Geometrical equivalence and action type geometrical equivalence of group representations

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(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 PósGraduação 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. - 2018. 84f.: il. Dissertação (mestrado) - Universidade Federal do Rio Grande do Norte, Centro de Ciências Exatas e da Terra, Programa de PósGraduação em Matemática Aplicada e Estatística. Natal, RN, 2018. 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 519.6. Elaborado por JOSENEIDE FERREIRA DANTAS - CRB-15/324.

(4) Dissertação de Mestrado sob o título. Geometrical Equivalence and Action Type Geome-. trical 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. 10. 1.1. Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. p. 10. 1.2. Thesis Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. p. 11. 2 Universal Algebra: Basic Denitions. p. 12. 2.1. Universal Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. p. 12. 2.2. Examples of Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . .. p. 24. 2.2.1. Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. p. 25. 2.2.2. Rings, Modules and Group Rings. p. 27. . . . . . . . . . . . . . . . . .. 3 Geometrical Equivalence of algebras. p. 29. 3.1. Galois Correspondence . . . . . . . . . . . . . . . . . . . . . . . . . . .. p. 29. 3.2. Quasi-identities and Quasi-varieties . . . . . . . . . . . . . . . . . . . .. p. 31. 3.3. Operators over Classes of Algebras. . . . . . . . . . . . . . . . . . . . .. p. 33. 3.4. Geometrical Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . .. p. 39. 4 Representation Theory and Geometrical Equivalence of Representations. p. 42. 4.1. Basic notions of Representation Theory . . . . . . . . . . . . . . . . . .. p. 42. 4.2. Galois Correspondence for Group Representations . . . . . . . . . . . .. p. 47. 4.3. Quasi-identities in. 4.4. Operators over Classes of Representations. . . . . . . . . . . . . . . . .. p. 52. 4.5. Geometrical Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . .. p. 57. REP − K. . . . . . . . . . . . . . . . . . . . . . . .. p. 50.

(11) 5 Action Type Geometrical Equivalence of Representations. p. 61. 5.1. Action Type Galois Correspondence . . . . . . . . . . . . . . . . . . . .. p. 61. 5.2. Action Type Quasi-identities . . . . . . . . . . . . . . . . . . . . . . . .. p. 63. 5.3. Operators. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. p. 67. 5.4. Action Type Geometrical Equivalence . . . . . . . . . . . . . . . . . . .. p. 75. 5.5. Action Type Geometrical Equivalence and Geometrical Equivalence of. V. and. Qr. Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. p. 79. 6 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 2.1.1. Let A be a non-empty set. If n is a non-negative integer, we dene:. (i) A0 = {∅}; (ii) An is the cartesian product of n copies of A, that is, an element a ∈ An has the form a = (a1 , a2 , ..., an ), where ai ∈ A, 1 ≤ i ≤ n. (iii) A mapping ω : An → 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. the (unique) element will be. ∅. in. A0 .. Hence,. ω ω. is completely determined by the image can be viewed as an element of. A,. ω(∅). of. whose name. constant.. Denition 2.1.2. 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 2.1.3. 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 bi ∈ B,. 1 ≤ i ≤ n, and every n-ary algebraic operation ω ∈ Ω, we have that ω(b1 , ..., bn ) ∈ B . If B is a subalgebra of A, we write B ≤ A.. Proposition 2.1.1. Let A be an algebra with signature Ω. If Bi ≤ A, then. T. Bi , 1 ≤ j ≤ n. Bi ≤ A,. i∈I. i ∈ I , where I is a set of indices.. Proof. Let bj ∈. T. and. ω∈Ω. be an. n-ary. algebraic operation. We have. i∈I. bj ∈ Bi , 1 ≤ j ≤ n, for every i ∈ I , and then ω(b1 , ..., bn ) ∈ Bi , for every i ∈ I . Hence T T ω(b1 , ..., bn ) ∈ Bi , and we conclude that Bi ≤ A.. that. i∈I. Denition 2.1.4. i∈I. Let A be an algebra with signature Ω, X ⊂ A and. BX = {B ≤ A | X ⊂ B}. The smallest algebra from BX is called the subalgebra generated by the set X . We denote this subalgebra by hXi.. Proposition 2.1.2. Let A be an algebra with signature Ω and X ⊂ A. Then we have that. hXi =. \. B.. B∈BX. Proof. By Proposition 2.1.1, we have that. T. B is a subalgebra of A. Now, since X ⊂ B , T for every B ∈ BX , we have that hXi ⊂ B, for every B ∈ BX . So, hXi ⊂ B. Moreover, B∈BX T T we have that B ⊂ hXi, because hXi is an element of BX . Therefore, hXi = B. B∈BX. B∈BX. B∈BX. In the following result, we describe the form of the elements of. Proposition 2.1.3. hXi.. Let A be an algebra with signature Ω and X ⊂ A. If.

(16) 14. M0 (X) = {cA |ω = c ∈ Ω, ρ(ω) = 0} ∪ X, Mk+1 (X) = {ω(a1 , ..., an )|ω ∈ Ω, ρ(ω) = n, ai ∈. k [. Mj (X), ∀1 ≤ i ≤ n},. j=0. then hXi = M (X), where M (X) =. ∞ S. Mj (X).. j=0. Proof. Firstly, we have to prove that M (X) is an algebra. Indeed, if a1 , ..., an ∈ M (X), there exist. j1 , ..., jn. it follows that. M (X) that. such that. ω(a1 , ..., an ) ∈ Mp+1 (X),. is an algebra. As. M0 (X) ⊂ hXi,. Mi (X) ⊂ hXi, where. ai ∈ Mji (X), ∀1 ≤ i ≤ n.. M (X). because. for every. contains. hXi. X,. So, if. ω∈Ω. p = M ax{ji |1 ≤ i ≤ n},. such that. we have that. ρ(ω) = n,. hXi ⊂ M (X).. is an algebra which contains. X.. and hence. Moreover, note. Now, suppose that. 1 ≤ i ≤ k . If b ∈ Mk+1 (X), we have that b = ω(a1 , ..., an ), k S ω ∈ Ω, ρ(ω) = n and ai ∈ Mj (X), ∀1 ≤ i ≤ n. Since Mj (X) ⊂ hXi, for for every. Mj (X) ⊂ hXi. hXi. is an algebra, thus we have. b = ω(a1 , ..., an ) ∈ hXi and then Mk+1 (X) ⊂ hXi. ∞ S M (X) = Mj (X) ⊂ hXi, and the proposition is proved.. It follows by induction that. every. 1 ≤ j ≤ k,. j=0 k S. it follows that. j=0 that. 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 2.1.5. Let A, B be two algebras with the same signature Ω. A mapping ϕ :. A → B is called a homomorphism if, for every a1 , ..., an ∈ A and ω ∈ Ω such that ρ(ω) = n, we have ϕ(ωA (a1 , ..., an )) = ωB (ϕ(a1 ), ..., ϕ(an )). A homomorphism if it is injective and. morphism.. isomorphism. is called. ϕ(cA ) = cB .. A∼ = B.. epimorphism. if it is a bijection. If. If there exists an isomorphism. and we denote it by then. ϕ:A→B. ϕ:A→B. Moreover, if. if it is surjective,. A = B,. then. , we say that. ϕ:A→B. A. is. ϕ. monomorphism is called. isomorphic. is a homomorphism and. endoto. B,. ω = cA ,. About composition of homomorphisms, we have the following result.. Proposition 2.1.4. 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 a1 , ..., an ∈ A and ω ∈ Ω such that ρ(ω) = n. It follows that (ψ ◦ ϕ)(ωA (a1 , ..., an )) = ψ(ϕ(ωA (a1 , ..., an ))) = ψ(ωB (ϕ(a1 ), ..., ϕ(an ))). Hence,. (ψ ◦ ϕ)(ωA (a1 , ..., an )) = ωC (ψ(ϕ(a1 )), ..., ψ(ϕ(an ))) = ωC ((ψ ◦ ϕ)(a1 ), ..., (ψ ◦ ϕ)(an )), and the result follows.. Denition 2.1.6. Let ϕ : A → B be a homomorphism. The kernel of ϕ is the set. Ker(ϕ) = {(a1 , a2 ) ∈ A2 |ϕ(a1 ) = ϕ(a2 )}. The image of ϕ is the set im(ϕ) = {ϕ(a) | a ∈ A}.. Remark 2.1.2:. b1 , ..., bn ∈ im(ϕ), for every. ω∈Ω. ϕ:A→B. If. there exist. such that. ϕ(ωA (a1 , ..., an )) ∈ im(ϕ).. is a homomorphism, then. a1 , ..., an ∈ A. ρ(ω) = n, Hence, if. such that. we have that. ϕ. im(ϕ) ≤ B ,. because for every. ϕ(ai ) = bi ,∀1 ≤ i ≤ n,. and then. ωB (b1 , ..., bn ) = ωB (ϕ(a1 ), ..., ϕ(an )) =. is a monomorphism, we have that. is an isomorphism. So, up to isomorphism, we can say that. A ≤ B.. ϕ : A → im(ϕ). Moreover,. Ker(ϕ) = {(a1 , a2 ) ∈ A2 |ϕ(a1 ) = ϕ(a2 )} = {(a1 , a2 ) ∈ A2 |a1 = a2 } = {(a, a)|a ∈ A}.. Proposition 2.1.5. Let ϕ : A → B be a homomorphism and X ⊂ A. Then ϕ(Mk (X)) =. Mk (ϕ(X)), for every k ∈ N.. Proof. Firstly, note that ϕ(cA ) = cB , for every ω = cA such that ρ(ω) = 0. So, ϕ(M0 (X)) = ϕ({cA |ω = c ∈ Ω, ρ(ω) = 0} ∪ X) ϕ(M0 (X)) = {cB |ω = c ∈ Ω, ρ(ω) = 0} ∪ ϕ(X) = M0 (ϕ(X)).. and then Now, suppose that we have that. ρ(ω) = n. and. ϕ(Mk (X)) = Mk (ϕ(X)),. for every. 1 ≤ i ≤ k.. j=0. b ∈ ϕ(Mk+1 (X)),. b = ϕ(a), where a ∈ Mk+1 (X), that is, a = ωA (a1 , ..., an ), with ω ∈ Ω, k k S S ai ∈ Mj (X), ∀1 ≤ i ≤ n. By the induction hypothesis, ϕ( Mj (X)) = j=0. k S. If. ϕ(Mj (X)) =. k S. Mj (ϕ(X)).. j=0 So,. j=0. b = ϕ(a) = ωB (ϕ(a1 ), ..., ϕ(an )) ∈ Mk+1 (ϕ(X)). and hence. ϕ(Mk+1 (X)) ⊂ Mk+1 (ϕ(X))..

