# nLab modular lattice

This is about a notion in order theory/logic. For an unrelated notion of a similar name in group theory/quadratic form-theory see at modular integral lattice.

(0,1)-category

(0,1)-topos

# Contents

## Idea

A modular lattice is a lattice where “opposite sides” of a “diamond” formed by four points $x \wedge y$, $x$, $y$, $x \vee y$ are “congruent”.

## Definition

A modular lattice is a lattice which satisfies a modular law, which we introduce after a few preliminaries.

In any lattice $L$, given two elements $x, y \in L$ with $x \leq y$, let $[x, y]$ denote the interval $\{z \colon x \leq z \leq y\}$. Then, given any two elements $a, b \in L$, there is an adjoint pair

$a \vee (-) \colon [a \wedge b, b] \stackrel{\leftarrow}{\to} [a, a \vee b] \colon (-) \wedge b$

where $a \vee (-)$ is left adjoint to $(-) \wedge b$. Indeed, for any $w \in [a \wedge b, b]$, we have a unit

$w \leq (a \vee w) \wedge b,$

whereas for any $z \in [a, a \vee b]$, we have dually a counit

$a \vee (z \wedge b) \leq z.$
###### Definition

A lattice $L$ is modular if for any $a, b \in L$, the adjoint pair

$a \vee (-) \dashv (-) \wedge b \colon [a, a \vee b] \to [a \wedge b, b]$

This is perhaps the most memorable definition for a category theorist: it is a precise expression of the slogan given in the Idea section.

It is immediate that the concept of modular lattice is self-dual, i.e., if $L$ is modular, then so is $L^{op}$.

## Alternative formulations

In the lattice-theoretic literature, modularity is usually formulated somewhat differently. Here are three alternative conditions on a lattice, all equivalent to that of Definition 1.

1. The modular law is the universal Horn sentence

$a \leq b \vdash (a \vee z) \wedge b = a \vee (z \wedge b).$
2. The modular identity is the universal equation

$(a \vee z) \wedge (a \vee b) = a \vee (z \wedge (a \vee b))$
3. Freyd’s modular law” (for lack of better term; see allegory) is the universal inequality

$(a \vee z) \wedge b \leq a \vee (z \wedge (a \vee b)).$

### Proofs of equivalence

#### Derivation of modular identity

To see that the modular identity follows from Definition 1, observe that for any $z \in L$ we have

$a \leq (a \vee z) \wedge (a \vee b) \leq a \vee b$

Let $w = (a \vee z) \wedge (a \vee b)$. Under $(-) \wedge b \colon [a, a \vee b] \to [a \wedge b, b]$, this element $w$ is sent to

$(a \vee z) \wedge (a \vee b) \wedge b = (a \vee z) \wedge b.$

Under Definition 1, this last element is sent back to $w$ by $a \vee (-)$. Therefore we have

$(a \vee z) \wedge (a \vee b) = w = a \vee ((a \vee z) \wedge b)$

and since this is true for all $a, b, z$, we can interchange $z$ and $b$ and rearrange by commutativity to get

$(a \vee z) \wedge (a \wedge b) = a \vee (z \wedge (a \vee b))$

which is the modular identity.

#### Modular law $\Leftrightarrow$ modular identity

To get the modular law from the modular identity, just use the fact that the hypothesis $a \leq b$ is equivalent to $a \vee b = b$, and use this to substitute $b$ for $a \vee b$ in the modular identity. Conversely, from the tautology $a \leq a \vee b$, we can instantiate the modular law to derive the modular identity.

#### Freyd’s modular law $\Leftrightarrow$ modular identity

From the tautology $(a \vee z) \wedge b \leq (a \vee z) \wedge (a \vee b)$, it is clear that Freyd’s modular law follows from the modular identity. Conversely, by substituting $a \vee b$ for $b$ in Freyd’s modular law, we derive the special case

$(a \vee z) \wedge (a \vee b) \leq a \vee (z \wedge (a \vee b))$

whereas the opposite inequality

$a \vee (z \wedge (a \vee b)) \leq (a \vee z) \wedge (a \vee b)$

holds in any lattice, so the modular identity follows from Freyd’s modular law.

