CONGRUENCE LATTICES OF SEMILATTICES ... - Semantic Scholar
CONGRUENCE LATTICES OF SEMILATTICES WITH OPERATORS JENNIFER HYNDMAN, J. B. NATION, AND JOY NISHIDA
We want to extend the duality Con(S, +, 0) ∼ =d Sub(S, ∧, 1) for finite semilattices, and more generally Con(S, +, 0) ∼ =d Sp (I(L)) to semilattices with operators (that preserve + and 0, i.e., endomorphisms added as operations). For simplicity, let us consider semilattices with one operator: S = hS, +, 0, gi, as the extension to a monoid of operators is straightforward. 1. Adjoints: the finite case We begin by recalling the general theory of adjoints on finite semilattices. A finite join semilattice with 0 is a lattice, with the naturally induced meet operation. Thus a finite lattice S can be regarded as a semilattice in two ways, either S = hS, +, 0i or S = hS, ∧, 1i. Given a (+, P0)-homomorphism g : S → T , define the adjoint h : T → S by h(t) = {s ∈ S : gs ≤ t} so that gs ≤ t iff s ≤ ht . Proof. The ⇒ direction is clear. For the resverse, assume s0 ≤ ht. P Then gs0 ≤ ght = {gs : gs ≤ t} ≤ t, as desired. As immediate consequences we get • ght ≤ t • hgs ≥ s We can think of h(t) as the largest element that maps below t, and in particular it must be the largest element of its ker g-class. Lemma 1. h(1) = 1 and h preserves meets: h(x ∧ y) = hx ∧ hy. Proof. The map h is order-preserving, so h(x ∧ y) ≤ hx ∧ hy. On the other hand, g(hx ∧ hy) ≤ ghx ≤ x and g(hx ∧ hy) ≤ ghy ≤ y, whence g(hx ∧ hy) ≤ x ∧ y. It follows that hx ∧ hy ≤ h(x ∧ y). Date: January 18, 2015. 1
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JENNIFER HYNDMAN, J. B. NATION, AND JOY NISHIDA
V Aside: Define k : S → T via k(s) = {t ∈ T : ht ≥ s}. Then we obtain t ≥ ks iff ht ≥ s iff t ≥ gs. Substituting t = gs and t = ks yields both inclusions: ks = gs. Let us denote the adjoint of g by h = gˆ. In this note, when there is only one operator, we continue to use just g and h. 2. Duality: the finite case Now we recall the duality theorem for congruence lattices of finite 0-semilattices without operators [3]. (Adjustments can be made for the case without 0.) Theorem 2. Let S = hS, +, 0i be a finite semilattice. Then Con(S, +, 0) ∼ =d Sub(S, ∧, 1) via the maps • σ(θ) = Uθ is the set of all maximal elements of θ-classes, V • ρ(U ) = θU is given by x θU y iff τ x = τ y, where τ x = {u ∈ U : u ≥ x}. It is convenient to use x ≤ y mod θ to mean x + y θ y. Note x θ y iff x ≤ y mod θ and y ≤ x mod θ. Lemma 3. Let hθ, U i be a pair with θ = ρ(U ) and U = σ(θ). • y ∈ Uθ iff x ≤ y mod θ implies x ≤ y. • x ≤ y mod θU iff y ≤ u implies x ≤ u for all u ∈ U . Given a congruence θ, let us show that Uθ = ρ(θ) is a meet semilattice with 1. Clearly 1 ∈ Uθ . Suppose u, v ∈ Uθ so that x ≤ u mod θ implies x ≤ u, and similarly for v. If x ≤ u ∧ v mod θ, then x ≤ u mod θ, whence x ≤ u. Likewise, x ≤ v, and thus x ≤ u ∧ v. Therefore u ∧ v ∈ Uθ . To see that a meet subsemilattice defines a congruence, let U be given. Suppose x ≤ y modulo the corresponding relation θU , i.e., y ≤ u ∈ U implies x ≤ u for all u ∈ U . Then easily y + t ≤ u ∈ U implies x + t ≤ u, and thus θU = ρ(U ) is a congruence. Finally, it is clear that ρ and σ are order-reversing bijections, completing the proof of the theorem. 3. Finite semilattices with operators Now Con(S, +, 0, g) is a sublattice of Con(S, +, 0), and Sub(S, ∧, 1, h) is a sublattice of Sub(S, ∧, 1). We just have to be sure we have the right sublattices. Lemma 4. Let hθ, U i be a pair with θ = ρ(U ) and U = σ(θ). Let g be an operator on hS, +, 0i and let h = gˆ be its adjoint. Then θ respects g iff U is closed under h.