(18) 16. a ∈ Mk+1 (ϕ(X)),. Moreover, if. we have that. a = ωB (b1 , ..., bn ),. where. bi ∈. k S. Mj (ϕ(X)),. j=0 for each. 1 ≤ i ≤ n. By the induction hypothesis, it follows that. k S. Mj (ϕ(X)) =. j=0 So, for each. 1 ≤ i ≤ n, there exists ai ∈ Mp (X) such that 1 ≤ p ≤ k. k S. ϕ(Mj (X)).. j=0 and. bi = ϕ(ai ). Thus,. a = ωB (ϕ(a1 ), ..., ϕ(an )) and. a = ϕ(ωA (a1 , ..., an )) ∈ ϕ(Mk+1 (X)). We conclude that. Mk (ϕ(X)),. Therefore,. Mk+1 (ϕ(X)) ⊂ ϕ(Mk+1 (X)).. ϕ(Mk+1 (X)) = Mk+1 (ϕ(X)), and it follows by induction that ϕ(Mk (X)) =. for every. Corollary 2.1.1. k ∈ N.. Let ϕ : A → B be a homomorphism. If A = hXi , then im(ϕ) = hϕ(X)i .. Proof. It follows immediately from the previous proposition: hϕ(X)i =. ∞ [. Mj (ϕ(X)) =. j=0. Remark 2.1.3: momorphisms. If. ∞ [. ϕ(Mj (X)) = ϕ(. j=0. A, B. ϕ:A→B. ∞ [. Mj (X)) = ϕ(hXi) = im(ϕ).. j=0. are two algebras with the same signature, the set of all howill be denoted by. Hom(A, B).. Q {Ai |i ∈ I} a family of algebras in signature Ω. If we consider X = Ai , the i∈I S elements of X are functions a : I → Ai , where a(i) ∈ Ai , ∀i ∈ I. If we denote a(i) = ai , Now, let. we can present each element of see, now, that. X. X. I∈i as a sequence. (ai )i∈I ,. is also an algebra with signature. Denition 2.1.7. where. ai ∈ Ai , ∀i ∈ I.. We will. Ω.. Let {Ai |i ∈ I} be a family of algebras with signature Ω. The set. i∈I. with the operations ω ∈ Ω dened by. (ω Q Ai (a1 , ..., an ))i = ωAi ((a1 )i , ..., (an )i ), i∈I. is called the Cartesian product of the family {Ai |i ∈ I}.. Remark 2.1.4: If Ai = A,∀i ∈ I, we denote. Q. Q i∈I. Ai = AI .. Ai.

(19) 17. Proposition 2.1.6. Let {Ai |i ∈ I} be a family of algebras with signature Ω. If j ∈ I is. xed, then the mapping. πj :. Y. Ai → Aj ,. i∈I. such that πj (a) = aj is a homomorphism.. Proof. Let a1 , ..., an ∈. Q. Ai. and. ω∈Ω. such that. ρ(ω) = n.. We have that. i∈I. πj ((ω(a1 , ..., an ))i∈I ) = (ω(a1 , ..., an ))j = ω((a1 )j , ..., (an )j ) = ω(πj (a1 ), ..., πj (an )), and. πj. is a homomorphism.. The mappings. πj :. Q. Ai → Aj. are called. projections. i∈I an important result: it is called Remak Theorem.. Theorem 2.1.1. of. Q. Ai. on. Aj .. Now, we have. i∈I. Let A be an algebra, {Ai |i ∈ I} be a family of algebras (A and Ai have. the same signature Ω) and ϕi : A → Ai be a family of homomorphisms. Then the mapping Q T ϕ:A→ Ai dened by (ϕ(a))i = ϕi (a) is a homomorphism and Ker(ϕ) = Ker(ϕi ). i∈I. i∈I. Proof. Let a1 , ..., an ∈ A and ω ∈ Ω such that ρ(ω) = n. Note that for every i ∈ I, (ϕ(ωA (a1 , ..., an )))i = ϕi (ωA (a1 , ..., an )) = ωAi (ϕi (a1 ), ..., ϕi (an )). On the other hand, for every. i ∈ I,. we have. (ω Q Ai (ϕ(a1 ), ..., ϕ(an )))i = ωAi ((ϕ(a1 ))i , ..., (ϕ(an ))i ) = ωAi (ϕi (a1 ), ..., ϕi (an )), i∈I. and then. ϕ(ωA (a1 , ..., an )) = ω(ϕ(a1 ), ..., ϕ(an )). We conclude that ϕ is a homomorphism.. Now,. Ker(ϕ) = {(a1 , a2 ) ∈ A2 |ϕ(a1 ) = ϕ(a2 )} = {(a1 , a2 ) ∈ A2 |(ϕ(a1 ))i = (ϕ(a2 ))i , ∀i ∈ I}, and then. Ker(ϕ) = {(a1 , a2 ) ∈ A2 |ϕi (a1 ) = ϕi (a2 ), ∀i ∈ I} \ = {(a1 , a2 ) ∈ A2 |(a1 , a2 ) ∈ Ker(ϕi ), ∀i ∈ I} = Ker(ϕi ). i∈I The theorem is proved.. Now, we have a result that connects Cartesian product and isomorphisms of algebras. For the proof see Theorem 4 in [9]..

(20) 18. Proposition 2.1.7. Let {Xj |j ∈ A} be a family of algebras with signature Ω and {Ai |i ∈ S I} be a family of pairwise disjoint index sets. If A = Ai , then i∈I. Y. Xj ∼ =. j∈A. Denition 2.1.8. Y Y ( Xj ). i∈I j∈Ai. Let A be an algebra with signature Ω. A binary relation T ⊂ A2 is. called a congruence if it is an equivalence relation on A and, for every ω ∈ Ω (ρ(ω) = n) and (ai , bi ) ∈ T , for every 1 ≤ i ≤ n, we have that. (ω(a1 , ..., an ), ω(b1 , ..., bn )) ∈ T.. Proposition 2.1.8. If Ti ⊂ A2 (i ∈ I) is a family of congruences, then T =. Proof. Firstly, we have to prove that T i ∈ I,. then. we have that. (a, a) ∈ T ,. is an equivalence relation. Note that. which means that. (a, b) ∈ Ti , ∀i ∈ I,. and so. an equivalence relation). It follows that. (a, b), (b, c) ∈ T , and then. then. (a, c) ∈ T.. ω ∈ Ω (ρ(ω) = n) i ∈ I,. Ti is a. i∈I. congruence.. every. T. (a, b), (b, c) ∈ Ti ,. Thus,. and. T. T. is reexive. Furthermore, if. (b, a) ∈ Ti. (b, a) ∈ T. for every. (a, a) ∈ Ti for. (this is true because each. and then. i∈I. (a, b) ∈ T ,. and so. T. Ti. is. is symmetric. Finally, if. (a, c) ∈ Ti. for every. i ∈ I,. is transitive, and so an equivalence relation. Now, for every. (ak , bk ) ∈ T ,. for every. 1 ≤ k ≤ n,. we have that. (ak , bk ) ∈ Ti ,. for all. and then. (ω(a1 , ..., an ), ω(b1 , ..., bn )) ∈ Ti , ∀i ∈ I Thus. (ω(a1 , ..., an ), ω(b1 , ..., bn )) ∈ T ,. Proposition 2.1.9. and we conclude that. T. is a congruence.. 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. ϕ(a) = ϕ(b), We conclude that. Ker(ϕ). (a, b) ∈ Ker(ϕ),. and then. then. ϕ(b) = ϕ(a).. is symmetric. Finally, if. ϕ(a) = ϕ(b) = ϕ(c),. Thus,. (b, a) ∈ Ker(ϕ).. (a, b), (b, c) ∈ Ker(ϕ),. which means that. (a, c) ∈ Ker(ϕ).. we have that.

