doi:10.7151/dmgaa.1190
ON SETS RELATED TO MAXIMAL CLONES
Yeni Susanti
Department of Mathematics Gadjah Mada University Yogyakarta Indonesia 55281
e-mail: inielsusan@yahoo.com
and Klaus Denecke
Institute of Mathematics Potsdam University Potsdam Germany
e-mail: kdenecke@rz.uni-potsdam.de
Abstract
For an arbitrary h-ary relation ρ we are interested to express n-clone P ol
nρ in terms of some subsets of the set of all n-ary operations O
n(A) on a finite set A, which are in general not clones but we can obtain P ol
nρ from these sets by using intersection and union. Therefore we specify the concept a function preserves a relation and moreover, we study the properties of this new concept and the connection between these sets and P ol
nρ. Particularly we study R
n,ka,bfor arbitrary partial order relations, equivalence relations and central relations.
Keywords: operations preserving relations, clones, semigroups.
2010 Mathematics Subject Classification: 08A30, 08A40.
1. Introduction
Let A be an arbitrary set and let O(A) be the set of all operations on the set
A. A clone on the set A is a subset of O(A) that are closed under superposition
and contains all projections. Recall that a projection e
nimaps every n-tuple
(a
1, . . . , a
n) ∈ A
nto a
i. Clones have been widely studied by many authors for
instance in [1, 2, 3, 4, 5, 6] and [10]. Among these clones, there are six well-
known clones preserving six classes of relations, namely affine relations, bounded
partial order relations, nontrivial equivalence relations, central relations, prime
permutations, and h-regularly generated relations. It is well-known that these clones are maximal according to Rosenberg’s characterization (see [3]). We are interested in the n-ary part of these clones which we call from now on n-clones.
Moreover, we specify the well-known concept of that ”a function preserves a relation” and show that we get the n-ary part of the maximal clones as union and intersection of sets of n-ary operations preserving (in our sense) a given relation. We show that these sets have particular interesting properties for other consideration in universal algebra.
First, we consider an arbitrary h-ary relation ρ and we want to represent P ol
nρ in terms of some subsets of O
n(A) which in general are not clones, but from these subsets we can obtain P ol
nρ by using intersection and union. For this aim, for arbitrary b ∈ A and a = (a
1, . . . , a
n) ∈ A
n, we define ρ
bk:= {(x
1, . . . , x
h−1) ∈ A
h−1|(x
1, . . . , x
k−1, b, x
k, . . . , x
h−1) ∈ ρ} and ρ
ak:= {((x
1,1, . . . , x
h−1,1), . . . , (x
1,n, . . . , x
h−1,n)) ∈ (A
h−1)
n| (x
1,i, . . . , x
k−1,i, a
i, x
k,i, . . . , x
h−1,i) ∈ ρ for every i ∈ {1, . . . , n}}. It is clear that ρ
akis a cartesian product of the h − 1 rela- tions ρ
ak1, . . . , ρ
akn, i.e., ρ
ak= ρ
ak1× · · · × ρ
akn. We say that f ∈ O
n(A) (a, b)
k- preserves ρ if and only if (f (x
1,1, . . . , x
1,n), . . . , f (x
h−1,1, . . . , x
h−1,n)) ∈ ρ
bkfor ev- ery ((x
1,1, . . . , x
h−1,1), . . . , (x
1,n, . . . , x
h−1,n)) ∈ ρ
ak. Then for every k ∈ {1, . . . , h}, we define R
a,bn,k:= {f ∈ O
n(A)|f (a, b)
k-preserves ρ}. We want to see the proper- ties of R
n,ka,b.
For further investigation, we recall the following concept of semigroup of n-ary operations on A. Let A be an arbitrary finite set and let O
n(A) be the set of all n-ary operations on A. On O
n(A), we define an operation + by f + g := f (g, . . . , g) for arbitrary f, g ∈ O
n(A), i.e., (f + g)(x) = f (g, . . . , g)(x) = f (g(x), . . . , g(x)) for every x ∈ A. To simplify the notation, we use x for (x, . . . , x) and thus we have (f + g)(x) = f (g(x)). It is easy to see that the operation + is associative giving a semigroup (O
n(A); +) (see [7, 9] and [10]). It is clear that if C is an n-clone on A, then (C; +) is a subsemigroup of (O
n(A); +). Moreover, for every f ∈ A
n, by π
fwe mean the unary operation on A such that π
f(x) = f (x) for every x ∈ A. We use c
nyfor the constant n-ary operation with value y.
Particularly for A = {0, 1}, we put C
4n:= {f ∈ O
n(A)|f (0) = 0, f (1) = 1},
¬C
4n:= {f ∈ O
n(A)|f (0) = 1, f (1) = 0}, K
0n:= {f ∈ O
n(A)|f (0) = f (1) = 0}
and K
1n:= {f ∈ O
n(A)|f (0) = f (1) = 1}. Clearly, C
4n, ¬C
4n, K
0nand K
1nare all disjoint and O
n(A) = C
4n∪ ¬C
4n∪ K
0n∪ K
1n. Moreover, the operation + has the following properties:
f + g =
g if f ∈ C
4n¬g if f ∈ ¬C
4nc
n0if f ∈ K
0nc
n1if f ∈ ¬K
0n(for more details see [9] and [10]).
2. Properties of R
n,ka,bIn this section, we study some properties of R
n,ka,bfor arbitrary h-ary relation ρ on A. The following propositions hold in O
n(A) for every n ≥ 1.
