Midpoint Algebras

Midpoint Algebras

[Peter Freyd]

I've defined the theory of midpoint algebras to be given by a single
binary operator satisfying the three equations;

            x|x = x                 Idempotence;
            x|y = y|x               Commutativity;
    (u|v)|(x|y) = (u|x)|(v|y)       Middle-two interchange;

and the Horn sentence:

          x|y = x|z  =>  y = z      Cancelation.

For a non-Euclidean example choose a subset of the circle consisting
of an oddly finite number of evenly placed points ("evenly placed"
meaning that the distances between any point and its two immediate
neighbors are the same). For any two points we take their midpoint to
be the unique point that's equally distant from the two of them. I'll
call this a cyclic midpoint algebra. There's one isomorphism type for
each odd natural number. And there aren't too many other finite
midpoint algebras:

1.
    Any non-empty finite midpoint algebra is isomorphic to a 
    product of cyclic midpoint algebras each of prime-power order.

(In particular, there are no finite midpoint algebras of even order.)

A little theorem that I find quite annoying:

2.
    The finite midpoint algebras are semantically complete with 
    respect to the Horn theory. That is, the universally quantified
    first-order Horn sentences on the predicate  x|y = z   that 
    hold for all cyclic midpoint algebras hold for all midpoint 
    algebras.

It's annoying because the finite examples are so unlike the motivating
example of Euclidean space. (Note the perversity: the midpoint of a
pair of adjacent points on a cyclic midpoint algebra is maximally far
from each of them.)

An algebra (for any theory) is said to be torsion-free if the only
finite subalgebras are those with just one element. Torsion freeness
is always given by a Horn theory. For midpoint algebras, torsion 
freeness can be specified with the sequence of Horn sentences of the
form:
          x = y|(y|(y|(...(y|(y|x))...)))   =>   x = y

I little theorem that I don't find at all annoying:

3.
    The real line is semantically complete with respect to the 
    universal quantified first-order theory of torsion-free 
    midpoint algebras.


Because:

1:
Given a non-empty finite midpoint algebra choose an element and label
it  0.  Define the halving function  h  by  hx = 0|x.  The cancelation
condition says that  h  is injective, therefore a permutation. Define
x+y  by  h(x+y) = x|y  and verify the abelian group axioms. The
decomposition theorem for finite abelian groups finishes the proof.

2:
Given an arbitrary non-empty midpoint algebra we may, as observed in
my first post on this subject, choose a  0  and obtain a larger 
algebra in which the halving map becomes an automorphism. (It may be 
explicitly constructed as the set whose elements are named by pairs
<x,n>  where  x  is an element in the given algebra and  n  is a 
natural number; <y,m>  names the same element iff  h^m(x) = h^n(y);
midpoints are defined by  <x,n>|<z,p>  = <h^p(x)|h^n(y),n+p>.) 

Define  x+y  as above to obtain a commutative monoid with the
cancelation property and apply the classical construction to obtain a
larger commutative monoid that is an abelian group. Note that this
group has unique division by two, hence we may view it as a module
over the ring, D, of dyadic rationals.

Given a Horn sentence with a counterexample in some midpoint algebra
we seek a counterexample in a cyclic midpoint algebra. The given
counterexample consists of a finite number of elements. If we embed
their ambient midpoint algebra into a  D-module, we may cut down
to the submodule generated therein by the image of those elements to 
obtain a finitely generated  D-module. One may easily verify that  D
is a principal ideal domain, hence this last  D-module decomposes as a
direct sum of cyclic  D-modules. The indecomposable cyclic modules
correspond to the prime-powers in  D, that is, there are finite cyclic
groups of odd (ordinary) prime power order (each automatically a  
D-module) and one infinite case,  D  itself, viewed as a  D-module.

For the finite set of elements to constitute a counterexample we
need to maintain one given non-equation, hence we can land in a cyclic
D-module. If it happens to be the infinite cyclic  D-module, 
factor out by a principal ideal generated by a sufficiently large
prime integer to maintain the non-equation.

3. 
Given a counterexample to a universally quantified first-order
sentence in a torsion-free midpoint algebra repeat the above
construction to obtain a counterexample in a finite direct sum of
cyclic  D-modules. Verify that it remains a counterexample when we 
factor out the torsion submodule -- that is, verify that the map from
the original torsion-free midpoint algebra remains an embedding. 

We now have a counterexample in a finite direct sum of copies of  D.
Enlarge to a finite direct sum of copies of the rational numbers. One
does not even need the axiom of choice to embed finite-dimensional
rational vector spaces into the one-dimensional real space.