Some Submaximal Elements of the Lattice of Finitary Operations on a Finite Set

1. Preambule

In 1941, Post presented the complete description of the countably many clones on two elements set. It turned out that, all such clones are finitely generated and the lattice of these clones is countable. The structure of the lattice of clones on finitely many (but more than 2) elements is more complex and is of the cardinality $2^{{\aleph }_0}$. For the k-element case, with $k\ge 3$, not much is known about the structure of the lattice of clones in spite of the efforts made by many researchers in this area. Therefore, every new piece of information is considered valuable. Indeed, it would be very interesting to know the clone lattice on the next level (below the maximal clones) and even a partial description will shed more light onto its structure. The description of this lattice will also help to find a simple basis for any set of operations on a finite set which is useful for the progress of logic and computer science. The complete description of all submaximal clones is known (below certain maximal clones) only for the 2-element case and the 3-element case (see [8,9,10]), however the result in ([8]) and many result in the literature on clones including those discussed in ([8,9,13,14,16]), require intensive knowledge of submaximal clones on arbitrary finite sets.

Recently, we have determined the eleven types of binary relations $\sigma$ such that the clones $Pol\sigma \cap Pol\rho$ are covered by $Pol\rho$, where $\rho$ is a fixed h-ary central relation ($h\ge 2$) on a given finite set (see [6]). Clone theory is considered to be very important because of its use to understand universal algebras and the problem of decidability in computer science. In fact, clone theory is also related to constraint satisfaction problem. The main problem is to decide whether a function on a finite set preserving some relations on this set belongs to a fixed clone on that set. For example, given a first order structure $(A;F;R)$ where A is a finite set; F the set of fundamental operations on A and R the set of all relations on A. It is known that $PolR$, the set of function on A preserving every relations on R, is a clone. Given a function $g\in PolR$, decide whether g belongs to the clone generated by the fundamental operations of $(A;F;R)$ is NP-complete. Clone theory is also useful to determine weak bases on sets of larger size (see [1,2,3]). Clone theory could also be useful for determining minimal bases to facilitate calculations in logics based on finite residuated lattices, which are tools of artificial intelligence. Our objective is to describe all submaximal clones determined by regular relations on finite set (because these submaximal clones are not studied in the literature for non-trivial regular relations on given finite set) to complete the Burle’s theorem (see [9], Page 161) . Since regular relation is difficult to handle, We begin this project by the cases of binary relations. A natural way to extend this research would be to consider relations of arity greater than 2.

2. Structure of the Paper

This paper is organized as follows : In Section 3 we give some necessary tools to understand this work and state the main result of this paper. In Section 4 we characterize the non-trivial equivalence relation $\sigma$ such that the clone $Pol\sigma \cap Pol\lambda$ is maximal below $Pol\lambda$. Section 5 is devoted to the case of bounded partial order. We prove that there is no submaximal clone. Section 6 studies the case of prime permutation. We show that for any prime permutation relation s on $E_k$, $Pol\lambda \cap Pol{s}^{\circ }$ is maximal below $Pol\lambda$ if and only if $\lambda$ is ${\theta }_s$-closed. Section 7 is reserved to the case of binary central relations. We prove that $Pol\sigma \cap Pol\lambda$ is maximal below $Pol\lambda$ if and only if $C_{\sigma }\cap {\left[x\right]}_{ϵ}$ is not empty where $C_{\sigma }$ is the set of central element of $\sigma$ and $ϵ$ is the equivalence relation characterizing $\lambda$.

3. Preliminaries

In this Section, we give some basic notions necessary to follow this work; for more details you can refer to ([7,9,11,12,13]). We begin with some generality on clone theory.

Let A be a fixed finite set with k elements, n and h be positive integers. An n-ary operation on A is a function $f:A^n\to A$. By ${\mathcal{O}}_A^{\left(n\right)}$ we denote the set of all n-ary operations on A, and by ${\mathcal{O}}_A$ the set $\underset{n\ge 1}{\cup }{\mathcal{O}}_A^{\left(n\right)}$ of all finitary operations on A. For $1\le i\le n$, the i-th projection on A is the operation ${\pi }_i^{\left(n\right)}:A^n\to A,(a_1,\dots ,a_n)\mapsto a_i$. For arbitrary positive integers m and n, there is a one-to-one correspondence between the functions $f:A^n\to A^m$ and the m-tuples $f=(f_1,\dots ,f_m)$ of functions $f_i:A^n\to A$ (for $i=1,\dots ,m$) via $f\mapsto f=(f_1,\dots ,f_m)$ with $f_i={\pi }_i^{\left(m\right)}\circ f$ for all $i=1,\dots ,m$. In particular, ${\pi }^{\left(n\right)}=({\pi }_1^{\left(n\right)},\dots ,{\pi }_n^{\left(n\right)})$ corresponds to the identity function $f:A^n\to A^n$. From now on, we will identify each function $f:A^n\to A^m$ with the corresponding m-tuples $f=(f_1,\dots ,f_m)\in {\left({\mathcal{O}}_A^{\left(n\right)}\right)}^m$ of n-ary operations. Using this convention, the composition of two functions $f=(f_1,\dots ,f_m):A^n\to A^m$ and $g=(g_1,\dots ,g_p):A^m\to A^p$ can be described as follows: $g\circ f=(g_1\circ f,\dots ,g_p\circ f)=(g_1(f_1,\dots ,f_m),\dots ,g_p(f_1,\dots ,f_m))$ where $g_i(f_1,\dots ,f_m)\left(a\right)=g_i(f_1\left(a\right),\dots ,f_m\left(a\right))$ for all $a\in A^n$ and $1\le i\le p$.

A clone on A is a subset $\mathcal{C}$ of ${\mathcal{O}}_A$ that contains all projections and is closed under composition; that is ${\pi }_i^{\left(n\right)}\in \mathcal{C}$ for all $n\ge 1$ and $1\le i\le n$, and $g\circ f\in {\mathcal{C}}^{\left(n\right)}$ whenever $g\in {\mathcal{C}}^{\left(m\right)}$ and $f\in {\left({\mathcal{C}}^{\left(n\right)}\right)}^m$ (for $m,n\ge 1$). The clones on A form a complete lattice ${\mathcal{L}}_A$ under inclusion. Moreover, for each set $F\subseteq {\mathcal{O}}_A$ of operations, there exists a smallest clone that contains F, which will be denoted by $\langle F\rangle$ and will be called clone generated by F. Clones can also be described via invariant relations. An h-ary relation $\rho$ on A is a subset of $A^h$. For an n-ary operation $f\in {\mathcal{O}}_A^{\left(n\right)}$ and an h-ary relation $\rho$ on A, we say that f preserves $\rho$ (or $\rho$ is invariant under f, or f is a polymorphism of $\rho$) if whenever f is applied coordinatewise to h-tuples from $\rho$, the resulting h-tuple belongs to $\rho$ i.e., for all $(a_{1,i},\dots ,a_{h,i})\in \rho ,i=1,\dots ,n$, $(f(a_{1,1},\dots ,a_{1,n}),f(a_{2,1},\dots ,a_{2,n})\dots ,f(a_{h,1},\dots ,a_{h,n}))\in \rho$. For any family $\mathcal{R}$ of (finitary) relations on A, the set $Pol\mathcal{R}$ of all operations $f\in {\mathcal{O}}_A$ that preserve each relation in R is easily seen to be a clone on A. Moreover, if A is finite, then it is a well-known fact that every clone on A is of the form $PolR$ for some family R of relations on A. If $R=\left\{\rho \right\}$, we write $Pol\rho$ for $Pol\left\{\rho \right\}$. Let $\rho \subseteq A^h$; for an integer $m>1$ and $a_i=(a_{1,i},\dots ,a_{m,i})\in A^m,1\le i\le h$, we will write $(a_1,\dots ,a_h)\in \rho$ if for all $j\in \{1,\dots ,m\},(a_{j,1},\dots ,a_{j,h})\in \rho$.

Since A is finite, it is well known that every clone on A other than ${\mathcal{O}}_A$ is contained in a maximal clone. We say that an h-ary relation $\rho$ on A is totally reflexive (reflexive for $h=2$) if $\rho$ contains the h-ary relation ${\iota }_A^h$ defined by

$${\iota }_A^h=\{(a_1,\dots ,a_h)\in A^h\mid \exists i,j\in \{1,\dots ,h\}:i\ne jand{a}_i=a_j\},$$

and is totally symmetric (symmetric if $h=2$) if $\rho$ is invariant under any permutation of its coordinates. If $\rho$ is totally reflexive and totally symmetric, we define the center of $\rho$, denoted by $C_{\rho }$, as follows:

$$C_{\rho }=\{a\in A\mid (a,a_2,\dots ,a_h)\in \rho forall{a}_2,\dots ,a_h\in A\}.$$

We say that $\rho$ is a central relation if $\rho$ is totally reflexive, totally symmetric and has a nonvoid center which is a proper subset of A. The elements of $C_{\rho }$ are called central elements of $\rho$. A nonvoid and proper subset of A is called unary central relation. Let $\rho$ be a binary relation on A, $\rho$ is an equivalence relation if $\rho$ is symmetric, reflexive and transitive; $\rho$ is non-trivial if $\rho \ne A^2$ and $\rho \ne \left\{\right(a,a)\mid a\in A\}$, $\rho$ is a bounded partial order if $\rho$ is reflexive, transitive, antisymmetric and there exist $\perp ,\top \in A$ such that for all $x\in A,(x,\top )\in \rho$ and $(\perp ,x)\in \rho$. $\rho$ is a prime permutation relationif $\rho =s^{\circ }$ the graph of the prime permutation s with the same length p (where p is prime).

For an integer $h\ge 3$, a family $T=\{{\theta }_1,\dots ,{\theta }_m\}$ ($m\ge 1$) of equivalence relations on A is called h-regular if each ${\theta }_i$ ($1\le i\le m$) has exactly h blocks, and for arbitrary blocks $B_i$ of ${\theta }_i$ ($1\le i\le m$), the intersection $B_1\cap B_2\cap \dots \cap B_m$ is nonempty. To each h-regular family $T=\{{\theta }_1,\dots ,{\theta }_m\}$ of equivalence relation on A, we associate an h-ary relation $\lambda$ on A as follows:

$$\lambda =\{(a_1,\dots ,a_h)\in A^h\mid (\forall i)(\exists j)j\ne i such that (a_i,a_j)\in {\theta }_i\}.$$

Relations of the form $\lambda$ are called h-regular (or h-regularly generated) relations. It is clear from the definition that h-regular relations are totally reflexive and totally symmetric, their arity h satisfies $3\le h\le \left|A\right|$, and $h=\left|A\right|$ holds if and only if T is the one-element family consisting of the equality relation. The characterization of maximal clones on finite sets given in [15] is well known. Therefore we have that whenever a relation $\gamma$ is a non-trivial equivalence relation, a bounded partial order, a binary central relation, a prime permutation relation or an h-regular relation with ($3\le h\le \left|A\right|$), the clone $Pol\gamma$ is a maximal clone. The binary diagonal relation on A is ${\Delta }_A=\{(a,a)\mid a\in A\}.$

We continue with some fundamental results on clone theory. The following result characterizes all maximal clones on finite sets.

Theorem 3.1. 

[15] For each finite set A with $\left|A\right|\ge 2$, the maximal clones on A are the clones of the form $Pol\rho$, where ρ is a relation of one of the following six types:

(1) 

a bounded partial order on A,

(2) 

a prime permutation on A,

(3) 

a prime affine relation on A,

(4) 

a non-trivial equivalence relation on A,

(5) 

a central relation on A,

(6) 

an h-regular relation on A.

For $h\in \mathbb{N}\setminus \left\{0\right\}$, $R_A^{\left(h\right)}$ denotes the set of h-ary relations on A. $R_A$ denotes the set of all finitary relations on A. For a set F of finitary operations on A, $Inv\left(F\right)$ denotes the set of relations preserved by every operation in F. A relational clone on A is a subset R of $R_A$ such that $Inv(PolR)=R$. The set of all relational clones with inclusion is a lattice called lattice of relational clones. For $R\subseteq R_A$, $\left[R\right]$ denotes the relational clone generated by R.

For instance, it is nice for us to give the following remark useful to justify some inclusions between clones.

Remark 3.2.([9], Theorem 2.6.2, 2.6.3 Page 132) Let $R\subseteq R_A$,

1. 

