Appendix C v2

Appendix C PhL 11.12.15 Here after stating a small Lemma we prove three theorems which, as is shown in Section C, can all be summarized in this one statement Alt(TS) = Alt(T^S) = Alt(TS^) = Alt(T^S^) . (C.5.4) or just Alt(TaSb) = Alt(TcSd) a,b,c,d ϵ {,^} (C.5.5) The theorems are intuitively obvious as noted below, but require C.1 The Rearrangement Theorem and two Lemmas 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) The first two sums are just reorderings of the third sum and so equal the third sum. This fact is true because the permutations P of {1,2...k} form a group. It is a fact that for any group G with k elements gi, ga {g1, g2, ....gk} = { gag1, gag2, ....gagk} = {g'1, g'2, ....g'k} = reordering of {g1, g2, ....gk} {g1, g2, ....gk}ga = { g1ga, g2ga, ....gkga} = {g"1, g"2, ....g"k} = reordering of {g1, g2, ....gk} and therefore Σi f(gagi) = Σif(giga) = Σif(gi) . 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.1.2) 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.1.3) Finally, consider P1P2{1,2...k} = P{1,2...k} = {i1,i2...ik} If P2 causes S2 position swaps and then P1 causes S1 more, then P does S2+S1 total swaps. Thus S(P1P2) = S(P1)+S(P2) (-1)S(PP) = (-1)S(P) (-1)S(P) = (-1)S(PP) (C.1.4) Lemma 1 Consider our usual notation (A.1.1) and (A.3.1) for creating a totally antisymmetric function f from function F, f(v1,v2,...vk) = ΣQ (-1)S(Q) F(vQ(1),vQ(2),...vQ(k)) Q = permutations of {1,2..k} . (C.1.5) Let P be any fixed permutation of {1,2...k}. We can apply the rearrangement theorem to (C.1.5) in two ways, first with Q → PQ and then second with Q → QP. The result is f(v1,v2,...vk) = ΣQ (-1)S(PQ) F(vPQ(1),vPQ(2),...vPQ(k)) = ΣQ (-1)S(QP) F(vQP(1),vQP(2),...vQP(k)) . Since (-1)S(PQ) = (-1)S(P)+S(Q) = (-1)S(P)(-1)S(Q) = (-1)S(QP) we then have f(v1,v2,...vk) = [ ΣQ (-1)S(Q) F(vQ(1),vQ(2),...vQ(k)) ] = (-1)S(P) [ ΣQ (-1)S(Q) F(vPQ(1),vPQ(2),...vPQ(k)) ] Lemma 1 = (-1)S(P) [ ΣQ (-1)S(Q) F(vQP(1),vQP(2),...vQP(k)) ] . (C.1.6) Lemma 2 In (C.1.5), f(v1,v2,...vk) = ΣQ(-1)S(Q) F(vQ(1),vQ(2),...vQ(k)) Q = permutations of {1,2..k} , (C.1.5) replace subscripts {1,2...k} with some permuted subscripts {P(1),P(2),...P(k)} = P{1,2..k} to get f (vP(1),vP(2),...vP(k)) = ΣQ(-1)S(Q) F(vQ(P(1)),vQ(P(2)),...vQ(P(k))) = ΣQ (-1)S(Q)F(vQP(1),vQP(2),...vQP(k)) = (-1)S(P) f(v1,v2,...vk) // from Lemma 1 line 3 = (-1)S(P) [ ΣQ (-1)S(Q) F(vQ(1),vQ(2),...vQ(k))] // from (C.1.2) Lemma 2 = ΣQ (-1)S(Q) F(vPQ(1),vPQ(2),...vPQ(k)) // from Lemma 1 line 2 (C.1.7) Any of these forms for f (vP(1),vP(2),...vP(k)) we regard as Lemma 2. C.2 Theorem One ΣP (-1)S(P) T(vP(1), vP(2) ....vP(k))S(vP(k+1), vP(k+2) ....vP(k+k')) = ΣP (-1)S(P) T^(vP(1), vP(2) ....vP(k))S(vP(k+1), vP(k+2) ....vP(k+k')) (C.2.1) Functions F belong to Lk while functions F^ belong to Λk Lk. Recall from (8.5.10) that T^(v1,v2....vk) = (1/k!) ΣQ (-1)S(Q)T(vQ(1),vQ(2)....vQ(k)) . (8.5.10) (C.2.2) The function T^ is an antisymmetrized version of the function T which in general has no particular symmetry relative to its arguments. Theorem (C.2.1) says that if we "pre-antisymmetrize" the function T by replacing it by T^, it makes no difference in the overall antisymmetrization of the product of functions. This certainly seems intuitively correct, but we nevertheless provide a formal proof. Using Lemma 2 we write (C.2.2) for permuted arguments as T^(vP(1),vP(2),...vP(k)) = (1/k!) ΣQ (-1)S(Q)T(vPQ(1),vPQ(2)....vPQ(k)) . (C.2.3) We start then with the second line of (C.2.1), ΣP (-1)S(P) T^(vP(1), vP(2) ....vP(k))S(vP(k+1), vP(k+2) ....vP(k+k')) = ΣP (-1)S(P) [ (1/k!) ΣQ (-1)S(Q)T(vPQ(1),vPQ(2)....vPQ(k))] S(vP(k+1), vP(k+2) ....vP(k+k')) = ΣQ [(1/k!) ΣP (-1)S(P) (-1)S(Q)T(vPQ(1),vPQ(2)....vPQ(k))S(vP(k+1), vP(k+2) ....vP(k+k'))] = ΣQ g(Q) (C.2.4) where g(Q) ≡ (1/k!)ΣP(-1)S(P)(-1)S(Q)T(vPQ(1),vPQ(2)...vPQ(k))S(vP(k+1), vP(k+2)...vP(k+k')) . (C.2.5) Below we shall show that, despite appearances, g(Q) does not depend on Q and we can then use Q = 1 (the identity permutation for Q) to evaluate g(Q) : g(Q) = g(1) = (1/k!) ΣP (-1)S(P)T(vP(1),vP(2)....vQP(k))S(vP(k+1), vP(k+2) ....vP(k+k')) (C.2.6) Then we find that ΣP (-1)S(P) T^(vP(1), vP(2) ....vP(k))S(vP(k+1), vP(k+2) ....vP(k+k')) = ΣQ g(Q) = ΣQ g(1) = g(1) ΣQ(1) = g(1) k! = k! g(1) = ΣP (-1)S(P)T(vP(1),vP(2)....vQP(k))S(vP(k+1), vP(k+2) ....vP(k+k')) (C.2.7) and so theorem (C.2.1) is proved. QED Loose End: To show that g(Q) does not depend on Q, we repeat from (C.2.5), g(Q) ≡ (1/k!)ΣP (-1)S(P) (-1)S(Q)T(vPQ(1),vPQ(2)....vPQ(k))S(vP(k+1), vP(k+2) ....vP(k+k')) . (C.2.5) Recall that P is a permutation of {1,2...k+k'} whereas Q is a permutation of {1,2..k}. We can extend the domain of permutation Q so it acts on {1,2...k+k'}, but it can only alter the first k integers. With this extension, we can think of both P and Q as permutations of {1,2...k+k'}. According to the rearrangement theorem (C.1.1) we can take P→ PX in (C.2.5) without changing the value of the sum. We select X = Q-1. Recall that S(Q) = S(Q-1) = S(X) in terms of swap count. Then, g(Q) = (1/k!)ΣP(-1)S(PX)(-1)S(X)T(vPXQ(1),vPXQ(2)....vPXQ(k))S(vPX(k+1), vPX(k+2) ....vPX(k+k')) . But (-1)S(PX) (-1)S(X) = [(-1)S(P) (-1)S(X)] (-1)S(X) = (-1)S(P), and XQ = 1 so then g(Q) ≡ (1/k!)