Algebraic Descent Through Closure Operators

Author: Thu-Le TRAN
Written by ChatGPT

1. Closure operators and equivalent forms of nuclearity

Let $(P,\leq)$ be a poset. A map $j:P\to P$ is a closure operator if it is monotone, extensive, and idempotent:
$$x\leq jx,\qquad x\leq y\Rightarrow jx\leq jy,\qquad j^2x=jx.$$
Write
$$P_j=\operatorname{Fix}(j)=\{x\in P:jx=x\}.$$

Let $\star:P\times P\to P$ be a commutative monotone operation.

Theorem 1

The following conditions are equivalent:
$$\tag{N} jx\star jy\leq j(x\star y),$$
$$\tag{J} j(jx\star jy)=j(x\star y),$$
$$\tag{A} j(x\star jy)=j(x\star y).$$

Proof. First, $(N)\Leftrightarrow(J)$. Since $x\leq jx$ and $y\leq jy$,
$$j(x\star y)\leq j(jx\star jy).$$
Under $(N)$,
$$j(jx\star jy)\leq j^2(x\star y)=j(x\star y),$$
hence $(J)$. Conversely, by extensivity and $(J)$,
$$jx\star jy\leq j(jx\star jy)=j(x\star y),$$
which gives $(N)$.

Next, $(J)\Leftrightarrow(A)$. Since
$$x\star y\leq x\star jy\leq jx\star jy,$$
applying $j$ and using $(J)$ gives
$$j(x\star y)\leq j(x\star jy)\leq j(jx\star jy)=j(x\star y),$$
hence $(A)$. Conversely, by commutativity and two applications of $(A)$,
$$j(jx\star jy)=j(jy\star jx)=j(jy\star x)=j(x\star jy)=j(x\star y).$$
Thus
$$(N)\Longleftrightarrow(J)\Longleftrightarrow(A).$$

A closure operator satisfying these equivalent conditions is called a nucleus for $\star$.

2. Descent of a commutative monoid

Let $(P,\star,e)$ be an ordered commutative monoid, with $\star$ monotone, and let $j$ be a nucleus for $\star$. Define on $P_j$
$$x\star_j y:=j(x\star y),\qquad e_j:=j(e).$$

Theorem 2

$(P_j,\star_j,e_j)$ is a commutative monoid.

Proof. Closure under $\star_j$ is immediate from idempotence. Commutativity follows directly from that of $\star$. For associativity, using $(A)$,
$$\begin{aligned} (x\star_j y)\star_j z &=j(j(x\star y)\star z)\\ &=j((x\star y)\star z)\\ &=j(x\star(y\star z))\\ &=j(x\star j(y\star z))\\ &=x\star_j(y\star_j z). \end{aligned}$$
For the identity, again by $(A)$,
$$x\star_j e_j=j(x\star je)=j(x\star e)=jx=x.$$
Hence the monoid structure descends from $P$ to $P_j$.

3. Descent of a commutative semiring

Let $(S,+,\cdot,0,1)$ be an ordered commutative semiring whose two operations are monotone. Let $j$ be a closure operator which is a nucleus for both $+$ and $\cdot$. Define
$$x+_j y:=j(x+y),\qquad x\cdot_jy:=j(xy),$$
and
$$0_j:=j0,\qquad 1_j:=j1.$$

Theorem 3

$(S_j,+_j,\cdot_j,0_j,1_j)$ is a commutative semiring.

Proof. By Theorem 2, both commutative monoid structures descend. It remains to prove distributivity and absorption. Using $(A)$ for multiplication, ambient distributivity, and $(J)$ for addition,
$$\begin{aligned} x\cdot_j(y+_jz) &=j(x\cdot j(y+z))\\ &=j(x(y+z))\\ &=j(xy+xz)\\ &=j(j(xy)+j(xz))\\ &=(x\cdot_jy)+_j(x\cdot_jz). \end{aligned}$$
Moreover,
$$x\cdot_j0_j=j(x\cdot j0)=j(x\cdot0)=j0=0_j.$$
Thus the complete commutative semiring structure descends to $S_j$.

4. Descent of an idempotent commutative semiring

Consider now an ordered idempotent commutative semiring
$$(S,\vee,\otimes,\bot,e),$$
where addition is the join:
$$x\vee x=x.$$
The key point is that no nuclearity assumption is needed for $\vee$.

Lemma 4

Every closure operator $j$ satisfies the nucleus condition for finite joins:
$$jx\vee jy\leq j(x\vee y).$$

Proof. Since $x\leq x\vee y$ and $y\leq x\vee y$, monotonicity gives
$$jx\leq j(x\vee y),\qquad jy\leq j(x\vee y).$$
Therefore
$$jx\vee jy\leq j(x\vee y).$$

Thus every closure operator is automatically a nucleus for the join operation.

Let now $j$ be a closure operator which is a nucleus only for $\otimes$. Define
$$x\vee_j y:=j(x\vee y),\qquad x\otimes_jy:=j(x\otimes y),$$
with
$$\bot_j:=j\bot,\qquad e_j:=je.$$

Theorem 4

$(S_j,\vee_j,\otimes_j,\bot_j,e_j)$ is an idempotent commutative semiring.

Proof. By Lemma 4, $j$ is automatically a nucleus for $\vee$, while nuclearity for $\otimes$ is assumed. Hence Theorem 3 applies. It remains only to check idempotence:
$$x\vee_jx=j(x\vee x)=jx=x.$$
Therefore the idempotent commutative semiring structure descends to $S_j$.

The essential principle can be summarized as
$$\text{closure}+\text{nuclearity} \quad\Longrightarrow\quad \text{algebraic descent to fixed points}.$$
For idempotent ordered structures, the join operation is special: its nuclearity is already forced by the closure axioms. Consequently, only the non-order-theoretic operation, such as $\otimes$, requires an additional nuclearity assumption.