Geometrical Equivalence and Action Type Geometrical Equivalence of Group Representations

Transcrição

1 Universidade Federal do Rio Grande do Norte Centro de Ciências Exatas e da Terra Programa de Pós-Graduação em Matemática Aplicada e Estatística Mestrado em Matemática Aplicada e Estatística Geometrical Equivalence and Action Type Geometrical Equivalence of Group Representations Josenildo Simões da Silva Natal-RN Março 2018

2 Josenildo Simões da Silva Geometrical Equivalence and Action Type Geometrical Equivalence of Group Representations Trabalho apresentado ao Programa de Pós- Graduação em Matemática Aplicada e Estatística da Universidade Federal do Rio Grande do Norte, em cumprimento com as exigências legais para obtenção do título de Mestre. Área de Concentração: Modelagem Matemática. Linha de Pesquisa: Matemática Computacional Orientador(a) Professor Doutor Arkady Tsurkov Universidade Federal do Rio Grande do Norte UFRN Programa de Pós-Graduação em Matemática Aplicada e Estatística PPGMAE Natal-RN Março 2018

3 Universidade Federal do Rio Grande do Norte - UFRN Sistema de Bibliotecas - SISBI Catalogação de Publicação na Fonte. UFRN - Biblioteca Setorial Prof. Ronaldo Xavier de Arruda - CCET Silva, Josenildo Simões da. Geometrical equivalence and action type geometrical equivalence of group representations / Josenildo Simões da Silva f.: il. Dissertação (mestrado) - Universidade Federal do Rio Grande do Norte, Centro de Ciências Exatas e da Terra, Programa de Pós- Graduação em Matemática Aplicada e Estatística. Natal, RN, Orientador: Arkady Tsurkov. 1. Matemática computacional - Dissertação. 2. Equivalência geométrica - Dissertação. 3. Equivalência geométrica de tipo de ação - Dissertação. 4. Representações de grupos - Dissertação. I. Tsurkov, Arkady. II. Título. RN/UF/CCET CDU Elaborado por JOSENEIDE FERREIRA DANTAS - CRB-15/324

4 Dissertação de Mestrado sob o título Geometrical Equivalence and Action Type Geometrical Equivalence of Group Representations apresentada por Josenildo Simões da Silva e aceita pelo Programa de Pós-Graduação em Matemática Aplicada e Estatística da Universidade Federal do Rio Grande do Norte, sendo aprovada por todos os membros da banca examinadora abaixo especicada: Professor Doutor Arkady Tsurkov Orientador(a) Departamento de Matemática Universidade Federal do Rio Grande do Norte Professor Doutor Mikhailo Dokuchaev Departamento de Matemática Universidade de São Paulo Professora Doutora Elena Aladova Departamento de Matemática Universidade Federal do Rio Grande do Norte Natal-RN, Quinta-feira, 22 de Março de 2018.

5 À minha família e amigos, que são a base de tudo.

6 Agradecimentos Agradeço a todos aqueles que ajudaram direta ou indiretamente para a construção desse trabalho, seja a ajuda em forma de almoços partilhados, conversas vespertinas com brigadeiro, dúvidas com problemas de Cálculo ou até mesmo palestras sobre Ciências Climáticas. Em especial, gostaria de agradecer à minha mãe, Sonia Maria da Silva, por todo o apoio nessa trajetória; à minha amada namorada Débora Elita de Sousa Silva, por todo afeto, carinho e consolo nos momentos difíceis; à minha querida amiga Kaliana Kaline de Oliveira, por toda doçura; aos professores do PPgMAE por todo o aprendizado e ao professor Arkady Tsurkov por toda a paciência e orientação.

7 A tarefa de viver é dura, mas fascinante. Ariano Suassuna

8 Equivalência Geométrica e Equivalência Geométrica Action Type de Representações de Grupos Autor: Josenildo Simões da Silva Orientador(a): Professor Doutor Arkady Tsurkov Resumo No presente trabalho, lidamos com conceitos de equivalência geométrica de álgebras universais. Concentramos nossa atenção na equivalência geométrica das representações de grupos. Neste caso, pode-se falar sobre equivalência geométrica (completa) e dois casos parciais: equivalência geométrica de grupos e equivalência geométrica de tipo de ação. Nosso trabalho tem dois objetivos principais. O primeiro é estudar equivalência geométrica de representações de grupos através de equivalências geométricas de grupo e de tipo de ação. O segundo objetivo é dar um exemplo que prova que não podemos concluir a equivalência geométrica de representações de grupos das correspondentes equivalência geométrica de tipo de ação e da equivalência geométrica de grupos. Palavras-chave: Equivalência geométrica, Representações de grupos, Equivalência geométrica de tipo de ação.

9 Geometrical Equivalence and Action Type Geometrical Equivalence of Group Representations Author: Josenildo Simões da Silva Advisor: Professor Doctor Arkady Tsurkov Abstract In the present work, we deal with concepts of geometrical equivalence of universal algebras. We concentrate our attention on geometrical equivalence of group representations. In this case one can speak about (full) geometrical equivalence and two partial cases: group geometrical equivalence and action-type geometrical equivalence. Our work has two main goals. The rst one is to study geometrical equivalence of group representations via group and action-type geometrical equivalences. The second goal is to give an example which proves that we can not conclude the geometrical equivalence of group representations from the corresponding action-type geometrical equivalence and group geometrical equivalence. Keywords: Geometrical Equivalence, Group Representations, Action Type Geometrical Equivalence.

10 Sumário 1 Introduction p Motivation p Thesis Overview p Universal Algebra: Basic Denitions p Universal Algebra p Examples of Algebras p Groups p Rings, Modules and Group Rings p Geometrical Equivalence of algebras p Galois Correspondence p Quasi-identities and Quasi-varieties p Operators over Classes of Algebras p Geometrical Equivalence p Representation Theory and Geometrical Equivalence of Representations p Basic notions of Representation Theory p Galois Correspondence for Group Representations p Quasi-identities in REP K p Operators over Classes of Representations p Geometrical Equivalence p. 57

11 5 Action Type Geometrical Equivalence of Representations p Action Type Galois Correspondence p Action Type Quasi-identities p Operators V and Q r p Action Type Geometrical Equivalence p Action Type Geometrical Equivalence and Geometrical Equivalence of Representations p Final Considerations p. 81 Referências p. 82

12 10 1 Introduction In the present work we consider concepts of geometrical equivalence of universal algebras and apply techniques from [18] and [23] for group representations. Speaking about geometrical equivalence of group representations we have on mind full geometrical equivalence and two partial cases: group geometrical equivalence and actiontype geometrical equivalence. One of the goal of our research is to study geometrical equivalence of group representations via group and action-type geometrical equivalences. Moreover, it is known (see [23], section 4.4) that two action type geometrically equivalent group representations whose corresponding groups are also geometrically equivalent are not necessarily full geometrically equivalent. But examples of such representations were not known. In this work we construct an example of such representation. 1.1 Motivation The considerations of databases and automatons have an important role in Applied Mathematics and Computer Science ([4], [6],[16]). For instance, it is well-known the equivalence problem for databases. It was rst posed by A.V. Aho, Y. Sagiv, J.D. Ullman in [1] and by C. Beeri, A. Mendelzon, Y. Sagiv, J. Ullmanin in [2]. B.I. Plotkin in [16] proposed a mathematical model of a database and gave a formal denition of the databases equivalence concept. The main peculiarity of this approach is that a database is treated as a certain algebraic structure, namely, as many-sorted algebra [9], [16]. The formal mathematical model of a database allows to solve formally the equivalence problem for databases. Nowadays, various solutions for the databases equivalence problem based on algebraic geometry approach were obtained in [11],[14], [20]. In this concern, considerations of geometrical equivalence of many-sorted algebras have a practical meaning. In the case of many sorted algebras we can consider full ge-

