Appendix A v4

This is a whole new Appendix A with new content selected from former appendices creation date 11.18.15 This was installed on 1.26.16, do not edit here. creation date 11.18.15 1 Appendix A: Permutation Support 1 A.1 Rearrangement Theorems and a Determinant Theorem 2 A.2 The Alt Operator 5 A.3 The Sym Operator 9 A.4 More on Alt and Sym and the decomposition of functions 12 A.5 Application to Tensors 14 (a) Alt Equations (translated from Section A.2 above) 14 (b) Sym Equations (translated from Section A.3 above) 15 (c) Alt/Sym and Other Equations (translated from Section A.4 above) 16 A.6 The permutation tensor ε 17 A.7 The wedge-product-of-vectors Alt equation 19 A.8 Application to Tensor Functions 20 (a) Alt Equations (translated from Section A.2 above) 21 (b) Sym Equations (translated from Section A.3 above) 22 (c) Alt/Sym and Other Equations (translated from Section A.4,A.6 and A.7 above) 23 (d) Alt/Sym when there are two sets of indices 24 A.9 The Ordered Sum Theorem 25 A.10 The Alt Operator in Dirac bra-ket notation 27 Appendix A: Permutation Support Summary Section A.1 describes our permutation notation, presents two rearrangement theorems for the permutation group, and then states the familiar determinant theorem det(M) = det(MT) in permutation notation. Section A.2 describes the action of a permutation operator on a generic function f(1,2...k), and states several theorems concerning multiple permutation operators. At the same time, the Alt operator is defined and various facts are proven concerning this operator. The notion of a totally antisymmetric generic function is directly related to the Alt operator. Section A.3 mimics Section A.2 for the Sym operator in place of the Alt operator. The notion of a totally symmetric generic function is directly related to the Sym operator. Section A.4 states some facts which concern both the Alt and the Sym operators together. Up to this point, the various facts and theorems have taken place in a "generic permutation space" which consists of functions of k arguments which are a permutation of 1,2...k, such as f(2,1,3...k). Section A.5 applies all the previous facts and theorems to the permutation space whose elements are the component index subscripts of a rank-k tensor, so f(1,2,3...k) = Tii...i. The results of Sections A.2, A.3 and A.4 are adapted to the tensor world in subsections (a),(b) and (c). Section A.6 deals with the permutation tensor εii...i and shows how it can provide an alternative to the permutation notation in some situations associated with the Alt operator. Section A.7 adapts the above generic results to the case f(1,2,....k) = (vj vj ..... vj) which is the tensor product of k vectors. Then the wedge product of k vectors is defined in terms of this application of the Alt operator, so that (vj ^ vj ^ .....^ vj) ≡ Alt(vj vj ..... vj). Section A.8 is similar to Section A.5, but the facts and theorems are applied not to tensors, but to "tensor functions", so here f(1,2...k) = T(vi,vi....vi). The permutation space is now the set of label subscripts on the k vector arguments of a tensor function. The results of Sections A.2, A.3 and A.4 are adapted to the tensor function world in subsections (a),(b) and (c). Section A.9 proves an obscure ordered permutation sum theorem that is used in (7.4.12). Section A.10 shows how to represent the Alt or Sym operators in Dirac bra-ket notation. A.1 Rearrangement Theorems and a Determinant Theorem Definition: A permutation P (of order k) reorders the list of integers [1,2,3...k] in some manner to give [i1,i2,i3...ik] . (A.1.1) Including the initial ordering [1,2,3...k], there are k! possible permutations. Fact: ΣP(1) = k! // there are k! equal terms in this sum (A.1.2) The permutation group rearrangement theorem states the following: ΣP f(QP) = ΣP f(PQ) = ΣP f(P) . (A.1.3) Here ΣP is a sum over all k! permutations of [1,2...k], 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...k] 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...k] P-1 exists for any P // just the inverse permutation. (A.1.4) 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] . (A.1.5) To show that [g'1, g'2, ....g'k] is a reordering of [g1, g2, ....gk], we have to show that no two elements of [g'1, g'2, ....g'k] 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, ....gk] enumerates the distinct group elements. Therefore Σi f(gagi) = Σif(giga) = Σif(gi) . (A.1.6) This is valid only if the sum is over all elements of the group, which in the rearrangement theorem (A.1.3) 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] [1,2...k] = P-1[i1,i2...ik] PP-1 = P-1P = 1 . (A.1.7) In the above, since [i1,i2...ik] is a permutation of [1,2...k], one can get from [1,2...k] to [i1,i2...ik] by making some number of swaps of the integers in [1,2...k]. Comment: We are following a Maple convention that [a,b,c...] is a "list" where order is significant, whereas {a,b,c...