Friday, June 30, 2017

On Triple Ratios: 5 of 6



S3 conjugations


          Define the “pivot” operator # on {1,2,3} this way:

                   a#b  =  c  if and only if
                   a=b=c=a     or a not= b not= c not= a
 
          That is, a#b is the same or the third.
The pivot a#b is “equally equal” to a and b. Here is its table:

#   1   2   3

1   1   3   2           
2   3   2   1           
3   2   1   3  
      
x#y pivots y around x in the triangle 1,2,3. Pivot generates the reflections symmetries of the triple; double reflections are rotations, so double pivot defines these rotations:

1+x    =   1#(2#x)  =  3#(1#x)  =  2#(3#x) ;
2+x    =   2#(1#x)  =  1#(3#x)  =  3#(2#x).

1+x moves x one step up the cycle 1 < 2 < 3 < 1.
2+x moves x one step down the cycle 1 < 2 < 3 < 1.


Here are function tables for the six permutations of the triple:

  x                        |         1       2       3
---------------------|--------------------------------
0 + x                    |         1       2       3
1 + x                    |         2       3       1
2 + x                    |         3       1       2 
1#x                      |         1       3       2  
2#x                      |         3       2       1  
3#x                      |         2       1       3  

This is the permutation group on three elements: S3. It is the same as the symmetries of the triangle. It contains three rotations (a+x) and three reflections  (a#x). Here is S3’s group table:
         
 a*b            |       0+     1+     2+     1#     2#     3#

 _________|_______________________________________

                   |
0+     |         0+     1+     2+     1#     2#     3#
1+     |         1+     2+     0+     3#     1#     2#
2+     |         2+     0+     1+     2#     3#     1#
1#     |         1#     2#     3#     0+     1+     2+              
2#     |         2#     3#     1#     2+     0+     1+
3#     |         3#     1#     2#     1+     2+     0+


Pivot has these laws:
Recall:                          a # a  =  a
Commutativity: a # b  =  b # a
Cancellation:               a # (a # b)  =   b    
Level associativity:    (a#b) # (c#d)  =  (a#c) # (b#d)    
Transposition:             a#b = c   if and only if     a = b#c
Self-distribution:         a # (b#c)   =   (a#b) # (a#c)

If R is a permutation in S3, then R distributes over #:
S3 symmetry:               R(a#b) =  Ra # Rb     
This is because pivot self-distributes, and every permutation in S3 derives from pivot.

          Pivot operates on triple ratios thus:

          a#(x1, x1, x3)        =       (xa#1, xa#2, xa#3)

          For any triple ratios x and y, and any index a:

          a#(x*y)       =       (a#x)*(a#y)
          a#(x/y)       =       (a#x)/(a#y)
          a#1             =       1
          a#0n            =       0a#n
          a#∞n           =       a#n
          a#-1n          =       -1a#n
          a#(x+ny)     =       (a#x)+a#n(a#y)


          (1#x)* (2#x)* (3#x)     =       1
          x * (1+(x))* (2+(x))     =       1

          Since (a;b;c) corresponds, in the 3 arithmetic, to the dual number ((a/c);(b/c)), then 3#(a;b;c) corresponds to its conjugate ((b/c);(a/c)) . Likewise, 2# corresponds, in the 2 arithmetic, to dual conjugation, and 1# is dual conjugation in the 1 arithmetic.
          So the permutations are conjugations within each of the three arithmetics. But they also turn the arithmetics into each other.


















Thursday, June 29, 2017

On Triple Ratios: 4 of 6



          The Lines at the Infinities


          Consider the “one-infinities”; ratios of the form (0;b;c). You can also write them as (a:b;c)* ∞1, where  1 equals (0;1;1). This applies:
          (x;y;z)+1(0;b;c)             =       (0; b; c)      if x is not zero
                                                          (0;0;0)        if x is zero
          So a 1-infinity plus any 1-finite ratio (i.e. with nonzero first term) absorbs it; and any two 1-infinities add to the indefinite ratio.
          (0;y;z)+2(0;b;c)             =       (0; by; cy+bz)     
                                                =       (0; 1; (c/b)+(z/y))
                                                =       (0; (b/c)[+](y/z); 1)
          (0;y;z)+3(0;b;c)             =       (0; bz+cy; cz)     
                                                =       (0; (b/c)+(y/z); 1)
                                                =       (0; 1; (c/b)[+](z/y))

          So +2 and +3 are addition and reduction on the line at one-infinity. Similarly, +3 and +1 are addition and reduction on the line at two-infinity, and +1 and +2 are addition and reduction on the line at three-infinity.