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?

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    $\begingroup$ How unfaithful are you ready to be? Is, for example, something like $L\twoheadrightarrow\{0<1\}\hookrightarrow\operatorname{Sp}(V)$ OK? $\endgroup$ Dec 7, 2022 at 3:46
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    $\begingroup$ It would be OK, but having a lattice homomorphism onto $\{0 < 1\}$ seems quite far from being a necessary condition, even though it's sufficient. $\endgroup$
    – Yegreg
    Dec 7, 2022 at 4:29
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    $\begingroup$ I wonder whether every lattice $L$ has a unique modular quotient $L^{\text{mod}}$ through which any map to a modular lattice factors? $\endgroup$ Dec 7, 2022 at 17:38
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    $\begingroup$ According to mathoverflow.net/questions/55515/… , the answer is yes. So we are reduced to answering your question for $L^{\text{mod}}$. $\endgroup$ Dec 7, 2022 at 19:25
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    $\begingroup$ A good test case would be to ask if a non-Desarguean projective plane has such a representation. en.wikipedia.org/wiki/Non-Desarguesian_plane $\endgroup$ Dec 7, 2022 at 20:15

1 Answer 1


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.


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