Proposition 1. Let A be an arbitrary finite set and let n ≥ 1, h ≥ 2 and 1 ≤ k ≤ h be natural numbers. Let ρ be an h-ary relation on A. Then the following propositions are true for every a, a
0∈ A
nand b, b
0, y ∈ A.
(i) If R
n,ka,b6= ∅, then ρ
ak= ∅ or ρ
bk6= ∅.
(ii) If ρ
bk⊆ ρ
bk0, then R
a,bn,k⊆ R
n,ka,b0. (iii) If ρ
ak⊆ ρ
ak0, then R
n,ka0,b⊆ R
a,bn,k.
(iv) Let 1 ≤ i ≤ n. If ρ
ak6= ∅, then R
n,ka,bcontains a projection e
niif and only if ρ
aki⊆ ρ
bk.
(v) R
n,ka,bcontains a constant operation c
nyif and only if (y, . . . , y) ∈ ρ
bk. (vi) If f ∈ R
n,kb,b0
and g ∈ R
n,ka,b, then f + g ∈ R
n,ka,b0.
Proof. (i) Let R
a,bn,k6= ∅ and let f ∈ R
n,ka,b. Assume that ρ
ak6= ∅. Then for ev- ery ((x
1,1, . . . , x
h−1,1), . . . , (x
1,n, . . . , x
h−1,n)) ∈ ρ
ak, we have (f (x
1,1, . . . , x
1,n), . . . , f (x
h−1,1, . . . , x
h−1,n)) ∈ ρ
bkand thus ρ
bk6= ∅.
(ii) Let ρ
bk⊆ ρ
bk0and let f ∈ R
n,ka,b. Then (f (x
1,1, . . . , x
1,n), . . . , f (x
h−1,1, . . . , x
h−1,n)) ∈ ρ
bkfor arbitrary ((x
1,1, . . . , x
h−1,1), . . . , (x
1,n, . . . , x
h−1,n)) ∈ ρ
ak. By assumption, (f (x
1,1, . . . , x
1,n), . . . , f (x
h−1,1, . . . , x
h−1,n)) ∈ ρ
bk0, i.e., f ∈ R
n,ka,b0and therefore R
n,ka,b⊆ R
n,ka,b0.
(iii) Let ρ
ak⊆ ρ
ak0and let f ∈ R
n,ka0,b. Then for every ((x
1,1, . . . , x
h−1,1), . . . , (x
1,n, . . . , x
h−1,n)) ∈ ρ
akwe have ((x
1,1, . . . , x
h−1,1), . . . , (x
1,n, . . . , x
h−1,n)) ∈ ρ
ak0and thus (f (x
1,1, . . . , x
1,n), . . . , f (x
h−1,1, . . . , x
h−1,n)) ∈ ρ
bkby assumption, i.e., f ∈ R
n,ka,b.
(iv) Let ρ
ak6= ∅. Let e
nibe in R
n,ka,band let (x
1,i, . . . , x
h−1,i) ∈ ρ
aki. By
assumption, we can find (x
1,j, . . . , x
h−1,j) ∈ ρ
akj, i 6= j = 1, . . . , n. Thus, we
obtain (x
1,i, . . . , x
h−1,i) = (e
ni(x
1,1, . . . , x
1,n), . . . , e
ni(x
h−1,1, . . . , x
h−1,n)) ∈ ρ
bkand
hence ρ
aki⊆ ρ
bk. Conversely, let ρ
aki⊆ ρ
bk. Then for every ((x
1,1, . . . , x
h−1,1), . . . ,
(x
1,n, . . . , x
h−1,n)) ∈ ρ
akwe have (e
ni(x
1,1, . . . , x
1,n), . . . , e
ni(x
h−1,1, . . . , x
h−1,n)) =
(x
1,i, . . . , x
h−1,i) ∈ ρ
aki⊆ ρ
bkand therefore e
ni∈ R
n,ka,b.
(v) Let c
nybe in R
n,ka,b. Then for every ((x
1,1, . . . , x
h−1,1), . . . , (x
1,n, . . . , x
h−1,n))
∈ ρ
akwe get (y, . . . , y) = (c
ny(x
1,1, . . . , x
1,n), . . . , c
ny(x
h−1,1, . . . , x
h−1,n)) ∈ ρ
bk. Conversely, let (y, . . . , y) ∈ ρ
bk. Then for every ((x
1,1, . . . , x
h−1,1), . . . , (x
1,n, . . . , x
h−1,n)) ∈ ρ
akwe obtain (c
ny(x
1,1, . . . , x
1,n), . . . , c
ny(x
h−1,1, . . . , x
h−1,n)) = (y, . . . , y)
∈ ρ
bk, i.e., c
ny∈ R
a,bn,k. (vi) Let f ∈ R
n,kb,b0
and g ∈ R
n,ka,b. Let ((x
1,1, . . . , x
h−1,1), . . . , (x
1,n, . . . , x
h−1,n))
∈ ρ
ak. Then we have (g(x
1,1, . . . , x
1,n), . . . , g(x
h−1,1, . . . , x
h−1,n)) ∈ ρ
bkand hence ((g(x
1,1, . . . , x
1,n), . . . , g(x
h−1,1, . . . , x
h−1,n)), . . . , (g(x
1,1, . . . , x
1,n), . . . , g(x
h−1,1, . . . , x
h−1,n))) ∈ ρ
bk. Thus ((f + g)(x
1,1, . . . , x
1,n), . . . , (f + g)(x
h−1,1, . . . , x
h−1,n))
= (f (g(x
1,1, . . . , x
1,n)), . . . , f (g(x
h−1,1, . . . , x
h−1,n))) ∈ ρ
bk0by assumption.
