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Note: This is a in the flavour of, say, "what does $X$ look like?" and "Is there a description of $Y$?" - so, hopefully, it is not too broad. Let's focus on the basics if necessary.

The Details:

Definition: A (meet/join) semilattice is a commutative, idempotent semigroup.

I prefer join semilattices (but they are isomorphic to meet semilattices.)

Definition (Green's Relations): Let $S$ be a semigroup and $S^1$ be $S$ with an one adjoined. Then

$$\begin{align} a\mathscr{L}b&\iff S^1a=S^1b,\\ a\mathscr{R}b&\iff aS^1=bS^1,\\ a\mathscr{J}b&\iff S^1aS^1=S^1bS^1,\\ a\mathscr{H}b&\iff (a\mathscr{L}b\land a\mathscr{R}b),\\ a\mathscr{D}b&\iff a(\mathscr{L}\circ\mathscr{R})b\\ &\iff a(\mathscr{R}\circ\mathscr{L})b, \end{align}$$

where $x(\rho\circ\sigma)y$ iff there is a $c$ such that $x\rho c$ and $c\sigma y$, where $\rho$ and $\sigma$ are equivalence relations.

The Question:

What can we say in general about Green's relations of a semilattice $L$?

Thoughts:

Due to commutativity, $\mathscr{L}=\mathscr{R}$; this implies $\mathscr{R}=\mathscr{H}$.

According to this (pdf), $\mathscr{J}$ is trivial.

Context:

I'm gathering relevant theorems for an idea I have for semilattices. I can't share what that idea is but I can tell you I'm looking at $XL$ for a certain type of $X\subseteq L$, where $L$ is a semilattice of a technical nature.

Shaun
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    $\mathscr{L} = \mathscr{R} = \mathscr{H} = \mathscr{J} = \mathscr{D}$ from commutativity – Jakobian Jan 17 '24 at 16:15
  • Thank you, @Jakobian. I'd accept that as an answer if you'd like. – Shaun Jan 17 '24 at 16:19
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    Another way to see that they are trivial is that maximal subgroups of a semilattice are one-point sets ${x}$, and equivalence class $[x]_\mathscr{H}$ is the maximal subgroup containing $x$, since $x$ is idempotent – Jakobian Jan 17 '24 at 16:20

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Since semilattices are commutative, $aS^1 = S^1a = S^1aS^1$. The last equality can be seen from $S^1aS^1 = a(S^1)^2 = aS^1$ since $S^1$ is a monoid.

Thus all the Green relations for semilattices are equal: $$\mathscr{L} = \mathscr{R} = \mathscr{J} = \mathscr{H} = \mathscr{D}.$$

Another easy way to see all of them are trivial, is that equivalence classes $[e]_\mathscr{H} = \{x : e\mathscr{H} x\}$ for an idempotent $e$ are precisely the maximal subgroups of $S$ with $e$ as the identity element (not to be confused with maximal subgroups in group theory, see wikipedia). For two distinct idempotents $e, f\in S$ those equivalence classes don't overlap (as a group cannot have two distinct identity elements), but all elements in a semilattice are idempotent. Hence all the Green relations are trivial.

Jakobian
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