ON ENDOMORPHISM MONOIDS OF PARTIAL ORDERS AND CENTRAL RELATIONS 1

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1 Novi Sad J. Math. Vol. 38, No. 1, 2008, ON ENDOMORPHISM MONOIDS OF PRTIL ORDERS ND CENTRL RELTIONS 1 Dragan Mašulović 2 bstract. In this paper we characterize pairs of Rosenberg s (ρ, σ) with the property that the endomorphism monoid of one of the s is properly contained in the endomorphism monoid of the other. We focus on the situations where one of the s is a partial order, or a central. MS Mathematics Subject Classification (2000): 0835, 0606 Key words and phrases: maximal clones, endomorphism monoids, Rosenberg s, partial orders, central s 1. Introduction In this paper we continue the work from [4, 5, 6, 7] on the description of all pairs (ρ, σ) of Rosenberg s satisfying End{ρ} End{σ}. Here, we focus on the situations where one of the s is a partial order, or a central. Results presented here, together with the results of [4, 5, 6, 7] cover all but two cases (see Table 1), bringing us, therefore, closer to the complete characterization. This line of research was motivated by the paper [1], where the completeness for some special structures (concrete near-rings) was investigated using techniques from clone theory. It appeared that unary parts (traces) of the maximal clones that contain the operation correspond to the maximal near-rings containing the identity map. Moreover, if for every two distinct unary parts M (1) i and M (1) j of such maximal clones we have M (1) i M (1) j, then every unary part is a maximal near ring. It is natural to ask what goes on in the general case, i.e. what the ship between any two traces of maximal clones on a finite set is. s it was expected, the width of this poset is doubly exponential, but it was rather surprising to find out that its height is equal to the size of the underlying set. Moreover, it turns out that the structure of this poset is quite rich. 2. Preliminaries O (n) Throughout the paper we assume that is a finite set and 3. Let denote the set of all n-ary operations on (so that O(1) = ) and let 1 Supported by the Grant Lattice methods and applications No /2007 of the Secretariat for Science and Technological Development of the utonomous Province Vojvodina 2 Department of Mathematics and Informatics, University of Novi Sad, Serbia, Trg Dositeja Obradovića 4, Novi Sad, masul@im.ns.ac.yu

2 112 D. Mašulović O = n 1 O(n) denote the set of all finitary operations on. For F O let F (n) = F O (n) be the set of all n-ary operations in F. set C O of finitary operations is a clone of operations on if it contains all projection maps πi n : n : (x 1,..., x n ) x i and is closed with respect to composition of functions in the following sense: whenever g C (n) and f 1,..., f n C (m) for some positive integers m and n then g(f 1,..., f n ) C (m), where the composition h = g(f 1,..., f n ) is defined by h(x 1,..., x m ) = g(f 1 (x 1,..., x m ),..., f n (x 1,..., x m )). For a clone C, the unary part C (1) of C, will be referred to as the trace of C. Let R (n) denote the set of all n-ary s on (including the empty ) and let R = n 1 R(n) ρ is a nonempty in R (n) denote the set of all finitary s on. If we say that its arity is n and write ar(ρ) = n. The arity of the empty is undefined. We say that an n-ary operation f preserves an h-ary ρ if the following holds: a 11 a 21 a h1 a 12 a 22 a h2 a 1n a 2n..,..,...,.. ρ implies a hn For a set Q of s and for a set F of operations let f(a 11, a 12,..., a 1n ) f(a 21, a 22,..., a 2n ).. ρ. f(a h1, a h2,..., a hn ) Pol Q = {f O : f preserves every ρ Q} Inv F = {ρ R : every f F preserves ρ}. Let Pol (n) Q = (Pol Q) O (n). For an h-ary θ h and a unary operation f it is convenient to write f(θ) = {(f(x 1 ),..., f(x h )) : (x 1,..., x h ) θ}. Then clearly f preserves θ if and only if f(θ) θ. It follows that Pol (1) Q is the endomorphism monoid of the al structure (, Q). Therefore instead of Pol (1) Q we simply write End Q. We shall omit the subscript in Pol Q, End Q and Inv F and write simply Pol Q, End Q and Inv F. If, however, F O B and Q R B for some B we shall keep the subscript and write Pol B Q, End B Q and Inv B F to indicate that we restrict our attention to operations and s on B. For a ρ of arity k, let ρ irr denote the irreflexive part of ρ, that is, the set of all (x 1,..., x k ) ρ such that x i x j for all i j. ρ is irreflexive if ρ = ρ irr. Let B. We say that f O (1) ρ R (k) irreflexively preserves an irreflexive B if whenever (b 1,..., b k ) ρ and (f(b 1 ),..., f(b k )) (B k ) irr then (f(b 1 ),..., f(b k )) ρ. We say that an irreflexive ρ is an irreflexive

