Boolean Semilattices

Clifford Bergman

Iowa State University

June 2015 Motivating Construction

Let G = hG, ·i be a groupoid (i.e., 1 binary operation) Form the complex algebra

G+ = hSb(G), ∩, ∪, ∼, ·, ∅, Gi

X · Y = { x · y : x ∈ X, y ∈ Y } “complex operation”

G+ is an expansion of a Boolean algebra X · ∅ = ∅ · X = ∅ (normality) X · (Y ∪ Z) = (X · Y ) ∪ (X · Z ) and (Y ∪ Z ) · X = (Y · X) ∪ (Z · X) (additivity) In fact, complete and atomic, and completely additive Boolean Groupoids

Deﬁnition A Boolean groupoid is an algebra B = hB, ∧, ∨, 0, ·, 0, 1i such that hB, ∧, ∨, 0, 0, 1i is a Boolean algebra x · 0 ≈ 0 · x ≈ 0 x · (y ∨ z) ≈ (x · y) ∨ (x · z) (y ∨ z) · x ≈ (y · x) ∨ (z · x).

BG = the variety of Boolean groupoids Crash Course in Universal Algebra

Algebra Structure with no relational parameters Variety Class of algebras closed under homomorphic image, subalgebra, and product Birkhoff’s Theorem K is a variety iff it is axiomatized by a set of equations Equation Universal, atomic sentence Variety Generated by K V(K) is the smallest variety containing K. V(K) = Mod(Eq(K)) Boolean Groupoids

BG = the variety of Boolean groupoids G a groupoid =⇒ G+ ∈ BG + Converse? Nope. |G |= 6 ℵ0

Is BG generated by complex algebras? Yes! Can retrieve G from G+:

a · b = c in G ⇐⇒ {a}·{b} = {c} in G+ Note that {a}, {b}, {c} are atoms of G+

Not all Boolean groupoids have atoms The product of two atoms might not be an atom

We need a more general approach Deﬁnition Let B be a Boolean groupoid.

B+ = hA, θi in which A is the set of atoms of B and θ = (x, y, z) ∈ A3 : z ≤ x · y

(a ternary relational structure)

For a groupoid G, let θ = { (x, y, x · y) }. + Then hG, θi = (G )+ Deﬁnition Let H = hH, ψi be a ternary relational structure H+ = hSb(H), ∩, ∪, ∼, ·, ∅, Hi in which X · Y = { z ∈ H :(∃x ∈ X)(∃y ∈ Y )(x, y, z) ∈ ψ }.

+ ∼ H a ternary relational structure =⇒ (H )+ = H B a complete and atomic Boolean algebra + ∼ =⇒ (B+) = B

There is a duality between the categories of ternary relational structures and complete, atomic Boolean algebras Canonical Extensions

Theorem (Jonsson-Tarski,´ 1951) Let B be a Boolean groupoid. There is a complete, atomic Boolean groupoid Bσ extending B such that ^ p · q = { a · b : a, b ∈ B, a ≥ p, b ≥ q }

for atoms p, q of Bσ.

σ Idea: (B )+ can serve as an approximation to a groupoid induced by B. Lemma (Jipsen) Every Boolean groupoid can be embedded into P+ for some partial groupoid P.

σ (B )+ P G

B Bσ P+ G+ Theorem Every Boolean groupoid lies in SH(G+) for some groupoid G.

Corollary BG = V { G+ : G a groupoid } V { G+ : G a groupoid } is axiomatized by the identities that “have to be true”

Will this hold for other classes of groupoids? Properties Reﬂected in the Complex Algebra

normality additivity ∴ monotonicity x ≤ y → x · z ≤ y · z p ≈ q is linear if each variable appears exactly once in p and in q + G p ≈ q ⇐⇒ G p ≈ q x · (y · z) ≈ (x · y) · z, x · y ≈ y · x, x · (x · y) ≈ y p ≈ q is semilinear if p has no repeated variables and each variable of q occurs in p + G p ≈ q ⇐⇒ G p ≤ q x ≈ x 2, x · y ≈ x Some Generic Questions

Let V be a variety of Boolean groupoids, K a class of groupoids or TRS. K+ = { G+ : G ∈ K }

Is V = V(K+)? Is V = SP(K+)? (“V is representable by K”) Is V(K+) ﬁnitely axiomatizable? Some Nice Axiomatizations

V { G+ : G a commutative groupoid } axiomatized by x · y ≈ y · x V { G+ : G an idempotent groupoid }: x ≤ x · x V { G+ : G a commutative, idempotent groupoid }: x · y ≈ y · x, x ≤ x · x

Together with axioms for Boolean algebras, normality, and additivity. (Jipsen)

So all are ﬁnitely based. Are these varieties represented by a class of groupoids? On the Other Hand. . .

What about the associative law?

Let Sg (CSg) be the variety of (commutative) semigroups.

Theorem (Jipsen) Neither V(Sg+) nor V(CSg+) is ﬁnitely based. On the Other, Other Hand. . .

Lz = left-zero semigroups (x · y ≈ x) V(Lz+) is ﬁnitely based and represented by Lz.

Rb = rectangular bands (x · (y · z) ≈ (x · y) · z, x ≈ x 2, x · y · z ≈ x · z) V(Rb+) is ﬁnitely based and represented by Rb. Boolean Semilattices

Let Sl denote the variety of semilattices, i.e. groupoids satisfying

x · (y · z) ≈ (x · y) · z x · y ≈ y · x x · x ≈ x.

