Examples of algebras satisfying (a+b)(c+d)=ac+bd

Question

Is there a known example of an algebra $(A, +, \cdot)$ with two binary commutative (see P.S below) and idempotent operations $+$ and $\cdot$ satisfying the identity $(a+b)(c+d)=ac+bd$?

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Actually I need to know is there any algebra $A$ having signature $\mathcal{L}$ such that for any $n$-ary $f\in \mathcal{L}$ and any $m$-ary $g\in\mathcal{L}$ we have $$ f(g(x_{11}, \ldots, x_{1m}), \ldots, g(x_{n1}, \ldots,x_{nm}))=g(f(x_{11}, \ldots,x_{n1}), \ldots, f(x_{1m},\ldots, x_{nm})) $$ Also I need to have all operations idempotent. As my motivation, I should say that for any variety of such algebras, we have the following interesting property: Let $F=F_V(x_1,\ldots,x_n)$ be a relative free algebra in $V$. Then the solution set of any system of equations in $F$ is a subalgebra of $F^k$ where $k$ is the number of indeterminate.

P.S. Instead of commutativity, consider the next two weaker conditions:

1. $(a+b)+(c+d)=(a+c)+(b+d)$.

2. $(ab)(cd)=(ac)(bd)$. \

P.S.2. It is good to use the phrase Medial instead of Commutative. So my question becomes: Is there a known example of an algebra $(A, +, \cdot)$ with two binary medial and idempotent operations satisfying the identity $(a+b)(c+d)=ac+bd$?

Answer 1

Assuming the properties of idempotency and commutativity, $a+b=(a+b)(a+b)=(a+b)(b+a)=ab+ba=ab+ab=ab$. So you will have essentially identical operations for + and *. I don't know what happens if you drop one or more of idempotency or commutativity.

Answer 2

There are large classes of algebras with two (or more) operations satisfying your identities. They are called modes. There is a monograph by Romanowska and Smith, called Modes, on this very topic. Here is a recent survey paper on the topic by Smith.

Answer 3

If you weaken your identities, then you end up with commuting one dimensional cellular automata. It turns out that commuting one dimensional cellular automata can essentially be represented as finite algebras $(A,+,\cdot)$ that satisfy the identity $(x\cdot y)+(y\cdot z)=(x+y)\cdot(y+z)$. Therefore, the algebras that you have mentioned will definitely produce commuting cellular automata.

Suppose $A$ is a finite set. Then $A^{\mathbb{Z}}$ becomes a compact space with the product topology. Let $\Phi:A^{\mathbb{Z}}\rightarrow A^{\mathbb{Z}}$ be the function defined by $\Phi(a_{r})_{r\in\mathbb{Z}}=(a_{r+1})_{r\in\mathbb{Z}}$. Then a one-dimensional cellular automaton with values in $A$ is a continuous function $G:A^{\mathbb{Z}}\rightarrow A^{\mathbb{Z}}$ such that $G\Phi=\Phi G$. It is well known that a function $G:A^{\mathbb{Z}}\rightarrow A^{\mathbb{Z}}$ is a cellular automaton if and only if there is some $n$ and function $F:A^{2n+1}\rightarrow A$ such that $G(a_{r})_{r\in\mathbb{Z}}=(F(a_{r-n},...,a_{r},...,a_{r+n}))_{r\in\mathbb{Z}}$. In other words, the cellular automaton function $G$ is determined by local coordinates and it is translation invariant. One may easily translate any cellular automata into a cellular automata of the form $G:A^{\mathbb{Z}}\rightarrow A^{\mathbb{Z}}$ such that there is a binary operation $+$ on $A$ such that $G(a_{r})_{r\in\mathbb{Z}}=(a_{r}+a_{r+1})_{r\in\mathbb{Z}}$.

