)] = Sym[(TsS)] = Sym[(TSs)] = Sym[(TsSs)] (C.6.7) The generalized results are then Alt[(T1)(T2) ...... (TN)] = Alt[(T1)a(T2)a ...... (TN)a] (C.6.13) where each ai can independently be a blank, (Ti) , or can be a ^ , (Ti)^. Similarly Sym[(T1)(T2) ...... (TN)] = Sym[(T1)a(T2)a ...... (TN)a] where each ai can independently be a blank, (Ti) , or can be an s, (Ti)s. (C.6.14) Comment: The final equations (C.6.13) and (C.6.14) are identical to (C.5.13) and (C.5.14) if we interpret the objects (like Alt[(T1)(T2) ...... (TN)] ) as abstract tensors. One simple interpretation of this fact is presented in the next section which shows that Section C.5 and Section C.6 treat the same objects in different bases, ********** Quiz Question: In the tensor function T(vi,vi, .... vi), the vector arguments are arbitrary, so they might not be linearly independent. How can one then be sure that |vI> is a complete basis? Suppose all k vectors vi are equal to e1. There is nothing wrong with the object e1 e1 ... e1 being a basis vector in Vk. [The space used in this section is Vk , not Lk. *********** R [T(vj,vj, .... vj)] = R [ Tjj....j] = ********************** C.5 Unified View of Tensors and Tensor Functions: Basis Change and Transformation In this section multiindex notations are shown in red to the right. (a) Basis In the bra-ket notation of quantum mechanics (Paul Dirac, 1939), a rank-k functional T is represented by the bra <T| which is an element of the dual space V*k. Meanwhile, the basis elements on the space Vk are written as kets, | vi,vi, .... vi > = |vi> |vi> ..... |vi> | vI> (C.5.1) where the ir are labels, not components. Each vi is a vector having n components, where n ≥ k. The tensor function T(vi,vi, .... vi) is then represented by the application of the functional <T| to vectors in Vk so that, <T | vi,vi, .... vi > = T(vi,vi, .... vi) . T(vI) = <T | vI > (C.5.2) Due to the tensor product (of vector spaces) construction of the "ket" shown in (C.5.1), the function shown in (C.5.2) is manifestly k-multilinear. Because Vk is a real Hilbert space (not complex), one has <a|b> = <b|a> for inner products, and then we can rewrite the function T(vi,vi, .... vi) in a more conventional (from a bra-ket viewpoint) form, T(vi,vi, .... vi) = < vi,vi, .... vi | T > . T(vI) = <vI | T > (C.5.3) The right side is the "projection" of a Hilbert Space "vector" | T > "onto the basis" < vi,vi, .... vi |. This is similar to a basic quantum mechanics k-particle wavefunction in the coordinate representation, ψ(r1, r2...rk) = <r1, r2...rk| ψ> (C.5.4) where ri is the position of particle i. Comments: 1. It happens that the Hilbert Space is complex for quantum mechanics so <a|b> = <b|a>*. 2. If the k particles are electrons or other half-integral spin particles which are in an "symmetric spin state", then the wavefunction (C.5.4) must be replaced by [Alt(ψ)](r1, r2...rk) in order to make it be totally antisymmetric in the coordinates ri, as required by "Fermi statistics" for half-integral spin particles. We mention this just to show that the Alt operator and the permutation group in general have important applications in quantum mechanics. 3. Mathematicians and physicists have different views concerning which side of an inner product <a|b> is which, see wiki. We are in the physics camp. For real Hilbert spaces both views are the same. The covariant tensor Tii....i is, we claim, this special case of (C.5.3), Tii....i = < ei,ei, .... ei | T > . TI = <eI | T > (C.5.5) From this one would conclude that Tii....i = T(ei,ei, .... ei) TI = T(eI) (C.5.6) in agreement with (6.2.1a). The contravariant form is then Tii....i = T(ei,ei, .... ei) = < ei,ei, .... ei | T > . TI = T(eI) = <eI| T> (C.5.7) One can say that the tensor Tii....i and the tensor function T(vi,vi, .... vi) are both representations of the same abstract tensor T/T in two different bases, |vI> and |eI>. Notice that <vI|vJ> = < vi|vj>< vi|vj> .... < vi|vj> = (vi vj)(vi vj) .... (vi vj) = δijδij...