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This will have the consequence: Theorem 5. If S is a finite semilattice with an operator g, and h = gˆ, then Con(S, +, 0, g) ∼ =d Sub(S, ∧, 1, h) . b then More generally, if G is a monoid of operators and H = G, Con(S, +, 0, G) ∼ =d Sub(S, ∧, 1, H) . So let us prove the lemma. Proof. Assume that θ is a semilattice congruence that respects g, and let z ∈ Uθ = σ(θ). To see that hz ∈ Uθ , assume s ≤ hz mod θ. Then gs ≤ ghz ≤ z mod θ. Since z ∈ Uθ , this implies gs ≤ z, whence s ≤ hz, as desired. Conversely, assume that U is h-closed, and that y ≤ u implies x ≤ u whenever u ∈ U , so that x ≤ y mod θU . Then gy ≤ u implies y ≤ hu ∈ U , wherefore x ≤ hu and gx ≤ u. Thus g respects θU . 4. Application: representing finite distributive lattices The duality gives us an easy representation of finite distributive lattices. This can also be deduced from a result of Tumanov [4], combined with Adaricheva and Nation [1]. Theorem 6. For every finite distributive lattice D, there is a finite semilattice with operators S such that D ∼ = Con(S, +, 0, G). Proof. As usual, we view D as the lattice of order ideals of an ordered set P . Because of the dual isomorphism involved, we work with the dual P d = Q = hQ, ≤i. Let v be a reverse linear extension of the order ≤ on Q, so that x ≤ y implies x w y, and form C0 = hQ, vi. Add a new top element 1 to C0 , forming C = C0 ∪ {1}. Since C is a chain, every subset will be a meet subsemilattice, and every order-preserving map will be meet-preserving. Now we add a set of operators H so that the sets I ∪ {1}, with I an order ideal of Q, will be exactly the H-closed subsets of C. For each covering relation a ≺ b in Q, note that b @ a, and define a function h ∈ H by if x @ b, x h(x) = a if x = b, 1 if x A b.
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JENNIFER HYNDMAN, J. B. NATION, AND JOY NISHIDA
It is easy to see that this does the trick! For Sub(C, ∧, 1, H) ∼ = O(Q), and then, keeping track of the dualities, Con(C, +, 0, G) ∼ = (O(Q))d ∼ = Fil(Q) O(Qd ) = O(P ) . 5. Adjoints: the general case To extend the theory to the general case, where S and T are notnecessarily-finite (0, +)-semilattices, we use (non-empty) ideals. Given a (+, 0)-homomorphism g : S → T , define the adjoint h : I(T ) → I(S) by h(J) = {s ∈ S : gs ∈ J}. It is straightforward to confirm that h(J) is an ideal of S. Moreover, gs ∈ J
iff s ∈ h(J) .
As an immediate consequence we get gh(J) ⊆ J for J ∈ I(T ). Now g(I) for I ∈ I(S) need not be an ideal, but it is an up-directed set, and the ideal it generates is g(I) = {x ∈ T : x ≤ g(z) for some z ∈ I}. With this minor adjustment, we have • gh(J) ≤ J for J ∈ I(T ), • hg(I) ≥ I for I ∈ I(S). Since h(J) is the ideal of all elements that map into J, it is the union of an up-directed set of ker g-classes. Recall that the union of a directed set of ideals is an ideal. Lemma 7. h(T ) = S and h preserves both arbitrary intersections and non-empty directed unions. It is possible however that h(0T ) > {0S }. 6. Duality without operators in general Now we recall the general version of the duality theorem for congruence lattices of 0-semilattices without operators [2]. For a complete lattice L, we use Sp (L) to denote the lattice of algebraic subsets of L, i.e., subsets that are closedVunder arbitrary meets and non-empty directed joins. Note that 1 = ∅ is in every algebraic subset, though 0 may not be. Theorem 8. Con(S, +, 0) ∼ =d Sp (I(S)) via the maps • σ(θ) = Uθ is the set of all θ-closed ideals of S.
