quasigroup midpoint algebras = Z[1/2]-modules
The goal of this blog post is to demonstrate that a quasigroup midpoint algebra induces a unique Z[1/2]-module up to isomorphism, and vice versa.
what is a quasigroup midpoint algebra
A midpoint algebra is a set and binary operation $(M, \mid)$ such that
$a \mid a = a$ (idempotent)
$a \mid b = b \mid a$ (commutative)
$(a \mid b) \mid (c \mid d) = (a \mid c) \mid (b \mid d)$ ("medial")
The medial axiom may look really arbitrary but it underpins two very important facts.
One comes up later, but an immediate consequence is that midpoint algebras are self-distributive, by letting b = a:
$(a \mid a) \mid (c \mid d) = (a \mid c) \mid (a \mid d)$ (medial)
$(a \mid a) \mid (c \mid d) = a \mid (c \mid d)$ (idempotent)
$a \mid (c \mid d) = (a \mid c) \mid (a \mid d)$ (equality is transitive)
In general, this is a very useful property.
Now, story time.
The algebraic structure that you will find online is known as a cancellative midpoint algebra.
Originally when I first published this, I screwed up and did not realize that if an algebra is "cancellative", this does not imply it is a quasigroup.
However, I have edited this post to invent an entirely new type of algebraic structure purely so I can demonstrate this neat result!
Anyway, to upgrade a magma to a quasigroup, all that is needed is that we introduce a new operation, division, and give it these properties:
$(a \mid b) / b = a$ (cancellation 1)
$(a / b) \mid b = a$ (cancellation 2)
Typically, quasigroups are defined with both a left- and right-division, but we can avoid this because our operation is commutative.
Something that will be useful is that we can prove $\mid$ also distributes over $/$:
$a \mid b = a \mid b$ (equality is reflexive)
$a \mid ((b / c) \mid c) = a \mid b$ (cancellation 2)
$(a \mid (b / c)) \mid (a \mid c) = a \mid b$ (self-distributive)
$a \mid (b / c) = (a \mid b) / (a \mid c)$ (divide both sides by $(a \mid c)$)
Now we have the basics to get into the proof.
motivation for what follows
We are motivated by the observation that any Z[1/2]-module induces a unique quasigroup midpoint algebra.
Indeed, let our module be denoted $(A, +)$.
(The letter A is suggestive of how this is just an abelian group equipped with a sensible "halving" operation.
Also the letter M was taken because M stands for Midpoint.)
Then we can define $a \mid b = \frac{a + b}{2}$ and $a / b = 2a - b$.
It is trivial to verify these obey each axiom of a quasigroup midpoint algebra listed above.
Our goal will be to uniquely reconstruct $(A, +)$ given $(M, \mid, /)$.
the proof
Given a quasigroup midpoint algebra $(M, \mid, /)$, choose an arbitrary element $z \in M$ and define:
$a + b = (a \mid b) / z$
$-a = z / a$
$\frac{a}{2} = z \mid a$
I call this element "z" because it will act as 0 (zero) in our constructed Z[1/2]-module.
It turns out that regardless of what z you pick, the result is always the same up to isomorphism.
This is because left-multiplication by $z' / z$ is a quasigroup midpoint algebra automorphism that sends $z$ to $z'$ for any $z'$ we desire.
This is part of why I verified above that $\mid$ distributes over both itself and division; this is sufficient to prove this is a valid endomorphism.
Then right-division by $z' / z$ is a two-sided inverse and thus proves it is in fact an automorphism.
However, we need right-division to be distributive and thus an endomorphism in order for this logic to work.
The proof is as follows:
$a \mid b = a \mid b$ (equality is transitive)
$a \mid b = (((a / c) \mid c) \mid ((b / c) \mid c))$ (cancellation 2)
$a \mid b = ((a / c) \mid (b / c)) \mid c$ (self-distributive)
$(a \mid b) / c = (a / c) \mid (b / c)$ (divide both sides by c)
Well, we also need to show it distributes over division, but that's easy enough.
$a / b = a / b$ (equality is transitive)
$a / b = ((a \mid c) / c) / ((b \mid c) / c)$ (cancellation 1)
$a / b = ((a \mid c) / (c \mid c)) / ((b \mid c) / (c \mid c))$ (idempotent)
$a / b = ((a / c) / (b / c)) \mid c$ (self-distributive)
$(a / b) / c = (a / c) / (b / c)$ (divide both sides by c)
With that out of the way, we can proceed without fear even though z was an arbitrary choice!
We can mindlessly verify quite a few abelian group axioms to start.
It is obvious that $a + b = b + a$.
A moment's thought reveals that $a + z = a$ by cancellation of z.
$a - a = z$ is slightly less trivial:
$a - a = (a \mid (z / a)) / z = z / z = z$
Note that $z / z = z$ follows from dividing both sides of $z \mid z = z$ by z.
Our penultimate obstacle is our greatest one yet: associativity of $+$.
Until now, we have not used mediality outside of its self-distributive consequences.
Essentially, everything above applies to what are known as "commutative quandles".
It is, in fact, impossible to prove the following fact without mediality, because there exist commutative quandles that are the result of a midpoint operation over commutative moufang loops, which are not associative.
$(z \mid a) \mid (b \mid c) = (z \mid a) \mid (b \mid c)$ (equality is reflexive)
$(z \mid a) \mid (b \mid c) = (z \mid a) \mid (c \mid b)$ (commutative)
$(z \mid a) \mid (b \mid c) = (z \mid c) \mid (a \mid b)$ (medial!!)
$a \mid ((b \mid c) / z) = c \mid ((a \mid b) / z)$ (divide both sizes by z, right-division distributing)
$(a \mid ((b \mid c) / z)) / z = (c \mid ((a \mid b) / z)) / z$ (divide both sides by z)
$a + (b + c) = (a + b) + c$ (definition)
And there we go!
All that is left to show is that we have a sensible action by $\frac{1}{2}$ on this abelian group, which is refreshingly simple:
$\frac{a}{2} + \frac{a}{2} = ((z \mid a) \mid (z \mid a)) / z = (z \mid a) / z = a$
as desired.
QED!
While this may be useless knowledge, I think it is funny how much structure a few rules can impose.
Hopefully you found it amusing.
Also, a reader suggested that I add a picture to help clarify the geometric meaning inherent in all of this.
Say no more.