δij // see (2.3.2) for basis {vr} with dual basis {vr} = δIJ . // orthonormal basis in the multiindex sense Since this result is general, it applies in particular to the basis |eI>, so <eI|eJ> = <vI|vJ> = δIJ . (C.5.8) (b) Basis change matrix The basis-change transformation matrix between the |vI> and |eI> bases is given by, MIJ ≡ < ei,ei, .... ei | vj,vj, .... vj > MIJ = <eI|vJ> (C.5.9) = < ei|vj>< ei|vj> .... < ei|vj> // inner products = (ei vj)(ei vj) .... (ei vj) = λi(vj)λi(vj) .... λi(vj) // see (2.11.3) = (vj)i (vj)i ...(vj)i // see (2.11.7) (C.5.10) = (vJ)I . // using a multiindex notation shown below (7.8.2) Entirely in multiindex notation, MIJ = <eI|vJ> = (vJ)I // mixed, see (2.1.6) line 2 or (C.5.11) MIJ = <eI|vJ> = (vJ)I . // pure covariant, see (2.1.6) line 4 The transpose is then, (MT)JI = MIJ = <eI|vJ> = <vJ| eI> // Hilbert Space is real (MT)JI = MIJ = <eI|vJ> = <vJ| eI> . (C.5.12) In the bra-ket notation completeness of an orthonormal basis is expressed this way: 1 = ΣJ |eJ><eJ| = ΣJ |eJ><eJ| = ΣJ |vJ><vJ| = ΣJ |vJ><vJ| (C.5.13) Proof: (example) Consider a general Vk tensor T : (1) |T> = 1|T> = ΣJ |eJ><eJ| T> = ΣJ TJ |eJ> // so basis |eJ> must be complete (2) |eI> = 1|eI> = ΣJ |eJ><eJ|eI> = ΣJ |eJ>δJI = |eI> // why orthonormal needed Therefore the up-tilt basis-change matrix M is real orthogonal, meaning MMT = 1 or MT = M-1 : (MMT)IK = ΣJ MIJ(MT)JK = ΣJ <eI|vJ><vJ| eK> = <eI| (ΣJ|vJ><vJ| )eK> = <eI | 1 | eK> = <eI | eK> = eI eK = δIK // see (2.11.2) or (C.5.14) MMT = 1 . // real orthogonal in the multi-index sense Note: In quantum mechanics with complex Vk, one gets instead MM† = 1 (unitary) . The connection then between the tensors and tensor functions is given by, TI = <eI| T> = <eI| 1 | T> = <eI| ΣJ |vJ><vJ| T> = ΣJ <eI|vJ><vJ| T> = = ΣJ MIJ T(vJ) . (C.5.15) Going the other direction, T(vI) = <vI | T > = <vI | 1 | T > = <vI | ΣJ |eJ><eJ| | T > = ΣJ <vI|eJ> <eJ|T > = ΣJ (MT)IJ TJ . (C.5.16) Example of (C.5.15): T(v1,v2) = Σjj (vj)i (vj)iTii Quiz Question: In the tensor function T(vi,vi, .... vi), the vector arguments are arbitrary, so they might not be linearly independent. How can one then be sure that |vI> is a complete basis? Suppose all k vectors vi are equal to e1. There is nothing wrong with the object e1 e1 ... e1 being a basis vector in Vk. [The space used in this section is Vk , not Lk. ] (c) Transformations of tensors and tensor functions Imagine taking the set of vectors {vi} and rotating them Under a Chapter 2 transformation from x-space to x'-space, arbitrary basis vectors ei transform to new basis vectors e'i = Rijej where R is the linearized form of a general transformation x' = F(x). We shall interpret this in bra-ket notation to say <e'i| = Rij <ej| . Then for a tensor product of the form (C.5.1) one has < e'i,e'i, .... e'i | = RijRij.....Rij < ej,ej, .... ej| . <e'I| = RIJ <eJ| (C.5.17) Closing the above "bra" equation from the right with "ket" |T> gives, < e'i,e'i, .... e'i | T> = RijRij.....Rij < ej,ej, .... ej| T> (C.5.18) According to (C.5.7) we interpret this to state, T'jj....j = RijRij.....Rij Tjj....j T'I = RIJ TJ (C.5.19) This is the usual rule for the transformation of a rank-k