example of polyadic algebra

Recall that the canonical example of a monadic algebra is that of a functional monadic algebra, which is a pair $(B,\exists )$ such that $B$ is the set of all functions from a non-empty set $X$ to a Boolean algebra $A$ such that, for each $f\in B$, the supremum and the infimum of $f(X)$ exist, and $\exists$ is a function on $B$ that maps each element $f$ to $f^{\exists }$, a constant element whose range is a singleton consisting of the supremum of $f(X)$.

The canonical example of a polyadic algebra is an extension (generalization) of a functional monadic algebra, known as the functional polyadic algebra. Instead of looking at functions from $X$ to $A$, we look at functions from $X^I$ (where $I$ is some set), the $I$-fold cartesian power of $X$, to $A$. In this entry, an element $x\in X^I$ is written as a sequence of elements of $A$: ${(x_i)}_{i\in I}$ where $x_i\in A$, or $(x_i)$ for short.

Before constructing the functional polyadic algebra based on the sets $X,I$ and the Boolean algebra $A$, we first introduce the following notations:

• for any $J\subseteq I$ and $x\in X^I$, define the subset (of $X^I$)

  $${[x]}_J:=\{y\in X^I\mid x_i=y_i\text{ for every }i\notin J\},$$

• for any function $\tau :I\to I$ and any $f:X^I\to A$, define the function $f_{\tau }$ from $X^I$ to $A$, given by

  $$f_{\tau }(x_i):=f(x_{\tau (i)}).$$

Now, let $B$ be the set of all functions from $X^I$ to $A$ such that

1. for every $f\in B$, every $J\subseteq I$ and every $x\in X^I$, the arbitrary join

   $$\bigvee f\left({[x]}_J\right)$$

   exists.

   Before stating the next condition, we introduce, for each $f\in B$, a function $f^{\exists J}:X^I\to A$ as follows:

   $$f^{\exists J}(x):=\bigvee f\left({[x]}_J\right).$$

   Now, we are ready for the next condition:

2. if $f\in B$, then $f^{\exists J}\in B$,

3. if $f\in B$, then $f_{\tau }\in B$ for $\tau :I\to I$.

Note that if $A$ were a complete Boolean algebra, we can take $B$ to be $A^{X^I}$, the set of all functions from $X^I$ to $A$.

Next, define $\exists :P(I)\to B^B$ by $\exists (J)(f)=f^{\exists J}$, and let $S$ be the semigroup of functions on $I$ (with functional compositions as multiplications), then we call the quadruple $(B,I,\exists ,S)$ the functional polyadic algebra for the triple $(A,X,I)$.

Remarks. Let $(B,I,\exists ,S)$ be the functional polyadic algebra for $(A,X,I)$.

• $(B,I,\exists ,S)$ is a polyadic algebra. The proof of this is not difficult, but involved, and can be found in the reference below.

• If $I$ is a singleton, then $(B,I,\exists ,S)$ can be identified with the functional monadic algebra $(B,\exists )$ for $(A,X)$, for $S$ is just $I$, and $X^I$ is just $X$.

• If $I$ is $\mathrm{\varnothing }$, then $(B,I,\exists ,S)$ can be identified with the Boolean algebra $A$, for $S=\mathrm{\varnothing }$ and $X^I$ is a singleton, and hence the set of functions from $X^I$ to $A$ is identified with $A$.