distributive lattice

A **distributive lattice** is a lattice in which join $\vee$ and meet $\wedge$ *distribute* over each other, in that for all $x,y,z$ in the lattice, the *distributivity laws* are satisfied:

- $x \vee (y \wedge z) = (x \vee y) \wedge (x \vee z)$,
- $x \wedge (y \vee z) = (x \wedge y) \vee (x \wedge z)$.

The nullary forms of distributivity hold in any lattice:

- $x \vee \top = \top$,
- $x \wedge \bot = \bot$.

Distributive lattices and lattice homomorphisms form a concrete category DistLat.

Any lattice that satisfies one of the two binary distributivity laws must also satisfy the other; isn't that nice? (This may safely be left as an exercise.) This convenience does *not* extend to infinitary distributivity, however.

As mentioned above, the theory of distributive lattices is self-dual, something that is proved in almost any account (or left as an exercise), but which is not manifestly obvious from the standard definition which chooses one of the two distributivity laws and goes from there. In this section we provide some other characterizations or axiomatizations which *are* manifestly self-dual.

Here is one such characterization:

A lattice is distributive if and only if the identity

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

is satisfied.

Again this may be left as a (somewhat mechanical) exercise.

Perhaps more useful in practice is the characterization in terms of “forbidden sublattices” due to Birkhoff. Namely, introduce the “pentagon” $N_5$ as the 5-element lattice $\{\bot \leq a, b, c \leq \top\}$ where $b \leq c$ and $a$ is incomparable to $b, c$, so that $\bot = a \wedge c$ and $a \vee b = \top$. Introduce the “thick diamond” $M_3$ as the 5-element lattice $\{\bot \leq a', b', c' \leq \top\}$ with $a', b', c'$ pairwise incomparable. Both $N_5$ and $M_3$ are self-dual. Birkhoff’s characterization is the following (manifestly self-dual) criterion.

A lattice $L$ is distributive if and only if there is no embedding of $N_5$ or $M_3$ into $L$ that preserves binary meets and binary joins.

This can be useful for determining distributivity or its failure, especially in cases where one can visualize a lattice via its Hasse diagram.

The necessity of the forbidden sublattice condition is clear in view of the fact that the cancellation law stated in the next result fails in $N_5$ and $M_3$. This result gives another self-dual axiomatization of distributive lattices.

A lattice $L$ is distributive if and only if the cancellation law holds: for all $y, z \in L$, we have $y = z$ whenever $x \wedge y = x \wedge z$ and $x \vee y = x \vee z$ (for some $x$).

“Only if”: if $x \wedge y = x \wedge z$ and $x \vee y = x \vee z$, then

$y = y \vee (x \wedge y) = y \vee (x \wedge z) = (y \vee x) \wedge (y \vee z) = (x \vee z) \wedge (y \vee z) = (x \wedge y) \vee z$

which implies $z \leq y$, and similarly we have $y \leq z$.

“If”: this is harder. Assuming the cancellation law for $L$, we first show $L$ is modular. Recall from modular lattice that for any lattice $L$ and $a, b \in L$, there is a covariant Galois connection

$([a \wedge b, b] \to [a, a \vee b]: x \mapsto a \vee x) \; \dashv \; ([a, a \vee b] \to [a \wedge b, b]: y \mapsto b \wedge y)$

and that $L$ is *modular* if this Galois connection is a Galois correspondence (or adjoint equivalence) for all $a, b$. Now, if $x \in [a \wedge b, b]$, then $a \vee x = a \vee (b \wedge (a \vee x))$ because $f = f g f$ for any Galois connection $f \dashv g$. From $a \wedge b \leq x \leq b$ we also have

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

and so by cancellation of the $a$'s, we conclude $x = b \wedge (a \vee x)$. Similarly (dually), for $y \in [a, a \vee b]$, we have $y = a \vee (b \wedge y)$. Hence $L$ is modular.

Now we show $L$ is distributive. Let $x, y, z \in L$ and consider the three elements

$\array{
u & \coloneqq & [x \wedge (y \vee z)] \vee (y \wedge z) & = & [x \vee (y \wedge z)] \wedge (y \vee z) \\
v & \coloneqq & [y \wedge (x \vee z)] \vee (x \wedge z) & = & [y \vee (x \wedge z)] \wedge (x \vee z) \\
w & \coloneqq & [z \wedge (x \vee y)] \vee (x \wedge y) & = & [z \vee (x \wedge y)] \wedge (x \vee y)
}$

where the non-definitional equalities follow from modularity. Using the first expressions, we compute

$\array{
u \vee v & = & [x \wedge (y \vee z)] \vee (y \wedge z) \vee [y \wedge (x \vee z)] \vee (x \wedge z) \\
& = & [x \wedge (y \vee z)] \vee [y \wedge (x \vee z)] \\
& = & [(x \wedge (y \vee z)) \vee y] \wedge (x \vee z) \\
& = & (x \vee y) \wedge (x \vee z) \wedge (y \vee z)
}$

where the third and fourth lines use modularity. By symmetry in the letters $x, y, z$, we also have $u \vee w = v \vee w = (x \vee y) \wedge (x \vee z) \wedge (y \vee z)$. Now the second expressions are dual to the first, so by duality we compute