Remark 2. By Proposition 1 (vi) it follows that R
n,kb,b
forms subsemigroup of (O
n(A); +) for every b ∈ A and for every 1 ≤ k ≤ h.
Recall that for every clone C, we call the n-ary part C ∩ O
n(A) an n-clone.
Proposition 3. Let A be an arbitrary finite set and let ρ be an arbitrary h-ary relation on A. The following assertions hold for every natural number n ≥ 1, a ∈ A
nand b ∈ A.
(i) If R
n,ka,bis an n-clone, then ρ
ak⊆ ρ
bk.
(ii) If π
f∈ R
1,kb,bfor every f ∈ R
n,ka,b, then R
n,ka,bforms a subsemigroup of (O
n(A); +).
(iii) For h = 2, if R
n,ka,bforms a subsemigroup of (O
n(A); +), then π
f∈ R
1,kb,bfor every f ∈ R
a,bn,k.
Proof. (i) If R
n,ka,bis an n-clone, then R
n,ka,bcontains all projections. Therefore by Proposition 1 (iv), ρ
ak⊆ ρ
bk.
(ii) Let f, g ∈ R
n,ka,b. For every ((x
1,1, . . . , x
h−1,1), . . . , (x
1,n, . . . , x
h−1,n)) ∈ ρ
akwe have (g(x
1,1, . . . , x
1,n), . . . , g(x
h−1,1, . . . , x
h−1,n)) ∈ ρ
bkand hence
((f + g)(x
1,1, . . . , x
1,n), . . . , (f + g)(x
h−1,1, . . . , x
h−1,n))
= (f (g(x
1,1, . . . , x
1,n)), . . . , f (g(x
h−1,1, . . . , x
h−1,n)))
= (π
f(g(x
1,1, . . . , x
1,n)), . . . , π
f(g(x
h−1,1, . . . , x
h−1,n))) ∈ ρ
bkby assumption. Therefore f + g ∈ R
a,bn,k, i.e., R
n,ka,bforms a subsemigroup of
(O
n(A); +).
(iii) Let h = 2. Let f ∈ R
n,ka,band let y ∈ ρ
bk. By Proposition 1 (v), c
ny∈ R
a,bn,kand hence f + c
ny∈ R
n,ka,bby assumption. Thus for every (x
1, . . . , x
n) ∈ ρ
akwe have (f + c
ny)(x
1, . . . , x
n) ∈ ρ
bk. Therefore π
f(y) = π
f(c
ny(x
1, . . . , x
n)) = f (c
ny(x
1, . . . , x
n)) = (f + c
ny)(x
1, . . . , x
n) ∈ ρ
bk, i.e., π
f∈ R
n,kb,b.
Theorem 4. Let A be an arbitrary finite set and let ρ be a binary relation on A.
For arbitrary n ≥ 1, a ∈ A
nand b ∈ A, it follows that R
a,bn,kforms a subsemigroup of (O
n(A); +) if and only if π
f∈ R
1,kb,bfor every f ∈ R
a,bn,k.
Proof. is clear by Proposition 3 (ii) and Proposition 3 (iii).
Corollary 5. Let A be an arbitrary finite set and let ρ be a binary relation on A.
For arbitrary a, b ∈ A, it follows that R
1,ka,bforms a subsemigroup of (O
1(A); ◦) if and only if R
1,ka,b⊆ R
1,kb,b.
Proof. It is clear by Theorem 4 and the fact that f = π
ffor n = 1.
Proposition 6. Let A be an arbitrary finite set and let ρ be an arbitrary h-ary re- lation on A. For every natural number n ≥ 1 it follows P ol
nρ ⊆ T
a∈An
S
b∈A
T
hk=1
R
n,ka,b.
Proof. Let f ∈ P ol
nρ and let a ∈ A
nbe arbitrary. We will show that f ∈ T
hk=1
R
n,ka,bfor some b ∈ A. Let ((x
1,1, . . . , x
h−1,1), . . . , (x
1,n, . . . , x
h−1,n)) ∈ ρ
ak, i.e., (x
1,i, . . . , x
k−1,i, a
i, x
k+1,i, . . . , x
h−1,1) ∈ ρ for every i ∈ {1, . . . , n}. Then (f (x
1,1, . . . , x
1,n), . . . , f (x
k−1,1, . . . , x
k−1,n), f (a), f (x
k+1,1, . . . , x
k+1,n), . . . , f (x
h−1,1, . . . , x
h−1,n)) ∈ ρ by assumption and therefore (f (x
1,1, . . . , x
1,n), . . . , f (x
h−1,1, . . . , x
h−1,n)) ∈ ρ
f (a)k, i.e., f ∈ R
n,ka,f (a).
Example 7. Let A be an arbitrary finite set and let n ≥ 1 be a natural number and let ρ be an h-ary relation on A. For every a ∈ A
nand b ∈ A, if ρ = {a ∈ A
h|a ∈ A}, then R
n,ka,bis an n-clone if and only if a = b if and only if R
n,ka,bforms a subsemigroup of (O
n(A); +). This can be explained as follows. It is clear that when R
a,bn,kis an n-clone, then R
n,ka,bforms a subsemigroup of (O
n(A); +) and moreover, contains all projections. Since ρ = {a ∈ A
h|a ∈ A}, it follows that ρ
akcontains only (a
1, . . . , a
n) ∈ ρ
n. Thus, a
i= e
ni(a) = b for every i = 1, . . . , n and hence a = b. Conversely, if a = b, then for every b ∈ A, ρ
bkcontains only b.