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• ρ(U ) = θU is given by x θU y iff for all J ∈ U we have x ∈ J ↔ y ∈ J. Lemma 9. Let hθ, U i be a pair with θ = ρ(U ) and U = σ(θ). • J ∈ Uθ iff xθy ∈ J implies x ∈ J. • x ≤ y mod θU iff y ∈ K ∈ U implies x ∈ K for all K ∈ U . Given a congruence θ, let us show that Uθ = ρ(θ) is an algebraic subsetTof S. Clearly S ∈ U . If Jα for α ∈ A are θ-closed ideals, then S so is α∈A Jα . Moreover, if A is up-directed, then α∈A Jα will be a θ-closed ideal as well. To see that an algebraic subset defines a congruence, let U be given. Suppose x ≤ y modulo the relation θU , i.e., y ∈ J ∈ U implies x ∈ J for all J ∈ U . Then easily y + t ∈ J ∈ U implies x + t ∈ J, and thus θU = ρ(U ) is a congruence. Finally, it is clear that ρ and σ are order-reversing bijections to the right sets, completing the proof of the theorem. 7. Why algebraic subsets Okay, so maybe it is not so clear. For any subset U ⊆ I(S), we can define the relation x ≤ y mod θU iff y ∈ K ∈ U implies x ∈ K for all K ∈ U . Define M(U ) to be the collection of all meets of subsets of U , and D(U ) to be the set of all joins of directed subsets of U . The following lemma is an elementary, if tedious, calculation. Lemma 10. For any subset U ⊆ IS, • MD(U ) ⊆ DM(U ), so the algebraic subset generated by U is DM(U ); • θU = θM(U ) = θDM(U ) . So we want to consider sets of ideals that have at least the structure of algebraic subsets; it remains to show that distinct algebraic subsets give distinct congruences. Lemma 11. Let U and V be algebraic subsets of I(S). Then θU ≤ θV iff V ≤ U . Proof. If V ≤ U , then clearly θU ≤ θV . For the converse, suppose V U , and choose an ideal K ∈ V − U . For each y ∈ K let Uy = T {J ∈ U : y ∈ J}. This isSan up-directed set of ideals with each Uy ∈ U . Since K ∈ / U , K 6= y∈K Uy . Thus there exists y0 ∈ K such that Uy0 6⊆ K. Choose x ∈ Uy0 −K. Since x ∈ Uy0 , we have y0 ∈ J ∈ U implies x ∈ J, so that x ≤ y0 mod θU . However, y0 ∈ K while x ∈ / K, so x y0 mod θV . Hence θU θV .
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JENNIFER HYNDMAN, J. B. NATION, AND JOY NISHIDA
8. Semilattices with operators Again Con(S, +, 0, g) is a sublattice of Con(S, +, 0), and the h-closed algebraic subsets Sp (S, h) form a sublattice of Sp (S). We want to be sure we have the right sublattices. The statement is the same, though the interpretation is more general. Lemma 12. Let hθ, U i be a pair with θ = ρ(U ) and U = σ(θ). Let g be an operator on hS, +, 0i and let h = gˆ be its adjoint. Then θ respects g iff U is closed under h. Consequently: Theorem 13. If S is a semilattice with an operator g, and h = gˆ, then Con(S, +, 0, g) ∼ =d Sp (I(S), h) . b then More generally, if G is a monoid of operators and H = G, Con(S, +, 0, G) ∼ =d Sp (I(S), H) . So we prove the lemma. Proof. Assume that θ is a semilattice congruence that respects g, and let J ∈ Uθ = σ(θ). To see that h(J) ∈ Uθ , assume xθy ∈ h(J). Then gxθgy ∈ gh(J). Since gh(J) ⊆ J, this implies gx ∈ J, whence x ≤ h(J), as desired. Conversely, assume that U is h-closed, and that y ∈ J implies x ∈ J whenever J ∈ U , so that x ≤ y mod θU . Then gy ∈ J with J ∈ U implies y ∈ h(J) ∈ U , wherefore x ∈ h(J) and gx ∈ J. Thus g respects θU . 9. The algebraic lattice version The ideals of a semilattice hS, +, 0i form an algebraic lattice I(S). Conversely, the compact elements of any algebraic lattice L form a join semilattice with 0 such that L ∼ = I(S). Thus the previous theorem can be interpreted as follows. Corollary 14. If S is a semilattice with a monoid G of operators, then there are an algebraic lattice L and a monoid of operations H that preserve arbitrary meets and directed non-empty joins, such that Con(S, +, 0, G) ∼ =d Sp (L, H) . For the converse, assume we are given an algebraic lattice L and a map h : L → L that preserves arbitrary meets and directed non-empty joins. Letting S denote the semilattice of compact elements of L, we
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want to define a (+, V 0)-preserving map g : S → S. Ostensibly we define g : S → L via gs = {j ∈ L : s ≤ jh}, so that gs ≤ j
iff s ≤ hj .
