Conference rusure::math

1239.0. "Quandles, their axioms and structure, and variations" by GUESS::DERAMO (Dan D'Eramo) Sat May 12 1990 11:25

Path: shlump.nac.dec.com!e2big.dec.com!decuac!haven!aplcen!uunet!cadence!holsz
From: [email protected] (Wlodzimierz Holsztynski)
Newsgroups: sci.math
Subject: quandles, their axioms and structure, and variations
Message-ID: <[email protected]>
Date: 11 May 90 03:39:52 GMT
Distribution: sci.math
Organization: Cadence Design Systems, Inc.
Lines: 226
 
Wlodzimierz Holsztynski				1990-April
 
		The Structure of Quandles
		-------------------------
 
In this post, and in (some :-) other in past, I try to take
sci.math seriously. It would be easier if we split sci.math
into 3:
 
	sci.math + sci.math.hist  +  sci.math.res
 
for general,  history  and  research.  Then each of the new groups
would enjoy more respect (certainly from me) than the present mess.
 
Ok, back to quandles. As Nathan Tenny has kindly informed us:
 
"A quandle is, by definition, a set with two binary operations (not
necessarily associative) satisfying:
 
         a + a = a * a = a
        (a + b) * b = (a * b) + b = a
        (a + b) + c = (a + c) + (b + c)
        (a * b) * c = (a * c) * (b * c)
 
This is from an article by David Joyce in 1982."
 
If you're still with me I'll bribe you into further reading
with the following example:
 
-------------------------------------------------------------------
EXAMPLE 0. Given an arbitrary set Q let  + = * = pr0,  i.e. define
 
	a + b = a * b = a	for every a,b from Q.
 
Then (Q,+,*) is a quandle.
-------------------------------------------------------------------
 
Now, please check with me my WHAT, then I'll provide the WHY.
 
THEOREM. If a system (Q,+,*) satisfies:
 
(i)	 a + a = a
(ii)	(a + b) * b = (a * b) + b = a
(iii)	(a + b) + c = (a + c) + (b + c)		/* the self-distributivity
						   or Thales condition  */
 
then (Q,+,*) is a quandle.
 
PROOF.	a*a = (a+a)*a = a	by (i) & (ii), proving the easy "half".
 
Next:
	    (a*b)*c =					 <use (ii) twice>
	(((((a*b)*c) + (b*c)) + c)      * c) * (b*c) =
							      <use (iii)>
	(((((a*b)*c) + c) + ((b*c) + c))* c) * (b*c) =
							 <use (ii) twice>
	  (((a*b)         +   b)        * c) * (b*c) =
							 <use (ii)>
	    (a				 * c) * (b*c)
 
This proves the "harder" half and completes the proof.
--------------
 
Now, HOW did I know? ('cos I knew the above without a proof, which
I needed only for confirmation). Easy. No mystery.  Given an element
 b  of a quandle Q let  Ab: Q --> Q  and  Mb: Q --> Q  be functions
defined by  Ab(x) = x + b  and  Mb(x) = x * b  for every x from Q.
Then axiom (ii)
 
        (a + b) * b = (a * b) + b = a
 
says that  Ab  and  Mb  are  each other's inverse. Thus knowing
{ Ab : b from Q }  allows you to know  { Mb : b from Q }. It does
not automatically guarantee that (i)-(iii) imply the self-distributivity
of * :	        (a * b) * c = (a * c) * (b * c)
 
(I was teasing you when I said "I knew", and when I said that I am serious)
but I felt that it was very likely, hence I went after it.  There should
be a general meta-theorem which would cover this special case. Then
we could say that we know.
 
----------------
 
THE STRUCTURE. Thus a quandle can be alternatively (but equivalently)
defined as a system  (Q,+)  with only one binary operation + such that
 +  is  idempotent (a+a=a)  and  self-distributive [(a+b)+c=(a+c)+(b+c)]
and such that all functions  Ab: Q --> Q  are bijections.  This would be
an economic definition to most of us except to algebraists. They like,
when they can, to define algebraic structures by equational identities,
as they do it for groups, rings, etc., where etc  does not include
fields.  Indeed, the cartesian product of equationally defined structures
(they're called universal algebras) is again a structure of the same type
(with operations defined coordinate-wise). But the cartesian product
of two fields is a ring which has zero-divisors.
 
Thus algebraists prefer rather 2 operations, and I am with them,
granted that we don't loose the picture.
 
On the other hand we can say that a quandle is a set Q together
with a bunch of bijections  Ab : Q --> Q,  one per each b from Q,
such that  b  is a fixed point of  Ab  and  the following ugly
commutativity law holds:
 
	Ac o Ab  =  AAc(b) o Ac
 
where  o  is the composition of functions. It says that instead
of preprocessing  Ac with an Ab you can post-process  Ac  with
an  Ab'  for another  b'  from Q, namely for  b' = (Ac)(b), and
the end result will be the same.
 
Yeah, the self-distributivity is a "weak" (and weird) form of
commutativity. 
 
