Order DOI 10.1007/s11083-010-9186-9
The Lattice of Compatible Quasiorders of Acyclic Monounary Algebras Danica Jakubíková-Studenovská · Reinhard Pöschel · Sándor Radeleczki
Received: 25 August 2010 / Accepted: 14 October 2010 © Springer Science+Business Media B.V. 2010
Abstract Acyclic monounary algebras are characterized by the property that any compatible partial order r can be extended to a compatible linear order. In the case of rooted monounary algebras A = (A, f ) we characterize the intersection of compatible linear extensions of r by several equivalent conditions and generalize these results to compatible quasiorders of A. We show that the lattice Quord A of compatible quasiorders is a disjoint union of semi-intervals whose maximal elements equal the intersection of their compatible quasilinear extensions. We also investigate algebraic properties of the lattices Quord A and Con A. Keywords Monounary algebra · Acyclic · Rooted · Quasiorder · Compatible linear extension Mathematics Subject Classifications (2010) Primary 08A60; Secondary 06B99 · 06F99 1 Introduction In [11] it was proved that a monounary algebra A = (A, f ) admits a compatible linear order if and only if it is acyclic, i.e. if each cycle of (the graph of) f has exactly
D. Jakubíková-Studenovská was supported by Slovak VEGA grant 2/0194/10. R. Pöschel and S. Radeleczki were supported by DFG grant 436 UNG 113/173/0-2. D. Jakubíková-Studenovská Institute of Mathematics, P. J. Šafárik University, Košice, Slovakia R. Pöschel (B) Institute of Algebra, TU Dresden, Germany e-mail:
[email protected] S. Radeleczki Institute of Mathematics, University of Miskolc, Miskolc, Hungary
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one element. Another remarkable property of an acyclic monounary algebra was shown in [14], namely that any compatible partial order r ⊆ A × A can be extended to a compatible linear order. It was also pointed out, that the intersection r¯ of all compatible linear extensions of r in general differs from r, and the question to characterize this intersection r¯ was raised. A solution for this question in case of acyclic monounary algebras was presented in [13] (see [15] for a generalization). The dimension theory of partially ordered sets (cf. [16]) is based on the notion of a realizer. Therefore, once we have a satisfactory description of the intersection of all compatible linear extensions, one can use the corresponding realizers in order to build a dimension theory for partially ordered (monounary) algebras along the lines of the classical development. In the present paper we often consider so-called rooted monounary algebras (where the graph of f has a root, cf. Section 2.4) and we characterize r¯ by several equivalent conditions. For instance, we prove that compatible partial orders ⊆ A × A satisfying the condition ∩ ker f = r ∩ ker f always form an interval in the lattice Quord A of all compatible quasiorders of a rooted algebra A, and that r¯ equals the greatest element in this interval and can be described by (see Theorem 4.3) r¯ = ∪ (x, y) ∈ A2 | ∃n ∈ N : f n x, f n y ∈ r \ . We generalize this construction to compatible quasiorders by showing that for any q ∈ Quord A the set { ∈ Quord A | ∩ ker f = q ∩ ker f } is a semi-interval (see what follows Definition 3.4), the maximal elements of which are equal to the intersections of their quasilinear extensions (cf. Theorem 4.6). In fact, we show that the form of this semi-interval is determined by its least element β = q ∩ ker f - which will be called the type of q. For instance, the semi-interval is an interval if and only if q has a so-called complete type (cf. Definition 3.6). We also investigate algebraic properties of the lattice Quord A and its sublattice Con A. Such properties may influence the shape of the above mentioned semiinterval constructions [9]. For instance, in the case of a finite acyclic and connected monounary algebra A it will be shown that Con A is always a strongly semimodular lattice (cf. Theorem 5.7). Moreover, Quord A shares some known properties of the lattice Quord(A) of all quasiorders on the set A (this lattice was studied intensively, see [2]). For a general approach concerning the structure of related lattices of monounary algebras we refer to [7].
2 Preliminaries 2.1 Quasiorders A quasiorder q on a set A is a reflexive and transitive relation q ⊆ A × A. The set of all quasiorders on A is denoted by Quord(A). The lattice Quord(A) is known to be algebraic, complemented, atomistic and dually atomistic (see e.g. [2]), and it contains as a complete sublattice the lattice Equ(A) of all equivalence relations on A. For q ∈ Quord(A) the inverse q−1 := {(y, x) | (x, y) ∈ q} is also a quasiorder and the relation q0 := q ∩ q−1
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is an equivalence on A. The q0 -equivalence class of an element a ∈ A will be denoted by [a]q0 := {b ∈ A | (a, b ) ∈ q0 }. Obviously, a quasiorder q is a partial order if and only if q0 = , and it is a linear order if and only if in addition q ∪ q−1 = ∇, where := {(x, x) | x ∈ A} and ∇ := A2 . It is well-known that q ∈ Quord(A) induces a natural partial order q/q0 on the factor set A/q0 := {[a]q0 | a ∈ A} given by [a]q0 , [b ]q0 ∈ q/q0 : ⇐⇒ ∃u ∈ [a]q0 ∃v ∈ [b ]q0 : (u, v) ∈ q . A straightforward argument shows that ([a]q0 , [b ]q0 ) ∈ q/q0 holds if and only if (x, y) ∈ q for all x ∈ [a]q0 and y ∈ [b ]q0 . 