tensor, as shown for example in the last line of (2.1.6). Under our transformation, the vi, being vectors, also transform as <v'i| = Rij <vj|, so that STOP! This is only valid if the vi vectors form a basis. < v'i,v'i, .... v'i | = RijRij.....Rij < vj,vj, .... vj| . <v'I| = RIJ <vJ| (C.5.20) Closing this bra equation with the ket |T> then gives < v'i,v'i, .... v'i |T> = RijRij.....Rij < vj,vj, .... vj| T> . (C.5.21) According to (C.5.3) we interpret this as saying T(v'i,v'i, .... v'i) = RijRij.....Rij T(vj,vj, .... vj) T(v'I) = RIJ T(vJ) (C.5.22) or in more detail T(Rijvj,Rijvj, .... .Rijvj) = RijRij.....Rij T(vj,vj, .... vj) . (C.5.23) The object on the left is some new function of the vectors vi which we can call T', so then T'(vj,vj, .... vj) = RijRij.....Rij T(vj,vj, .... vj) T'(vI) = RIJ T(vJ) (C.5.24) and so the tensor function T is transformed into some new tensor function T'. If the function T is k-multilinear, then so is T' . And if T is totally antisymmetric, then so is T' . ************ Defining Tii....i ≡ [T(e)] ii....i, the expansion of a general rank-k tensor T in terms of contravariant coefficients can be written T = Σii....i Tii....i (eiei... ei) . (2.10.14)  The coefficients can be projected out according to (eiei... ei) T = Tii...i (2.10.15) with an appropriate generalization of the dot product to Vk , (v1v2...vk) (u1u2...uk) ≡ Σii....i (v1v2...vk)ii....i (u1u2...uk)ii....i = Σii....i (v1)i(v2)i... (vk)i (u1)i(u2)i... (uk)i = (v1 u1) (v2 u2) .... (vk uk) . (2.10.15) In ordinary multiindex notation, I ≡ i1, i2...ik eI ≡ eiei... ei eI ≡ eiei... ei TI ≡ Tii...i (2.10.16) the above equations can be compactly written as T = ΣI TI eI (2.10.14) (2.10.17) eI T = TI . (2.10.15) (2.10.18) ************ A rank-k tensor T in Vk has this general expansion on the er basis, T = Σii....i Tii....i (ei ei ..... ei) . (5.2.1) In the notation of (2.10.2) or (2.10.14) we identify Tii....i = [T(e)]ii....i . As expected, [T]jj...j = Σii....i Tii....i (ei ei ..... ei)jj...j = Σii....i Tii....i (δij δij .... δij) = Tjj...j . (5.2.2) The coefficients Tii....i can be projected out from T as in (2.10.15), (eiei... ei) T = Tii...i (2.10.15) (5.2.2) where the dot product of two "vectors" in Vk is defined in (2.10.15). Using the notion of a multiindex I (an ordinary multiindex), I ≡ {i1, i2, .....ik} // each is ranges 1,2....n (5.2.2) and a shorthand notation for the basis vectors eI ≡ ei ei ..... ei eI ≡ ei ei ..... ei (5.2.3) the general rank-k tensor T in Vk can be expanded in the following compact restatement of (5.2.1), T = ΣI TI eI . (5.2.4) and the coefficients TI can be projected out as in Section eI T = TI . (2.10.18) (5.2.5) w As noted in ******************** Scalar and Vector under transformations This is a confusing subject, addressed in detail in Appendix D. The functional λi a vector in the dual space V*, but is not displayed in a bold font the way ei is . Whereas ei has components (ei)i, the object λi has no components. In that sense it is a scalar quantity. Imagine a version of Picture A (2.1.1) where we transform from x-space with basis vectors ei to some x'-space with basis vectors e'i where (e'i)a = Rab(ei)b. Under such a component transformation, a dot product like α v transforms as a scalar. So, whereas v and α transform as vectors, the function α(v) transforms as a scalar field, just as in (2.2.6) and (2.1.15) (e). That is to say, α(v) = α v = α' v' = α'(v') . (2.11.13a) However, with respect to the vector transformation v'i = Rijvj the function α(v) transforms as a vector, (v'i)a = Rab (vi)b . (D.4.8) (2.11.13b) The