Join of two elements

Let $a,b$ be two elements with atoms. The pair $(a,b)$ satisfies the following universal property: $(a,b)$ supports both $a$ and $b$ , and if $y$ is another element with this property, then $y$ supports $(a,b)$ .

We demonstrate the existence of an element denoted $a\land b$ , with a dual property: $a\land b$ is supported both by $a$ and by $b$ , and if $y$ is another element with this property, then $y$ is supported by $a\land b$ . The construction does not require any assumptions on the symmetry.

Recall that an element $x$ supports an element $y$ if for any permutation $\pi$ of the atoms, $\pi(x)=x$ implies $\pi(y)=y$ . Equivalently, there is an equivariant mapping which maps $x$ to $y$ . If $x$ supports $y$ and $y$ supports $x$ , then $x$ and $y$ are equivariantly equivalent. The element $a\land b$ satisfying the above universal property is unique, up to equivariant equivalence.

Example. If $a,b,c$ are distinct atoms, then $(a,b)\land (b,c)$ is equivariantly equivalent to $b$ . On the other hand, $\set{a,b}\land\set{b,c}$ is equivariantly equivalent to $\emptyset$ .

We now present the construction of $a\land b$ in general. It is vaguely inspired by the construction of a pushout using coproducts and coequalizers.

Let $a$ and $b$ be two elements. Let $A$ denote the orbit of $a$ and $B$ denote the orbit of $b$ . There is a smallest equivariant relation $\sim$ on the disjoint union $A\coprod B$ such that $a\sim b$ . Let $a\land b$ denote the $\sim$ -equivalence class of the element $a$ .

Claim. $a\land b$ is supported both by $a$ and by $b$ , and if $y$ is another element with this property, then $y$ is supported by $(a\land b)$ .

Proof. Since $\sim$ is equivariant, it defines an equivariant mapping $f$ from $A\coprod B$ to $P_{fs}(A\coprod B)$ , which assigns to an element its equivalence class. In particular, $f(a)=a\land b$ is supported by $a$ . Similarly, $a\land b$ is supported by $b$ .

Now, let $y$ be any element which is supported both by $a$ and by $b$ . Let $Y$ be the orbit of $y$ . Then there is a unique equivariant mapping $f:A\to Y$ such that $f(a)=y$ , and a unique equivariant mapping $g:B\to Y$ such that $g(b)=y$ . Consider the equivalence relation $\approx$ on $A\times B$ such that $u\approx v$ for $u\in A,v\in B$ iff $f(u)=g(v)$ .

Observe that there is an equivariant mapping $h$ from the set of $\approx$ -equivalence classes to $Y$ , and that this mapping maps the $\approx$ -equivalence class of $a$ to $y$ .

This relation is equivariant and since $a\approx b$ , it follows that it is coarser than the relation $\sim$ . In particular, there is an equivariant mapping which maps the $\sim$ -equivalence class of an element $u$ to its $\approx$ -equivalence class.

It follows that there is an equivariant mapping which maps the element $a\land b$ , i.e., the $\sim$ -equivalence class of $a$ , to $y$ .

Atompress

Join of two elements

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Computation with atoms