lowering of a semilattice

Question

In these Lecture Notes the notion of lowering a semilattice is introduced, there it is stated:

Sometimes "broken" elements need to be looked at and computed with. Now any semilattice can have an "extra" 1 joined "at the top" of the partial ordering. [...] we make the adjunction by "lowering" a copy of the whole semilattice with the aid of a new unary operation.

AXIOM II: Lowering

p* < 1

0* = 0

(p v q)* = p* v q*

p* = q* => p = q

So as I understand it, simply a top element is added to the semilattice, but at the beginning of the lecture notes a semilattice is already defined as a system with a top element $\langle S, 0, 1, \le, \lor \rangle$, so this lowering operation make no sense to me. Does it correspond to something like in my picture:

This make not much sense to me. The exercises ask then for "the simplest example of a semilattice that has a lowering operation", for me this would be $2^* = \{ 0, 1, 1^*\}$, i.e. the semilattice $2 = \{0,1\}$ adjoined with a new top element. Is this right?

Answer 1

It’s not very clear, but I think that he means that a semilattice $L$ has a lowering operation $*$ if there is a map $*:L\to L:p\mapsto p^*$ such that

• $p^*<1$ for all $p\in L$,
• $0^*=0$,
• $(p\lor q)^*=p^*\lor q^*$ for all $p,q\in L$, and
• if $p^*=q^*$, then $p=q$.

This amounts to saying that $*$ is an injective homomorphism from $L$ to $L\setminus\{1\}$ that fixes $0$. It follows that if $L$ has a lowering operation, $L$ must be infinite. The simplest example would appear to be $$L=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$$ with the usual ordering, so that $x\lor y=\max\{x,y\}$; the map

$$*:L\to L:x\mapsto\begin{cases} 0,&\text{if }x=0\\\\ \frac1{n+1},&\text{if }x=\frac1n \end{cases}$$

is a lowering operation.

This operation in effect leaves the original $1$ as an ‘extra’ element above a copy $L^*$ of the original lattice $L$: the effect is obtained not by adding an extra $1$ above $L$, but by pushing $L$ down inside itself. In that too-brief discussion he’s contrasting these two ways of getting an extra element at the top: you can simply add one, or, if your lattice allows, you can push the lattice down off its original top element.