Appendix C

Appendix C PhL 11.12.15 Notes added 1.14.16. In this early version of Appendix C, I used subscripts and ^ to represent tensor functions which were elements of Lk (non-dual, wedge) and Λk(dual, wedge) . This now seems odd since all my tensor symbols below are script, indicating they are all dual. I was writing Λk Lk but that just makes no sense to me now. Maybe I had different definitions at that time. But in all the below, I have things like T(vP(1), vP(2) ....vP(k)) which is a tensor function in the dual world, so not in Lk. So I will just put this doc into my collection of old appendix C docs. I think the key idea is that I later learned to do all these things "generically" and then apply those generic results separately to tensors and tensor functions. 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 Lemma If f(v1,v2,...vk) = ΣQ(-1)S(Q) F(vQ(1),vQ(2),...vQ(k)) Q = permutations of {1,2..k} (C.1.1) then if P is some particular permutation of {1,2..k}, one can write f (vP(1),vP(2),...vP(k)) = ΣQ(-1)S(Q) F( vP(Q(1)),vP(Q(2)),...vP(Q(k))) . (C.1.2) In the subscripts of the v arguments of F, notice that the summation index (here Q) goes on the right. The sum in (C.1.2) is then F(vP(1),vP(2),...vP(k)) – F(vP(2),vP(1),...vP(k)) + other signed permutations , fulfilling the intention of the notation which is to antisymmetrize F with respect to the arguments. Since P(Q(i)) = (PQ)(i) ≡ PQ(i), we rewrite the above claim as f (vP(1),vP(2),...vP(k)) = ΣQ(-1)S(Q) F( vPQ(1),vPQ(2),...vPQ(k)) . // The Lemma (C.1.3) In the notation of (8.5.11) one can write (C.1.1) as f(v1,v2,...vk) = [Alt(F)](v1,v2,...vk) or just f = Alt(F). (C.1.4) and then the Lemma states that f(vP(1),vP(2),...vP(k)) = [Alt(F)](vP(1),vP(2),...vP(k)) . (C.1.5) Comment: One is tempted to say that, since v1 → vQ(1) in the (C.1.1), one should have vP(1) → vQ(P(1)) in (C.1.2). This is wrong and gives an invalid result which differs from the intended antisymmetrization. What about the rearrangement theorem? f (vP(1),vP(2),...vP(k)) = ΣQ(-1)S(Q) F( vPQ(1),vPQ(2),...vPQ(k)) . // The Lemma Replace Q → P-1Q = XQ to get f (vP(1),vP(2),...vP(k)) = ΣQ(-1)S(XQ) F( vPXQ(1),vPXQ(2),...vPXQ(k)) . // The Lemma = ΣQ(-1)S(Q)(-1)S(P) F( vP(1),vP(2),...vP(k)) = (-1)S(P) ΣQ(-1)S(Q) F( vP(1),vP(2),...vP(k)) = (-1)S(P) k! F( vP(1),vP(2),...vP(k)) ????? This says every term is the same, but that is wrong. 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 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. According to tiny theorem (C.1.2) one may then write (C.2.2) 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')) = (1/k!) ΣQ [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 To show that g(Q) does not depend on Q, consider 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 ***, which says ΣP h(P) = ΣP h(PX) for any fixed permutation X, we can take P→ PX in (C.2.5) without changing the value of the sum. We select X = Q-1. Note that S(Q) = S(Q-1) = S(X) in terms of swap count. Then, 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')) = (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 QX = 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. 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...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 (3.2.1) is proved. QED 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", S → B, and rename the labels, Alt(Ta[k]Bg[k"]) = Alt(Td[k]Bh[k"]) a,g,d,h ϵ {,^} (C.5.6) Now in (C.5.6) set Bg[k"] = Sb[k']Rc[k"] g,b,c ϵ {,^} Bh[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) Then (C.5.6) becomes, 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](T1)a[k] .....(T1)a[k] ] = Alt[(T1)b[k](T1)b[k] .....(T1)b[k] ] ai,bi ϵ {,^} (C.5.10) where ki is the rank of Ti (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(T1)a .....(T1)a ] = Alt[(T1)b(T1)b .....(T1)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.