} is a "set" where order is not significant. The swap count S(P) Any two permutations of [1,2,...k] can be linked by a number of pairwise swaps of the integers. For example, if we have P[1,2,...k] = [i1,i2,...ik], one can get from the first integer sequence to the second by doing some number S(P) of pairwise swaps. The integer S(P) is not unique, but whether it is an even or an odd integer is unique, so the factor (-1)S(P) is unique to a particular P (we leave it to the reader to prove this fact) . Sometimes (-1)S(P) is called the parity of permutation P. Example: [1,2,3] → [2,1,3] S(P) = 1 (-1)S(P) = -1 [1,2,3] → [1,3,2] → [2,3,1] → [2,1,3] S(P) = 3 (-1)S(P) = -1 (A.1.8) 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) . (A.1.9) 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) so (-1)S(PP) = (-1)S(P) (-1)S(P) = (-1)S(PP) . (A.1.10) From the above these trivial corollaries follow : (-1)S(PP) = 1, (-1)S(PQ) = (-1)S(QP). (-1)S(P) (-1)S(P) = 1 (-1)S(P) = (-1)S(Q) (-1)S(PQ) (A.1.11) Fact: ΣP(-1)S(P) = 0 (A.1.12) Proof : By the rearrangement theorem (A.1.3) and then (A.1.11) we know that, for any permutation Q, ΣP(-1)S(P) = ΣP(-1)S(QP) = (-1)S(Q)ΣP(-1)S(P) . Select a Q which has (-1)S(Q) = -1. Then ΣP(-1)S(P) = - ΣP(-1)S(P) ΣP(-1)S(P) = 0. QED Another Rearrangement Theorem Another version of the rearrangement theorem is the following, ΣQ f(Q) = ΣQ f(Q-1) . (A.1.13) Again, this is just a reordering of the sum. Consider, {g1-1, g2-1, ....gk-1} = {g1', g2', ....gk'} = reordering of {g1, g2, ....gk} To show that {g1', g2', ....gk'} is a reordering of {g1, g2, ....gk} 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, ....gk} enumerates the distinct group elements. Therefore, Σi f(gi) = Σif(gi-1) (A.1.14) 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) (A.1.15) where dg is called the invariant Haar measure. For SO(3) it is dg = dφd(cosθ)dψ (Euler angles). A Determinant Theorem The determinant of a k x k matrix M can be written in several equivalent ways, det(Mab) = Σii...i εii...iM1iM2i ...Mki = ΣP (-1)S(P) M1P(1)M2P(2) ...MkP(k) = [ M11M22....Mkk + signed permutations of the second index ] . (A.1.16) The permutation tensor εii...i is described in (A.6.1) but is not important right here. It is well known that det(MT) = det(M) (T=transpose, swap rows and columns). The above det(Mab) expressions are therefore also valid if one replaces Mab →MTab = Mba : det(Mab) = Σii...i εii...iMi1Mi2 ...Mik = ΣP (-1)S(P) MP(1)1MP(2)2 ...MP(k)k = [ M11M22....Mkk + signed permutations of the first index ] . (A.1.17) Our Determinant Theorem (not much of a theorem) then states, ΣP (-1)S(P) M1P(1)M2P(2) ...MkP(k) = ΣP (-1)S(P) MP(1)1MP(2)2 ...MP(k)k (A.1.18) or det(M**) = det(MT**) . Basically this says a determinant of a matrix is unchanged if one swaps the rows with the columns. Our main interest is getting this statement expressed in the ΣP notation. Raising the second index to make M be a mixed rank-2 tensor gives ΣP (-1)S(P) M1P(1)M2P(2) ...MkP(k) = ΣP (-1)S(P) MP(1)1MP(2)2 ...MP(k)k (A.1.19) or det(M**) = det(MT**) A.2 The Alt Operator The generic Alt operator acts on a function f of the integers [1,2....k] to create a new function, g = Alt(f), as follows: g(1,2...k) = [Alt(f)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) (A.2.1) where f is any function and P are the k! permutations of the set of integers [1,2....k]. Informally we write the above as g(1,2...k) = (1/k!) [ f(1,2...k) - f(2,1...k) + other signed permutations ] Examples: g(1,2) = [Alt(f)](1,2) = (1/2) [ f(1,2) - f(2,1) ] (A.2.2) g(1,2,3) = [Alt(f)](1,2,3) = (1/6) [ f(1,2,3) - f(1,3,2) + f(3,1,2) - f(3,2,1) + f(2,3,1) - f(2,1,3) ] . Now, let R be some permutation of [1,2....k]. We write R[1,2....k] = [R(1),R(2).....R(k)] (A.2.3) where for example R(1) gives the integer into which 1 is converted by the permutation R. We can apply the operator R to a function of 1,2...k in this manner, R f(1,2,...k) = f( R(1),R(2).....R(k) ) . (A.2.4) Fact: Any permutation R is a linear operator, so R( Σiaifi) = Σiai(Rfi) (A.2.5) Proof: Let h(1,2,...k) ≡ Σiaifi(1,2,...k). Then R h(1,2,...k) = h( R(1),R(2).....R(k) ) // (A.2.4) applied to h = Σiaifi( R(1),R(2).....R(k) ) // definition of h = Σiai R fi(1,2,...k) . // (A.2.4) applied to fi QED Now suppose R = QP, the product of two permutations Q and P. Then starting with (A.2.3), (QP) f(1,2,...k) = f( (QP)(1),(QP)(2).....(QP)(k) ) = f( Q(P(1), Q(P(2)).... Q(P(k) ) ≡ f( QP(1), QP(2)... .QP(k) ) . (A.2.6) But from (A.2.4), (QP) f(1,2,...k) = Q { P f(1,2,...k)} = Q f( P(1),P(2).....P(k) ) (A.2.7) Therefore we have shown that Q f( P(1),P(2).....P(k) ) = f( QP(1), QP(2)... .QP(k) ) . (A.2.8) Definition: A function f(1,2..k) is totally antisymmetric if it changes sign when any two arguments are swapped. (A.2.9) Examples: f(1,2) = - f(2,1) f is totally antisymmetric f(1,2,3) = -f(2,1,3) f(1,2,3) = -f(3,2,1) f is totally antisymmetric f(1,2,3) = -f(1,3,1) Fact: f(1,2..k) totally antisymmetric P f(1,2,3..k) = (-1)S(P) f(1,2,3..k) (A.2.10) Proof: [] S(P) is the number of pairwise swaps going from [1,2,3...k] to P[1,2,3...k] = [i1,i2,i3...in]. If f is totally antisymmetric by the definition above, each such swap causes a minus sign, and the product of these minus signs is then (-1)S(P). [] If