Again note s ≤ ghs. We claim that if s ∈ S, then gs ∈ S, i.e., gs is compact. W Suppose gs ≤ A. Then _ _ _ s ≤ hgs ≤ h( A) = h( F) finite F ⊆A
and the last is a directed set, so s ≤ h(∨F ) for some finite F ⊆ A. W This implies gs ≤ F , as desired. Another routine calculation shows that Sp (L, H) is a complete sublattice of Sp (L). 10. More distributive representations As an application, we can slightly extend Theorem 6. Theorem 15. Let P be a countable ordered set with the property that ↓ x is finite for every x ∈ P , and let D be the lattice of order ideals O(P ). Then there is a semilattice with operators S such that D ∼ = Con(S, +, 0, G). Lemma 16. An ordered set P has a linear extension to ω if and only if P is countable and has the property that ↓ x is finite for every x ∈ P . Proof. Assume that P has the property, and index its elements as x0 , x1 , x2 , . . . . We can then construct the linear extension C recursively: Given cj for j < n, let cn = xm where xm is the element of least index that is minimal in P − {c0 , . . . , cn−1 }. This respects order, and the property insures that every element is eventually chosen. The reverse direction is clear. To prove the theorem, let P be an ordered set with the property, and let Q = P d . W.l.o.g. P is infinite. Then Q has a reverse linear extension C0 ∼ = ω; form C = C0 ∪ {1} ∼ = ω + 1. Every subset S ⊆ C with 1 ∈ S is algebraic. Again we add a set of operators H so that the sets I ∪ {1}, with I an order ideal of Q, will be exactly the H-closed subsets of C. For each covering relation a ≺ b in Q, note that b @ a, and define a function h ∈ H exactly as before: if x @ b, x h(x) = a if x = b, 1 if x A b.
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JENNIFER HYNDMAN, J. B. NATION, AND JOY NISHIDA
Since C ∼ = ω + 1, these maps preserve arbitrary meets and (directed) joins. Thus we obtain Sp (C, H) ∼ = O(Q), and keeping track of dualities as in the finite case, Con(C, +, 0, G) ∼ = O(P ), as desired. 11. Representing O(P ) + 1 as Con(S, +, 0, G) In this section, we show that for any ordered set P , we can represent the linear sum 1 + Fil(P ) as Sp (C, H) for an algebraic chain C with suitable operators. W.l.o.g. P is infinite. Let α be the initial ordinal of cardinality |P |. Let L be the set of limit ordinals in α, including 0 as a limit ordinal, and let N = P \ L be the set of non-limit ordinals. Add a new top element T to α, forming an algebraic chain C isomorphic to the ordinal α + 1. Meets are trivial in α + 1, since every set has a least element, while the join of a set is either 0, or its largest element, or some limit ordinal in L, or T . For every pair of limit ordinals i, j ∈ L let gi,j be defined by ( j if x ≤ i, gi,j (x) = T if x > i. In particular, gi,j (i) = j. Note that these operations preserve arbitrary meets and joins in C. If we let H0 = {gi,j : i, j ∈ L}, then the algebraic subsets Sp (C, H0 ) are {T } and all sets of the form {T } ∪ L ∪ S with S an arbitrary subset of N . Now the intention is to add operations, so that the algebraic subsets of (C, H) are {T } and sets {T } ∪ L ∪ U that are in one-to-one correspondence with the order filters of P . To this end, set up a bijection f : N → P . For any pair of non-limit ordinals i, j ∈ N let if x < i, 0 hi,j (x) = j if x = i, T if x > i. In particular, hi,j (i) = j, and again, because i is a non-limit ordinal, the operations preserve arbitrary meets and joins in C. It remains only to choose an appropriate subset H1 of these operations. Using the bijection f , let H1 = {hi,j : f (i) ≤P f (j)}. Then form H = H0 ∪ H1 . By construction, the H-closed subsets of C are {T } and {T } ∪ L ∪ U with f −1 (U ) an order filter in P . Before stating the theorem, we recall that for any ordinal β, O(β) ∼ = d d β + 1 and O(β d ) ∼ 1 + β . In particular, β or β could be used as the = ordered set P .
