Appendix A

creation date 11.4.15 Warning: This old Appendix A contains items that don't appear anywhere else! In particular, A.3 and the first A.6 and A.7. Do not throw this doc out!!! 1.14.16 It used to be crucial stuff. Appendix A: Supporting Theorems 1 A.1 Antisymmetric Tensor Theorem 1 A.2 Antisymmetric Function Theorem 2 A.3 Ordered Sum Theorem 3 A.4 Sum of permutation signs vanishes 4 A.5 Orthogonality of the Alt and Sym Projection Operators 5 A.6 Theorem involving Kronecker deltas 5 A.6 Some Facts about Determinants 6 A.7 Theorem on raising and lowering P 7 Appendix A: Supporting Theorems The following theorem appears in definition (A.2.1) of Alt and (A.2.11). A.1 Antisymmetric Tensor Theorem The tensor Yii...i ≡ ΣP (-1)S(P) RP(i)P(i)....P(i) is totally antisymmetric. (A.1.1) Note: The sum shown here is defined in ** to be [alt(R)]ii...i so this theorem also claims, Y = alt(R) = a totally antisymmetric tensor for any R (A.1.2) Proof: We shall show that Yii...i = - Yii...i and then the same argument trivially applies to any pair of indices. Start with, Yii...i = ΣP (-1)S(P) RP(i)P(i)....P(i) . (A.1.3) In this sum, P represents a permutation of the integers {1,2...k}, and P(ir) ≡ iP(r) so that P really acts on the subscript r. S(P) is the number of swaps involved in permutation P. For example, if we define the specific permutation P' by P'{1,2...k} = {2,1...k}, then S(P') = 1. Since in this case i2 = P'(ii), we can write i1 = P(i2) = PP'(i1). Similarly, i2 = P(i1) = PP'(i2). For any other subscript like i3 we have i3 = P(i3) = PP'(i3). The above sum can then be written Yii...i = ΣP (-1)S(P) RPP'(i)PP'(i)....PP'(i) . (A.1.4) Since S(PP') = S(P) + S(P') for any P and P', and since S(P') = 1, we know S(P) = S(PP') - 1, so then Yii...i = ΣP (-1)S(PP')-1 RPP'(i)PP'(i)....PP'(i) = (-1) ΣP (-1)S(PP') RPP'(i)PP'(i)....PP'(i) . (A.1.5) Finally we can use the rearrangement theorem of group theory applied to the permutation group (see PL ***). This theorem says that, ΣP f(QP) = ΣP f(PQ) = ΣP f(P) (A.1.6) where Q is any fixed permutation. The first two sums are just reorderings of the third sum and so equal the third sum. Selecting Q = P' we have ΣP f(PP') = ΣP f(P) so then Yii...i = (-1) ΣP (-1)S(P) RP(i)P(i)....P(i) (A.1.7) = - Yii...i QED (A.1.8) Corollary: Tensor Y12...k = ΣP (-1)S(P) RP(1)P(2)....P(k) is totally antisymmetric. (A.1.9) Proof: This is an application of Theorem 1 where we take in = n. The theorem and corollary are fairly obvious when one writes Yii...i = ΣP (-1)S(P) RP(i)P(i)....P(i) = Rii...i + signed permutations (A..1.10) or Y12...k = ΣP (-1)S(P) RP(1)P(2)....P(k) = R12...k + signed permutations (A.1.11) Examples: Yiii = Riii - Riii + Riii - Riii + Riii - Riii Y123 = R123 - R132 + R312 - R321 + R231 - R213 (A.1.12) ________________________________________________________________________________________________ The following theorem should probably have all v's on the right. It then seems to be just a repeat of what I proved above, but here it is for a special tensor function. This does not seem useful or correct on 1.14.16. A.2 Antisymmetric Function Theorem The function F(vi,vi.....vi) ≡ ΣP (-1)S(P) (fi)P(j)(fi)P(j)...