What is the mental model for (modular) lattices?

Question

When we do computations* in abstract commutative rings, we use a mental model all the time: the arithmetic of integers. After the foundations are developed, we don't really care about the specific ring axioms, especially when handling complex expressions. For abstract boolean algebras, we think of propositional logic and truth values. For fields, the mental model will usually be the rational or real numbers. For abelian categories, we usually think of the category of modules over a ring. For partial orders, already the choice of the symbol $\leq$ indicates that we usually think of the common less-than-or-equal relation of numbers.

Now what about lattices? Which mental model do most professionals** use for computations in abstract lattices?

More formally perhaps: What is a "small" class of "concrete" lattices ("models") such that an equation in the language of lattices holds in all models if and only if it holds in all lattices? The word "concrete" shall exclucde free lattices, which otherwise would provide a trivial answer.

For the purpose of this question, all lattices will be bounded (have smallest and largest element). I ask this because even though I can do computations in lattices, right now they are a bit "mechanic". Every step has to be done by hand, using the lattice axioms. This is much more tedious than computations in, say, rings, where I can just rely on my experience with integers.

For distributive lattices we can always use the mental model of sets with $\vee$ = union, $\wedge$ = intersection. Formally, this is justified by Birkhoff's representation theorem. (This theorem holds for every finite distributive lattice, but every valid expression lies in a finite distributive sublattice, so that is sufficient.)

Unfortunately, since not every lattice is distributive, this means that sets with $\vee$ = union, $\wedge$ = intersection are not an adequate mental model for lattices. They are not sufficient to distinguish the expressions $x \wedge (y \vee z)$ and $(x \wedge y) \vee (x \wedge z)$.

The prime example of a modular lattice is the lattice of submodules of a module, and I believe this was also the reason for the terminology. However, from the mathoverflow question Representation theorem for modular lattices (a question I completely forgot!) the lattice of submodules satisfies further identities, such as the Arguesian law. This indicates that submodules of a module are probably not a good mental model for modular lattices.

I know that a lattice is modular (resp., distributive) iff it does not contain a copy of $N_5$ (resp., $M_3$ or $N_5$), but I am afraid that this doesn't help with computations.

I saw here and in some papers that the notation $ab = a \wedge b$ and $a+b = a \vee b$ is used. However, this means that I need to become comfortable with identities such as $a + a = a$ and $a + ab = a$, which is a bit weird since I have to "override" my mental model for the rules of $+$ and $\cdot$.

*With computations, I mean specifcally mean deductions in equational logic. So, even though the ring of integers has all sorts of special properties (principal ideal domain, characteristic $0$, ...), the ring-theoretic equations satisfied by $\mathbb{Z}$ are only those satisfied in every commutative ring. This is why $\mathbb{Z}$ is an adequate model for general commutative rings. Formally: if $p,q \in \mathbb{Z}[T_1,\dotsc,T_n]$ are two polynomials such that $p(z_1,\dotsc,z_n) = q(z_1,\dotsc,z_n)$ for all $z_1,\dotsc,z_n \in \mathbb{Z}$, then $p = q$ as polynomials.

**I have formulated it this way to prevent this question from being opinion-based and then perhaps getting closed. While I am looking forward to hearing about your personal preferences for mental models in the comments, I am looking for answers preferably from mathematicians who have professionally worked with lattices before and therefore know what is "standard" in the field.

Answer 1

Here's a partial answer. Every algebraic lattice is isomorphic to the congruence lattice of some algebra (Gratzer/Schmidt). In particular, every finite lattice is isomorphic to the congruence lattice of some algebra, and if memory serves the equational theory of algebraic lattices is no larger than the equational theory of lattices itself. So at least in a wide variety of circumstances, we can get away with thinking of a lattice as a lattice of congruences.

Answer 2

The quantum logic program (see the wiki page) aimed at proving that any orthomodular orthocomplemented lattice was a sublattice of the lattice of closed subspaces in a (generalized) Hilbert space, but unfortunately, there seems to be no conclusive proof.

Answer 3

For modular lattices this is too much to ask, since the word problem is unsolvable even for free modular lattices. Though you can get pretty far if you just think about modules. It would be hard to stumble across non-Desarguesian planes if you didn't know about them already.

For general lattices, Noah's suggestion is a good one, and universal algebraicists contribute to general lattice theory pretty regularly. If you don't want to picture a whole algebra, every lattice can be represented as a lattice of equivalence relations that are 4-permutable. Modular lattices can be represented as 3-permutable equivalence relations. (You can find the precise result here: https://math.hawaii.edu/~jb/math618/os4uh.pdf)

Another formulation is that can also embed every lattice in the lattice of closed sets for some closure operator. (The algebraic lattice result can be thought as a strengthening of this to imposing a restriction on the type of closure operator.)

The word problem for free lattices is solvable, so you could also look at that directly. Freese, Jezek and Nation have a book on free lattices that give the result.