P = any pairwise swap, (-1)S(P) = -1, so f(1,2..k) is then totally antisymmetric. Fact: The function g(1,2..k) ≡ [Alt(f)](1,2...k) is totally antisymmetric in its arguments. (A.2.11) Proof: Let Q be some permutation of [1,2....k]. Then apply Q to the function g(1,2..k), Q g(1,2..k) = Q { (1/k!) ΣP (-1)S(P)f(P(1),P(2)...P(k)) } // definition of g = (1/k!) ΣP (-1)S(P) Q f(P(1),P(2)...P(k)) // (A.2.5), Q is linear = (1/k!) ΣP (-1)S(P) f(QP(1),QP(2)...QP(k)) // (A.2.8) = (-1)S(Q) (1/k!) ΣP (-1)S(QP) f(QP(1),QP(2)...QP(k)) // (A.1.11) = (-1)S(Q) (1/k!) ΣP (-1)S(P) f(P(1),P(2)...P(k)) // (A.1.3), rearrang. thm. = (-1)S(Q) g(1,2..k) // definition of g By (A.2.10) it follows that g(1,2,..k) is totally antisymmetric. QED Fact: Alt is a linear operator, so Alt(Σiaifi) = Σiai Alt(fi) . (A.2.12) Proof: Let h(1,2,...k) ≡ Σiaifi(1,2,...k). Then [Alt(h)](1,2,...k) = (1/k!) ΣP (-1)S(P)h( P(1),P(2)...P(k) ) // (A.2.1) def of Alt(h) = (1/k!) ΣP (-1)S(P){ Σiaifi(P(1),P(2)...P(k)) } // definition of h = Σiai [ (1/k!) ΣP (-1)S(P) fi(P(1),P(2)...P(k)) ] // reorder sums = Σiai Alt(fi) // (A.2.1) def of Alt(fi) Fact: Alt is a projection operator, so Alt(Alt(f)) = Alt(f) . (A.2.13) Comment: This is why (1/k!) is included in the definition of Alt. Proof: By (A.2.11) we know that Alt is a totally antisymmetric function, and therefore from (A.2.10), P [Alt(f)](1,2...k)] = (-1)S(P)[Alt(f)](1,2...k)] . (A.2.14) Next, consider that [Alt(f)](P(1),P(2)...P(k) ) = P [Alt(f)](1,2...k) // (A.2.4) applied with f→ Alt(f), R→P = (-1)S(P)[Alt(f)](1,2...k) . // (A.2.14) (A.2.15) Now examine Alt(Alt(f)) : [Alt(Alt(f))](1,2...k) = (1/k!) ΣP (-1)S(P)[Alt(f)]( P(1),P(2)...P(k) ) = (1/k!) ΣP (-1)S(P){(-1)S(P)[Alt(f)](1,2...k)]} // (A.2.15) = (1/k!) ΣP[Alt(f)](1,2...k)]} // (-1)S(P)(-1)S(P) = 1 = [Alt(f)](1,2...k)] {(1/k!)ΣP(1)} // reorder factors = [Alt(f)](1,2...k)] {1} . // (A.1.2) QED Fact: If f is a totally antisymmetric function, then Alt(f) = f . (A.2.16) Proof: Alt(f)(1,2...k) = (1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) // definition of Alt(f) (A.2.1) = (1/k!) ΣP (-1)S(P) P f(1,2,...k) // (A.2.4) with R→P = (1/k!) ΣP (-1)S(P) (-1)S(P) f(1,2,...k) // (A.2.10) = (1/k!) ΣP f(1,2,...k) // (A.1.11) = (1/k!) f(1,2,...k) { ΣP (1) } // reorder = f(1,2,...k) // (A.1.2) Fact: If f is totally antisymmetric, then f(1,2,3...k) =( -1)k-1 f(2, 3,... k-1, k, 1) forward cyclic (A.2.17) f(1,2,3...k) = (-1)k-1 f(k, 1, 2, 3,... k-1) backward cyclic Proof: Let B be the particular permutation which does this: B[1,2,3..k-1,k] = B[2,3,...k-1,k,1] (Backward cyclic). One then has S(B) = k-1 because it takes k-1 swaps to move the 1 from one end to the other. If f is totally antisymmetric, then according to (A.2.10) one has B f(1,2,3..k) = (-1)S(B) f(1,2,3..k) so then f(2,3,... k,1) = B f(1,2,3..k) = (-1)S(B) f(1,2,3..k) = (-1)k-1 f(1,2,3...k) . On the other hand, if F[1,2,3..k-1,k] = [k, 1, 2, 3,... k-1] (Forward cyclic), S(F) = k-1 for the same reason, and then f(k, 1, 2, 3,... k-1) = F f(1,2,3..k) = (-1)S(F) f(1,2,3..k) = (-1)k-1 f(1,2,3...k) . Example: A very commonly used fact is that, for k = 3, (-1)k-1 = (-1)2 = 1 and so f(1,2,3) = f(2,3,1) = f(3,1,2) f totally antisymmetric (A.2.18) Suppose now that function f has k arguments and function g has k' arguments. We then define a meaning for the tensor product in our generic function space as follows Definition: Tensor product: (fg)(1,2,....k+k') ≡ f(1,2...k) g(k+1,k+2....k+k') (A.2.19) A.3 The Sym Operator This section is an obvious copy, paste and edit job on the previous section. We omit what would be (A.3.3) through (A.3.8) since they would be the same as (A.2.3) through (A.2.8). The changes are mainly these: antisymmetric → symmetric (-1)P → 1 Alt → Sym . The Sym operator acts on a function of the integers [1,2....k] to create a new function, g = Sym(f), as follows: g(1,2...k) = [Sym(f)](1,2...k) ≡ (1/k!) ΣPf( P(1),P(2)...P(k) ) (A.3.1) where f is any function and P are the k! permutations of the set of integers [1,2....k]. Informally we write the above as g(1,2...k) = (1/k!) [ g(1,2...k) + g(2,1...k) + other permutations ] Examples: g(1,2) = [Sym(f)](1,2) = (1/2) [ f(1,2) + f(2,1) ] (A.3.2) g(1,2,3) = [Sym(f)](1,2,3) = (1/6) [ f(1,2,3) + f(1,3,2) + f(3,1,2) + f(3,2,1) + f(2,3,1) + f(2,1,3) ] Definition: A function f(1,2..k) is totally symmetric if it is unchanged when any two arguments are swapped. (A.3.9) Examples: f(1,2) = f(2,1) f is totally symmetric f(1,2,3) = f(2,1,3) f(1,2,3) = f(3,2,1) f is totally symmetric f(1,2,3) = f(1,3,1) Fact: f(1,2..k) totally symmetric P f(1,2,3..k) = f(1,2,3..k) (A.3.10) Proof: [] S(P) is the number of pairwise swaps going from [1,2,3...k] to P[1,2,3...k] = [i1,i2,i3...in]. If f is totally symmetric by the definition above, each such swap causes a plus sign, and the product of these plus signs is then 1. [] If P = any pairwise swap, (-1)S(P) = 1, so f(1,2..k) is then totally symmetric. Fact: The function g(1,2..k) ≡ [Sym(f)](1,2...k) is totally symmetric in its arguments. (A.3.11) Proof: Let Q be some permutation of [1,2....k]. Then apply Q to the function g(1,2..k), Q