2.2 Monounary Algebras and Compatible Relations For a unary mapping f : A → A we often write for simplicity f x and f n x instead of f (x) and f n (x) (where f 0 (x) := x, f n+1 (x) := f ( f n (x)), n ∈ N). A mapping f : A → A preserves a binary relation ⊆ A × A (or f is an endomorphism of , is invariant for or compatible with f , notation f ), if ∀x, y ∈ A : (x, y) ∈ =⇒ ( f x, f y) ∈ . For a monounary algebra A = (A, f ) we consider the following particular sets of compatible relations: Refl A := ⊆ A2 | reflexive relation, f Quord A := ⊆ A2 | quasiorder, f Con A := ⊆ A2 | equivalence relation, f Pord A := ⊆ A2 | partial order, f . It is known that Refl A is a completely distributive algebraic lattice, Quord A is a complete sublattice of Quord(A), and Con A is a complete sublattice of Equ(A). This follows from the facts that the infimum of a system {αi | i ∈ I} of elements in each of these lattices is equal to the set theoretical intersection {αi | i ∈ I}; the supremum of {αi | i ∈ I} in Refl A is {αi | i ∈ I}, and in all other mentioned lattices it is equal to ( {αi | i ∈ I})tra where α tra stands for the transitive closure of a binary relation α ⊆ A2 . Pord A is an order ideal of Quord A, since r ∈ Pord A and q ⊆ r imply q ∈ Pord A for any q ∈ Quord A. If q ∈ Quord(A, f ) then q0 ∈ Con(A, f ) and A = (A, f ) has a natural homomorphic image A/q0 := (A/q0 , fˆ), where fˆ is defined by fˆ([x]q0 ) := [ f (x)]q0 . It is known that the partial order q/q0 is also compatible: q/q0 ∈ Pord(A/q0 , fˆ). 2.3 Types Now consider a monounary algebra A = (A, f ) and the relation ker f := {(x, y) ∈ A2 | f x = f y}. Since (Refl A, ∩, ∪) is a complete, completely distributive lattice and ker f ∈ Con A ⊆ Refl A, we obtain that θker f = (1 , 2 ) ∈ (Refl A)2 | 1 ∩ ker f = 2 ∩ ker f
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is a complete congruence of Refl A. Therefore, for any ∈ Refl A its congruence class []θker f is an interval [min , max ] in the lattice Refl A with α ∈ Refl A | (, α) ∈ θker f , max := α ∈ Refl A | (, α) ∈ θker f . min := It is easy to check that min = ∩ ker f and max = {α ∈ Refl A | α ∩ ker f = min } = (min )max , and that {min | ∈ Refl A} and {max | ∈ Refl A} are complete sublattices of Refl A both isomorphic to Refl A/θker f ∼ = ker f ]Refl A (see [3, 4]). As any reflexive relation β ⊆ ker f is trivially compatible with f and β ∩ ker f = β, all these classes have the form
β, β max Refl A = { ∈ Refl A | β = ∩ ker f } for some ⊆ β ⊆ ker f . For any relation ∈ Refl A, the least element β = ∩ ker f = min of its θker f -class is called the type of . If is a quasiorder, then its type β is also a quasiorder. The set of all quasiorder types, i.e. types of quasiorders of A, is denoted by TA . Remark Since the elements of TA are just quasiorders of the set A contained in ker f , TA coincides with the principal ideal ker f ]Quord(A) in Quord(A). As mentioned above, every q ∈ Quord(A) with q ⊆ ker f is compatible ((a, b ) ∈ q implies ( f a, f b ) ∈ ⊆ q). Because ker f ∈ Quord A, the principal ideals TA = ker f ]Quord(A) in Quord(A) and ker f ]Quord(A) in Quord A coincide. A further justification for our terminology “quasiorder type” will be given in Definition 3.4. The lattice TA can be described completely as follows. Lemma 2.1 For a monounary algebra A = (A, f ) the lattice TA of its quasiorder types is isomorphic to the direct product of full quasiorder lattices: TA ∼ Quord(B). = B∈A/ ker f
Proof Each β ∈ TA is determined by the family (β B ) B∈A/ ker f of all restrictions β B := β ∩ B2 to the blocks B ∈ A/ ker f of ker f . Then we have β = B∈A/ ker f β B ∈ Quord A for each choice of β B ∈ Quord(B). Since any direct product of atomistic (dually atomistic, complemented) lattices is also an atomistic (dually atomistic, complemented) lattice and since TA is a principal ideal in the algebraic lattice Quord A, we obtain: Corollary 2.2 TA = ker f ]Quord A is an atomistic, dually atomistic and complemented algebraic lattice. 2.4 Rooted Algebras A monounary algebra A = (A, f ) is acyclic, if f n a = a for some a ∈ A and positive n ∈ N+ implies f a = a (i.e. there are no cycles except fixed points). A monounary algebra A is called connected if for any x, y ∈ A there exist some n, m ∈ N such that f n x = f m y holds (see e.g. [7]).
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For a monounary algebra A = (A, f ) the following conditions are equivalent: (*) ∀x, y ∈ A∃n ∈ N : f n x = f n y. (**) A is acyclic, connected and has a fixed point. Since for any finite acyclic monounary algebra (A, f ) the operation f : A → A has at least one fixed point, it follows that finite monounary algebras satisfy condition (*) if and only if they are acyclic and connected. We call an algebra satisfying (*) also a rooted algebra since the unique fixed point is the “root” of the graph of f (which is a directed tree). 3 The Shape of Quord A Let A = (A, f ) be a monounary algebra. In this section we make a look inside the lattice Quord A. By Definitions 3.4 and 3.5, Quord A can be partitioned into semiintervals, the lattice structure of their least elements, i.e. of TA , is completely clarified by Lemma 2.1. Moreover we study under which conditions these semi-intervals become intervals. With Lord A we shall denote the set of all compatible linear orders of an algebra A. Definition 3.1 For ∈ Refl A let us define the binary relation
:= ∪ (x, y) ∈ A2 | ∃n ∈ N : f n x, f n y ∈ \ . Lemma 3.2
= β max ∩ ∩ ker f = β, ⊆ max and β (i) Let ∈ Refl A and β = ∩ ker f . Then f . ker (ii) The mapping → is a closure operator on Refl A.
= R (i.e. R is closed). (iii) If R ∈ Lord A then R Proof (i) It is easy to see that has the same type β as . Thus ∩ ker f = β and ⊆ max max max max
= . Clearly, β ⊆ β ∩ ker f . As (a, b ) ∈ β ∩ ker f \ implies β
, we get ( f n a, f n b ) ∈ β max ∩ ker f \ = β \ for some n ∈ N, i.e. (a, b ) ∈ β max f .
= β ∩ ker β (ii) directly follows from the definition of .