Thus e
ni(b) = b, i.e., e
ni∈ R
n,kb,b
for every 1 ≤ i ≤ n. Now, for every f, g ∈ R
n,kb,b
, we have (f + g)(b) = f (b) = b that implies f + g ∈ R
n,kb,b
. Hence, R
n,kb,b
is an
n-clone. Furthermore, let R
n,ka,bform a subsemigroup of (O
n(A); +). For every
f ∈ R
n,ka,b, it follows that f (a) = b. Thus, if g
1, . . . , g
n∈ R
a,bn,k, then we have
f (g
1, . . . , g
n)(a) = f (g
1, . . . , g
1)(a) = (f + g
1)(a) = b and thus R
a,bn,kis an n-clone.
3. Properties of R
n,ka,bfor some particular relations
In this section, we study some properties of R
n,ka,bfor some particular relations, i.e., partial order relation, equivalence relation and central relation. Instead of R
a,bn,k, we use P
a,bn,kfor partial order relation ≤. The following propositions are true for arbitrary partial order relation ≤ on A.
Proposition 8. Let (A; ≤) be an arbitrary finite partially ordered set and let n ≥ 1 be a natural number. Then the following properties hold for every a, a
0∈ A
n, b, b
0∈ A and k = 1, 2.
(i) P
a,bn,k6= ∅.
(ii) P
a,bn,k⊆ P
a,bn,k0if and only if b ∈ ≤
bk0. (iii) If ≤
ak⊆ ≤
ak0, then P
an,k0,b⊆ P
a,bn,k.
(iv) Let 1 ≤ i ≤ n. P
a,bn,kcontains the projection e
niif and only if a
i∈ ≤
bk. (v) P
a,bn,kcontains the constant operation c
nyif and only if y ∈ ≤
bk. Moreover,
P
a,bn,kcontains exactly | ≤
bk| constant operations.
(vi) If g ∈ P
a,bn,kand f ∈ P
n,kb,b0
, then f + g ∈ P
a,bn,k0.
Proof. (i) Since b ∈ ≤
bkfor every b ∈ A and k = 1, 2, then by Proposition 1 (v), c
nb∈ P
a,bn,k.
(ii) Let P
a,bn,k⊆ P
a,bn,k0. Since b ∈ ≤
bkthen by (i) c
nb∈ P
a,bn,k⊆ P
a,bn,k0and hence for every (x
1, . . . , x
n) ∈ ≤
akwe have b = c
nb(x
1, . . . , x
n) ∈ ≤
bk0. The opposite direction is clear by Proposition 1 (ii).
(iii), (iv), (v) and (vi) are clear respectively by (iii), (iv), (v) and (vi) of Proposition 1.
If A has the least element and the greatest element, then we have the following properties.
Proposition 9. Let (A; ≤) be an arbitrary finite partially ordered set and let n ≥ 1 be a natural number. If A has the least and the greatest element and V
A
is the least element and W
A
is the greatest element in A, then for every a ∈ A
nand b ∈ A the following propositions are true.
(i) P
a,bn,1= O
n(A) if and only if b = V
A
(P
a,bn,2= O
n(A) if and only if b = W
A
).
(ii) P
a,bn,1= {c
nb} if and only if a = V
A
and b = W
A
(P
a,bn,2= {c
nb} if and only if a = W
A
and b = V
A
).
Proof. We prove for k = 2 and similar way for k = 1.
(i) Let P
a,bn,2= O
n(A). Then c
ny∈ P
a,bn,2for all y ∈ A and hence for all (x
1, . . . , x
n) ∈
≤
a2we obtain y = c
ny(x
1, . . . , x
n) ∈ ≤
b2for every y ∈ A, i.e., b = W
A
. Con- versely, let b = W
A
and let f ∈ O
n(A). Then for all (x
1, . . . , x
n) ∈ ≤
a2, we have f (x
1, . . . , x
n) ∈ ≤
b2, i.e., O
n(A) = P
a,bn,2.
(ii) Let P
a,bn,2= {c
nb}. Assume a 6= W
A
. If b = W
A
, then by (i), P
a,bn,2= O
n(A), a contradiction. If b 6= W
A
, then consider an n-ary operation f ∈ O
n(A) satisfying f (x
1, . . . , x
n) = b for every (x
1, . . . , x
n) 6= W
A
and f ( W
A
) = W
A
. This f is not equal to c
nband is in P
a,bn,2. Hence P
a,bn,26= {c
nb}, a contradiction. Assume now b 6=
V
A
. Then by Proposition 8 (v), {c
nVA
, c
nb} ⊆ P
a,bn,2, a contradiction. Conversely, let f ∈ P
a,bn,2= P
Wn,2A,V
A
and let (x
1, . . . , x
n) ∈ A
n. Since (x
1, . . . , x
n) ∈ ≤
W
A
2
, then f (x
1, . . . , x
n) ∈ ≤
V
A
2
, i.e., f (x
1, . . . , x
n) = V
A
, i.e., f = c
nVA
. Hence P
a,bn,2= {c
nb}.
