## Monday, November 18, 2013

### Conjugates of Fields; 1 of 6; Conjugation

Conjugates of Fields

Or: Parallel Arithmetics

1.     Conjugation
4.     Modular Fermat Fields
5.     Logarithmic and Exponential Fields
6.     Other Conjugate Fields

1. Conjugation

Given functions f(x) and g(x), if g-1(x) exists, then we can define the conjugate of f by g:

g[f](x)    =    g(f(g-1(x))) .

This resembles matrix conjugation in coordinate transformations; also conjugation in group theory. As there, conjugation tends to preserve function relations.

Theorem: Conjugation is a homomorphism. That is:
g[f] o g[h]  =  g[ f o h ]
g[f -1]  =  g[ f ] -1
g[identity] = identity
g[f[h]]  =  (gof)[h]
If k is a constant, then g[k] = g(k)
Composition, inverse and identity are preserved by conjugation. So are all equational identities,  up to tranformation of constants. Conjugates of fields are fields, conjugates of vector spaces are vector spaces, conjugates of matrix rings are matrix rings, and so on.

Given an invertible real function g(x), define these "conjugate field" operations:
a g[+] b    =    g ( g-1(a)  +  g-1(b) )
a g[-] b    =    g ( g-1(a)  -  g-1(b) )
a g[*] b    =    g ( g-1(a)  *  g-1(b) )
a g[/] b    =    g ( g-1(a)  /  g-1(b) )
g[-] a    =    g ( - g-1(a) )
g[1/] a    =    g ( 1/ g-1(a) )

The definition and properties of conjugation imply the following three theorems.

Slide Rule Theorem:
a g[+] b  =  c       if and only if          g-1(a) + g-1(b)  =  g-1(c)
a g[-] b  =  c        if and only if          g-1(a) - g-1(b)  =  g-1(c)
a g[*] b  =  c       if and only if          g-1(a) * g-1(b)  =  g-1(c)
a g[/] b  =  c        if and only if           g-1(a) / g-1(b)  =  g-1(c)
g[-] b  =  c       if and only if               -   g-1(b)    =  g-1(c)
g[1/] b  =  c        if and only if             1  /  g-1(b)   =  g-1(c)

I call this the Slide Rule theorem because this is how a generalized slide rule works. For instance, * = exp[+], so we can find * using a logarithmic scale.

Generalized Distribution Theorem:
g(a + b)   =   g(a)  g[+]  g(b)
g(a - b)   =   g(a)  g[-]  g(b)
g(a * b)   =   g(a)  g[*]  g(b)
g(a / b)   =   g(a)  g[/]  g(b)
g(-b)      =   g[-]  g(b)
g(1/b)     =   g[1/]  g(b)

I dedicate this theorem with gratitude to my students, who have tried in vain over many years to apply a defective version of this theorem; one with a "+" written where there should be a "g[+]". Due to that error, I never gave those students any credit - until now.

Conjugate Field Theorem:

a g[+] b  =  b g[+] a     ;           a g[*] b  =  b g[*] a

(a g[+] b) g[+] c  =  a g[+] (b g[+] c)
(a g[*] b) g[*] c  =  a g[*] (b g[*] c)

a g[+] g(0)  =  a                       ;                       a g[*] g[1]  =  a
a g[-] a  =  g(0)                        ;                       a g[/] a  =  g(1)

a  g[-]  b    =   a  g[+]  g[-](b)
a  g[/]  b    =   a  g[*]  g[1/](b)

a  g[*]  (b g[+] c)   =   (a g[*] b)  g[+]  (a g[*] c)

These are the field axioms: commutativity, associativity, identities, inverses, and distibutivity. Thus the conjugate of a field is a field; g[+] is a field under g[*] with identities g(0) and g(1).