If $f\in PolR$, then $f\in Pol\left(\right[R\left]\right)$.

2. 

If σ and ${\sigma }^{\prime }$ are relations such that ${\sigma }^{\prime }\in \left[\left\{\sigma \right\}\right]$, then $Pol\sigma \subseteq Pol{\sigma }^{\prime }$.

For two clones $\mathcal{C}$ and $\mathcal{D}$ on A, we say that $\mathcal{C}$ is maximal in $\mathcal{D}$ if $\mathcal{D}$ covers $\mathcal{C}$ in ${\mathcal{L}}_A$, we also say that $\mathcal{C}$ is submaximal if $\mathcal{C}$ is maximal in a clone $\mathcal{D}$ and $\mathcal{D}$ is a maximal clone on A. For a maximal clone $\mathcal{D}$, there are two types of clones $\mathcal{C}$ being maximal in $\mathcal{D}$: $\mathcal{C}$ is meet-reducible if $\mathcal{C}=\mathcal{D}\cap \mathcal{F}$ for a maximal clone $\mathcal{F}$ distinct from $\mathcal{D}$ (but not necessarily unique) and $\mathcal{C}$ is meet-irreducible if it is not meet-reducible.

Definition 3.3.([9], Page 126) Let $h\in \mathbb{N}\setminus \left\{0\right\}$. An h-ary relation $\rho \in R_A^{\left(h\right)}$ is called diagonal relation if there exists an equivalence relation ε on $\{1,\dots ,h\}$ such that $\rho :=\{(a_1,\dots ,a_h)\in A^h\mid (i,j)\in \epsilon ⟹a_i=a_j\}$.

The set of all diagonal relations on A is denoted by $D_A$ and ${\displaystyle D_A=\{\varnothing \}\cup \bigcup _{h\ge 1}{D}_A^{\left(h\right)}}$, where $D_A^{\left(h\right)}$ is the set of all h-ary diagonal relations on A. In particular, $A^h$ and ${\delta }^h=\{(x,x,\dots ,x)\in A^h\mid x\in A\}$ are diagonal relations on A. Fore more details, see [9].

From now on, we assume that we are working on the set $E_k=\{0,1,\dots ,k-1\}$. For any integer $2\le h\le k$, we denote by ${\iota }_k^h$ the set ${\iota }_{E_k}^h$. It is well known (From Theorem 3.1, relation of type $\left(6\right)$) that the Słupecki clone $Pol{\iota }_k^k$ is a maximal clone. Subclones of $Pol{\iota }_k^k$ are known (see [9], Page 161, Burle’s Theorem). If $\sigma$ is an equivalence relation, we denote by ${\left[a\right]}_{\sigma }$ the $\sigma$-class of a.

Remark 3.4 

For any h-regular relation on $E_k$, we have $3\le h\le k$. The case $h=k$ is given by the Słupecki clone $Pol{\iota }_k^k$. Submaximal clones of Słupecki clone are known.

In this paper, we consider the h-ary ($3\le h<k$) regular relation $\lambda$ on $E_k$ defined by the h-ary regular family $T=\{{\theta }_1,\dots ,{\theta }_m\}$, where $m\ge 1$ and every non-trivial equivalence relation ${\theta }_i,1\le i\le m$ contains h blocks.

Now we define some types of binary relations useful to express our results. Let k and h be two integers such that $h,k\ge 3$. For a prime permutation s of order p on $E_k$, we denote by ${\theta }_s$ the equivalence relation consisting of pairs $(a,b)\in E_k^2$ with $a=s^i\left(b\right)$ for some $0\le i<p$.

Definition 3.5 

Let λ be an h-ary regular relation on $E_k$ and σ be a non-trivial equivalence relation on $E_k$.

1. 

We say that λ is σ-closed if $(a_1,\dots ,a_h)\in \lambda$ whenever $(u_1,\dots ,u_h)\in \lambda$ for some $u_1,\dots ,u_h\in E_k$ with $(a_i,u_i)\in \sigma$ for all $1\le i\le h$.

2. 

There is a transversal T for the σ-classes means that there exist $u_1,\dots ,u_t\in E_k$ such that $(u_i,u_j)\notin \sigma$ for all $1\le i<j\le t,(u_{i_1},u_{i_2},\dots ,u_{i_h})\in \lambda$ for all $1\le i_1\dots ,i_h\le t$ and $T=\{u_1,\dots ,u_t\}$.

Let $\epsilon$ be the binary relation defined on $E_k$ by

$$\epsilon =\{(a,b)\in E_k^2\mid (a,b,a_3,\dots ,a_h)\in \lambda ,\forall a_3,\dots ,a_h\in E_k\}$$

and the h-ary relation ${\gamma }_h$ defined by :

$${\gamma }_h=\{(a_1,\dots ,a_h)\in E_k^h\mid \exists v_1,\dots ,v_h\in E_k,(v_1,\dots ,v_h)\in \lambda ,\mathit{and}$$

$$\forall 1\le i\le h,\forall j_1,\dots ,j_{h-2}\in \{1,\dots ,h\}\setminus \left\{i\right\},(v_i,a_i,a_{j_1},\dots ,a_{j_{h-2}})\in \lambda \}$$

We will denote by $\underline{h}$, the set $\{1,\dots ,h\}$ and $S_h$, the set of all permutations on $\{1,\dots ,h\}$, $h\ge 2$.

Now we state the main result of this paper.

Theorem 3.6 

Let λ be an h-ary regular relation on $E_k$ determined by the h-ary regular family $T=\{{\theta }_1,\dots ,{\theta }_m\}$ and σ be a binary relation on $E_k$ such that $Pol\sigma$ is a maximal clone. Then $Pol\sigma \cap Pol\lambda$ is a submaximal clone of $Pol\lambda$ if and only if one of the following conditions is satisfied:

I. 

σ is a non-trivial equivalence relation and λ is σ-closed;

II. 

σ is a non-trivial equivalence relation, $\lambda ={\gamma }_h$ and for all $x,y\in E_k$,

${\left[x\right]}_{\sigma }\cap {\left[y\right]}_{\epsilon }\ne \varnothing$;

III. 

σ is a prime permutation and λ is ${\theta }_{\sigma }$-closed;

IV. 

σ is a binary central relation and for all $x\in E_k$, $C_{\sigma }\cap {\left[x\right]}_{\epsilon }\ne \varnothing$.

Before the Proof of Theorem 3.6, we give some examples to fix some ideas.

Example 3.7 

Let $E_{10}=\{0,1,2,3,4,5,6,7,8,9\}$. Let ${\theta }_1$ be the equivalence relation with blocks $\{0,1,2\},$$\{3,4,5\},\{6,7,8,9\}$ and ${\theta }_2$ the equivalence relation with blocks $\{0,3,6\},\{1,4,7\},\{2,5,8,9\}$. Then $T=\{{\theta }_1,{\theta }_2\}$ is an 3-ary regular family.

1. 

Set $\theta ={\theta }_1\cap {\theta }_2$. We have $\theta ={\Delta }_{E_4}\cup \{(8,9),(8,9)\}$ and λ is θ-closed. Thus $Pol\lambda \cap Pol\theta$ is maximal below $Pol\lambda$.

2. 

Let $\epsilon =\{(a,b)\in E_{10}^2\mid (a,b,c)\in \lambda ,\forall c\in E_{10}\}$ and σ be the equivalence relation with blocks $\{0,1,2,3,4,5,6,7\},\{8,9\}$. It is easy to show that $\epsilon ={\theta }_1\cap {\theta }_2=\theta$. Furthermore, $\forall x,y\in E_{10}$, ${\left[x\right]}_{\epsilon }\cap {\left[y\right]}_{\sigma }\ne \varnothing$. Hence $Pol\lambda \cap Pol\sigma$ is maximal below $Pol\lambda$.

3. 

Let $\pi =\left(01\right)\left(23\right)\left(45\right)\left(67\right)\left(89\right)$. π is a prime permutation on $E_{10}$ of order 2 and ${\theta }_{\pi }$ is the equivalence relation with blocks $\{0,1\},\{2,3\},\{4,5\},\{6,7\},\{8,9\}$. Set $T_1:=\left\{{\theta }_{\pi }\right\}$. Then the regular relation ${\lambda }_{T_1}$ is ${\theta }_{\pi }$-closed. So $Pol{\lambda }_{T_1}\cap Pol{\pi }^{\circ }$ is maximal below $Pol{\lambda }_{T_1}$.

The Proof of Theorem 3.6 follows from propositions 4.1, 5.1, 6.2 and 7.1.

Firstly we look at the non-trivial equivalence case.

4. Non-Trivial Equivalence Relation

Let $\lambda$ be an h-ary regular relation $(3\le h<k)$ on $E_k$ determined by the h-regular family $T=\{{\theta }_1,\dots ,{\theta }_m\}$, $m\ge 1$ and $\sigma$ be a non-trivial equivalence relation on $E_k$ with t classes ($t\ge 2$).

Here we state the main result of this section.

Proposition 4.1 

Let λ be an h-ary regular relation $(3\le h<k)$ on $E_k$ determined by the h-regular family $T=\{{\theta }_1,\dots ,{\theta }_m\}$, $m\ge 1$ and σ be a non-trivial equivalence relation on $E_k$ with t classes ($t\ge 2$). Then $Pol\lambda \cap Pol\sigma$ is maximal below $Pol\lambda$ if and only if (λ is σ-closed) or ($\lambda ={\gamma }_h$ and for all $x,y\in E_k$, ${\left[x\right]}_{\sigma }\cap {\left[y\right]}_{\epsilon }\ne \varnothing$).

The necessary condition of Proposition 4.1 is given by Lemmas 4.2, and 4.4.

Lemma 4.2 

Let λ be an h-ary regular relation $(3\le h<k)$ on $E_k$ determined by the h-regular family $T=\{{\theta }_1,\dots ,{\theta }_m\}$, $m\ge 1$ and σ be a non-trivial equivalence relation on $E_k$ with t classes ($t\ge 2$) such that λ is σ-closed. Then $Pol\lambda \cap Pol\sigma$ is maximal below $Pol\lambda$.

Before the Proof of Lemma 4.2, we will give some useful properties of $\sigma$ in Lemma 4.3. For $a,b\in E_k^n$, we write $(a,b)\in \sigma$ if $(a_i,b_i)\in \sigma$ for all $1\le i\le n$, where $a:=(a_1,\dots ,a_n)$ and $b:=(b_1,\dots ,b_n)$. Let $g\in Pol\lambda \setminus Pol\sigma$ be an n-ary operation, then there exist $(a_1,b_1),\dots ,(a_n,b_n)\in \sigma$ such that $(g(a_1,\dots ,a_n),g(b_1,\dots ,b_n))\notin \sigma$.

Lemma 4.3 

Let $m\ge 1$ be an integer, λ be an h-ary regular relation $(3\le h<k)$ on $E_k$ determined by the h-regular family $T=\{{\theta }_1,\dots ,{\theta }_m\}$, $m\ge 1$ and σ be a non-trivial equivalence relation on $E_k$ with t classes ($t\ge 2$), let $g\in Pol\lambda \setminus Pol\sigma$ be an n-ary operation . Then for all $e,d\in E_k^m$ such that $e\ne d$ and $(e,d)\in \sigma$, there exists an m-ary operation $f_{ed}\in \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle$ such that $(f_{ed}\left(e\right),f_{ed}\left(d\right))\notin \sigma$.

Proof. 

We distinguish two cases: (i) $m=1$ and (ii) $m>1$.

Let $g\in Pol\lambda \setminus Pol\sigma$ be an n-ary operation. Then there exist $(a_1,b_1),\dots ,(a_n,b_n)\in \sigma$ such that $(g(a_1,\dots ,a_n),g(b_1,\dots ,b_n))\notin \sigma$. For $m=1$, let $e,d\in E_k$ such that $e\ne d$ and $(e,d)\in \sigma$, we will construct a unary operation $f_{ed}\in \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle$ such that $(f_{ed}\left(e\right),f_{ed}\left(d\right))\notin \sigma$. For $1\le i\le n$, consider the unary operation $f_{ed}^i$ defined on $E_k$ by $f_{ed}^i\left(x\right)=a_i$ if $x=e$ and $f_{ed}^i\left(x\right)=b_i$ otherwise. Since $(a_i,b_i)\in \sigma$ and $Im{f}_{ed}^i=\{a_i,b_i\}$, then $f_{ed}^i\in Pol\sigma$. Furthermore, $\lambda$ is totally reflexive, thus $f_{ed}^i\in Pol\lambda$ and $f_{ed}^i\in Pol\lambda \cap Pol\sigma$. Let $f_{ed}$ be the unary operation defined on $E_k$ by : $f_{ed}\left(x\right)=g\left(f_{ed}^1\left(x\right),\dots ,f_{ed}^n\left(x\right)\right)$, $\forall x\in E_k$. We have $f_{ed}\in \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle$ (because $f_{ed}^i\in Pol\lambda \cap Pol\sigma$) and $(f_{ed}\left(e\right),f_{ed}\left(d\right))=(g(a_1,\dots ,a_n),g(b_1,\dots ,b_n))$$\notin \sigma$. Hence, $f_{ed}$ is the operation wanted.