\ . Then ∃n ∈ N : ( f n a, f n b ) ∈ R \ . As (a, b ) ∈ R−1 would (iii) Let (a, b ) ∈ R
= R. imply ( f n a, f n b ) ∈ R ∩ R−1 = , we conclude (a, b ) ∈ R. Thus R Remark It is easy to see that the mapping → max = ( ∩ ker f )max is also a closure operator on Refl A (containing according to Lemma 3.2(i)). The next proposition shows when these closure operators coincide. Proposition 3.3 A is rooted if and only if = max for all ∈ Refl A. = Proof Let A be rooted, ∈ Refl A of type β = ∩ ker f . In order to prove
since β
⊆ max it is enough to show β max ⊆ β ⊆ max = β max . Take (a, b ) ∈ β max ,
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a = b . Then, by Section 2.4(*), there exists an m ∈ N with ( f m a, f m b ) ∈ ker f \ ,
. Conversely, thus ( f m a, f m b ) ∈ (β max ∩ ker f ) \ = β \ . This implies (a, b ) ∈ β max f ) ∪ assume = for all ∈ Refl A and consider the relation α := (A × A \ ker . Let (a, b ) ∈ α. Then ( f a, f b ) ∈ α, otherwise ( f a, f b ) ∈ ker f \ would imply f , a contradiction. Con( f n a, f n b ) ∈ ker f \ for some n ∈ N, thus (a, b ) ∈ ker f , we obtain α ∩ ker f = . Hence max = sequently α ∈ Refl A. Since ker f ⊆ ker f = A × A = ∇.
= yields α = , whence we get ker α max ⊇ α. Then max = This result implies that A is rooted. Definition 3.4 For β ∈ TA let Q(β) := {q ∈ Quord A | q ∩ ker f = β}. Q(β) = [β, β max ]Refl A ∩ Quord A. Hence we have the disjoint union Quord A = Q(β).
Clearly
β∈TA
We shall see that these sets Q(β) are semi-intervals, i.e. a union of intervals (in the quasiorder lattice) with the same least element β as shown in Fig. 1. More precisely we have: Proposition 3.5 Let β ∈ TA and let {qi | i ∈ I} be the set of all maximal quasiorders in Q(β). Then
⊆ i∈I qi ⊆ β max . (a) Q(β) = i∈I [β, qi ]Quord A , and β (b) If A is a rooted algebra, then β max = i∈I qi .
Proof (a) Since each q ∈ Q(β) contains β by definition, we have [β, q]Quord A ⊆ Q(β). For the first part of (a) it remains to prove that each p ∈ Q(β) is contained in a maximal element of Q(β). As the union of an ascending chain of compatible quasiorders is again a compatible quasiorder, by Zorn’s Lemma there is a maximal element in {q ∈ Q(β) | p ⊆ q}. Since by definition every element of
Fig. 1 The semi-interval Q(β)
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⊆ Q(β) is contained in β max , we get i∈I qi ⊆ β max . It remains to prove β n0 n0
i∈I qi . Take (a, b ) ∈ β \ , then there exist n0 ∈ N such that ( f a, f b ) ∈ β \ . Consider q := (β ∪ {( f n a, f n b ) | n ∈ {0, 1, . . . , n0 − 1})tra . Then β ⊆ q ∈ Quord(A, f ). We are going to show that q has type β (then q ∈ Q(β) is contained in a maximal qi ∈ Q(β), hence (a, b ) ∈ i∈I qi , and the proof is finished).
has type β (cf. Lemma 3.2(ii)) and β ⊆ q, it is sufficient to show q ⊆ Since β
. Let (x0 , y0 ) ∈ q, i.e. there is a sequence x0 , x1 , . . . , xr = y0 (r ∈ N) such that β either (xi , xi+1 ) ∈ β or (xi , xi+1 ) = ( f mi a, f mi b ) for some mi ∈ {0, 1, . . . , n0 − 1}.
by transitivity of β. If the latter case does not appear we have (x0 , y0 ) ∈ β ⊆ β Otherwise, let m be the minimum of all mi for which (xi , xi+1 ) = ( f mi a, f mi b ). Let k := n0 − m (note that k ≥ 1 since all mi ≤ n0 − 1). Then for the sequence f k x0 , f k x1 , . . . , f k xr = f k y0 we have f k xi = f k xi+1 (if (xi , xi+1 ) ∈ β or mi > m) or ( f k xi , f k xi+1 ) = ( f n0 a, f n0 b ) (if mi = m). Since f n0 a = f n0 b , there can be only one pair (xi , xi+1 ) with ( f k xi , f k xi+1 ) = ( f n0 a, f n0 b ), therefore we have ( f k x0 , f k y0 ) =
. ( f k xi1 , f k xi1 +1 ) = ( f n0 a, f n0 b ) ∈ β \ , consequently, (x0 , y0 ) ∈ β (b) directly follows from Proposition 3.3 and (a). Remark Let Max Y denotes the set of maximal elements in a set Y ⊆ Quord A. Then, for any types β1 , β2 ∈ TA we have Max(Max Q(β1 ) ∧ Max Q(β2 )) = Max Q(β1 ∩ β2 ).
is transitive and Q(β) is Now we shall study those quasiorder types β for which β −1 an interval. Recall that β0 := β ∩ β ∈ Con A (cf. Section 2.1). Definition 3.6 A type β ∈ TA is called complete if for any x, y ∈ A and s ∈ N+ we have s f x, f s y ∈ β0 \ =⇒ [x]ker f ⊆ [x]β0 . (3.1) Remark 3.7 From the definition it follows immediately: (i) For any type β ∈ TA we have β complete ⇐⇒ β −1 complete ⇐⇒ β0 = β ∩ β −1 complete. (ii) If ker f ⊆ q ∈ Quord A then q has a complete type. (iii) If q ∈ Pord(A, f ) then q has a complete type (because then β0 = and β0 \ = ∅). Proposition 3.8 Let A be a monounary algebra and β ∈ TA . Then
is transitive if and only if β is complete. (a) β (b) If Q(β) is an interval, then β is a complete type. Proof s s
be transitive. Then also (β
)0 = (β (a) “=⇒”: Let β 0 ) is transitive. Let ( f x, f y) ∈ β0 \ for some s ∈ N+ and x, y ∈ A. Then (x, y) ∈ β 0 (cf. Definition 3.1) and
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(y, x) ∈ β 0 by symmetry of β0 . Let z ∈ [x]ker f , i.e. f z = f x. Thus ( f s z, f s y) = ( f s x, f s y) ∈ β0 \ and we get (z, y) ∈ β 0 . By transitivity of β 0 we have (z, x) ∈ β 0 ∩ ker f = β0 , i.e. z ∈ [x]β0 . Hence we obtain [x]ker f ⊆ [x]β0 .
