Complex scalings

Given an additive domain V, let AddOK(V) stand for {(V:f:(:V|)): f respects addition}. For any sub-collection, A, of AddOK(V), let Scaling(A) stand for {f in AddOK(V): for all a, b in A, f&on;a&on;b = a&on;b&on;f}, let Linear(A) stand for {g in AddOK(V): for all z in Scaling(A), z&on;g = g&on;z}, and let antiLinear(A) stand for {g&on;a: g in Linear(A), a in a}. I shall be particularly interested in the case where A = antiLinear(A) and every member of AddOK(V) is a sum whose terms are drawn only from Linear(A) and antiLinear(A).

The definitions immediately tell us:

and we may infer

Linear(A) is closed under addition

since z&on;(g+f) = z&on;g + z&on;f = g&on;z + f&on;z = (g+f)&on;z whenever g and f are in Linear(A) and z is in Scaling(A)

Linear(A) is closed under composition

since z&on;g = g&on;z and z&on;f = f&on;z do imply z&on;f&on;g = f&on;z&on;g = f&on;g&on;z

Linear(A) subsumes {f&on;a: f in antiLinear(A), a in A}

since f is g&on;b for some g in Linear(A) and b in A, whence f&on;a is g&on;b&on;a with g in Linear(A) and b, a in A, whence b&on;a in Linear(A); so f&on;a is a composite of members of Linear(A)

Given two members of antiLinear(A), can we expect their composite to be in Linear(A) ? It's of form g&on;a&on;f&on;b with g, f in Linear(A), a, b in A. Can we be sure a&on;f is in antiLinear(A), given a in A and f in Linear(A) ? If g&on;b is in Linear(A) for some b in antiLinear(A), can we infer that g is antiLinear ?

If A is closed under addition and subsumes {a&on;b&on;c: a, b, c in A}, I'll say that it presages {(V:f:(:V|)) shadow(A) = {a, b&on;c, a+b&on;c: a, b, c in A}. This is equal to A if it is closed under composition and addition; I am interested in it when A is closed under addition but not composition (e.g. A = {antilinears}). Likewise, define S(A) = {(V;f|(:V|)): f respects addition and commutes with all members of {b&on;c: b, c in A}} and note that this always subsumes {real scalings}. Now, for any z in S(A), I know z commutes with all members of {b&on;c: b, c in A}: when I compose it with just one member of A, what happens ?

Given a collection, A, of mappings (V:|(:V|)) which respect addition, I'll say that A shadows {mapping (V:f|(:V|)): f respect addition} precisely if this last is equal to {a, b&on;c, a + b&on;c: a, b, c in A}. The mappings in A will be described as antilinear, those in {b&on;c: b, c in A} as linear; mappings which respect addition and commute with all linears will be described as scalings. I shall need to show that, for every scaling z there is a scaling s for which: a in A implies a&on;z = s&on;a; s and z will then be described as conjugate to one another. If A is closed under composition and addition, it's equal to any collection it shadows.

To go from real scalings to (at least) complex ones, I need an abelian family of mappings (V:|(:V|)), among which composition needs to be closed and cancellable though possibly not complete, to construe as the non-zero scalings; these shall include the positive real scalings, necessarily as a sub-family. Linear maps are those which respect addition and commute with all these scalings. We can then express our family as a product of two families; the positive real scalings and a family of phases; the inverse automorphism of the latter, combined with the identity automorphism of the former, gives an authomorphism of the scalings known as conjugation - preserve scale, invert phase.

Antilinear maps are those which respect addition but when composed before a scaling yield the same result as composing after the conjugate of that scaling. I expect every mapping which respects addition to be expressible as a sum of terms, each of which is either linear or antilinear (i.e. as a linear plus an antilinear, at least when V has an additive identity to allow us the constant zero map, which is both linear and antilinear; absent this, we need to be able to leave out one kind of map, hence the wording given).

I may concievably want to allow a non-abelian family of phases, but at what point does this degenerate into treating the rotations of my additive domain as scalar multiplication ? Even in the abelian case, I'm not sure I can define conjugation without somehow establishing a unit-sized scaling for each phase, to do which without depending on conjugation I seem to need the exp function (hence differentiation), though I may be able to play games with the square root of -1, in so far as I can get my hands on it.

Does respect for addition enable me to define differentiation ? Not without multiplicative completeness (paddock-style, i.e. except for 0).

Written by Eddy.
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