# Geometric interpretation of distributivity

I'm studying distributivity in the lattice theory, consequently, I'm after any ideas that might help to develop some intuition.

In elementary school level algebra, distributivity property

(a+b) c = ac + bc


is pictured geometrically as the two adjacent rectangles: a x c and b x c. (Geometric interpretation follows from linearity of inner product space, I guess).

Next, for Boolean algebras (which are lattices) there are the two interpretations: Venn diagram, and cube.

On Venn diagram we draw union of sets a and b, then intersect the result with c. Then, we compare it with a intersect with c unioned with b intersect with c. One can convince yourself that the result is the same, although it is less intuitive than in the previous case of adjacent rectangles.

When I do the same "operational" drawing on (4 dimensional) Boolean cube image, the result is again the same, but the procedure is even less convincing. Why (a v b) ^ c is the same as (a ^ c) v (b ^ c), is it due to some properties of the geometric shape of the cube?

In a similar fashion I drew couple of cases on graph paper (representing N^2 lattice). Again, why (a v b) ^ c is the same as (a ^ c) v (b ^ c) is not obvious at all.

Both Boolean cube and N^2 lattice are Cartesian products of 1-dimensional lattices which are total orders. IIRC, distributivity carries over to Cartesian products, so one just have to analyze 1-dimensional picture only. So we have to see that

min(max(a,b),c) = max(min(c,a),min(c,b))


holds in every of the following cases:

a < b < c
a < c < b
c < a < b


(where transpositions of a and b are ignored because of commutativity)

Again, this case analysis is not exactly geometrically evident interpretation...

Group theory promoted non-commutativity property into a whole new topic; by analogy, had anybody try to introduce "distributor", that is [a,b,c] (this is reminiscent of standard group commutator notation) such that

(a v b) ^ c  = (a ^ c) v (b ^ c) v [a,b,c]


or there is a reason why this wouldn't work?

• You have interchanged some maxs and mins and it is a bit hard to pick out the questions from your text. – user11235 Apr 21 '11 at 20:02
• The identity with min and max was inconsistent -- fixed, thank you. – Tegiri Nenashi Apr 21 '11 at 21:42

## 2 Answers

There is some interesting "geometry" lurking within the study of Boolean algebras: as realized by Marshall Stone, every Boolean algebra can be realized as an algebra of sets (that is, as a subalgebra of a power set); more interestingly, this subalgebra consists of clopen (closed and open) sets of a topological space often called the spectrum of the Boolean algebra, and this is where "geometry" comes in. If we view the concept of Boolean algebra as essentially equivalent to that of Boolean ring, then the construction is a special case of the spectrum of a commutative ring, much studied by algebraic geometers.

One way of "explaining" the distributivity law of a power set $P(X)$ (which would automatically be inherited by sublattices of $P(X)$) starts by associating to each subset $A \subseteq X$ its characteristic function $\chi_A: X \to \{0, 1\}$, defined by $\chi_A(x) = 1$ iff $x \in A$. The power set is thereby isomorphic (as Boolean algebra) to the set of functions $f: X \to \{0, 1\}$. The lattice operations on the latter are defined pointwise, i.e., $(f \wedge g) = f(x) \wedge g(x)$ and $(f \vee g)(x) = f(x) \vee g(x)$, so it suffices to check that distributivity holds in the lattice $\{0, 1\}$.

This is somewhat trivial; if you like, you can just think of $\{0, 1\}$ as a Boolean ring instead (which you might like since you are comfortable with distributivity in ring theory). Multiplication in a Boolean ring (a commutative ring in which every element is idempotent) coincides with the meet operation; addition corresponds to symmetric difference, and the join operation is defined in a Boolean ring by

$$x \vee y = x + y + xy$$

and I invite you to prove that distributivity in the lattice-theoretic sense follows from distributivity in the ring sense plus idempotence, and this involves no case-splitting.

I am not sure what your question is concerning "distributor".

You ask "Why $(a \vee b) \wedge c$ is the same as $(a \wedge c) \vee (b \wedge c)$, is it due to some properties of the geometric shape of the cube?" It's because there are no $N_5$ or $M_3$ sublattices in the cube, or in any distributive lattice for that matter. (Remember, a necessary and sufficient condition for a lattice to be distributive is that it contain no sublattices isomorphic to $N_5$ or $M_3$. Looking at a Hasse diagram and seeing why $N_5$ and $M_3$ break distributivity seems fairly geometric to me.)

In response to your second question, commutators have been defined on congruence lattices of general algebras and extensively studied. They behave nicely when such lattices are modular (See Freese & McKenzie).

I initially didn't think your notion of commutator (or "distributor") would get you very much, but I'm having second thoughts... as you know, in any lattice we have $(a\vee b) \wedge c \geq (a\wedge c) \vee (b\wedge c)$, so perhaps you could define your term [a,b,c] to be the largest element such that $[a,b,c] \wedge (a\vee b) \wedge c \leq (a\wedge c) \vee (b\wedge c)$. Note, however, it's possible that there will be more than one maximal element with this property, in which case [a,b,c] would not be well-defined. It might be interesting to determine the class of lattices in which this term is well-defined; cf. residuated lattices (in which given x and y there is unique largest z such that $x\wedge z \leq y$).

Since you're "after any ideas that might help to develop some intuition," it might also help to think about the "cancellativity" property: a lattice is distributive iff for all $a, b, c$, the equalities $a\wedge c = b\wedge c$ and $a\vee c = b\vee c$ together imply $a=b$ (cf. cancellativity in integral domains). Also, in distributive lattices, an element is meet (join) prime iff it is meet (join) irreducible (cf. UFD's).

• Previously I commented that, although the commutator is defined on congruence lattices, this is not much of a limitation, since every algebraic lattice is the congruence lattice of some algebra (Gratzer-Schmidt 1963). However, this is wrong since the definition of commutator requires that we know the algebra explicitly. – William DeMeo Apr 26 '11 at 23:37