# Models of a certain (weird) equational theory

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?

• Do you mean that you want $x\ast x=x$ to be never true, or just false for at least one $x$? – Captain Lama Mar 13 '20 at 20:49
• Just false for at least one $x$ is sufficient. – User7819 Mar 13 '20 at 20:51
• Then you can take for instance any non-trivial commutative group of exponent $3$. – Captain Lama Mar 13 '20 at 20:58
• $x * y = -x$ should work – Daniel Schepler Mar 13 '20 at 21:26
• @CaptainLama That should be an answer! – Noah Schweber Mar 13 '20 at 21:29

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).

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.

• +1, this is neat - although per Captain Lama's comment it's also massive overkill. – Noah Schweber Mar 13 '20 at 21:30

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.)