13 11 ometrical equivalence and partial geometrical equivalences, i.e. geometrical equivalences of the corresponding sorts of many-sorted algebra(see [16], [24]). The studies of relations between these equivalences is important for Computer Science. Investigation presented in this thesis is an example of such studies. Group representations are a particular case of many sorted algebras (see [18], [25]). In particular, in our work we analyze the relations between the geometrical equivalence and action type geometrical equivalence of group representations. 1.2 Thesis Overview This work consists of six chapters. In Chapter 2 we give all necessary denitions and results from universal algebras. In particular, we consider such algebraic structures as groups, rings and modules. In Chapter 3 we include concepts and results about geometrical equivalence of universal algebras. We begin with closure operators and the denition of quasi-identities, which will give us a logical relation between geometrically equivalent algebras. Then we dene operators over classes of algebras and consider their properties. Finally, we give the denition of geometrical equivalence and various criteria of geometrical equivalence of algebras. In Chapter 4 we apply the notions from Chapter 3 to study the geometrical equivalence of group representations. We start with basic denitions and facts from Representation Theory. Then we specify the notions of closure operators, of operators over classes of algebras and of geometrical equivalence from Chapter 3 for the case of group representation. The main aim of this chapter is to dene the geometrical equivalence of group representations. In Chapter 5 we consider the action type geometrical equivalence of group representations. In this case we deal with only one part of representation, which is related to module. Our main goal is to present an example which shows that the action type geometrical equivalence of two representations and the geometrical equivalence of their groups do not imply full geometrical equivalence. Conclusions are presented in Chapter 6.

14 12 2 Universal Algebra: Basic Denitions This chapter contains the basis for our work. It describes, according to [7] and [22], the basic concepts of Universal Algebra. Moreover we will present some particular cases of algebras which will be useful in the next chapters. 2.1 Universal Algebra Denition Let A be a non-empty set. If n is a non-negative integer, we dene: (i) A 0 = { }; (ii) A n is the cartesian product of n copies of A, that is, an element a A n has the form a = (a 1, a 2,..., a n ), where a i A, 1 i n. (iii) A mapping ω : A n A is called an n-ary algebraic operation over A. The number n is called arity (or rank) of ω, and it will be denoted by ρ(ω) = n. The operation ω is called nullary, unitary, binary or ternary if its arity is 0, 1, 2 or 3, respectively. Remark 2.1.1: A nullary operation ω is completely determined by the image ω( ) of the (unique) element in A 0. Hence, ω can be viewed as an element of A, whose name will be constant. Denition An (universal) algebra is a pair (A, Ω), where A is a non-empty set and Ω is a collection of algebraic operations over A. The collection Ω is called signature (or type) of the algebra A.

15 13 Denition Let A be an algebra with signature Ω and B A. We say that B is a subalgebra of A if it is closed over the operations from Ω, that is, for every b i B, 1 i n, and every n-ary algebraic operation ω Ω, we have that ω(b 1,..., b n ) B. If B is a subalgebra of A, we write B A. Proposition Let A be an algebra with signature Ω. If B i A, then i I B i A, i I, where I is a set of indices. Proof. Let b j B i, 1 j n and ω Ω be an n-ary algebraic operation. We have i I that b j B i, 1 j n, for every i I, and then ω(b 1,..., b n ) B i, for every i I. Hence ω(b 1,..., b n ) B i, and we conclude that B i A. i I i I Denition Let A be an algebra with signature Ω, X A and B X = {B A X B}. The smallest algebra from B X is called the subalgebra generated by the set X. We denote this subalgebra by X. Proposition Let A be an algebra with signature Ω and X A. Then we have that X = B B X B. Proof. By Proposition 2.1.1, we have that B is a subalgebra of A. Now, since X B, B B X for every B B X, we have that X B, for every B B X. So, X B. Moreover, B B X we have that B X, because X is an element of B X. Therefore, X = B. B B X B B X In the following result, we describe the form of the elements of X. Proposition Let A be an algebra with signature Ω and X A. If

16 14 M 0 (X) = {c A ω = c Ω, ρ(ω) = 0} X, k M k+1 (X) = {ω(a 1,..., a n ) ω Ω, ρ(ω) = n, a i M j (X), 1 i n}, then X = M(X), where M(X) = M j (X). j=0 Proof. Firstly, we have to prove that M(X) is an algebra. Indeed, if a 1,..., a n M(X), there exist j 1,..., j n such that a i M ji (X), 1 i n. So, if p = Max{j i 1 i n}, it follows that ω(a 1,..., a n ) M p+1 (X), for every ω Ω such that ρ(ω) = n, and hence M(X) is an algebra. As M(X) contains X, we have that X M(X). Moreover, note that M 0 (X) X, because X is an algebra which contains X. Now, suppose that M i (X) X, for every 1 i k. If b M k+1 (X), we have that b = ω(a 1,..., a n ), where ω Ω, ρ(ω) = n and a i k M j (X), 1 i n. Since M j (X) X, for every 1 j k, it follows that j=0 j=0 k M j (X) X. X is an algebra, thus we have j=0 that b = ω(a 1,..., a n ) X and then M k+1 (X) X. It follows by induction that M(X) = M j (X) X, and the proposition is proved. j=0 Now, we have a very important denition of a special kind of mapping, which preserves the algebraic structure between two algebras with the same signature. Denition Let A, B be two algebras with the same signature Ω. A mapping ϕ : A B is called a homomorphism if, for every a 1,..., a n A and ω Ω such that ρ(ω) = n, we have ϕ(ω A (a 1,..., a n )) = ω B (ϕ(a 1 ),..., ϕ(a n )). A homomorphism ϕ : A B is called epimorphism if it is surjective, monomorphism if it is injective and isomorphism if it is a bijection. If A = B, then ϕ is called endomorphism. If there exists an isomorphism ϕ : A B, we say that A is isomorphic to B, and we denote it by A = B. Moreover, if ϕ : A B is a homomorphism and ω = c A, then ϕ(c A ) = c B. About composition of homomorphisms, we have the following result. Proposition Let A, B, C be algebras with the same signature Ω. If ϕ : A B and ψ : B C are homomorphisms, then the composition ψ ϕ : A C is a homomorphism.