3 On endomorphism monoids of partial orders and central s 113 invariant of F O (1) on B if every operation in F irreflexively preserves ρ. The set of all irreflexive invariants of F on B will be denoted by Irr B F. If the underlying set is finite and has at least three elements, then the lattice of clones has cardinality 2 ℵ 0. However, one can show that the lattice of clones on a finite set has a finite number of coatoms, called maximal clones, and that every clone distinct from O is contained in one of the maximal clones. One of the most influential results in clone theory is the explicite characterization of the maximal clones, obtained by I. G. Rosenberg as the culmination of the work of many mathematicians. It is usually stated in terms of the following six classes of finitary s on (the so-called Rosenberg s). (R1) Bounded partial orders. These are partial orders on with a least and a greatest element. (R2) Nontrivial equivalence s. These are equivalence s on distinct from := {(x, x) : x } and 2. (R3) Permutational s. These are s of the form {(x, π(x)) : x }, where π is a fixpoint-free permutation of with all cycles of the same length p, where p is a prime. (R4) ffine s. For a binary operation on let λ := {(x, y, u, v) 4 : u v = x y}. ρ is called affine if there is an elementary abelian p-group (,,, 0) on such that ρ = λ. Suppose now that is an elementary abelian p-group. Then it is wellknown that f Pol{λ } if and only if f(x 1 y 1,..., x n y n ) = f(x 1,..., x n ) f(y 1,..., y n ) f(0,..., 0) for all x i, y i. In case f is unary, this condition becomes f(x y) = f(x) f(y) f(0). (R5) Central s. ll unary s are central s. For central s ρ of arity h 2 the definition is as follows: ρ is said to be totally symmetric if (x 1,..., x h ) ρ implies (x π(1),..., x π(h) ) ρ for all permutations π, and it is said to be totally reflexive if (x 1,..., x h ) ρ whenever there are i j such that x i = x j. n element c is central if (c, x 2,..., x h ) ρ for all x 2,..., x h. Finally, ρ h is called central if it is totally reflexive, totally symmetric and has a central element. (R6) h-regular s. Let Θ = {θ 1,..., θ m } be a family of equivalence s on. We say that Θ is an h-regular family if every θ i has precisely h blocks, and additionally, if B i is an arbitrary block of θ i for i {1,..., m}, then m i=1 B i.

4 114 D. Mašulović n h-ary ρ h is h-regular if h 3 and there is an h-regular family Θ such that (x 1,..., x h ) ρ if and only if for all θ Θ there are distinct i, j with x i θx j. Note that regular s are totally reflexive and totally symmetric. Theorem 2.1. (Rosenberg [8]) clone M of operations on a finite set is maximal if and only if there is a ρ from one of the classes (R1) (R6) such that M = Pol{ρ}. Every central ρ can be written as C ρ R ρ T ρ, where C ρ is the central part of ρ which consists of all the tuples of distinct elements containing at least one central element, R ρ is the reflexive part of ρ which consists of all the tuples (x 1,..., x k ) such that there are i j with x i = x j, and T ρ is the tail of ρ and consists of all the tuples (x 1,..., x k ) such that x 1,..., x k are distinct non-central elements. The center of ρ will be denoted by Z(ρ). This is the set of all central elements of ρ. For a ρ by dom(ρ) we denote the domain of ρ, that is, the set D such that ρ D ar(ρ). For most s in this paper the domain is obvious. However, for the tail T ρ of a central ρ we set dom(t ρ ) = \ Z(ρ). There is another way to look at regular s. Given a finite set, 3, and an h-regular family Θ = {θ 1,..., θ m } on, let R Θ = {(x 1,..., x h ) : ( θ Θ)( i j)x i θx j } denote the corresponding h-regular. We define the elementary (h, m)- Ψ h,m on {1,..., h} m in the following way: Ψ h,m = {( a 1 1.,..., a 1 m a h 1. a h m ) : ( i {1,..., m})( j k)a j i = ak i Note that the elementary (h, m)- is the h-regular on {1,..., h} m defined by the h-regular family Θ = {θ1,..., θm}, where {( b 1 b θi = ) },. 1. : b 1 i = b 2 i. b 1 m b 2 m Then, there exists a surjective mapping λ : {1,..., h} m such that R Θ = {(x 1,..., x h ) : (λ(x 1 ),..., λ(x h )) Ψ h,m }. Conversely, for every surjective mapping λ : {1,..., h} m the is an h-regular. {(x 1,..., x h ) : (λ(x 1 ),..., λ(x h )) Ψ h,m } }.

