Big Problem #1

Big Problem #1 PhL 11.16.15 1.15.16 This matter is dealt with now in Appendix C (C.1.14). Consider, f^[z] = (1/k!) ΣQ (-1)σS(Q) f[Q(z)] where z = {1,2...k}. The Big Problem #1 is this: What happens if you apply a permutation P of {1,2...k+k'} to the above object? Certainly this object "makes sense", P(z) = P(1,2...k) = (P(1),P(2)....P(k)) In my matrix theory, I think of P as a (k+k')x(k+k') matrix, so then P(z) seems to be a size mismatch since the thing z is like a vector with only k components. I would have to think of z = (z,0) to make this matrix idea match. But forget matrix. Just think of P(z) as a list of k integers in the range 1..k+k' . But P(z) does not appear in Big Problem #1. What is the meaning of f[Q(z)] which does appear? f[Q(z)] = f[Q(1), Q(2), ....Q(k)] = seems well-defined OK, then here is the $64 question: what is the meaning of this P f[Q(z)] = P f[Q(1), Q(2), ....Q(k)] = ?? One seemingly logical approach is to just have P act on each item in the list, so 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))) But we pause again to think about the meaning of PQ(1). If I think of P(Q(1)) then I know that Q(1) is some integer n in the set z, and then P(n) is completely well-defined. But eventually I want to use a rearrangement theorem in the space 1,2..k+k' and that is going to require that I interpret P(Q(1)) as (PQ)(1) and more generally P(Q(z)) = (PQ)(z). But how do you think about (PQ)(z) ? This requires thinking about object (PQ). I need to think about this object (PQ) because I am later going to try to use a rearrangement theorem. P and Q are not in the same world really, so I am unclear what to say. I can talk about (PQ'). Here is a matrix view of things P = ξ = Pξ = = well defined Q' = Q'ξ = = (PQ') = = (PQ')ξ = = well defined = So OK, suppose I make this identification z as a list = as a vector Then (PQ') = but I only care what happens to the z list items. So then (PQ')(z) = P11Q (z) I am not getting anywhere here, am flailing with no success (a very common situation). ********************************************** Then using this we seem to get P f^[z] = P { (1/k!) ΣQ (-1)σS(Q) f[Q(z)] } = (1/k!) ΣQ (-1)σS(Q) P f[Q(z)] = (1/k!) ΣQ (-1)σS(Q) f[PQ(z)] Now consider the following candidate theorem, ΣP (-1)σS(P) f[P(z)] F[P(Z)] = ΣP (-1)σS(P f^[P(z)] F[P(Z)]