. We can assume “⇐=”: Suppose β is complete. Let (a, b ), (b , c) ∈ β that both pairs are not in and a = c. Then there exist m, n ∈ N with ( f m a, f m b ), ( f n b , f n c) ∈ β \ . Observe that m = n = 0 implies
. Therefore, to prove (a, c) ∈ β
, we need (a, b ), (b , c) ∈ β whence (a, c) ∈ β ⊆ β to consider only the following three cases: Case 1 m = n. If m > n, then β ⊆ ker f implies ( f m a, f m c) = ( f m a, f m b ) ∈
. The case n > m is similar. β \ , i.e. (a, c) ∈ β Case 2 m = n > 0 and f n a = f n c. In this case ( f n a, f n b ), ( f n b , f n c) ∈ β and
. f n a = f n c implies ( f n a, f n c) ∈ β \ , i.e. (a, c) ∈ β Case 3 m = n > 0 and f n a = f n c. In this case there exists a least t ∈ N with f t a = f t c. Clearly, t ≤ n and t > 0 follows from a = c. Further, ( f n a, f n b ), ( f n b , f n c) ∈ β \ and f n a = f n c imply that ( f n a, f n b ) ∈ β0 \ . Now let x := f t−1 a, y := f t−1 b , z := f t−1 c and s := n − (t − 1). Then s > 0 and (x, z) ∈ ker f . Moreover s + t − 1 = n implies ( f s x, f s y) = ( f n a, f n b ) ∈ β0 \ . As z ∈ [x]ker f , by Eq. 3.1 we get z ∈ [x]β0 , i.e., ( f t−1 a, f t−1 c) ∈ β0 \ ⊆ β \ ,
. Summarizing, β
is transitive. consequently (a, c) ∈ β (b) Assume that Q(β) is an interval. Then Max Q(β) contains a single element q, f ⊆ β max ∩ ker f = β ,
⊆ q. Thus β
⊆ q ∩ ker and Proposition 3.5(a) implies β
whence β = q ∩ ker f . Since ker f is a complete type, ker f is transitive, and
is also transitive. Thus β is a complete type according to (a). hence β Corollary 3.9 (a) Let A be a rooted algebra. Then the following conditions are equivalent: (i) β max is transitive, (ii) Q(β) is an interval, (iii) β is a complete type.
∈ Pord A. (b) If β ∈ TA ∩ Pord A then β (c) If A is a rooted algebra and ∈ Pord A then = max is again a partial order. Proof (a) (i) ⇐⇒ (ii): Clearly, if Q(β) is an interval, by Proposition 3.5(b), β max must coincide with the largest element, which is transitive. Conversely, if β max is transitive then β max ∈ Q(β) and it is the largest element of Q(β).
Propositions 3.3 and 3.8. (i) ⇐⇒ (iii): Directly follows from β max = β (b) If β ∈ TA ∩ Pord A then it is a complete type (see Remark 3.7(iii)) and
is transitive by Proposition 3.8. It remains to prove antitherefore β
. If a = b , then there exist m, n ∈ N with symmetry: Let (a, b ), (b , a) ∈ β ( f m a, f m b ), ( f n b , f n a) ∈ β \ . It follows that ( f k a, f k b ), ( f k b , f k a) ∈ β with
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k = max{m, n} and the antisymmetry of β leads to the contradiction f k a = f k b . Therefore it follows a = b .
. (c) Follows from (b) since β = ∩ ker f ∈ Pord A and max = β max = β Corollary 3.10 If A is rooted then Pord A =
{Q(β) | β ∈ TA partial order}.
Proof Because Quord A = {Q(β) | β ∈ TA }, we have Pord A ⊆ {Q(β) | β ∈ TA , Q(β) ∩ Pord A = ∅}. By Remark 3.7 and Corollary 3.9, Q(β) ∩ Pord A = ∅ im
] ⊆ Pord A. Hence Pord A = {Q(β) | β ∈ TA is a partial order}. plies Q(β) = [β, β Example 3.11 Let A = (A, f ) be the 6-element monounary algebra with A = {a0 , a1 , a2 , b 1 , b 2 , c2 } and f a0 = f a1 = f b 1 = a0 , f a2 = f c2 = a1 , f b 2 = b 1 . The graph of f with the kernel classes is in the left part of Fig. 2. An example for a type is the relation β = ∪ {(a1 , b 1 ), (b 1 , a1 ), (a2 , c2 )}. The semi-interval Q(β) is given (together with β max ) in the middle of the figure. This semi-interval is the union of two intervals: Q(β) = [β, q1 ] ∪ [β, q2 ]. The directed graphs of the ∧-irreducible quasiorders are described in large circles (omitting all loops). For the algebra A we mention the following numbers (obtained by computer): (i) quasiorders: | Quord A| = 6784 (but | Refl A| = 34222500); (ii) types: |TA | = | Quord(3)| · | Quord(2)| · | Quord(1)| = 29 · 4 · 1 = 116 (cf. Lemma 2.1), among them 12 piecewise linear and 57 partial orders; (iii) linear orders: | Lord A| = 12; (iv) quasilinear quasiorders: | QLord A| = 211; (v) partial orders: | Pord A| = 3, 325.
Fig. 2 The algebra (A, f ) and the semi-interval Q(β) ⊆ Quord(A, f )
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4 Linear and Quasilinear Extensions As already mentioned in the introduction, every ∈ Pord A has a compatible linear extension if and only if A is acyclic [14]. The nontrivial part (“⇐”) of this result also follows from Theorem 4.3 below for rooted algebras. A type β ∈ TA is called piecewise linear if β| B := β ∩ B2 is a linear order on B for each equivalence class B ∈ A/ ker f . Lemma 4.1 Let A = (A, f ) be a rooted algebra and β ∈ TA a piecewise linear type.