g(1,2..k) = Q { (1/k!) ΣP f(P(1),P(2)...P(k)) } // definition of g = (1/k!) ΣP Q f(P(1),P(2)...P(k)) // (A.2.5), Q is linear = (1/k!) ΣP f(QP(1),QP(2)...QP(k)) // (A.2.8) = (1/k!) ΣP f(P(1),P(2)...P(k)) // (A.1.3), rearrang. thm. = g(1,2..k) // definition of g By (A.3.10) it follows that g(1,2,..k) is totally symmetric. QED Fact: Sym is a linear operator, so Sym(Σiaifi) = Σiai Sym(fi) (A.3.12) Proof: Let h(1,2,...k) ≡ Σiaifi(1,2,...k). Then [Sym(h)](1,2,...k) = (1/k!) ΣP h( P(1),P(2)...P(k) ) // (A.3.1) def of Sym(h) = (1/k!) { Σiaifi(P(1),P(2)...P(k)) } // definition of h = Σiai [ (1/k!) fi(P(1),P(2)...P(k)) ] // reorder sums = Σiai Sym(fi) // (A.3.1) def of Sym(fi) Fact: Sym is a projection operator, so Sym(Sym(f)) = Sym(f) . (A.3.13) Comment: This is why (1/k!) is included in the definition of Sym. Proof: By (A.3.11) we know that Sym is a totally symmetric function, and therefore from (A.3.10), P [Sym(f)](1,2...k)] = [Sym(f)](1,2...k)] . (A.3.14) Next, consider that [Sym(f)](P(1),P(2)...P(k) ) = P [Sym(f)](1,2...k)] // (A.2.4) applied with f→ Sym(f), R→P = [Sym(f)](1,2...k)] . // (A.3.14) (A.3.15) Now examine Sym(Sym(f)) : [Sym(Sym(f))](1,2...k) = (1/k!) ΣP [Sym(f)]( P(1),P(2)...P(k) ) = (1/k!) ΣP {[Sym(f)](1,2...k)]} // (A.3.15) = [Sym(f)](1,2...k)] {(1/k!)ΣP(1)} // reorder = [Sym(f)](1,2...k)] {1} // (A.1.2) QED Fact: If f is a totally symmetric function, then Sym(f) = f . (A.3.16) Proof: Sym(f)(1,2...k) = (1/k!) ΣP f( P(1),P(2)...P(k) ) // definition of Sym(f) (A.3.1) = (1/k!) ΣP P f(1,2,...k) // (A.2.4) with R→P = (1/k!) ΣP f(1,2,...k) // (A.3.10) = (1/k!) f(1,2,...k) { ΣP (1) } // reorder = f(1,2,...k) // (A.1.2) Fact: If f is totally symmetric , then f(1,2,3...k) = f(2, 3,... k-1, k, 1) forward cyclic (A.3.17) f(1,2,3...k) = f(k, 1, 2, 3,... k-1) backward cyclic This is just a special case of (A.3.10) which says Q f(1,2,3..k) = f(1,2,3..k) for any Q, so it certainly true for F = forward cyclic or B = backward cyclic permutations. Example: f(1,2,3) = f(2,3,1) = f(3,1,2) f totally symmetric (A.3.18) For comparison, recall (A.2.18) which said f(1,2,3) = f(2,3,1) = f(3,1,2) f totally antisymmetric (A.2.18) A.4 More on Alt and Sym and the decomposition of functions Fact: The projection operators Alt and Sym are orthogonal, so Alt(Sym(f)) = Sym(Alt(f)) = 0 . (A.4.1) Proof left: Alt(Sym(f)) = Alt( {(1/k!) ΣPf( P(1),P(2)...P(k) ) } = (1/k!) ΣP [Alt(f)]( P(1),P(2)...P(k) ) ] // (A.2.12), Alt is linear = (1/k!) ΣP (-1)S(P)[Alt(f)](1,2...k) // (A.2.15) = {(1/k!) [Alt(f)](1,2...k)} {ΣP (-1)S(P)} // reorder = {(1/k!) [Alt(f)](1,2...k)} {0} // (A.1.12) = 0 Proof right: Sym(Alt(f)) = Sym( {(1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) } = (1/k!) ΣP (-1)S(P)[Sym(f)]( P(1),P(2)...P(k) ) ] // (A.3.12), Sym is linear = (1/k!) ΣP (-1)S(P)[Sym(f)](1,2...k) // (A.3.15) = {(1/k!) [Sym(f)](1,2...k)} {ΣP (-1)S(P)} // reorder = {(1/k!) [Sym(f)](1,2...k)} {0} // (A.1.12) = 0 We can define a third projection operator this way, Else() ≡ 1 - Alt() - Sym() // projection operator Else(f) = f - Alt(f) - Sym(f) . // applied to f(1,2,3...k) (A.4.2) One can then decompose an arbitrary function f into three pieces, f = Alt(f) + Sym(f) + Else(f) = a + s + e // a = a(1,2....k) etc (A.4.3) where the "else" piece is whatever is left over, which is to say, e ≡ f - a - s. Then consider, Alt(f) = Alt(a + s + e) = Alt(a) + Alt(s) + Alt(e) // (A.2.12), Alt is linear = Alt(Alt(f)) + Alt(Sym(f)) + Alt(Else(f)) // (7.5.14) = Alt(f) + 0 + Alt(Else(f)) // (A.2.13) and (A.3.1) Alt(Else(f)) = 0 (A.4.4) Sym(f) = Sym(a + s + e) = Sym(a) + Sym(s) + Sym(e) // (A.2.12), Alt is linear = Sym(Alt(f)) + Sym(Sym(f)) + Sym(Else(f)) // (7.5.14) = 0 + Sym(f) + Sym(Else(f)) // (A.3.13) and (A.3.1) Sym(Else(f)) = 0 (A.4.5) This verifies that the "else" piece e of a function has neither a totally antisymmetric nor a totally symmetric component. Example 1: For a function f(1,2) one has from (A.2.2) and (A.3.2), a(1,2) = (1/2) [ f(1,2) - f(2,1)] s(1,2) = (1/2) [ f(1,2) + f(2,1) ] e(1,2) = f(1,2) - a(1,2) - s(1,2) = 0 (A.4.6) so the leftover else piece e(1,2) is null. Example 2: On the other hand, for a function f(1,2,3) one has from (A.2.2) and (A.3.2) a(1,2,3) =(1/6) [ f(1,2,3) - f(1,3,2) + f(3,1,2) - f(3,2,1) + f(2,3,1) - f(2,1,3) ] // (A.2.2) s(1,2,3) =(1/6) [ f(1,2,3) + f(1,3,2) + f(3,1,2) + f(3,2,1) + f(2,3,1) + f(2,1,3) ] // (A.3.2) e(1,2,3) = f(1,2,3) - (1/6) [ f(1,2,3) - f(1,3,2) + f(3,1,2) - f(3,2,1) + f(2,3,1) - f(2,1,3) ] - (1/6) [ f(1,2,3) + f(1,3,2) + f(3,1,2) + f(3,2,1) + f(2,3,1) + f(2,1,3) ] = f(1,2,3) - (1/3) [ f(1,2,3) + f(3,1,2) + f(2,3,1) ] = (2/3) f(1,2,3) - (1/3) [ f(3,1,2) + f(2,3,1) ] (A.4.7) so in this case the leftover piece e(1,2,3) is not null. In the case that f is either totally antisymmetric or totally symmetric, we know from (A.2.18) and (A.3.18) that all cyclic permutations of f are the same (for k = odd). In these cases, we can see explicitly from (A.4.7) that e(1,2,3) = 0, as expected. A.5 Application to Tensors We now restate the "generic" results of Sections A.2, A.3 and A.4 for this special case: f(1,2...k) = Tii...i . // a "tensor" T ϵ Vk . (A.5.1) Here T is any rank-k tensor (either in the weak or strong sense mentioned in ***). This f seems perhaps an odd looking "function", but one can consider it to be just an evaluation of this more respectable mapping, f(a,b,c,...q) = Tiii...i a,b,c... ϵ {1,2...k} f: {1,2...k}k → Vk . (A.5.2) This technical mapping issue is not important because we are just regarding Tii...i as a "carrier" of the labels 1,2,3..k, from the point of view of doing permutations. The actual indices like i1 could be arbitrary objects (labeled pancakes) as far as the permutation theorems are concerned, but in our applications we have in mind that i1 is an integer in the range 1,2....n where n = dim(V) and n is unrelated to the tensor rank k. In Section A.5 we shall instead apply our results to "tensor functions", f(1,2,3...k) = T(vi, vi, ....vi) Again, from a permutation point of view, T(vi, vi, ....vi) is just a carrier of the labels 1,2...k. The permutation theorems don't care whether or not vi happens to be a vector in V labeled by i1, or even whether or not vi happens to be an argument of a function T. Here then are some Section A.2,A.3,A.4 results translated according to f(1,2,3...k) = Tii...i . For some of the translations, we show the actual equation from above, then its translation. For others we just state the translated result. In all the results below, one can always specialize to the case i1,i2...ik → 1,2,...k. The resulting equations are then as if our mapping were f(1,2...k) = T12...k. Note then that iP(r) → i(r) in a superscript. (a) Alt Equations (translated from Section A.2 above) The basic Alt definition of (A.2.1) g(1,2...k) = [Alt(f)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) (A.2.1) becomes, Gii...i = [Alt(F)]ii...i = (1/k!) ΣP (-1)S(P) Fii...i (A.5.3) G = Alt(F) . // definition of Alt acting on a tensor Examples: Gii = [Alt(F)]ii = (1/2) [ Fii - Fii ] Giii = [Alt(F)]iii = (1/6) [ Fiii - Fiii + the other four terms ] (A.5.4) In practice, we might more easily write Gabc = [Alt(F)]abc = (1/6) (Fabc - Facb + Fcab - Fcba + Fbca - Fbac) but when it comes time to prove permutation-related theorems, we use indices like iii . Continuing on, R[1,2....k] = [R(1),R(2).....R(k)] of (A.2.3) becomes, with R→ P, P Tii...i = Tii...i (A.2.3) (A.5.5) Fact: Any permutation P is a linear operator, so P( ΣiaiTiii...i) = Σiai(PTiii...i) (A.2.5) (A.5.6) Definition: A tensor Tii...i is totally antisymmetric if it changes sign when any two superscripts are swapped. (A.2.9) (A.5.7) Example: Tii...i = - Tii...i or Tabc = -Tbac Fact: Tii...i totally antisymmetric P Tii...i = (-1)S(P) Tii...i , where P is any permutation of [1,2..k]. (A.2.10) (A.5.8) Fact: The function Tii...i ≡ [Alt(F)]ii...i is totally antisymmetric in its indices. (A.2.11) (A.5.9) Fact: Alt is a linear operator, so Alt (ΣjajTjii...i) = Σjaj [Alt(Tj)]ii...i. (A.2.12) (A.5.10) Fact: Alt is a projection operator, so Alt(Alt(T)) = Alt(T) . (A.2.13) (A.5.11) Fact: If T is a totally antisymmetric rank-k tensor, then Alt(T) = T . (A.2.16) (A.5.12) (b) Sym Equations (translated from Section A.3 above) The basic Sym definition of (A.3.1) g(1,2...k) = [Sym (f)](1,2...k) ≡ (1/k!) ΣP f( P(1),P(2)...P(k) ) (A.3.1) becomes, Gii...i = [Sym (F)]ii...i = (1/k!) ΣP Fii...i (A.5.12) G = Sym (F) . // definition of Sym acting on a tensor Examples: Gii = [Sym(F)]ii = (1/2) [ Fii + Fii ] Giii = [Sym(F)]iii = (1/6) [ Fiii + Fiii + the other four terms ] (A.5.13) Gabc = [Sym(F)]abc = (1/6) (Fabc + Facb + Fcab+ Fcba + Fbca + Fbac) Definition: A tensor Tii...i is totally symmetric if it is unchanged when any two superscripts are swapped. (A.3.9) (A.5.14) Example: Tii...i = Tii...i or Tabc = Tbac = Tacb Fact: Tii...i totally symmetric P Tii...i = Tii...i, where P is any permutation of [1,2..k]. (A.3.10) (A.5.15) Fact: The function Tii...i ≡ [Sym(F)]ii...i is totally symmetric in its indices. (A.2.11) (A.5.16) Fact: Sym is a linear operator, so Sym (ΣjajTjii...i) = Σjaj [Sym(Tj)]ii...i. (A.3.12) (A.5.17) Fact: Sym is a projection operator, so Sym (Sym (T)) = Sym (T) . (A.3.13) (A.5.18) Fact: If T is a totally symmetric rank-k tensor, then Sym(T) = T . (A.3.16) (A.5.19) (c) Alt/Sym and Other Equations (translated from Section A.4 above) Fact: The projection operators Alt and Sym are orthogonal, so Alt(Sym(f)) = Sym(Alt(f)) = 0 . (A.4.1) (A.5.20) Fact: A tensor Tii...i can be decomposed in the following manner: (A.5.21) Tii...i = Aii...i + Sii...i + Eii...i (A.4.3) Alt(A) = A Sym(A) = 0 Alt(E) = 0 (A.4.4) Alt(S) = 0 Sym(S) = S Sym(E) = 0 (A.4.5) where A is totally antisymmetric, S is totally symmetric, and E is whatever is left over. Definition: Tensor product: (TS)ii...i ≡ Tii...i Sii...i, where the ranks of tensors T,S are k,k'. (A.2.19) (A.5.22) A.6 The permutation tensor ε The permutation tensor ε of rank k is written εii...i where each subscript must be an element of {1,2...k}. The values of the tensor are these: ε12..k = +1 εii...i changes sign if any two indices are swapped . Therefore, if two or more indices are the same, εii...i = 0 . εii...i ≡ εii...i (A.6.1) The