Appendix C v4

Appendix C v4 PhL 11.17.15 C.1 The Rearrangement Theorem 1 C.2 Theorem One 3 C.3 Theorem Two 5 C.4 Theorem Three 8 C.5 Applications of the Theorems to Tensors: Theorem Generalizations 9 C.6 Applications of the Theorems to Tensor Functions 11 C.7 A Unified View of Tensors and Tensor Functions 13 After a statement of the very powerful "rearrangement theorem", the next three sections use that theorem to prove certain other theorems (One, Two and Three) where we have attempted to abstract as much as possible the "permutational nature" of the objects involved. Then in Sections C.5 and C.6 these theorems are applied to tensors and tensor functions, resulting in certain simple statements involving the Alt and Sym operators in the tensor and tensor function worlds. Section C.7 then shows that in fact these two worlds are related by a real orthogonal transformation. C.1 The Rearrangement Theorem First we restate the permutation group rearrangement theorem from Appendix A ΣP f(QP) = ΣP f(PQ) = ΣP f(P) . (A.3.6) (C.1.1) Here ΣP is a sum over all n! permutations of {1,2...n}, and Q is any one of these permutations. The first two sums are just reorderings or rearrangements of the third sum and so equal the third sum. Proof: This theorem is true because the permutations P of {1,2...n} form a group G : P1P2 = P3 ϵ G // closure (P1P2)P3 = P1(P2P3) // associative P = I // identity exists, permutation that does nothing to {1,2...n} P-1 exists for any P // just the inverse permutation. (C.1.2) It is a fact that for any group G with n elements gi, ga {g1, g2, ....gn} = { gag1, gag2, ....gagn} = {g'1, g'2, ....g'n} = reordering of {g1, g2, ....gn} {g1, g2, ....gn}ga = { g1ga, g2ga, ....gnga} = {g"1, g"2, ....g"n} = reordering of {g1, g2, ....gn} . (C.1.3) To show that {g'1, g'2, ....g'n} is a reordering of {g1, g2, ....gn}, we have to show that no two elements of {g'1, g'2, ....g'n} are the same. Suppose for example g'1 = g'2 . That would imply gag1 = gag2. Since ga-1 exists in a group for any ga, apply ga-1 to both sides to get ga-1gag1 = ga-1gag2 or g1 = g2. But that contradicts the basic starting point that {g1, g2, ....gn} enumerates the distinct group elements. Therefore Σi f(gagi) = Σif(giga) = Σif(gi) . (C.1.4) This is valid only if the sum is over all elements of the group, which in (C.1.1) means the sum ΣP must be over all permutations P. In any group, if g exists, so does g-1, and it is just some element of the group. For the permutation group P-1 exists and is in fact the permutation which reverses the permutation of P : P{1,2...k} = {i1,i2...ik} {i1,i2...ik} = P-1{1,2...k} PP-1 = P-1P = 1 (C.1.5) It seems clear that the number of position swaps to get from {1,2...k} to {i1,i2...ik} is the same as it is going the other direction, so S(P-1) = S(P) . (C.1.6) Finally, consider P1P2{1,2...k} = P{1,2...k} = {i1,i2...ik} . P = P1P2 If P2 causes S2 position swaps and then P1 causes S1 more, then P does S2+S1 total swaps. Thus S(P) = S(P1P2) = S(P1) + S(P2) and (-1)S(PP) = (-1)S(P) (-1)S(P) = (-1)S(PP) (C.1.7) Another