ΣP (-1)S(P) T(vP(1),vP(2)....vP(k))S(vPX(k+1), vPX(k+2) ....vPX(k+k')) Now we claim that vPX(k+1) = vP(k+1) . The reason is that, since our extended permutation Q acts only on the first k integers of {1,2...k+k'}, then Q-1 = X also only acts on these first k indices, since Q-1 is just some other element of the permutation group of these integers. Thus X(k+1) = (k+1) and similarly for the other higher argument subscripts. Then g(Q) ≡ (1/k!)ΣP (-1)S(P) T(vP(1),vP(2)....vP(k))S(vP(k+1), vP(k+2) ....vP(k+k')) and, as claimed, g(Q) does not depend on Q. C.3 Theorem Two ΣP (-1)S(P) T(vP(1), vP(2) ....vP(k))S(vP(k+1), vP(k+2) ....vP(k+k')) = ΣP (-1)S(P) T(vP(1), vP(2) ....vP(k))S^(vP(k+1), vP(k+2) ....vP(k+k')) (C.3.1) This theorem is similar to (C.2.1) where we now do → ^ on the second function S. We briefly mimic the steps from Section C.2, omitting most of the words. The summation S^ index is R instead of Q. S^(vk+1,vk+2....vk+k') = (1/k'!) ΣR (-1)S(R)S(vR(k+1),vR(k+2)....vR(k+k')) . (C.3.2) S^(vP(k+1), vP(k+2) ....vP(k+k')) = (1/k'!) ΣR (-1)S(R)S(vPR(k+1),vPR(k+2)....vPR(k+k')) (C.3.3) ΣP (-1)S(P) T(vP(1), vP(2) ....vP(k))S^(vP(k+1), vP(k+2) ....vP(k+k')) = ΣP (-1)S(P) T(vP(1), vP(2) ....vP(k)) [(1/k'!) ΣR (-1)S(R)S(vPR(k+1),vPR(k+2)....vPR(k+k')) ] = ΣR g(R) (C.3.4) g(R) ≡ (1/k'!)ΣP(-1)S(P)(-1)S(R)T(vP(1),vP(2)...vP(k))S(vPR(k+1), vPR(k+2)...vPR(k+k')) . (C.3.5) To show g(R) = independent of R, take P→ PY where Y = R-1 (YR=1) and use the rearrangement theorem to rewrite g(R), (1/k'!) ΣP (-1)S(PY) (-1)S(R)T(vPY(1),vPY(2)....vPY(k))S(vPYR(k+1), vPYR(k+2) ....vPYR(k+k')) = (1/k'!) ΣP (-1)S(P)T(vPY(1),vPY(2)....vPY(k))S(vP(k+1), vP(k+2) ....vP(k+k')) = (1/k'!) ΣP (-1)S(P)T(vP(1),vP(2)....vP(k))S(vP(k+1), vP(k+2) ....vP(k+k')) = g(1) . (C.3.6) In the last line vPY(1) = vP(1) because Y only acts on the upper index set {k+1...k+k'}. Therefore ΣP (-1)S(P) T(vP(1), vP(2) ....vP(k))S^(vP(k+1), vP(k+2) ....vP(k+k')) = ΣR g(R) = ΣR g(1) = g(1) ΣR(1) = g(1) k'! = k'! g(1) = ΣP (-1)S(P)T(vP(1),vP(2)....vP(k))S(vP(k+1), vP(k+2) ....vP(k+k')) (C.3.7) so theorem (C.3.1) is proved. QED C.4 Theorem Three ΣP (-1)S(P) T(vP(1), vP(2) ....vP(k))S(vP(k+1), vP(k+2) ....vP(k+k')) = ΣP (-1)S(P) T^(vP(1), vP(2) ....vP(k))S^(vP(k+1), vP(k+2) ....vP(k+k')) (C.4.1) Now we do → ^ in both functions at once. Start with the second line and install (C.2.3) for T^ and (C.3.3) for S^ to get ΣP (-1)S(P) T^(vP(1), vP(2) ....vP(k))S^(vP(k+1), vP(k+2) ....vP(k+k')) = ΣP (-1)S(P)[ (1/k!) ΣQ (-1)S(Q)T(vPQ(1),vPQ(2)....vPQ(k))] [(1/k'!) ΣR (-1)S(R)S(vPR(k+1),vPR(k+2)....vPR(k+k'))] = ΣQ,R g(Q,R) (C.4.2) where g(Q,R) = (1/k!)(1/k'!)