Models of a certain (weird) equational theory

Question

Consider the following (single-sorted) equational/algebraic theory with one binary operation symbol $\ast$ whose axioms are as follows: $$(x \ast x) \ast (x \ast x) = x$$ $$(x \ast y) \ast (x \ast y) = (x \ast x) \ast (y \ast y).$$

I am interested in models of this theory where $\ast$ is NOT idempotent, i.e. where $x \ast x = x$ is not true for every $x$ in the model. So far, I have come up with the following toy model of this theory, where $\ast$ is not idempotent: the carrier is $\{0, 1\}$, and the binary operation $\ast$ is defined as follows: $$0 \ast 0 = 1,$$ $$1 \ast 1 = 0,$$ $$0 \ast 1 = 0,$$ $$1 \ast 0 = 1.$$

My question is, are there any more 'natural' models of this theory where $\ast$ is NOT idempotent, i.e. are there any non-idempotent binary operations satisfying the above axioms that have been previously studied in mathematics?

Answer 1

Let me describe how to produce typical models of this equational theory using a different but equivalent language.

First, number the two given axioms:

Axiom $(1)$. $(x*x)*(x*x)=x$
Axiom $(2)$. $(x*y)*(x*y)=(x*x)*(y*y)$

Let $\sigma(x)=x*x$ be the squaring map with respect to $*$, and let $x\odot y=\sigma(x*y)$. Axiom (1) asserts exactly that $\sigma$ is a permutation of exponent $2$, while Axiom (2) asserts exactly that $\sigma$ commutes with $*$. Since $\sigma$ also commutes with itself, it will then commute with $\odot$, which is a composition of $\sigma$ and $*$. Since $x\odot y$ is defined to be $\sigma(x*y)$, and $\sigma$ has exponent $2$, we can recover $*$ from $\sigma$ and $\odot$ by $x*y=\sigma(\sigma(x*y))=\sigma(x\odot y)$.

Altogether, this shows that we can convert between the $*$-language and the $\odot,\sigma$-language using these definitions:

$\sigma(x):=x*x$.

$x\odot y:=\sigma(x*y)$.

$x*y:=\sigma(x\odot y)$.

Now, in order to translate theories, we observe that an algebra $\langle A; *\rangle$ in the language $\{*\}$ satisfies Axioms (1) and (2) iff the corresponding algebra $\langle A; \odot, \sigma\rangle$ in the language $\{\sigma,\odot\}$ satisfies

Axiom $(1)'$. the binary operation of $\langle A; \odot\rangle$ is idempotent, and
Axiom $(2)'$. $\sigma$ is an exponent-2 automorphism of $\langle A; \odot\rangle$.

That is, up to a change of language, a model of the original axioms is simply an idempotent binary algebra equipped with an exponent-$2$ automorphism.

Examples.

It is not hard to characterize the examples where $\sigma$ is trivial (i.e., $\sigma$ is the identity function). Any such algebra is obtained from an idempotent binary algebra $\langle A; \odot\rangle$ by setting $x*y=x\odot y$.

It is not hard to characterize the examples where $\odot$ is trivial (i.e., $\odot$ is one of the projections $x\odot y = x$ for all $x, y$ or $x\odot y = y$ for all $x, y$). In this case, for any set $A$ let $\sigma: A\to A$ be any permutation of exponent $2$ ($\sigma^2(x)=x$). Then $x*y:=\sigma(x)$ or $x*y:=\sigma(y)$ are both operations on $A$ satisfying Axioms (1) and (2). The example in the question statement is of this type.

Let $\mathbb A = \langle A; \odot\rangle$ be any idempotent binary algebra. Let $\mathbb B = \mathbb A\times \mathbb A$. Let $\sigma: \mathbb B\to \mathbb B: (b,c)\mapsto (c,b)$ be the automorphism of switching coordinates. A change of language converts $\langle B; \odot, \sigma\rangle$ into a model of Axioms (1) and (2).

Let $M$ be an $R$-module. Suppose that $r,s\in R$ commute with each other and $s^2=1$. Then $x\odot y:=rx+(1-r)y$ is idempotent and $\sigma(x)=sx$ is an exponent 2 automorphism of $\langle M; \odot\rangle$, so if we equip $M$ with only the operation $x*y=\sigma(x\odot y) = srx+s(1-r)y$, then $\langle M; *\rangle$ will satisfy Axioms (1) and (2).

Answer 2

There is such a theory that was actually going to be the topic of my dissertation. Let $R$ be a root system, let $x,y$ be roots, and let $s_x$ be the reflection in the hyperplane normal to $x$. Then we may define $$x*y = s_x(y)$$ Then there does not exist any $x$ for which $x*x=x$. In fact, for all $x$ we have $$x*x = -x$$ We therefore have $$(x*x)*(x*x) = - (-x) = x$$ Note we have $$x*(-y)=-(x*y)$$ and $$(x*x)*y = x*y$$

We also have $$(x*y)*(x*y)=-(x*y)$$ and $$(x*x)*(y*y) = x*(-y) = -(x*y)$$ so $$(x*x)*(y*y)=(x*y)*(x*y)$$

The two axioms necessary to prove these things are

1. $x*(x*y)=y$

2. $x*(y*z)=(x*y)*(x*z)$

There is a third axiom that ensures that the resulting algebra is a root system, but it is a bit more of a pain to state.

Answer 3

This answer expands on a comment of Captain Lama; if they post an answer of their own I'll delete this one, and I've made it community-wiki so I don't get reputation for their work.

Note that the second condition is an immediate consequence of associativity and commutativity. So any commutative semigroup satisfying $x^4=x$ will satisfy your theory - for example, the group $\mathbb{Z}/3\mathbb{Z}$.

(Of course, there are structures satisfying your theory which are not commutative semigroups, but commutative semigroups are relatively simple things to think about.)