) is a linear order. Then β max ( = β
is a partial order. It remains to show linearity: Proof In view of Corollary 3.9(b), β
(in particular a = b ). According to Section 2.4(*) choose m ∈ N Assume (a, b ) ∈ /β
). such that ( f m a, f m b ) ∈ ker f \ . Clearly ( f m a, f m b ) ∈ / β (otherwise (a, b ) ∈ β m m Thus, because β| B is a linear order for each block B of ker f , we have ( f b , f a) ∈
. β \ , consequently (b , a) ∈ β Remark 4.2 For a rooted algebra A it follows that any R ∈ Lord A is completely determined by its (piecewise linear) type: (R ∩ ker f )max = R. The intersection of all compatible linear extensions of a compatible partial order was characterized in [13]. In the particular case of rooted monounary algebras we get another description which clearly shows the role of types for partial orders. A generalization of this to quasiorders is given in Theorem 4.6(ii). Theorem 4.3 Let A be a rooted algebra, r ∈ Pord A of type β ∈ TA . Then
= β max = β
{R ∈ Lord A | r ⊆ R},
i.e., the intersection of all compatible linear extensions of r is β max (in particular there always exists a compatible linear extension). Proof
\ . Then ( f n a, f n b ) ∈ β \ ⊆ R for some “⊆” Let r ⊆ R ∈ Lord A and (a, b ) ∈ β n n ∈ N. If (b , a) ∈ R then ( f b , f n a) ∈ R, thus ( f n a, f n b ) ∈ by antisymmetry of R, a contradiction. Therefore (b , a) ∈ / R, i.e. (a, b ) ∈ R by linearity. “⊇” By the classical theorem of Dushnik&Miller, the partial order β is the intersection of linear orders L ∈ Lord(A): β = j j∈J L j . Thus β = β ∩ ker f = (L ∩ ker f ). Clearly, L ∩ ker f is a piecewise linear type, therefore R j := j j j∈J (L j ∩ ker f )max ∈ Lord A by Lemma 4.1. It is easy to prove that in a rooted al j = j; therefore r ⊆ β max = (L j ∩ ker f )max = gebra we have j∈J j∈J j∈J {R ∈ Lord A | r ⊆ R}. j∈J R j ⊇ For the next theorem we need the following lemma, which might be of its own interest. Recall that q \ q−1 = q \ q0 for any q ∈ Quord A.
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Lemma 4.4 Let β be a type of A = (A, f ). (a) For q ∈ Q(β), let (a, b ) ∈ / q but ( f a, f b ) ∈ q \ q−1 . Then (q ∪ {(a, b )})tra is of type β, too. (b) Let q ∈ Max Q(β). Then ( f n a, f n b ) ∈ q \ q−1 implies (a, b ) ∈ q \ q−1 . Proof (a) Clearly, β ⊆ (q ∪ {(a, b )})tra ∩ ker f . Thus let (x, y) ∈ (q ∪ {(a, b )})tra ∩ ker f , i.e. f x = f y and there is a sequence x = x0 , x1 , . . . , xm = y such that (xi−1 , xi ) ∈ q ∪ {(a, b )} for i ∈ {1, . . . , m}. If (a, b ) does not appear among that pairs then (x, y) ∈ q ∩ ker f = β (because q is transitive). Otherwise, since f q, we apply f to that sequence and get ( f xi−1 , f xi ) ∈ q ∪ {( f a, f b )} ⊆ q. Since f x0 = f x = f y = f xm by assumption, we have a circle of elements of q and by transitivity of q we get ( f b , f a) ∈ q in contradiction to ( f a, f b ) ∈ / q−1 , i.e. this case can be excluded. Therefore (x, y) ∈ β. (b) directly follows from (a), since for maximal q the conditions of (a) would imply the contradiction q = (q ∪ {(a, b )})tra , therefore ( f a, f b ) ∈ q \ q−1 must imply (a, b ) ∈ q \ q−1 and by induction also ( f n a, f n b ) ∈ q \ q−1 implies (a, b ) ∈ q \ q−1 . Definition 4.5 A quasiorder q ∈ Quord(A) is called quasilinear if every two elements are comparable with respect to q, i.e. q ∪ q−1 = ∇. Note that q is quasilinear iff q/q0 is a linear order on the factor set A/q0 (where q0 = q ∩ q−1 , see Section 2.1). For an algebra A we define QLord A := {q ∈ Quord A | q is quasilinear}. Theorem 4.6 Let A be a rooted algebra. (i) For any q ∈ Quord A we have {λ ∈ QLord A | q ⊆ λ, λ0 = q0 } = q0 ∪ (x, y) ∈ A2 | ∃ n ∈ N : ( f n x, f n y) ∈ q \ q0 . (ii) For any q ∈ Max Q(β), β ∈ TA we have q= {λ ∈ QLord A | q ⊆ λ} = {λ ∈ QLord A | q ⊆ λ, λ0 = q0 }, i.e., each quasiorder which is maximal in Q(β), is the intersection of its quasilinear extensions. Proof (i) Since A/q0 is also rooted, Theorem 4.3 gives that 0 = q/q { ∈ Lord(A/q0 ) | q/q0 ⊆ }. (◦ ) Since any λ ∈ Quord A with q ⊆ λ and λ0 = q0 is of the form λ = {(x, y) | ([x]q0 , [y]q0 ) ∈ } (for some ∈ Lord(A/q0 ) with q/q0 ⊆ ), the lifting of (◦ ) in A2 easily gives the desired formula.
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(ii) If q ∈ Max Q(β), then the right side of (i) equals q by Lemma 4.4(b). Thus q ⊆ (i) {λ ∈ QLord A | q ⊆ λ} ⊆ {λ ∈ QLord A | q ⊆ λ, λ0 = q0 } = q. Proposition 4.7 In a rooted algebra A, for any complete type β ∈ TA we have β max =
λ ∈ QLord A | β ⊆ λ = λ .