(fi)P(j) is totally antisymmetric in its arguments. (A.2.1) We shall show that F(vi,vi.....vi) = - F(vi,vi.....vi) and then the same argument trivially applies to any pair of indices. The proof is very similar to the proof of Theorem A.1 above. Start with F(vi,vi.....vi) = ΣP (-1)S(P) (fi)P(j)(fi)P(j)...(fi)P(j) . (A.2.2) Define permutation P' exactly as in Section A.2, except here permutations act on the j indices. Then for example j1 = P'(j2) and so on, allowing the above equation to be rewritten as F(vi,vi.....vi) = ΣP (-1)S(P) (fi)PP'(j)(fi)PP'(j)...(fi)PP'(j) = (-1) ΣP (-1)S(PP') (fi)PP'(j)(fi)PP'(j)...(fi)PP'(j) (A.2.3) = (-1) ΣP (-1)S(P) (fi)P(j)(fi)P(j)...(fi)P(j) // rearrangement theorem (A.2.4) = - F(vi,vi.....vi) QED (A.2.5) _______________________________________________________________________________________________ This theorem I think still needs to be put somewhere in wedge doc. It is not in the newest Appendix A. Maybe it is no longer needed, but I need to keep in mind that it is sitting right here. 1.14.16 A.3 Ordered Sum Theorem (ΣP [ΣP(i)<P(i)<...<P(i)]) fii...i = Σi<i<...<i [ΣP fP(i)P(i)...P(i)] . (A.3.1) Rather than present a formal proof, we look at the two simplest cases and the general case is then obvious. k = 2: First, consider Q ≡ Σi≠i fii = [Σi<i + Σi>i] fii = [Σi<i + Σi<i] fii= (ΣP[ΣP(i)<P(i)]) fii . (A.3.2) On the other hand, Q = [Σi<i + Σi<i] fii = Σi<i fii + Σi<i fii = Σi<i fii + Σi<i fii //dummy swap i1↔i2 in 2nd term = Σi<i [ fii + fii] = Σi<i [ΣP fP(i)P(i)] . (A.3.3) Thus we have proven the Theorem for k = 2, (ΣP [ΣP(i)P(i)]) fii = Σi<i [ΣP fP(i)P(i)] (A.3.4) k = 3: First, consider Q ≡ Σi≠i≠i fiii = (Σi<i<i + Σi<i<i + Σi<i<i + Σi<i<i + Σi<i<i + Σi<i<i) fiii = (ΣP [ΣP(i)<P(i)<P(i)]) fiii . (A.3.5) On the other hand we can rename the summation indices in all but the first sum to get Q = (Σi<i<i + Σi<i<i + Σi<i<i + Σi<i<i + Σi<i<i + Σi<i<i) fiii as is 2↔3 1↔3 = Σi<i<i fiii + Σi<i<i fiii + Σi<i<i fiii + 3 more sums = Σi<i<i [ fiii + fiii + fiii + 3 more terms ] = Σi<i<i [ΣP fP(i)P(i)P(i)]. (A.3.6) Thus we have proven the Theorem for k =3, (ΣP [ΣP(i)<P(i)<P(i)]) fiii = Σi<i<i [ΣP fP(i)P(i)P(i)] . (A.3.7) The argument for a k-fold sum proceeds in the same manner, and we end up with (ΣP [ΣP(i)<P(i)<...<P(i)]) fii...i = Σi<i<...<i [ΣP fP(i)P(i)...P(i)] . (A.3.1) ________________________________________________________________________________________________ This theorem is (A.1.12) of the newest Appendix A A.4 Sum of permutation signs vanishes ΣP (-1)S(P) = 0 (A.4.1) Proof: Suppose in (A.1.1) we take Rii...i = 1 . (A.4.2) Then (A.1.1) says Yii...i ≡ ΣP (-1)S(P) RP(i)P(i)....P(i) = ΣP (-1)S(P). (A.4.3) Theorem (A.1.1) says that Yii...i is totally antisymmetric for any R. But the right side of (A.4.3) does not depend on the indices, so we have for example Yii...i = ΣP (-1)S(P) Yii...i = ΣP (-1)S(P) (A.4.4) Then ΣP (-1)S(P) = Yii...i = - Yii...i = - ΣP (-1)S(P) (A.4.5) and we