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Theorem 17. For any ordered set P , there is an semilattice with operators such that O(P ) + 1 ∼ = Con(S, +, 0, G). Corollary 18. For any ordinal β, the lattices β + 1 and 2 + β d can be represented as congruence lattices of semilattices with operators. References [1] K. Adaricheva and J. Nation, Lattices of quasi-equational theories as congruence lattices of semilattices with operators, parts I and II, International Journal of Algebra and Computation 22 (2012), N7. [2] S. Fajtlowicz and J. Schmidt, B´ezout families, join congruences and meetirreducible ideals, Lattice Theory (Proc. Colloq., Szeged, 1974), pp. 51–76, Colloq. Math. Soc. Janos Bolyai, Vol 14, North Holland, Amsterdam, 1976. [3] R. Freese and J. Nation, Congruence lattices of semilattices, Pacific J. of Math. 49 (1973), 51–58. [4] V. Tumanov, Finite distributive lattices of quasivarieties, Algebra and Logic 22 (1983), 119-129. Department of Mathematics, University of Northern British Columbia, Prince George, BC, Canada E-mail address: [email protected] Department of Mathematics, University of Hawaii, Honolulu, HI 96822, USA E-mail address: [email protected] Department of Mathematics, University of Hawaii, Honolulu, HI 96822, USA E-mail address: [email protected]
We want to extend the duality Con(S, +, 0) ∼ =d Sub(S, ∧, 1) for finite semilattices, and more generally Con(S, +, 0) ∼ =d Sp (I(L)) to semilattices with operators (that preserve + and 0, i.e., endomorphisms added as operations). For simplicity, let us consider semilattices with one operator: S = hS, +, 0, gi, as the extension to a monoid of operators is straightforward. 1. Adjoints: the finite case We begin by recalling the general theory of adjoints on finite semilattices. A finite join semilattice with 0 is a lattice, with the naturally induced meet operation. Thus a finite lattice S can be regarded as a semilattice in two ways, either S = hS, +, 0i or S = hS, ∧, 1i. Given a (+, P0)-homomorphism g : S → T , define the adjoint h : T → S by h(t) = {s ∈ S : gs ≤ t} so that gs ≤ t iff s ≤ ht . Proof. The ⇒ direction is clear. For the resverse, assume s0 ≤ ht. P Then gs0 ≤ ght = {gs : gs ≤ t} ≤ t, as desired. As immediate consequences we get • ght ≤ t • hgs ≥ s We can think of h(t) as the largest element that maps below t, and in particular it must be the largest element of its ker g-class. Lemma 1. h(1) = 1 and h preserves meets: h(x ∧ y) = hx ∧ hy. Proof. The map h is order-preserving, so h(x ∧ y) ≤ hx ∧ hy. On the other hand, g(hx ∧ hy) ≤ ghx ≤ x and g(hx ∧ hy) ≤ ghy ≤ y, whence g(hx ∧ hy) ≤ x ∧ y. It follows that hx ∧ hy ≤ h(x ∧ y). Date: January 18, 2015. 1
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JENNIFER HYNDMAN, J. B. NATION, AND JOY NISHIDA
V Aside: Define k : S → T via k(s) = {t ∈ T : ht ≥ s}. Then we obtain t ≥ ks iff ht ≥ s iff t ≥ gs. Substituting t = gs and t = ks yields both inclusions: ks = gs. Let us denote the adjoint of g by h = gˆ. In this note, when there is only one operator, we continue to use just g and h. 2. Duality: the finite case Now we recall the duality theorem for congruence lattices of finite 0-semilattices without operators [3]. (Adjustments can be made for the case without 0.) Theorem 2. Let S = hS, +, 0i be a finite semilattice. Then Con(S, +, 0) ∼ =d Sub(S, ∧, 1) via the maps • σ(θ) = Uθ is the set of all maximal elements of θ-classes, V • ρ(U ) = θU is given by x θU y iff τ x = τ y, where τ x = {u ∈ U : u ≥ x}. It is convenient to use x ≤ y mod θ to mean x + y θ y. Note x θ y iff x ≤ y mod θ and y ≤ x mod θ. Lemma 3. Let hθ, U i be a pair with θ = ρ(U ) and U = σ(θ). • y ∈ Uθ iff x ≤ y mod θ implies x ≤ y. • x ≤ y mod θU iff y ≤ u implies x ≤ u for all u ∈ U . Given a congruence θ, let us show that Uθ = ρ(θ) is a meet semilattice with 1. Clearly 1 ∈ Uθ . Suppose u, v ∈ Uθ so that x ≤ u mod θ implies x ≤ u, and similarly for v. If x ≤ u ∧ v mod θ, then x ≤ u mod θ, whence x ≤ u. Likewise, x ≤ v, and thus x ≤ u ∧ v. Therefore u ∧ v ∈ Uθ . To see that a meet subsemilattice defines a congruence, let U be given. Suppose x ≤ y modulo the corresponding relation θU , i.e., y ≤ u ∈ U implies x ≤ u for all u ∈ U . Then easily y + t ≤ u ∈ U implies x + t ≤ u, and thus θU = ρ(U ) is a congruence. Finally, it is clear that ρ and σ are order-reversing bijections, completing the proof of the theorem. 3. Finite semilattices with operators Now Con(S, +, 0, g) is a sublattice of Con(S, +, 0), and Sub(S, ∧, 1, h) is a sublattice of Sub(S, ∧, 1). We just have to be sure we have the right sublattices. Lemma 4. Let hθ, U i be a pair with θ = ρ(U ) and U = σ(θ). Let g be an operator on hS, +, 0i and let h = gˆ be its adjoint. Then θ respects g iff U is closed under h.