ΣP (-1)S(P)(-1)S(Q)(-1)S(R) * T(vPQ(1),vPQ(2)....vPQ(k))S(vPR(k+1),vPR(k+2)....vPR(k+k')) . We now show g(Q,R) is independent of both Q and R. First, use the rearrangement theorem with P→PX with X=Q-1 to get = (1/k!)(1/k'!)ΣP (-1)S(Q)(-1)S(R)T(vPXQ(1),vPXQ(2)....vPXQ(k))S(vPXR(k+1),vPXR(k+2)....vPXR(k+k')) g(Q,R) = (1/k!)(1/k'!)ΣP (-1)S(PX)(-1)S(Q)(-1)S(R) T(vPXQ(1),vPXQ(2)....vPXQ(k))S(vPXR(k+1),vPXR(k+2)....vPXR(k+k')) = (1/k!)(1/k'!)ΣP (-1)S(P)(-1)S(R) T(vP(1),vP(2)....vP(k))S(vPXR(k+1),vPXR(k+2)....vPXR(k+k')) Now XR = RX because X=Q-1 only affects the lower indices, while R only affects the higher ones. Therefore vPXR(k+1) = vPRX(k+1). But X has no effect on an upper index like (k+1) so = vPR(k+1). So, g(Q,R) = (1/k!)(1/k'!)ΣP (-1)S(P)(-1)S(R) T(vP(1),vP(2)....vP(k))S(vPR(k+1),vPR(k+2)....vPR(k+k')) and Q has left the playing field. Next, use the rearrangement theorem again this time with P→PY with Y = R-1 to get g(Q,R) = (1/k!)(1/k'!)ΣP (-1)S(PY)(-1)S(R) T(vPY(1),vPY(2)....vPY(k))S(vPYR(k+1),vPYR(k+2)....vPYR(k+k')) = (1/k!)(1/k'!)ΣP (-1)S(P)T(vP(1),vP(2)....vP(k))S(vP(k+1),vP(k+2)....vP(k+k')) = g(1,1) In the left factor vPY(1) = vP(1) because Y = R-1 has no effect on lower indices like 1,2..k. So now g(Q,R) is independent of both Q and R so we replace it with g(1,1). Then ΣP (-1)S(P) T^(vP(1), vP(2) ....vP(k))S^(vP(k+1), vP(k+2) ....vP(k+k')) = ΣQ,R g(Q,R) = ΣQ,R g(1,1) = g(1,1)[ΣQ(1)][ΣR(1)] = g(1,1) k! k'! = k! k'! g(1,1) = ΣP (-1)S(P)T(vP(1),vP(2)....vP(k))S(vP(k+1),vP(k+2)....vP(k+k')) QED. C.5 Theorems in compact Alt notation It is painful to develop further "theorems" using the cumbersome notation above. Here we adopt a shorthand notation with the idea that, if we wanted, we could write out each line in the full notation used above. Recall the definition of Alt for functions, [Alt(X)](v1,v2....vk) ≡ (1/k!) ΣP (-1)S(P)X(vP(1),vP(2)....vP(k)) . (8.5.11) Using this definition, we can translate Theorem (C.2.1), ΣP (-1)S(P) T(vP(1), vP(2) ....vP(k))S(vP(k+1), vP(k+2) ....vP(k+k')) = ΣP (-1)S(P) T^(vP(1), vP(2) ....vP(k))S(vP(k+1), vP(k+2) ....vP(k+k')), (C.2.1) into the compact notation shown on the first line below, Alt(T[k]S[k']) = Alt(T^[k]S[k']) (C.2.1) (C.5.1) Alt(T[k]S[k']) = Alt(T[k]S^[k']) (C.3.1) (C.5.2) Alt(T[k]S[k']) = Alt(T^[k]S^[k']) (C.4.1) (C.5.3) and similarly for the last two lines. For example T[k] means that T[k] ϵ Lk and has k vector arguments, while S^[k'] ϵ Λk and has k' vector arguments. The arguments themselves are in the implied sequential groups: {1,2...k} for the first tensor, {k+1,k+2....k+k'} for the second, and so on for more tensors. If there were a third tensor of type [k"], its index group would be {k+k'+1,k+k'+2, ...k+k'+k"}. Combining the above three