#### Modular identity $\Rightarrow$ definition 1

Finally, we derive the adjoint equivalence of Definition 1 from the modular identity. One half of the adjoint equivalence states that if $a \leq z \leq a \vee b$, then $z = a \vee (z \wedge b)$; if this holds, then the other half follows because it is the dual statement. If $a \leq z \leq a \vee b$, then

$z = (a \vee b) \wedge z = (a \vee b) \wedge (a \vee z)$

just by the laws of a lattice. By the modular identity (again switching $b$ and $z$), the right side equals $a \vee (b \wedge (a \vee z))$. But since $a \vee z = z$, this equals $a \vee (b \wedge z) = a \vee (z \wedge b)$, as was to be shown.

## Examples

• Every distributive lattice, e.g., a Heyting algebra, is modular. Indeed, if $a \leq b$ in a distributive lattice, we have

$(a \vee z) \wedge b = (a \wedge b) \vee (z \wedge b) = a \vee (z \wedge b)$

which proves the modular law.

• For any Mal'cev variety or Mal’cev algebraic theory, the lattice of internal equivalence relations of an algebra is a modular lattice. The equivalence classes often arise as cosets of kernels; for example, for a vector space $V$, equivalence relations correspond to subspaces of $V$, and form a modular lattice. Other examples include the lattice of normal subgroups of a group, the lattice of two-sided ideals of a ring, etc.

• In fact, any lattice of commuting equivalence relations on a set is a modular lattice (being a suballegory of the allegory of sets, one in which composition provides the join).

• Every abstract projective plane gives rise to a modular lattice $L$ whose underlying set is the disjoint union

$\{0\} \cup \{1\} \cup \{points\} \cup \{lines\}$

where $0$ is taken as bottom, $1$ as top, the points are atoms, and the lines are coatoms, ordered by the incidence relation. The projective plane need not be Desarguesian.

• Young–Fibonacci lattice

## Characterization

The smallest non-modular lattice has 5 elements and is called the pentagon, denoted $N_5$. It can be described as the lattice $\{\bot, a, b, c, \top\}$ where $b \leq c$ and $a$ is incomparable with $b$ and $c$.

###### Theorem

(Dedekind) A lattice $L$ is modular if and only if there is no injective function $f \colon N_5 \to L$ that preserves meets and joins.

(Notice we are leaving out the condition of preservation of the top and bottom elements.) The direction $\Rightarrow$ is easy enough: $L$ being modular is incompatible with an injective $f: N_5 \to L$ preserving meets and joins, since we contradict the modular law in $L$ by applying $f$ to $(b \vee a) \wedge c = \top \wedge c = c$ and $b \vee (a \wedge c) = b \vee \bot = b$.

This is reminiscent of forbidden minor characterizations of certain classes of graphs; see graph minor. There is a similar “forbidden sublattice” characterization of distributive lattices – see this comment by Tom Leinster at the $n$-Category Café.

## Free modular lattices

Free modular lattices tend to be complicated. Dedekind showed that the free modular lattice on 3 elements has 28 elements; its Hasse diagram can be seen in these lecture notes by J.B. Nation (chapter 9, page 100).

N.B.: this notion of lattice is meant with respect to the signature $(\wedge, \vee)$; if we include top and bottom constants in the signature, then the free modular lattice on three elements has 30 elements. A compelling illustration (in gif form) which exhibits triality of this lattice is given in this $n$-Category Café post, as part of a larger discussion which explores the connection with linear representations of the quiver $D_4$ (the Coxeter diagram of $SO(8)$).

For $n \geq 4$, the free modular lattice generated by $n$ elements is infinite and in fact has an undecidable word problem (Freese, Herrmann).