17 15 Proof. Let a 1,..., a n A and ω Ω such that ρ(ω) = n. It follows that (ψ ϕ)(ω A (a 1,..., a n )) = ψ(ϕ(ω A (a 1,..., a n ))) = ψ(ω B (ϕ(a 1 ),..., ϕ(a n ))). Hence, (ψ ϕ)(ω A (a 1,..., a n )) = ω C (ψ(ϕ(a 1 )),..., ψ(ϕ(a n ))) = ω C ((ψ ϕ)(a 1 ),..., (ψ ϕ)(a n )), and the result follows. Denition Let ϕ : A B be a homomorphism. The kernel of ϕ is the set Ker(ϕ) = {(a 1, a 2 ) A 2 ϕ(a 1 ) = ϕ(a 2 )}. The image of ϕ is the set im(ϕ) = {ϕ(a) a A}. Remark 2.1.2: If ϕ : A B is a homomorphism, then im(ϕ) B, because for every b 1,..., b n im(ϕ), there exist a 1,..., a n A such that ϕ(a i ) = b i, 1 i n, and then for every ω Ω such that ρ(ω) = n, we have that ω B (b 1,..., b n ) = ω B (ϕ(a 1 ),..., ϕ(a n )) = ϕ(ω A (a 1,..., a n )) im(ϕ). Hence, if ϕ is a monomorphism, we have that ϕ : A im(ϕ) is an isomorphism. So, up to isomorphism, we can say that A B. Moreover, Ker(ϕ) = {(a 1, a 2 ) A 2 ϕ(a 1 ) = ϕ(a 2 )} = {(a 1, a 2 ) A 2 a 1 = a 2 } = {(a, a) a A}. Proposition Let ϕ : A B be a homomorphism and X A. Then ϕ(m k (X)) = M k (ϕ(x)), for every k N. Proof. Firstly, note that ϕ(c A ) = c B, for every ω = c A such that ρ(ω) = 0. So, ϕ(m 0 (X)) = ϕ({c A ω = c Ω, ρ(ω) = 0} X) and then ϕ(m 0 (X)) = {c B ω = c Ω, ρ(ω) = 0} ϕ(x) = M 0 (ϕ(x)). Now, suppose that ϕ(m k (X)) = M k (ϕ(x)), for every 1 i k. If b ϕ(m k+1 (X)), we have that b = ϕ(a), where a M k+1 (X), that is, a = ω A (a 1,..., a n ), with ω Ω, ρ(ω) = n and a i k M j (X), 1 i n. By the induction hypothesis, ϕ( k M j (X)) = j=0 k ϕ(m j (X)) = k M j (ϕ(x)). So, j=0 j=0 b = ϕ(a) = ω B (ϕ(a 1 ),..., ϕ(a n )) M k+1 (ϕ(x)) and hence ϕ(m k+1 (X)) M k+1 (ϕ(x)). j=0

18 16 Moreover, if a M k+1 (ϕ(x)), we have that a = ω B (b 1,..., b n ), where b i k M j (ϕ(x)), for each 1 i n. By the induction hypothesis, it follows that k M j (ϕ(x)) = k ϕ(m j (X)). So, for each 1 i n, there exists a i M p (X) such that 1 p k and b i = ϕ(a i ). Thus, a = ω B (ϕ(a 1 ),..., ϕ(a n )) and a = ϕ(ω A (a 1,..., a n )) ϕ(m k+1 (X)). Therefore, M k+1 (ϕ(x)) ϕ(m k+1 (X)). We conclude that ϕ(m k+1 (X)) = M k+1 (ϕ(x)), and it follows by induction that ϕ(m k (X)) = M k (ϕ(x)), for every k N. Corollary Let ϕ : A B be a homomorphism. If A = X, then im(ϕ) = ϕ(x). j=0 j=0 j=0 Proof. It follows immediately from the previous proposition: ϕ(x) = M j (ϕ(x)) = ϕ(m j (X)) = ϕ( M j (X)) = ϕ( X ) = im(ϕ). j=0 j=0 j=0 Remark 2.1.3: If A, B are two algebras with the same signature, the set of all homomorphisms ϕ : A B will be denoted by Hom(A, B). Now, let {A i i I} a family of algebras in signature Ω. If we consider X = A i, the i I elements of X are functions a : I A i, where a(i) A i, i I. If we denote a(i) = a i, I i we can present each element of X as a sequence (a i ) i I, where a i A i, i I. We will see, now, that X is also an algebra with signature Ω. Denition Let {A i i I} be a family of algebras with signature Ω. The set A i i I with the operations ω Ω dened by (ω A i (a 1,..., a n )) i = ω Ai ((a 1 ) i,..., (a n ) i ), i I is called the Cartesian product of the family {A i i I}. Remark 2.1.4: If A i = A, i I, we denote i I A i = A I.

19 17 Proposition Let {A i i I} be a family of algebras with signature Ω. If j I is xed, then the mapping π j : A i A j, i I such that π j (a) = a j is a homomorphism. Proof. Let a 1,..., a n i I A i and ω Ω such that ρ(ω) = n. We have that π j ((ω(a 1,..., a n )) i I ) = (ω(a 1,..., a n )) j = ω((a 1 ) j,..., (a n ) j ) = ω(π j (a 1 ),..., π j (a n )), and π j is a homomorphism. The mappings π j : A i A j are called projections of A i on A j. Now, we have i I i I an important result: it is called Remak Theorem. Theorem Let A be an algebra, {A i i I} be a family of algebras (A and A i have the same signature Ω) and ϕ i : A A i be a family of homomorphisms. Then the mapping ϕ : A A i dened by (ϕ(a)) i = ϕ i (a) is a homomorphism and Ker(ϕ) = Ker(ϕ i ). i I i I Proof. Let a 1,..., a n A and ω Ω such that ρ(ω) = n. Note that for every i I, (ϕ(ω A (a 1,..., a n ))) i = ϕ i (ω A (a 1,..., a n )) = ω Ai (ϕ i (a 1 ),..., ϕ i (a n )). On the other hand, for every i I, we have (ω A i (ϕ(a 1 ),..., ϕ(a n ))) i = ω Ai ((ϕ(a 1 )) i,..., (ϕ(a n )) i ) = ω Ai (ϕ i (a 1 ),..., ϕ i (a n )), i I and then ϕ(ω A (a 1,..., a n )) = ω(ϕ(a 1 ),..., ϕ(a n )). We conclude that ϕ is a homomorphism. Now, Ker(ϕ) = {(a 1, a 2 ) A 2 ϕ(a 1 ) = ϕ(a 2 )} = {(a 1, a 2 ) A 2 (ϕ(a 1 )) i = (ϕ(a 2 )) i, i I}, and then Ker(ϕ) = {(a 1, a 2 ) A 2 ϕ i (a 1 ) = ϕ i (a 2 ), i I} The theorem is proved. = {(a 1, a 2 ) A 2 (a 1, a 2 ) Ker(ϕ i ), i I} = i I Ker(ϕ i ). Now, we have a result that connects Cartesian product and isomorphisms of algebras. For the proof see Theorem 4 in [9].

20 18 Proposition Let {X j j A} be a family of algebras with signature Ω and {A i i I} be a family of pairwise disjoint index sets. If A = A i, then i I X j = ( j A i I j A i X j ). Denition Let A be an algebra with signature Ω. A binary relation T A 2 is called a congruence if it is an equivalence relation on A and, for every ω Ω (ρ(ω) = n) and (a i, b i ) T, for every 1 i n, we have that (ω(a 1,..., a n ), ω(b 1,..., b n )) T. Proposition If T i A 2 (i I) is a family of congruences, then T = i I T i is a congruence. Proof. Firstly, we have to prove that T is an equivalence relation. Note that (a, a) T i for every i I, then (a, a) T, which means that T is reexive. Furthermore, if (a, b) T, we have that (a, b) T i, i I, and so (b, a) T i (this is true because each T i is an equivalence relation). It follows that (b, a) T and then T is symmetric. Finally, if (a, b), (b, c) T, then (a, b), (b, c) T i, for every i I and so (a, c) T i for every i I, and then (a, c) T. Thus, T is transitive, and so an equivalence relation. Now, for every ω Ω (ρ(ω) = n) and (a k, b k ) T, for every 1 k n, we have that (a k, b k ) T i, for all i I, and then (ω(a 1,..., a n ), ω(b 1,..., b n )) T i, i I Thus (ω(a 1,..., a n ), ω(b 1,..., b n )) T, and we conclude that T is a congruence. Proposition Let A, B be algebras with the same signature Ω and ϕ : A B a homomorphism. Then Ker(ϕ) is a congruence. Proof. Firstly, note that, for every a A ϕ(a) = ϕ(a), and then (a, a) Ker(ϕ). Hence, Ker(ϕ) is reexive. Moreover, if (a, b) Ker(ϕ), then ϕ(a) = ϕ(b), and then ϕ(b) = ϕ(a). Thus, (b, a) Ker(ϕ). We conclude that Ker(ϕ) is symmetric. Finally, if (a, b), (b, c) Ker(ϕ), we have that ϕ(a) = ϕ(b) = ϕ(c), which means that (a, c) Ker(ϕ).