Let $m>1$, $e,d\in E_k^m$ such that $e\ne d$ and $(e,d)\in \sigma$, then there exists $1\le i\le m$ such that $e_i\ne d_i$ and $(e_i,d_i)\in \sigma$, we can construct $f_{e_i{d}_i}$ as in Case$m=1$. Set $f_{ed}=f_{e_i{d}_i}\circ {\pi }_i^{\left(m\right)}$. We have $f_{ed}\in \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle$ and $(f_{ed}\left(e\right),f_{ed}\left(d\right)=(g(a_1,\dots ,a_n),g(b_1,\dots ,b_n))\notin \sigma$. Hence $f_{ed}$ is the m-ary operation wanted. □

Proof. (Proof of Lemma 4.2). Let $g^n\in Pol\lambda \setminus \left(Pol\sigma \cap Pol\lambda \right)$ be an n-ary operation. Using the operation constructed in Lemma 4.3 we will show that $\langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle =Pol\lambda$.

We have $\langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle \subseteq Pol\lambda$. It remains to prove that $Pol\lambda \subseteq \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle$. Let $u\in Pol\lambda$ be an m-ary operation on $E_k$. Let us show that $u\in \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle$. For $e,d\in E_k^m$ such that $e\ne d$ and $(e,d)\in \sigma$, there exists an m-ary operation $f_{ed}\in \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle$ such that $(f_{ed}\left(e\right),f_{ed}\left(d\right))\notin \sigma$ (by Lemma 4.3). Set $S=\{f_{ed}\mid e,d\in E_k^m,(e,d)\in \sigma ande\ne d\}$, denoted for reason of simple notation by $S=\{f_i\mid 1\le i\le \ell \}$ ( $Card\left(S\right)=\ell$) and define the mapping $ext:E_k^m⟶E_k^{m+\ell }$, by $ext\left(x\right)=(\mathbf{x},f_1\left(x\right),\dots ,f_{\ell }\left(x\right))$, for all $x\in E_k^m$.

Let $x,y\in E_k^m$ such that $x\ne y$, we have $(ext(x),ext(y\left)\right)\notin \sigma$. In fact, if $(ext(x),ext(y\left)\right)\in \sigma$, then $(x,y)\in \sigma ,x\ne y$ and $(f_{xy}\left(x\right),f_{xy}\left(y\right))\in \sigma$. Thus, $f_{xy}\in S$; contradiction(due to Lemma 4.3).

Now we define the operation H on the range of ext by $H(ext(x\left)\right)=u\left(x\right)$. We choose and fix $T=\{e_1,\dots ,e_t\}$ such that $(e_i,e_j)\notin \sigma$, $\forall 1\le i<j\le t$ (due to $\sigma$ contains t blocks). Define the unary operation $\alpha$ on $E_k$ by $\alpha \left(a\right)=i$ if and only if $(a,e_i)\in \sigma$. Thus for all $a\in E_k$,$(a,e_{\alpha \left(a\right)})\in \sigma$. We have $ext\left(E_k^m\right)E_k^{m+\ell }$, so we can construct an extension $\tilde{H}$ of H on $E_k^{m+\ell }$ as follow:

$\tilde{H}\left(y\right)=\left\{\begin{array}{cc}u\left(u\right)\hfill & if\exists u\in E_k^m:(ext\left(u\right),y)\in \sigma ,\hfill \\ u\left(e_{\alpha \left(y_1\right)},\dots ,e_{\alpha \left(y_m\right)}\right)\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$

Let us show that $\tilde{H}\in Pol\lambda \cap Pol\sigma$. Firstly, we show that $\tilde{H}\in Pol\sigma$. Let $a=(a_1,\dots ,a_{m+\ell })$ and $b=(b_1,\dots ,b_{m+\ell })$ such that $(a,b)\in \sigma$. We distinguish two cases:

Case 1: There exists $u\in E_k^m$ such that $(ext(u),a)\in \sigma$. Then $(ext(u),b)\in \sigma$ (because $(a,b)\in \sigma$ and $\sigma$ is transitive); hence $(\tilde{H}\left(a\right),\tilde{H}\left(b\right))=(u\left(u\right),u\left(u\right))\in \sigma$.

Case 2: For all $u\in E_k^m$, $\left(ext\right(a),u)\notin \sigma$, then for all $u\in E_k^m$, $\left(ext\right(u),b)\notin \sigma$ ( because $(\mathbf{a},\mathbf{b})\in \sigma$ and $\sigma$ is transitive). For all $1\le i\le m+l$, $e_{\alpha \left(a_i\right)}=e_{\alpha \left(b_i\right)}$ (since $(a_i,b_i)\in \sigma$).

Therefore, $(\tilde{H}\left(a\right),\tilde{H}\left(b\right))=\left(u\left(e_{\alpha \left(a_1\right)},\dots ,e_{\alpha \left(a_{m+\ell }\right)}\right),u\left(e_{\alpha \left(b_1\right)},\dots ,e_{\alpha \left(b_{m+\ell }\right)}\right)\right)\in \sigma$. Thus $\tilde{H}\in Pol\sigma$.

Secondly, we show that $\tilde{H}\in Pol\lambda$.

Let $a_1=(a_{1,1},\dots ,a_{1,h}),\dots ,a_m=(a_{m,1},\dots ,a_{m,h}),\dots ,a_{m+l}=(a_{m+\ell ,\ell },\dots ,a_{m+\ell ,h})\in \lambda$, then for all $y\in E_k^{m+\ell }$, there exists $v\in E_k^m$ such that $\tilde{H}\left(y\right)=u\left(v\right)$ and $(y_1,\dots ,y_m,v)\in \sigma$ (by the definition of $\tilde{H}$).

For all $1\le j\le h$, set $d_j=(a_{1,j},\dots ,a_{m,j},\dots ,a_{m+l,j})$, $b_j=\left(e_{\alpha \left(a_{1,j}\right)},\dots ,e_{\alpha \left(a_{m,j}\right)},\dots ,e_{\alpha \left(a_{m+\ell ,j}\right)}\right)$, $d_j^{\prime }=(a_{1,j},\dots ,a_{m,j})$ and $b_j^{\prime }=\left((e_{\alpha \left(a_{1,j}\right)},\dots ,e_{\alpha \left(a_{m,j}\right)}\right)$. We have $(d_j,b_j),(d_j^{\prime },b_j^{\prime })\in \sigma$, $1\le j\le h$.

Since $d_j=(a_{1,j},\dots ,a_{m,j},\dots ,a_{m+\ell ,j})\in E^{m+l}$, there exists $v_j=(v_{1,j},\dots ,v_{m,j})\in E_k^m$ such that $\tilde{H}\left(d_j\right)=u\left(v_j\right)$ and $\left((a_{1,j},\dots ,a_{m,j}),(v_{1,j},\dots ,v_{m,j})\right)\in \sigma$. Thus $(\tilde{H}\left(d_1\right),\dots ,\tilde{H}\left(d_h\right))=(u\left(v_1\right),\dots ,u\left(v_h\right))$.

For all $1\le i\le m$ and $1\le j\le h$, $(v_{i,j},e_{\alpha \left(a_{i,j}\right)})\in \sigma$ and $\left(e_{\alpha \left(a_{i,1}\right)},e_{\alpha \left(a_{i,2}\right)},\dots ,e_{\alpha \left(a_{i,h}\right)}\right)\in \lambda$; since $\lambda$ is $\sigma$-closed, we have $(v_{i,1},\dots ,v_{i,h})\in \lambda$, $\forall 1\le i\le m$.

Therefore $\left(u(v_{1,1},\dots ,v_{m,1}),u(v_{1,2},\dots ,v_{m,2}),(u(v_{1,h},\dots ,v_{m,h})\right)=(u\left(v_1\right),\dots ,u\left(v_h\right))=$$(\tilde{H}\left({\mathbf{d}}_1\right),\dots ,\tilde{H}\left({\mathbf{d}}_h\right))\in \lambda$ (because $u\in Pol\lambda$). Thus $\tilde{H}\in Pol\lambda$. It follows that $\tilde{H}\in Pol\lambda \cap Pol\sigma$ and for all $x\in E_k^m$, $\tilde{H}\left(x,f_{m+\ell }\left(x\right),\dots ,f_{m+l}\left(x\right)\right)=\tilde{H}\left(ext\left(x\right)\right)=\tilde{H}\circ ext\left(x\right)=u\left(x\right)$. Thus $u=\tilde{H}\circ ext\in \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle$. Therefore $Pol\lambda \subseteq \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle$. We conclude that $\langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle =Pol\lambda$ and $Pol\lambda \cap Pol\sigma$ is maximal below $Pol\lambda$. □

Lemma 4.4 

Let λ be an h-ary regular relation $(3\le h<k)$ on $E_k$ determined by the h-regular family $T=\{{\theta }_1,\dots ,{\theta }_m\}$, $m\ge 1$ and σ be a non-trivial equivalence relation on $E_k$ with t classes ($t\ge 2$) such that $\lambda ={\gamma }_h$ and for all $x,y\in E_k$, ${\left[x\right]}_{\sigma }\cap {\left[y\right]}_{\epsilon }\ne \varnothing$. Then $Pol\lambda \cap Pol\sigma$ is maximal below $Pol\lambda$.

Before the Proof of Lemma 4.4, we recall some notations. The binary relation $\epsilon$ is defined by

$$\epsilon =\{(a,b)\in E_k^2\mid (a,b,a_3,\dots ,a_h)\in \lambda ,\forall a_3,\dots ,a_h\in E_k\}.$$

Note that $T=\{u_1,\dots ,u_t\}$ is a transversal for the $\sigma$-classes such that $T^h\subseteq \lambda$ and the h-ary relation ${\gamma }_h$ is defined by :

${\gamma }_h=\{(a_1,\dots ,a_h)\in E_k^h\mid \exists v_1,\dots ,v_h\in E_k:(v_1,\dots ,v_h)\in \lambda ,\mathit{and}\forall 1\le i\le h,\forall j_1,\dots ,j_{h-2}\in \{1,\dots ,h\}\setminus \left\{i\right\},(v_i,a_i,a_{j_1},\dots ,a_{j_{h-2}})\in \lambda \}$.

For all $x,y\in E_k$, we choose and fix ${\alpha }_{xy}\in {\left[x\right]}_{\sigma }\cap {\left[y\right]}_{\epsilon }$ (because ${\left[x\right]}_{\sigma }\cap {\left[y\right]}_{\epsilon }\ne \varnothing$).

Proof. 

Let $g^n\in Pol\lambda \setminus \left(Pol\sigma \cap Pol\lambda \right)$ be an n-ary operation. Using the operation constructed in Lemma 4.3, we will show that $\langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle =Pol\lambda$. We have $\langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle \subseteq Pol\lambda$. It remains to prove that $Pol\lambda \subseteq \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle$. Let $u\in Pol\lambda$ be an m-ary operation on $E_k$. We will show that $u\in \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}$. Using the notations given in the proof of Lemma 4.2, we define the operation $\tilde{H}$ as follows:

$\tilde{H}\left(y\right)=\left\{\begin{array}{cc}{\alpha }_{u\left(x\right)u(y_1,\dots ,y_m)}\hfill & if\exists x\in E_k^m:(ext\left(x\right),y)\in \sigma ,\hfill \\ {\alpha }_{u_{1}u(y_1,\dots ,y_m)}\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$

Let us show that $\tilde{H}\in Pol\lambda \cap Pol\sigma$. Firstly, we show that $\tilde{H}\in Pol\sigma$. Let $a=(a_1,\dots ,a_{m+\ell })$ and $b=(b_1,\dots ,b_{m+\ell })$ such that $(\mathbf{a},\mathbf{b})\in \sigma$. We distinguish two cases:

Case 1: There exits $x\in E_k^m$ such that $(ext(x),a)\in \sigma$. Then $(ext(x),b)\in \sigma$ (because $(\mathbf{a},\mathbf{b})\in \sigma$ and $\sigma$ is transitive). We obtain

$(\tilde{H}\left(a\right),\tilde{H}\left(b\right))=({\alpha }_{u\left(x\right)u(a_1,\dots ,a_m)},{\alpha }_{u\left(x\right)u(b_1,\dots ,b_m)})\in \sigma$.