equations we get, Alt(T[k]S[k']) = Alt(T^[k]S[k']) = Alt(T[k]S^[k']) = Alt(T^[k]S^[k']) . (C.5.4) We combine these into a single equation as follows Alt(Ta[k]Sb[k']) = Alt(Tc[k]Sd[k']) a,b,c,d ϵ {,^} (C.5.5) where a,b,c,d can each take either value or ^. Now take k' → k'+k", S → B, and rename the labels, Alt(Ta[k]Bg[k'+k"]) = Alt(Td[k]Bh[k'+k"]) a,g,d,h ϵ {,^} (C.5.6) Here k'+k" is the assumed rank of tensors B and B^. Now in (C.5.6) set Bg[k'+k"] = Sb[k']Rc[k"] g,b,c ϵ {,^} Bh[k'+k"] = Se[k']Rf[k"] h,e,f ϵ {,^} . (C.5.7) . which says merely that we are assuming the functions Bg and Bh have the following factored forms, Bg(vP(k+1), vP(k+2) ....vP(k+k")) = Sb(vP(k+1), vP(k+2) ....vP(k+k'))Rc(vP(k+k'+1), vP(k+k'+2) ....vP(k+k'+k")) Bh(vP(k+1), vP(k+2) ....vP(k+k")) = Se(vP(k+1), vP(k+2) ....vP(k+k'))Rf(vP(k+k'+1), vP(k+k'+2) ....vP(k+k'+k")) . (C.5.8) Inserting (C.5.7) into (C.5.6) gives, Alt(Ta[k]Sb[k']Rc[k"]) = Alt(Td[k]Se[k']Rf[k"]) a,b,c,d,e,f ϵ {,^} (C.5.9) where each label a,b,c,d,e,f can independently take either value in {,^}. This is our desired new theorem, which could be derived "in longhand" using the methods of earlier sections of this Appendix. The reader can see that this result can be generalized as follows, Alt[(T1)a[k](T2)a[k] .....(TN)a[k] ] = Alt[(T1)b[k](T2)b[k] .....(TN)b[k] ] ai,bi ϵ {,^} (C.5.10) where ki is the rank of Ti (rank = number of vector arguments), and where each of the ai and each of the bi can independently be set to either or ^. Since the rank k1 can be implied by the name T1, we can restate the above in less cluttered notation Alt[(T1)a(T2)a .....(TN)a ] = Alt[(T1)b(T2)b .....(TN)b ] ai,bi ϵ {,^} (C.5.11) Again, the interpretation of this theorem is simple: it does not matter if one or more tensors are pre-antisymmetrized ( → ^) if one is going to antisymmetrize the entire product with Alt. Tensor Product Notation Using the definitions shown in (8.12.7,8,9) for tensor products of functions, wherever we have an abutted product of functions, a symbol can be inserted. For example, T(v1),v2)....vk)S(vk+1), vk+2 ....vk+k') = (TS)(v1,v2, .....vk+k') or T[k]T[k'] = (TS)[k+k'] Then (C.5.5) Alt(Ta[k]Sb[k']) = Alt(Tc[k]Sd[k']) a,b,c,d ϵ {,^} (C.5.5) becomes Alt[(TaSb)[k+k']] = Alt[(TcSd)[k+k']] a,b,c,d ϵ {,^} Similarly, (C.5.9), Alt(Ta[k]Sb[k']Rc[k"]) = Alt(Td[k]Se[k']Rf[k"]) a,b,c,d,e,f ϵ {,^} (C.5.9) becomes Alt[(TaSbRc)[k+k'+k"]] = Alt[(TdSeRf)[k+k'+k"]] Displaying the ranks is not necessary, so we write Alt(TaSb) = Alt(TcSd) a,b,c,d ϵ {,^} (C.5.12) Alt(TaSbRc) = Alt(TeSeRf) a,b,c,d,e,f ϵ {,^} (C.5.13) Alt[(T1)a(T2)a .....(TN)a ] = Alt[(T1)b(T2)b .....(TN)b], ai,bi ϵ {,^} (C.5.14)