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This will have the consequence: Theorem 5. If S is a finite semilattice with an operator g, and h = gˆ, then Con(S, +, 0, g) ∼ =d Sub(S, ∧, 1, h) . b then More generally, if G is a monoid of operators and H = G, Con(S, +, 0, G) ∼ =d Sub(S, ∧, 1, H) . So let us prove the lemma. Proof. Assume that θ is a semilattice congruence that respects g, and let z ∈ Uθ = σ(θ). To see that hz ∈ Uθ , assume s ≤ hz mod θ. Then gs ≤ ghz ≤ z mod θ. Since z ∈ Uθ , this implies gs ≤ z, whence s ≤ hz, as desired. Conversely, assume that U is h-closed, and that y ≤ u implies x ≤ u whenever u ∈ U , so that x ≤ y mod θU . Then gy ≤ u implies y ≤ hu ∈ U , wherefore x ≤ hu and gx ≤ u. Thus g respects θU . 4. Application: representing finite distributive lattices The duality gives us an easy representation of finite distributive lattices. This can also be deduced from a result of Tumanov [4], combined with Adaricheva and Nation [1]. Theorem 6. For every finite distributive lattice D, there is a finite semilattice with operators S such that D ∼ = Con(S, +, 0, G). Proof. As usual, we view D as the lattice of order ideals of an ordered set P . Because of the dual isomorphism involved, we work with the dual P d = Q = hQ, ≤i. Let v be a reverse linear extension of the order ≤ on Q, so that x ≤ y implies x w y, and form C0 = hQ, vi. Add a new top element 1 to C0 , forming C = C0 ∪ {1}. Since C is a chain, every subset will be a meet subsemilattice, and every order-preserving map will be meet-preserving. Now we add a set of operators H so that the sets I ∪ {1}, with I an order ideal of Q, will be exactly the H-closed subsets of C. For each covering relation a ≺ b in Q, note that b @ a, and define a function h ∈ H by if x @ b, x h(x) = a if x = b, 1 if x A b.
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JENNIFER HYNDMAN, J. B. NATION, AND JOY NISHIDA
It is easy to see that this does the trick! For Sub(C, ∧, 1, H) ∼ = O(Q), and then, keeping track of the dualities, Con(C, +, 0, G) ∼ = (O(Q))d ∼ = Fil(Q) O(Qd ) = O(P ) . 5. Adjoints: the general case To extend the theory to the general case, where S and T are notnecessarily-finite (0, +)-semilattices, we use (non-empty) ideals. Given a (+, 0)-homomorphism g : S → T , define the adjoint h : I(T ) → I(S) by h(J) = {s ∈ S : gs ∈ J}. It is straightforward to confirm that h(J) is an ideal of S. Moreover, gs ∈ J
iff s ∈ h(J) .
As an immediate consequence we get gh(J) ⊆ J for J ∈ I(T ). Now g(I) for I ∈ I(S) need not be an ideal, but it is an up-directed set, and the ideal it generates is g(I) = {x ∈ T : x ≤ g(z) for some z ∈ I}. With this minor adjustment, we have • gh(J) ≤ J for J ∈ I(T ), • hg(I) ≥ I for I ∈ I(S). Since h(J) is the ideal of all elements that map into J, it is the union of an up-directed set of ker g-classes. Recall that the union of a directed set of ideals is an ideal. Lemma 7. h(T ) = S and h preserves both arbitrary intersections and non-empty directed unions. It is possible however that h(0T ) > {0S }. 6. Duality without operators in general Now we recall the general version of the duality theorem for congruence lattices of 0-semilattices without operators [2]. For a complete lattice L, we use Sp (L) to denote the lattice of algebraic subsets of L, i.e., subsets that are closedVunder arbitrary meets and non-empty directed joins. Note that 1 = ∅ is in every algebraic subset, though 0 may not be. Theorem 8. Con(S, +, 0) ∼ =d Sp (I(S)) via the maps • σ(θ) = Uθ is the set of all θ-closed ideals of S.