Theorem 10. Let (A; ≤) be an arbitrary finite partially ordered set and let n ≥ 1 be a natural number. For arbitrary a ∈ A
nand b ∈ A the following propositions are equivalent for k = 1, 2.
(i) P
a,bn,kis an n-clone.
(ii) P
a,bn,k= P
n,kb,b
.
(iii) a ∈ ≤
bkand P
a,bn,k⊆ P
n,kb,b
.
Proof. (i)⇒(ii) Let P
a,bn,kbe an n-clone. Then by Proposition 3 (i), ≤
ak⊆
≤
bkand thus by Proposition 8 (iii), P
nb,b
⊆ P
a,bn. Now, let f ∈ P
a,bnand let (y
1, . . . , y
n) ∈ ≤
bk, i.e., y
i∈ ≤
bkfor every i = 1, 2, . . . , n. Then by Proposition 8 (v), c
nyi∈ P
a,bn,kand hence f (c
ny1, . . . , c
nyn) ∈ P
a,bn,kfor 1 ≤ i ≤ n. Therefore, for ev- ery (x
1, . . . , x
n) ∈ ≤
ak, f (c
ny1, . . . , c
nyn)(x
1, . . . , x
n) ∈ ≤
bkand hence f (y
1, . . . , y
n) = f (c
ny1(x
1, . . . , x
n), . . . , c
nyn(x
1, . . . , x
n)) = f (c
ny1, . . . , c
nyn)(x
1, . . . , x
n) ∈ ≤
bk, i.e., f ∈ P
n,kb,b
. Thus P
a,bn,k⊆ P
n,kb,b
and hence P
a,bn,k= P
n,kb,b
. (ii)⇒(iii) By Proposition 8 (iv), P
n,kb,b
contains all projections. Therefore
P
a,bn= P
b,bncontains all projections and hence by Proposition 8 (iv), a
i∈ ≤
bkfor
every i ∈ {1, . . . , n}, i.e., a ∈ ≤
bk.
(iii)⇒(i) By assumption and Proposition 8 (iv), P
a,bncontains all projections.
Moreover, let f, g
1, . . . , g
nbe in P
a,bn⊆ P
nb,b
. Then, g
i(x
1, . . . , x
n) ∈ ≤
bkfor every (x
1, . . . , x
n) ∈ ≤
akand i = 1, 2, . . . , n. Thus, (g
1(x
1, . . . , x
n), . . . , g
n(x
1, . . . , x
n)) ∈
≤
bk. Hence for every (x
1, . . . , x
n) ∈ ≤
ak, f (g
1, . . . , g
n)(x
1, . . . , x
n) = f (g
1(x
1, . . . , x
n), . . . , g
n(x
1, . . . , x
n)) ∈ ≤
bk. Therefore, f (g
1, . . . , g
n) ∈ P
a,bn,kand hence P
a,bn,kis an n-clone.
Theorem 11. Let (A; ≤) be an arbitrary finite partially ordered set and let n ≥ 1 be a natural number. For every a ∈ A
n, b ∈ A it follows that P ol
n≤
= T
a∈An
S
b∈A
(P
a,bn,1∩ P
a,bn,2).
Proof. (⊆) is clear by Proposition 6.
(⊇) Let f ∈ T
a∈An
S
b∈A
(P
a,bn,1∩ P
a,bn,2). Let (u
i, v
i) ∈ ≤, i = 1, 2, . . . , n. Now, take a = u. By assumption, for this a ∈ A
n, we can find b ∈ A such that f ∈ P
a,bn,1∩ P
a,bn,2. Therefore f (u) ∈ ≤
b2and f (v) ∈ ≤
b1, i.e., (f (u), f (v)) ∈ ≤.
Hence f ∈ P ol
nρ
A.
Example 12. Let A = {0, 1}. By Proposition 9, P
1,0n,2= {c
n0} and P
n,10,1
= {c
n1}. Moreover, P
a,0n,1= O
n({0, 1}) = P
a,1n,2. Therefore P
a,0n,1∩ P
a,0n,2= P
a,0n,2and P
a,1n,1∩ P
a,1n,2= P
a,1n,1and hence S
b∈A
(P
a,bn,1∩ P
a,bn,2) = P
a,1n,1∪ P
a,0n,2and we get P ol
n≤
= T
a∈An
S
b∈A
(P
a,bn,1∩ P
a,bn,2) = T
a∈An
(P
a,1n,1∪ P
a,0n,2) by Theorem 11. Now consider all operations on O
2({0, 1}) as follows
f
1f
2f
3f
4f
5f
6f
7f
8f
9f
10f
11f
12f
13f
14f
15f
16(0, 0) 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
(0, 1) 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1
(1, 0) 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
(1, 1) 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1.
Remark 13. From above example, we have f
1= c
20, f
16= c
21and we obtain K
02= {c
20, f
2, f
3, f
4}, C
42= {f
5, f
6, f
7, f
8}, ¬C
42= {f
9, f
10, f
11, f
12} and K
12= {f
13, f
14, f
15, c
21}. By Proposition 9, P
(1,1),02,2= {c
20} and P
(0,0),12,1= {c
21}. Moreover, it is easy to see that
P
(0,1),12,1= {f
7, f
8, f
15, c
21} P
(0,0),02,2= K
0n∪ C
42P
(1,0),12,1= {f
6, f
8, f
14, c
n1} P
(0,1),02,2= {c
20, f
2, f
5, f
6}
P
(1,1),12,1= C
42∪ K
12P
(1,0),02,2= {c
20, f
3, f
5, f
7}
and hence we obtain P ol
2≤= C
42∪ {c
20, c
21}.