Case 2: For all $x\in E_k^m$, $(ext(x),a)\notin \sigma$, then for all $x\in E_k^m$, $(ext(x),b)\notin \sigma$ (because $(a,b)\in \sigma$ and $\sigma$ is transitive).

We have $(\tilde{H}\left(a\right),\tilde{H}\left(b\right))=({\alpha }_{u_{1}u(a_1,\dots ,a_m)},{\alpha }_{u_{1}u(b_1,\dots ,b_m)})\in \sigma$. Thus $\tilde{H}\in Pol\sigma$.

Secondly, we show that $\tilde{H}\in Pol\lambda$.

Let $a_1=(a_{1,1},\dots ,a_{1,h}),\dots ,a_m=(a_{m,1},\dots ,a_{m,h}),\dots ,a_{m+l}=(a_{m+\ell ,\ell },\dots ,a_{m+\ell ,h})\in \lambda$. For all $1\le j\le h$, set $d_j=(a_{1,j},\dots ,a_{m,j},\dots ,a_{m+l,j})$ and $b_j=\left(a_{1,j},\dots ,a_{m,j}\right)$. We will show that $(\tilde{H}\left(d_1\right),\dots ,\tilde{H}\left(d_h\right))\in \lambda$. We look at the following two cases:

Case 1: For $1\le j\le h$, $\exists x_j=(x_{1,j},\dots ,x_{m,j})\in E_k^m$ such that $(ext\left(x_j\right),d_j)\in \sigma$. Then, $\tilde{H}\left(d_j\right)={\alpha }_{u(x_{1,j},\dots ,x_{m,j})u(a_{1,j},\dots ,a_{m,j})}={\alpha }_{u\left(x_j\right)u\left(b_j\right)}$. For all $1\le j\le h$, set $v_j=u\left(b_j\right)$. We have $(v_1,\dots ,v_h)\in \lambda$ (because $u\in Pol\lambda$). Since $\lambda ={\gamma }_h$, we obtain $(\tilde{H}\left(d_1\right),\dots ,\tilde{H}\left(d_h\right))=({\alpha }_{u\left(x_1\right)u\left(b_1\right)},\dots ,{\alpha }_{u\left(x_h\right)u(b_h})\in \lambda$ (due to the fact that for $1\le i\le h$, $({\alpha }_{u\left(x_i\right)u\left(b_i\right)},v_i)\in \epsilon$).

Case 2: There exists $1\le j\le h$ such that $\forall x\in E_k^m$, $(ext\left(x\right),d_j)\notin \sigma$. Then $\tilde{H}\left(d_j\right)={\alpha }_{u_{1}u\left(b_j\right)}$.

We obtain $(\tilde{H}\left(d_1\right),\dots ,\tilde{H}\left(d_h\right))=({\alpha }_{u\left(x_1\right)u\left(b_1\right)},\dots ,{\alpha }_{u_{1}u\left(b_j\right)},\dots ,{\alpha }_{u\left(x_h\right)u\left(b_h\right)})$. Since $(u\left(b_1\right),\dots ,u\left(b_h\right))$$\in \lambda$ and $({\alpha }_{u\left(x_1\right)u\left(b_1\right)},u\left(b_1\right))\in \epsilon$, we conclude that

$({\alpha }_{u\left(x_1\right)u\left(b_1\right)},u\left(b_2\right),\dots ,u\left(b_h\right))\in \lambda$. Repeating the same argument, we obtain by induction that the resulting h tuple $(\tilde{H}\left(d_1\right),\dots ,\tilde{H}\left(d_h\right))$ is in $\lambda$. Thus, $\tilde{H}\in Pol\lambda$.

For all $x\in E_k^m$, $\tilde{H}\left(x,f_{m+\ell }\left(x\right),\dots ,f_{m+l}\left(x\right)\right)=\tilde{H}\left(ext\left(x\right)\right)=\tilde{H}\circ ext\left(x\right)=u\left(x\right)$. Thus, $u=\tilde{H}\circ ext\in \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle$. Therefore $Pol\lambda \subseteq \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle$. We conclude that $\langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle =Pol\lambda$ and $Pol\lambda \cap Pol\sigma$ is maximal below $Pol\lambda$. □

The sufficient condition of Proposition 4.1 is obtained by the following proposition.

Proposition . 

Let λ be an h-ary regular relation on $E_k$$(3\le h<k)$ determined by the h-regular family $T=\{{\theta }_1,\dots ,{\theta }_m\}$, and σ be a non-trivial equivalence relation on $E_k$ with t classes ($t\ge 2$)such that $Pol\lambda \cap Pol\sigma$ is maximal below $Pol\lambda$. Then, λ is σ-closed or $\lambda ={\gamma }_h$ and for all $x,y\in E_k,{\left[x\right]}_{\sigma }\cap {\left[y\right]}_{\epsilon }\ne \varnothing$.

It’s proof is obtained from the following Lemmas.

Let $\lambda$ be an h-ary regular relation $(3\le h<k)$ on $E_k$ determined by the h-regular family $T=\{{\theta }_1,\dots ,{\theta }_m\}$, $m\ge 1$ and $\sigma$ be a non-trivial equivalence relation on $E_k$ with t classes ($t\ge 2$). We set :

${\sigma }_j=\{(a_1,\dots ,a_h)\in E_k^h\mid \exists u_1\in {\left[a_1\right]}_{\sigma },\dots u_j\in {\left[a_j\right]}_{\sigma },(u_1,\dots ,u_j,a_{j+1},\dots ,a_h)\in \lambda \}$, $1\le j\le h$. We have $\lambda \subseteq {\sigma }_j\subseteq E_k^h$ and $Pol\lambda \cap Pol\sigma \subseteq Pol\lambda \cap Pol{\sigma }_j$ for $1\le j\le h$ (due to ${\sigma }_j\in \left[\{\lambda ,\sigma \}\right]$).

Let $(a,b)\in E_k^2\setminus \sigma$ and $(e,d)\in \sigma$ such that $e\ne d$. Consider the unary operation $f_1$ defined on $E_k$ by

$$f_1\left(t\right)=\left\{\begin{array}{cc}a\hfill & \mathrm{if} t=e,\hfill \\ b\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Since $(e,d)\in \sigma$ and $(f_1\left(e\right),f_1\left(d\right))=(a,b)\notin \sigma$, then $f_1\notin Pol\sigma$; but $f_1\in Pol\lambda \cap Pol{\sigma }_j$ ( because $Im\left(f_1\right)=\{a,b\},{\{a,b\}}^h\subseteq \lambda \subseteq {\sigma }_j$, in fact $\lambda$ and ${\sigma }_j$ are totally reflexive). Therefore, for $1\le j\le h$, $Pol\lambda \cap Pol\sigma ⊊Pol\lambda \cap Pol{\sigma }_j$. Since ${\sigma }_h$ is totally reflexive, totally symmetric and $\lambda \subseteq {\sigma }_h\subseteq E_k^h$, we obtain the following three cases: $\left(1\right)\lambda ={\sigma }_h$, $\left(2\right)\lambda ⊊{\sigma }_h⊊E_k^h$ and $\left(3\right){\sigma }_h=E_k^h$.

We begin with case $\left(1\right)\lambda ={\sigma }_h$. The following lemma shows that $\lambda$ is $\sigma$-closed.

Lemma 4.6 

Under the assumptions of Proposition 4.5 and $\lambda ={\sigma }_h$, we have λ is σ-closed.

Proof. 

Assume that $\lambda ={\sigma }_h$. It follows from definition that $\lambda$ is $\sigma$-closed. □

We continue with case $\left(2\right)\lambda ⊊{\sigma }_h⊊E_k^h$. The next lemma proves that this case can not occur.

Lemma 4.7 

Under the assumptions of Proposition 4.5, the case $\left(2\right)\lambda ⊊{\sigma }_h⊊E_k^h$ is impossible.

Proof. 

Assume that $\lambda ⊊{\sigma }_h⊊E_k^h$. Let $(a_1,\dots ,a_h)\in E_k^h\setminus {\sigma }_h$ and $(u_1,\dots ,u_h)\in {\sigma }_h\setminus \lambda$. There exists an equivalence relation ${\theta }_1$ in the h-regular family T associated to $\lambda$ such that for $1\le i<j\le h,(u_i,u_j)$ is not an element of ${\theta }_1$. Consider the unary operation g defined on $E_k$ by $g\left(x\right)=a_i$ if and only if $x\in {\left[u_i\right]}_{{\theta }_1}$. Hence g preserves $\lambda$ (because g restricted to each ${\theta }_1$-class is constant) and does not preserve ${\sigma }_h$ (because $(u_1,\dots ,u_h)\in {\sigma }_h$ and $(g\left(u_1\right),\dots ,g\left(u_h\right))=(a_1,\dots ,a_h)\notin {\sigma }_h$). Hence $Pol\lambda \cap Pol\sigma ⊊Pol{\sigma }_h\cap Pol\lambda ⊊Pol\lambda$. Thus $Pol\lambda \cap Pol\sigma$ is not maximal below $Pol\lambda$. □

Now we finish our investigation with case $\left(3\right){\sigma }_h=E_k^h$. We recall that $\sigma$ has t classes $(t\ge 2)$. For $i\ge 2$, we denote by ${\xi }_i$ the i-ary relation defined on $E_k$ by: ${\xi }_i=\{(a_1,\dots ,a_i)\in E_k^i\mid \exists a_1^{\prime }\in {\left[a_1\right]}_{\sigma },\dots ,a_i^{\prime }\in {\left[a_i\right]}_{\sigma },{\{a_1^{\prime },\dots ,a_i^{\prime }\}}^h\subseteq \lambda \}$.

We have ${\sigma }_h={\xi }_h=E_k^h$ and ${\xi }_t$ satisfies one of the following two conditions : $\left(3.1\right){\xi }_t=E_k^t$, $\left(3.2\right){\xi }_t\ne E_k^t$.

In the case ${\xi }_t\ne E_k^t$, we denote by n the least integer N such that ${\xi }_N\ne E_k^N$. Since ${\xi }_h=E_k^h$, we have $n>h$. The next result shows that only the case $\left(3.1\right)$ is possible.

Lemma 4.8 

Under the assumptions of Proposition 4.5 and ${\sigma }_h=E_k^h$, we have ${\xi }_t=E_k^t$.

Proof. 

Assume that ${\xi }_t\ne E_k^t$. The minimality of n yields that ${\xi }_{n-1}=E_k^{n-1}$. It is easy to check that ${\xi }_n$ is totally reflexive and totally symmetric. Furthermore, we have $Pol\lambda \cap Pol\sigma ⊊Pol\lambda \cap Pol{\xi }_n\subseteq Pol\lambda$ (due to $\sigma$ binary, $n>h$, ${\xi }_n$ totally reflexive and totally symmetric). Let $(a_1,\dots ,a_h)\in E_k^h\setminus \lambda$ and $(u_1,\dots ,u_h)\in \lambda \setminus {\iota }_k^h$. The unary operation f defined on $E_k$ by

$$f\left(t\right)=\left\{\begin{array}{cc}{a}_i\hfill & \mathrm{if}t=u_i,1<i\le h,\hfill \\ a_1\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

preserves ${\xi }_n$ (because $n>h$, $Im\left(f\right)=\{a_1,\dots ,a_h\}$, ${\xi }_n$ is totally symmetric and totally reflexive ) and does not preserve $\lambda$ ( because $(u_1,\dots ,u_h)\in \lambda$ and $(f\left(u_1\right),\dots ,f\left(u_h\right))=(a_1,\dots ,a_h)\notin \lambda$. Thus $Pol\lambda \ne Pol{\xi }_n$ and we have $Pol\lambda \cap Pol\sigma ⊊Pol\lambda \cap Pol{\xi }_n⊊Pol\lambda$; contradicting the maximality of $Pol\lambda \cap Pol\sigma$ in $Pol\lambda$. Hence, ${\xi }_t=E_k^t$. □

Now we assume that ${\xi }_t=E_k^t$. Therefore there exist $u_1,\dots ,u_t\in E_k$ such that $(u_i,u_j)\notin \sigma$, $1\le i<j\le t$ and ${\{u_1,\dots ,u_t\}}^h\subseteq \lambda$. We set $W=\{u_1,\dots ,u_t\}$; W is a transversal of $\sigma$ and $\lambda$.

