Which lattices have non-trivial linear representations? Suppose we have a a bounded lattice $L$. We might ask: does there exist a non-trivial linear representation of $L$, i.e. a lattice homomorphism $\rho: L \to \text{Sp}(V)$, where $V$ is a non-zero finite dimensional vector space, and $\text{Sp}(V)$ is the lattice of subspaces of $V$, and we require $\rho(0) = \{ 0\}$ and $\rho(1) = V$.
One simple necessary condition is the existence of a function $d: L \to \mathbb{N}$, satisfying:

*

*$d(0) = 0$, and $d(1) = n > 0$

*$d(a) \leq d(b)$ whenever $a \leq b$

*$d(a) + d(b) = d(a\lor b) + d(a\land b)$
Is the existence of such a $d$ sufficient?
I've encountered various related results about faithful representations (i.e. sublattices of $\text{Sp}(V)$). E.g. those lattices need to be modular and arguesian. However, for representations that are not required to be faithful, we might get away with non-modularity for example, by just quotienting out the non-modular part of the lattice (e.g. collapsing the vertical edge in the non-modular pentagon $N_5$). I haven't found any literature on what sort of conditions might remain if we don't require faithfulness. Are there any known results regarding this?
 A: Thanks to David E Speyer for the helpful pointers.
I can now answer the first question (is the existence of the function $d$ sufficient?) in the negative.
A function $d$ with the given properties is an isotone valuation as defined in point 6, Chapter V of Birkhoff's "Lattice theory". As in Theorem 9 in the same chapter, we can take a quotient of the lattice, on which the induced $d$ is strictly increasing (called a metric lattice in Birkhoff), and which is therefore a modular lattice.
So the existence of such a $d$ (modulo some finiteness condition) is equivalent to having a non-trivial modular quotient.
Based on this, we might suspect that the existence of such a $d$ is not sufficient to have a non-trivial linear representation, since sublattices of $\text{Sp}(V)$ have further properties (e.g. being arguesian).
As a concrete counter-example, we can take the incidence lattice $L$ of a finite non-Desarguesian projective plane. Such an $L$ is modular (the function $d$ can be taken to be $d(point) = 1$, $d(line) = 2$), non-arguesean, and as it turns out also simple (i.e. without any smaller non-trivial quotients). It therefore cannot have a non-trivial arguesean quotient.
The fact that such a lattice $L$ is simple follows from Corollary 1 in Chapter V of Birkhoff:

A modular lattice of finite length is ‘‘simple’’ (i.e., without
proper congruence relations) if and only if all its prime quotients are projective.

I believe the fact that there's a line passing through any pair of points is sufficient in this case to show that all prime quotients are projective.
Birkhoff, Garrett, Lattice theory, New York: American Mathematical Society (AMS). v, 155 p. (1940). ZBL0063.00402.