21 19 Therefore, Ker(ϕ) is transitive and we conclude that Ker(ϕ) is an equivalence relation. Now, for every ω Ω (ρ(ω) = n) and (a i, b i ) Ker(ϕ), we have ϕ(ω A (a 1,..., a n )) = ω B (ϕ(a 1 ),..., ϕ(a n )), and also ω B (ϕ(a 1 ),..., ϕ(a n )) = ω B (ϕ(b 1 ),..., ϕ(b n )). It follows that ϕ(ω A (a 1,..., a n )) = ω B (ϕ(b 1 ),..., ϕ(b n )) = ϕ(ω A (b 1,..., b n )), and we conclude that Ker(ϕ) is a congruence. Denition Let A be an algebra with signature Ω and T A 2 be a congruence.the quotient algebra is the set A/T with operations ω A/T ([a 1 ] T,..., [a n ] T ) = [ω A (a 1,..., a n )] T, where ω Ω (ρ(ω) = n) and a i A, for every 1 i n. For every x A, [x] T equivalence class of x by T. is an This denition is correct, because if [a i ] T = [b i ] T, for every 1 i n, then (a i, b i ) T for every 1 i n. Therefore (ω A (a 1,..., a n ), ω A (b 1,..., b n )) T and [ω A (a 1,..., a n )] T = [ω A (b 1,..., b n )] T. Proposition Let A/T be a quotient algebra with signature Ω. The mapping τ : A A/T dened by τ(a) = [a] T is an epimorphism and Ker(τ) = T. Proof. Let ω Ω (ρ(ω) = n) and a i A for every 1 i n. By the denition of τ, we have τ(ω A (a 1,..., a n )) = [ω A (a 1,..., a n )] T. Since [ω A (a 1,..., a n )] T = ω A/T ([a 1 ] T,..., [a n ] T ), it follows that τ(ω A (a 1,..., a n )) = ω A/T (τ(a 1 ),..., τ(a n )), and so τ is a homomorphism. Moreover, if b A/T, there exists a A such that b = [a] T = τ(a), which means that τ is an epimorphism. Finally, note that Ker(τ) = {(a 1, a 2 ) A 2 τ(a 1 ) = τ(a 2 )} = {(a 1, a 2 ) A 2 [a 1 ] T = [a 2 ] T }. Since {(a 1, a 2 ) A 2 [a 1 ] T = [a 2 ] T } = {(a 1, a 2 ) A 2 (a 1, a 2 ) T } = T, we conclude that Ker(τ) = T. The epimorphism τ above is called the natural epimorphism. Now, we will see a very important result, called First Isomorphism Theorem (see [22], Theorem 6.12):

22 20 Theorem Let A, B be algebras with signature Ω. If ϕ : A B is a homomorphism, then there exists an isomorphism ψ : A/Ker(ϕ) im(ϕ) such that ϕ = ψτ, where τ : A A/Ker(ϕ) is the natural epimorphism. Proof. Firstly, consider the mapping ψ : A/Ker(ϕ) im(ϕ) dened by ψ([a] Ker(ϕ) ) = ϕ(a). Note that [a 1 ] Ker(ϕ) = [a 2 ] Ker(ϕ) if and only if (a 1, a 2 ) Ker(ϕ), which takes place if and only if ψ([a 1 ] Ker(ϕ) ) = ϕ(a 1 ) = ϕ(a 2 ) = ψ([a 2 ] Ker(ϕ) ). Then ψ is well-dened and injective. If b im(ϕ), there exists a A such that b = ϕ(a) = ψ([a] Ker(ϕ) ). Therefore ψ is surjective. Now, if ω Ω (ρ(ω) = n) and a i A, for every 1 i n, we have that ψ(ω A/Ker(ϕ) ([a 1 ] Ker(ϕ),..., [a n ] Ker(ϕ) )) = ψ([ω A (a 1,..., a n )] Ker(ϕ) ) = ϕ(ω A (a 1,..., a n )). Since ϕ is a homomorphism, we have ϕ(ω A (a 1,..., a n )) = ω B (ϕ(a 1 ),..., ϕ(a n )) = ω B (ψ([a 1 ] Ker(ϕ) ),..., ψ([a n ] Ker(ϕ) )), and then ψ is a homomorphism, and so an isomorphism. Finally, note that for every a A, it is true that ϕ(a) = ψ([a] Ker(ϕ) ) = ψ(τ(a)), and the proof is complete. Now, we will consider the following sets: M 0 (X) = {ω ω Ω, ρ(ω) = 0} X, k Mk+1(X) = {ωb 1 b 2...b n ω Ω, ρ(ω) = n > 0, b 1, b 2,..., b n Mj (X), 1 i 0 n; b i0 Mk (X)} and j=0 M (X) = Mj (X), j=0 for an arbitrary set X and a signature Ω. For each ω Ω such that ρ(ω) = 0 dene the correspondent constant in M (X) by c M (X), and ω M (X)(b 1, b 2,..., b n ) = ωb 1 b 2...b n,

23 21 for ω Ω (ρ(ω) = n), b 1, b 2,..., b n M (X), we have that M (X) is an algebra with signature Ω. It is called the algebra of terms on the alphabet X in the signature Ω. With this denition, it is clear that if ω 1 (b 1, b 2,..., b n ) = ω 2 (m 1, m 2,..., m k ), then ω 1 = ω 2, ρ(ω 1 ) = n = k = ρ(ω 2 ) and b i = m i, 1 i n. The following proposition gives us some interesting properties of M (X). Proposition Let X be an arbitrary set and Ω a signature. The algebra M (X) has the following properties: (i) If i j, then M i (X) M j (X) = ; (ii) M (X) = X. Proof. (i) Firstly, note that X M i (X) =, for i > 0. Now, suppose that i 0 > 0 and a M 0 (X) M i 0 (X). Since a M i 0 (X), we have that a = ω(b 1, b 2,..., b n ), where ρ(ω) = n > 0. On the other hand, a M 0 (X), and it follows that ρ(ω) = 0. With this contradiction, we conclude that M 0 (X) M i (X) =, whenever i > 0. Now, suppose that Mi (X) Mk (X) =, for every 0 < i k. If ω 1 (b 1, b 2,..., b n1 ) M k+1(x), ω 2 (t 1, t 2,..., t n2 ) M i (X), 0 < i k, and ω 1 (b 1, b 2,..., b n1 ) = ω 2 (t 1, t 2,..., t n2 ), it follows that ω 1 = ω 2, n 1 = n 2, and b p = t p, for all 1 p n. By the denition of Mk+1 (X), there exists 1 j n such that b j Mk (X). On the other hand, by the denition of Mi (X), it follows that t j Mj 0 (X), where j 0 < i k. By the hypothesis, we have that b j induction. t j, it is a contradiction. The result follows by (ii) Since X M (X) and M (X) is an algebra, we have that X M (X). Now, if B is an algebra which contains X, that is, B B X, it follows that {ω ω Ω, ρ(ω) = 0} B, because B is an algebra. So, M 0 (X) B. Now, suppose that M i (X) B, for every 1 i k. If m 1, m 2,..., m n k j=0 M j (X), we have that