Proof
⊆ λ = λ, consequently the left side is contained “⊆” Let β ⊆ λ = λ. Then β max = β in the right one. “⊇” From Theorem 4.6(ii) (and Corollary 3.9) we get β max = {λ ∈ QLord A | max β max ⊆ λ, λ0 = β0max }. We ⊆ λ, λ0 = β0max implies β ⊆ λ = shall show that β max
λ, consequently β ⊇ {λ ∈ QLord A | β ⊆ λ = λ} and the proof is finished. λ = λ. Clearly λ ⊆ λ. Let Thus assume β max ⊆ λ, λ0 = β0max , it remains to show n n
(a, b ) ∈ λ. Then there is n ∈ N with ( f a, f b ) ∈ λ \ . Since λ is quasilinear, we have (a, b ) ∈ λ (and we are done) or (b , a) ∈ λ. In the latter case we conclude max
0 = β
0 ⊆ ( f n b , f n a) ∈ λ \ , thus ( f n a, f n b ) ∈ λ0 \ , hence (a, b ) ∈ λ0 = β =β 0 λ and we are also done. A posteriori we see that Theorem 4.3 in turn can be obtained from Theorem 4.6(ii) and Proposition 4.7, since λ0 = is equivalent to λ ∈ Lord A. Remark 4.8 By Theorem 4.6 we know that the maximal elements in each semiinterval Q(β) are intersections of quasilinear quasiorders. It arises the question which intersections of quasilinear quasiorders provide the maximal elements of Q(β). Thus let {λi | i ∈ I} ⊆ QLord A be a set of quasilinear quasiorders and let β := q ∩ ker f be the type of the intersection q := i∈I λi . Then q is a maximal element in the semi-interval Q(β) if and only if there exists a subset I ⊆ I such that λ = λ and for every proper subset J ⊂ I we have β = i∈I i i∈I i i∈J λi ∩ ker f .
5 Properties of the Lattices Quord A and Con A In this section we investigate properties of the lattices Quord A and Con A for rooted monounary algebras (for simplicity we mostly consider finite algebras). We will show that these lattices are glued sums of intervals in Quord(A) and Equ(A), respectively. Recall that Quord A is a complete sublattice of Quord(A), and Con A is a complete sublattice of Equ(A). For any lattice L = (L, ≤) and x, y ∈ L, we write x ≺L y if x is covered by y (i.e. x is a lower neighbour of y). For a monounary algebra A = (A, f ) and q ∈ Quord A we define the relations f (q) := {( f x, f y) | (x, y) ∈ q}, f (q) := ∪ f (q)tra , and f −1 (q) := (x, y) ∈ A2 | ( f x, f y) ∈ q .
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Lemma 5.1 For q, q1 , q2 ∈ Quord A and ε ∈ Con A we have: f (ε), f −1 (ε) ∈ Con A, (i) f (q), f −1 (q) ∈ Quord A and −1 −1 (ii) f (q) ⊆ q ⊆ f ( f (q)) ⊆ f (q), (iii) f (q1 ) ⊆ q2 ⇐⇒ q1 ⊆ f −1 (q2 ), (iv) For any α ∈ Quord(A) with f (q) ⊆ α ⊆ f −1 ( f (q)) we have α ∈ Quord A (and therefore, α ∈ Con A whenever α ∈ Equ(A)). We omit the straightforward proof of this lemma. Remark 5.2 Condition Lemma 5.1(iv) says in particular that, for q ∈ Quord A, the interval [ f (q), f −1 ( f (q))]Quord(A) in Quord(A) coincides with the interval −1 [ f (q), f ( f (q))]Quord A in Quord A (analogously for q ∈ Con A, Equ(A) and Con(A)). An interpretation for Lemma 5.1 in terms of polarities will be given as an application below. Let q(a, b ) and q f (a, b ) denote the least quasiorder on A and the least compatible quasiorder of (A, f ), respectively, containing the pair (a, b ) ∈ A2 . Clearly q(a, b ) = ∪ {(a, b )}, and it is easy to check that q f (a, b ) = {( f n a, f n b ) | n ∈ N}tra ∪ . Notice, that for any q1 , q2 ∈ Quord(A) with q1 ≺ q2 , there exists an (a, b ) ∈ A2 such that q2 = q1 ∨ q(a, b ) = (q1 ∪ {(a, b )})tra . Proposition 5.3 Let A = (A, f ) be a f inite rooted monounary algebra and q1 , q2 ∈ Quord A. If q1 ≺ q2 in Quord A (in Con A, resp.), then q1 , q2 ∈ [ f (q2 ), f −1 ( f (q2 ))], and q1 ≺ q2 holds in Quord(A) (in Equ(A), resp.), too. Proof (i) If q1 ≺ q2 holds in Quord A, then there exists (a, b ) ∈ q2 \ q1 such that q2 = q1 ∨ q f (a, b ). Hence q2 = (q1 ∪ q f (a, b ))tra = (q1 ∪ {( f n a, f n b ) | n ∈ N})tra . As A is rooted, it satisfies Section 2.4(*) and there exists a minimal number k > 0 such / q1 . This implies c = d and that ( f k a, f k b ) ∈ q1 . Hence (c, d) := ( f k−1 a, f k−1 b ) ∈ (c, d) ∈ q2 \ q1 . Since q1 < q1 ∨ q f (c, d) ≤ q2 and q1 ≺ q2 , we get q2 = q1 ∨ q f (c, d). Clearly, ( f c, f d) = ( f k a, f k b ) ∈ q1 , whence we obtain ( f n c, f n d) ∈ q1 , for all n > 1. Thus we get q2 = q1 ∨ q f (c, d) = (q1 ∪ {( f n c, f n d) | n ∈ N})tra = (q1 ∪ {(c, d)})tra . Next we prove f (q2 ) ⊆ q1 : Indeed, take (x, y) ∈ q2 . Then there exist z0 , z1 , . . . , zs ∈ A with x = z0 , y = zs , such that for each i ∈ {1, . . . , s} either (zi−1 , zi ) ∈ q1 or (zi−1 , zi ) = (c, d) holds. Then ( f zi−1 , f zi ) ∈ q1 , for all i ∈ {1, . . . , s}. This yields ( f x, f y) = ( f z0 , f zs ) ∈ q1 , so we get f (q2 ) = {( f (x), f (y)) | (x, y) ∈ q2 }tra ∪ ⊆ q1 . −1 Thus q1 , q2 ∈ [ f (q2 ), f ( f (q2 ))] by Lemma 5.1(ii). As by Remark 5.2 f (q2 ) and f (q2 )) determine the same intervals in Quord A and Quord(A), q1 ≺ q2 holds f −1 ( in Quord(A), too. Finally, take q1 , q2 ∈ Con A with q1 ≺ q2 in Con A. Now, since Quord A is finite and Con A is a sublattice of it, there exists a q ∈ Quord A that satisfies q1 ≤ q ≺ q2 in Quord A. Then, obviously, q1 ≤ q−1 ≺ q2 , and the congruence q0 = q ∩ q−1 satisfies q1 ≤ q0 < q2 in Con A. Since q1 ≺ q2 , we get q1 = q0 . As q ≺ q2 and q−1 ≺ q2 hold in Quord A, according to the above proof we get f (q2 ) ⊆ q and f (q2 ) ⊆ q−1 . Hence f (q2 ) ⊆ q0 = q1 . Clearly, f (q2 ) ⊆ q1 and Lemma 5.1(ii)
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imply q1 , q2 ∈ [ f (q2 ), f −1 ( f (q2 ))]. Then q1 ≺ q2 also holds in Equ(A), according to Remark 5.2. Polarities and Compatible Tolerances Let P = (P, ≤) be a poset. A pair (K, H) of maps K : P → P, H : P → P is called a polarity (or strong adjunction) on P if for any x, y ∈ P, we have K(x) ≤ x and K(x) ≤ y ⇐⇒ x ≤ H(y)
(5.1)
This yields also x ≤ H(x) for all x ∈ P. If (K, H) is a polarity on a lattice L = (L, ≤), then K is a join-homomorphism and H is a meet-homomorphism of L, i.e. for all x 1 , x2 ∈ L K(x1 ∨ x2 ) = K(x1 ) ∨ K(x2 ),
H(x1 ∧ x2 ) = H(x1 ) ∧ H(x2 )
(5.2)
Polarities on a lattice L can be characterized also as pairs (K, H) satisfying Eq. 5.2 and K(x) ≤ x ≤ H(x)
(5.3)
K(H(x)) ≤ x ≤ H(K(x))
(5.4)
for all x ∈ L. In the case of a finite lattice L, there is a one-to-one correspondence between its compatible tolerances and polarities given in the next Lemma 5.4. A compatible tolerance of L is a reflexive and symmetric relation τ ⊆ L2 compatible with the lattice operations. A block of τ is a subset B ⊆ L maximal with respect to the property B2 ⊆ τ . If L is finite, then each block B is an interval [u, v]L = {x ∈ L | u ≤ x ≤ v} in L (cf. [3]). Lemma 5.4 Let L = (L, ∨, ∧) be a f inite lattice. (i) Let (K, H) be a polarity on L. Then the relation τ ⊆ L2 def ined by τ :=
[K(c), H(K(c))]2L
(5.5)
c∈L
is a compatible tolerance on L and the blocks of τ are intervals of the form [K(c), H(K(c))]L for some c ∈ L. (ii) Let τ be a compatible tolerance on L. Then (K, H) given by K(x) :=
{y ∈ L | (x, y) ∈ τ },
H(x) :=
{y ∈ L | (x, y) ∈ τ }
(5.6)
is a polarity, and we have τ = {(x, y) | K(x ∨ y) ≤ x ∧ y} where τ corresponds to (K, H) according to (i). Proof (i) and (ii) are known in the literature (see e.g. [6] or [8, 5.3.25]), it remains to prove the assertion concerning the blocks of τ : Let L be finite and B a block of τ . Then B = [u, v] for some u, v ∈ L. As (u, v) ∈ τ , by Eq. 5.5 there is a c ∈ L with K(c) ≤ u ≤ v ≤ H(K(c)) and (x, y) ∈ τ for all x, y ∈ [K(c), H(K(c))]. So, by the maximality property of a tolerance block it follows B = [u, v] = [K(c), H(K(c))].
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Application According to Lemma 5.1(i)–(iii) the mappings K : q → f (q), k : ε → f (ε),
H : q → f −1 (q) (q ∈ Quord A) and their restrictions h : ε → f −1 (ε) (ε ∈ Con A)
are polarities on the lattices Quord A and Con A, respectively. Consequently, (K, H) and (k, h) satisfy also Eqs. 5.3 and 5.4. Now we can apply Lemma 5.4(i): The following relations T ⊆ (Quord A)2 and S ⊆ (Con A)2 are compatible tolerances
2 T := f (q)) Quord A (5.7) f (q), f −1 ( q∈Quord A
S :=
ε∈Con A
f (ε), f −1 ( f (ε))
2 Con A
.