5 On endomorphism monoids of partial orders and central s 115 The complete characterization of all mappings that preserve regular s can be found in [3]. We present this result without proof. Denote by x (i) the i-th coordinate of the tuple x {1,..., h} m. Let f be an n-ary function on the set. We define the function f i : n {1,..., h} in the following way: f i (x 1,..., x n ) := λ i (f(x 1,..., x n )), where λ i (x) = λ(x) (i) = π i λ(x). Proposition 2.2. ([3]) n n-ary function f on a set preserves an h-regular R Θ if and only if for each i either f i has at most h1 distinct values or there exist a permutation s on {1,..., h}, a j {1,..., n} and a v {1,..., m} such that f i (x 1,..., x n ) = s(λ v (x j )). The results we have obtained up to now, including the results obtained in this paper, are summarized in Table 1. The entries in this table are to be interpreted in the following way: we write if End ρ End σ for every pair (ρ, σ) of s of the indicated type; we write if there is a complete characterization of the situation End ρ End σ; we write? if there is no definite answer, but the reference below the question mark contains some partial results. 3. Bounded partial orders Let (, ) be a partially ordered set. Recall that S is a retract of if there is an idempotent monotonous map f : such that f() = S. We shall say that S is a pseudoretract of if there is a (not necessarily idempotent) monotonous map f : such that f() = S. Thus, pseudoretracts of are nothing but substructures of that are at the same time homomorphic images of. If ρ S k and θ k are arbitrary s, we say that (S, ρ) is a pseudoretract of (, θ) if there is a monotonous map f : such that f() = S and f(θ) = ρ. We start with the characterization of the ship between bounded partial orders and regular s. In we obtained the following partial result: Proposition 3.1. ( Proposition 4.25) Let be a bounded partial order. If R Θ is an h-regular defined by Θ = {θ 1,..., θ m } where m 2, then End( ) End R Θ. Therefore, if End( ) End R Θ then Θ = {θ}. We shall now complete the characterization. Let θ be an equivalence on and let S be a subset of. We write S/θ to denote S/(θ S 2 ), the set of equivalence classes of the restriction of θ to S.

6 116 D. Mašulović ρ σ Bounded partial order Equivalence Permutational ffine Unary central k ary central, k 2 h regular Bounded partial order Prop. 3.4 Prop. 3.2 Equivalence [4] Permutational ffine Unary central k ary central, k 2 Prop. 4.6 Prop h regular?? [4] Table 1: Summary of some partial results for End ρ End σ

7 On endomorphism monoids of partial orders and central s 117 Proposition 3.2. Let (, ) be a partially ordered set, let Θ be an h-regular family, h 3, and R Θ the regular generated by the regular family Θ. Then End( ) End R Θ if and only if Θ = {θ} and for every pseudoretract S of, if S/θ = h and (S, ρ) is a pseudoretract of (, θ) then ρ = θ S. Proof. ( ) Let End( ) End R Θ. Then by Proposition 3.1 we have Θ = {θ} for some equivalence θ. Let (S, ρ) be a pseudoretract of (, θ) and let S/θ = h. Then there is a monotonous map f : such that f() = S and f(θ) = ρ. Let B 1,..., B h be the blocks of θ. Since f End( ) End R Θ and f()/θ = S/θ = h it follows from Proposition 2.2 that there exists a permutation α : {1,..., h} {1,..., h} such that f(b i ) B α(i) for all i. So, f End θ and hence ρ = f(θ) θ S. Let us show that f(θ) = θ S. Take any (u, v) θ S. Then u, v B α(j) for some j. Since f(b i ) B α(i) for all i, from u, v S = f() it follows that there exist p, q B j such that (f(p), f(q)) = (u, v). Therefore, ρ = f(θ) = θ S. ( ) ssume that Θ = {θ} and that for every pseudoretract S of, if S/θ = h and (S, ρ) is a pseudoretract of (, θ) then ρ = θ S. Take any f End( ). If f()/θ < h then f obviously preserves R Θ, so assume that f()/θ = h. Let S = f() and ρ = f(θ). Then (S, ρ) is a pseudoretract of (, θ) whence ρ = θ S. So, f preserves θ and consequently f End R Θ since Θ = {θ}. s for the ship between endomorphism monoids of bounded parital orders and central s, in and we obtained the following: Proposition 3.3. (Proposition 4.10, Proposition 4.19 ) Let (, ) be a partially ordered set. If ρ is a unary central then End( ) End ρ. If ρ is a binary central then End( ) End ρ if and only if is not a chain and ρ = ( ) ( ) 1. We shall now consider central s of arity h 3. For = D let Cen h (D) = {(x 1,..., x h ) ( h ) irr : there exists an i such that x i D}. Proposition 3.4. Let be a bounded partial order and σ a central of arity h 3. Then End( ) End σ if and only if T σ Irr \Z(σ) End( ), and for every pseudoretract S of, if S \ Z(σ) h and (S, D) is a pseudoretract of (, Z(σ)) then Cen h S\Z(σ)(D) T σ. Proof. ( ) Let B = \Z(σ). Suppose that T σ / Irr B End( ). Then there exist f End( ) and (b 1,..., b h ) T σ such that (f(b 1 ),..., f(b h )) (B h ) irr but (f(b 1 ),..., f(b h )) / T σ. Let us show that in this case f does not preserve σ. It is obvious that (b 1,..., b h ) T σ σ. On the other hand, (f(b 1 ),..., f(b h )) / σ since the tuple consists of distinct noncentral elements but it does not belong to T σ. Therefore, f End( ) \ End σ.