24 22 ω(m 1, m 2,..., m n ) B, for every ω Ω such that ρ(ω) = n, because B is an algebra k and Mj (X) B. Thus, Mk+1 (X) B, and by induction we conclude that j=0 M (X) B. As B B X was chosen abitrarily, it follows that M (X) B = B B X X. Therefore, M (X) = X. Proposition Let A be an algebra with signature Ω and X an arbitrary set. For every mapping f : X A, there exists a unique homomorphism ϕ : M (X) A such that ϕ X = f. Proof. Dene ϕ : M (X) A by ϕ(x) = f(x), x X, ϕ(c) = c A, c Ω, ρ(ω) = 0. If ϕ(t) is dened for t k Mi (X), we dene ϕ in Mk+1 (X) by i=0 ϕ(ωt 1 t 2...t n ) = ω A (ϕ(t 1 ),..., ϕ(t n )), ω Ω, ρ(ω) = n. Now, we have to check that ϕ is a homomorphism. Indeed, if ω Ω is an operation such that ρ(ω) = n and t 1,..., t n M (X), we have that there exist k N and i 0 {1, 2,..., n} such that t 1,..., t n k Mi (X) and t i0 Mk (X). So, i=0 ϕ(ω M (X)(t 1...t n )) = ϕ(ωt 1...t n ) = ω A (ϕ(t 1 ),..., ϕ(t n )). Thus, ϕ is a homomorphism. Now, if there exists another homomorphism ψ : M (X) A such that ψ(x) = f(x), x X, then we have ψ(x) = f(x) = ϕ(x), x X and ψ(c) = c A = ϕ(c), c Ω, ρ(ω) = 0, which means that ψ M 0 (X) = ϕ M 0 (X). Now, suppose that ψ k = ϕ k. Let Mi (X) Mi (X) t = ωt 1...t n Mk+1 (X), where ρ(ω) = n, t 1,..., t n k Mi (X) and there exists i 0 {1, 2,..., n} such that t i0 Mk (X). So, we obtain i=0 t = ωt 1...t n = ω M (X)(t 1,..., t n ), i=0 i=0

25 23 ψ(t) = ψ(ω M (X) (t 1,..., t n )) = ω A (ψ(t 1 ),..., ψ(t n )), and nally ψ(t) = ω A (ψ(t 1 ),..., ψ(t n )) = ω A (ϕ(t 1 ),..., ϕ(t n )) = ϕ(ω M (X) (t 1,..., t n )) = ϕ(t). Hence, by induction we conclude that ψ = ϕ. Proposition Let A, B be two algebras with signature Ω, ϕ : M (X) B a homomorphism and ψ : A B an epimorphism. There exists a homomorphism χ : M (X) A such that ϕ = ψχ. Proof. As ψ is an epimorphism, for every x X there exists a x A such that ψ(a x ) = ϕ(x). So, dening f : X A by f(x) = a x, for any x X, we have by the previous proposition a (unique) homomorphism χ : M (X) A such that χ(x) = f(x), x X. Now, note that ψ(χ(x)) = ψ(f(x)) = ψ(a x ) = ϕ(x), x X, Then, dening g : X B by g(x) = ϕ(x) = (ψχ)(x), x X, we conclude by the previous proposition that ϕ = ψχ. Now we will see the denition of fullling for an identity, and the denition of variety of algebras. Denition Let X = {x 1, x 2,..., x n } and f(x 1,..., x n ), g(x 1,..., x n ) M (X). We say that an algebra A fullls the identity f(x 1,..., x n ) = g(x 1,..., x n ), if for every ϕ Hom(M (X), A) we have ϕ(f(x 1,..., x n )) = ϕ(g(x 1,..., x n )). Denition Let J be a set of identities. The variety dened by J is the class of all algebras with the same signature Ω which fulll all identities from J. We write V ar(j ).

26 24 Remark 2.1.5: The set of quasi-identities J can be empty. We will nish this section with the denition of a free algebra and some important results about it. The proof for those results can be found in [3]. Denition Let Θ be a variety of algebras with signature Ω, W Θ and X W. We say that W is a free algebra of the variety Θ generated by the set X of free generators if for every A Θ and for every mapping f : X A there exists a unique homomorphism ϕ : W A such that ϕ X = f. This algebra is denoted by W = W Θ (X). Theorem Let Θ be a variety of algebras with signature Ω such that there exists A Θ which contains more than one element ( A > 1). For every cardinal α there exists a free algebra W Θ (X) such that X = α. Remark 2.1.6: In Theorem 2.1.3, W Θ (X) = M (X)/J Θ (X), where J Θ (X) is the set of all identities f = g (f, g M (X)) which are fullled in every algebra of Θ. It is easy to prove that J Θ (X) is a congruence in M (X). Proposition Let Θ be a variety of algebras with signature Ω, W Θ and X W. (i) W Θ (X) = X ; (ii) W Θ (X) = W Θ (Y ) if X = Y. Proposition Let Θ be a variety of algebras with signature Ω. For every A Θ, there exists W Θ (X) such that A = W Θ (X)/T, where T is a congruence in W Θ (X). The following Proposition is similar to Proposition : Proposition Let Θ be a variety of algebras with signature Ω, A, B Θ, W Θ (X) a free algebra in Θ, ϕ : W Θ (X) B a homomorphism and ψ : A B an epimorphism. There exists a homomorphism χ : W Θ (X) A such that ϕ = ψχ. 2.2 Examples of Algebras In this section, we consider various examples of universal algebras which will be useful in the subsequent considerations. The interested reader can check [5], [8] and [10] for more details about these structures.

27 Groups Denition A group is a set G with three algebraic operations (i) A Binary operation. : G G G (we denote.(a, b) = ab, for every a, b G), (ii) An unary operation G a a 1 G, (iii) A constant unit 1 G G, which fulll the following identities: (a) For every a, b, c G, a(bc) = (ab)c. (b) For every a G, a1 G = 1 G a = a. (c) For every a G, aa 1 = a 1 a = 1 G. A group G is called abelian if ab = ba, for every a, b G. Let G be a group. The order of a G is the smallest n N such that a n = 1 G. The order of a will be denoted by a. Let X be an arbitrary set. We can consider words over X by the following way: Denition Words over X are sequences w = a 1 a 2...a n, where a i X, i {1, 2,...n}. If n = 0, we have the empty word, which will be denoted by ε. If we denote X = { ã a X} (where ã is just a symbol obtained from a and ), it is clear that X X =. Put X = X X and, for b X, dene b = { ã, if b = a X, a, if b = ã X. We say that an expression w is a group word if it has the form w = a 1 a 2...a n, where a i X, for every i {1, 2,...n}. A group word w = a 1 a 2...a n (a i X ) is called reduced

