Consider a generic binary operation $B$. It is possible for $B$ to obey "set-independent" functional equations, that is functional equation that do not make any reference to an underlying domain and co-domain sets which $B$ acts upon. For example if $B(B(x,y),z) = B(x,B(y,z)$ then we say that $B$ is an associative operator. Another classical example is if $B(x,y) = B(y,x)$ then we say that $B$ is a commutative operator.
From here on out we will assume our "set-independent" functional equations always have a finite amount of functional nesting in them.
A question I was curious about:
Is it always possible to find an algebraic representation of $B$ over $\mathbb{C}^n$ that respects a collection of "set-independent" functional equations?
More Detail:
What that means, given an operator $B$ that obeys one or more "set-independent" functional equations (ex: associativity, commutativity etc...), is it always possible to find an integer $n$ and a multivariate polynomial $b: \mathbb{C}^{n} \times \mathbb{C}^n \rightarrow \mathbb{C}^n$ such that $b$ obeys the same functional equations?
A Concrete Example:
How do you algebraically represent an associative operator? That is an operator $B(x,y)$ which obeys $B(B(x,y),z) = B(x,B(y,z))$
It turns out the map $(x,y) \rightarrow x*y$ is associative and meets our criteria, but looking at it you would be quick to state "that's not very satisfying, what about associative operators that are NOT commutative?" I.E. when we find such algebraic representations we want to find ones that obey NO additional functional equations than they must.
Over $\mathbb{C}^1$ this appear to be impossible but over $\mathbb{C}^4$ we have $2\times2$ matrix multiplication which satisfies exactly this, associativity, and no other assumptions such as commutativity.
Where this question gets difficult:
Suppose we look at a binary operation that obeys some more arbitrary functional equation, it seems like its very hard to find "pure" algebraic solutions that don't assume more relations.
As a concrete example say we look at: $B(a,B(B(b,c),d)) = B(B(B(a,b),c),d)$ (at this point the parenthesis are not intuitive to look at so I put a pictorial diagram below to help make it more palatable.)
This functional equation, I believe, is a MUCH weaker statement than associativity, so finding an algebraic expression that obeys this but not associativity/other set-independent relations has been quite difficult.