fact that α(v) is both a scalar and a vector is discussed in Appendix D.4. When functional α is expanded on the basis functionals λi, the coefficients αi are the components of the the vector α which transforms as a vector under a component transformation. These coefficients αi are projected out by (2.11.9). ******* um = (un um) un = δnm un = um // not very interesting (um(u))n = δnm um = (un um) un = gnm un // a fact already known in (2.4.3) (um(u))n = gnm um = (en um) en = Rnm en (um(e))n = Rnm um = (en um) en = Rnm en (um(e))n = Rnm um = (un um) un = gnm un // a fact already known in (2.4.3) (um(u))n = gnm um = (un um) un = δnm un = um // not very interesting (um(u))n = δnm um = (en um) en = Rnm en (um(e))n = Rnm um = (en um) en = Rnm en (um(e))n = Rnm em = (un em) un = Rmn un (em(u))n = Rmn em = (un em) un = Rmn un (em(u))n = Rmn em = (en em) en = δnm en = em // not very interesting (em(e))n = δnm em = (en em) en = g'nm en // a fact already known in (2.3.3) (em(e))n = g'nm em = (un em) un = Rmn un (em(u))n = Rmn em = (un em) un = Rmn un (em(u))n = Rmn em = (en em) en = g'nm en // a fact already known in (2.3.3) (em(e))n = g'nm em = (en em) en = δnm en = em // not very interesting (em(e))n = δnm ***** Now look back at (2.3.4) which says this (en)i = Rni (en)i = Rni . (7.18.1)' (2.3.4) Which components of en and en are implied here? Looking at (2.6.4) we see that the more precise equations are these, (en(u))i = Rni (en(u))i = Rni . (2.6.7) col 3 row 1 col 4 row 2 ********* Old Section 2.6: 2.6 A change in notation : Picture E Due to the way expansions work, as shown in (2.5.1), we shall now cosmetically modify Picture A noted above, so that x-space becomes u-space (right side) and x'-space becomes e-space (left side). The metric tensors will then be called g(u) on the right, and g(e) on the left (E is used because B,C,D are already used up in Ref ** ). (2.1.1) (2.6.1) The equations of all earlier sections of Chapter 2 can be mapped from Picture A to Picture E as follows: X → X(u) where X is any x-space object X' → X(e) where X' is any x'-space object . However, the R matrix elements stay the same, since they are neither x-space nor x'-space objects. Examples: Rab → Rab g→ g(u) dx'a = Rabdxb → dx(e)a = Rabdx(u)b g' →g(e) V'a = RabVb → V(e)a = RabV(u)b V = Vn un → V = V(u)n un V = V'n en → V = V(e)n en . The four expansions of (2.5.1) are now (showing the n sum explicitly) V = Σn V(u)n un where un V = V(u)n V = Σn V(u)n un where un V = V(u)n V = Σn V(e)n en where en V = V(e)n V = Σn V(e)n en where en V = V(e)n from (2.5.1) (2.6.2) We can now make the following observations: 1. When a vector V is expanded on the un, the expansion coefficients are V(u)n . 2. When a vector V is expanded on the un, the expansion coefficients are V(u)n . 3. When a vector V is expanded on the en, the expansion coefficients are V(e)n . 4. When a vector V is expanded on the en, the expansion coefficients are V(e)n . (2.6.3) Notice that there is a distinction for example between V(u)n and V(e)n . Here the index n labels the vector component, but these two components are not the same. The number V(u)n is the component of V in the un basis, whereas V(e)n is the component of the same vector V but in the en basis. Since the bases are different, the components are different. We can apply the four equations shown on the right of (2.6.2) sequentially to the basis vectors V = um, um, em, em to obtain a set of 16 equations. The very first equation would be un um = um(u)n. One can then look up the dot product to find that un um = δnm and then