28 26 if a i ã i+1, that is, if it does not contain a subword aã, where a X. We will assume that the empty word ε is reduced. The set of all reduced words from X will be denoted by F (X). If b X and w = a 1 a 2...a n F (X), we dene w ε = w and b, if n = 0, ab, if n = 1 and b ã, w b = ε, if n = 1 and b = ã, a 1 a 2...a n 1, if n > 1 and b = ã n, a 1 a 2...a n b, if n > 1 and b ã n. The multiplication b w is dened analogously. Moreover, for elements w 1 = a 1 a 2...a n and w 2 = b 1 b 2...b n in F (X), w 1 w 2 can be dened recursively by w 1 w 2 = [(a 1 a 2...a n ) b 1 ] (b 2...b n ). So, we can consider the following algebraic operations over F (X): (i) (.) 1 : F (X) F (X) is such that, for w = a 1 a 2...a n F (X), ε, if n = 0, w 1 = ã, if n = 1, ã n...ã 2 ã 1, if n > 1. (ii) : F (X) F (X) F (X) is such that w 1 w 2 = w 1 w 2. With these two operations, it is easy to verify that F (X) becomes a group with unit 1 G = ε. Denition Let G be a group and X an arbitrary set. We say that G is the free group generated by X if G = F (X). Remark 2.2.1: It is possible to prove that this denition of free group is a particular case of denition We will nish this subsection with a result which is called the Third Isomorphism Theorem. We will use it in Chapter 5. The interested reader can nd its proof in [8], see Theorem

29 27 Theorem Let G be a group. If H G, A G and H A, then (G/H)/(A/H) = G/A Rings, Modules and Group Rings Denition A ring is a set R which has two binary operations + and (to simplify the notation, we will denote a b by ab) satisfying the following properties: (i) (R, +) is an abelian group. The identity element of + will be denoted by 0 and for a R, the inverse of a with respect to + will be denoted by a. (ii) The operation is associative. (iii) The operation is distributive with respect to +. In other words, for every a, b, c R, a(b + c) = ab + ac and (a + b)c = ac + bc; If has an identity element 1 R then R is called ring with unit. A commutative ring with unit R is called a eld if every element a 0 has an inverse a 1 R with respect to. Denition Let R be a ring with unity. A (left) R-module M is an abelian group (M, +) with additional operation : R M M such that, for all r, s R and m, n M, we have: (i) (r + s) m = r m + s m; (ii) r (m + n) = r m + r n; (iii) (rs) m = r (s m); (iv) 1 R m = m. If the ring R is a eld, then R-module M is called a vector space. Denition A (left) R-module M is called free if there exists a set E M such that:

30 28 (i) M = { n i=1 r ix i n N, r i R, x i E, i {1, 2,..., n}} (M = E ); (ii) If r i R, x i E (i {1, 2,..., n}) and n i=1 r ix i = 0, then r 1 = r 2 =... = r n = 0 R. The set E is called the basis of R-module M. Denition Let G be a group and R a ring with unity. The group ring RG is the set of all sums α = g G a g g, where a g R, for all g G and the set {a g 0} is nite. Let α = g G a gg and β = g G b gg be elements of RG. We dene α + β = g G(a g + b g )g, and, if β = h G b hh, αβ = ( a g g)( b h h) = g G h G a g b h gh. g G,h G To simplify the notation above, we put αβ = u G C uu, where C u = gh=u a g b h. It is not dicult to see that, with these operations, RG becomes a ring with unity ɛ = 1 R e (e is the unity element of the group G). The mapping ν : G RG dened by ν(g) = h G a hh, where a g = 0 R, if h g and a g = 1 R, is an embedding from G to RG. So, we can consider G as a subset of RG. The material presented here will be important in the next chapters. Particularly, this section will be important in the studying of geometrical equivalence of representations, which will take place in Chapter 4.

31 29 3 Geometrical Equivalence of algebras In the present chapter we consider some concepts of universal algebraic geometry over a xed variety of algebras Θ (see [12], [13], [15] and [17]). Firstly, we begin with denitions of closure operators and their properties. 3.1 Galois Correspondence First of all, we need to introduce the concept of Galois correspondence and closure operators (see, for instance, [21]). Denition Let A, B be sets ordered by inclusion. A Galois correspondence on the pair (A, B) is a pair of maps (F, G), F : A B and G : B A, with the following properties: (i) S T A F (S) F (T ); (ii) S T B G(S) G(T ); (iii) T G(F (T )), for every T A; (iv) S F (G(S)) for every S B. Let S A and T B. Denition Elements cl(s) = G(F (S)) and cl(t ) = F (G(T )) are called the Galois closures of S and T, respectively. Denition Elements S A and T B are called Galois closed if cl(s) = S and cl(t ) = T.

32 30 Let Θ be a variety of algebras and a set X be given. In this chapter, we assume that X is always nite. Denition Let W Θ (X) be the free algebra generated by the set X. Equations over W Θ (X) are elements of W Θ (X) W Θ (X), i.e., pairs of theform (w 1, w 2 ), where w 1, w 2 W Θ (X). We denote (w 1, w 2 ) by w 1 = w 2. Denition The solution of equation w 1 = w 2 in H Θ is a point µ of the space Hom(W Θ (X), H) such that µ(w 1 ) = µ(w 2 ), i.e., (w 1, w 2 ) Ker(µ). We can dene maps on W Θ (X) W Θ (X) and on Hom(W Θ (X), H), which we denote by the same symbol. For a set T W Θ (X) W Θ (X), we put: T H = {µ Hom(W Θ (X), H) T Ker(µ)}. It is called an algebraic set. Now let a set A Hom(W Θ (X), H) be given. We can consider the set: A H = µ A Ker(µ). This set is a congruence on W Θ (X). The next proposition describes some properties of maps. Proposition Let H Θ. For S, T W Θ (X) W Θ (X) and A, B Hom(W Θ (X), H), we have the following properties: (i) S T S H T H ; (ii) A B A H B H ; (iii) T T H ; (iv) A A H. Proof. (i) If µ T H, then T Ker(µ). Since S T, we have that S Ker(µ). So, µ S H and then T H S H.

33 31 (ii) Since A B, it follows that µ B Ker(µ) µ A Ker(µ) and so B H A H. (iii) If µ T H, we have that T Ker(µ) and then T µ T H Ker(µ) = T H. (iv) For every µ 0 A, we have that µ A Ker(µ) Ker(µ 0). Thus, µ 0 A H and A A H. Proposition says that the maps dened above provide us a Galois correspondence between the set of equations T W Θ (X) W Θ (X) and the set of points µ Hom(W Θ (X), H). So, by Denition we can speak about the H-closure for a set of equations T W Θ (X) W Θ (X), which can be dened by T H = (T H) H = Ker(µ), µ T H and about the H-closure for a set of points A Hom(W Θ (X), H): A H = (A H) H = {µ Hom(W Θ (X), H) A H Ker(µ)}. According to Denition 3.1.3, a set of equations T W Θ (X) W Θ (X) is H-closed if T H A Hom(W Θ(X), H) is H-closed if A H following result takes place. = A. The Proposition Let H Θ. For every set of equations T W Θ (X) W Θ (X), T H is the smallest H-closed congruence containing the set T. Proof. We have that T H such that T R, we have that R H T H is an H-closed congruence. Moreover, if R is an H-closed congruence and then T H R H = R. 3.2 Quasi-identities and Quasi-varieties Denition Let Θ be a variety of algebras. A quasi-identity is a logic formula which has the form