Generally, for A = {0 ≤ 1} and for every n ≥ 1, by applying Proposition 8, we have some properties on P
a,bn,1and P
a,bn,2as follows
(i) c
n1∈ P
a,bn,1and c
n0∈ P
a,bn,2since 1 ∈ ≤
b1and 0 ∈ ≤
b2.
(ii) If a 6= 0, then P
a,bn,1∩ C
4n6= ∅. This fact holds since 1 ∈ ≤
b1which implies e
ni∈ C
4nis contained in P
a,bn,1by Proposition 8 (iv) for some i such that a
i6= 0. Similarly, if a 6= 1, then P
a,bn,2∩ C
4n6= ∅.
(iii) If b = 1, then P
a,bn,1∩ (K
0n∪ ¬C
4n) = ∅. It is clear that 1 ∈ ≤
a1. But for all f ∈ K
0n∪ ¬C
4n, we have f (1) = 0 6∈ ≤
11and hence f 6∈ P
a,bn,1. Similarly if b = 0, then P
a,bn,2∩ (K
1n∪ ¬C
4n) = ∅.
(iv) P ol
n≤= C ∪ {c
n0, c
n1} for some C ⊆ C
4n.
Now, we come to the properties of R
a,bn,kfor an arbitrary equivalence relation θ on A. By symmetry property of θ, it follows that θ
1b= θ
b2and θ
1a= θ
a2for every b ∈ A and a ∈ A
nand these imply R
n,1a,b= R
a,bn,2. Therefore, we can omit the number k.
Moreover, since θ
b1= θ
b2is actually an equivalence class that contains b, then we use [b]
θinstead θ
band we use E
a,bn,θinstead of R
n,ka,bsince we might have various equivalence relations on A. Thus, by f ∈ E
a,bn,θwe mean an n-ary operation such that (f (x
1, . . . , x
n), b) ∈ θ for every (x
1, . . . , x
n) satisfying (x
i, a
i) ∈ θ for every i = 1, . . . , n.
Proposition 14. Let A be an arbitrary finite set and n ≥ 1 be an arbitrary natural number. For an arbitrary equivalence relation θ
A6= A × A on A, the following properties are true for arbitrary a, a
0∈ A
nand b, b
0∈ A.
(i) E
a,bn,θ6= ∅.
(ii) [b]
θ= [b
0]
θif and only if E
a,bn,θ∩ E
a,bn,θ06= ∅ if and only if E
a,bn,θ= E
a,bn,θ0. (iii) [a]
θn= [a
0]
θnif and only if E
a,bn,θ= E
an,θ0,b.
(iv) If [a]
θn6= [a
0]
θn, then E
a,bn,θ∩ E
an,θ0,b06= ∅.
(v) Let 1 ≤ i ≤ n. E
a,bn,θcontains a projection e
niif and only if [b]
θ= [a
i]
θ. (vi) E
a,bn,θcontains a constant operation c
nyif and only if y ∈ [b]
θ. Moreover,
E
a,bn,θcontains precisely |[b]
θ| constant operations.
(vii) If f ∈ E
n,θb,b0
and g ∈ E
a,bn,θ, then f + g ∈ E
a,bn,θ0.
Proof. (i) By reflexivity of θ and Proposition 1 (v), c
nb∈ E
a,bn,θ, i.e., E
a,bn,θ6= ∅.
(ii) By Proposition 1 (ii), if [b]
θ= [b
0]
θ, then E
a,bn,θ= E
a,bn,θ0. Thus if [b]
θ= [b
0]
θ, then E
a,bn,θ∩ E
n,θa,b06= ∅ since E
a,bn,θ6= ∅ by (i). Conversely, let E
a,bn,θ∩ E
a,bn,θ06= ∅ and let f ∈ E
a,bn,θ∩ E
a,bn,θ0. Then for every (x
1, . . . , x
n) ∈ [a]
θnwe have f (x
1, . . . , x
n) ∈ [b]
θ∩ [b
0]
θand hence [b]
θ= [b
0]θ.
(iii) By Proposition 1 (iii), if [a]
θn= [a
0]
θn, then E
a,bn,θ= E
an,θ0,b. Conversely, let E
a,bn,θ= E
an,θ0,b. Assume that [a]
θn6= [a
0]
θn, i.e., a 6∈ [a
0]
θnand a
06∈ [a]
θn. Since θ
A6= A × A, then there is b
0∈ A such that [b]
θ6= [b
0]
θ. Now consider an n-ary operation f on A such that f (x
1, . . . , x
n) = b for all (x
1, . . . , x
n) ∈ [a]
θnand f (a
0) = b
0. Then it is clear that f ∈ E
a,bn,θbut f 6∈ E
an,θ0,band hence E
a,bn,θ6= E
an,θ0,b, a contradiction.
(iv) Let [a]
θn6= [a
0]
θn. Consider an f ∈ O
n(A) such that f (x
1, . . . , x
n) = b for every (x
1, . . . , x
n) ∈ [a]
θnand f (x
1, . . . , x
n) = b
0for every (x
1, . . . , x
n) ∈ [a
0]
θn. It is clear that f ∈ E
a,bn,θ∩ E
an,θ0,band hence E
a,bn,θ∩ E
an,θ0,b06= ∅.