8 118 D. Mašulović Suppose now that there is a pseudoretract (S, D) of (, Z(σ)) such that S \ Z(σ) h but Cen h S\Z(σ)(D) T σ. Let f be a monotonous map such that f() = S and f(z(σ)) = D, and take any (b 1,..., b h ) Cen h S\Z(σ)(D) \ T σ. Note that (b 1,..., b h ) / σ since the tuple consists of distinct noncentral elements but it does not belong to T σ. lso, there is an i such that b i D, say b 1 D. Since f() = S and f(z(σ)) = D there exist c Z(σ) and a 2,..., a h such that f(c) = b 1 and f(a i ) = b i for all i 2. Now, (c, a 2,..., a h ) σ but (f(c), f(a 2 ),..., f(a h )) = (b 1, b 2,..., b h ) / σ, so f End( ) \ End σ. ( ) Suppose that End( ) End σ and take any f End( ) \ End σ. Then there is a (a 1,..., a h ) σ such that (b 1,..., b h ) / σ, where b i = f(a i ) for all i. Therefore, all the b i s are distinct noncentral elements and consequently all the a i s are distinct. If (a 1,..., a h ) T σ then T σ / Irr \Z(σ) End( ) since f End( ), all the b i s are distinct noncentral elements but (b 1,..., b h ) / T σ. ssume now that (a 1,..., a h ) C σ. Let S = f() and D = f(z(σ)). Then clearly (S, D) is a pseudoretract of (, Z(σ)), but (b 1,..., b h ) Cen h S\Z(σ)(D) \ T σ. If σ is a central with T σ = we have the following simpler version of the above result: Proposition 3.5. Let σ be a central of arity h 3 such that T σ =. Then End( ) End σ if and only if for every pseudoretract S of, if S \ Z(σ) h and (S, D) is a pseudoretract of (, Z(σ)) then D Z(σ). Proof. ( ) Let End( ) End σ, let S be a pseudoretract of such that S \ Z(σ) h and let (S, D) be a pseudoretract of (, Z(σ)). Then there is a monotonous map f such that f() = S and f(z(σ)) = D. Since f End( ) End σ and f() \ Z(σ) h it follows that f(z(σ)) Z(σ). Therefore, D Z(σ). ( ) Take any f End( ), let S = f() and D = f(z(σ)). Then (S, D) is a pseudoretract of (, Z(σ)). If f() \ Z(σ) < h then it is easy to see that f preserves σ. If, however, f() \ Z(σ) h then according to the assumption D Z(σ) since (S, D) is a pseudoretract of (, Z(σ)). Therefore, f(z(σ)) Z(σ) and consequently f preserves σ. 4. Central s In we solved the following special case of the problem of comparing endomorphism monoids of central s: Proposition 4.1. Let ρ and σ be distinct central s. (a) If σ is a unary central then End ρ End σ [7, Lemma 3.2]. (b) Suppose that ar(σ) 2 and T ρ =. Then End ρ End σ if and only if ar(ρ) < ar(σ), Z(ρ) = Z(σ) and T σ = [7, Proposition 4.1].