These are the identities of associativity, commutativity, and idempotence Deﬁnition A Boolean semilattice is a Boolean groupoid satisfying

x · (y · z) ≈ (x · y) · z x · y ≈ y · x x ≤ x · x

The variety is denoted BSl. Easy to check that V(Sl+) ⊆ BSl. One wishes that these are equal. They are not.

Example Let H = {a, b}. Deﬁne TRS on H with θ = (a, a, a), (a, b, b), (b, a, b), (b, b, a), (b, b, b) . H+ ∈ BSl + + but Sl x ∧ (y · 1) ≤ x · y while H fails this identity Open Questions

+ 1 Is V(Sl ) finitely based? Is the equational theory decidable? + 2 Is either BSl or V(Sl ) generated by its finite members? 3 Is there a finitely axiomatizable class K of TRS such that BSl = V(K+)? Algebraic Theory of BSl

Let B ∈ BSl, x ∈ B. ↓x = x · 1

Theorem ‘↓’ yields a closure operator on B: x ≤ ↓x = ↓ ↓ x and x ≤ y =⇒ ↓x ≤ ↓y

x is closed if x = ↓x For a semilattice S and X ∈ S+

↓X = X · S = { y ∈ S :(∃x ∈ X) y ≤ x } the downset generated by X. Congruence Ideals

Let θ ∈ Con(B). I = 0/θ = { x ∈ B :(0, x) ∈ θ } is an ideal of the Boolean algebra

x ∈ I =⇒ x θ 0 =⇒ ↓x θ ↓0 = 0 =⇒ ↓x ∈ I

Conversely I an ideal closed under ↓ implies I = 0/θ for some θ ∈ Con(B)

“Congruence ideals” Consequences

Let B be a Boolean semilattice

1 Let a ∈ B. The smallest congruence ideal containing a is (↓a]. 2 BSl has equationally deﬁnable principal congruences (EDPC) 3 B is subdirectly irreducible iff it has a smallest nonzero closed element 4 B is simple iff x > 0 =⇒ ↓x = 1 Thus, for S ∈ Sl, S+ is SI iff S has a least element S+ is simple iff S is trivial 5 B subdirectly irreducible implies Bσ subdirectly irreducible B simple implies Bσ simple Discriminator Algebras

Let A be a set. The discriminator on A is ( z if x = y dA(x, y, z) = x if x 6= y.

An algebra A is a discriminator algebra if dA is a term of A.

Discriminator algebras are simple every subalgebra is a discriminator alg generate arithmetical varieties Theorem Every simple BSl is a discriminator algebra, with term

d(x, y, z) = x ∧ ↓(x ⊕ y) ∨ z ∧ ↓(x ⊕ y)0 Theorem The subvariety of BSl generated by all simple algebras is axiomatized, relative to BSl, by

↓(↓x)0 ≈ (↓x)0

This is the largest discriminator subvariety of BSl.

Problem: Is this variety representable by a nice family of TRS? An Interesting Subvariety

Tot = hH, H3i : H any set . Every member of Tot+ is a discriminator algebra. + V(Tot ) is axiomatized (relative to BSlD) by x · y · 1 ≈ x · y. Linear Semilattices

Recall BSl x ≤ x · x. What about IBSl = { B ∈ BSl : B x ≈ x · x }?

+ Let S ∈ Sl. S x ≈ x · x iff S is linearly ordered Proof: X · X ⊆ X =⇒ X is a subsemilattice So S has the property that every subset is a subsemilattice. Thus S is linearly ordered Theorem Let LS denote the class of linearly ordered semilattices. Then IBSl = SP(LS+). In IBSl, x · y = (x ∧ ↓y) ∨ (y ∧ ↓x)

Thus IBSl is term-equivalent to its closure-reduct

This variety of modal algebras is S4,3 Consequences of EDPC

For a class K,

Hω(K) = { B/θ : B ∈ K, θ a compact congruence }

If K ⊆ BSl, then B/θ = B/(c] for a closed element c.

A is splitting (relative to V) if subdirectly irreducible and there is a largest subvariety excluding A. Denoted V/A, deﬁned by a single equation. Theorem (Blok-Pigozzi) If V has EDPC then every ﬁnite subdirectly irreducible algebra is splitting. In this case

V/A = { B ∈ V : A ∈/ SHω(B) } . 2ℵ0 subvarieties

For n ∈ ω,

a1 a2 a3 an

Yn =

+ Each Yn is subdirectly irreducible, hence splitting.

Easy to check:

+ + n 6= m =⇒ Ym ∈/ SH(Yn )

+ + Since Yn is ﬁnite, H(Yn ) = Hω(Yn ) + Let T ⊆ ω and VT = V { Yn : n ∈ T }. Then

+ + m ∈/ T =⇒ Ym ∈/ SH Yn : n ∈ T =⇒ + + Yn : n ∈ T ⊆ BSl/Ym

+ Splitting, so BSl/Ym is a variety. Therefore

+ + VT = V Yn : n ∈ T ⊆ BSl/Ym

+ so Ym ∈/ VT . + m ∈/ T =⇒ Ym ∈/ VT

Thus S 6= T =⇒ VS 6= VT .

Hence BSl has uncountably many subvarieties Wrap-up

The variety of Boolean semilattices is a very natural, ﬁnitely axiomatizable variety Has a rich arithmetic structure Numerous open problems on the border of all the components of BLAST Thanks for listening!