tensor εii...i has kk components, but only k! of those components are non-zero. One arrives at k! by allowing k values for i1, then only (k-1) values for i2, and so on. The tensor εii...i is totally antisymmetric by (A.5.7) since any index swap causes a minus sign. Fact: Apart from scale, the εii...i tensor is the only totally antisymmetric tensor one can construct. (A.6.2) Proof: From the definition of εii...i, we see that if Aii...i is a arbitrary totally antisymmetric tensor, then one can write Aii...i = [ A12...k] εii...i . (A.6.3) Here εii...i does the bookkeeping for swaps of index pairs. The scale factor is then A12...k . Use of the ε tensor We noted already that all our permutation results can be specialized to ir → r. For example, [Alt(T)]ii...i = (1/k!) ΣP (-1)S(P) Tii...i (A.5.3) then becomes [Alt(T)]12...k = (1/k!) ΣP (-1)S(P) TP(1)P(2)...P(k) . (A.6.4) Now we make the following claim, ΣP (-1)S(P) TP(1)P(2)...P(k) = Σii...i εii...i Tii...i, ir = 1,2...k . (A.6.5) Each of the ir sums runs from 1 to k. Notice that each side has k! non-vanishing terms in its sum. Suppose P[1,2,3...k] = [i1, i2, i3 ...ik]. Then we claim that the parity of the permutation is given by (-1)S(P) = εii...i . (A.6.6) To see why this is so, start off with the identity permutation P = 1 which has (-1)S(P) = (-1)0 = 1. In this case P[1,2...k] = [1,2...k] and conveniently ε123...k = 1, so both sides of ** agree. Now swap 1↔2 and then the left side is (-1)1 = -1 and the right side is ε213...k = - ε123...k = - 1 and again both sides agree. Now swap 2↔3. The left side is (-1)2 and the right side is - ε132...k = ε123...k = 1, and again both sides agree. In this way one can exhaust all permutations P and the equation is always true. On the left side of (A.6.5) the permutations are enumerated by P, while on the right they are enumerated by i1, i2, i3 ...ik which is restricted by the ε tensor to be a permutation of 1,2,3...k. Basically the notation on each side of (A.6.5) is describing the same instructions for forming the sum. Example with k = 3 (A.6.7) ΣP (-1)S(P) TP(1)P(2)P(3) = T123 - T213 + T231 - T321 + T312 - T132 . The only simple way to form this sum is to keep doing swaps. We show in red the pair that will be swapped to make the next term on the right. Compare then to Σiii3 εiii Tiii . To enumerate the terms, we use the 3! = 6 non-zero values of εiii in the same order as above Σiii3 εiii Tiii = ε123 T123 + ε213 T213 + ε231 T231 + ε321T321 + ε312 T312 + ε132 T132 = (1) T123 + (-1) T213 + (1) T231 + (-1) T321 + (1) T312 + (-1)T132 = T123 - T213 + T231 - T321 + T312 - T132 Here the signs of the ε factors alternate as shown because each one is obtained by an index pair swap on the preceding term. Application Consider, Tii...i = (vi vi ..... vi) . We can specialize this to say T12...k = (v1 v2 ..... v3) and then apply P using (A.2.4) with R→P to get TP(1)P(2)...P(k) = (vP(1) vP(2) ..... vP(k)) . Then (A.6.5) ΣP (-1)S(P) TP(1)P(2)...P(k) = Σii...i εii...i Tii...i, ir = 1,2...k (A.6.5) becomes ΣP (-1)S(P) (vP(1) vP(2) ..... vP(k)) = Σii...i εii...i (vi vi ..... vi), ir = 1,2...k (A.6.8) A.7 The wedge-product-of-vectors Alt equation Here we consider a new application for our generic function f[1,2...k], namely, f[1,2,....k] = (vj vj ..... vj) . (A.7.1) Here the js label the generic objects vj and is some generic operator. Then, consider the generic Alt definition, [Alt(f)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) . (A.2.1) Formally speaking, the left side would have to be written something like this, [Alt(f)](1,2...k) = [Alt((vj vj ..... vj))](1,2...k) = Alt(vj vj ..... vj) . (A.7.2) On the first line the asterisks are place holders which will get the arguments in the argument list. This lets us make a formal association f → (vj vj ..... vj) for a function without arguments. On the right side of (A.2.1) we have f( P(1),P(2)...P(k) ) = (vj vj ..... vj) . (A.7.3) We end up then with this statement Alt(vj vj ..... vj) = (1/k!) ΣP (-1)S(P) (vj vj ..... vj) . (A.7.4) If it happens that means the tensor product, and if the vj happen to be vectors in V, then the above expression happens to be our definition (7.1.2) for the wedge product of k vectors: (vj ^ vj ^ .....^ vj) = Alt(vj vj ..... vj) . (A.7.5) As noted earlier, we can always specialize replacing jr → r. Then f[1,2,....k] = (v1 v2 ..... vk) . Alt(v1 v2 ..... vk) = (1/k!) ΣP (-1)S(P) (vP(1) vP(2) ..... vP(k)) (A.7.6) (v1 ^ v2 ^ .....