9 On endomorphism monoids of partial orders and central s 119 This proposition handles the cases where at least one of the central s is unary, or T ρ =. In this section we consider the remaining case where ar(ρ) 2, ar(σ) 2 and T ρ. The main problem in the analysis of central s comes from the fact that the tail of a central can be arbitrary (actually, any totally symmetric irreflexive ). Hence, if ρ and σ are central s and End ρ End σ then T σ should be a sort of an invariant of T ρ, but this does not mean that T σ Inv End T ρ. The paper [6] raises an intriguing point in the analysis of the structure of endomorphism monoids of central s. We are now going to make the approach of [6] explicite and describe the tools in a fashion of Fraïssé-type analysis of first-order structures that will enable us to precisely formulate the feeling that T σ is a sort of an invariant of T ρ. Let ρ be a k-ary and let h > k. We write (a 1,..., a h ) ρ (b 1,..., b h ) to denote that all a 1,..., a h are distinct, all b 1,..., b h are distinct, {a 1,..., a h, b 1,..., b h } dom(ρ), and whenever i 1,..., i k {1,..., h} are k distinct indices such that (a i1,..., a ik ) ρ, then (b i1,..., b ik ) ρ. We say that σ is a Fraïssé filter over ρ if ar(σ) > ar(ρ), and if (a 1,..., a h ) σ and (a 1,..., a h ) (b 1,..., b h ) σ. ρ (b 1,..., b h ) dom(σ) h then Recall that a structure is homogeneous if every isomorphism between two finite substructures extends to an automorphism of the structure. Recently, P. Cameron and J. Nešetřil introduced a notion of homomorphism-homogeneity of structures as a straightforward generalization of homogeneity [2]. Let (, ρ) be a al structure. local homomorphism of (, ρ) is a homomorphism f : (S, ρ S ) (T, ρ T ), where S, T. We say that the structure (, ρ) is homomorphism-homogeneous if every local homorphism extends to an endomorphism, that is, if for every homorphism f : (S, ρ S ) (T, ρ T ), where S, T, there is an endomorphism g End(, ρ) such that f = g S. We say that a ρ is homomorphism-homogeneous if (, ρ) is homomorphism-homogeneous. Lemma 4.2. Let ρ be a homomorphism-homogeneous of arity k 2 and let σ be a totally reflexive of arity h > k with dom(ρ) = dom(σ). The following are equivalent (1) End ρ End σ, (2) σ is a Fraïssé filter over ρ, (3) σ irr Irr End ρ. Proof. (1) (2): Suppose End ρ End σ. Take any (a 1,..., a h ) σ and ρ suppose (a 1,..., a h ) (b 1,..., b h ). Let S = {a 1,..., a h }, T = {b 1,..., b h } and f : S T : a i b i. Then f is a local homomorphism of (, ρ), and by

10 120 D. Mašulović homomorphism-homogeneity of ρ it extends to an endomorphism g End ρ. So, g End σ, and consequently (g(a 1 ),..., g(a h )) σ. But g(a i ) = f(a i ) = b i, for all i. Therefore, (b 1,..., b h ) σ. (2) (3): Let σ be a Fraïssé filter over ρ and let us show that every g End ρ irreflexively preserves σ irr. Take any g End ρ and any (a 1,..., a h ) σ irr such that the g(a i ) s are all distinct. Since g End ρ it follows that ρ (g(a 1 ),..., g(a h )), whence (g(a 1 ),..., g(a h )) σ irr since σ is (a 1,..., a h ) a Fraïssé filter over ρ. (3) (1): ssume that σ irr Irr End ρ, take any g End ρ and let us show that g End σ. Take any (a 1,..., a h ) σ and let b i = g(a i ). If there are i j such that b i = b j then (b 1,..., b h ) σ since σ is totally reflexive. Otherwise, all the b i s are distinct. Since σ irr Irr End ρ and g End ρ we obtain (b 1,..., b h ) σ irr σ. Therefore, g End σ. Lemma 4.3. Every central is homomorphism-homogeneous. Proof. Let ρ be a central and f : (S, ρ S ) (T, ρ T ), where S, T, a local homomorphism of (, ρ). Choose an arbitrary central element c of ρ and define an extension g : of f by { f(x), x S, g(x) = c, otherwise. Then it is easy to see that g preserves ρ. Take any (a 1,..., a k ) ρ. If f(a i ) = f(a j ) for some i j then trivially (f(a 1 ),..., f(a k )) ρ. If all the f(a i ) s are distinct, then all the a i s are also distinct. If there is a j such that a j / S, then f(a j ) = c and (f(a 1 ),..., f(a k )) ρ. Finally, if {a 1,..., a k } S then from the fact that f is a local homomorphism of (, ρ) it follows that (f(a 1 ),..., f(a k )) ρ. Corollary 4.4. Let ρ be a central of arity k 2 and let σ be a totally reflexive of arity h. Then End ρ End σ if and only if h > k and σ is a Fraïssé filter over ρ. Proof. In view of the previous two lemmas it suffices to show that h > k. But from [4, Proposition 5.3] we know that if ρ and σ are totally reflexive and totally symmetric distinct s such that σ contains at least one tuple (b 1,..., b h ) of distinct elements and σ h, where h = ar(σ), then End ρ End σ implies ar(ρ) < ar(σ). lthough rather general, the above statement can be used to derive certain information about central s. Corollary 4.5. Let ρ and σ be central s, k = ar(ρ) 2, h = ar(σ) 2. If End ρ End σ then h > k and Z(ρ) Z(σ).

