TITLE: Are all n-ary operators simply compositions of binary operators?
QUESTION [17 upvotes]: Take for example $A \times B \cdot C$ = $(A \times B) \cdot C$ where $A, B, C$ are 3-component real vectors.
We can define a 3-nary operator $\times - \cdot$ that is a composition of the two common binary operators $\times$ and $\cdot$.
The same thing happens with most functions (operators) - the way we calculate them is by doing smaller binary problems and adding together.
Every time I try to come up with an $(n > 2)$-ary operator my mind automatically looks for binary operators.
So, the question is, do there exist operators (of some weird kind in some branch of math) that cannot be decomposed into 2-ary and 1-ary operators?
Thanks.

REPLY [9 votes]: Here is another point of view, which I learned from Richard Garner. 
Suppose we have an algebraic theory $\mathbb{T}$ – so a fixed set of finitary operations and a set of equations that these satisfy. Let $\mathbb{T}\textbf{-Alg}$ be the category of all $\mathbb{T}$-algebras and their homomorphisms. We define a natural $n$-ary operation to be a family of functions $\theta_X : X^n \to X$, indexed by $\mathbb{T}$-algebras $X$, such that for all homomorphisms $f : X \to Y$, the operation $\theta$ commutes with $f$, in the sense that
$$\theta \circ (f \times \cdots \times f) = f \circ \theta$$
where $f \times \cdots \times f : X^n \to Y^n$ is the function that applies $f$ to each component separately.
Now, it is a fact that any algebraic theory $\mathbb{T}$ admits a notion of ‘free $\mathbb{T}$-algebra on a set of generators’. Formally, this is defined as the functor $F : \textbf{Set} \to \mathbb{T}\textbf{-Alg}$ which is left adjoint to the forgetful functor $U : \mathbb{T}\textbf{-Alg} \to \textbf{Set}$; in simpler terms, the free $\mathbb{T}$-algebra on a set $S$ is the $\mathbb{T}$-algebra $F S$ together with a function $\eta_S : S \to F S$, called the ‘insertion of generators’, such that for all functions $f : S \to X$, where $X$ is a $\mathbb{T}$-algebra, there is a unique $\mathbb{T}$-algebra homomorphism $h : F S \to X$ such that $h \circ \eta_S = f$. In this general language, your question can be interpreted as follows: is every natural $n$-ary operation a composite of operations defined in $\mathbb{T}$?
It turns out the answer is yes! By some simple abstract nonsense involving the Yoneda lemma, it can be shown that there is a canonical bijection between the set of natural $n$-ary operations and the set of elements of $F \{ e_1, \ldots, e_n \}$. This means that every natural $n$-ary operation corresponds to some string of symbols formed by $e_1, \ldots, e_n$ and the operations of the theory $\mathbb{T}$. But this precisely means that the $n$-ary operation $\theta_X : X^n \to X$ can be written as a composite of the operations of $\mathbb{T}$ together with the canonical projection operators $\pi_j : X^n \to X^m$. For example, if we take $\mathbb{T}$ to be the theory of groups, and $\theta_X (x_1, \ldots, x_n) = x_1 {x_2}^{-1} x_3 \cdots {x_n}^{(-1)^n}$, then the operation $\theta$ corresponds to the word $e_1 {e_2}^{-1} e_3 \cdots {e_n}^{(-1)^n}$ in the free group generated by $\{ e_1, \ldots, e_n \}$, and this is clearly a composition of the group multiplication and the inversion.
Of course, there are algebraic theories $\mathbb{T}$ where there are no constants, unary or binary operations whatsoever. For instance, $\mathbb{T}$ could be the theory of heaps, which has a single ternary operation. In this context it makes no sense to ask whether that ternary operation can be rewritten as a binary operation of $\mathbb{T}$, because there simply aren't any binary operations in $\mathbb{T}$ at all. But it turns out that when we augment $\mathbb{T}$ with one constant (and no axioms!), $\mathbb{T}$ becomes the theory of groups, and we know that the theory of groups is can be presented using one constant, one unary operation, and one binary operation. So one might ask: does every algebraic theory $\mathbb{T}$ admit a presentation using only operations of arity at most $2$? Unfortunately, I do not know the answer to this.