(21) 19. Therefore,. Ker(ϕ). Now, for every. is transitive and we conclude that. ω ∈ Ω (ρ(ω) = n). and. Ker(ϕ). (ai , bi ) ∈ Ker(ϕ),. is an equivalence relation.. we have. ϕ(ωA (a1 , ..., an )) = ωB (ϕ(a1 ), ..., ϕ(an )), and also. ωB (ϕ(a1 ), ..., ϕ(an )) = ωB (ϕ(b1 ), ..., ϕ(bn )). It follows that that. Ker(ϕ). ϕ(ωA (a1 , ..., an )) = ωB (ϕ(b1 ), ..., ϕ(bn )) = ϕ(ωA (b1 , ..., bn )), and we conclude. is a congruence.. Denition 2.1.9. Let A be an algebra with signature Ω and T ⊂ A2 be a congruence.The. quotient algebra is the set A/T with operations. ωA/T ([a1 ]T , ..., [an ]T ) = [ωA (a1 , ..., an )]T , where ω ∈ Ω (ρ(ω) = n) and ai ∈ A, for every 1 ≤ i ≤ n. For every x ∈ A, [x]T is an equivalence class of x by T . This denition is correct, because if for every. 1 ≤ i ≤ n.. Therefore. [ai ]T = [bi ]T ,. for every. 1 ≤ i ≤ n,. (ωA (a1 , ..., an ), ωA (b1 , ..., bn )) ∈ T. and. then. (ai , bi ) ∈ T. [ωA (a1 , ..., an )]T =. [ωA (b1 , ..., bn )]T .. Proposition 2.1.10. 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 ai ∈ A for every 1 ≤ i ≤ n. By the denition of τ , we have. τ (ωA (a1 , ..., an )) = [ωA (a1 , ..., an )]T .. Since. [ωA (a1 , ..., an )]T = ωA/T ([a1 ]T , ..., [an ]T ),. it. follows that. τ (ωA (a1 , ..., an )) = ωA/T (τ (a1 ), ..., τ (an )), and so. τ. is a homomorphism. Moreover, if. [a]T = τ (a),. which means that. τ. b ∈ A/T ,. there exists. a∈A. such that. b=. is an epimorphism. Finally, note that. Ker(τ ) = {(a1 , a2 ) ∈ A2 |τ (a1 ) = τ (a2 )} = {(a1 , a2 ) ∈ A2 |[a1 ]T = [a2 ]T }. Since. {(a1 , a2 ) ∈ A2 |[a1 ]T = [a2 ]T } = {(a1 , a2 ) ∈ A2 |(a1 , a2 ) ∈ 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 2.1.2. 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. if and only if injective. If. [a1 ]Ker(ϕ) = [a2 ]Ker(ϕ). if and only if. (a1 , a2 ) ∈ Ker(ϕ),. ψ([a1 ]Ker(ϕ) ) = ϕ(a1 ) = ϕ(a2 ) = ψ([a2 ]Ker(ϕ) ).. b ∈ im(ϕ),. is surjective. Now, if. there exists. a∈A. ω ∈ Ω (ρ(ω) = n). such that. and. ai ∈ A,. Then. ψ. which takes place. is well-dened and. b = ϕ(a) = ψ([a]Ker(ϕ) ). for every. 1 ≤ i ≤ n,. Therefore. ψ. we have that. ψ(ωA/Ker(ϕ) ([a1 ]Ker(ϕ) , ..., [an ]Ker(ϕ) )) = ψ([ωA (a1 , ..., an )]Ker(ϕ) ) = ϕ(ωA (a1 , ..., an )). Since. ϕ. is a homomorphism, we have. ϕ(ωA (a1 , ..., an )) = ωB (ϕ(a1 ), ..., ϕ(an )) = ωB (ψ([a1 ]Ker(ϕ) ), ..., ψ([an ]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:. M0∗ (X) = {ω | ω ∈ Ω, ρ(ω) = 0} ∪ X,. ∗ Mk+1 (X). = {ωb1 b2 ...bn | ω ∈ Ω, ρ(ω) = n > 0, b1 , b2 , ..., bn ∈. k [. Mj∗ (X), ∃ 1 ≤ i0 ≤ n; bi0 ∈ Mk∗ (X)}. j=0 and. M ∗ (X) =. ∞ [. Mj∗ (X),. j=0 for an arbitrary set. X. and a signature. correspondent constant in. ∗. M (X). by. Ω.. For each. cM ∗ (X) ,. ω∈Ω. such that. and. ωM ∗ (X) (b1 , b2 , ..., bn ) = ωb1 b2 ...bn ,. ρ(ω) = 0. dene the.

(23) 21. for. ω ∈ Ω (ρ(ω) = n), b1 , b2 , ..., bn ∈ M ∗ (X),. signature. Ω.. It is called the. we have that. M ∗ (X). is an algebra with. algebra of terms on the alphabet X in the signature Ω.. ω1 (b1 , b2 , ..., bn ) = ω2 (m1 , m2 , ..., mk ),. this denition, it is clear that if. then. With. ω1 = ω2 ,. ρ(ω1 ) = n = k = ρ(ω2 ) and bi = mi , ∀ 1 ≤ i ≤ n. The following proposition gives us some interesting properties of. Proposition 2.1.11. M ∗ (X).. Let X be an arbitrary set and Ω a signature. The algebra M ∗ (X). has the following properties: (i) If i 6= j, then Mi∗ (X) ∩ Mj∗ (X) = ∅; (ii) M ∗ (X) = hXi .. Proof. (i) Firstly, note that. a∈. M0∗ (X). ∩. ρ(ω) = n > 0.. X ∩ Mi∗ (X) = ∅,. Mi∗0 (X). Since. a∈. On the other hand,. a∈. this contradiction, we conclude that suppose that. for. i > 0.. Now, suppose that. Mi∗0 (X), we have that. Mi∗ (X) ∩ Mk∗ (X) = ∅,. a = ω(b1 , b2 , ..., bn ),. M0∗ (X), and it follows that. M0∗ (X) ∩ Mi∗ (X) = ∅, for every. i0 > 0. 0 < i ≤ k.. and. where. ρ(ω) = 0.. With. i > 0.. Now,. whenever. If. ∗ ω1 (b1 , b2 , ..., bn1 ) ∈ Mk+1 (X),. ω2 (t1 , t2 , ..., tn2 ) ∈ Mi∗ (X), 0 < i ≤ k, and. ω1 (b1 , b2 , ..., bn1 ) = ω2 (t1 , t2 , ..., tn2 ), it follows that of. ∗ Mk+1 (X),. ω1 = ω2 , n1 = n2 ,. there exists. bp = tp ,. and. 1 ≤ j ≤ n. such that. ∗ by the denition of Mi (X), it follows that the hypothesis, we have that. for all. bj 6= tj ,. 1 ≤ p ≤ n.. bj ∈ Mk∗ (X).. tj ∈. By the denition. On the other hand,. Mj∗0 (X), where. j0 < i ≤ k.. By. it is a contradiction. The result follows by. induction. (ii) Since if. B. X ⊂ M ∗ (X). and. M ∗ (X). is an algebra, we have that. is an algebra which contains. Ω, ρ(ω) = 0} ⊂ B , Mi∗ (X) ⊂ B ,. because. for every. B. X,. that is,. B ∈ BX ,. it follows that. ∗ is an algebra. So, M0 (X). 1 ≤ i ≤ k.. If. hXi ⊂ M ∗ (X).. {ω | ω ∈. ⊂ B. Now, suppose k S m1 , m2 , ..., mn ∈ Mj∗ (X), we have j=0. Now,. that that.

