Proc. Indian Acad. Sci. (Math. Sci.) Vol. 122, No. 3, August 2012, pp. 339–350. c Indian Academy of Sciences Quotient semigroups and extension semigroups RONG XING1, CHANGGUO WEI2,∗ and SHUDONG LIU3 1Department of Mathematics, Teachers College of Qingdao University, Qingdao 266071, People’s Republic of China 2School of Mathematical Sciences, Ocean University of China, Qingdao 266100, People’s Republic of China 3School of Mathematical Sciences, Qufu Normal University, Qufu, Shandong 273165, People’s Republic of China ∗ Corresponding author E-mail: [email protected]; [email protected]; [email protected] MS received 2 March 2011; revised 23 May 2011 Abstract. We discuss properties of quotient semigroup of abelian semigroup from ∗ the viewpoint of C -algebra and apply them to a survey of extension semigroups. Cer- tain interrelations among some equivalence relations of extensions are also considered. Keywords. Quotient semigroup; extension semigroup; congruence relation. 1. Introduction Abelian groups and semigroups play an important role in the classification of C∗-algebras and their extensions. Especially, equipped with addition, the sets of equivalence classes of extensions become quotient semigroups. Therefore, the theory of quotient semigroups is essential for studying C∗-algebras and their extensions, though it is not a new object from the viewpoint of pure algebra. But the existing theory of semigroups in algebra is not suitable to being applied to C∗-algebra because of lack of details. In order to understand and apply C∗-algebra extension theory and KK-theory, it is crucial to study the theory of quotient semigroups from the viewpoint of C∗-algebra. Classifications of C∗-algebras and their extensions are two main aspects in the field of operator algebra since 1970’s. One can see [1] for details of their developments (see also [2,5–8,12–15]). In the recent past, interest in extension algebras has resulted in classifi- cation of such algebras [10,11,17–19]. Unlike classification of extensions, Ext-groups in KK-theory provide little information for classification of extension algebras, and so the isomorphism equivalence of extensions has not been studied. Besides extension semigroups, the semigroups of Murray–von Neumann equivalence classes of projections have already been applied to classifying C∗-algebras. Therefore, we need to study systematically the questions of semigroups relating to C∗-algebras and their extensions. In §2, we discuss some fundamental questions of quotient semigroups relating to classi- fying C∗-algebras. In §3, we give a survey of extension semigroups. Most of them are not new. We discuss certain interrelations among some equivalence relations of C∗-algebra extensions. 339 340 Rong Xing, Changguo Wei and Shudong Liu 2. Quotient semigroups In this section, we discuss several fundamental questions on semigroups. These questions are important in studying extension semigroups of C∗-algebras, but they are not con- sidered systematically in existing literature. We also give some examples to make these questions clear. Let us firstly recall some definitions on semigroups. Suppose that S is a semigroup. Then an equivalence relation ∼ on S is said to be congruent if a ∼ b and c ∼ d implies ac ∼ bd for any elements a, b, c and d in S.LetS/∼ be the quotient set and π : S → S/ ∼ the quotient mapping which maps a to a˜, where a˜ is the equivalence class of S containing a. The following results are easy to prove (see also [4]). PROPOSITION 2.1 Let S be a semigroup and ∼ an equivalence relation on S. Then the following statements are equivalent: (1) The relation ∼ is a congruence relation; (2) If a ∼ b, then we have ac ∼ bc and ca ∼ cb for each element c ∈ S; (3) The set S/∼ becomes a semigroup, which is called quotient semigroup, equipped with the multiplication defined by a˜ · b˜ = ab. PROPOSITION 2.2 If the quotient semigroup S/∼ admits a unit element E, then ker π = E and ker π is a sub-semigroup, where π : S → S/∼ is the quotient homomorphism. In the rest of this section, we only discuss the properties of quotient semigroup of abelian semigroup. The semigroups we considered in the following are always abelian if they are not specialized. Similar to the construction of quotient group and