$\textbf{Proposition}$ Suppose that $F,G:A^{\mathbb{Z}}\rightarrow A^{\mathbb{Z}}$ are cellular automata such that $F(a_{r})_{r\in\mathbb{Z}}=(a_{r}+a_{r+1})_{r\in\mathbb{Z}},G(a_{r})_{r\in\mathbb{Z}}=(a_{r}\cdot a_{r+1})_{r\in\mathbb{Z}}$. Then $FG=GF$ if and only if the algebra $(A,+,\cdot)$ satisfies the identity $(x+y)\cdot(y+z)=(x\cdot y)+(y\cdot z)$.

$\textbf{Proof}$ This result follows from the observation that $$FG(a_{r})_{r\in\mathbb{Z}}=f((a_{r}\cdot a_{r+1})_{r\in\mathbb{Z}})= ((a_{r}\cdot a_{r+1})+(a_{r+1}\cdot a_{r+2}))_{r\in\mathbb{Z}}$$ and $$GF(a_{r})_{r\in\mathbb{Z}}=((a_{r}+a_{r+1})\cdot(a_{r+1}+a_{r+2})).$$ $\mathbf{QED}$

The reader is referred to the paper [1] for more information on commuting cellular automata and algebras that satisfy the identity $(x\cdot y)+(y\cdot z)=(x+y)\cdot(y+z)$.

1. Moore, Cristopher; Boykett, Timothy. Commuting cellular automata. Complex Systems 11 (1997), no. 1, 55–64.

Answer 4

This answer is similar to the other answer that I gave but here I claim that the identity $(a\cdot b)+(c\cdot d)=(a+c)\cdot(b+d)$ is equivalent to whether two 2-dimensional cellular automata commute.

Suppose that $A$ is a finite set and $\cdot,+$ are binary operations on $A$. Then we may define cellular automata $F,G:A^{\mathbb{Z}^{2}}\rightarrow A^{\mathbb{Z}^{2}}$ by letting $F(a_{r,s})_{r,s\in\mathbb{Z}}=(a_{r,s}+a_{r+1,s})_{r,s\in\mathbb{Z}}$ and $G(a_{r,s})_{r,s\in\mathbb{Z}}=(a_{r,s}\cdot a_{r,s+1})_{r,s\in\mathbb{Z}}$. Then it can easily be shown that $F\circ G=G\circ F$ if and only if the identity $(w\cdot x)+(y\cdot z)=(w+y)\cdot(x+ z)$ holds.

$\textbf{A Generalization}$ (edited November 15, 2014).

It turns out the more general identity asked in the question involving operations of higher arity still reduces to the notion of whether $``$orthogonal$"$ cellular automata commute with each other.

Let $A$ be a set. Let $s:A^{m}\rightarrow A,t:A^{n}\rightarrow A$ be functions.

Define maps $G,H:A^{\mathbb{Z}^{n+m}}\rightarrow A^{\mathbb{Z}^{n+m}}$ by letting

$$\large G(x_{i_{1},...,i_{n+m}})_{i_{1},...,i_{n+m}\in\mathbb{Z}}$$ $$\large =s(x_{i_{1}+1,...,i_{m},i_{m+1},...,i_{m+n}},...,x_{i_{1},...,i_{m}+1,i_{m+1},...,i_{m+n}})$$

and

$$\large H((x_{i_{1},...,i_{n+m}})_{i_{1},...,i_{n+m}\in\mathbb{Z}})$$

$$\large=(t(x_{i_{1},...,i_{m},i_{m+1}+1,...,i_{m+n}},...,x_{i_{1},...,i_{m},i_{m+1},...,i_{m+n}+1})_{i_{1},...,i_{m+n}\in\mathbb{Z}}.$$

Then the cellular automata mappings $G$ and $H$ commute (i.e. $G\circ H=H\circ G$) if and only if the operations $s$ and $t$ satisfy the entropic identity: $$s(t(z_{1,1},...,z_{1,n}),...,t(z_{m,1},...,z_{m,n})) =t(s(z_{1,1},...,z_{m,1}),...,s(z_{1,n},...,z_{m,n})).$$

I am unsure if the relation between modes and entropic algebras and commutative cellular automata which act on orthogonal dimensions has been studied before.