conclude that ΣP (-1)S(P) = 0. QED Example: Yiii = Riii - Riii + Riii - Riii + Riii - Riii There are three + signs and three - signs. ________________________________________________________________________________________________ These theorems appear in latest App A as (A.4.1). A.5 Orthogonality of the Alt and Sym Projection Operators Alt(Sym(X)) = Sym(Alt(X) = 0 (A.5.1) The Alt and Sym operators are defined in ** and ** as [Alt(X)]ii...i ≡ ΣP (-1)S(P) XP(i)P(i)...P(i) (A.5.2) [Sym(X)]ii...i ≡ ΣP XP(i)P(i)...P(i) (A.5.3) Proof that Alt(Sym(X)) = 0 Alt(Sym(X)) = Alt( ΣP XP(i)P(i)...P(i)) = ΣP [Alt(XP(i)P(i)...P(i))] = ΣP [ ΣP' (-1)S(P')XP(P'i)P(P'i)...P(P'i)] = ( )2 ΣP'(-1)S(P'){ ΣP [XP(P'i)P(P'i)...P(P'i)] } = ( )2 [ΣP'(-1)S(P')]{ ΣP [XP(i)P(i)...P(i)] } // rearrangement theorem (A.1.6) = 0 // because [ΣP'(-1)S(P')] = 0 by (A.3.1) QED Sym(Alt(T)) = Sym (ΣP(-1)S(P)XP(i)P(i)...P(i))= ΣP(-1)S(P)[Sym(XP(i)P(i)...P(i))] = ΣP(-1)S(P) { ΣP' XP(P'i)P(P'i)...P(P'i)} = ( )2 [ΣP(-1)S(P)]{ ΣP' [XP'(i)P'(i)...P'(i)] } // rearrangement theorem (A.1.6) = 0 // because [ΣP(-1)S(P)] = 0 by (A.3.1) QED _______________________________________________________________________________________________ This theorem has yet to be placed anywhere, and I don't know if it is still needed, I suspect yes. A.6 Theorem involving Kronecker deltas ΣP (-1)S(P) δP(j)i δP(j)i... δP(j)i = ΣP (-1)S(P) δjP(i) δjP(i)...δjP(i) (A.6.1) Define Q ≡ P-1 which is the inverse of the permutation P of the k integers {1,2...k}. Then, δP(j)i = 1 P(j1) = i1 j1 = P-1(i1) = Q(i1) δjQ(i) = 1 (A.6.2) so that δP(j)i = δjQ(i) . (A.6.3) Then ΣP (-1)S(P) δP(j)i δP(j)i... δP(j)i = ΣP (-1)S(P) δjQ(i) δjQ(i)...δjQ(i) . (A.6.4) But it is a general fact that, if Q = P-1, then ΣP (-1)S(P) f(Qx) = ΣQ (-1)S(Q) f(Qx) , (A.6.5) because S(P) = S(P-1) = number of swaps in either P or P-1, and summing over all P is the same as summing over all inverse permutations P-1 (either sum exhausts the permutation group). Therefore, replacing dummy Q by P on the right, ΣP (-1)S(P) f(Qx) = ΣP (-1)S(P) f(Px) (A.6.6) so the right side of (A.6.4) gets Q replaced by P so that ΣP (-1)S(P) δP(j)i δP(j)i... δP(j)i = ΣP (-1)S(P) δjP(i) δjP(i)...δjP(i) which is the claim of this theorem. QED _______________________________________________________________________________________________ This stuff now appears in Appendix A Section A.1. A.6 Some Facts about Determinants A standard form of the determinant of a kxk matrix M is the following (see *** ) det(M) = Σii...i εii...iM1iM2i ...Mki = ΣP (-1)S(P) M1P(1)M2P(2) ...MkP(k) = [ M11M22....Mkk + signed permutations of the second index ] (A.6.1) It is easy to show from the first line that if we swap any pair of "first indices" on two of the factors in M1iM2i ...Mki, the determinant changes sign. This is just the well-known rule that a determinant changes sign if any two rows are swapped. Because of this fact, we may write εjj...jdet(M) = Σii...i εii...iMjiMji ...Mji = ΣP (-1)S(P)MjP(1)MjP(2) ...MjP(k) . (A.6.2) If two jr are the same, both sides vanish (determinant has two