one gets the result that um(u)n = δnm . The next equation is un um = um(u)n and we look up this dot product to find un um = gnm and so um(u)n = gnm. Rather than do all these calculations, since the dot products are already listed in (2.5.4), we can just read off the 16 results we want. This then produces the rightmost column of equations in (2.5.4) above, which we transcribe here: (um(u))n = δmn (um(u))n = g(u)mn (em(u))n = Rmn (em(u))n = Rmn (um(u))n = g(u)mn (um(u))n = δmn (em(u))n = Rmn (em(u))n = Rmn (um(e))n = Rnm (um(e))n = Rnm (em(e))n = δmn (em(e))n = g(e)mn (um(e))n = Rnm (um(e))n = Rnm (em(e))n = g(e)mn (em(e))n = δmn . (2.6.4) We have replaced g→g(u) and g'→g(e) and have made cosmetic changes such as δnm = δmn as well as g(u)mn = g(u)nm since all metric tensors are symmetric. Equations (2.6.4) give the contravariant (up) and covariant (down) components of all four basis vector types evaluated in both the u and e bases. Notice in (2.6.4) [ col 3 row 3] that (em(e))n = δmn . According to the Comment at the end Section 2.3, this equation applies for an arbitrary set of basis functions em(x). The equation is eminently reasonable. Suppose we expand em = [ Σn (em(e))n en] . We can see that we must have (em(e))n = δmn. Expanding a basis vector on its own basis yields a single term in the sum. Being true for any basis, (em(e))n = δmn must also be true for the u basis, and we see as well that (um(u))n = δmn in (2.6.4) [ col 2 row 2]. Let's now review the paradox (2.1) presented at the start of Chapter 2, but in our "improved" notation. We first restate the paradox in covariant notation: Paradox. Let v = Σnvnen be the expansion of a vector v onto a basis en. Since em is a vector, use this same expansion to expand v = em to get em = Σn(em)n en. This implies that (em)n = δmn. Thus, the only possible basis vectors em allowed in the universe are axis-aligned unit vectors! (2.6.5) Now we go through the same paradox presentation with more precise notation; Paradox? Let v = Σnv(e)nen be the expansion of a vector v onto a basis en. Since em is a vector, use this same expansion to expand v = em to get em = Σn(em(e)n) en. This implies that (em(e))n = δmn. This agrees with (2.6.4) column 3 row 3. There is no paradox. Without clear labeling, one assumes that by default the paradox statement (em)n = δmn means (em(u))n = δmn and this seems to rule out general basis vectors. In reality, (em)n = δmn means (em(e))n = δmn which is blatantly true as noted above, and this still allows us to have arbitrary basis vector components (em(u))n = Rmn as indicated by (2.6.4) col 3 row 1. (2.6.6) Conclusion: One must be careful when dealing with components of vectors and tensors to understand the space to which a vector or tensor belongs. For example, in Picture A we had a vector V'a(x') belonging to x'-space and we are now writing that as (V(e))a (x(e)) belonging to e-space using Picture E. When V is a basis vector like en (n is a label not an index), we have (en')a(x') → (en(e))a (x(e)). Now look back at (2.3.4) which says this (en)i = Rni (en)i = Rni . (7.18.1)' (2.3.4) Which components of en and en are implied here? Looking at (2.6.4) we see that the more precise equations are these, (en(u))i = Rni (en(u))i = Rni . (2.6.7) col 3 row 1 col 4 row 2 We now restate the transformation rules for rank-1 and rank-2 tensors in our (2.6.1) Picture E context: [V(e)]a = Rab[V(u)]b from (2.1.5) [M(e)]ab = Raa' Rbb' [M(u)]a'b' from (2.1.6) (2.6.8) The transformation of a vector field would be written, [V(e)(x(e))]a = Rab(x(u))[V(u)(x(u))]b . (2.6.9) ******