11 On endomorphism monoids of partial orders and central s 121 Proof. We already know that End ρ End σ implies h > k, [4, Proposition 5.3]. Take any c Z(ρ) and a 2,..., a h. If {c, a 2,..., a h } < h then (c, a 2,..., a h ) σ since σ is totally reflexive. If {a 2,..., a h } Z(ρ) then (c, a 2,..., a h ) σ since the tuple contains a central element of σ. So, assume that {c, a 2,..., a h } = h and {a 2,..., a h } Z(ρ) =. Take any d Z(σ). Then all d, a 2,..., a h are distinct and (d, a 2,..., a h ) σ. Moreover, (d, a 2,..., a h ) ρ (c, a 2,..., a h ) since c is a central element of ρ. Now, from End ρ End σ we get by Corollary 4.4 that σ is a Fraïssé filter over ρ. This together with (d, a 2,..., a h ) σ and (d, a 2,..., a h ) ρ (c, a 2,..., a h ) yields (c, a 2,..., a h ) σ. Let θ be a k-ary and let h > k. We say that d dom(θ) is an h-central element for θ if there exists a B dom(θ) such that B = h 1, d / B and {d} (B k1 ) irr θ. We also say that B is a set of witnesses for d. Proposition 4.6. Let ρ and σ be central s, k = ar(ρ) 2, h = ar(σ) 2, and T ρ. Then End ρ End σ if and only if Z(ρ) Z(σ), h > k, if d dom(t σ ) is an h-central element of T ρ witnessed by {b 2,..., b h } dom(t σ ) then (d, b 2,..., b h ) T σ, and (C σ T σ ) dom(tρ ) is a Fraïssé filter over T ρ. Proof. ( ) Suppose End ρ End σ. Then Z(ρ) Z(σ) and h > k according to Corollary 4.5. Let d dom(t σ ) be an h-central element of T ρ witnessed by {b 2,..., b h } dom(t σ ). Take any c Z(ρ) Z(σ) and consider d, x = c f(x) = x, x {b 2,..., b h } c, otherwise. Then f End ρ. Therefore, f End σ which together with (c, b 2,..., b h ) σ yields (d, b 2,..., b h ) σ. Since d, b 2,..., b h are distinct noncentral elements, we obtain (d, b 2,..., b h ) T σ. Take any (a 1,..., a h ) (C σ T σ ) dom(tρ ) and let (a 1,..., a h ) T ρ (b 1,..., b h ). Take any c Z(ρ) Z(σ) and consider { b i, x = a i, i {1,..., h} f(x) = c, otherwise. Then f End ρ. Therefore, f End σ which together with (a 1,..., a h ) σ yields (b 1,..., b h ) σ. But {b 1,..., b h } dom(t ρ ) so (b 1,..., b h ) (C σ T σ ) dom(tρ).

12 122 D. Mašulović ( ) Suppose End ρ End σ and let h > k and Z(ρ) Z(σ). Take any f End ρ \ End σ and a tuple (a 1,..., a h ) σ such that (f(a 1 ),..., f(a h )) / σ. Let b i = f(a i ), i = 1,..., h. Then {b 1,..., b h } Z(σ) =, all the b i s are distinct and consequently all the a i s are distinct. If there is an i such that a i Z(ρ), then from f End ρ it follows that b i is an h-central element of T ρ witnessed by {b 1,..., b h } \ {b i }. Since {b 1,..., b h } \ Z(σ) we get a contradiction with the third requirement. If {a 1,..., a h } Z(ρ) =, then from f End ρ it follows that (a 1,..., a h ) T ρ (b 1,..., b h ). Since (a 1,..., a h ) belongs to (C σ T σ ) dom(tρ ) and (b 1,..., b h ) does not, we have that (C σ T σ ) dom(tρ) is not a Fraïssé filter over T ρ which contradicts the fourth requirement. Corollary 4.7. Let ρ and σ be central s, k = ar(ρ) 2, h = ar(σ) 2, and T ρ. dditionally, assume Z(ρ) = Z(σ) and let B = dom(t ρ ) = dom(t σ ) = \ Z(ρ). Then End ρ End σ if and only if h > k, if d B is an h-central element of T ρ witnessed by {b 2,..., b h } B then (d, b 2,..., b h ) T σ, and T σ is a Fraïssé filter over T ρ. In case T σ = and Z(ρ) = Z(σ) we obtain an even simpler characterization: Proposition 4.8. Let ρ and σ be central s, k = ar(ρ) 2, h = ar(σ) 2, and let T σ = and Z(ρ) = Z(σ). Then End ρ End σ if and only if h > k and T ρ has no h-central elements. Proof. ( ) ssume that End ρ End σ and let h > k. Take any f End ρ \ End σ and a tuple (a 1,..., a h ) σ such that (f(a 1 ),..., f(a h )) / σ. Let b i = f(a i ), i = 1,..., h. Then {b 1,..., b h } Z(σ) =, all the b i s are distinct and consequently all the a i s are distinct. Since T σ =, from the fact that all the a i s are distinct it follows that at least one of the a i s is in Z(σ), say, a 1 Z(σ). But then from f End ρ it follows that b 1 is an h-central element of T ρ witnessed by {b 2,..., b h } ( ) Suppose that h > k and that T ρ has an h-central element d witnessed by {b 2,..., b h }. Then {d, b 2,..., b h } \ Z(σ) whence follows that (d, b 2,..., b h ) / σ since T σ =. Take any c Z(σ) and consider d, x = c f(x) = x, x {b 2,..., b h } c, otherwise. Then f End ρ since d is an h-central element of T ρ witnessed by {b 2,..., b h }, and f / End σ since (c, b 2,..., b h ) belongs to σ and (d, b 2,..., b h ) does not. We shall say that a central ρ is End-maximal if End ρ End σ for every central σ ρ.