version of the rearrangement theorem is the following, ΣQ f(Q) = ΣQ f(Q-1) (C.1.8) Again, this is just a reordering of the sum. Consider, {g1-1, g2-1, ....gn-1} = {g1', g2', ....gn'} = reordering of {g1, g2, ....gn} To show that {g1', g2', ....gn'} is a reordering of {g1, g2, ....gn} we have to show that no two elements are the same. Suppose for example that g1' = g2' . That would say g1-1 = g2-1 which in turn says g1 = g2, but that contradicts the basic starting point that {g1, g2, ....gn} enumerates the distinct group elements. Therefore, Σi f(gi) = Σif(gi-1) (C.1.9) Comment: For continuous groups (like the rotation group SO(3)) , the rearrangement theorems become ∫dg f(gag) = ∫dg f(gga) = ∫dg f(g) ∫dg f(g) = ∫dg f(g-1) (C.1.10) where dg is called the invariant Haar measure. For SO(3) it is dg = dφd(cosθ)dψ (Euler angles). C.2 Theorem One Consider the following set of k+k' integers, {1,2....k, k+1,k+2....k+k'} = {1,2....k+k' } (C.2.1) Partition this list into a low and high half by defining z ≡ 1,2....k Z = k+1,k+2....k+k' (C.2.2) Then {1,2....k+k'} = {z,Z} (C.2.3) Now let Q be a permutation of the lower integers {1,2...k} = z. There are k! possible permutations, so we know that ΣQ (1) = k! (C.2.4) We can extend the meaning of Q so it applies to the entire set of integers {1,2....k+k'} merely by stating that this extended Q' does not alter the higher integers. Then Q'(z) = Q(z) = z' = some permutation of the lower integers (C.2.5a) Q'(Z) = Z // since Q has no effect on the higher integers (C.2.5b) Q'(z, Z) = {Q'(z), Q'(Z)} = {Q(z), Z} *** (C.2.5c) Now imagine we have a function f of the lower integers and a function F of the higher ones, f[z] = f[1,2....k] F[Z] = F[k+1,k+2....k+k'] (C.2.6) We use [] in place of () merely to improve clarity below. Think of the integers as generic labels on the functions. Here are two applications we shall consider later on, f[z] = f[1,2....k] = Tii...i = components of a rank-k tensor f[z] = f[1,2....k] = T(vi,vi, .... vi) = a rank-k tensor function (C.2.7) Now let P be a general permutation of {1,2....k+k'} = {z,Z}. P(z,Z) = {P(z), P(Z)} (C.2.8) Notice that QP and PQ are undefined since P and Q operate in different spaces, but Q'P and PQ' are both defined since both permutations Q' and P operate in the space of {1,2....k+k'}. Recall now the meaning of S(Q) as the number of swaps required to go from z to Q(z) . This is the same as the number of swaps required to go from {z,Z} to Q'{z,Z} = {Q(z),Z}. Therefore S(Q) = S(Q') (C.2.9) We shall now prove the following rather obscure looking but very powerful theorem : Theorem One ΣP (-1)σS(P) f[P(z)] F[P(Z)] = ΣP (-1)σS(P) f^[P(z)] F[P(Z)] where (C.2.10) f^[z] ≡ (1/k!) ΣQ (-1)σS(Q) f[Q(z)] σ = 1 or 0 The purpose of σ is to state the theorem with and without the (-1)S(P) factor. Proof: Our first task is to process the second line f^[z] = (1/k!) ΣQ (-1)σS(Q) f[Q(z)] = (1/k!) ΣQ (-1)σS(Q') f[Q'(z)] // (C.2.9) and (C.2.5a) (C.2.11) Apply permutation P to the above equation, P f^[z] = P { (1/k!) ΣQ (-1)σS(Q') f[Q'(z)]} = (1/k!) ΣQ (-1)σS(Q') P f[Q'(z)] . (C.2.12) The object appearing on the right is this, P f [Q'(z)] = P f [ Q'(1), Q'(2) ....Q'(k) ] = f [ P(Q'(1)), P(Q'(2))....P(Q'(k)) ] = f [P(Q'(z)] = f [(PQ')(z)] ≡ f [PQ'(z)] . (C.2.13) Putting (C.2.13) into (C.2.12) gives P f^[z] = f^[P(z)] = (1/k!) ΣQ (-1)σS(Q') f [PQ'(z)] . (C.2.14) Now rewrite the claimed theorem (C.2.10) as LHS = RHS. We shall process the RHS and show that it is equal to the LHS : RHS = ΣP (-1)σS(P) f^[P(z)] F[P(Z)] = ΣP (-1)σS(P) { (1/k!) ΣQ (-1)σS(Q') f[PQ'(z)] } F[P(Z)] // (C.2.14) for f^[P(z)] = (1/k!) ΣQ ΣP (-1)σS(PQ') f[PQ'(z)]} F[P(Z)] // reorder and (C.1.7) = (1/k!) ΣQ ΣP(-1)σS(PQ') f[PQ'(z)]} F[PQ'(Z)] // Q'(Z) = Z from (C.2.5b) = (1/k!) ΣQ ΣP(-1)σS(P) f[P(z)]} F[P(Z)] // ΣP rearrangement theorem (C.1.1) = ΣP(-1)σS(P) f[P(z)]} F[P(Z)] { (1/k!) ΣQ (1)} // reorder = ΣP(-1)σS(P) f[P(z)]} F[P(Z)] { 1 } // ΣQ (1) = k! from (C.2.4) = LHS QED C.3 Theorem Two This section is a copy, paste and edit version of Section C.2. Equations that are the same are in italics. Consider the following set of k+k' integers, {1,2....k, k+1,k+2....k+k'} = {1,2....k+k' } (C.2.1) Partition this list into a low and high half by defining z ≡ 1,2....k Z = k+1,k+2....k+k' (C.2.2) Then {1,2....k+k'} = {z,Z} (C.2.3) Now let R be a permutation of the upper integers {k+1,k+2....k+k'} = Z. There are k'! possible permutations, so we know that ΣR (1) = k'! (C.3.4) We can extend the meaning of R so it applies to the entire set of integers {1,2....k+k'} merely by stating that this extended R' does not alter the lower integers. Then R'(Z) = R(Z) = Z'' = some permutation of the upper integers (C.3.5a) R'(z) = z // since R has no effect on the lower integers (C.3.5b) R'(z, Z) = {R'(z), R'(Z)} = {z, R(Z)} *** (C.3.5c) Now imagine we have a function f of the lower integers and a function F of the higher ones, f[z] = f[1,2....k] F[Z] = F[k+1,k+2....k+k'] (C.2.6) We use [] in place of () merely to improve clarity below. Think of the integers as generic labels on the functions. Here are two applications we shall consider later on, F[Z] = F [k+1,k+2....k+k'] = Sii...i = components of a rank-k' tensor F[Z] = F [k+1,k+2....k+k'] = S(vi,vi, .... vi) = a rank-k' tensor function (C.3.7) Now let P be a general permutation of {1,2....k+k'} = {z,Z}. P(z,Z) = {P(z), P(Z)} (C.2.8) Notice that RP and PR are undefined since P and R operate in different spaces, but R'P and PR' are both defined since both permutations R' and P operate in the space of {1,2....k+k'}. Recall now the meaning of S(R) as the number of swaps required to go from Z to R(Z) . This is the same as the number of swaps required to go from {z,Z} to R'{z,Z} = {z,R(Z)}. Therefore S(R) = S(R') (C.3.9) We shall now prove the following equally obscure looking but very powerful theorem : Theorem Two ΣP (-1)σS(P) f[P(z)] F[P(Z)] = ΣP (-1)σS(P) f[P(z)] F^[P(Z)] where (C.3.10) F^[Z] ≡ (1/k'!) ΣR (-1)σS(R) F[R(Z)] σ = 1 or 0 The purpose of σ is to state the theorem with and without the (-1)S(P) factor. Proof: Our first task is to process the