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 $X^{I}$)

 $[x]_{J}:=\{y\in X^{I}\mid x_{i}=y_{i}\mbox{ for every }i\notin J\},$
• 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. 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. 2.

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

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

## References

Title example of polyadic algebra ExampleOfPolyadicAlgebra 2013-03-22 17:53:20 2013-03-22 17:53:20 CWoo (3771) CWoo (3771) 15 CWoo (3771) Example msc 03G15 functional polyadic algebra