13 On endomorphism monoids of partial orders and central s 123 Proposition 4.9. central ρ on is End-maximal if and only if = ar(ρ) Z(ρ). Proof. ( ) Let ρ be a central which is not End-maximal. Then there is a central σ such that End ρ End σ. This further implies ar(ρ) < ar(σ) and Z(ρ) Z(σ). Since σ has at least ar(σ) noncentral elements, we have that ar(σ) Z(σ), so ar(ρ) Z(ρ) < ar(σ) Z(σ). ( ) Let ρ be a central of arity k such that k Z(ρ) < and let ρ = R ρ C ρ T ρ be the decomposition of ρ into its reflexive part, central part and the tail. Let h = Z(ρ) and define R h, C h and T h as follows: R h = {(x 1,..., x h ) h : x i = x j for some i j}, C h = {(x 1,..., x h ) ( h ) irr : x i Z(ρ) for some i}, T h = {(x 1,..., x h ) ( h ) irr : there is an i such that (x i, x j2,..., x jk ) T ρ whenever j 2,..., j k {1,..., h} \ {i} are k 1 distinct indices}. Let us show that σ = C h R h T h is a central such that End ρ End σ. The σ is clearly totally reflexive and totally symmetric, and has at least all of Z(ρ) as its central elements. To show that σ is a central, we have to show that σ h. Let \Z(ρ) = {b 1,..., b h }. Since each b i is a noncentral element, for every i there exist d 2,..., d k such that (b i, d 2,..., d k ) / ρ. But then {d 2,..., d k } \ Z(ρ), so d 2 = b j2,..., d k = b jk for some k 1 distinct indices j 2,..., j k {1,..., h} \ {i}. Thus, for every i there exist k 1 indices j 2,..., j k {1,..., h} \ {i} such that (b i, b j2,..., b jk ) / T ρ. Therefore, (b 1,..., b h ) / T h and hence (b 1,..., b h ) / σ. Clearly, ar(σ) > ar(ρ) and Z(σ) = Z(ρ). Let us show that End ρ End σ. Take any f End ρ, (a 1,..., a h ) σ and let d i = f(a i ), i {1,..., h}. If there is an i such that d i Z(σ) or there are i j such that d i = d j, then (d 1,..., d h ) σ. So, assume now that d 1,..., d h are distinct noncentral elements. Then the a i s are also all distinct. If there is an i such that a i Z(ρ) then for every k 1 distinct indices j 2,..., j k {1,..., h} \ {i} we have (a i, a j2,..., a jk ) ρ. Since f End ρ, we get (d i, d j2,..., d jk ) ρ. Moreover, (d i, d j2,..., d jk ) T ρ since the d i s are distinct noncentral elements. Therefore, (d 1,..., d h ) T h σ. If the a i s are distinct noncentral elements, then (a 1,..., a h ) T h whence follows that there is an i such that (a i, a j2,..., a jk ) T ρ whenever j 2,..., j k {1,..., h} \ {i} are k 1 distinct indices. Since f End ρ, we get (d i, d j2,..., d jk ) ρ. Moreover, (d i, d j2,..., d jk ) T ρ since the d i s are distinct noncentral elements. Therefore, (d 1,..., d h ) T h σ. There exist pairs (ρ, R Θ ) such that ρ is a central of arity k 2, R Θ is an h-regular and End ρ End R Θ. First of all, if Θ = { } where = {(x, x) : x } then End R Θ = and clearly End ρ End R Θ. In case Θ { }, we have shown in [7, Propositions 4.6 and 4.7] that End ρ End R Θ