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• ρ(U ) = θU is given by x θU y iff for all J ∈ U we have x ∈ J ↔ y ∈ J. Lemma 9. Let hθ, U i be a pair with θ = ρ(U ) and U = σ(θ). • J ∈ Uθ iff xθy ∈ J implies x ∈ J. • x ≤ y mod θU iff y ∈ K ∈ U implies x ∈ K for all K ∈ U . Given a congruence θ, let us show that Uθ = ρ(θ) is an algebraic subsetTof S. Clearly S ∈ U . If Jα for α ∈ A are θ-closed ideals, then S so is α∈A Jα . Moreover, if A is up-directed, then α∈A Jα will be a θ-closed ideal as well. To see that an algebraic subset defines a congruence, let U be given. Suppose x ≤ y modulo the relation θU , i.e., y ∈ J ∈ U implies x ∈ J for all J ∈ U . Then easily y + t ∈ J ∈ U implies x + t ∈ J, and thus θU = ρ(U ) is a congruence. Finally, it is clear that ρ and σ are order-reversing bijections to the right sets, completing the proof of the theorem. 7. Why algebraic subsets Okay, so maybe it is not so clear. For any subset U ⊆ I(S), we can define the relation x ≤ y mod θU iff y ∈ K ∈ U implies x ∈ K for all K ∈ U . Define M(U ) to be the collection of all meets of subsets of U , and D(U ) to be the set of all joins of directed subsets of U . The following lemma is an elementary, if tedious, calculation. Lemma 10. For any subset U ⊆ IS, • MD(U ) ⊆ DM(U ), so the algebraic subset generated by U is DM(U ); • θU = θM(U ) = θDM(U ) . So we want to consider sets of ideals that have at least the structure of algebraic subsets; it remains to show that distinct algebraic subsets give distinct congruences. Lemma 11. Let U and V be algebraic subsets of I(S). Then θU ≤ θV iff V ≤ U . Proof. If V ≤ U , then clearly θU ≤ θV . For the converse, suppose V U , and choose an ideal K ∈ V − U . For each y ∈ K let Uy = T {J ∈ U : y ∈ J}. This isSan up-directed set of ideals with each Uy ∈ U . Since K ∈ / U , K 6= y∈K Uy . Thus there exists y0 ∈ K such that Uy0 6⊆ K. Choose x ∈ Uy0 −K. Since x ∈ Uy0 , we have y0 ∈ J ∈ U implies x ∈ J, so that x ≤ y0 mod θU . However, y0 ∈ K while x ∈ / K, so x y0 mod θV . Hence θU θV .
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JENNIFER HYNDMAN, J. B. NATION, AND JOY NISHIDA
8. Semilattices with operators Again Con(S, +, 0, g) is a sublattice of Con(S, +, 0), and the h-closed algebraic subsets Sp (S, h) form a sublattice of Sp (S). We want to be sure we have the right sublattices. The statement is the same, though the interpretation is more general. Lemma 12. Let hθ, U i be a pair with θ = ρ(U ) and U = σ(θ). Let g be an operator on hS, +, 0i and let h = gˆ be its adjoint. Then θ respects g iff U is closed under h. Consequently: Theorem 13. If S is a semilattice with an operator g, and h = gˆ, then Con(S, +, 0, g) ∼ =d Sp (I(S), h) . b then More generally, if G is a monoid of operators and H = G, Con(S, +, 0, G) ∼ =d Sp (I(S), H) . So we prove the lemma. Proof. Assume that θ is a semilattice congruence that respects g, and let J ∈ Uθ = σ(θ). To see that h(J) ∈ Uθ , assume xθy ∈ h(J). Then gxθgy ∈ gh(J). Since gh(J) ⊆ J, this implies gx ∈ J, whence x ≤ h(J), as desired. Conversely, assume that U is h-closed, and that y ∈ J implies x ∈ J whenever J ∈ U , so that x ≤ y mod θU . Then gy ∈ J with J ∈ U implies y ∈ h(J) ∈ U , wherefore x ∈ h(J) and gx ∈ J. Thus g respects θU . 9. The algebraic lattice version The ideals of a semilattice hS, +, 0i form an algebraic lattice I(S). Conversely, the compact elements of any algebraic lattice L form a join semilattice with 0 such that L ∼ = I(S). Thus the previous theorem can be interpreted as follows. Corollary 14. If S is a semilattice with a monoid G of operators, then there are an algebraic lattice L and a monoid of operations H that preserve arbitrary meets and directed non-empty joins, such that Con(S, +, 0, G) ∼ =d Sp (L, H) . For the converse, assume we are given an algebraic lattice L and a map h : L → L that preserves arbitrary meets and directed non-empty joins. Letting S denote the semilattice of compact elements of L, we
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want to define a (+, V 0)-preserving map g : S → S. Ostensibly we define g : S → L via gs = {j ∈ L : s ≤ jh}, so that gs ≤ j
iff s ≤ hj .