$u \wedge v = u \wedge w = v \wedge w = (x \wedge y) \vee (x \wedge z) \vee (y \wedge z).$

Now by cancellation of the $u$'s, we may conclude $v = w$, but in that case we obtain

$(x \vee y) \wedge (x \vee z) \wedge (y \vee z) = v \vee w = v \wedge w = (x \wedge y) \vee (x \wedge z) \vee (y \wedge z)$

While the expressions for $u, v, w$ in the preceding proof may look as though they come out of thin air, the underlying idea is that the sublattice of $L$ generated by $x, y, z$ is the image of a lattice map $F(3) \to L$ out of the free modular lattice $F(3)$ on three elements. The only obstruction to distributivity in $F(3)$ is the presence of an $M_3$-sublattice appearing in the center of its Hasse diagram. The middle elements of that sublattice correspond to the formal expressions for $u, v, w$ given above, and the proof shows that under the cancellation law, we have $u = v = w$ in $L$, making the thick diamond collapse to a point in $L$ and removing the obstruction to distributivity.

From Proposition , it is not very hard to deduce Birkhoff’s theorem. The presence of a copy of $M_3$ or $N_5$ in a non-distributive lattice $L$ is deduced from a failure of the cancellation law where we have three elements $x, y, z$ with $x \wedge y = x \wedge z$, $x \vee y = x \vee z$, and $y \neq z$. If $y, z$ are comparable, say $y \leq z$, then the set $\{x \wedge y, x, y, z, x \vee y\}$ forms an $N_5$. If $y, z$ are incomparable, then we have either $y \vee z \lt x \vee y$, or $y \wedge z \gt x \wedge y$, or both $y \vee z = x \vee y$ and $y \wedge z = x \wedge y$; in the first two cases we get an $N_5$ (e.g., $\{x \wedge y, x, y, y \vee z, x \vee y\}$ for the first case), and in the third case the set $\{x \wedge y, x, y, z, x \vee y\}$ forms an $M_3$.

Any Boolean algebra, and even any Heyting algebra, is a distributive lattice.

Any linear order is a distributive lattice.

An integral domain is a Prüfer domain? iff its lattice of ideals is distributive. The classical example is $\mathbb{Z}$; equivalently, the (opposite of the) multiplicative monoid $\mathbb{N}$ ordered by divisibility, with $1$ at the bottom and $0$ at the top.

The lattice of Young diagrams ordered by inclusion is distributive.

A distributive lattice that is complete (not necessarily completely distributive) may be **infinitely distributive** or said to satisfiy the **infinite distributive law** :

$x \wedge (\bigvee_i y_i) = \bigvee_i (x\wedge y_i)$

This property is sufficient to give the lattice Heyting algebra stucture where the implication $a\Rightarrow b$ (or exponential object $b^a$) is:

$(u \Rightarrow v) = \bigvee_{x \wedge u \leq v} x$

Note that this property does not imply the dual **co-infinitely distributive** property:

$x \vee (\bigwedge_i y_i) = \bigwedge_i (x\vee y_i)$

Instead this dual gives the lattice co-Heyting structure where the co-implication or “subtraction” ($\backslash$) is

$(u \backslash v) = \bigwedge_{u \leq v \vee x} x$

If a lattice has both properties, as in a completely distributive lattice, then it has bi-Heyting structure (both Heyting and co-Heyting) and the two exponentials are equal.

$(u \Rightarrow v) = \bigvee_{x \wedge u \leq v} x \qquad = \qquad (u \backslash v) = \bigwedge_{u \leq v \vee x} x$

Since a finite distributive lattice is completely distributive it is a bi-Heyting lattice, as shown above.

Let $FinDistLat$ be the category of finite distributive lattices and lattice homomorphisms, and let $FinPoset$ be the category of finite posets and order-preserving functions. These are contravariantly equivalent, thanks to the presence of an ambimorphic object:

**Proposition.** The opposite category of $FinDistLat$ is equivalent to $FinPoset$:

$FinDistLat^{op} \simeq FinPoset
\,.$

This equivalence is given by the hom-functor

$[-,2] \;\colon\; FinDistLat^{op} \stackrel{\simeq}{\to} FinPoset$

where $2$ is the 2-element distributive lattice, and in the other direction by

$[-,2] \;\colon\; FinPoset^{op} \stackrel{\simeq}{\to} FinDistLat$

where $2 = \{0,1\}$ is the 2-element poset with $0 \lt 1$.

This baby form of *Birkhoff duality* is (in one form or another) mentioned in many places; the formulation in terms of hom-functors may be found in

- Gavin C. Wraith,
*Using the generic interval*, Cah. Top. Géom. Diff. Cat.**XXXIV**4 (1993) pp.259-266. (pdf)

Every distributive lattice, regarded as a category (a (0,1)-category), is a *coherent category*. Conversely, the notion of coherent category may be understood as a categorification of the notion of distributive lattice. A different categorification is the notion of distributive category.

The completely distributive algebraic lattices (the frames of opens of Alexandroff locales ) form a reflective subcategory of that of all distributive lattices. The reflector is called *canonical extension*.

Last revised on October 14, 2016 at 14:44:58. See the history of this page for a list of all contributions to it.