^ vk) = Alt(v1 v2 ..... vk) . (A.7.7) A.8 Application to Tensor Functions We now restate the "generic" results of Sections A.2, A.3 and A.4 for this special case: f(1,2...k) = T(vi,vi....vi) // a "tensor function" T ϵ V*k . (A.8.1) We apologize for copy, paste and edit, but things really are exactly parallel to the tensor discussion above. Here T is any rank-k tensor function. This f seems perhaps an odd looking "function", but one can consider it to be just an evaluation of this more respectable mapping, f(a,b,c,...q) = T(vi,vi....vi) a,b,c... ϵ {1,2...k} f: {1,2...k}k → Vf*k . (A.8.2) This technical mapping issue is not important because we are just regarding T(vi,vi....vi) as a "carrier" of the labels 1,2,3..k, from the point of view of doing permutations. The actual indices like i1 could be arbitrary objects (labeled pancakes) as far as the permutation theorems are concerned, but in our applications we have in mind that i1 is an integer in the range 1,2....n where n = dim(V) and n is unrelated to the tensor rank k. Here then are some Section A.2,A.3,A.4 results translated according to f(1,2,3...k) = T(vi,vi....vi) . For some of the translations, we show the actual equation from above, then its translation. For others we just state the translated result. In all the results below, one can always specialize to the case i1,i2...ik → 1,2,...k. The resulting equations are then as if our mapping were f(1,2...k) = T(v1,v2....vk). Note then that iP(r) → i(r) in a subscript. (a) Alt Equations (translated from Section A.2 above) The basic Alt definition of (A.2.1) g(1,2...k) = [Alt(f)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) (A.2.1) becomes, G(vi,vi....vi) = [Alt(F)](vi,vi....vi) = (1/k!) ΣP (-1)S(P)F(vi,vi....vi) (A.8.3) G = Alt(F) // definition of Alt acting on a tensor Examples: G(vi,vi) = [Alt(F)](vi,vi) = (1/2)[ F(vi,vi) - F(vi,vi) ] (A.8.4) G(vi,vi,vi) = [Alt(F)](vi,vi,vi) = (1/6)[ F(vi,vi,vi) - F(vi,vi,vi+ the other four terms ] In practice, we might more easily write G(va,vb,vc) = [Alt(F)](va,vb,vc) = (1/6) [ F(va,vb,vc) - F(vb,va,vc) + the other four terms ] but when it comes time to prove permutation-related theorems, we use subscripts like i1. Continuing on, R[1,2....k] = [R(1),R(2).....R(k)] of (A.2.3) becomes, with R→ P, P T(vi,vi....vi) = T(vi,vi....vi) . (A.2.3) (A.8.5) Fact: Any permutation P is a linear operator, so P( ΣrarTr(vi,vi....vi)) = Σrar(PTr(vi,vi....vi)) (A.2.5) (A.8.6) Definition: A tensor function T(vi,vi....vi) is totally antisymmetric if it changes sign when any two arguments are swapped. (A.2.9) (A.8.7) Example: T(vi,vi....vi) = - T(vi,vi....vi) or T(va,vb....vq) = - T(vb,va....vq) Fact: T(vi,vi....vi) totally antisymmetric P T(vi,vi....vi) = (-1)S(P) T(vi,vi....vi) , where P is any permutation of [1,2..k]. (A.2.10) (A.8.8) Fact: The function T(vi,vi....vi) ≡ [Alt(F)]T(vi,vi....vi) is totally antisymmetric in its labels. (A.2.11) (A.8.9) Fact: Alt is a linear operator, so Alt (ΣjajTj(vi,vi....vi)) = Σjaj [Alt(Tj)](vi,vi....vi). (A.2.12) (A.8.10) Fact: Alt is a projection operator, so Alt(Alt(T)) = Alt(T) . (A.2.13) (A.8.11) Fact: If T is a totally antisymmetric rank-k tensor, then Alt(T) = T . (A.2.16) (A.8.12) (b) Sym Equations (translated from Section A.3 above) The basic Sym definition of (A.3.1) g(1,2...k) = [Sym (f)](1,2...k) ≡ (1/k!) ΣP f( P(1),P(2)...P(k) ) (A.3.1) becomes, G(vi,vi....vi) = [Sym(F)](vi,vi....vi) = (1/k!) ΣP F(vi,vi....vi) (A.8.12) G = Sym(F) // definition of Alt acting on a tensor Examples: G(vi,vi) = [Sym(F)](vi,vi) = (1/2)[ F(vi,vi) + F(vi,vi) ] (A.8.13) G(vi,vi,vi) = [Sym(F)](vi,vi,vi) = (1/6)[F(vi,vi,vi)+F(vi,vi,vi+ the other four terms ] In practice, we might more easily write G(va,vb,vc) = [Sym(F)](va,vb,vc) = (1/6) [ F(va,vb,vc) + F(vb,va,vc) + the other four terms ] but when it comes time to prove permutation-related theorems, we use subscripts like i1. Definition: A tensor function T(vi,vi....vi) is totally symmetric if it is unchanged when any two arguments are swapped. (A.3.9) (A.8.14) Example: T(vi,vi....vi) = T(vi,vi....vi) or T(va,vb....vq) = T(vb,va....vq) Fact: T(vi,vi....vi) totally symmetric P T(vi,vi....vi) = T(vi,vi....vi), where P is any permutation of [1,2..k]. (A.3.10) (A.8.15) Fact: The function T(vi,vi....vi) ≡ [Sym(F)]T(vi,vi....vi) is totally symmetric in its labels. (A.2.11) (A.8.16) Fact: Sym is a linear operator, so Sym(ΣjajTj(vi,vi....vi)) = Σjaj [Sym(Tj)](vi,vi....vi). (A.3.12) (A.8.17) Fact: Sym is a projection operator, so Sym (Sym (T)) = Sym (T) . (A.3.13) (A.8.18) Fact: If T is a totally symmetric rank-k tensor, then Sym(T) = T . (A.2.16) (A.8.19) (c) Alt/Sym and Other Equations (translated from Section A.4,A.6 and A.7 above) Fact: The projection operators Alt and Sym are orthogonal, so Alt(Sym(T)) = Sym(Alt(T)) = 0 . (A.4.1) (A.8.20) Fact: A tensor function T(vi,vi....vi) can be decomposed in the following manner: (A.8.21) T(vi,vi....vi) = A(vi,vi....vi) + S(vi,vi....vi) + E(vi,vi....vi) (A.4.3) Alt(A) = A Sym(A) = 0 Alt(E) = 0 (A.4.4) Alt(S) = 0 Sym(S) = S Sym(E) = 0 (A.4.5) where A is totally antisymmetric, S is totally symmetric, and E is whatever is left over. Definition: Tensor product: (TS)(vi,vi....vi) ≡ T(vi,vi....vi) S(vi,vi....vi), where the ranks of tensors T,S are k,k'. (A.2.19) (A.8.22) The following are based on Section A.6 and concern use of the ε tensor with tensor functions. If A is totally antisymmetric, then A(vi,vi....vi) = [ A(v1,v2....vk)] εii...i . (A.6.3) (A.8.22) The tensor function [Alt(T)](v1,v2....vk) can be expressed as, ΣP (-1)S(P) T(vP(1),vP(2)....vP(k)) = Σii...i εii...i T(vi,vi....vi) (A.6.5) (A.8.23) Let T(vi,vi....vi) = (αi αi ..... αi)(vi,vi....vi) so T(v1,v2....vk) = (α1 α2 ..... αk)(v1,v2....vk) Then (A.8.23) gives this way to write (αj ^ αj ^ .....