Again note s ≤ ghs. We claim that if s ∈ S, then gs ∈ S, i.e., gs is compact. W Suppose gs ≤ A. Then _ _ _ s ≤ hgs ≤ h( A) = h( F) finite F ⊆A
and the last is a directed set, so s ≤ h(∨F ) for some finite F ⊆ A. W This implies gs ≤ F , as desired. Another routine calculation shows that Sp (L, H) is a complete sublattice of Sp (L). 10. More distributive representations As an application, we can slightly extend Theorem 6. Theorem 15. Let P be a countable ordered set with the property that ↓ x is finite for every x ∈ P , and let D be the lattice of order ideals O(P ). Then there is a semilattice with operators S such that D ∼ = Con(S, +, 0, G). Lemma 16. An ordered set P has a linear extension to ω if and only if P is countable and has the property that ↓ x is finite for every x ∈ P . Proof. Assume that P has the property, and index its elements as x0 , x1 , x2 , . . . . We can then construct the linear extension C recursively: Given cj for j < n, let cn = xm where xm is the element of least index that is minimal in P − {c0 , . . . , cn−1 }. This respects order, and the property insures that every element is eventually chosen. The reverse direction is clear. To prove the theorem, let P be an ordered set with the property, and let Q = P d . W.l.o.g. P is infinite. Then Q has a reverse linear extension C0 ∼ = ω; form C = C0 ∪ {1} ∼ = ω + 1. Every subset S ⊆ C with 1 ∈ S is algebraic. Again we add a set of operators H so that the sets I ∪ {1}, with I an order ideal of Q, will be exactly the H-closed subsets of C. For each covering relation a ≺ b in Q, note that b @ a, and define a function h ∈ H exactly as before: if x @ b, x h(x) = a if x = b, 1 if x A b.
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JENNIFER HYNDMAN, J. B. NATION, AND JOY NISHIDA
Since C ∼ = ω + 1, these maps preserve arbitrary meets and (directed) joins. Thus we obtain Sp (C, H) ∼ = O(Q), and keeping track of dualities as in the finite case, Con(C, +, 0, G) ∼ = O(P ), as desired. 11. Representing O(P ) + 1 as Con(S, +, 0, G) In this section, we show that for any ordered set P , we can represent the linear sum 1 + Fil(P ) as Sp (C, H) for an algebraic chain C with suitable operators. W.l.o.g. P is infinite. Let α be the initial ordinal of cardinality |P |. Let L be the set of limit ordinals in α, including 0 as a limit ordinal, and let N = P \ L be the set of non-limit ordinals. Add a new top element T to α, forming an algebraic chain C isomorphic to the ordinal α + 1. Meets are trivial in α + 1, since every set has a least element, while the join of a set is either 0, or its largest element, or some limit ordinal in L, or T . For every pair of limit ordinals i, j ∈ L let gi,j be defined by ( j if x ≤ i, gi,j (x) = T if x > i. In particular, gi,j (i) = j. Note that these operations preserve arbitrary meets and joins in C. If we let H0 = {gi,j : i, j ∈ L}, then the algebraic subsets Sp (C, H0 ) are {T } and all sets of the form {T } ∪ L ∪ S with S an arbitrary subset of N . Now the intention is to add operations, so that the algebraic subsets of (C, H) are {T } and sets {T } ∪ L ∪ U that are in one-to-one correspondence with the order filters of P . To this end, set up a bijection f : N → P . For any pair of non-limit ordinals i, j ∈ N let if x < i, 0 hi,j (x) = j if x = i, T if x > i. In particular, hi,j (i) = j, and again, because i is a non-limit ordinal, the operations preserve arbitrary meets and joins in C. It remains only to choose an appropriate subset H1 of these operations. Using the bijection f , let H1 = {hi,j : f (i) ≤P f (j)}. Then form H = H0 ∪ H1 . By construction, the H-closed subsets of C are {T } and {T } ∪ L ∪ U with f −1 (U ) an order filter in P . Before stating the theorem, we recall that for any ordinal β, O(β) ∼ = d d β + 1 and O(β d ) ∼ 1 + β . In particular, β or β could be used as the = ordered set P .
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Theorem 17. For any ordered set P , there is an semilattice with operators such that O(P ) + 1 ∼ = Con(S, +, 0, G). Corollary 18. For any ordinal β, the lattices β + 1 and 2 + β d can be represented as congruence lattices of semilattices with operators. References [1] K. Adaricheva and J. Nation, Lattices of quasi-equational theories as congruence lattices of semilattices with operators, parts I and II, International Journal of Algebra and Computation 22 (2012), N7. [2] S. Fajtlowicz and J. Schmidt, B´ezout families, join congruences and meetirreducible ideals, Lattice Theory (Proc. Colloq., Szeged, 1974), pp. 51–76, Colloq. Math. Soc. Janos Bolyai, Vol 14, North Holland, Amsterdam, 1976. [3] R. Freese and J. Nation, Congruence lattices of semilattices, Pacific J. of Math. 49 (1973), 51–58. [4] V. Tumanov, Finite distributive lattices of quasivarieties, Algebra and Logic 22 (1983), 119-129. Department of Mathematics, University of Northern British Columbia, Prince George, BC, Canada E-mail address: [email protected] Department of Mathematics, University of Hawaii, Honolulu, HI 96822, USA E-mail address: [email protected] Department of Mathematics, University of Hawaii, Honolulu, HI 96822, USA E-mail address: [email protected]
