By general results for every set $X$ there is a free bounded lattice $L(X)$ on $X$. I would like to understand the element structure of this lattice. The cases $X=\emptyset$, $X=\{x\}$ and $X=\{x,y\}$ are quite easy. But for $X=\{x,y,z\}$ we get an infinite lattice. But what are the elements explicitly? Is there any normal form available? Or, is there any natural representation of $L(X)$? Compare this to the free group on two generators, which might be quite abstract, but it can be explicitly realized as a certain subgroup of $\mathrm{SL}(2,\mathbb{Z})$, generated by two matrices (see Ping-pong lemma). So is there a natural and non-abstract example of a bounded lattice which contains the free bounded lattice on three generators? Notice that the recursive description at Wikipedia doesn't answer these questions.


Solution 1:

First question:
But what are the elements explicitly? Is there any normal form available?

Yes.
Let $X$ be the set of generators of a free lattice. The elements of the free lattice over $X$ are the terms in the language modulo the lattice identities. Up to commutativity and associativity, such terms can be expressed in these forms:

$x_1, x_2, \ldots$ (generators=elements of $X$=rank 1 terms)

$s = (s_1\vee\cdots\vee s_k)$ (finite joins of lower complexity terms)

$t = (t_1\wedge\cdots\wedge t_{\ell})$ (finite meets of lower complexity terms)

Each such term has a rank, which is just its length. (That is, it is the number of characters needed to write it down, including parentheses.) Within a finite join or meet, I use only outer parentheses.

Whitman proved that each element of the free lattice over $X$ is representable by a minimal rank lattice term, which is unique up to commutativity and associativity. This is called its canonical form. You can decide if a term is written in canonical form using this theorem:

Thm.

(1) Any element of $X$ is in canonical form.

(2) A join $s = (s_1\vee\cdots\vee s_k)$ is in canonical form iff
(2.1) each $s_i$ is in canonical form,
(2.2) each $s_i$ is either a generator (= an element of $X$), or is a formal meet,
(2.3) the $s_i$'s are pairwise incomparable, and
(2.4) if $s_i = \bigwedge s_{i_j}$, then $s_{i_j}\not\leq t$ for any $j$.

(3) A meet $t = (t_1\wedge\cdots\wedge t_k)$ is in canonical form iff the duals of (2.1)-(2.4) hold.

This is Whitman's result for the unbounded case. If you want the free lattice in the variety of bounded lattices (with bounds named by constants), just add an extra top element $1$ and bottom element $0$ to the above description.

Second question:
Or, is there any natural representation of L(X)?

I don't know a satisfying answer off the top of my head. But every lattice is embeddable in a lattice of equivalence relations on some set. It follows that the free lattice over $X$ is isomorphic to a sublattice of a lattice $\textrm{Eq}(Z)$, of equivalence relations on some set $Z$, generated by $|X|$-many "sufficiently independent" equivalence relations. Perhaps it would not be too hard to determine what "sufficiently independent" means.