For $j\in \underline{h}$, we set ${\displaystyle {\delta }_j=\bigcap _{s\in S_h}{\left({\sigma }_j\right)}_s}$. It is easy to check that ${\delta }_j$ is totally symmetric and totally reflexive. We have ${\delta }_h=E_k^h$. For $1\le j\le h-1$, $\lambda \subseteq {\sigma }_j\subseteq E_k^h$ and we distinguish the following cases: $\left(4.1\right)\lambda ={\delta }_j$, $\left(4.2\right)\lambda ⊊{\delta }_j⊊E_k^h$ and $\left(4.3\right){\delta }_j=E_k^h$.

Now, we study the subcase $\left(4.2\right)\lambda ⊊{\delta }_j⊊E_k^h$. The following lemma shows that this case can not occur.

Lemma 4.9 

Under the assumptions of Proposition 4.5 and there is a transversal W of σ-classes such that $W^h\subseteq \lambda$, there is no $1\le j\le h-1$ such that $\lambda ⊊{\delta }_j⊊E_k^h$.

Proof. 

Assume that there exists $1\le j\le h-1$ such that $\lambda ⊊{\delta }_j⊊E_k^h$. Then a similar argument as in the Proof of Lemma 4.7 shows that $Pol\lambda \cap Pol\sigma ⊊Pol\lambda \cap Pol{\delta }_j⊊Pol\lambda$ and we obtain a contradiction. □

We continue our discussion with subcase $\left(4.1\right)\lambda ={\delta }_j$ for some $1\le j\le h-1$. We can see in the following lemma that it is also impossible.

Lemma 4.10 

Under the assumptions of Proposition 4.5 and there is a transversal W of σ-classes such that $W^h\subseteq \lambda$, there is no $1\le j\le h-1$ such that $\lambda ={\delta }_j$.

Proof. 

Assume that there is $1\le j\le h-1$ such that $\lambda ={\delta }_j$. Since ${\delta }_i\subseteq {\delta }_j$, for all $1\le i\le j\le h-1$ and $\lambda \subseteq {\delta }_i$, for all $1\le i\le h-1$, then ${\displaystyle \lambda \subseteq \bigcap _{i=1}^{h-1}{\delta }_i={\delta }_1}$.

Thus, $\lambda ={\delta }_1$ (due to $\lambda \subseteq {\delta }_1$). Recall that ${\displaystyle {\delta }_1=\bigcap _{s\in S_h}{\left({\sigma }_1\right)}_s}$ and

$${\sigma }_1=\{(a_1,\dots ,a_h)\in E_k^h\mid \exists u\in E_k,(a_1,u)\in \sigma \wedge (u,a_2,\dots ,a_h)\in \lambda \}.$$

Thus ${\sigma }_1\ne E_k^h$. We have the following possibilities: $\left(i\right){\sigma }_1=\lambda$ and $\left(ii\right)\lambda ⊊{\sigma }_1⊊E_k^h$.

Assume that $\left(i\right)\lambda ={\sigma }_1$ holds. Let $(a_1,\dots ,a_h)\in E_k^h\setminus \lambda$; since $W=\{u_1,\dots ,u_t\}$ is a transversal, without loss of generality we can affirm that $(u_1,\dots ,u_h)\in \lambda$ and $(a_1,u_1),\dots ,(a_h,u_h)\in \sigma (★)$. Thus, $(a_1,u_2,\dots ,u_h)\in {\sigma }_1=\lambda$. By induction and the totally symmetry of $\lambda$, we obtain $(a_1,a_2,\dots ,a_h)\in \lambda$; contradicting the choice of $(a_1,a_2,\dots ,a_h)$. Therefore $\left(ii\right)$$\lambda ⊊{\sigma }_1⊊E_k^h$ holds. Since ${\displaystyle \lambda =\bigcap _{s\in S_h}{\left({\sigma }_1\right)}_s}$, we have $Pol{\sigma }_1\subseteq Pol\lambda$; in addition ${\sigma }_1\in \left[\{\sigma ,\lambda \}\right]$, so $Pol\lambda \cap Pol\sigma \subseteq Pol{\sigma }_1$. It follows that $Pol\lambda \cap Pol\sigma \subseteq Pol{\sigma }_1\subseteq Pol\lambda$. Let $(a,b)\in E_k^2\setminus \sigma$ and $(u,v)\in \sigma$ such that $u\ne v$. Let $f_1$ be defined on $E_k$ by $f\left(t\right)=a$ if $t=u$ and $f\left(t\right)=b$ otherwise. Then the unary operation $f_1$ preserves ${\sigma }_1$ (because ${\sigma }_1$ is totally reflexive and $Im\left(f_1\right)=\{a,b\}$) and does not preserve $\sigma$ (due to $(u,v)\in \sigma$ and $(f_1\left(u\right),f_1\left(v\right))=(a,b)\notin \sigma$. Therefore $Pol\lambda \cap Pol\sigma ⊊Pol{\sigma }_1$. A similar argument as in the proof of Lemma 4.7 shows that $Pol\lambda \cap Pol{\sigma }_1⊊Pol\lambda$. Thus, $Pol\lambda \cap Pol\sigma ⊊Pol{\sigma }_1\cap Pol\lambda ⊊Pol\lambda$; contradicting the maximality of $Pol\lambda \cap Pol\sigma$ in $Pol\lambda$. □

From Lemmas 4.6, 4.7, 4.8, 4.9, and 4.10, we conclude that for all $1\le j\le h-1$, ${\delta }_j=E_k^h$. Therefore ${\displaystyle {\delta }_1=E_k^h=\bigcap _{s\in S_h}{\left({\sigma }_1\right)}_s}$. Hence $E_k^h={\sigma }_1={\left({\sigma }_1\right)}_s$ for all $s\in S_h$.

Now we assume that for all $(a_1,\dots ,a_h)\in E_k^h$, $\exists u\in {\left[a_1\right]}_{\sigma }$ such that $(u,a_2,\dots ,a_h)\in \lambda$ and there is a transversal $W=\{u_1,\dots ,u_t\}$ for $\sigma$ and $\lambda$ such that $W^h\subseteq \lambda$.

For $h\le l\le k$, we set ${\rho }_l=\{(a_1,\dots ,a_l)\in E_k^l\mid \exists u\in E_k,\left\{u\right\}\times {\{a_1,\dots ,a_l\}}^{h-1}\subseteq \lambda \}$.

Clearly $\lambda \subseteq {\rho }_h\subseteq E_k^h$, and every ${\rho }_l$, $h\le l\le k$, is totally symmetric. Since $\sigma$ is a binary relation and $h>2$, we have $Pol\sigma \cap Pol\lambda Pol\lambda \cap Pol{\rho }_h$. We distinguish the following three cases : $\left(1\right)\lambda ={\rho }_h$, $\left(2\right)\lambda {\rho }_h{E}_k^h$ and $\left(3\right){\rho }_h=E_k^h$.

Now, we study the subcase $\left(2\right)\lambda {\rho }_h{E}_k^h$. We show in the next lemma that $\left(2\right)$ is impossible.

Lemma 4.11 

Under the assumptions of Proposition 4.5, ${\sigma }_1=\lambda$ and there exists a transversal $W=\{u_1,\dots ,u_t\}$ of σ-classes such that $W^h\subseteq \lambda$, subcase $\lambda ⊊{\rho }_h⊊E_k^h$ is impossible.

Proof. 

Use same argument as in the proof of Lemma 4.7. □

We continue with subcase $\left(3\right){\rho }_h=E_k^h$. It is also shown that this case is impossible.

Lemma 4.12 

Under the assumptions of Proposition 4.5, ${\sigma }_1=\lambda$ and there exists a transversal $W=\{u_1,\dots ,u_t\}$ of σ-classes such that $W^h\subseteq \lambda$, subcase ${\rho }_h=E_k^h$ is also impossible.

Proof. 

Assume that ${\rho }_h=E_k^h$. We will show that ${\rho }_k=E_k^k$.

Suppose that ${\rho }_k\ne E_k^k$. Let n be the least integer such that ${\rho }_n\ne E_k^n$. Then $n>h$ (because ${\rho }_h=E_k^h$). Since $n,h>2$, we have $Pol\sigma \cap Pol\lambda Pol{\rho }_h\cap Pol\lambda$. Furthermore, ${\rho }_n$ is totally reflexive (due to ${\rho }_{n-1}=E_k^{n-1}$). Let $(a_1,\dots ,a_h)\in E_k^h\setminus \lambda$ and $(u_1,\dots ,u_h)\in \lambda \setminus {\iota }_k^h$. The unary operation f defined on $E_k$ by :

$$f\left(t\right)=\left\{\begin{array}{cc}{a}_i\hfill & \mathrm{if}t=u_i,1<i\le h,\hfill \\ a_1\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

preserves ${\rho }_n$ (because $n>h$ and ${\rho }_n$ is totally reflexive) and does not preserve $\lambda$ (because $(u_1,\dots ,u_h)\in \lambda$ and $(f\left(u_1\right),\dots ,f\left(u_h\right))=(a_1,\dots ,a_h)\notin \lambda$). Hence $Pol\sigma \cap Pol\lambda Pol{\rho }_h\cap Pol\lambda Pol\lambda$, contradicting the maximality of $Pol\lambda \cap Pol\sigma$ in $Pol\lambda$. Thus ${\rho }_k=E_k^k$. Set $E_k=\{a_1,\dots ,a_k\}$. Then $(a_1,\dots ,a_k)\in E_k^k={\rho }_k$, so there exists $u\in E_k$ such that $\left\{u\right\}\times {\{a_1,\dots ,a_k\}}^{h-1}\subseteq \lambda$. Thus u is a central element of $\lambda$; contradiction. □

Now we finish our discussion with subcase $\left(3\right)\lambda ={\rho }_h$.

For $h\le l\le k$, we set

${\gamma }_l=\{(a_1,\dots ,a_l)\in E_k^l\mid \exists v_1,\dots ,v_l\in E_k:{\{v_1,\dots ,v_l\}}^h\subseteq \lambda ,\forall 1\le i\le l,\forall j_1,\dots ,j_{h-2}\in \{1,\dots ,l\}\setminus \left\{i\right\},(v_i,a_i,a_{j_1},\dots ,a_{j_{h-2}})\in \lambda \}$.

Clearly $\lambda \subseteq {\gamma }_h\subseteq E_k^h$, and every ${\gamma }_l$, $h\le l\le k$, is totally symmetric. Since $\sigma$ is a binary relation and $h>2$, we have $Pol\sigma \cap Pol\lambda Pol\lambda \cap Pol{\gamma }_h$. ${\gamma }_h$ satisfies one of the following three cases : $\left(1\right)\lambda ={\gamma }_h$, $\left(2\right)\lambda {\gamma }_h{E}_k^h$ and $\left(3\right){\gamma }_h=E_k^h$.

Firstly, we show that the subcase $\left(2\right)\lambda {\gamma }_h{E}_k^h$ is impossible.

Lemma 4.13 

Under the assumptions of Proposition 4.5, ${\sigma }_1=\lambda$, there exists a transversal $W=\{u_1,\dots ,u_t\}$ of σ-classes such that $W^h\subseteq \lambda$ and ${\rho }_h=\lambda$, subcase $\lambda {\gamma }_h{E}_k^h$ is impossible.

Proof. 

Assume that $\lambda {\gamma }_h{E}_k^h$. Using similar argument as in the proof of Lemma 4.7 we obtain the conclusion. □

Secondly, we prove also that subcase ${\gamma }_h=E_k^h$ can not occur.

Lemma 4.14 

Under the assumptions of Proposition 4.5, ${\sigma }_1=\lambda$, there exists a transversal $W=\{u_1,\dots ,u_t\}$ of σ-classes such that $W^h\subseteq \lambda$ and ${\rho }_h=\lambda$, subcase ${\gamma }_h=E_k^h$ is impossible.

Proof. 

Assume that ${\gamma }_h=E_k^h$. We will show that ${\gamma }_k=E_k^k$.

