For neutral elements in a lattice, the definition is typically stated as follows:
An element a of a lattice L is neutral, iff every triple {a, x, y} generates a distributive sublattice of L.
Which now leads to my question: What is this "distributive sublattice generated by {a, x, y}" and how do you compute it? I have not been able to find a proper explanation on this.
I am familiar with the concept of the free lattice on $n$ generators. From my understanding, you can iteratively generate it by starting with the generators, computing the meet and join pairwise (including each generator with itself), then doing the same with the results, and so on. This yields an infinite lattice on 3 or more generators.
I am also familiar with the free distributive lattice on 3 generators: https://www.researchgate.net/figure/The-free-distributive-lattice-on-three-generators-Free-D-3_fig10_299594715
So the way I infer the "sublattice generated by {a, x, y}" works, is that you generate it like the free lattice, but you disregard duplicates? As in, for each $x \in L$, you only generate the lowest rank expression that yields it. This would make it so that the generated lattice is finite and a sublattice of $L$. I have not been able to find confirmation on this though, so my interpretation may be entirely incorrect.
Assuming this interpretation is correct, am I correct in the assumption, that the free distributive lattice is the largest potential sublattice generated by {a, x, y} that is also distributive? By extension all distributive lattices generated by {a, x, y} would have to be a sublattice of the free distributive lattice.
No, that's not the right interpretation.
In a loose way, one could even say it's quite the opposite, since the sub-lattice of $L$ generated by $\{a,x,y\}$ is the smallest sub-lattice of $L$ which contains $a,x,y$, in the sense that if $K \leq L$ and $a,x,y \in L$, then the sub-lattice of $L$ generated by $\{a,x,y\}$ is a sub-lattice of $K$ too.
On the other hand, you can think of the free lattice generate by a set of elements as the largest lattice which is generated by those elements.
The difference is that in the free lattice generated by $\{a,x,y\}$, it is assumed that there is no non-trivial equational relation between these elements. In particular $u \nleq v$ for each $u,v$ in the set, and also $u \wedge v \neq w$ and $u \vee v \neq w$ for any $u,v,w$ in the set.
Now it may happen that $u \leq v$ in $L$ or $u \wedge v = w$ or $u \vee v = w$ (or other non-trivial equations), and so that must also be the case in sub-lattice of $L$ generated by those elements.
Now how do you compute the sub-lattice generated by $\{a,x,y\}$?
If you are acquainted with other algebraic structures (such as groups), you can notice that it is the same here.
Start with the set $X_0=\{a,x,y\}$, and compute, given $X_i$, $$X_{i+1} = \{ u \wedge v, u \vee v : u,v \in X_i \},$$ where the elements $u\wedge v$ and $u \vee v$ are computed in $L$.
When $X_i = X_{i+1}$, then $X_i$ is the sub-lattice of $L$ generated by $X_0 = \{a,x,y\}$.
It turns out to be much simpler than it looks.