Does this “reverse distributivity” ever occur: $a \circ (b\times c) = (a \times b) \circ (a \times c)$?

Question

I was reading a book containing a typo to the effect that they defined the distributive property as: $$ a \circ (b\times c) = (a \times b) \circ (a \times c) \tag{*}\label{*} $$

which is wrong of course. I will call the property (*) "reverse distributivity" for now. It got me wondering:

Are there any examples of structures with this "reverse distributivity"? What can we say about such a structure? And are there names for these things?

Some findings so far:

If we assume the existence of neutral elements, then things quickly degenerate. Assume that $(M, \circ, 1_\circ, \times, 1_\times)$ is an algebraic structure with two binary operators satisfying (*), and where $1_\circ$ and $1_\times$ are neutral elements. Then we have: $$ 1_\circ = (1_\circ \times 1_\times) \circ (1_\circ \times 1_\times) \stackrel{\eqref{*}} = 1_\circ \circ (1_\times \times 1_\times) = 1_\times $$ so the identity elements are in fact equal. Let $1 := 1_\times = 1_\circ$. Then, for any $a,b\in M$: $$ a \times b = 1 \circ (a \times b) \stackrel{\eqref{*}} = (1 \times a ) \circ (1 \times b) = a \circ b $$ so in fact the two compositions are the same. In this case (*) becomes $$ a \circ (b\circ c) = (a \circ b) \circ (a \circ c) $$ which seems to be known as self-distributivity and shows up in a number of places (e.g. group conjugation and logical implication).

But if we want two (different) compositions that satisfy (*), then this shows that they at least cannot both have neutral elements. (If we only assume the existence of $1_\times$, then we can show that $a \circ a = a \circ 1_\times$ for all $a$). I haven't gone much further than this.

Answer 1

I will treat the case where $\times$ forms a monoid with unit element $1$. From monoids, one can go in two directions: towards group-like structures or towards lattice-like structures. The group-like cases are all trivial, but we can find many non-degenerate lattice-like examples.

So consider a monoid $(M,\times,1)$ and assume that we have an operation $\circ$ that satisfies the reverse distributivity law $a\circ(b \times c) = (a \times b)\circ(a \times c)$ for any $a,b,c \in M$. It follows then that have $b\circ c = (1 \times b)\circ(1\times c)=1\circ(b\times c)$ for all $b,c \in M$.

Define the function $f: M \rightarrow M$ as $f(x) = 1\circ x$. Then reverse distributivity becomes $f(a \times b \times c) = a\circ (b \times c) = (a \times b) \circ (a \times c) = f(a \times b \times a \times c)$. Vice versa, any function $f: M \rightarrow M$ satisfying $f(a \times b \times c) = f(a\times b \times a \times c)$ for all $a,b,c \in M$ gives rise to a reverse distributive operation via $a\circ b = f(a \times b)$. Therefore:

1. Any potential $f$ must satisfy $f(a) = f(a \times a)$ for all $a \in M$. This follows by setting $b,c = 1$ in the equation above.

2. If $M$ is a group, then all such functions $f$ are in fact constant. To see this, notice that $f(a\times b \times c) = f(a \times b \times a \times c)$, then set $c=1$ and $b=a^{-1}$ to conclude $f(1) = f(a \times a^{-1}) = f(a \times a^{-1} \times a) = f(a)$ for all $a \in M$.

3. If $M$ is an idempotent commutative monoid (semilattice), then the equality $f(a \times b \times c) = f(a \times a \times b \times c) = f(a \times b \times a \times c)$ holds for any function $f: M \rightarrow M$. This observation yields plenty of non-degenerate examples. For example, consider the Boolean algebra $\{\text{true},\text{false}\}$ of truth values, and set $\times$ to be disjunction $\vee$. Define $a \not\vee b = \neg(a \vee b)$. The pair $(\vee,\not\vee)$ satisfies reverse distributivity, i.e. $a \not\vee (b \vee c) = (a \vee b) \not\vee (a \vee c)$. Similarly for $\wedge$.

If the structure $(M,\times)$ is a general semigroup, then these observations fail. Interestingly, by analyzing all 14 non-constant binary operations, we can show that all suitable pairs of operations $(\times, \circ)$ over truth values have $a \circ b = f(a \times b)$ for some $f$. However, there is a three-element structure $(S,\times,\circ)$ with $(\times,\circ)$ satisfying reverse distributivity that does not have $a \circ b = f(a \times b)$ for any function $f$ and with $a\times b$ non-constant. E.g. Consider the two operations below.

x|012    o|012
-----    -----
0|121    0|011
1|211    1|011
2|111    2|011

Since $b \times c \neq 0$, we have $a \circ (b \times c) = 1 = 1 \circ 1 = (a \times b) \circ (a \times c)$, but not $a \circ b = f(a \times b)$, since that would give $0 = 0 \circ 0 = f(0 \times 0) = f(1) = f(1 \times 1) = 1 \circ 1 = 1$.

edit: While I'm sure that someone, somewhere has already named these structures, I doubt that there's a widely-agreed-upon name for them, even among people who actively study these structures (especially given that we don't have such a name even for the self-distributive case, and even when equipped with additional axioms. Viz. racks, crystals, automorphic sets, etc.).

Answer 2

For a yet trivial example in which $\circ$ and $\times$ denote different operations (and $\circ$ has a neutral element, and additionally, they're both associative and commutative), consider $M = \{0,1\}$, with $\circ$ being the usual product, and $\times$ a null product (that is, $x \times y = 0$, for all $x$ and $y$).

Then the equation $$x \circ (y \times x) = (x \times y) \circ (x \times z)$$ reduces to $$x \circ 0 = 0 \circ 0,$$ which is true.

It is not a very interesting example, but the answer to your question is yes, there exist such structures.
The example you deduced from the existence of identities for both operations (both operations being equal and self-distributive) is also perfectly legitimate.

There are many more others (some without any of the trivial aspects the above examples show), but concerning names, usually these structures are only named if they are found to be interesting in some context.
As a rule, the answer to the question "Are there such algebras satisfying these axioms" is yes, at least if the axioms are equalities. In these cases, there's always at least the trivial example, that is, the one-element algebra with whatever are the operations.