^ αj) : ΣP (-1)S(P) (αP(1) αP(2) ..... αP(k)) (A.6.8) (A.8.24) = Σii...i εii...i (αi αi ..... αi), ir = 1,2...k The following is based on Section A.7. (αj ^ αj ^ .....^ αj) = Alt(αj αj ..... αj) (A.7.5) (A.8.25) (d) Alt/Sym when there are two sets of indices It is not uncommon to encounter objects like the following (Xjj...j)ii...i where the jr are labels and the ir are tensor component indices. An example would be the components of a wedge product of k vectors (vj^ vj^ .....^ vj)ii...i . In this situation, we have to clarify which of the two sets of indices is being acted upon by the Alt operator. We might do this as follows, using AltI and AltJ , AltI[(Xjj...j)ii...i] = (1/k!) ΣP (-1)S(P) (Xjj...j)ii...i AltJ[(Xjj...j)ii...i] = (1/k!) ΣP (-1)S(P) (Xjj...j)ii...i In general, the above two objects are different. Now recall Fact (A.5.12) from above, Fact: If T is a totally antisymmetric rank-k tensor, then Alt(T) = T . (A.2.16) (A.5.12) If it happens that (Xjj...j)ii...i is totally antisymmetric in the ir , then AltI[(Xjj...j)ii...i] = (Xjj...j)ii...i If it happens that (Xjj...j)ii...i is totally antisymmetric in the jr , then AltJ[(Xjj...j)ii...i] = (Xjj...j)ii...i If it happens that (Xjj...j)ii...i is totally antisymmetric separately in the ir and the jr, then we have AltI[(Xjj...j)ii...i] = AltJ[(Xjj...j)ii...i] since both are equal to (Xjj...j)ii...i Recalling Fact (A.5.19) Fact: If T is a totally symmetric rank-k tensor, then Sym(T) = T . (A.3.16) (A.5.19) we conclude a similar fact for SymI and SymJ . We then summarize these results Fact: If (Xjj...j)ii...i is totally antisymmetric in both sets of indices, then AltI[(Xjj...j)ii...i] = AltJ[(Xjj...j)ii...i] (A.8.26) Fact: If (Xjj...j)ii...i is totally symmetric in both sets of indices, then SymI[(Xjj...j)ii...i] = SymJ[(Xjj...j)ii...i] (A.8.27) Example: According to (7.2.9) the object (vj^ vj^ .....^ vj)ii...i is totally antisymmetric in both sets of indices. Therefore, AltI (vj^ vj^ .....^ vj)ii...i = AltJ (vj^ vj^ .....^ vj)ii...i (A.8.28) A.9 The Ordered Sum Theorem The ordered sum theorem states that, (ΣP [ΣP(i)<P(i)<...<P(i)]) fii...i = Σi<i<...<i [ΣP fP(i)P(i)...P(i)] . (A.9.1) 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.9.2) 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.9.3) Thus we have proven the Theorem for k = 2, (ΣP [ΣP(i)P(i)]) fii = Σi<i [ΣP fP(i)P(i)] (A.9.4) 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.9.5) 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.9.6) 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.9.7) 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.9.1) A.10 The Alt Operator in Dirac bra-ket notation We won't make use of what follows, but it seems a reasonable way to incorporate operators like Alt and Sym into the Dirac bra-ket notation. In the above discussion, when we say g = Alt(f), what we mean is that, g(1,2...k) = Alt(f)(1,2...k) = (1/k!) ΣP(-1)P f(P(1),P(2)....P(k) ) . (A.10.1) The stripped down statement g = Alt(f) is somewhat vague because there are no visible labels for Alt to act upon. We can give g = Alt f a more concrete meaning using the Dirac bra-ket notation. Write, g(1,2...k) = <g | 1,2...k> <g | = functional (A.10.2) Alt(f)(1,2...k) = < Alt f |1,2...k > < Alt f | = functional (A.10.3) Here |1,2...k > is a ket in some generic permutation space Gk . Next we define Alt |1,2...k> ≡ (1/k!) ΣP(-1)P |P(1),P(2)...P(k)> (A.10.4) < Alt f | 1,2...k> ≡ <f | Alt 1,2...k> . (A.10.5) In (A.10.5) we are pretending that Alt is a symmetric (self-adjoint) operator AltT = Alt which can be swung for free from the bra space to the ket space. It is just notation. Then we can interpret the stripped statement g = Alt(f) to mean < g | = < Alt f | . (A.10.6) When both sides are closed with the ket |1,2,...k> we get the intended result, < g | 1,2,...k> = < Alt f |1,2...k> = <f | Alt 1,2...k> = <f | [(1/k!) ΣP(-1)P |P(1),P(2)...P(k)>] = (1/k!) ΣP(-1)P <f | P(1),P(2)...P(k)> (A.10.7) which then says g(1,2..k) = (1/k!) ΣP(-1)P f(P(1),P(2)....P(k) ) . (A.10.1) (A.10.8) In our two applications we have tensor: | 1,2...k> = | ei, ei ...ei> <T| 1,2...k> = <T|ei, ei ...ei> = Tii...i (A.10.9) tensor function: | 1,2...k> = | vi, vi ...vi> <T| 1,2...k> = <T|vi, vi ...vi> = T(vi, vi ...vi) . (A.10.10) Example: Prove that Alt(Alt(f)) = Alt(f) in Dirac notation. That is, show < Alt Alt f | = <Alt f |. Proof: Let | i1,i2...ik> be an arbitrary permutation of | 1,2...k> ( an arbitrary ket in Gk ) < Alt Alt f | i1,i2...ik> = < Alt f | Alt i1,i2...ik> = < Alt f | (1/k!) ΣP(-1)P iP(1),iP(2)...iP(k)> = <f | Alt (1/k!) ΣP(-1)P iP(1),iP(2)...iP(k)> = (1/k!) ΣP(-1)P <f | Alt iP(1),iP(2)...iP(k)> = (1/k!) ΣP(-1)P <f | (1/k!) ΣQ (-1)Q iQP(1),iQP(2)...iQP(k)> = (1/k!)2 ΣP <f | ΣQ (-1)QP iQP(1),iQP(2)...iQP(k)> = (1/k!)2 ΣP <f | ΣQ (-1)Q iQ(1),iQ(2)...iQ(k)> // rearrangement theorem (A.1.3) = (1/k!)2 (k!) <f | ΣQ (-1)Q iQ(1),iQ(2)...iQ(k)> // ΣP(1) = k! = <f | Alt i1,i2...ik> = <Alt f | i1,i2...ik> (A.10.11) Since this is true for all kets in Gk we conclude that < Alt Alt f | = <Alt f |. QED