(v), (vi) and (vii) are clear by Proposition 1 (iv), (v) and (vi).
Theorem 15. Let A be an arbitrary finite set and n ≥ 1 be an arbitrary natural number and let θ 6= A × A be an arbitrary equivalence relation on A. Then for arbitrary a ∈ A
nand b ∈ A it follows that E
a,bn,θis an n-clone if and only if [a]
θn= [b]
θn.
Proof. If E
a,bn,θis an n-clone, then by Proposition 3 (i), a ∈ [b]
θn, i.e., [a]
θn= [b]
θn. Conversely, let [a]
θn= [b]
θn. By Proposition 1 (iv), E
a,bn,θcontains all projections. Now, let f, g
1, . . . , g
n∈ E
a,bn,θand (x
1, . . . , x
n) ∈ [a]
θnbe arbitrary.
Then g
1(x
1, . . . , x
n), . . . , g
n(x
1, . . . , x
n) ∈ [b]
θand therefore (g
1(x
1, . . . , x
n), . . . , g
n(x
1, . . . , x
n)) ∈ [b]
θn= [a]
θn. Thus, f (g
1, . . . , g
n)(x
1, . . . , x
n) = f (g
1(x
1, . . . , x
n), . . . , g
n(x
1, . . . , x
n)) ∈ [b]
θ, i.e., f (g
1, . . . , g
n) ∈ E
a,bn,θ.
Now, if we define E
an,θ:= {f ∈ O
n(A)|f (x
1, . . . , x
n) ∈ [f (a)]
θfor every (x
1, . . . , x
n) ∈ [a]
θn}, then we have the following proposition.
Proposition 16. Let A be an arbitrary finite set and n ≥ 1 be an arbitrary natural number. For an arbitrary equivalence relation θ 6= A × A on A and an arbitrary a ∈ A
nit follows that E
an,θ= S
b∈A
E
a,bn,θ. Proof. (⊆) It is clear by definition.
(⊇) Let f ∈ S
b∈A
E
a,bn,θ. Then, we can find b ∈ A such that f ∈ E
a,bn,θand thus for every (x
1, . . . , x
n) ∈ [a]
θn, we have f (x
1, . . . , x
n), f (a) ∈ [b]
θ, i.e.,
f (x
1, . . . , x
n) ∈ [f (a)]
θ, i.e., f ∈ E
an,θ. Hence S
b∈A
E
a,bn,θ⊆ E
n,θaand therefore E
an,θ= S
b∈A
E
a,bn,θ.
Theorem 17. Let A be an arbitrary finite set and n ≥ 1 be an arbitrary natural number. For an arbitrary equivalence relation θ 6= A × A on A, it follows that P ol
nθ = T
a∈An
S
b∈A
E
a,bn,θ= T
a∈An
E
an,θ.
Proof. (⊆) is clear by Proposition 6 and by Proposition 16.
(⊇) Let f ∈ T
a∈An
E
an,θ. For every (x
i, y
i) ∈ θ, i = 1, . . . , n we have (x
1, . . . , x
n) ∈ [(y
1, . . . , y
n)]
θn. By assumption, we know that f ∈ E
(yn,θ1,...,yn)
. Therefore f (x
1, . . . , x
n) ∈ [f (y
1, . . . , y
n)]
θand thus (f (x
1, . . . , x
n), f (y
1, . . . , y
n))
∈ θ, i.e., f ∈ P ol
nθ.
Recall that an h-ary relation ζ on A is called a central relation if ζ satisfies these three properties: (i) totally symmetric, i.e., if (a
1, . . . , a
h) ∈ ζ, then (a
σ(1), . . . , a
σ(h)) ∈ ζ for all permutations σ on {1, . . . , h} (ii) totally reflexive, i.e., κ
hA⊆ ζ for κ
hA:= {(a
1. . . , a
h)|∃ i ∃j (i 6= j ∧ a
i= a
j)} and (iii) there exists ∅ 6= C ⊆ A such that (c, a
2, . . . , a
h) ∈ ζ for every c ∈ C and for all a
2, . . . , a
h∈ A. We call the set C as the central of ζ. Now, we consider an arbitrary h-ary central relation ζ, h ≥ 2. We use h ≤ |A| since otherwise we would have ζ = A
hand the center of ζ would be trivial. Moreover, by totally symmetry property of ζ, we have ζ
kb= ζ
lband ζ
ka= ζ
lafor every a ∈ A
n, b ∈ A and k 6= l ∈ {1, . . . , h}. Therefore, we can again omit the number k and since we might have many central relation on A, we then use C
a,bn,ζinstead of R
n,ka,b. Without lost of generality, we use implicitly k = h, i.e., by f ∈ C
a,bn,ζwe mean n-ary operation satisfying (f (x
1,1, . . . , x
1,n), . . . , f (x
h−1,1, . . . , x
h−1,n), b) ∈ ζ for every (x
1,i, . . . , x
h−1,i, a
i) ∈ ζ, i = 1, . . . , n. By the third property of ζ, i.e., for every (x
1, . . . , x
h−1) ∈ A
h−1and for all c ∈ C, it follows (x
1, . . . , x
h−1, c) ∈ ζ and we have the following simple properties.
Proposition 18. Let ζ be an h-ary relation on A with central C. For every b ∈ A and a ∈ A
nwe have the following properties.
(i) ζ
b⊆ ζ
cfor every c ∈ C.
(ii) ζ
a⊆ ζ
cfor every c ∈ C
n.