14 124 D. Mašulović implies k < h, Θ = {θ}, θ has a single nontrivial block B and Z(ρ) B. Moreover, if T ρ =, we were able to provide a complete characterization [7, Proposition 4.8]: if T ρ = and Θ { } then End ρ End R Θ if and only if k < h, Θ = {θ}, θ has a single nontrivial block B and Z(ρ) = B. Here we shall deal with the remaining case of T ρ and thus finish the characterization. Proposition Let ρ be a central of arity k 2 such that T ρ and let R Θ be an h-regular, where Θ { }. Then End ρ End R Θ if and only if: (1) k < h; (2) Θ = {θ}, where /θ = { B, {b 2 },..., {b h } } and B 2, i.e. B is a single nontrivial block of θ; (3) Z(ρ) B; (4) if b i is an (h 1)-central element of T ρ for some i {2,..., h}, then {b 2,..., b h } \ {b i } cannot be a set of witnesses; and T ρ (5) for all a 2,..., a h dom(t ρ ), if (a 2,..., a h ) (b 2,..., b h ) then (a 2,..., a h ) is a permutation of (b 2,..., b h ). Proof. ( ) Suppose End ρ End R Θ. Then we know from [7, Propositions 4.6 and 4.7] that k < h, Θ = {θ}, θ has a single nontrivial block B and Z(ρ) B. Suppose now that (4) does not hold. Then there is an i {2,..., h} such that b i is an (h1)-central element of T ρ with the set of witnesses {b 2,..., b h }\{b i }. For the sake of simplicity, let b 2 be an (h 1)-central element of T ρ with the set of witnesses {b 3,..., b h }. Take any c Z(ρ), b 1 B \ {c} and consider c, x = b 1 b 2, x = c f(x) = x, x {b 3,..., b h } c, otherwise. Then f End ρ \ End R Θ which contradicts End ρ End R Θ. Finally, assume that (5) does not hold. Then there exist a 2,..., a h dom(t ρ ) such that (a 2,..., a h ) T ρ (b 2,..., b h ) and (a 2,..., a h ) is not a permutation of (b 2,..., b h ). Then {a 2,..., a h } B since all the a i s are distinct. If {a 2,..., a h } B 2 take an arbitrary a 1 \ {a 2,..., a h }. If, however, {a 2,..., a h } B = 1 take an arbitrary a 1 B\{a 2,..., a h }. Take any c Z(ρ) and consider c, x = a 1 f(x) = b i, x = a i, i {2,..., h} c, otherwise. Then f End ρ \ End R Θ which contradicts End ρ End R Θ. ( ) Suppose End ρ End R Θ, but (1), (2) and (3) are true. Take any f End ρ \ End R Θ and any (a 1,..., a h ) R Θ such that (f(a 1 ),..., f(a h )) / R Θ. Then there is a b 1 B such that (f(a 1 ),..., f(a h )) is a permutation of

15 On endomorphism monoids of partial orders and central s 125 (b 1,..., b h ). Since s we are working with are totally symmetric, without loss of generality we can assume that b i = f(a i ), i {1,..., h}. If there is an i 2 such that a i Z(ρ) then b i is an (h 1)-central element for T ρ witnessed by {b 2,..., b h } \ {b i }. If, however, {a 2,..., a h } Z(ρ) = then from (a 1,..., a h ) R Θ it follows that {a 2,..., a h } B, while f End ρ implies (a 2,..., a h ) T ρ (b 2,..., b h ). References [1] ichinger, E., Mašulović, D., Pöschel, R., Wilson, J. S., Completeness for concrete near-rings. Journal of lgebra 279 (2004), [2] Cameron, P. J., Nešetřil, J., Homomorphism-homogeneous al structures. Combinatorics. Probability and Computing 15 (2006), [3] Lau, D., Bestimmung der Ordnung maximaler Klassen von Funktionen der k- wertigen Logik. Z. Math. Logik u. Grundl. Math. 24 (1978), [4] Pech, M., Mašulović, D., On the height of the poset of endomorphism monoids of regular s. (accetped for publication in Multiple-Valued Logic) Pech, M., Mašulović, D., On traces of maximal clones. Novi Sad J. Math. Vol. 35 No. 1 (2005), [6] Ponjavić, M., On the structure of the poset of endomorphism monoids of central s. Contributions to General lgebra 16 (2005), Ponjavić, M., Mašulović, D., On chains and antichains in the partially ordered set of traces of maximal clones. Contributions to General lgebra 15 (2004), [8] Rosenberg, I. G., Über die funktionale Vollständigkeit in den mehrwertigen Logiken. Rozpr. ČSV Řada Mat. Přir. Věd. Praha 80 4 (1970), Received by the editors November 12, 2007