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.