By Proposition 1, we have the following properties for an arbitrary h-ary central relation ζ on A.
Proposition 19. Let A be an arbitrary finite set and let n ≥ 1, h ≥ 2 be natural
numbers. Let ζ be an h-ary central relation on A with C as the central. Then the
following propositions are true for every a, a
0∈ A
nand b, b
0, y ∈ A.
(i) C
a,bn,ζ6= ∅.
(ii) If ζ
b⊆ ζ
b0, then C
a,bn,ζ⊆ C
a,bn,ζ0. (iii) If ζ
a⊆ ζ
a0, then C
an,ζ0,b⊆ C
a,bn,ζ.
(iv) Let 1 ≤ i ≤ n. C
a,bn,ζcontains a projection e
niif and only if ζ
ai⊆ ζ
b. (v) C
a,bn,ζcontains the constant operation c
nyif and only if (y, . . . , y) ∈ ζ
b. (vi) If f ∈ C
n,ζb,b0
and g ∈ C
a,bn,ζ, then f + g ∈ C
a,bn,ζ0.
As a consequence of Proposition 18 and Proposition 19, we have
Proposition 20. Let A be an arbitrary finite set and let n ≥ 1, h ≥ 2 be natural numbers. Let ζ be an h-ary central relation on A with C as the central. Then the following propositions are true for every a ∈ A
nand b ∈ A.
(i) C
a,bn,ζ⊆ C
a,cn,ζfor every c ∈ C.
(ii) C
c,bn,ζ⊆ C
a,bn,ζfor every c ∈ C
n.
(iii) If b ∈ C, then C
a,bn,ζcontains all projections, contains all constant operations and moreover is an n-clone.
(iv) If C
an,ζ0,bcontains all projections for all a
0∈ A
n, then b ∈ C.
(v) If h = 2 and C
a,bn,ζcontains all constant operations, then b ∈ C.
(vi) If h ≥ 3, then C
a,bn,ζcontains all constant operations.
Proof. (i) By Proposition 18 (i) and Proposition 19 (ii).
(ii) By Proposition 18 (ii) and Proposition 19 (iii).
(iii) By Proposition 18 (i), Proposition 19 (iv) and Proposition 19 (v). More- over, since for every x
1, . . . , x
h−1∈ A it follows that (x
1, . . . , x
h−1) ∈ ζ
b, then for every (x
1,1, . . . , x
h−1,1), . . . , (x
1,n, . . . , x
h−1,n) ∈ ζ
aand for every f, g
1, . . . , g
n∈ C
a,bn,ζ, we have (f (g
1, . . . , g
n)(x
1,1, . . . , x
1,n), . . . , f (g
1, . . . , g
n)(x
h−1,1, . . . , x
h−1,n))
∈ ζ
b, i.e., f (g
1, . . . , g
n) ∈ C
a,bn,ζ. Thus C
a,bn,ζis an n-clone.
(iv) Let (x
1, . . . , x
h−1) ∈ A
h−1be arbitrary. Then (x
1, . . . , x
h−1, x
h−1) ∈ ζ.
By assumption, for every a
0∈ A
nsuch that a
i0= x
h−1, we have that C
an,ζ0,bcontains all projections and hence by Proposition 19 (iv), ζ
ai0⊆ ζ
b. Therefore,
(x
1, . . . , x
h−1) ∈ ζ
xh−1= ζ
ai0⊆ ζ
b. Thus (x
1, . . . , x
h−1, b) ∈ ζ and hence b ∈ C
since (x
1, . . . , x
h−1) is arbitrary.
(v) Let h = 2 and let y ∈ A. By assumption, C
a,bn,ζcontains c
ny. Therefore, by Proposition 19 (v), y ∈ ζ
b, i.e., (y, b) ∈ ζ. Since y is arbitrary we have b ∈ C.
(vi) By totally reflexive property of ζ it follows that for every (y, . . . , y) ∈ A
h−1, h ≥ 3 we have (y, . . . , y, b) ∈ ζ, i.e., c
ny∈ C
a,bn,ζby Proposition 19 (v).
Proposition 21. Let A be an arbitrary finite set and let n ≥ 1, h ≥ 2 be natural numbers. Let ζ be an h-ary central relation on A with C as the central. Then for every a ∈ A
nand b ∈ A, C
a,bn,ζcontains all constant operations if and only if b ∈ C or h ≥ 3.
Proof. If C
a,bn,ζcontains all constant operations and h < 3, i.e., h = 2, then b ∈ C by Proposition 20 (v). The converse is clear by Proposition 20 (iii) and Proposition 20 (vi).
Proposition 22. Let A be an arbitrary finite set and let n ≥ 1, h ≥ 2 be natural numbers. Let ζ be an h-ary central relation on A with C as the central. For b ∈ A, the following propositions are equivalent.
(i) C
a,bn,ζcontains all projections for all a ∈ A
n. (ii) C
a,bn,ζis an n-clone for all a ∈ A
n.
(iii) b ∈ C.
Proof. (i)⇒(iii) is clear by Proposition 20 (iv).
(iii)⇒(ii) is clear by Proposition 20 (iii).
(ii)⇒(i) is obvious by definition.
The following property is clear by Proposition 6.
Proposition 23. Let A be an arbitrary finite set and let n ≥ 1, h ≥ 2 be natural numbers. Let ζ be an h-ary central relation on A with C as the central. Then P ol
nζ ⊆ T
a∈An
S
b∈A