identical rows). Otherwise, the εjj...j on the left implements the fact just stated above about swapping "first indices" on the M factors. If M happens to be a covariant "up-tilt" matrix, then (A.7.1 εjj...j det(M) = Σii...i εii...iMjiMji ...Mji = ΣP (-1)S(P)MjP(1)MjP(2)...MjP(k) (A.6.3) Since det(M) = det(MT), we can write (A.7.1) as εjj...jdet(M) = Σii...i εii...iMijMij ...Mij = ΣP (-1)S(P)MP(1)jMP(2)j ...MP(k)j . (A.6.4) Again, if M is an uptilt matrix, this reads εjj...jdet(M) = Σii...i εii...iMijMij ...Mij = ΣP (-1)S(P)MP(1)jMP(2)j ...MP(k)j . (A.6.5) _______________________________________________________________________________________________ This has not been placed anywhere, and I don't know if it is needed. A.7 Theorem on raising and lowering P ΣP (-1)S(P) (vj)P(i) (vj)P(i) .... (vj)P(i) = ΣP(-1)S(P) (vP(j))i (vP(j))i ... (vP(j))i (A.7.1) The theorem claims one can move the P operator from the ir indices to the jr indices without changing the sum. It is the index subscripts that P permutes, for example, P(i1) ≡ iP(1), and the P are permutations of {1,2..k}. The actual indices ir and labels jr can be arbitrary positive integers. For either expansion the P = 1 term in is (vj)i(vj)i ....(vj)i so the claim seems at least feasible. Example: For k = 2 one has first line = ΣP (-1)S(P) (vj)P(i)(vj)P(i) = (vj)i(vj)i - (vj)i(vj)i second line = ΣP(-1)S(P) (vP(j))i(vP(j))i = (vj)i(vj)i - (vj)i (vj)i so the theorem is valid for k = 2. Proof: Define matrix M to be Mab = (va)b (A.7.2) Warning: Then we need to show that ΣP (-1)S(P) MjP(i)MjP(i) .... MjP(i) = ΣP(-1)S(P) MP(j)iMP(j)i ... MP(j)i . (A.7.3) We shall now examine the two sides of this equation. We know from (A.1.1) that the LHS is totally antisymmetric in the ir indices and can therefore be written LHS = ΣP (-1)S(P) MjP(i)MjP(i) .... MjP(i) = εii...i { ΣP (-1)S(P) MjP(1)MjP(2) .... MjP(k)} . (A.7.4) But (A.6.3) says that {...} = εjj...j det(M), so we find that LHS = εii...i εjj...j det(M) . (A.7.5) This shows that the left side of (A.7.1) is in fact totally antisymmetric with respect to both ir and jr. Meanwhile, we know from (A.1.1) that the RHS of (A.7.3) is totally antisymmetric in the jr indices and can therefore be written RHS = ΣP(-1)S(P) MP(j)iMP(j)i ... MP(j)i = εjj...j { ΣP (-1)S(P) MP(1)iMP(2)i ... MP(k)i } . (A.7.6) But (A.6.5) with j→ i says that {...} = εii...i det(M), so we find that RHS = εjj...j εii...i det(M) (A.7.7) Since LHS = RHS, we have proven (A.7.3) and therefore (A.7.1). QED Corollary 1: Theorem (A.7.1) is valid if we select {j1,j2...jk} = {1,2,..k}. In this case one gets ΣP (-1)S(P) (v1)P(i) (v2)P(i) .... (vk)P(i) = ΣP(-1)S(P) (vP(1))i (vP(2)i ... (vP(k))i = εii...i det(M) where Mab = (va)b (A.7.8) Corollary 2: Application to Kronecker deltas, ΣP (-1)S(P) δjP(i) δjP(i)...δjP(i) = ΣP (-1)S(P) δP(j)i δP(j)i... δP(j)i (A.7.9) Although this is not quite a special case of (A.7.1), the proof is identical if one replaces (A.7.2) by Mab = δab . (A.7.10) _______________________________________________________________________________________________