Mitchell Faulk June 12, 2014 Equivalence Relations and Quotient Spaces 1 Equivalence Relations Recall that a relation on a set X is a subset S of the cartesian product X × X. Problem Set 2 has several examples of relations, so I will not spend time on examples here. Instead, I would like to discuss special types of relations, called equivalence relations, and how to deal with these special relations in mathematical constructions. As an example, I will introduce the notion of a quotient vector space. Let me start by discussing an equivalent formulation of a relation on a set X, namely, as a map X × X ! f0; 1g. We can think of a relation S on X as defining a map RS : X × X ! f0; 1g where 1; (x; y) 2 S R (x; y) = S 0; (x; y) 2= S In the case that RS(x; y) = 1, we say that x is related to y and we write this statement as xRSy. In the other case, the elements x and y are unrelated. The previous paragraph implies that we have a map R : frelations on Xg −! Set(X × X; f0; 1g) S 7! RS where Set(X×X; f0; 1g) is the collection of maps X×X ! f0; 1g. One can show that this map R is a bijection, and thus we have an alternate, yet equivalent way of thinking about a relation on a set X, namely as a map from X × X ! f0; 1g. Exercise 1.1. Show that the map R is a bijection by giving an inverse. Remark 1.2. More generally, for any set D, we may construct a bijection fsubsets of Dg Set(D; f0; 1g): The set of subsets of D is usually denoted P(D), and is called the power set of D. If D has finite cardinality jDj, the previous observation implies that the cardinality of P(D) is jP(D)j = 2jDj: This is somewhat of a common technique in mathematics: to view one con- struction as another equivalent construction. Having several different view- points can be somewhat of an advantage, as each different viewpoint allows for 1 a different approach to a proof. On the other hand, having too many different viewpoints can make certain things confusing, since it might not be clear which viewpoint is best in which situation. Moreover, constantly switching viewpoints can be confusing in itself. Nevertheless, I tend to believe in the slogan \the more viewpoints, the merrier." In the spirit of the slogan, we now give an equivalent formulation for an equivalence relation on a set X. In the end, we will see that giving an equivalence relation on X is the same as specifying a partition of the set X. Recall that a relation is called an equivalence relation if it is reflexive, sym- metric, and transitive. (These three terms are defined in Problem Set 2.) If ∼ is an equivalence relation on a set X, then for any element x 2 X, we may introduce the equivalence class of x, denoted [x] and defined by [x] := fy 2 X : x ∼ yg: Note that [x] is a subset of the set X, namely, the subset consisting of all elements of X that are related to x under the equivalence relation. We then write X= ∼ for the collection of all equivalence classes, that is, X= ∼:= f[x]: x 2 Xg: We have a natural surjective map π : X ! (X= ∼) which sends an element x to its equivalence class [x], that is, π(x) = [x]: Example 1.3. For a positive integer n, define a relation ∼ on Z by the rule x ∼ y if there is an integer k such that x = y + kn. (Morally, one should think of this relation as saying that x is related to y if x and y have the same remainder mod n.) It follows directly from the definition that the equivalence class of x 2 Z is given by [x] = fx + kn : k 2 Zg = f: : : ; x − 2n; x − n; x; x + n; x + 2n; : : :g: We let Zn denote the set of equivalence classes under this relation. It is a rewarding exercise to show that the cardinality of the set Zn is finite and equal to n. Exercise 1.4. Show that the cardinality of Zn is n. The previous example hints at a phenomenon that occurs in general: the set of equivalence classes form a partition of the parent set. (The term partition is defined in Problem Set 2.) More precisely, we have the following proposition, whose proof we leave as an exercise. Proposition 1.5. For an equivalence relation on a set X, the set of equivalence classes form a partition of the set X. 2 Reformulated in another way, Proposition 1.5 says that we have a map fequivalence relations on Xg −! fpartitions of Xg which sends an equivalence relation to the partition given by its equivalence classes. One could hope that this map is a bijection, which would mean that giving an equivalence relation on a set X is the same as giving a partition of a set X. And in fact, this is the case, as the following theorem asserts, which I let you prove in Problem Set 2. Theorem 1.6. Let X be a set. Then there are mutually inverse maps fequivalence relations on Xg fpartitions of Xg: 2 Well-defined maps Let us motivate this section with a question. Suppose that we have a set X with an equivalence relation ∼. Suppose also that we have a map f : X ! Y from X into another set Y . We can ask: When does the map f give rise to a map fe :(X= ∼) ! Y from the set of equivalence classes of X into Y ? More precisely, we are interested in when we can define fe by the rule fe([x]) = f(x): To see why this is even a question at all, let us consider an example. Let f : Z ! Z be the identity map. We can ask: Does f induce a map fe : Z2 ! Z? Because we are using such a nice map, namely the identity map, one would guess that the answer to this question is yes. However, this is certainly not the case! Indeed, note that f maps 1 7! 1 and 3 7! 3. However, in Z2, the elements 1 and 3 represent the same equivalence class, meaning that [1] = [3] in Z2. Thus, if fe were a function, we would have that 1 = f(1) = fe([1]) = fe([3]) = f(3) = 3; which is clearly false. So f does not induce a map fe in this case. If we examine what went wrong in the previous example, we are led naturally to introduce the following notion. Definition 2.1. Let X be a set with an equivalence relation ∼ and let f : X ! Y be a function from X into another set Y . We say that f induces a well-defined map fe :(X= ∼) −! Y if and only if the following property is satisfied by the map f: whenever x and x0 are elements of X such that x ∼ x0, we have f(x) = f(x0). If this property is satisfied, then the map fe is defined by the rule fe([x]) = f(x) for x 2 X. 3 Exercise 2.2. Let Zn denote the integers mod n. Let + : Z × Z ! Z denote the addition map on Z. For an integer y 2 Z, the mapping x 7! x + y defines a map Ry : Z ! Z. Show that the map f := π ◦ Ry : Z ! Zn induces a well-defined map fe : Zn ! Zn. This shows that the binary operation given by ⊕ : Zn × Zn −! Zn ([x]; [y]) 7! [x] ⊕ [y] := [x + y] is well-defined in the first slot. To show that it is well-defined in the second slot, you should show that the map g := π ◦ Ly : Z ! Zn induces a well-defined map ge : Zn ! Zn, where Ly : Z ! Z is the map defined by Ly(x) = y + x. The upshot of this exercise is that we can equip Zn with a binary operation ⊕ : Zn ×Zn ! Zn induced by addition on Z. As a consequence of this, it is easy to see that (Zn; ⊕; [0]) is a group. 3 Quotient vector spaces Let V be a vector space over the field k and let U be a subspace of V . From this data, we will construct a new vector space V=U called the quotient space whose vectors are equivalence classes of vectors from V and whose operations of addition and scalar multiplication are induced by the corresponding operations on V . The idea of the quotient space construction is to collapse the subspace U to a single vector. That is, we should view all elements of U as representing the same vector, that is, as equivalent vectors under some appropriately defined equivalence relation. More precisely, define the following relation on the set V : We say that v ∼ w if there is a vector u 2 U such that v = w + u. We leave it as an exercise to show that ∼ indeed defines an equivalence relation. Exercise 3.1. Show that ∼ is an equivalence relation. We let V=U denote the set of equivalence classes of vectors under this equiv- alence relation, that is, V=U = f[v]: v 2 V g: This means that the set V=U has as its elements subset of V . Our goal is to endow this collection of subsets with the structure of a vector space. First, we might ask ourselves, what is the zero vector in this vector space? The following exercise hints at the answer to this question.