A Theorem on Permutation Sums

A Theorem on Permutation Sums PhL 9.9.15 I think the theorem is this: Σabc... Xabc... ΣP (-1)P YP(a)P(b)... = Σabc... Yabc... ΣP (-1)P XP(a)P(b)... where X and Y are objects having n subscripts and P is a permutation of the n subscripts. How would one prove something like this? I could restate the theorem this way Σabc...ΣP (-1)P [ Xabc... YP(a)P(b)... - XP(a)P(b)...Yabc...] = 0 Still not obvious to me. I would like to get this into some group notation as in Appendix B of method of Lagrange. There I claim this: Plan A Σabc...Fabc... = ΣP' FP'(1)P'(2)P'(3)... so our Σabc can be written as a permutation sum with no signs. Now restate the original theorem with only 3 subscripts just to make things simpler. Σabc Xabc ΣP (-1)P YP(a)P(b)P(c) = Σabc Yabc ΣP (-1)P XP(a)P(b)P(c) The right side can then be rewritten, = ΣP' YP'(a)P'(b)P'(c) ΣP (-1)P XP(P'(1))P(P'(2))P(P'(3)) = ΣP'ΣP (-1)P YP'(a)P'(b)P'(c) XP(P'(1))P(P'(2))P(P'(3)  I think there is a better way: Plan B. Change to i,j,k as sum indices and we want to show Σijk Xijk ΣP (-1)P YP(i)P(j)P(k) = Σijk Yijk ΣP (-1)P XP(i)P(j)P(k) Go back to the basic ideas of Appendix B z0 ≡ . (B.1.1) a ≡ = A = A z0 (B.1.2) Write theorem as Σa1a2a3 Xa1a2a3 ΣP (-1)P YP(a1)P(a2)P(a3) = Σa1a2a3 Ya1a2a3 ΣP (-1)P XP(a1)P(a2)P(a3) Now rewrite this in permutation vector notation as Σa Xa ΣP (-1)S YPa = ? = Σa Ya ΣP (-1)S XPa (*) where P is a permutation matrix acting on the vector a within the subscript. What exactly is integer S here? It is the parity of permutation P. It is the number of swaps it takes to go from z0 to Pzo. Now look again at Σa f(a) = Σa f(Ca) . // permutation sum rearrangement theorem (B.2.4) where you can just "throw in" a permutation matrix C and not change the sum. Write LHS of * Σa Xa ΣP (-1)S YPa = ΣP (-1)S (Σa XaYPa) Then add an arbitrary permutation matrix C so replace a → Ca everywhere in the sum = ΣP (-1)S (Σa XCaYPCa) Now in this term select C = P-1 so we then get = ΣP (-1)S (Σa XP-1aYa) ** Now I think we need another theorem which would say ΣP (-1)S f(P-1a) = ΣP' (-1)S' f(P'a) The ΣP is over all permutations, and if P' = P-1, then sum over P' is a rearrangement of the same sum. And the swap parity of P is the same as that of P-1, so that S' = S. Thus, our theorem really says ΣP (-1)S f(P-1a) = ΣP (-1)S f(Pa) Then we can write ** as = ΣP (-1)S (Σa XPaYa) Summary: Want so show that Σabc... Xabc... ΣP (-1)S YP(a)P(b)... = Σabc... Yabc... ΣP (-1)S XP(a)P(b)... Step 1. Claim that in either sum, only terms with a ≠ b ≠ c contribute. Consider in the first term Qabc... ≡ ΣP (-1)S YP(a)P(b)... = ΣP (-1)S+1 YP(b)P(a)P(c)... Suppose we had a = b. Then we get Qaac... ≡ ΣP (-1)S YP(a)P(a)... = ΣP (-1)S+1 YP(a)P(a)P(c)... = 0 This is 0 because any expression that equals minus itself is 0. Step 2. Use the permutation vector and matrix notation of Appendix B to write the claimed theorem as Σa Xa ΣP (-1)S YPa = Σa Ya ΣP (-1)S XPa. ? Now we regard P as a permutation matrix which acts on the index vector a. This matrix P is not a function of a. We move the P sums all the way to the left to get ΣP (-1)S ( Σa Xa YPa ) (*) = ΣP (-1)S ( Σa Ya XPa ) ? Step 3. The parenthetical quantity has the form Σa f(a). The rearrangement theorem says that Σa f(a) = Σa f(Ca) for any permutation matrix C Select C = P-1 so that LHS(*) = ΣP (-1)S ( Σa Xa YPa ) = ΣP (-1)S ( Σa XP-1a YPP-1a ) = ΣP (-1)S ( Σa XP-1a Ya ) Step 4. Use the following theorem ΣP (-1)S f(P-1a) = ΣP (-1)S f(Pa) to continue the above to get LHS(*) = [ ΣP (-1)S ( Σa XPa Ya ) But RHS(*) is this, RHS(*) = ΣP (-1)S ( Σa Ya XPa ) This we have proven that LHS = RHS.