second line F^[Z] = (1/k'!) ΣR (-1)σS(R) F[R(Z)] = (1/k'!) ΣR (-1)σS(R') F[R'(Z)] / (C.3.9) and (C.3.5a) (C.3.11) Apply permutation P to the above equation, P F^[Z] = P {(1/k'!) ΣR (-1)σS(R') F[R'(Z)]} = (1/k'!) ΣR (-1)σS(R') P F[R'(Z)] . (C.2.12) The object appearing on the right is this, P F[R'(Z)] = P F [ R'(k+1), R'(k+1) ....R'(k+k') ] = F [ P(R'(k+1)), P(R'(k+2)) ... P(R'(k+k')) ] = F [P(R'(Z)] = F [(PR')(Z)] ≡ F [PR'(Z)] . (C.3.13) Putting (C.3.13) into (C.3.12) gives P F^[Z] = F^[P(Z)] = (1/k'!) ΣR (-1)σS(R') F [PR'(Z)] (C.3.14) Now rewrite the claimed theorem (C.3.10) as LHS = RHS. We shall process the RHS and show that it is equal to the LHS : RHS = ΣP (-1)σS(P) f[P(z)] F^[P(Z)] = ΣP (-1)σS(P) f[P(z)]{(1/k'!) ΣR (-1)σS(R') F [PR'(Z)]} // (C.3.14) for F^[P(Z)] = (1/k'!)ΣR ΣP(-1)σS(PR') f[P(z)] F [PR'(Z)] // reorder and (C.1.7) = (1/k'!)ΣR ΣP(-1)σS(PR') f[PR'(z)] F [PR'(Z)] // R'(z) = Z from (C.3.5b) = (1/k'!)ΣR ΣP(-1)σS(P) f[P(z)] F [P(Z)] // ΣP rearrangement theorem (C.1.1) = ΣP(-1)σS(P) f[P(z)] F [P(Z)] { (1/k'!) ΣR (1) } // reorder = ΣP(-1)σS(P) f[P(z)] F [P(Z)] {1 } // ΣR (1) = k'! from (C.3.4) = LHS (C.3.10) Alternate Proof of Theorem 2 An alternate proof of Theorem Two is two start with Theorem One and just make these changes z ↔ Z f↔F k↔k' Q→R Here is Theorem One ΣP (-1)σS(P) f[P(z)] F[P(Z)] = ΣP (-1)σS(P) f^[P(z)] F[P(Z)] where f^[z] ≡ (1/k!) ΣQ (-1)σS(Q) f[Q(z)] σ = 1 or 0 (C.2.10) and here is Theorem One with the above changes applied, ΣP (-1)σS(P) F[P(Z)] f[P(z)] = ΣP (-1)σS(P) F^[P(Z)] f[P(z)] where F^[z] ≡ (1/k'!) ΣR (-1)σS(R) F[R(Z)] σ = 1 or 0 . (C.2.10)swap This is the same as Theorem Two which we quote from above, ΣP (-1)σS(P) f[P(z)] F[P(Z)] = ΣP (-1)σS(P) f[P(z)] F^[P(Z)] where F^[Z] ≡ (1/k'!) ΣR (-1)σS(R) F[R(Z)] σ = 1 or 0 (C.3.10) We went ahead with the detailed proof for two reasons. First, the swap proof might not be convincing to the reader. Second and more importantly, the detailed proof provides steps which are crucial to proving Theorem Three below. C.4 Theorem Three Now both functions have a ^ subscript. ΣP (-1)σS(P) f[P(z)] F[P(Z)] = ΣP (-1)σS(P) f^[P(z)] F^[P(Z)] where f^[z] ≡ (1/k!) ΣQ (-1)σS(Q) f[Q(z)] (C.4.1) F^[Z] ≡ (1/k'!) ΣR (-1)σS(R) F[R(Z)] σ = 1 or 0 The purpose of σ is to state the theorem with and without the (-1)S(P) factor. This theorem will involve both R and Q, as well as R' and Q' from earlier sections. Note that R'Q' = Q'R' (C.4.2) because Q' acts only on the lower integers in (1,2...k+k') while R' acts only on the upper integers. Recall these results from previous sections, f^[P(z)] = (1/k!) ΣQ (-1)σS(Q') f [PQ'(z)] (C.2.14) F^[P(Z)] = (1/k'!) ΣR (-1)σS(R') F [PR'(Z)] (C.3.14) Now rewrite the claimed theorem (C.4.1) as LHS = RHS. We shall process the RHS and show that it is equal to the LHS : RHS = ΣP (-1)σS(P) f^[P(z)] F^[P(Z)] = ΣP (-1)σS(P){ (1/k!) ΣQ (-1)σS(Q') f [PQ'(z)]}{ (1/k'!) ΣR (-1)σS(R') F [PR'(Z)]} = (1/k!)(1/k'!)ΣQΣR ΣP (-1)σS(PR'Q') f [PQ'(z)] F [PR'(Z)] // reorder and (C.1.7) = (1/k!)(1/k'!)