(24) 22. ω(m1 , m2 , ..., mn ) ∈ B, for every ω ∈ Ω such that ρ(ω) = n, because B is an algebra k S ∗ (X) ⊂ B , and by induction we conclude that and Mj∗ (X) ⊂ B. Thus, Mk+1 j=0 T M ∗ (X) ⊂ B . As B ∈ BX was chosen abitrarily, it follows that M ∗ (X) ⊂ B= hXi.. Therefore,. Proposition 2.1.12. B∈BX. M ∗ (X) = hXi.. 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) = cA , ∀c ∈ Ω, ρ(ω) = 0. If. ϕ(t). is dened for. t∈. k S. Mi∗ (X),. we dene. ϕ. in. ∗ Mk+1 (X). by. i=0. ϕ(ωt1 t2 ...tn ) = ωA (ϕ(t1 ), ..., ϕ(tn )), ω ∈ Ω, ρ(ω) = n. Now, we have to check that. ϕ. is a homomorphism. Indeed, if. ω∈Ω. is an operation such. ρ(ω) = n and t1 , ..., tn ∈ M ∗ (X), we have that there exist k ∈ N and i0 ∈ {1, 2, ..., n} k S such that t1 , ..., tn ∈ Mi∗ (X) and ti0 ∈ Mk∗ (X). So,. that. i=0. ϕ(ωM ∗ (X) (t1 ...tn )) = ϕ(ωt1 ...tn ) = ωA (ϕ(t1 ), ..., ϕ(tn )). 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) = cA = ϕ(c), ∀c ∈ Ω, ρ(ω) = 0, which means that. ψ|M0∗ (X) = ϕ|M0∗ (X) .. Now, suppose that. ψ| Sk. Mi∗ (X). i=0. t = ωt1 ...tn ∈ {1, 2, ..., n}. ∗ Mk+1 (X), where. such that. ρ(ω) = n, t1 , ..., tn ∈. ti0 ∈ Mk∗ (X).. k S. = ϕ| Sk. Mi∗ (X). i=0. Mi∗ (X) and there exists i=0. So, we obtain. t = ωt1 ...tn = ωM ∗ (X) (t1 , ..., tn ),. . Let. i0 ∈.

(25) 23. ψ(t) = ψ(ωM ∗(X) (t1 , ..., tn )) = ωA (ψ(t1 ), ..., ψ(tn )), and nally. ψ(t) = ωA (ψ(t1 ), ..., ψ(tn )) = ωA (ϕ(t1 ), ..., ϕ(tn )) = ϕ(ωM ∗(X) (t1 , ..., tn )) = ϕ(t). Hence, by induction we conclude that. Proposition 2.1.13. ψ = ϕ.. 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 ϕ(x).. is an epimorphism, for every. So, dening. f : X → A. by. x∈X. f (x) = ax ,. proposition a (unique) homomorphism. there exists. for any. χ : M ∗ (X) → A. ax ∈ A. x ∈ X,. such that. ψ(ax ) =. we have by the previous. such that. χ(x) = f (x), ∀ x ∈ X .. Now, note that. ψ(χ(x)) = ψ(f (x)) = ψ(ax ) = 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 2.1.10. Let X = {x1 , x2 , ..., xn } and f (x1 , ..., xn ), g(x1 , ..., xn ) ∈ M ∗ (X). We. say that an algebra A fullls the identity. f (x1 , ..., xn ) = g(x1 , ..., xn ), if for every ϕ ∈ Hom(M ∗ (X), A) we have. ϕ(f (x1 , ..., xn )) = ϕ(g(x1 , ..., xn )).. Denition 2.1.11. 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 2.1.12. 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 = WΘ (X).. Theorem 2.1.3. 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 WΘ (X) such that |X| = α.. Remark 2.1.6: In Theorem 2.1.3, WΘ (X) = M ∗ (X)/JΘ (X), where JΘ (X) is the set of all identities prove that. f = g (f, g ∈ M ∗ (X)). JΘ (X). which are fullled in every algebra of. is a congruence in. Proposition 2.1.14. Θ.. It is easy to. M ∗ (X).. Let Θ be a variety of algebras with signature Ω, W ∈ Θ and X ⊂ W .. (i) WΘ (X) = hXi; (ii) WΘ (X) ∼ = WΘ (Y ) if |X| = |Y |.. Proposition 2.1.15. Let Θ be a variety of algebras with signature Ω. For every A ∈ Θ, there exists WΘ (X) such that A ∼ = WΘ (X)/T , where T is a congruence in WΘ (X). The following Proposition is similar to Proposition 2.1.13:. Proposition 2.1.16. Let Θ be a variety of algebras with signature Ω, A, B ∈ Θ , WΘ (X). a free algebra in Θ, ϕ : WΘ (X) → B a homomorphism and ψ : A → B an epimorphism. There exists a homomorphism χ : WΘ (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) 25. 2.2.1. Groups. Denition 2.2.1. 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 3 a 7→ a−1 ∈ G, (iii) A constant unit 1G ∈ G, which fulll the following identities: (a) For every a, b, c ∈ G, a(bc) = (ab)c. (b) For every a ∈ G, a1G = 1G a = a. (c) For every a ∈ G, aa−1 = a−1 a = 1G .. A group Let. G a. order of. Let. X. G. is called. abelian. be a group. The. if. order. will be denoted by. ab = ba, of. a∈G. for every. a, b ∈ G.. is the smallest. n ∈ N∗. an = 1G .. The. |a|.. be an arbitrary set. We can consider words over. Denition 2.2.2. such that. X. by the following way:. Words over X are sequences. w = a1 a2 ...an , where ai ∈ X, ∀ i ∈ {1, 2, ...n}. If n = 0, we have the empty word, which will be denoted by ε.. ˜ ={a X ˜ | a ∈ X} (where a ˜ is just a symbol obtained ˜ = ∅. Put X ∗ = X ∪ X ˜ and, for b ∈ X ∗ , dene X ∩X ( ˜, if b = a ∈ X, ˜b = a ˜ a, if b = a ˜ ∈ X.. If we denote is clear that. We say that an expression. ai ∈ X ∗ ,. for every. w. is a. i ∈ {1, 2, ...n}.. group word. A group word. if it has the form. from. a. and. w = a1 a2 ...an ,. w = a1 a2 ...an (ai ∈ X ∗ ). is called. ∼),. it. where. reduced.

(28) 26. if. ai 6= a ˜i+1 ,. that is, if it does not contain. that the empty word by. ε. a subword. a˜ a,. where. a ∈ X ∗.. We will assume. X. will be denoted. is reduced. The set of all reduced words from. F (X). If. b ∈ X∗. and. w = a1 a2 ...an ∈ F (X),. we dene. w◦ε=w. and.   b, if n = 0,      ˜,   ab, if n = 1 and b 6= a w◦b= ε, if n = 1 and b = a ˜,     a1 a2 ...an−1 , if n > 1 and b = a ñ ,     a1 a2 ...an b, if n > 1 and b 6= a ñ . The multiplication. w2 = b1 b2 ...bn. in. b◦w. is dened analogously. Moreover, for elements. F (X), w1 ◦ w2. w1 = a1 a2 ...an. and. can be dened recursively by. w1 ◦ w2 = [(a1 a2 ...an ) ◦ b1 ] ◦ (b2 ...bn ). So, we can consider the following algebraic operations over. (i). (.)−1 : F (X) → F (X). is such that, for. w−1 =. (ii). · : F (X) × F (X) → F (X).     ε,. F (X):. w = a1 a2 ...an ∈ F (X), if. n = 0,. a ˜, if n = 1,    a ñ ...˜ a2 a ˜1 , if n > 1.. is such that. w1 · w2 = w1 ◦ w2 .. With these two operations, it is easy to verify that. F (X). becomes a group with unit. 1G = ε.. Denition 2.2.3. 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 2.1.12. 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 4.2.3..

(29) 27. Theorem 2.2.1. Let G be a group. If H E G, A E G and H ≤ A, then. (G/H)/(A/H) ∼ = G/A.. 2.2.2. Rings, Modules and Group Rings. Denition 2.2.4. 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 6= 0. has an inverse. a−1 ∈ R. with respect. to ·.. Denition 2.2.5. 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) 1R · m = m. If the ring. R. is a eld, then. Denition 2.2.6 that:. R-module M. is called a. vector space.. A (left) R-module M is called free if there exists a set E ⊆ M such.

(30) 28. (i) M = {. Pn. i=1 ri xi |. n ∈ N, ri ∈ R, xi ∈ E, ∀i ∈ {1, 2, ..., n}} (M = hEi);. (ii) If ri ∈ R , xi ∈ E (i ∈ {1, 2, ..., n}) and. Pn. i=1 ri xi. = 0, then. r1 = r2 = ... = rn = 0R . The set E is called the basis of R-module M .. Denition 2.2.7. Let G be a group and R a ring with unity. The group ring RG is the. set of all sums. X. α=. ag g,. g∈G. where ag ∈ R, for all g ∈ G and the set {ag 6= 0} is nite. Let. α=. P. g∈G. ag g. and. β=. P. g∈G bg g be elements of. α+β =. X. RG.. We dene. (ag + bg )g,. g∈G and, if. β=. P. h∈G bh h,. X. X X bh h) = ag g)( αβ = ( g∈G. g∈G,h∈G. h∈G. To simplify the notation above, we put. αβ =. Cu =. ag bh gh.. P. X. u∈G. Cu u,. where. ag b h .. gh=u. It is not dicult to see that, with these operations,. RG. becomes a ring with unity. = 1R e (e is the unity element of the group G). The mapping ν : G → RG dened by P ν(g) = h∈G ah h, where ag = 0R , if h 6= g and ag = 1R , 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 3.1.1. 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. Denition 3.1.2. T ⊂ B. Elements cl(S) = G(F (S)) and cl(T ) = F (G(T )) are called the Galois. closures of S and T , respectively.. Denition 3.1.3 cl(T ) = T .. Elements S ⊂ A and T ⊂ B are called Galois closed if cl(S) = S and.