(5.8)
Glued Tolerances and Sums Let τ be a compatible tolerance of a lattice L. Then we consider the factor lattice L/τ defined on the blocks of τ as follows: B1 , B2 ∈ L/τ , B1 ≤ B2 ⇐⇒ B1 ∧ B2 = {x ∧ y | x ∈ B1 , y ∈ B2 } ⊆ B1 [1]. The relation τ is called a glued tolerance, if B1 ≺L/τ B2 implies B1 ∩ B2 = ∅ for B1 , B2 ∈ L/τ . If L is finite, then any glued tolerance τ of L contains all covering pairs x ≺L y ; moreover, this property characterizes glued tolerance relations. Therefore, the compatible tolerance generated by all covering pairs, i.e. by the set {(x, y) ∈ L2 | x ≺L y} is the least glued tolerance of L, called the skeleton of L (cf. [3]) which will be denoted by (L). We say that a lattice L is glued by lattices Li (i ∈ I), or that L is a glued sum of the lattices Li , if there exists a glued tolerance τ ⊆ L2 such that the blocks of τ are just the lattices Li , i.e. L/τ = {Li | i ∈ I} (see [5, 10]). In what follows we mainly consider finite rooted algebras (i.e. acyclic connected monounary algebras, cf. Section 2.4(*)), and we collect some properties of the tolerances T and S defined in Eqs. 5.7 and 5.8. Proposition 5.5 Let A = (A, f ) be a f inite rooted monounary algebra. Then we have: T and S are glued tolerances on Quord A and Con A, respectively, and S = f (q), T ∩ (Con A)2 . Each block of T and S is an interval of the form [ f −1 ( f (q))]Quord(A) (with q ∈ Quord A ) and [ f (ε), f −1 ( f (ε))]Equ(A) (with ε ∈ Con A ), respectively. (ii) Quord A is glued by some intervals of Quord(A), and Con A is glued by some intervals of Equ(A). (i)
Proof Let q1 ≺ q2 be a covering pair in Quord A, respectively in Con A. Then, by using the calculations of the proof of Proposition 5.3 and formulas 5.7 and 5.8, we get (q1 , q2 ) ∈ T, respectively (q1 , q2 ) ∈ S. In view of Remark 5.2, the blocks of T and S are full intervals in the lattices Quord(A) and Equ(A), respectively. The inclusion S ⊆ T ∩ (Con A)2 is also clear. Now we prove T ∩ (Con A)2 ⊆ S: Take (ε1 , ε2 ) ∈ T ∩ (Con A)2 . Then ε1 ∨ ε2 ∈ Con A, and f (q) ⊆ εi ⊆ f −1 ( f (q)), for some q ∈ Quord A (i ∈ {1, 2}). Since (K, H) is a polarity on Quord A, we get f (εi ) ⊆ f (q), and hence f (ε1 ∨ ε2 ) = f (ε1 ) ∨ f (ε2 ) ⊆ f (q) ⊆ εi . As relation (Eq. 5.4) implies ε1 ∨ ε2 ⊆ f −1 ( f (ε1 ∨ ε2 )), we get f (ε1 ∨ ε2 ) ⊆ εi ⊆ f −1 ( f (ε1 ∨ ε2 )) (i ∈ {1, 2}). Thus (ε1 , ε2 ) ∈ S. (ii) is an obvious consequence of (i).
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For a complete lattice L and any x ∈ L we define x+ := {y ∈ L | y ≺ x}. If L is a finite lattice, then by the definition of the skeleton (L) we have (x+ , x) ∈ (L). Applying Proposition 5.5, we obtain: Corollary 5.6 Let A be a f inite rooted monounary algebra. Then any q ∈ Quord A and ε ∈ Con A satisfy f (q) ≤ q+ ≤ f −1 ( f (q)) in Quord A and f (ε) ≤ ε+ ≤ f −1 ( f (ε)) in Con A. A finite lattice L is called lower locally distributive if, for any x ∈ L, the interval [x+ , x] is a distributive lattice. In [2] it is proved that for any finite set A and any partial order r ⊆ A2 , the interval [, r] in Quord(A) is lower locally distributive. A lattice L is (upper) semimodular if, for any a, b ∈ L, a ∧ b ≺ a implies b ≺ a ∨ b . An element j = 0L is ∨-irreducible if j = x ∨ y implies either j = x or j = y (for x, y ∈ L). Let J(L) denote the set of ∨- irreducible elements of L. In a lattice L of finite height any j ∈ J(L) has a unique lower cover j∗ ≺ j. A semimodular lattice L of finite height is called strongly semimodular if, for any j ∈ J(L) and x ∈ L, j ≤ j∗ ∨ x implies j ≤ x. Atomistic, algebraic, semimodular lattices are called geometric lattices. Geometric lattices of finite height are strongly semimodular (see e.g. [12]). In [10] it is proved that a finite lattice L is strongly semimodular if and only if L is glued by geometric lattices. Theorem 5.7 Let A be a f inite rooted monounary algebra. Then Con A is a strongly semimodular lattice, and any compatible partial order r of A the interval [, r] in Quord A is a lower locally distributive lattice. Proof Define q⊕ := { p ∈ Quord(A) | p ≺ q} for any q ∈ Quord A . Then q⊕ ≤ { p ∈ Quord A | p ≺ q} =: q+ , and hence [q+ , q] is a sublattice of the interval [q⊕ , q] in Quord(A). Take an arbitrary q ≤ r. Since, in view of [2], the interval [, r] in Quord(A) is lower locally distributive, [q⊕ , q] is a distributive lattice. Therefore, [q+ , q] in Quord A is also distributive for each q ≤ r, and this means that [, r] in Quord A is a lower locally distributive lattice. Further, since any interval of Equ(A) is a geometric lattice, in view of Proposition 5.5, Con A is glued by geometric lattices. As Con A is finite, it is a strongly semimodular lattice, according to [10]. Corollary 5.8 Let A be a f inite rooted monounary algebra. Then the relation S (cf. Eq. 5.8) is just the skeleton tolerance of Con A, i.e. S = (Con A), and its blocks are isomorphic to f inite direct products of f inite partition lattices. Proof As S is a glued tolerance on Con A, (Con A) ⊆ S. To prove the converse inclusion, take ε1 , ε2 ∈ Con A with (ε1 , ε2 ) ∈ S. Then there is a block B of S such that ε1 , ε2 ∈ B, and B = [α, β] ⊆ Con A for some α, β ∈ Con A, α ≤ β. In view of Proposition 5.5(ii) this interval of Con A is the same as the interval I := [α, β]Equ(A) in Equ(A). Since I is a geometric lattice, it is dually atomistic, too. As the dual atoms of I are dual atoms of [α, β]Con A , they are those ξ ∈ Con A that satisfy α ≤ ξ ≺ β in Con A. Hence α = {ξ ∈ Con A | α ≤ ξ ≺ β}, and since (ξ, β) ∈ (Con A) for each ξ ≺ β, we get (α, β) = {(ξ, β) | α ≤ ξ ≺ β} ∈ (Con A). The blocks of (Con A) are convex sets, thus α ≤ εi ≤ β (i ∈ {1, 2}) implies (ε1 , ε2 ) ∈ (Con A). This proves S = (Con A). Since any block of S is an interval in the finite lattice Equ(A), the
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second assertion follows from the fact that any interval of Equ(A) is isomorphic to a direct product of partition lattices. Acknowledgement The authors thank the referee for helpful hints and suggestions that considerably improved the presentation of the results.
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