ΣQΣR ΣP (-1)σS(PR'Q') f [PQ'R'(z)] F [PR'Q'(Z)] // Q'(Z) = Z from (C.2.5b) // R'(z) = Z from (C.3.5b) = (1/k!)(1/k'!)ΣQΣR ΣP (-1)σS(PR'Q') f [PR'Q'(z)] F [PR'Q'(Z)] // (C.4.2) R'Q' = Q'R' = (1/k!)(1/k'!)ΣQΣR ΣP (-1)σS(P[R'Q']) f [P[R'Q'](z)] F [P[R'Q'](Z)] = (1/k!)(1/k'!)ΣQΣR ΣP (-1)σS(P) f [P(z)] F [P(Z)] // ΣP rearrangement theorem (C.1.1) = ΣP (-1)σS(P) f [P(z)] F [P(Z)] { (1/k!)ΣQ(1) }{ (1/k'!)ΣR(1) } // reorder = ΣP (-1)σS(P) f [P(z)] F [P(Z)] { 1 }{ 1 } // (C.2.4) and (C.3.4) = LHS (C.4.1) QED C.5 Applications of the Theorems to Tensors: Theorem Generalizations Here are the three theorems collected from above : 1. ΣP (-1)σS(P) f[P(z)] F[P(Z)] = ΣP (-1)σS(P) f^[P(z)] F[P(Z)] (C.2.10) 2. ΣP (-1)σS(P) f[P(z)] F[P(Z)] = ΣP (-1)σS(P) f[P(z)] F^[P(Z)] (C.3.10) 3. ΣP (-1)σS(P) f[P(z)] F[P(Z)] = ΣP (-1)σS(P) f^[P(z)] F^[P(Z)] (C.4.1) (C.5.1) where f^[z] ≡ (1/k!) ΣQ (-1)σS(Q) f[Q(z)] z = 1,2,...k F^[Z] ≡ (1/k'!) ΣR (-1)σS(R) F[R(Z)] σ = 1 or 0 Z = k+1,k+2..k+k' 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.2.7) 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.2.10) 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. Generalization of the three theorems The theorems derived above can be generalized in the following manner. Suppose for example we have a set of integers 1,2,3......(k1+k2+...+kN) = 1,2,3....κN. Instead of partitioning Z into 2 groups Z = (z,z') as done above, we partition the integers into N groups Z = (z1, z2....zN) as follows: κ1 = k1 // "cumulative ranks", as in (7.11.6) κ2 = k1+ k2 κ3 = k1+ k2 + k3 ... κN = k1 + k2 + ... + kN = Σi=1N ki . (7.11.6) Z = (z1, z2....zN) (C.5.8) z1 = 1,2,3...κ1 // the partitions z2 = κ1+1,κ2+2,...κ2 z2 = κ2+1,κ2+2,...κ3 ... zN = κN-1, κN-1 + 1, ...κN . And instead of functions f and F, we have functions f1, f2....fN. Whereas for N = 2 we had 22-1 = 3 theorems, for general N there will be 2N - 1 theorems. If define L ≡ ΣP (-1)σS(P)f1[P(z1)]f2[P(z2)] ... fN[P(zN)] // Left side of theorems (C.5.9) then here are those theorems: ( exercise for the reader, use induction or brute force ) 1. L = ΣP (-1)σS(P)(f1)^[P(z1)]f2[P(z2)] ... fN[P(zN)] 2. L = ΣP (-1)σS(P)f1[P(z1)](f2)^[P(z2)] ... fN[P(zN)] 3. L = ΣP (-1)σS(P)(f1)^[P(z1)](f2)^[P(z2)] ... fN[P(zN)] 4. L = ΣP(-1)σS(P)f1[P(z1)]f2[P(z2)](f3)^[P(z3)] ...fN[P(zN)] ...... (2N-1). L = ΣP(-1)σS(P)(f1)^[P(z1)](f2)^[P(z2)](f3)^[P(z3)] ...(fN)^[P(zN)] (C.5.10) Translated to Alt/Sym, notation, we then find for the case N = 3 Alt[(TSR)] = Alt[(T^SR)] = Alt[(TS^R)] = Alt[(TSR^)] = Alt[(T^S^R)] = Alt[(T^SR^)] = Alt[(TS^R^)] = Alt[(T^S^R^)] (C.5.11) Sym[(TSR)] = Sym[(TsSR)] = Sym[(TSsR)] = Sym[(TSRs)] = Sym[(TsSsR)] = Sym[(TsSRs)] = Sym[(TSsRs)] = Sym[(TsSsRs)] (C.5.12) In general one can write Alt[(T1)(T2) ...... (TN)] = Alt[(T1)a(T2)a ...... (TN)a] (C.5.13) where each ai can independently be a blank, (Ti) , or can be a ^ , (Ti)^. This then is the ultimate statement that arbitrary pre-antisymmetrizing of one or more tensors in a totally antisymmetric product makes no difference. Similarly Sym[(T1)(T2) ...... (TN)] = Sym[(T1)a(T2)a ...... (TN)a] (C.5.14) where