(32) 30. Let. X. Θ. be a variety of algebras and a set. X. be given. In this chapter, we assume that. is always nite.. Denition 3.1.4. Let WΘ (X) be the free algebra generated by the set X . Equations over. WΘ (X) are elements of WΘ (X)×WΘ (X), i.e., pairs of thef orm (w1 , w2 ), where w1 , w2 ∈ WΘ (X). We denote (w1 , w2 ) by w1 = w2 .. Denition 3.1.5. The solution of equation w1 = w2 in H ∈ Θ is a point µ of the space. Hom(WΘ (X), H) such that µ(w1 ) = µ(w2 ), i.e., (w1 , w2 ) ∈ Ker(µ). We can dene maps on by the same symbol . 0. WΘ (X) × WΘ (X). . For a set. and on. Hom(WΘ (X), H),. T ⊂ WΘ (X) × WΘ (X),. which we denote. we put:. TH0 = {µ ∈ Hom(WΘ (X), H) | T ⊂ Ker(µ)}. It is called an algebraic set. Now let a set. A ⊂ Hom(WΘ (X), H). be given. We can. consider the set:. A0H =. \. Ker(µ).. µ∈A This set is a congruence on. WΘ (X).. The next proposition describes some properties of. 0 maps .. Proposition 3.1.1. Let H ∈ Θ. For S, T ⊂ WΘ (X) × WΘ (X) and A, B ⊂ Hom(WΘ (X), H),. we have the following properties: 0 (i) S ⊂ T ⇒ SH ⊃ TH0 ; 0 (ii) A ⊂ B ⇒ A0H ⊃ BH ;. (iii) T ⊂ TH00 ; (iv) A ⊂ A00H .. Proof. (i) If. 0 µ ∈ TH0 , then T ⊂ Ker(µ). Since S ⊂ T , we have that S ⊂ Ker(µ). So, µ ∈ SH. and then. 0 TH0 ⊂ SH ..

(33) 31. (ii) Since. (iii) If. A ⊂ B,. µ ∈ TH0 ,. T. µ∈B. Ker(µ) ⊂. T ⊂ Ker(µ). we have that. µ0 ∈ A,. (iv) For every. A⊂. it follows that. we have that. T. T. and then. µ∈A. µ∈A. Ker(µ). T ⊂. T. and so. 0 µ∈TH. 0 BH ⊂ A0H .. Ker(µ) = TH00 .. Ker(µ) ⊂ Ker(µ0 ).. Thus,. µ0 ∈ A00H. and. A00H .. Proposition 3.1.1 says that the maps pondence between the set of equations. µ ∈ Hom(WΘ (X), H). of equations. 0. dened above provide us a Galois corres-. T ⊂ WΘ (X) × WΘ (X). and the set of points. So, by Denition 3.1.2 we can speak about the. T ⊂ WΘ (X) × WΘ (X),. H -closure. for a set. which can be dened by. TH00 = (TH0 )0H =. \. Ker(µ),. 0 µ∈TH. and about the. H -closure. for a set of points. A ⊂ Hom(WΘ (X), H):. A00H = (A0H )0H = {µ ∈ Hom(WΘ (X), H) | A0H ⊂ Ker(µ)}. According to Denition 3.1.3, a set of equations if. TH00 = T ,. and a set of points. T ⊂ WΘ (X) × WΘ (X). A ⊂ Hom(WΘ (X), H). is. H -closed. if. is. H -closed. A00H = A.. The. following result takes place.. Proposition 3.1.2. Let H ∈ Θ. For every set of equations T ⊂ WΘ (X) × WΘ (X), TH00. is the smallest H -closed congruence containing the set T .. Proof. We have that such that. 3.2. T ⊂ R,. TH00. is an. H -closed congruence. Moreover, if R is an H -closed congruence. we have that. 0 RH ⊂ TH0. and then. 00 TH00 ⊂ RH = R.. Quasi-identities and Quasi-varieties. Denition 3.2.1. Let Θ be a variety of algebras. A quasi-identity is a logic formula. which has the form.

(34) 32. (. n ^. (wi0 = wi00 )) ⇒ (w00 = w000 ),. i=1. where. wi0 , wi00. ∈ WΘ (X), 0 ≤ i ≤ n and |X| = m, m, n ∈ N.. Denition 3.2.2. Let Θ be a variety of algebras and H ∈ Θ. We say that H fullls. a quasi-identity. (. n ^. (wi0 = wi00 )) ⇒ (w00 = w000 ). i=1. if for all α ∈ Hom(WΘ (X), H) such that α(wi0 ) = α(wi00 ), 1 ≤ i ≤ n, we have that V α(w00 ) = α(w000 ). We write H (( ni=1 (wi0 = wi00 )) ⇒ (w00 = w000 )). We can generalize the denition of a quasi-identity above and consider innite quasiidentities, i.e., the logic formula which have the form. (. ^ (wi0 = wi00 )) ⇒ (w00 = w000 ), i∈I. where. I. is a set of indices. In this case,. 0 such that α(wi ). α ∈ Hom(WΘ (X), H) Now, let. X. V H (( i∈I (wi0 = wi00 )) ⇒ (w00 = w000 )) =. α(wi00 ), for any Θ.. be a class of algebras in a variety. i∈. if for every. I , we have α(w00 ). We denote by. q − id(X). all quasi-identities which are fullled on all algebras of this class. Moreover, if of quasi-identities in a variety of algebras algebras in the variety by the class denote by and. Θ. X ⊂ Θ by. we denote by. that fulll all quasi-identities of. is dened by the set of algebras. q − V ar(X).. q − V ar(X). Θ,. If. X = {H} (H ∈ Θ),. q − V ar(H).. Q.. q − V ar(Q). the set of. Q. is a set. the set of all. The quasi-variety generated. q − V ar(q − id(X)),. we denote. = α(w000 ).. q − id(X). by. which we. q − id(H). In the following proposition, we will see an interesting. relation between quasi-identities and the maps . 0. .. Proposition 3.2.1 ^ H (( (wi0 = wi00 )) ⇒ (w00 = w000 )), i∈I. where. {wi0 , wi00. | i ∈ I} ∪. {w00 , w000 }. ∈ WΘ (X) if and only if. (w00 , w000 ) ∈ {(wi0 , wi00 ) | i ∈ I}00H .. Proof. (⇒). For every. µ0 ∈ Hom(WΘ (X), H). such that. (wi0 , wi00 ) ∈ Ker(µ0 ),. µ0 ∈ {(wi0 , wi00 ) | i ∈ I}0H , we have that (w00 , w000 ) ∈ Ker(µ0 ), T 0 00 00 µ∈{(w0 ,w00 ) | i∈I}0 Ker(µ) = {(wi , wi ) | i ∈ I}H . i. i. H. and then. i.e.,. (w00 , w000 ) ∈.

(35) 33. (⇐). We have that. α ∈ Hom(WΘ (X), H) α∈R. (w00 , w000 ) ∈. is such that. 3.3. µ∈R ker(µ), (wi0 , wi00 ) ∈. where. Ker(α),. (w00 , w000 ) ∈ µ∈R Ker(µ) ⊂ Ker(α). V H (( i∈I (wi0 = wi00 )) ⇒ (w00 = w000 )). T. and then. have that. T. R = {(wi0 , wi00 ) | i ∈ I}0H . for all. Thus,. i ∈ I,. If. it follows that. (w00 , w000 ) ∈ Ker(α).. So, we. 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 3.3.1. 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 iΘh.. Denition 3.3.2. An operator over classes of algebras is a map. U :iΘh→iΘh. We say that an operator algebras. U. if. over classes of algebras is. X, we have UUX = UX. A class of algebras X. UX = X.. An operator. operator is called operator. U. U. is called. operator of extension. is called. Finally, if. U. monotone. if. X ⊂ UX,. if. closed. is called. if, for every class of. closed over the operator. X1 ⊂ X2 ⇒ UX1 ⊂ UX2 .. An. X.. An. for every class of algebras. operator of extension over the class of algebras X. U1 , U2 , ..., Un. if. X ⊂ UX.. are operators over classes of algebras, we denote by. {U1 , U2 , ..., Un }(X) the minimal class of algebras containing the class. U1 , U2 , ..., Un . also by. The operator. {U1 , U2 , ..., Un }(X),. iΘh→iΘh. dened by. X. which is closed over all operators. X 7→ {U1 , U2 , ..., Un }(X). will be denoted. 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 3.3.3. Let Θ be a variety of algebras and X be a class of algebras in iΘh.. (i) H ∈ S(X) if H ≤ G, where G ∈ X..

