Appendix C v3

Appendix C v3 PhL 11.16.15 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) 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) C.2 Theorem One ΣP (-1)S(P) f(vP(1), vP(2) ....vP(k))g(vP(k+1), vP(k+2) ....vP(k+k')) = ΣP (-1)S(P) f^(vP(1), vP(2) ....vP(k))g(vP(k+1), vP(k+2) ....vP(k+k')) (C.2.1) If we define Alt(h) = (1/k!) ΣP (-1)S(P) h(vP(1), vP(2) ....vP(k)) then Theorem One says this Alt(fg) = Alt(f^g) It looks like a convolution theorem!!! What does that theorem say? Spectral end of Section 3 a(g) = ∫dg1 b(g1-1g)c(g1) Aσkk' = Σk" Bσkk" Cσk"k' This is for a continuous group. For a discrete group I would guess a(g) = Σg1 b(g1-1g)c(g1) and for the permutation group a(P) = ΣQ b(Q-1P)c(Q) Use the rearrangement theorem with Q→PQ, such that Q-1 → (PQ)-1 = Q-1P-1 we get a(P) = ΣQ b(Q-1P-1P)c(PQ) = ΣQ b(Q-1)c(PQ) I should have also stated this rule ΣQ f(Q) = ΣQ f(Q-1) because Q-1 rearranges the group. How do I know that? {g1-1, g2-1, ....gn-1} = {g'1, g'2, ....g'k} = reordering of {g1, g2, ....gk} Where is this theorem stated and what is the theorem called? Suppose we had g'1 = g'2 . Then we would have g1-1 = g2-1 which would mean g1 = g2. Done! So then I have another form a(P) = ΣQ b(Q-1)c(PQ) = ΣQ b(Q)c(PQ-1) So far then we have a(P) = ΣQ b(Q-1P)c(Q) a(P) = ΣQ b(Q-1)c(PQ) a(P) = ΣQ b(Q)c(PQ-1) Now on this third form do rearrangement with Q→QP to get so Q-1 → (QP)-1 = P-1Q-1 a(P) = ΣQ b(QP)c(PP-1Q-1) = ΣQ b(QP)c(Q-1) So here is our full list a(P) = ΣQ b(Q-1P)c(Q) a(P) = ΣQ b(Q-1)c(PQ) a(P) = ΣQ b(Q)c(PQ-1) a(P) = ΣQ b(QP)c(Q-1) Now what does "diagonalization" look like? What exactly is our theorem of interest with this convolution sum form? What is the projection? What are the "representations" of the permutation group? What are the analogs of the Djmm'(g) of the rotation group? The permutation group is non-abelian since PQ ≠ QP. What is a matrix representation of this group? I did that in my Appendix B of Lagrange doc, to wit a ≡ = A = A z0 (B.1.2) where A is an nxn matrix which has 1's in the right places to create this permutation vector a. For example, a = Azo ↔ = . So do I have a general method of constructing the matrices? Consider, = A a = Azo Here A is the permutation group element, whereas a and z0 are inert carriers. But you can think of a representing A relative to z0. Then we should have A = a z0-1 = z0-1 What exactly is z0-1 ? Vectors don't have inverses! Not sure this path is useful right now. What is the general theorem I am trying to prove? Here is one form, ΣP (-1)S(P) b(vP(1), vP(2) ....vP(k))c(vP(k+1), vP(k+2) ....vP(k+k')) = ΣP (-1)S(P) b^(vP(1), vP(2) ....vP(k))c(vP(k+1), vP(k+2) ....vP(k+k')) (C.2.1) where b^(v1,v2....vk) = (1/k!) ΣQ (-1)S(Q)b(vQ(1),vQ(2)....vQ(k)) . (8.5.10) (C.2.2) Does this fit into my convolution mold or not? Write it out in more detail. First write, b^(vP(1),vP(2)....vP(k)) = (1/k!) ΣQ (-1)S(Q)b(vPQ(1),vPQ(2)....vPQ(k)) . (8.5.10) Then we have ΣP (-1)S(P) b(vP(1), vP(2) ....vP(k))c(vP(k+1), vP(k+2) ....vP(k+k')) = ΣP (-1)S(P) (1/k!) ΣQ (-1)S(Q)b(vPQ(1),vPQ(2)....vPQ(k))c(vP(k+1), vP(k+2) ....vP(k+k')) We can extend Q from {1,2..k} to {1,2...k+k'} by stating that Q acts only on the lower integers. That is to say, Q{1,2...k, k+1,k+2...k+k') = {Q(1),Q(2)2...Q(k), k+1,k+2...k+k') Now