34 32 n ( (w i = w i )) (w 0 = w 0), i=1 where w i, w i W Θ (X), 0 i n and X = m, m, n N. Denition Let Θ be a variety of algebras and H Θ. We say that H fullls a quasi-identity n ( (w i = w i )) (w 0 = w 0) i=1 if for all α Hom(W Θ (X), H) such that α(w i) = α(w i ), 1 i n, we have that α(w 0) = α(w 0). We write H (( n i=1 (w i = w i )) (w 0 = w 0)). We can generalize the denition of a quasi-identity above and consider innite quasiidentities, i.e., the logic formula which have the form ( i I (w i = w i )) (w 0 = w 0), where I is a set of indices. In this case, H (( i I (w i = w i )) (w 0 = w 0)) if for every α Hom(W Θ (X), H) such that α(w i) = α(w i ), for any i I, we have α(w 0) = α(w 0). Now, let X be a class of algebras in a variety Θ. We denote by q id(x) the set of all quasi-identities which are fullled on all algebras of this class. Moreover, if Q is a set of quasi-identities in a variety of algebras Θ, we denote by q V ar(q) the set of all algebras in the variety Θ that fulll all quasi-identities of Q. The quasi-variety generated by the class X Θ is dened by the set of algebras q V ar(q id(x)), which we denote by q V ar(x). If X = {H} (H Θ), we denote q id(x) by q id(h) and q V ar(x) by q V ar(h). In the following proposition, we will see an interesting relation between quasi-identities and the maps. Proposition H (( i I (w i = w i )) (w 0 = w 0)), where {w i, w i i I} {w 0, w 0} W Θ (X) if and only if (w 0, w 0) {(w i, w i ) i I} H. Proof. ( ) For every µ 0 Hom(W Θ (X), H) such that (w i, w i ) Ker(µ 0 ), µ 0 {(w i, w i ) i I} H, we have that (w 0, w 0) Ker(µ 0 ), and then (w 0, w 0) µ {(w i,w i ) Ker(µ) = {(w i, w i I} i ) i I} H. H i.e.,

35 33 ( ) We have that (w 0, w 0) µ R ker(µ), where R = {(w i, w i ) i I} H. If α Hom(W Θ (X), H) is such that (w i, w i ) Ker(α), for all i I, it follows that α R and then (w 0, w 0) µ R Ker(µ) Ker(α). Thus, (w 0, w 0) Ker(α). So, we have that H (( i I (w i = w i )) (w 0 = w 0)). 3.3 Operators over Classes of Algebras In this section, we give the denition of some operators whose properties will be very useful for our theory. But rstly we need to know what an operator over classes of algebras is. Denition Let Θ be a variety of algebras. A class of algebras X Θ is called an abstract class if, for every H X, we have that every isomorphic copy of H is contained in X. The family of all the abstract classes from Θ is denoted by Θ. Denition An operator over classes of algebras is a map U : Θ Θ. We say that an operator U over classes of algebras is closed if, for every class of algebras X, we have UUX = UX. A class of algebras X is called closed over the operator U if UX = X. An operator U is called monotone if X 1 X 2 UX 1 UX 2. An operator is called operator of extension if X UX, for every class of algebras X. An operator U is called operator of extension over the class of algebras X if X UX. Finally, if U 1, U 2,..., U n are operators over classes of algebras, we denote by {U 1, U 2,..., U n }(X) the minimal class of algebras containing the class X which is closed over all operators U 1, U 2,..., U n. The operator Θ Θ dened by X {U 1, U 2,..., U n }(X) will be denoted also by {U 1, U 2,..., U n }(X), in order to simplify the notation. Now, we will dene some important operators which are closely related with the geometrical equivalence of algebras, and after that we consider some of their properties. Denition Let Θ be a variety of algebras and X be a class of algebras in Θ. (i) H S(X) if H G, where G X.

36 34 (ii) H C(X) if H = i I G i, where G i X, for every i I. (iii) H L(X) if, for every subalgebra H 0 H nitely generated, we have H 0 X. Remark 3.2.1: If U 1, U 2 composition U 1 U 2 by U 1 U 2. are operators over classes of algebras, we will denote the Proposition Let Θ be a variety of algebras. The operators S,C, and L have the following properties: (i) S,C and L are monotone operators; (ii) S and C are operators of extension; (iii) L is an operator of extension over the classes of algebras which are closed under the operator S, that is, if X Θ is such that S(X) = X, then X L(X); (iv) SL(X) = L(X), for every X Θ ; (v) S,C, and L are closed operators; (vi) CS(X) SC(X), for every X Θ ; (vii) CL(X) LSC(X), for every X Θ ; (viii) {L,S,C} (X) = LSC(X), for every X Θ. Proof. (i) Let X 1, X 2 Θ such that X 1 X 2. If H S(X 1 ), we have that H G, where G X 1 X 2. Then H S(X 2 ) and S is monotone. If H C(X 1 ), we have that H = G i, where G i X 1 X 2, for every i I. Then i I H C(X 2 ) and we conclude that C is monotone. Now, if H L(X 1 ), for every nitely generated subalgebra H 0 H, we have that H 0 X 1 X 2. Then H L(X 2 ) and L is monotone. (ii) Let H X. Since H is a subalgebra of itself, we have that H S(X), and S is an operator of extension. Now, for any index j, we have that H = H I, where I = {j}. Thus H C(X), and C is an operator of extension.

37 35 (iii) Let X Θ such that S(X) = X. If H X and H 0 H is a nitely generated subalgebra, we have that H 0 S(X) = X. Hence H L(X), and X L(X). (iv) Let X Θ. Since S is an operator of extension, we have that L(X) SL(X). Now, if H SL(X), there exists G L(X) such that H G. If H 0 H is a nitely generated subalgebra, then H 0 H G. By the denition of the operator L we have H 0 X, and so H L(X). We conclude that SL(X) L(X) and SL(X) = L(X). (v) Let X Θ. By (ii), we have that S(X) SS(X). Now, if H SS(X), then there exists G S(X) such that H G. Since G S(X), there exists A X such that G A. So, we have H G A and H S(X). Thus, SS(X) S(X), and S is closed. Moreover, by (ii), we have that C(X) CC(X). Now, let H CC(X). So, H = H i, where H i C(X), for all i I. For every i I i I, there exists a set of indices J i such that H i = A j, where A j X for every j J i j J i, i I, and J i1 J i2 =, for i 1 i 2. So, we have H = ( A j ), and by i I j J i Proposition 2.1.7, it follows that H = A j C(X), where J = J i. So, as C(X) j J i I is an abstract class, we have that H C(X), and then CC(X) C(X). We conclude that C is closed. For the operator L, by (iv) we have SL(X) = L(X), that is, L(X) is closed under S. So, by (iii), it follows that L(X) LL(X). Now, if H LL(X) and H 0 H is nitely generated, we have that H 0 L(X). As H 0 is a nitely generated subalgebra of itself, it follows that H 0 X. Thus, H L(X). We conclude that LL(X) L(X) and hence L is closed. (vi) Let G CS(X). We have that G = i I G i, where G i S(X), for all i I. So, for every i I there exists A i X such that G i A i. If π i : G G i are projections and ν i : G i A i are inclusion mappings, we dene ϕ i : G A i by ϕ i (g) = ν i (π i (g)) = π i (g) (this last equality takes place because π i (g) G i A i, for any i I ). By the Remak Theorem we have a homomorphism ϕ : G i I A i dened by σ i (ϕ(g)) = ϕ i (g) = π i (g), for any i I, where σ i : j I A j A i are projections. Thus, Ker(ϕ) = i I Ker(ϕ i ) = {(g 1, g 2 ) G G π i (g 1 ) = π i (g 2 ), i I}, and Ker(ϕ) = {(g, g) g G} G G. So, ϕ is an embedding, and G SC(X).

