]> Conjugate

# Conjugate

Every ringlet has a prime subringlet generated by repeated addition, that models the natural numbers. From this we can obtain as much of the positive reals, within the ringlet, as it contains a representation for; it can then be interesting to consider a transformation of the ringlet under which the product of any non-zero value with its transformed value is positive.

I'm mostly concerned with the case of ringlets of character 0, where the ringlet homomorphism that embeds the naturals in our ringlet is monic; distinct naturals are represented by distinct members of the prime subringlet, hence the same shall be true of such reals as we get. While I'll try not to restrict unnecessarily to this case, the relevance of positives gets mangled in a ring with cyclic addition. In particular, character zero lets us identify the maximal strictly ordered sub-ringlet of our ringlet, to use as the real values in it, with the positives among these taking centre stage.

### Reals within the ringlet

Where the ringlet has multiplicative inverses for the non-zero members of its prime subringlet, we can extend it to an image of the positive ratios or, when the ringlet is additively complete, the rationals. There may even be a ringlet homomorphism from the positives or even a ring homomorphism from the reals into our ringlet. If our ringlet has character 0, its image of the positive naturals has a strict order and we can look for the maximal strictly-ordered sub-ringlet of our ringlet, whose order is consistent with that on the prime subringlet. By any of these means we may identify a sub-ringlet of values worthy to be termed real and, within these, a sub-ringlet of positive values, on which our ringlet's arithmetic has some chance of respecting order. It is usual and natural to denote the real values within a ringlet by the usual denotations for the real number that corresponds to it.

Since the prime subringlet lies in the centre of every ringlet, we can expect that, for the ringlet's multiplication, all of the ringlet's real values commute with all values; so the positives within the ringlet form a subringlet of its centre. (Likewise, when the ringlet is a ring, its reals form a subring of its centre.) Every automorphism or coautomorphism of the ringlet is an automorphism of its addition and preserves 1, hence the prime subring; it is thus reasonable to expect that it also preserves the rest of the ringlet's reals.

Note that a ringlet is a module over any central subringlet, including its reals or positives. This may let us analyse the ring in terms of the module structure, e.g. expressing it in terms of a well-chosen basis, which may help with some of the analysis below.

Because the reals within a ringlet are always generated from the multiplicative identity, 1, which is preserved by every ring automorphism, all of a ringlet's reals are also preserved by every automorphism. Consequently, every automorphism of the ring can be understood in terms of its action on the ring modulo scaling by the positive real values in the ring. Furthermore, the real values of any ringlet lie in its multiplication's centre.

## Definition

I'll describe a self-inverse coautomorphism, c, of a ringlet as a conjugation precisely if, for every non-zero value r, r.c(r) is positive (and, in particular, real). Naturally, r = 0 gives zero (which is also real) for r.c(r). (When the multiplication is commutative, being a coautomorphism is the same as being an automorphism. Having it be a coautomorphism in general ensures that r.s.c(r.s) = r.s.c(s).c(r) has s.c(s) real, so we can swap r past it to get s.c(s).r.c(r), also real, consistent with being r.s times its image under c.) In particular, (: r.c(r) ←r :) is a homomorphism from the non-zero multiplication of our ringlet to its restriction to the ringlet's {positives}. (It's fatuously also a cohomomorphism, as the multiplication on its image is commutative.)

Those values r for which r.c(r) = 1 shall then be of particular interest. In particular, within the ringlet, they form a subgroup of the multiplication: one of them is 1 (c is a coautomorphism, so c(1) = 1, so 1.c(1) = 1) and each of them has an inverse (c is the inversion mapping).

In so far as the ringlet's positives include square roots for the r.c(r) of at least some values r (notably including every positive), we can define the ({positives}: norm :) on our ringlet for which r.c(r) = power(2, norm(r)) and thus obtain a mapping on our non-zero values (: r/norm(r) ←r :) that maps each to a value whose norm is 1; the multiplication on these values then characterises our ringlet's multiplication modulo positive scaling. However, any r for which r.c(r) has no square root get left out of (: norm |) and thus don't get represented by this, so the most interesting case needs enough square roots.

When we have a conjugation on our ringlet (including the case of the fatuous identity conjugation on any sub-ringlet of the reals), we can use it to define antilinearity and related properties for mappings between the ringlet's modules; when we use this to define hermitian forms, it makes possible a positive (or, when desired, real) metric on modules of the ringlet.