Suppose that ${\gamma }_k\ne E_k^k$. Let n be the least integer such that ${\gamma }_n\ne E_k^n$. Then $n>h$ ( because ${\gamma }_h=E_k^h$) and ${\gamma }_n$ is totally reflexive. Using a similar argument as in the proof of Lemma 4.7, we obtain $Pol\sigma \cap Pol\lambda Pol{\gamma }_n\cap Pol\lambda Pol\lambda$, contradicting the maximality of $Pol\sigma \cap Pol\lambda$ in $Pol\lambda$. So, ${\gamma }_k=E_k^k$.

Now we show that this fact yields $\lambda =E_k^h$. Since ${\gamma }_k=E_k^k$, then for every $a_0,\dots ,a_{h-1}\in E_k$ there exist certain $v_0,\dots ,v_{h-1}\in E_k$ with

$(v_0,v_1,\dots ,v_{h-1})\in \lambda \left(4.1\right)$

and

$\forall 0\le i\le h-1,\forall j_1,\dots ,j_{h-2}\in \{0,\dots ,h-1\}\setminus \left\{i\right\},(a_i,v_i,a_{j_1},\dots ,a_{j_{h-2}})\in \lambda \left(4.2\right)$.

By induction, we will show that

$\forall l\ge 0\forall a_0,\dots ,a_{l-1}\in E_k\mid (a_0,\dots ,a_{l-1},v_l,v_{t+1},\dots ,v_{h-1})\in \lambda \left(4.3\right)$.

For $l=0$, $\left(4.3\right)$ follows from $\left(4.1\right)$. Assume $\left(4.3\right)$ holds for $1\le l\le h-2$. Then $(a_0,\dots ,a_{l-1},v_l,v_{t+1},\dots ,$$v_{h-1})\in \lambda$ from $\left(4.3\right)$. Choosing $u=v_l$, we can see that $(a_0,\dots ,a_{l-1},a_l,v_{l+1},\dots ,v_{h-1})\in {\rho }_h=\lambda$. By $\left(4.3\right)$ and $l=h$, we get $\lambda =E_k^h$ which is a contradiction. □

Now we end this discussion with subcase $\left(1\right)\lambda ={\gamma }_h$.

Let $\epsilon$ be the binary relation defined on $E_k$ by

$\epsilon =\{(a,b)\in E_k^2\mid (a,b,a_3,\dots ,a_h)\in \lambda ,\forall a_3,\dots ,a_h\in E_k\}$. The following proposition show that $\epsilon$ is an equivalence relation and the next lemma gives the link between the equivalence classes of $\epsilon$ and $\sigma$.

Proposition 4.15.([9], page 205-206 ) Under the assumptions of Proposition 4.5 and ${\gamma }_h=\lambda$, ε is an equivalence relation on $E_k$. Furthermore, for all $a,b,a_1,\dots ,a_{h-1}\in E_k$ such that $(a,b)\in \epsilon$, we have

$(a,a_1,\dots ,a_{h-1})\in \lambda ⟺(b,a_1,\dots ,a_{h-1})\in \lambda$.

Lemma 4.16 

Under the assumptions of Proposition 4.5, ${\sigma }_1=\lambda$, there exists a transversal $W=\{u_1,\dots ,u_t\}$ of σ-classes such that $W^h\subseteq \lambda$ and ${\gamma }_h=\lambda$, then for all $x,y\in E_k$, ${\left[x\right]}_{\sigma }\cap {\left[y\right]}_{\epsilon }\ne \varnothing$.

Proof. 

Assume that ${\gamma }_h=\lambda$. We set ${\alpha }_h=\{(a_1,\dots ,a_h)\in E_k^h\mid (a_1,a_2)\in \sigma \}$.

Assume that ${\alpha }_h\subseteq \lambda$. Let $(a_1,\dots ,a_h)\in E_k^h$, there exists $v\in E_k$ such that $(a_1,v)\in \sigma$ and $(v,a_2,\dots ,a_h)\in \lambda$. Since ${\alpha }_h\subseteq \lambda$, we have $\left\{v\right\}\times {\{a_1,\dots ,a_h\}}^{h-1}\subseteq \lambda$. Thus $(a_1,\dots ,a_h)\in \lambda$ (because $\lambda ={\rho }_h$). So, $E_k^h=\lambda$; contradiction. We conclude that ${\alpha }_h⊈\lambda$. Let $(e_1,\dots ,e_h)\in E_k^h\setminus \lambda$ such that $(e_1,e_2)\in \sigma$. We set ${\beta }_h=\{(a_1,\dots ,a_h)\in E_k^h\mid \exists v\in {\left[a_1\right]}_{\sigma },(v,a_2,a_{j_1},\dots ,a_{j_{h-2}})\in \lambda \forall j_1,\dots ,j_{h-2}\in \{1,\dots ,h\}\}$.

Clearly, we have $\lambda \subseteq {\beta }_h\subseteq E_k^h$. $(e_1,\dots ,e_h)\in {\beta }_h\setminus \lambda$, taking $v=e_2$. Hence $\lambda ⊊{\beta }_h\subseteq E_k^h$.

If $\lambda ⊊{\beta }_h⊊E_k^h$, then using a similar argument as in the proof of Lemma 4.7 we obtain a contradiction. So, ${\beta }_h=E_k^h$. By induction on $l\ge h$ and the fact that ${\beta }_{h+1}$ is totally reflexive, we can show that ${\beta }_k=E_k^k$.

Assume that $E_k=\{a_1,\dots ,a_k\}$. Then $(a_1,\dots ,a_k)\in E_k^k={\beta }_k$ and there exists $v\in {\left[a_1\right]}_{\sigma }$ such that $(v,a_2,a_{j_1},\dots ,a_{j_{h-2}})\in \lambda$, $\forall j_1,\dots ,j_{h-2}\in \{1,\dots ,k\}$. Thus $(v,a_2)\in \epsilon$ and ${\left[a_1\right]}_{\sigma }\cap {\left[a_2\right]}_{\epsilon }\ne \varnothing$. □

Now we are ready to give the proof of Proposition 4.5 and 4.1.

Proof. (Proof of Proposition 4.5) Combining Lemmas 4.6-4.16 and Proposition 4.15, we obtain the result. □

Proof. (Proof of Proposition 4.1) The necessary condition of Proposition 4.1 is given by Lemmas 4.2 and 4.4 and the sufficient condition is obtained by Proposition 4.5. □

Secondly, we investigate the bounded partial order case.

5. Bounded Partial Order

Let $\lambda$ be an h-ary regular relation $(3\le h<k)$ on $E_k$ and $\sigma$ be a bounded partial order on $E_k$ with greatest element ⊤ and least element ⊥. In this section, we will show that $Pol\lambda \cap Pol\sigma$ is not maximal below $Pol\lambda$. We recall the relation ${\sigma }_1$ defined below. ${\sigma }_1=\{(a_1,\dots ,a_h)\in E_k^h\mid \exists u\in E_k,(a_1,u)\in \sigma \wedge (u,a_2,\dots ,a_h)\in \lambda \}$. We have $\lambda \subseteq {\sigma }_1\subseteq E_k^h$ and $Pol\lambda \cap Pol\sigma ⊊Pol\lambda \cap Pol{\sigma }_1\subseteq Pol\lambda$.

Now we state the main result of this section.

Proposition 5.1 

Let λ be an h-ary regular relation on $E_k$ ($3\le h<k$). If σ is a bounded partial order on $E_k$ with greatest element ⊤ and least element ⊥, then $Pol\lambda \cap Pol\sigma$ is not maximal below $Pol\lambda$.

The proof will be shared in Lemmas 5.2, 5.3 and 5.4. Comparing $\lambda$ and ${\sigma }_1$, we have the following cases: $\left(a\right)\lambda ={\sigma }_1$, $\left(b\right)\lambda ⊊{\sigma }_1⊊E_k^h$ and $\left(c\right){\sigma }_1=E_k^h$. The next lemma shows that the case $\left(a\right)$ is impossible.

Lemma 5.2 

If $Pol\lambda \cap Pol\sigma$ is maximal below $Pol\lambda$, then the case $\lambda ={\sigma }_1$ is impossible.

Proof. 

Assume that $\lambda ={\sigma }_1$. Let $a_2,\dots ,a_h\in E_k$, we have $(a_2,\top )\in \sigma$ and $(\top ,\top ,\dots ,a_h)\in \lambda$ (due to ⊤ greatest element of $\sigma$ and $\lambda$ totally reflexive). Thus $(a_2,\top ,\dots ,a_h)\in {\sigma }_1=\lambda$. Hence ⊤ is a central element of $\lambda$; contradiction.

□

Now we look at the case $\left(c\right)$. The following lemma shows that it is impossible.

Lemma 5.3 

If $Pol\lambda \cap Pol\sigma$ is maximal below $Pol\lambda$, then the case ${\sigma }_1=E_k^h$ is impossible.

Proof. 

Assume that ${\sigma }_1=E_k^h$. Let $a_2,\dots ,a_h\in E_k$; we have $(\top ,a_2,\dots ,a_h)\in E_k^h={\sigma }_1$, there exists $u\in E_k$ such that $(\top ,u)\in \sigma$ and $(u,a_2,\dots ,a_h)\in \lambda$. Hence $u=\top$ (due to ⊤ is the greatest element of $\sigma$) and $(\top ,a_2,\dots ,a_h)\in \lambda$ . Therefore ⊤ is a central element of $\lambda$; contradiction.

□

Finally we show that the case $\left(b\right)$ is also impossible in the next lemma.

Lemma 5.4 

If $Pol\lambda \cap Pol\sigma$ is maximal below $Pol\lambda$, then the case $\lambda ⊊{\sigma }_1⊊E_k^h$ is impossible.

Proof. 

Assume that $\lambda ⊊{\sigma }_1⊊E_k^h$. Let $(a_1,\dots ,a_h)\in E_k^h\setminus {\sigma }_1$ and $(u_1,\dots ,u_h)\in {\sigma }_1\setminus \lambda$. Consider again the unary operation g defined on $E_k$ by $g\left(x\right)=a_i$ if and only if $x\in {\left[u_i\right]}_{{\theta }_1}$ where ${\theta }_1\in T$ and T is the h-regular family associated to $\lambda$. g preserves $\lambda$ and does not preserve ${\sigma }_1$, since $(u_1,\dots ,u_h)\in {\sigma }_1$ and $(g\left(u_1\right),\dots ,g\left(u_h\right))=(a_1,\dots ,a_h)\notin {\sigma }_1$ and g restricted to each ${\theta }_1$-class is constant. Hence $Pol\lambda \cap Pol\sigma ⊊Pol{\sigma }_1\cap Pol\lambda ⊊Pol\lambda$, contradicting the maximality of $Pol\lambda \cap Pol\sigma$ below $Pol\lambda$. □

Proof. ( Proof of Proposition 5.1) Using Lemma 5.2, Lemma 5.3 and Lemma 5.4, we observe that $Pol\sigma \cap Pol\lambda$ is not maximal below $Pol\lambda$. □

Thirdly, we investigate the case of prime permutation relation.

6. Prime Permutation Relation

In this section, s is a fixed point free permutation on $E_k$ with $s^p=id$ ( p prime). We denote by $s^{\circ }$, the graph of s. In other words, $s^{\circ }=\{(a,b)\in E_k^2\mid b=s\left(a\right)\}$. Let ${\theta }_s$ be the equivalence relation on $E_k$ consisting of all pairs $(a,b)\in E_k^2$ with $a=s^i\left(b\right)$ for some $0\le i<p$.

Definition 6.1 

Let $k\ge 3$, λ be an h-ary regular relation $(3\le h<k)$ and s be a fixed point free permutation on $E_k$ with $s^p=id$ ( p prime ). An equivalence relation θ is said transversal to s if $s\in Pol\theta$ and ${\theta }_s\cap \theta ={\Delta }_{E_k}$ where ${\Delta }_{E_k}$ is the equality relation on $E_k$.

Here we state the main result of this section.

Proposition 6.2 

Let λ be an h-ary regular relation on $E_k$ and s be a fixed point free permutation on $E_k$ with $s^p=id$ ( p prime). Then $Pol\lambda \cap Pol{s}^{\circ }$ is maximal below $Pol\lambda$ if and only if λ is ${\theta }_s$-closed.

Before the proof of this proposition, we recall the following characterization of the relational clone generated by the prime permutation s and the regular relation $\lambda$.