(36) 34. (ii) H ∈ C(X) if H =. Q. i∈I. Gi , where Gi ∈ X, for every i ∈ I .. (iii) H ∈ L(X) if, for every subalgebra H0 ≤ H nitely generated, we have H0 ∈ X.. Remark 3.2.1: If U1 , U2 composition. U1 ◦ U 2. Proposition 3.3.1. by. are operators over classes of algebras, we will denote the. U1 U2 .. 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 ∈ iΘh is such that S(X) = X, then X ⊂ L(X); (iv) SL(X) = L(X), for every X ∈ iΘh; (v) S ,C , and L are closed operators; (vi) CS(X) ⊂ SC(X), for every X ∈ iΘh; (vii) CL(X) ⊂ LSC(X), for every X ∈ iΘh; (viii) {L,S ,C} (X) = LSC(X), for every X ∈ iΘh.. Proof. (i) Let. X1 , X2 ∈ iΘh. G ∈ X1 ⊂ X2 . If. H ∈ C(X1 ),. H ∈ C(X2 ) Now, if. Then. X1 ⊂ X2 .. H ∈ S(X2 ). we have that. and. H=. H ∈ L(X1 ),. H ∈ X.. Q. If. S. H ∈ S(X1 ),. H ≤ G,. where. is monotone.. Gi ,. where. Then. Since. H. C. Gi ∈ X1 ⊂ X2 ,. for every. i ∈ I.. Then. is monotone.. for every nitely generated subalgebra. H ∈ L(X2 ). and. L. H ∈ C(X),. and. is a subalgebra of itself, we have that. C. H0 ≤ H ,. we have that. is monotone.. operator of extension. Now, for any index Thus. we have that. i∈I. and we conclude that. H0 ∈ X1 ⊂ X2 . (ii) Let. such that. j,. we have that. is an operator of extension.. H ∈ S(X),. H = HI,. and. where. S. is an. I = {j}..

(37) 35. X ∈ iΘh. (iii) Let. such that. S(X) = X.. subalgebra, we have that (iv) Let if. X ∈ iΘh. Since S. H ∈ SL(X),. H ∈ X. If. H0 ∈ S(X) = X.. H0 ≤ H. and. Hence. is a nitely generated. H ∈ L(X),. and. is an operator of extension, we have that. there exists. G ∈ L(X). such that. H ≤ G.. If. X ⊂ L(X).. L(X) ⊂ SL(X). Now, H0 ≤ H. is a nitely. generated subalgebra, then. H0 ≤ H ≤ G. By the denition of the operator L we have. H0 ∈ X,. We conclude that. (v) Let. H ∈ L(X).. and so. X ∈ iΘh.. (ii),. By. G ∈ S(X). exists. G ≤ A.. So, we have. closed. Moreover, by. S(X) ⊂ SS(X).. we have that. H ≤ G.. such that. H ≤G≤A (ii),. H ∈ CC(X).. SL(X) ⊂ L(X). Since. and. we have that. Now, if. G ∈ S(X),. H ∈ S(X).. and. H ∈ SS(X),. there exists. Thus,. SL(X) = L(X). then there. A∈X. SS(X) ⊂ S(X),. such that and. S. is. C(X) ⊂ CC(X).. Q. Hi ∈ C(X), for all i ∈ I . For every Q i∈ Hi = Aj , where Aj ∈ X for every j∈Ji Q Q j ∈ Ji , i ∈ I , and Ji1 ∩ Ji2 = ∅, for i1 6= i2 . So, we have H = ( Aj ), and by i∈I j∈Ji Q S 0 Proposition 2.1.7, it follows that H ∼ Aj ∈ C(X), where J = Ji . So, as C(X) = Hi , where i∈I I , there exists a set of indices Ji such that. Now, let. So,. H =. j∈J 0 is an abstract class, we have that that. C. So, by. (iii),. L,. by. (iv). we have. it follows that. of itself, it follows that and hence. L. CC(X) ⊂ C(X).. We conclude. there exists. νi : Gi → Ai. H ∈ L(X).. L(X). H ∈ LL(X). is closed under and. We conclude that. LL(X) ⊂ L(X). G = Ai ∈. Q. Gi , where Gi ∈ S(X), for all i i∈I X such that Gi ≤ Ai . If πi : G →. are inclusion mappings, we dene. ∈ I Gi. ϕi : G → Ai. (this last equality takes place because. j∈I. Ker(ϕ) =. T. Ker(ϕi ) = {(g1 , g2 ) ∈ G × G | πi (g1 ) = πi (g2 ), ∀i ∈ I},. i∈I. Ker(ϕ) = {(g, g) | g ∈ G} ⊂ G × G. ϕ. is. is a nitely generated subalgebra. projections. Thus,. So,. H0 ≤ H. .. are by. πi (g) ∈ Gi ⊂ Ai , Q i ∈ I ). By the Remak Theorem we have a homomorphism ϕ : G → Ai i∈I Q by σi (ϕ(g)) = ϕi (g) = πi (g), for any i ∈ I , where σi : Aj → Ai are. ϕi (g) = νi (πi (g)) = πi (g). dened. Thus,. We have that. i ∈ I. projections and. for any. Now, if. H0 ∈ L(X). As H0. H0 ∈ X.. that is,. is closed.. G ∈ CS(X).. So, for every. SL(X) = L(X),. L(X) ⊂ LL(X).. nitely generated, we have that. (vi) Let. i∈I and then. is closed.. For the operator. S.. H ∈ C(X),. is an embedding, and. G ∈ SC(X).. and.

(38) 36. X ∈ iΘh.. (vii) Let. i∈I. . Let. For every. , let. 0. Gi. are projections. Since we have that. i ∈ I , πi (g). So, for every. 0. πi (g) ∈ Gi ,. G=. we have that. be the subalgebra of. g ∈ G0 ,. If. Q. Gi , where Gi ∈ L(X), for every i∈I be a nitely generated subalgebra with generators g1 , g2 , ..., gk .. G0 ≤ G. i∈I. i ∈ I.. Thus,. G ∈ CL(X),. πi : G → Gi. where any. If. for all. g. Gi. generated by. Gi ∈ L(X),. we have that. is generated by the elements. is generated by the elements. i∈I. and. πi (g1 ), πi (g2 ), ..., πi (gk ),. G0 ⊂. Q. 0. Gi .. So,. 0. Gi ∈ X,. for. g1 , g2 , ..., gk .. πi (g1 ), πi (g2 ), ..., πi (gk ).. G0 ∈ SC(X).. It follows that. i∈I. G ∈ LSC(X), X ∈ iΘh.. (viii) Let. and then By. (v). and. SLSC(X) = LSC(X). Now, by. (vii),. CL(X) ⊂ LSC(X). (iv),. respectively, we have that. The class. it follows that. LSC(X). LLSC(X) = LSC(X). and. S.. is closed under the operator. C.. is closed under the operators. CLSC(X) ⊂ LSCSC(X).. By. (vi),. L. and. 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). Now, note that. X ⊂ {L,S ,C} (X) ⊂ LSC(X) By. (i), LSC(X) ⊂ LSC {L,S ,C} (X) ⊂ LSCLSC(X). and. 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 operators and the operators. Proposition 3.3.2. L, S. and. C. H -closure. .. Let Θ be a variety of algebras and T ⊂ WΘ (X) × WΘ (X) a congru-. ence. T is H -closed if and only if WΘ (X)/T ∈ SC(H)..

(39) 37. Proof. (⇒) Let T. be. H -closed.. So, we have that. \. TH00 =. Ker(α) = T.. 0 α∈TH. TH0 = {αi | i ∈ I}.. Denote. αi (w),. Dene the mapping. πi : H I → H. where. [w1 ]T = [w2 ]T ,. µ : WΘ (X)/T → H I. are projections. We have that. µ. πi (µ([w]T )) =. by. is well-dened, because if. we have that. (w1 , w2 ) ∈ T =. \. Ker(αi ),. which means that. αi (w1 ) = αi (w2 ),. for all. i ∈ I.. i∈I Therefore,. µ([w1 ]T ) = µ([w2 ]T ).. πi (µ([w1 ]T )) = πi (µ([w2 ]T )), and. (w1 , w2 ) ∈ Ker(αi ),. Now, if. i ∈ I,. for all. for every. (w1 , w2 ) ∈. i ∈ I.. \. µ([w1 ]T ) = µ([w2 ]T ), which means that. we have. αi (w1 ) = αi (w2 ),. for all. i ∈ I,. It follows that. Ker(αi ) = T. and. [w1 ]T = [w2 ]T .. i∈I We conclude that. (⇐) Let. Let. µ. is an embedding. Thus,. WΘ (X)/T ∈ SC(H).. πi : H I → H (i ∈ I). epimorphism.. Then,. WΘ (X)/T ∈ SC(H).. Then we have an embedding. µ : WΘ (X)/T ,→ H I .. τ : WΘ (X) → WΘ (X)/T be the natural T πi µτ ∈ Hom(WΘ (X), H). We claim that: T = Ker(πi µτ ) . Let be projections and. i∈I. (w1 , w2 ) ∈ T .. We have that. πi (µ(τ (w1 ))) = πi (µ(τ (w2 ))), T (w1 , w2 ) ∈ Ker(πi µτ ), we i∈I we conclude that and. {πi µτ | i ∈ I} ⊂ TH0 T ⊂ TH00 ,. for all have. i ∈ I.. µ(τ (w1 )) = µ(τ (w2 )), and then T that T ⊂ Ker(πi µτ ). Now, if. and hence It follows. i∈I for every. πi (µ(τ (w1 ))) = πi (µ(τ (w2 ))),. µ(τ (w1 )) = µ(τ (w2 )).. [w1 ]T = [w2 ]T .. Since. [w1 ]T = [w2 ]T. As. µ. is an embedding, we have. and. τ (w1 ) = τ (w2 ). (w1 , w2 ) ∈ T , and the claim is proved. Finally, we have that T T Ker(πi µτ ) ⊃ TH00 . We conclude that TH00 ⊂ Ker(πi µτ ) = T .. Thus and. i∈I. we have. Corollary 3.3.1. i ∈ I,. TH00 = T ,. i∈I and therefore. T. is. H -closed.. 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 G0 ≤ G be a nitely generated subalgebra. There exists an epimorphism τ : WΘ (X) → G0 , where X is nite (because G0 is nitely generated). If T = Ker(τ ), then WΘ (X)/T ∼ = G0 . Since G0 ≤ G ⊂ C(G), we have that G0 ∈ SC(G) and WΘ (X)/T ∼ = G0 ∈ SC(G).. Thus,. T. is. G-closed.. By the assumption,. T. is. H -closed.. So, it follows.