## Existence

There's no guarantee that we can define any conjugations on a given ring (much less induce a norm on all of it); and I need to examine whether the conjugation is necessarily unique. First, let's see how far we can get towards constructing a conjugation from what we know.

We can define an equivalence on the ringlet's non-zero values by r ~ s iff x.r = y.s for some positive x, y. (We can do the same with reals in place of positives, but this identifies additive inverses; in particular, it identifies +1 with −1, which I don't want to do.) Since positive multiplication commutes, this respects the multiplication, so we can consider the multiplication of the ringlet reduced modulo this equivalence.

Suppose the ringlet's multiplication is cancellable and consider r.s ~ t.s; this says that there are some positive x, y for which x.r.s = y.t.s in the ringlet, where we can cancel s to get x.r = y.t and infer r ~ t. As reals commute multiplicatively, the s.r ~ s.t case can be written s.r.x = s.t.y to cancel the s and infer r ~ t again. So our equivalence preserves cancellability of multiplication.

Even when the ringlet's multiplication isn't complete, we may induce completion by reducing modulo positive scalings. For example, when we work with the complex whole numbers, using i to denote one of its square roots of −1, we don't have even the simple ratios of whole numbers that we need to complete multiplication in our positives (which are just the positive naturals). None the less, for whole numbers n, m, p, q, when asked to complete (n +i.m).z = p +i.q, we can use z = (n −i.m).(p +i.q) = (n.p +m.q) +i.(n.q −m.p), which gives us (n +i.m).z = (n.n +m.m).(p +i.q) with n.n +m.m positive hence (n +i.m).z ~ p +i.q as required. So we may induce a complete multiplication by reducing an incomplete one modulo positive scaling.

When the reduced multiplication is complete, we get an inversion relation from it, that relates each value of our original ringlet to those that multiply by it to get a positive product (i.e. a product equivalent to 1). We can, indeed, define this relation in any case, regardless of the reasoning that has lead me to it; any conjugation of our ringlet must be a mapping subsumed by this relation. For one to exist, every value of our ringlet must be both a left value and a right value of this relation. That requires, for each r, some s with r.s real and non-zero and some t with t.r real and non-zero. This suffices (just multiply s and t by suitable reals) to ensure each non-zero r has an s and t for which r.s and t.r are positive.

In a ring of linear automorphisms of a vector space over {reals}, the reals (of the ringlet) are simply the real multiples of the identity linear map; since these (aside from zero) are non-singular, no degenerate linear map has a non-zero real multiple; so there definitely are rings in which no such conjugation is possible.

As long as the relation (: r←s; r.s is positive :) does have every value of our ring as a left and a right value, we have some hope of finding a conjugation as a sub-relation of it. For the conjugation to be unique, we need this relation to be a mapping when read modulo positive scalings (i.e. the equivalence discussed above). Being a self-inverse coautomorphism (and, in particular, preserving the reals) is then a strong condition that has a good chance of constraining the conjugation to be unique and may indeed suffice to constrain there to be one.

So suppose we have a ringlet R, with positive sub-ringlet P in R's centre, and we know every non-zero value of R has a multiple in P. If the non-zero multiplication in R is cancellable, r.n and r.m in P, for values r, n, m, implies some positive p, q for which r.n.p = r.m.q whence n.p = m.q and n ~ m. So the candidats to be r's conjugate are equivalent modulo positive scaling; we just need some way to chose the candidate that's the same size as r. We get a pair of rays, {p.r: p in P} and {s: s.r in P}, between which we need to chose an isomorphism that preserves P-scaling. When we do this with the complex numbers, a value and its conjugate also sum to a real, which suffices to select which member of the ray to use. Can we do the same more generally ?

Consider non-singular linear automorphisms of the lattice {lists ({integer}: |3)}. For any three points in the lattice, we can map the three generators – [1, 0, 0], [0, 1, 0] and [0, 0, 1] – to the three given points of the lattice; if the plane through the points doesn't pass through the origin, the result is non-singular. Its determinant will be a whole number; and the construction that would normally give its inverse, but with division by the determinant skipped, will give us an automorphism that composes with it to give that determinant, as a scaling. Given or take sign, we now have a positive. For example,

•  1 0 1 1 1 0 0 1 1
·  1 1 −1 −1 1 1 1 −1 1
=  2 0 0 0 2 0 0 0 2
is positive;

(and the multiplication commutes) but no positive scalings of the two factors will give a real sum. So we can't use scale to make sum real as our rule to make a conjugation; but the fact that we can't may be the hint that this ring doesn't have a conjugation.

Written by Eddy.