Proposition 6.3.([16], Proposition 3.14) Let $k\ge 3$, λ be an h- ary regular relation $(3\le h<k)$ and s be a fixed point free permutation on $E_k$ with $s^p=id$ ( p prime ). The relational algebra $\left[\{s^{\circ },\lambda \}\right]$ contains one of the following relations:

(1) 

a non-trivial equivalence relation which is either ${\theta }_s$-closed or transversal to s;

(2) 

an affine relation determined by an elementary abelian p-group $(\mathit{k};+)$ such that there exists an element $c\in \mathit{k}$ with $s\left(x\right)=x+c$ for all $x\in \mathit{k}$;

(3) 

a central relation;

(4) 

a ${\theta }_s$-closed regular relation.

The next lemma gives the sufficient condition in Proposition 6.2.

Lemma 6.4 

Let $k\ge 3$, λ be an h-ary regular relation $(3\le h<k)$ and s be a fixed point free permutation on $E_k$ with $s^p=id$ ( p prime ). If $Pol\lambda \cap Pol{s}^{\circ }$ is maximal below $Pol\lambda$, then λ is ${\theta }_s$-closed.

Proof. 

Assume that $Pol\lambda \cap Pol{s}^{\circ }$ is maximal below $Pol\lambda$. From Proposition 6.3, the relational algebra $\left[\{s^{\circ },\lambda \}\right]$ contains a relation $\gamma$ satisfying $\left(1\right),\left(2\right),\left(3\right)$ or $\left(4\right)$. Thus $Pol\lambda \cap Pol{s}^{\circ }\subseteq Pol\lambda \cap Pol\gamma \subseteq Pol\lambda$. From Rosenberg’s classification theorem (see Theorem 3.1) $Pol\gamma$ is a maximal clone. Assume that $\gamma \ne \lambda$, then $Pol\lambda \cap Pol\gamma ⊊Pol\lambda$ ( because $Pol\lambda$ and $Pol\gamma$ are two different maximal clones). Thus $Pol\lambda \cap Pol{s}^{\circ }\subseteq Pol\lambda \cap Pol\gamma ⊊Pol\lambda$. Let $a\in E_k$. Consider the constant unary operation $C_a$ with value a. It is easy to see that $C_a$ preserves $\lambda$ and $\gamma$ and $C_a$ does not preserve $s^{\circ }$. Hence $Pol\lambda \cap Pol{s}^{\circ }⊊Pol\lambda \cap Pol\gamma ⊊Pol\lambda$, contradicting the choice of s. Therefore $\gamma =\lambda$ and we conclude that $\lambda$ is ${\theta }_s$-closed. □

The converse of the equivalence in Proposition 6.2 follows from the following result due to Rosenberg and Szendrei.

Proposition 6.5.([16], Proposition 4.3) Let $k\ge 3$, s be a fixed point free permutation on $E_k$ with $s^p=id$ (p prime) and λ be an h-ary ${\theta }_s$-closed regular relation ($3\le h<k$). The relational subalgebras of $\left[\{s^{\circ },\lambda \}\right]$ form a 4-element boolean lattice consisting of $\left[\{s^{\circ },\lambda \}\right]$, $\left[\left\{s^{\circ }\right\}\right]$, $\left[\right\{\lambda \left\}\right]$ and $\left[\left\{{\Delta }_{E_k}\right\}\right]$.

The next Corollary gives the maximality of $Pol\lambda \cap Pol{s}^{\circ }$ in $Pol\lambda$.

Corollary 6.6 

Let λ be an h-ary regular relation on $E_k$ ($3\le h<k$) and s be a prime permutation relation on $E_k$ such that λ is ${\theta }_s$-closed. Then $Pol{s}^{\circ }\cap Pol\lambda$ is maximal below $Pol\lambda$.

Proof. 

It follows from Proposition 6.5. □

Proof. (Proof of Proposition 6.2) It follows from Lemma 6.4 and Corollary 6.6. □

Fourthly, we focus our attention on binary central relation case.

7. Binary Central Relation

In this section, we give a necessary and sufficient condition on a binary central relation $\sigma$ such that the clone $Pol\sigma \cap Pol\lambda$ is covered by $Pol\lambda$, where $\lambda$ is an h-ary regular relation on $E_k$ ($3\le h<k$).

Let $\sigma$ be a binary central relation on $E_k$. For $l\ge 2$, we set

${\omega }_l=\{(a_1,\dots ,a_l)\in E_k^l\mid \exists u\in E_k,(a_1,u),\dots ,(a_l,u)\in \sigma \mathit{and}\left\{u\right\}\times {\left\{a_1,\dots ,a_l\right\}}^{h-1}\subseteq \lambda \}$. From definition ${\omega }_l$ is totally symmetric for $2\le l$. We consider again the equivalence relation $\epsilon$ defined on $E_k$ by

$$\epsilon =\{(a,b)\in E_k^2\mid (a,b,a_3,\dots ,a_h)\in \lambda ,\forall a_3,\dots ,a_h\in E_k\}.$$

Here we state the main result of this section.

Proposition 7.1 

Let $k\ge 3$, λ be an h-ary regular relation on $E_k$ ($3\le h<k$) and σ be a binary central relation on $E_k$. Then $Pol\lambda \cap Pol\sigma$ is maximal below $Pol\lambda$ if and only if for all $x\in E_k$, $C_{\sigma }\cap {\left[x\right]}_{\epsilon }\ne \varnothing$.

The necessary condition of Proposition 6.7 is stated in the following Lemma.

Lemma 7.2 

Let λ be an h-ary regular relation on $E_k$ ($3\le h<k$) determined by the h-regular family $T=\{{\theta }_1,\dots ,{\theta }_m\}$, $m\ge 1$ and σ be a binary central relation on $E_k$ such that $\forall x\in E_k$, $C_{\sigma }\cap {\left[x\right]}_{\epsilon }\ne \varnothing$. Then $Pol\lambda \cap Pol\sigma$ is a submaximal clone of $Pol\lambda$.

Proof. 

Assume that $\forall x\in E_k$, $C_{\sigma }\cap {\left[x\right]}_{\epsilon }\ne \varnothing$. For all $x\in E_k$, we choose and fix $c_x\in C_{\sigma }\cap {\left[x\right]}_{\epsilon }$. Let $g^n\in Pol\lambda \setminus \left(Pol\sigma \cap Pol\lambda \right)$ be an n-ary operation. Using the operation H constructed in the proof of Lemma 4.3, we will show that $\langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle =Pol\lambda$.

Clearly, we have $\langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle \subseteq Pol\lambda$. It remains to prove that $Pol\lambda \subseteq \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle$. Let $u\in Pol\lambda$ be an m-ary operation on $E_k$. We will show that $u\in \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}$. We extend H to $\tilde{H}$ defined by:

$$\tilde{H}\left(y\right)=\left\{\begin{array}{cc}u\left(x\right)\hfill & \mathrm{if} y=ext\left(x\right),\hfill \\ c_{u(y_1,\dots ,y_m})\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Let’s show that $\tilde{H}\in Pol\lambda \cap Pol\sigma$. Firstly, we show that $\tilde{H}\in Pol\sigma$.

Let $a=(a_1,\dots ,a_{m+\ell })$ and $b=(b_1,\dots ,b_{m+\ell })$ such that $(a,b)\in \sigma$. We distinguish two cases:

Case 1: If $a=ext\left(x\right)$ and $b=ext\left(y\right)$, then $x=y$ (From Lemma 4.3). Hence $(\tilde{H}\left(a\right),\tilde{H}\left(b\right))=(u\left(x\right),u\left(x\right))\in \sigma$.

Case 2: For all $x\in E_k^m,a\ne ext\left(x\right)$ or For all $x\in E_k^m,b\ne ext\left(x\right)$; without loss of generality, we suppose that For all $x\in E_k^m,a\ne ext\left(x\right)$. Then $(\tilde{H}\left(a\right),\tilde{H}\left(b\right))=(c_{u(a_1,\dots ,a_m)},\tilde{H}\left(b\right))\in \sigma$ (because $c_{u(a_1,\dots ,a_m)}\in C_{\sigma }$). Thus $\tilde{H}\in Pol\sigma$.

Secondly, we show that $\tilde{H}\in Pol\lambda$.

Let $a_1=(a_{1,1},\dots ,a_{1,h}),\dots ,a_m=(a_{m,1},\dots ,a_{m,h}),\dots ,a_{m+l}=(a_{m+\ell ,\ell },\dots ,a_{m+\ell ,h})\in \lambda$. For all $1\le j\le h$, set $d_j=(a_{1,j},\dots ,a_{m,j},\dots ,a_{m+l,j})$ and $b_j=\left(a_{1,j},\dots ,a_{m,j}\right)$. We will show that $(\tilde{H}\left(d_1\right),\dots ,\tilde{H}\left(d_h\right))\in \lambda$. We distinguish again two cases:

Case 1:$\forall 1\le j\le h$, $d_j=ext\left(b_j\right)$, then we obtain

$(\tilde{H}\left(d_1\right),\dots ,\tilde{H}\left(d_h\right))=(u\left(b_1\right),\dots ,u\left(b_h\right))\in \lambda$ (due to $(b_1,\dots ,b_h)\in \lambda$ and $u\in Pol\lambda$).

Case 2: There exists $1\le j\le h$ such that $\tilde{H}\left(d_j\right)=c_{u\left(b_j\right)}$, then we have $(\tilde{H}\left(d_1\right),\dots ,\tilde{H}\left(d_h\right))=(\tilde{H}\left(d_1\right),\dots ,c_{u\left(b_j\right)},\dots ,\tilde{H}\left(d_h\right))\in \lambda$ because $(c_{u\left(b_j\right)},u\left(b_j\right))\in \epsilon$, $\lambda$ is totally symmetric, $(u\left(b_1\right),\dots ,u\left(b_h\right))$$\in \lambda$ and we can replace gradually $u\left(b_k\right)$ by $c_{u(b_k})$ until obtained the desire tuple. Thus $\tilde{H}\in Pol\lambda$. For all $x\in E_k^m$, $\tilde{H}\left(x,f_{m+\ell }\left(x\right),\dots ,f_{m+l}\left(x\right)\right)=\tilde{H}\left(ext\left(x\right)\right)=\tilde{H}\circ ext\left(x\right)=u\left(x\right)$. Thus, $u=\tilde{H}\circ ext\in \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle$. Therefore $Pol\lambda \subseteq \langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle$. We conclude that $\langle (Pol\lambda \cap Pol\sigma )\cup \left\{g\right\}\rangle =Pol\lambda$. Thus $Pol\lambda \cap Pol\sigma$ is maximal below $Pol\lambda$. □

The sufficient condition in Proposition 6.7 is given by the following statement.

Proposition 7.3 

Let λ be an h-ary regular relation $(3\le h<k)$ on $E_k$ and σ be a binary central relation on $E_k$. If $Pol\lambda \cap Pol\sigma$ is maximal below $Pol\lambda$, then $\forall x\in E_k$, $C_{\sigma }\cap {\left[x\right]}_{\epsilon }\ne \varnothing$.

The proof of Proposition 7.3 is investigated in the following Lemmas.

Let $\lambda$ be an h-ary regular relation $(3\le h<k)$ on $E_k$ and $\sigma$ be a binary central relation on $E_k$ such that $Pol\lambda \cap Pol\sigma$ is maximal below $Pol\lambda$. Recall that ${\omega }_2=\{(a_1,a_2)\in E_k^2\mid \exists u\in E_k,(a_1,u),(a_2,u)\in \sigma \mathit{and}\left\{u\right\}\times {\left\{a_1,a_2\right\}}^{h-1}\subseteq \lambda \}$.

Clearly $\sigma \subseteq {\omega }_2\subseteq E_k^2$ and $Pol\lambda \cap Pol\sigma \subseteq Pol\lambda \cap Pol{\omega }_2\subseteq Pol\lambda$ (due to $\lambda$ totally reflexive and ${\omega }_2\in \left[\lambda ,\sigma \right]$). We look at the following three subcases: $\left(1\right)\sigma {\omega }_2{E}_k^2$, $\left(2\right){\omega }_2=\sigma$ and $\left(3\right){\omega }_2=E_k^2$.

Firstly, we show in the next lemma that the subcase $\left(1\right)\sigma {\omega }_2{E}_k^2$ can not occur.

