Tensor Alt Theorems

Tensor Alt Theorems PhL 11.15.15 1.15.16 All this material now exists in App A or App C. S.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) Two Lemmas Consider the equation f = Alt(F) written in components fii...i = ΣQ (-1)S(Q) FQ(i)Q(i)...Q(i) (C.1.5) We can apply the rearrangement theorem to in two ways, first with Q → PQ and then second with Q → QP. The results are fii...i = ΣQ (-1)S(PQ) FPQ(i)PQ(i)...PQ(i) = (-1)P ΣQ (-1)S(Q) FPQ(i)PQ(i)...PQ(i) (C.1.6) fii...i = ΣQ (-1)S(QP) FQP(i)QP(i)...QP(i) = (-1)PΣQ (-1)S(Q) FQP(i)QP(i)...QP(i) Therefore Lemma 1: ΣQ (-1)S(Q) FPQ(i)PQ(i)...PQ(i) = ΣQ (-1)S(Q) FQP(i)QP(i)...QP(i) (C.1.7) Now in (C.1.5) replace ir → P(ir) to get Lemma 2: fP(i)P(i)....P(i) = ΣQ (-1)S(Q) FQP(i)QP(i)...QP(i) = ΣQ (-1)S(Q) FPQ(i)PQ(i)...PQ(i) (C.1.8) where on the last line we have used (C.1.7). C.2 Theorem One Alt(Alt(T)S) = Alt(TS) (C.2.1) Interpretation: If we pre-antisymmetrize T prior to antisymmetrizing TS, it makes no difference. This certainly seems intuitively reasonable, but here is a formal brute force proof. Assume rank(T) = k and rank(S) = k'. Recall from (7.5.3) that, for a tensor X of rank r one has, [Alt(X)]ii...i ≡ (1/k!) ΣP (-1)S(P) XP(i)P(i)...P(i) (7.5.3) (C.2.2) Now apply (C.2.2) to X = Alt(T)S to get an expression for the left side of (C.2.1), [Alt(Alt(T)S)]ii...i = (1/(k+k')!) ΣP (-1)S(P)[Alt(T) S]P(i)P(i)....P(i) (C.2.3) We know from the usual outer product notation that [Alt(T) S]P(i)P(i)...P(i) = [Alt(T)]P(i)P(i)....P(i) SP(i)P(i)....P(i) (C.2.4) Meanwhile, using X = T in (C.2.2) gives, [Alt(T)]ii...i ≡ (1/k!) ΣQ (-1)S(Q) TQ(i)Q(i)...Q(i) . (7.5.3) (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'}. Taking ir → P(ir) in (C.2.5) gives [Alt(T)]P(i)P(i)....P(i) = (1/k!) ΣQ (-1)S(Q) TQP(i)QP(i)...QP(i) . (C.2.6) Use Lemma 2 (C.1.8) with f = Alt(T) and F = T to restate the above as [Alt(T)]P(i)P(i)....P(i) = [ (1/k!) ΣQ (-1)S(Q) TPQ(i)PQ(i)...PQ(i)] (C.2.7) Install (C.2.7) into (C.2.4) to get [Alt(T) S]P(i)P(i)...P(i) = [(1/k!) ΣQ (-1)S(Q) TPQ(i)PQ(i)...PQ(i)] SP(i)P(i)....P(i) (C.2.9) Then install (C.2.9) into (C.2.3) to get [Alt(Alt(T)S)]ii...i = (1/(k+k')!) ΣP (-1)S(P) * [ (1/k!) ΣQ(-1)S(Q) TPQ(i)PQ(i)...PQ(i)] SP(i)P(i)....P(i) (C.2.10) Rewrite this as [Alt(Alt(T)S)]ii...i = ΣQ g(Q) (C.2.11) where g(Q) ≡ (1/(k+k')!) ΣP (-1)S(P)(1/k!) (-1)S(Q) TPQ(i)PQ(i)...PQ(i) SP(i)P(i)....P(i) = (1/(k+k')!)(1/k!) ΣP (-1)S(PQ) TPQ(i)PQ(i)...PQ(i) SP(i)P(i)....P(i) . (C.2.12) We shall now show that, despite appearances, g(Q) does not really depend on Q. On g(Q) use the rearrangement theorem with P→PX with X = Q-1 so XQ = QX = 1. This gives g(Q) = (1/(k+k')!)(1/k!) ΣP (-1)S(PXQ) TPXQ(i)PXQ(i)...PXQ(i) SPX(i)PX(i)....PX(i) = (1/(k+k')!)(1/k!) ΣP (-1)S(P) TP(i)P(i)...P(i) SPX(i)PX(i)....PX(i). (C.2.13) Since Q has no effect on the upper indexes in the set {i1....k+k'}, we know that Q-1 also has no effect, since Q-1 is just some other element of the set of permutations of these first indices. Therefore X(ik+1) = Q-1(ik+1) = ik+1 (C.2.14) and the same for the other higher indices. Thus (C.2.13) becomes, g(Q) = (1/(k+k')!)(1/k!) ΣP (-1)S(P) TP(i)P(i)...P(i) SP(i)P(i)....P(i) (C.2.15) which shows g(Q) in fact has no dependence on Q, so we can write g(Q) = g(1) (C.2.16) where 1 is the identity permutation. Then from (C.2.12) one finds, g(1) = (1/(k+k')!)(1/k!) ΣP (-1)S(P) TP(i)P(i)...P(i) SP(i)P(i)....P(i) (C.2.17) Then from (C.2.11), [Alt(Alt(T)S)]ii...i = ΣQ g(Q ) = k! g(1) = (1/(k+k')!) ΣP (-1)S(P) TP(i)P(i)...P(i) SP(i)P(i)....P(i) = (1/(k+k')!) ΣP (-1)S(P) (TS)P(i)P(i)...P(i) . (C.2.18) But this we recognize as (C.2.2) applied to X = TS, so therefore [Alt(Alt(T)S)]ii...i = Alt(TS)ii...i (C.2.19) or Alt(Alt(T)S) = Alt(TS) QED Why is Spivak's proof so short, or does he ever do this Theorem? I think this is too simple a theorem for him to even state.