define z = {1,2...k} Z = {k+1,k+2....k+k'} Then we write the above as Q{z,Z} = {Q(z),Z) Then Q(z) = Q(z) Q(Z) = Z Now make these further definitions, b(vP(1), vP(2) ....vP(k)) = b( vP(z)) z = zlo = {1,2...k} c(vP(k+1), vP(k+2) ....vP(k+k')) = c(vP(Z)) Z = zhi = {k+1,k+2....k+k'} Then for example vP(Z) = vP(QZ) = vPQ(Z) Then my desired theorem can be written this way ΣP (-1)S(P) b( vP(z)) c(vP(Z)) = (1/k!) ΣQ ΣP (-1)S(PQ) b( vPQ(z))c(vP(Z)) In the second line, replace vP(Z) = vPQ(Z) as noted above to get RHS = (1/k!) ΣQ ΣP (-1)S(PQ) b( vPQ(z))c(vPQ(Z)) Now use the rearrangement theorem with P→PQ-1 [ so PQ → P ] RHS = (1/k!) ΣQ ΣP (-1)S(P) b(vP(z))c(vP(Z)) = ΣP (-1)S(P) b( vP(z))c(vP(Z)) [ (1/k!) ΣQ(1) ] = ΣP (-1)S(P) b( vP(z))c(vP(Z)) = LHS Suppose we did not have the (-1)P factor? We could still prove this theorem, ΣP b( vP(z)) c(vP(Z)) = (1/k!) ΣQ ΣP b( vPQ(z))c(vP(Z)) This would be the corresponding theorem for symmetrization! Hurray! How would this then fit into the component index world? Theorem there says ΣP (-1)S(P) TP(i)P(i)...P(i) SP(i)P(i)....P(i) = ΣP (-1)S(P) T^P(i)P(i)...P(i) SP(i)P(i)....P(i) where T^P(i)P(i)...P(i) = (1/k!) ΣQ (-1)Q TPQ(i)PQ(i)...PQ(i) Now write TP(i)P(i)...P(i) = TP(i) SP(i)P(i)....P(i) = SP(i) TQ(i)Q(i)...Q(i) = TQ(i) TPQ(i)PQ(i)...PQ(i) = TPQ(i) Then our theorem claims that ΣP (-1)S(P) TP(i) SP(i) = ΣP (-1)S(P) { (1/k!) ΣQ (-1)Q TPQ(i)} SP(i) or ΣP (-1)S(P) TP(i) SP(i) = (1/k!) ΣQ (-1)S(PQ) TPQ(i) SP(i) Now we know that PQ(z) = P(z) and therefore TPQ(i) = TP(i) and therefore our theorem says ΣP (-1)S(P) TP(i) SP(i) = (1/k!) ΣQ (-1)S(PQ) TPQ(i) SPQ(i) Then use the rearrangement theorem on the right to get ΣP (-1)S(P) TP(i) SP(i) = (1/k!) ΣQ (-1)S(P) TP(i) SP(i) QED Now how to fit this into the general template? b( v(z)) = Ti It is really just a function of z. Desired general theorem ΣP (-1)S(P) b( vP(z)) c(vP(Z)) = (1/k!) ΣQ ΣP (-1)S(PQ) b( vPQ(z))c(vP(Z)) could be written this way ΣP (-1)S(P) b(P(z)) c(P(Z)) = (1/k!) ΣQ ΣP (-1)S(PQ) b[PQ(z)]c[P(Z)] Then our two applications would be b(z) = B(vz) b(z) = Ti ************************************************************* Appendix C 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'}. We can then apply P to *** to get, P(Q'(z,Z) = (PQ')(z,Z) = PQ'(z,Z) = (PQ'(z), PQ'(Z)) = (PQ(z), PQ'(Z)) // = (PQ(z), P(Q'(Z)) // = (PQ(z), P(Z)) // (C.2.9) 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.2.10) 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 f^[z] ≡ (1/k!) ΣQ (-1)σS(Q) f[Q(z)] σ = 1 or 0 (C.2.1) 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.2.10) and (C.2.5a) (C.2.2) 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.2.3) 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.2.4) Putting (C.2.4) into (C.2.3) gives P f^[z] = (1/k!) ΣQ (-1)σS(Q') f [PQ'(z)] (C.2.5) Now rewrite the claimed theorem (C.2.1) 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)] } F[P(Z)] // (C.2.5) = (1/k!) ΣQ ΣP (-1)σS(PQ') f[PQ'(z)]} F[P(Z)] // reorder and (****) = (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 = 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 (****) = LHS QED Maybe it is OK !!!! ???? ************************************ So everything now hinges on Lemma 1. I have to show that f^[P(z)] = ΣQ (-1)σS(Q') f[Q'P(z)] = ΣQ (-1)σS(Q') f[PQ'(z)] Start with f^[z] = ΣQ (-1)σS(Q) f[Q(z)] Let P be a permutation of {1,2...k}. Apply the rearrangement theorem to *** with Q→ QP to get f^[z] = ΣQ (-1)σS(QP) f[QP(z)] = (-1)σS(P) ΣQ (-1)σS(Q) f[QP(z)] Now instead apply the rearrangement theorem to ** with Q → PQ f^[z] = ΣQ (-1)σS(PQ) f[PQ(z)] = (-1)σS(P) ΣQ (-1)σS(Q) f[PQ(z)] Therefore we have shown that ΣQ (-1)σS(Q) f[QP(z)] = ΣQ (-1)σS(Q) f[PQ(z)] ** where Q and P are permutations of z = {1,2....k} . Now the Big Problem: how do I extend the above to P' and Q' so that, ΣQ (-1)σS(Q) f[Q'P'(z)] = ΣQ (-1)σS(Q) f[P'Q'(z)] where P' = (P,P") Q' = (Q, 1) ? Q'P'(z) = Q' (P(z),P"(z)) = (Q'P(z),Q'P"(z)) = ??? Example: P'{1,2,3,4} = {4,2,3,1} Q'{4,2,3,1} = { Q'(4),Q'(2),Q'(3),Q'(1)} = { 4, Q'(2), 3, Q'(1) } Need a better notation. How about Z = z,z' = low, high sets Consider f^[Z] = ΣQ(-1)S(Q) f(Q(Z)) Q = full permutations now I could then show using the full rearrangement theorem in the larger space that f^[P(Z)] = ΣQ(-1)S(Q) f(QP(Z)) = ΣQ(-1)S(Q) f(PQ(Z)) Now let q be a permutation of just {1,2...k}. Then write Q(z,z') = ( Q(z),Q(z')) = (Q(z), z') P(z,z') = ( P(z),P(z')) = nothing else to be said. Q = (q, 1) P = (p, p') QP(Z) = ( QP(z), QP(z') ) = nothing else can be said PQ(Z) = ( PQ(z), PQ(z') ) = ( PQ(z), P(z') ) When ΣQ is over all permutations of the full set, there are (k+k')! terms, I will never be able to reduce that to some kind of Σq. ******************* start over I know that, ΣQ (-1)σS(Q) f[QP(z)] = ΣQ (-1)σS(Q) f[PQ(z)] ** where Q and P are permutations of z = {1,2....k} . How do I extend the above to P' and Q' so that, ΣQ (-1)σS(Q) f[Q'P'(z)] = ΣQ (-1)σS(Q) f[P'Q'(z)] where P' = (P,P2) Q' = (Q, 1) ? One way would be to show that QP(z) = Q'P'(z) (1) P'Q'(z) = PQ(z) (2) I don't think these are true. For example Q'P'(z) = ? Here P'(z) could generate arbitrary integers mixed into both high and low. Q'P'(z) = Q' { P'(1),P'(2).....P'(k) } = what ?? This is a fundamental issue. What do you MEAN by such an expression! P'{1,2,3,4} = {4,2,3,1} Q'{4,2,3,1} = { Q'(4),Q'(2),Q'(3),Q'(1)} = { 4, Q'(2), 3, Q'(1) } ?? The notation is just not clean! Suppose Q'{1,2,3,4} = {2,1,3,4) This says Q'{1,2,3,4} = {Q'(1),Q'(2),Q'(3),Q'(4)} = {2,1,3,4} Then what is this Q'{4,2,3,1} = ?? Does Q' act on the position, or does it act on the actual integer?? I think it acts on the integer, so Q'{4,2,3,1} = { Q'(4),Q'(2),Q'(3),Q'(1)} = { 4, Q'(2), 3, Q'(1) } = { 4, 1, 3, 2} One way out would be to reinterpret, f^[z] = ΣQ (-1)σS(Q) f[Q(z)] If permutations really act on the "integers" and not the position, then Pf^[z] = f^[(P(z)] = ΣQ (-1)σS(Q) f[Q(P(z))] **********************************8 Go back to the actual problems, 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.2.1) Applying this rule to {i1, i2, i3, i4} = {1,3,2,4} would give Q{1,3,2,4} = {iQ(1),iQ(2),iQ(3),iQ(4)} = What then is this: QP{1,2,3...k} = Q{i1, i2, ....ik} = { Q(i1),Q(i2)....Q(i3)} Here is an important question. Can one say that Q(ir) = iQ(r) ? If this is valid, then we can say QP{1,2,3...k} = { Q(i1),Q(i2)....Q(i3)} = { iQ(1), iQ(2).... iQ(3)} = {ij1, ij2, ...ijk } Let's now look at an example. P{1,2,3,4} = {4,2,3,1} Q{1,2,3,4} = {1,3,2,4} QP{1,2,3,4} = Q{4,2,3,1} = ??? The notation is just not clean! Suppose Q'{1,2,3,4} = {2,1,3,4)