Lemma 7.4 

Under the assumptions of Proposition 7.3, the subcase $\sigma {\omega }_2{E}_k^2$ is impossible.

Proof. 

Let $c\in C_{\sigma }$. Since $\sigma ⊊{\omega }_2⊊E_k^2$, then ${\omega }_2$ is a binary central relation with $c\in C_{{\beta }_2}$. Therefore $\sigma$ and ${\omega }_2$ are two different central relations. Thus $Pol\lambda \cap Pol{\sigma }_2⊊Pol\lambda \cap Pol{\omega }_2⊊Pol\lambda$ (because $Pol\lambda$ and $Pol{\omega }_2$ are two different maximal clones and $\sigma ⊊{\omega }_2$) , contradicting the maximality of $Pol\sigma \cap Pol\lambda$ below $Pol\lambda$. □

Secondly, we show that subcase ${\omega }_2=\sigma$ is also impossible.

Lemma 7.5 

Under the assumptions of Proposition7.3, the subcase ${\omega }_2=\sigma$ is also impossible.

Proof. 

Assume that ${\omega }_2=\sigma$. In this case we have $h=3$ (because $\lambda$ is totally reflexive ). For $l\ge 3$, we set

${\theta }_l^{\prime }=\{(a_1,\dots ,a_l)\in E_k^l\mid \exists u\in E_k,(a_1,u)\in \sigma and\forall j_1,\dots ,j_{h-2}\in \{1,\dots ,l\},$$(u,a_2,a_{j_1},\dots ,a_{jh-2})\in \lambda \}$.

Clearly $\lambda \subseteq {\theta }_3^{\prime }\subseteq E_k^3$, hence ${\theta }_3^{\prime }$ is totally reflexive and $Pol\lambda \cap Pol\sigma ⊊Pol\lambda \cap Pol{\theta }_3^{\prime }\subseteq Pol\lambda$(due to ${\theta }_3^{\prime }\in \left[\lambda ,\sigma \right]$).

Let $c\in C_{\sigma }$. Since c is not a central element of $\lambda$, there exist $a_2,a_3\in E_k$ such that $(c,a_2,a_3)\notin \lambda$. So $(c,a_2)\in \sigma$, $(a_2,a_2,c),(a_2,a_2,a_3)\in \lambda$. Hence $(c,a_2,a_3)\in {\theta }_3^{\prime }\setminus \lambda$. Thus $\lambda ⊊{\theta }_3^{\prime }\subseteq E_k^3$.

If $\lambda ⊊{\theta }_3^{\prime }⊊E_k^3$, then using a similar argument as in the proof of Lemma 4.7, we obtain $Pol\lambda \cap Pol\sigma ⊊Pol\lambda \cap Pol{\theta }_3^{\prime }⊊Pol\lambda$; contradicting the maximality of $Pol\lambda \cap Pol\sigma$ below $Pol\lambda$. Hence ${\theta }_3^{\prime }=E_k^3$.

Now we will show that ${\theta }_k^{\prime }=E_k^k$. We assume that ${\theta }_k^{\prime }\ne E_k^k$. Let n be the least integer N such that ${\theta }_N^{\prime }\ne E_k^N$. Then $3<n\le k$ and ${\theta }_n^{\prime }$ is totally reflexive (due to ${\theta }_{n-1}^{\prime }=E_k^{n-1}$). A similar argument as in the proof of Lemma 4.8 yields that $Pol\lambda \cap Pol\sigma ⊊Pol\lambda \cap Pol{\theta }_n^{\prime }⊊Pol\lambda$(due to ${\theta }_n^{\prime }$ is not a diagonal relation); contradicting the maximality of $Pol\lambda \cap Pol\sigma$ below $Pol\lambda$. Thus ${\theta }_k^{\prime }=E_k^k$.

Furthermore, We show that $\epsilon \subseteq \sigma$. Let $(a,b)\in \epsilon$. Since ${\theta }_k^{\prime }=E_k^k$, there exists $u\in E_k$ such that $(u,a)\in \epsilon$ and $(b,u)\in \sigma$. Let $c\in C_{\sigma }$. We have $(a,c),(b,c)\in \sigma$ and $(a,b,c)\in \lambda$ (because $(a,b)\in \epsilon$). Hence $(a,b)\in {\omega }_2=\sigma$. Therefore $\epsilon \subseteq \sigma$.

Let $(x,y)\in E_k^2$. We assume that $E_k=\left\{x,y,a_3,\dots ,a_k\right\}$. Therefore $(x,y,a_3,\dots ,a_k)\in E_k^k={\theta }_k^{\prime }$. So there exists $u\in E_k$ such that $(x,u)\in \sigma$ and $(u,y,v_1,\dots ,v_{h-2})\in \lambda$ for all $v_1,\dots ,v_{h-2}\in E_k$. Thus $(y,u)\in \epsilon$. Hence $(y,u)\in \sigma$ (because $\epsilon \subseteq \sigma$). Thus $(x,u),(y,u)\in \sigma$ and $(x,y,u)\in \lambda$ (because $(y,u)\in \epsilon$). Therefore $(x,y)\in {\omega }_2$, and $\sigma ={\omega }_2=E_k^2$; contradiction. □

We finish with subcase $\left(3\right){\omega }_2=E_k^2$. The following lemma shows that every $\epsilon$-block contains a central element of $\sigma$.

Lemma 7.6 

Under the assumptions of Proposition 7.3 and ${\omega }_2=E_k^2$, then for all $x\in E_k$, $C_{\sigma }\cap {\left[x\right]}_{\epsilon }\ne \varnothing$.

Proof. 

Assume that ${\omega }_2=E_k^2$. We will show that ${\omega }_{h-1}=E_k^{h-1}$. If ${\omega }_{h-1}\ne E_k^{h-1}$, then we denote by n the least integer N such that ${\omega }_N\ne E_k^N$. Then $2<n<h-1$. Since ${\omega }_n$ is totally reflexive and $n>2$, for $(a,b)\in \sigma$ such that $a\ne b$ and $(u,v)\notin \sigma$, the unary operation f defined by $f\left(a\right)=u$ and $f\left(x\right)=v$ if $x\ne a$, preserves $w_n$ and $\lambda$; and does not preserve $\sigma$. Hence $Pol\lambda \cap Pol\sigma ⊊Pol\lambda \cap Pol{\omega }_n$; it remains the next inequality to conclude. Let $a,b\in \sigma$ such that $a\ne b$, we fix $a_1=(a,a,\dots ,a),a_2=(a,b,\dots ,a),\dots ,a_n=(a,a,\dots ,b)$, n tuples of $E_k^n$. Let $(u_1,\dots ,u_n)\in E_k^n$. The n-ary operation g defined on $E_k$ by $g\left(a_i\right)=u_i$ for $1\le i\le n$ and $g\left(x\right)=u_1$ otherwise, preserves $\lambda$ (due to $\lambda$ totally reflexive and $h>n$) and does not preserve ${\omega }_n$ (due to $a_i\in {\omega }_n$, $1\le i\le n$ and $g(a_1,\dots ,a_n)=(u_1,\dots ,u_n)\notin E_k^n)$; therefore $Pol\lambda \cap Pol{\omega }_n⊊Pol\lambda$; contradicting the maximality of $Pol\lambda \cap Pol\sigma$ in $Pol\lambda$. Thus ${\omega }_{h-1}=E_k^{h-1}$.

Further we show that ${\omega }_h=E_k^h$. It is easy to see that ${\omega }_h\cap \lambda$ is an h-ary totally reflexive and totally symmetric relation contains in $\lambda$ and $Pol\lambda \cap Pol\sigma \subseteq Pol\lambda \cap Pol\left({\omega }_h\cap \lambda \right)$. We discuss two subcases: $\left(i\right)$${\omega }_h\cap \lambda =\lambda$, $\left(ii\right)$${\omega }_h\cap \lambda ⊊\lambda$. We proceed first with case $\left(ii\right)$. Since $\lambda \cap {\omega }_h$ is totally reflexive, we have $Pol\lambda \cap Pol\sigma ⊊Pol\lambda \cap Pol\left({\omega }_h\cap \lambda \right)$. Choosing $(u_1,\dots ,u_h)\in \lambda \setminus {\omega }_h$ and using an operation similar to g below, we obtain that $Pol\lambda \cap Pol\left({\omega }_h\cap \lambda \right)⊊Pol\lambda$, contradicting the maximality. Hence case $\left(ii\right)$ can not occur. Now we suppose that $\left(i\right)$ holds. It means that $\lambda$ is a subset of ${\omega }_h$. Thus three subcases can occur: $\left(i\right)$$\lambda ={\omega }_h$, $\left(ii\right)$$\lambda ⊊{\omega }_h⊊E_k^h$ and $\left(iii\right)$${\omega }_h=E_k^h$.

We begin with subcase $\left(iii\right)$${\omega }_h=E_k^h$. Using a similar argument as above in the proof of the fact that ${\omega }_{h-1}=E_k^{h-1}$, we can observe that ${\omega }_k=E_k^k$. Therefore $\lambda$ has a central element which is a contradiction.

Now we continue with subcase $\left(ii\right)$$\lambda ⊊{\omega }_h⊊E_k^h$. A similar argument as in the proof of Lemma 4.7 shows that this case is also impossible.

We finish with subcase $\left(i\right)$$\lambda ={\omega }_h$.

We consider the relation

${\phi }_l=\{(a_1,\dots ,a_l)\in E_k^l\mid \exists u\in E_k,(a_1,u),\dots ,(a_l,u)\in \sigma and\forall j_1,\dots ,j_{h-2}\in \{1,\dots ,l\},$$(u,a_2,a_{j_1},\dots ,a_{jh-2})\in \lambda \}$ for $l\ge 2$.

We observe that $\lambda$ is a subset of ${\phi }_h$. Hence, we obtain two possibilities: $\left(i\right)$$\lambda \subseteq {\phi }_h⊊E_k^h$ and $\left(ii\right)$$E_k^h={\phi }_h$.

Firstly, we discuss case $\left(i\right)$$\lambda \subseteq {\phi }_h⊊E_k^h$. Let $c\in C_{\sigma }$. Since c is not a central element of $\lambda$, there exist $a_2,\dots ,a_h\in E_k$ such that $(c,a_2,\dots ,a_h)\notin \lambda$. Hence, $(a_2,c,a_3,\dots ,a_h)\notin \lambda$ (because $\lambda$ is totally symmetric). We have $(a_i,c)\in \sigma$ for $2\le i\le h$ and $(c,c,a_{j_1},\dots ,a_{j_{h-2}})\in \lambda$ for all $j_1,\dots ,j_{h-2}\in \{1,\dots ,h\}$. So $(a_2,c,\dots ,a_h)\in {\phi }_h\setminus \lambda$. Using the h tuple $(a_2,c,\dots ,a_h)$ and the operation g defined in the proof of Lemma 4.7, we obtain $Pol\lambda \cap Pol\sigma ⊊Pol\lambda \cap Pol{\phi }_h⊊Pol\lambda$, which is a contradiction.

Secondly we investigate case ${\phi }_h=E_k^h$. A similar argument as in the proof of Lemma 4.7 shows that ${\phi }_k=E_k^k$.

We assume that $E_k=\{a_1,\dots ,a_k\}$. Then $(a_1,\dots ,a_k)\in E_k^k={\phi }_k$, so there exists $u\in E_k$ such that $(a_1,u),\dots ,(a_k,u)\in \sigma$ and $(u,a_2,a_{j_1},\dots ,a_{j_{h-2}})\in \lambda$ for all $j_1,\dots ,j_{h-2}\in \{1,\dots ,k\}$. Hence $u\in C_{\sigma }$ and $(u,a_2)\in \epsilon$. Thus $\forall x\in E_k,C_{\sigma }\cap {\left[x\right]}_{\epsilon }\ne \varnothing$. □

Proof. (Proof of Proposition 7.3) It follows from the combination of Lemma 7.4, Lemma 7.5 and Lemma 7.6. □

Proof. (Proof of Proposition 6.7) It follows from the combination of Proposition 6.7, Lemma 7.2 and Proposition 7.3. □

8. Conclusion and Further Research

In this work, we have characterized all binary relations $\sigma$ such that the clone $Pol\lambda \cap Pol\sigma$ is a submaximal clone of $Pol\lambda$ for a fixed regular relation $\lambda$ on a k-element set ($k>2$). The complete characterization of submaximal clone in the lattice of clone on finite set is still opened and we will continue to explore it in our future work.