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In mathematics, in the area of order theory, a free lattice is the free object corresponding to a lattice. As free objects, they have the universal property.

Formal definition

Any set X may be used to generate the free semilattice FX. The free semilattice is defined to consist of all of the finite subsets of X, with the semilattice operation given by ordinary set union. The free semilattice has the universal property. The universal morphism is (FX,η), where η is the unit map η:X→FX which takes x∈X to the singleton set {x}. The universal property is then as follows: given any map f:X→L from X to some arbitrary semilattice L, there exists a unique semilattice homomorphism ${\tilde {f}}:FX\to L$ such that $f={\tilde {f}}\circ \eta$. The map ${\tilde {f}}$ may be explicitly written down; it is given by

$S\in FX\mapsto \bigvee \left\{f(s)\vert s\in S\right\}$

Here, $\bigvee$ denotes the semilattice operation in L. This construction may be promoted from semilattices to latticesTemplate:Clarify; by construction the map ${\tilde {f}}$ will have the same properties as the lattice.

The symbol F is then a functor from the category of sets to the category of lattices and lattice homomorphisms. The functor F is left adjoint to the forgetful functor from lattices to their underlying sets. The free lattice is a free object.

Word problem

Example computation of x∧z ~ x∧z∧(x∨y)

x∧z∧(x∨y) ≤~ x∧z by 5. since x∧z ≤~ x∧z by 1. since x∧z = x∧z

x∧z ≤~ x∧z∧(x∨y) by 7. since x∧z ≤~ x∧z and x∧z ≤~ x∨y by 1. since x∧z = x∧z by 6. since x∧z ≤~ x by 5. since x ≤~ x by 1. since x = x

The word problem for free lattices has some interesting aspects. Consider the case of bounded lattices, i.e. algebraic structures with the two binary operations ∨ and ∧ and the two constants (nullary operations) 0 and 1. The set of all well-formed expressions that can be formulated using these operations on elements from a given set of generators X will be called W(X). This set of words contains many expressions that turn out to denote equal values in every lattice. For example, if a is some element of X, then a∨1 = 1 and a∧1 =a. The word problem for free bounded lattices is the problem of determining which of these elements of W(X) denote the same element in the free bounded lattice FX, and hence in every bounded lattice.

The word problem may be resolved as follows. A relation ≤_~ on W(X) may be defined inductively by setting w ≤_~ v if and only if one of the following holds:

1.   w = v (this can be restricted to the case where w and v are elements of S),
2.   w = 0,
3.   v = 1,
4.   w = w₁ ∨ w₂ and both w₁≤_~v and w₂≤_~v hold,
5.   w = w₁ ∧ w₂ and either w₁≤_~v or w₂≤_~v holds,
6.   v = v₁ ∨ v₂ and either w≤_~v₁ or w≤_~v₂ holds,
7.   v = v₁ ∧ v₂ and both w≤_~v₁ and w≤_~v₂ hold.

This defines a preorder ≤_~ on W(X), so an equivalence relation can be defined by w~v when w≤_~v and v≤_~w. One may then show that the partially ordered quotient space W(X)/~ is the free bounded lattice FX. The equivalence classes of W(X)/~ are the sets of all words w and v with w≤_~v and v≤_~w. Two well-formed words v and w in W(X) denote the same value in every bounded lattice if and only if w≤_~v and v≤_~w; the latter conditions can be effectively decided using the above inductive definition. The table shows an example computation to show that the words x∧z and x∧z∧(x∨y) denote the same value in every bounded lattice. The case of lattices that are not bounded is treated similarly, omitting rules 2. and 3. in the above construction.

The solution of the word problem on free lattices has several interesting corollaries. One is that the free lattice of a three-element set of generators is infinite. In fact, one can even show that every free lattice on three generators contains a sublattice which is free for a set of four generators. By induction, this eventually yields a sublattice free on countably many generators. This property is reminiscent of SQ-universality in groups.

The proof that the free lattice in three generators is infinite proceeds by inductively defining

p_n+1 = x ∨ (y ∧ (z ∨ (x ∧ (y ∨ (z ∧ p_n)))))

where x, y, and z are the three generators, and p₀=x. One then shows, using the inductive relations of the word problem, that p_n+1 is strictly greater than p_n, and therefore all infinitely many words p_n evaluate to different values in the free lattice FX.

The complete free lattice

Another corollary is that the complete free lattice "does not exist", in the sense that it is instead a proper class. The proof of this follows from the word problem as well. To define a complete lattice in terms of relations, it does not suffice to use the finitary relations of meet and join; one must also have infinitary relations defining the meet and join of infinite subsets. For example, the infinitary relation corresponding to "join" may be defined as

$\operatorname {sup} _{N}:(f:N\to FX)$

Here, f is a map from the elements of a cardinal N to FX; the operator $\operatorname {sup} _{N}$ denotes the supremum, in that it takes the image of f to its join. This is, of course, identical to "join" when N is a finite number; the point of this definition is to define join as a relation, even when N is an infinite cardinal.

The axioms of the pre-ordering of the word problem may be adjoined by the two infinitary operators corresponding to meet and join. After doing so, one then extends the definition of $p_{n}$ to an ordinally indexed $p_{\alpha }$ given by

$p_{\alpha }=\operatorname {sup} \{p_{\beta }\vert \beta <\alpha \}$

when $\alpha$ is a limit ordinal. Then, as before, one may show that $p_{\alpha +1}$ is strictly greater than $p_{\alpha }$. Thus, there are at least as many elements in the complete free lattice as there are ordinals, and thus, the complete free lattice cannot exist as a set, and must therefore be a proper class.

References

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• Peter T. Johnstone, Stone Spaces, Cambridge Studies in Advanced Mathematics 3, Cambridge University Press, Cambridge, 1982. (ISBN 0-521-23893-5) (See chapter 1)