quotient linear space, quotient semi- group may be induced by the equivalence relation which is induced by sub-semigroup. S1 Let (S, +) be an abelian semigroup and S1 a sub-semigroup of S. Define a relation ∼ S1 on S by a ∼ b if and only if there exist elements x1, x2 ∈ S1 such that a + x1 = b + x2. Remarks. S1 S1 (1) The relation ∼ is a congruence relation on S.LetS/S1 = S/∼. Then we call it the quotient semigroup induced by S1. S (2) It should be pointed out, however, that a ∼1 b is not equivalent to that and there exists an element x ∈ S1 such that a + x = b or b + x = a. See Example 2. There are some essential questions we have to consider. Suppose that ∼ is a congruence relation on S. Question 1. Does S/∼ admit a unit element? If S/∼ has a unit element E,isE a sub- semigroup? What is the relation between E and ker π? What about S/S1? Quotient semigroups and extension semigroups 341 Question 2. When is the congruence relation ∼ induced by a sub-semigroup S1, that is, when do we have S/∼=S/S1? Question 3. If S/∼ is a group, does S/ ker π = S/∼ hold? Does the element in S/∼ have the form a ker π? Question 4. In S/S1, when does ker π = S1 hold? Conclusion 1. (1) In general, S/∼ does not admit a unit element (see Example 1). S/∼ admits a unit ele- ment if and only if there exists an equivalence class E such that E is a sub-semigroup and a + e ∼ a for any a ∈ S and e ∈ E. (2) If E is the unit element of S/∼, then E is a sub-semigroup and E = ker π. (3) The quotient semigroup S/S1 induced by a sub-semigroup S1 admits a unit element E, that is, E = ker π. One can see that S1 ⊂ E, but the converse does not hold in general (see Example 2). Example 1. Let S = N ⊕ N and define (a, b) ∼ (c, d) ⇔ b = d. Then ∼ is a congruence relation and S/∼={N ⊕ 1, N ⊕ 2,...}. Hence S/∼ does not admit a unit element. This congruence relation can not be induced by any sub-semigroup of S. Example 2. Let S = N and S1 ={6, 8, 10,...}. Since 1+(2n +8) = (2n −1)+10, 2n + ˜ ˜ 8 ∈ S1 and 10 ∈ S1, 1 = 2n − 1 (n ∈ N).Wealsohave2 = 2n (n ∈ N). Therefore ˜ ˜ S/S1 ={1, 2}. (1) It is easy to see that ker π ={2, 4, 6,...},sokerπ = S1. (2) We have 2 ∼ 4. However, there is not an element x ∈ S1 such that 2 + x = 4or 4 + x = 2. (3) One can also see that 2 ∼ 6 and 6 ∈ S1,but2∈/ S1. PROPOSITION 2.3 If there is a sub-semigroup S1 such that S/∼=S/S1, then S/S1 = S/ ker π. S ker π S Proof. It is equivalent to show that a ∼1 b if and only if a ∼ b.Ifa ∼1 b, then there exist ker π elements x1, x2 ∈ S1 such that a + x1 = b + x2, and hence a ∼ b (since S1 ⊂ ker π). π ker∼ , ∈ π + = Conversely, if a b, then there exist elements x1 x2 ker such that a x1 + π( ) = π( ) ∼S1 b x2, and hence a b . It follows that a b. S1 Conclusion 2. If ∼ is induced by a sub-semigroup S1, i.e. ∼⇔∼, then S1 ⊂ ker π and S/∼=S/S1 = S/ ker π by Proposition 2.3. Hence ker π is the maximal sub-semigroup with S/∼=S/S1. From the above discussion, if S/∼ admits a unit element, the unit element must be ker π ker π ker π, and a ∼ b implies a ∼ b. The remaining question is that when are ∼ and ∼ equivalent. 342 Rong Xing, Changguo Wei and Shudong Liu Conclusion 3. ker π (1) If S/∼ is a group, then a ∼ b ⇔ a ∼ b,sowehaveS/∼=S/ ker π. (2) It is not necessary that each element of S/∼ has the form a ker π. See Example 3 below. Proof. To prove Conclusion 3(1), it is easy to see that π(a) = π(b) from a ∼ b. Suppose π(c) is the inverse of π(a) in S/∼. Then π(a) + π(c) = π(b) + π(c) is the unit element ker π of S/∼. Hence a + c, b + c ∈ ker π. Since a + (b + c) = b + (a + c), a ∼ b. Thus S/∼=S/ ker π if S/∼ is a group. Example 3. Let S = N ⊕ Z and ∼ the congruence relation induced by the sub-semigroup N⊕0. Then S/∼ has the unit E and E = ker π = N⊕0. Hence S/∼ is a group. Consider the equivalence class A ={n ⊕ 5 : n ∈ N} of a = (3, 5) ∈ A. Then aE ={(3 + n, 5) : n ∈ N},butaE = A. Conclusion 4. In order to find the condition for ker π = S1,wegivethefollow- ing definition. Let S be an abelian semigroup and S1, S2 be sub-semigroups of S. Put S2 − S1 ={a ∈ S :∃xi ∈ Si , s.t. a + x1 = x2}. Then S2 − S1 is a sub-semigroup of S. We have the following results for Question 4. Theorem 2.4. Let π : S → S/S1 be the quotient homomorphism. Then ker π = S1 − S1 and hence ker π = S1 if and only if S1 = S1 − S1.