38 36 (vii) Let X Θ. If G CL(X), we have that G = G i, where G i L(X), for every i I i I. Let G 0 G be a nitely generated subalgebra with generators g 1, g 2,..., g k. For every i I, let G i be the subalgebra of G i generated by π i (g 1 ), π i (g 2 ),..., π i (g k ), where π i : G G i are projections. Since G i L(X), we have that G i X, for any i I. If g G 0, we have that g is generated by the elements g 1, g 2,..., g k. So, for every i I, π i (g) is generated by the elements π i (g 1 ), π i (g 2 ),..., π i (g k ). Thus, π i (g) G i, for all i I and G 0 G i. So, G 0 SC(X). It follows that i I G LSC(X), and then CL(X) LSC(X). (viii) Let X Θ. By (v) and (iv), respectively, we have that LLSC(X) = LSC(X) and SLSC(X) = LSC(X). The class LSC(X) is closed under the operators L and S. Now, by (vii), it follows that CLSC(X) LSCSC(X). By (vi), we have LSCSC(X) LSSCC(X) = LSC(X). The last equality takes place because S and C are closed operators. Hence, CLSC(X) LSC(X). Since C is an operator of extension, we have LSC(X) CLSC(X). Thus CLSC(X) = LSC(X) and therefore LSC(X) is closed under the operator C. Now, note that X {L,S,C} (X) LSC(X) By (i), and LSC(X) LSC {L,S,C} (X) LSCLSC(X) LSC(X) {L,S,C} (X) LSC(X). We conclude that {L,S,C} (X) = LSC(X). We nish this section with some results that ilustrate a relation between the H-closure operators and the operators L, S and C. Proposition Let Θ be a variety of algebras and T W Θ (X) W Θ (X) a congruence. T is H-closed if and only if W Θ (X)/T SC(H).

39 37 Proof. ( ) Let T be H-closed. So, we have that T H = Ker(α) = T. α T H Denote T H = {α i i I}. Dene the mapping µ : W Θ (X)/T H I by π i (µ([w] T )) = α i (w), where π i : H I H are projections. We have that µ is well-dened, because if [w 1 ] T = [w 2 ] T, we have that (w 1, w 2 ) T = i I Ker(α i ), which means that α i (w 1 ) = α i (w 2 ), for all i I. Therefore, µ([w 1 ] T ) = µ([w 2 ] T ). Now, if µ([w 1 ] T ) = µ([w 2 ] T ), we have π i (µ([w 1 ] T )) = π i (µ([w 2 ] T )), for all i I, which means that α i (w 1 ) = α i (w 2 ), for all i I, and (w 1, w 2 ) Ker(α i ), for every i I. It follows that (w 1, w 2 ) i I Ker(α i ) = T and [w 1 ] T = [w 2 ] T. We conclude that µ is an embedding. Thus, W Θ (X)/T SC(H). ( ) Let W Θ (X)/T SC(H). Then we have an embedding µ : W Θ (X)/T H I. Let π i : H I H (i I) be projections and τ : W Θ (X) W Θ (X)/T be the natural epimorphism. Then, π i µτ Hom(W Θ (X), H). We claim that: T = i I Ker(π i µτ). Let (w 1, w 2 ) T. We have that [w 1 ] T = [w 2 ] T and hence µ(τ(w 1 )) = µ(τ(w 2 )), and then π i (µ(τ(w 1 ))) = π i (µ(τ(w 2 ))), for all i I. It follows that T i I Ker(π i µτ). Now, if (w 1, w 2 ) i I Ker(π i µτ), we have π i (µ(τ(w 1 ))) = π i (µ(τ(w 2 ))), for every i I, and we conclude that µ(τ(w 1 )) = µ(τ(w 2 )). As µ is an embedding, we have τ(w 1 ) = τ(w 2 ) and [w 1 ] T = [w 2 ] T. Thus (w 1, w 2 ) T, and the claim is proved. Finally, we have that {π i µτ i I} T H and Ker(π i µτ) T H. We conclude that T H Ker(π i µτ) = T. i I i I Since T T H, we have T H = T, and therefore T is H-closed. Corollary Let Θ be a variety of algebras and G, H Θ. Every G-closed congruence is an H-closed congruence if and only if G LSC(H). Proof. ( ) Let G 0 G be a nitely generated subalgebra. There exists an epimorphism τ : W Θ (X) G 0, where X is nite (because G 0 is nitely generated). If T = Ker(τ), then W Θ (X)/T = G 0. Since G 0 G C(G), we have that G 0 SC(G) and W Θ (X)/T = G 0 SC(G). Thus, T is G-closed. By the assumption, T is H-closed. So, it follows

40 38 that W Θ (X)/T SC(H) and hence G 0 = WΘ (X)/T SC(H). We conclude that G LSC(H). ( ) Let G LSC(H) and T W Θ (X) W Θ (X) be a G-closed congruence. We have by the previous proposition that W Θ (X)/T SC(G). So there exists an embedding µ : W Θ (X)/T G I (I is a set of indices). Let π i : G I G (i I) be projections. Since X is nite and W Θ (X)/T = τ(w Θ (X)) = τ( X ), where τ : W Θ (X) W Θ (X)/T is the natural epimorphism, we have by Corollary that im(π i (µ(w Θ (X)/T ))) = im(π i (µ(τ( X )))) = π i (µ(τ(x))), for all i I, and then im(π i µ) is a nitely generated subalgebra of G, for every i I. So, for every i I, im(π i µ) SC(H), because G LSC(H). Thus, there exist embeddings ν i : im(π i µ) H J i, where J i is a set of indices. Now, dene λ : W Θ (X)/T H J i by i I σ i (λ([w] T )) = ν i (π i (µ([w] T ))), where σ i : H J i H J i are projections. If λ([w 1 ] T ) = i I λ([w 2 ] T ), then and σ i (λ([w 1 ] T )) = σ i (λ([w 2 ] T )), for all i I, ν i (π i (µ([w 1 ] T ))) = ν i (π i (µ([w 2 ] T ))), for all i I. Since ν i is an embedding, for any i I, it follows that π i (µ([w 1 ] T )) = π i (µ([w 2 ] T )), for all i I, and µ([w 1 ] T ) = µ([w 2 ] T ). As µ is an embedding, we have that [w 1 ] T = [w 2 ] T. Therefore, λ is an embedding. We conclude that W Θ (X)/T SC(H), and hence T is H-closed. Corollary LSC(H) q V ar(h). Proof. Let H 0 LSC(H). So, we have that every H 0 -closed congruence T is H-closed. If H (( i I(w i = w i )) (w 0 = w 0)), we have that (w 0, w 0) {(w i, w i ) i I} H. Once {(w i, w i ) i I} H 0 is H 0 -closed, then it is H-closed. By Proposition 3.1.2, (w 0, w 0) {(w i, w i ) i I} H {(w i, w i ) i I} H 0, and the result follows.