each ai can independently be a blank, (Ti) , or can be an s, (Ti)s. This then is the ultimate statement that arbitrary pre-symmetrizing of one or more tensors in a totally symmetric product makes no difference. There are several "Alt theorems" that can be derived from the above statements. For example Theorem 1: Alt( Alt(A^ B^) C^) = Alt(A^ B^ C^) = Alt(A^ Alt(B^ C^)) Proof: Define T ≡ A^ B^ T^ = Alt(A^B^) , then: Alt( Alt(A^ B^) C^) = Alt(T^ C^) = Alt(T C^) // (C.5.7) = Alt((A^ B^) C^) = Alt(A^B^C^) // (2.8.22) is associative Define S ≡ B^ C^ S^ = Alt(B^C^), then: Alt(A^ Alt(B^ C^)) = Alt(A^ S^) = Alt(A^ S) // (C.5.7) = Alt(A^ (B^ C^)) = Alt(A^B^C^) // (2.8.22) is associative Theorem 2: A^ ^ B^ ^ C^ = constant * Alt(A^ B^ C^) I already know from (7.11.3) that A^ ^ B^ ^ C^ = ΣI(TSR)I eI STOP Question: Can you write (ei^ ei ......^ ei) = Alt(ei ei ...... ei) Well remember [Alt(X)]ii...i ≡ (1/k!) ΣP (-1)S(P) XP(i)P(i)...P(i) (7.5.3) So try X = ej ej ...... ej Xii....i = δij δij....δij Then [Alt(ej ej ...... ej)]ii...i = (1/k!) ΣP (-1)S(P) δP(i)j δP(i)j....δP(i)j STOP. Recall, 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))i(vP(j))i......(vP(j))i = (1/k!) ΣP (-1)S(P) (vj)P(i)(vj)P(i)......(vj)P(i) // A.6.1 Meanwhile. Alt(vj vj ..... vj)ii....i = (1/k!) ΣP (-1)S(P) (vj)P(i)(vj)P(i)......(vj)P(i) So I have omitted this major idea for vectors. 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.2.7) 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.2.10) 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, C.7 A Unified View of Tensors and Tensor Functions: Basis Change In this section multiindex notations are shown in red to the right. 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.7.1) 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.7.2) 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.7.3) 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.7.4) 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.7.4) 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.7.3), Tii....i = < ei,ei, .... ei | T > . <eI | T > (C.7.5) From this one would conclude that Tii....i = T(ei,ei, .... ei) TI = T(eI) (C.7.6) 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.7.7) 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 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.7.8) The basis-change transformation matrix between the |vI> and |eI> is given by, MIJ ≡ < ei,ei, .... ei | vj,vj, .... vj > MIJ = <eI|vJ> (C.7.9) = < 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.7.10) = (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.7.11) 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.7.12) 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.7.13) 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.7.14) 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.7.15) 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.7.16) 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? This is an example of a "fuzzy question". 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.