(40) 38. that. WΘ (X)/T ∈ SC(H). and hence. G0 ∼ = WΘ (X)/T ∈ SC(H).. G∈. We conclude that. LSC(H).. (⇐). Let. G ∈ LSC(H). and. T ⊂ WΘ (X) × WΘ (X). have by the previous proposition that. µ : WΘ (X)/T → GI X. is nite and. be a. WΘ (X)/T ∈ SC(G).. (I is a set of indices). Let. G-closed. So there exists an embedding. πi : GI → G (i ∈ I ). WΘ (X)/T = τ (WΘ (X)) = τ (hXi),. where. congruence. We. be projections. Since. τ : WΘ (X) → WΘ (X)/T. is the. natural epimorphism, we have by Corollary 2.1.1 that. im(πi (µ(WΘ (X)/T ))) = im(πi (µ(τ (hXi)))) = hπi (µ(τ (X)))i , and then. im(πi µ). is a nitely generated subalgebra of. i ∈ I , im(πi µ) ∈ SC(H), im(πi µ) → H Ji ,. where. Ji. because. G ∈ LSC(H).. G,. i ∈ I,. for all. for every. i ∈ I.. So, for every. Thus, there exist embeddings. is a set of indices. Now, dene. Q. λ : WΘ (X)/T →. νi :. H Ji. by. i∈I. σi (λ([w]T )) = νi (πi (µ([w]T ))), λ([w2 ]T ),. where. σi :. Q. H Ji → H Ji. λ([w1 ]T ) =. are projections. If. i∈I. then. σi (λ([w1 ]T )) = σi (λ([w2 ]T )),. for all. i ∈ I,. and. νi (πi (µ([w1 ]T ))) = νi (πi (µ([w2 ]T ))), Since. νi. is an embedding, for any. i ∈ I,. µ([w1 ]T ) = µ([w2 ]T ).. As. µ. i ∈ I.. it follows that. πi (µ([w1 ]T )) = πi (µ([w2 ]T )), and. for all. for all. i ∈ I,. is an embedding, we have that. [w1 ]T = [w2 ]T .. WΘ (X)/T ∈ SC(H),. T. is an embedding. We conclude that. and hence. is. Therefore,. λ. H -closed.. Corollary 3.3.2 LSC(H) ⊂ q − V ar(H). Proof. Let H0 ∈ LSC(H). So, we have that every H0 -closed congruence T If. H ((. V. (wi0 = wi00 )) ⇒ (w00 = w000 )),. we have that. is. (w00 , w000 ) ∈ {(wi0 , wi00 ) | i ∈ I}00H .. i∈I. {(wi0 , wi00 ) | i ∈ I}00H0. is. H0 -closed,. then it is. {(wi0 , wi00 ) | i ∈ I}00H ⊂ {(wi0 , wi00 ) | i ∈ I}00H0 ,. H -closed.. H -closed.. By Proposition 3.1.2,. and the result follows.. Once. (w00 , w000 ) ∈.

(41) 39. 3.4. Geometrical Equivalence. In this section, we will dene the most important concept of this chapter: the geometrical equivalence of algebras.. Denition 3.4.1. Two algebras H1 , H2 ∈ Θ are called geometrically equivalent if TH00 1 =. TH00 2 , for every T ⊂ WΘ (X) × WΘ (X) and every nite set X . If two algebras H1 and H2 are geometrically equivalent we will write H1 ∼ H2 . Next propositions give us criteria and properties of geometrically equivalent algebras.. Proposition 3.4.1. Let Θ be a variety of algebras and H1 , H2 ∈ Θ. Then H1 ∼ H2 if and. only if every H1 -closed congruence is H2 -closed and vice versa.. Proof. (⇒) Let T ⊂ WΘ (X) × WΘ (X) be a H1 -closed congruence. By the assumption, T = TH00 1 = TH00 2 (⇐). and we conclude that. For every. T. is. H2 -closed. The other case is proved analogously.. T ⊂ WΘ (X) × WΘ (X),. By the assumption,. TH00 1. is. H2 -closed.. prove the converse inclusion. So,. Proposition 3.4.2. we have that. Since. T ⊂ TH00 1 ,. TH00 1. is a. we have. H1 -closed. TH00 2 ⊂ TH00 1 .. congruence.. Similarly, we. TH00 1 = TH00 2 .. Let Θ be a variety of algebras and H1 , H2 ∈ Θ. If H1 ∼ H2 , then. q − id(H1 ) = q − id(H2 ).. Proof.. (wi0 = wi00 )) ⇒ (w00 = w000 )), then (w00 , w000 ) ∈ {(wi0 , wi00 ) | i ∈ I}00H1 = i∈I V 00 0 00 {(wi , wi ) | i ∈ I}H2 . Thus, H2 (( (wi0 = wi00 ) ⇒ (w00 = w000 )) and hence q − id(H1 ) ⊂ If. H1 ((. q − id(H2 ).. V. i∈I The converse inclusion can be proved similarly.. Proposition 3.4.3. Let Θ be a variety of algebras and H1 , H2 ∈ Θ. Then H1 ∼ H2 if and. only if H1 and H2 have the same innite quasi-identities.. Proof. (⇒) It follows from the last proposition. (⇐). Let. T ⊂ WΘ (X) × WΘ (X) H1 ((. and. ^. (wi0 ,wi00 )∈T. (w00 , w000 ) ∈ TH00 1 .. We have that. (wi0 = wi00 )) ⇒ (w00 = w000 ))..

(42) 40. From the assumption, it follows that. ^. H2 ((. (wi0 = wi00 )) ⇒ (w00 = w000 )),. (wi0 ,wi00 )∈T and. (w00 , w000 ) ∈ TH00 2 .. Then. TH00 1 ⊂ TH00 2 .. Analogously, we can show that. TH00 2 ⊂ TH00 1 ,. So,. TH00 1 = TH00 2 . An algebra. Hom(H1 , H2 ). H1. is said to be. approximated. by the algebra. H2. if there exists. A ⊂. such that. \. Ker(α) = {(h, h) | h ∈ H1 } ⊂ H1 × H1 .. α∈A With the denition above, we have the following result:. Proposition 3.4.4. Let Θ be a variety of algebras and H1 , H2 ∈ Θ. H1 is approximated. by H2 if and only if there exists an embedding µ : H1 ,→ H2I (I is a set of indices).. Proof. (⇒) There exists A ⊂ Hom(H1 , H2 ) such that Let. A = {αi | i ∈ I}(I. πi : H I → H. is a set of indices). Dene. are projections. We have that. µ. T. Ker(α) = {(h, h) | h ∈ H1 }.. α∈A. µ : H1 → H2I. by. πi (µ(h)) = αi (h), where. is an embedding, because if. µ(h1 ) = µ(h2 ),. we have. αi (h1 ) = πi (µ(h1 )) = πi (µ(h2 )) = αi (h2 ), and then. (⇐). (h1 , h2 ) ∈ Ker(αi ),. i ∈ I.. Thus. (h1 , h2 ) ∈. T. i ∈ I,. h1 = h2 .. Let us assume that there exists an embedding. be projections. Put if. for any. for all. µ : H1 ,→ H2I .. Let. πi : H I → H. αi = πi µ and A = {αi | i ∈ I}. We have that A ⊂ Hom(H1 , H2 ) and,. Ker(αi ),. it follows that. i∈I. πi (µ(h1 )) = αi (h1 ) = αi (h2 ) = πi (µ(h2 )), and then. µ(h1 ) = µ(h2 ).. is approximated by. Proposition 3.4.5. Since. µ. is an embedding, we have. for all. i ∈ I,. h1 = h2 .. We conclude that. H1. H2 .. Let Θ be a variety of algebras and H1 , H2 ∈ Θ. H1 ∼ H2 if and only. if LSC(H1 ) = LSC(H2 )..