-------------------
 
Now let's make algebraists happy and forget about the bijectivity
of maps  Ab, at least for a while. Thus let's consider systems
(Q,+) such that the binary operation is idempotent (see (i) above)
and self-distributive on the right (axiom (iii) above). I'll call
such (Q,+) a soloid.  We are going to define a universal associative
soloid.
 
Start with an arbitrary set D. Let  U(D)  be the set of all
(finite) strings of elements of D, excluding the empty one.
 
Let  S(D)  be the set of all strings of U(D) such that each
character (element of D) appears at most one time.  Thus  S(D) is
finite when  D  is finite.
 
Anyway, there is a map (retraction)  r : U(D) --> S(D), where  r(F)
is the expression obtained from F by removing all characters which are
followed later in F by its own copy.  Finally, define + in S(D):
 
	F + G  =  r(FG)		  for all non-empty  F,G  from  S(D).
 
It's obvious (give me a break :-) that  (S(D),+)  is a soloid.
It is the free associative soloid, universal in the well known sense:
 
	for every function  p : D --> Q  into any associative
	soloid Q there exists a unique solomorphism (i.e. homomorphism
	of soloids)  P : S(D) --> Q  such that  P|D = p.
 
COROLLARY. Every finitely generated associative soloid is finite.
 
---------------------------------------------------------------------
 
EXAMPLE 1. Let Q be the field of rational numbers (or any module over
the Z[1/2] ring, i.e. an abelian group with a unique division by 2).
Define  a*b = (a+b)/2  to be the arithmetic mean.  Then  (Q,*) is
a commutative (i.e. * is commutative) but not associative soloid.
Furthermore, let a#b = 2a-b. Then  (Q,#,*) is a quandle. The triple
(Q,*,#) is also a quandle (you can always switch the operations to
get a new quandle from a given one). The two are not isomorphic
since # is not commutative.
 
In 1961/62 I've introduce the following systems (Q, s: QxQ-->Q);
I was motivated to axiomatizatize a portion of geometry, the mid-point
of course:
 
(1) s(a,a) = a
(2) s(a,b) = s(b,a)
(3) s(s(a,b),s(c,d)) = s(s(a,c),s(b,d))
(4) if s(a,c)=s(b,c) then a=b
 
Then I observed that (1) and (3) imply the following Thalesian
property, as I liked to call it then:
 
(iii') s(s(a,b),c) = s(s(a,c),s(b,c))
 
and asked if the (self-distr.) condition (iii') can replace (3).  Actually,
I thought about (iii') first but needed the stronger (3) to prove that
every (Q,s) which satisfies (1)-(4) admits essentially unique isomorphic
embedding into a system (Q',s'), generated by the image of (Q,s), which in
addition to (1)-(4) satisfies
 
(5) for every a,c from Q' there exists b such that  s(a,c)=b.
 
In other words such special soloids (Q,s) embedd into quandles. I've also
made the (easy) observation that the systems which satisfy (1)-(5), i.e.
the abelian quandles, are isomorphic, as soloids (or quandles), to
Z[1/2]-modules. The abelian groups with every element of finite odd
order have provided me with a nice class of examples.
 
On the other hand, Siemion Fajtlowicz has given an example of a
finite abelian quandle which fails to satisfy (3).
 
On the third :-) hand, Jose Slawny considered Finsler diff.
mfld's, with s(a,b) being defined only locally as the mid point
on the geodesic connecting a and b. Then he proved that if
cond's (1),(2),(iii'),(4) are satisfied then (3) holds - everything
locally.
---------------------------------------------------------------------
 
Axioms (1)-(4) above are independent, none is redundant. In particular
we may have  non-commutative  systems which satisfy (1),(3),(4) and
even the symmetric version of (4). Let Q be a Z[1/3]-module. Then
define:
		s(a,b) = (1/3)*a + (2/3)*b
 
Then again we have a quandle; we can define
 
		q(a,b) = 3*a - 2*b	for every  a,b  from Q;
 
then
	q(s(a,b),b) = s(q(a,b),b) = a	for every  a,b  from Q.
 
---------------------------------------------------------------------
 
More generally, I was interested in (a trivial question) of
an axiomatization of affine spaces (and convex sets). Let  A  be an
arbitrary commutative ring with 1, we can define an A-modaff (Q,m)
where  m : QxQxA --> Q  satisfies:
 
(A0)  m(a,b;0) = a
(A1)  m(b,a;t) = m(a,b,1-t)
(A3)  m(m(a,b;t),m(c,d;t);u) = m(m(a,c;u),m(b,d;u);t)
(A4)  if m(a,c;t) = m(b,c;t)	then	a=b
 
for all  a,b,c,d  from Q  and t,u from A.
---------------------------------------------------------------------
 
There are more open questions that I am able to type (I'm not
a touch typist, rather a key hunter). Thus I'll quit for the time
being. I know that it's not even a start. Thus now you can start.
 
-- Wlodek