(43) 41. Proof. (⇒) If H1 ∼ H2 , we have by Corollary 3.3.1 that H1 ∈ LSC(H2 ). The operators C, S. and. every So,. L are monotone, so LSC(H1 ) ⊂ LSC(LSC(H2 )). Since {LSC}(X) = LSC(X), for. X ∈ iΘh, then LSC(H1 ) ⊂ LSC(H2 ). Similarly, we can prove the converse inclusion.. LSC(H1 ) = LSC(H2 ). (⇐). Let. H1 ∈ C(H1 ). LSC(H1 ) = LSC(H2 ). and. C(H1 ) ⊂ SC(H1 ).. As. C. Since. and. S. S. are operators of extension, we have that. is a closed operator and. L. is an operator of. extension over the classes of algebras which are closed under the operator. H1 ∈ SC(H1 ) ⊂ LSC(H1 ) = LSC(H2 ). WΘ (X). is. H2 -closed,. conclude that. and. TH00 2 ⊂ TH00 1 .. So, every. H1 -closed. congruence. Analogously, we can prove that. S , it follows that T ⊂ WΘ (X) × TH00 1 ⊂ TH00 2 .. We. TH00 1 = TH00 2 .. Remark 3.4.1:. Note that, if every nitely generated subalgebra. H0 ≤ H1. can be. approximated by. H2 , we have an embedding µ : H0 ,→ H2I , that is, H0 ∈ SC(H2 ) and hence. H1 ∈ LSC(H2 ).. Conversely, if. H0 ≤ H1 , Therefore,. it follows that. H0. H1 ∈ LSC(H2 ),. H0 ∈ SC(H2 ),. can be approximated by. Proposition 3.4.6. for every nitely generated subalgebra. that is, there exists an embedding. µ : H0 ,→ H2I .. H2 .. Let Θ be a variety of algebras and H1 , H2 ∈ Θ. H1 ∼ H2 if and only. if every nitely generated subalgebra H0 ≤ H1 can be approximated by H2 and vice versa.. Proof. (⇒) Let H1 ∼ H2 . By the previous proposition, we have that LSC(H1 ) = LSC(H2 ). Since. H1 ∈ LSC(H1 ),. LSC(H1 ), (⇐). we have that. H1 ∈ LSC(H2 ).. H2 ∈. and the result follows by Remark 3.4.1.. By Remark 3.4.1, we have that. H1 -closed. Similarly, we can prove that. congruence is. H2 -closed. H1 ∈ LSC(H2 ). and vice versa. Thus. and. H2 ∈ LSC(H1 ),. so every. H1 ∼ H2 .. Now, it is time to apply concepts and results from this chapter for the studying of the algebraic geometry over variety of group representations..

(44) 42. 4. Representation Theory and Geometrical Equivalence of Representations. In this chapter we apply the general notions of universal algebraic geometry to the case of group representations (see [18]). We start from the basic concepts of Representation Theory (see [19]).. 4.1. Basic notions of Representation Theory. For the sake of convenience we present three approaches to the notion of a group representation.. Denition 4.1.1. Let K be a commutative eld, G a group and V a K -vector space.. A representation of G on V is a group homomorphism ρ : G → AutK V (Notation:. ρ = (V, G)). We suppose that elements of AutK V are applied on elements of V on the right.. Denition 4.1.2. Let G be a group and V a K -vector space. An action of G on V is a. mapping (v, g) 7→ v ◦ g from V × G to V such that (i) (ii). v 7→ v ◦ g. is an automorphism of. V , ∀ g ∈ G;. v ◦ (g1 g2 ) = (v ◦ g1 ) ◦ g2 , ∀ v ∈ V, ∀ g1 , g2 ∈ G.. Let. (V, G). be a representation,. v. be a vector from. put. (v)ρ(g) = v ◦ g, we obtain an action of. G. on. V,. because. V. and. g. be an element of. G.. If we.

(45) 43. (i). ∀ g ∈ G, ρ(g) ∈ AutK V ;. (ii) For every. g1 , g2 ∈ G,. we have. ρ(g1 g2 ) = ρ(g1 )ρ(g2 ),. v ∈V,. and so, for every. v ◦ (g1 g2 ) = (v)ρ(g1 g2 ) = (v)ρ(g1 )ρ(g2 ) = (v ◦ g1 )ρ(g2 ) = (v ◦ g1 ) ◦ g2 . Thus. ◦. every. g∈G. V.. is, indeed, an action of. on. V.. ρ(g) : V → V. the mapping. Moreover, if. G. g1 , g2 ∈ G. and. v ∈V,. Conversely, if dened by. ◦. is an action of. (v)ρ(g) = v ◦ g. G. on. V,. then for. is an automorphism of. then we have. (v)ρ(g1 g2 ) = v ◦ (g1 g2 ) = (v ◦ g1 ) ◦ g2 = ((v)ρ(g1 ))ρ(g2 ) = (v)ρ(g1 )ρ(g2 ). So. ρ(g1 g2 ) = ρ(g1 )ρ(g2 ), and ρ is a representation of G. By these considerations, we have. the following result:. Proposition 4.1.1 (V, G). is a representation if and only if there exists an action of G. on V .. (V, G). Now, let. be a representation and. K. a eld. For every. v∈V. and. u=. P. ag g ∈. g∈G. KG,. we put. v◦u=v◦. X. ag g =. a. KG-module.. ρ(g) : V → V the element. Then it is a. dened by. v ∈V.. V. KG-module.. becomes a. K -module,. and so for each. (v)ρ(g) = v ◦ g ,. Since the elements of. by the axioms of module that, for every. ag (v ◦ g).. g∈G. g∈G It is not dicult to see that. X. where. K. ◦. Conversely, suppose that. g ∈ G,. V. is. we can consider map. is the action of the group element. commute with each element. g ∈ G,. g. on. it follows. α, β ∈ K. (vα + uβ)ρ(g) = (vα + uβ) ◦ g = (vα) ◦ g + (uβ) ◦ g = (vg) ◦ α + (ug) ◦ β and Hence,. ρ(g). is a linear mapping. Again by the axioms of module, we have. (v)ρ(g1 )ρ(g2 ). the image. (vα + uβ)ρ(g) = ((v)ρ(g))α + ((u)ρ(g))β, ∀v, u ∈ V.. Thus,. ρ : G → EndK V. ρ(G) ⊂ EndK V. We conclude that. ρ. Proposition 4.1.2 are equivalent:. is a homomorphism.. EndK V. is a semigroup, but. has only invertible elements. In other words,. is a representation of. G.. (v)ρ(g1 g2 ) =. ρ(G) ⊂ AutK V .. We have the following proposition:. Let V be a K -vector space and G a group. The following statements.

(46) 44. (i) (V, G) is a representation. (ii) There exists an action of G on V . (iii) V is a KG-module. In what follows, we will always use the item (iii) from the proposition above as the denition for a group representation:. Denition 4.1.3. The pair (V, G) is a representation if V is a KG-module.. Here, we will denote the multiplication of elements of variety of representations of groups over a xed eld. Denition 4.1.4. K. V. by elements of. will be denoted by. G. by. ◦.. The. REP − K .. Let (V, G) ∈ REP −K. (V0 , G0 ) is called a subrepresentation if V0 ≤ V ,. G0 ≤ G and V0 is a KG0 -submodule. We write (V0 , G0 ) ≤ (V, G).. Denition 4.1.5. The pair (V, G) is called nitely generated if the group G is nitely. generated and V is nitely generated as a KG-module.. Denition 4.1.6. Let (V, G), (W, H) ∈ REP − K. A homomorphism (α, β): (V, G) →. (W, H) is a pair of homomorphisms α : V → W and β : G → H such that α(v ◦ g) = α(v) ◦ β(g), ∀ v ∈ V, g ∈ G. A homomorphism. (α, β). is a monomorphism (epimorphism, isomorphism) if. α. and. β. are. monomorphisms (epimorphisms, isomorphisms). Now, we have an immediate consequence of the denition above:. Proposition 4.1.3. Let (V1 , G1 ), (V2 , G2 ), (V3 , G3 ) ∈ REP − K . If (α, β) : (V1 , G1 ) →. (V2 , G2 ) and (µ, ν) : (V2 , G2 ) → (V3 , G3 ) are homomorphisms, then the composition (µ ◦ α, ν ◦ β) : (V1 , G1 ) → (V3 , G3 ) is also a homomorphism.. Proof. Indeed, for every v ∈ V1 and g ∈ G1 , since (α, β) is a homomorphism, we have (µ ◦ α)(v ◦ g) = µ(α(v ◦ g)) = µ(α(v) ◦ β(g)),.