1 Quotient spaces

This are notes for the math 370 class of Januari 25.

Definition 1.1 Let S be a set. An equivalence relation is a relation which holds between certain elements of S. Notation: a b. It should satisfy: ∼ (i) a a for all a S (reflexivity), ∼ ∈ (ii) if a b then b a (symmetry), ∼ ∼ (iii) if a b and b c then a c (transitivity). ∼ ∼ ∼ The set of all elements equivalent to a fixed element a is called the equivalence class of a;itis denoted by Ca.

Example 1.2 (i) S = students in the class .Wesayx y if x and y have the same age. (ii) Let S be as in (i), but{ now we say two students} are equivalent∼ if they speak the same languages. (iii) Let S be the set of triangles in the plane. We say two of them are equivalent if they are congruent.

Proposition 1.3 Let S be a set and be an equivalence relation on S.LetCa and Cb be two ∼ equivalence classes. Then either Ca = Cb or Ca Cb = . ∩ ∅

Proof. Suppose that Ca and Cb have an element in common, call this element d.Thend Ca ∈ and d Cb. The definition of equivalence class now implies that a d and b d, hence a b by transitivity∈ of the equivalence relation. ∼ ∼ ∼ Now let x be an arbitrary element of Cb.Thenb x.Sincea b,wehavea x (transitivity). ∼ ∼ ∼ This implies x Ca. This shows that Cb Ca. The inclusion Ca Cb follows by reversing the ∈ ⊂ ⊂ role of a and b.SoCa = Cb. 

In Example (i) the equivalence classes are: Ci = all student of i years . Here is another example of an equivalence relation. Let V be a vector space{ over a field F and let} W be a subspace of V .WeuseW to define an equivalence relation on the elements of V as follows:

v w if v w W. ∼ − ∈

This defines an equivalence relation since v v =0V W ,sov v.Ifv w then v w W . So w v = (v w) W ,andw v.If−v w and∈ w u,then∼ v w∼and w u −are in∈ W . Therefore− v −u =(− v ∈w) (w u)isanelementof∼ ∼ W . ∼ − − − − − − Let us consider what the equivalence classes look like in an example. Let V = R2 and W =< (0, 1) >= (0,a) a R .Letv =(x0,y0). Then w =(x1,y1) v if v w =(x0 x1,y0 y1) W , { | ∈ } ∼ − − − ∈ i.e. x0 = x1. We find that Cv = (x0,a) a R . We can do the same computation for W 0 =< { | ∈ } (1, 1) >. In this case the equivalence class Cv = (x0 + a, y0 + a) a R . (Check!) In general, we have the following results. { | ∈ }

Proposition 1.4 Let W be a subspace of a vector space V ,andlet be as above. Let C be an equivalence class and v C.Then ∼ ∈ C = v + W := v + w w W . { | ∈ } Proof. By definition C = u u v = u u v W . For such a u we have u v = w W ,so { | ∼ } { | − ∈ } − ∈ u = v + w. 

1 Definition 1.5 Let C1 = v1 + W and C2 = v2 + W . We define an addition of equivalence classes by C1 + C2 is equal to the equivalence class (v1 + v2)+W .Forascalara F , we define the ∈ equivalence class of a C1 as av1 + W . · Proposition 1.6 These operations are well-defined, i.e. they are independent of the choices.

Proof. Let v0 C1 and v0 C2. We need to show that the class containing v0 + v0 is equal to the 1 ∈ 2 ∈ 1 2 class containing v1 + v2. The vectors v1 and v0 are both in C1, therefore v1 v0 = w1 W .Also 1 − 1 ∈ v2 v0 = w2 W .Sov1 + v2 = v0 + w1 + v0 + w2 =(v0 + v0 )+(w1 + w2) (v0 + v0 )+W .The − 2 ∈ 1 2 1 2 ∈ 1 2 vector v1 + v2 is both in (v1 + v2)+W andin(v10 + v20 )+W . The two classes have a non-trivial intersection, therefore they are equal. The proof that scalar multiplication is also well-defined is similar.  Proposition 1.7 The addition and scalar multiplication defined above makes the set of equivalence classes into a vector space. This vector space is denoted by V/W and is called the quotients space.

Proof. The zero element 0 =0+W . The negative element of a class v + W is ( v)+W .The V/W − verification of the 8 axioms is straight forward.  Proposition 1.8 Let W be a subspace of V .Supposedim(V )=n and dim(W )=m.Then dim(V/W)=n + m.

Proof. Let B =(w1,...,wm)beabasisofW . We can extend it to a basis (w1,...,wm,v1,...,vn m) − of V .LetCi = vi + W for i =1,...,n m. We are going to show that C1,...Cn m form a basis for V/W. − − First we have to check that the Ci are linearly independent. Suppose

c1 C1 + cn m +Cn m =0, · ··· − · − here the ci are scalars. There are elements xi W such that ∈

(c1v1 + x1)+...(cn mvn m + xn m)=x0. − − −

Therefore c1v1 + cn mvn m W .But(w1,...,wm,v1,...,vn m)isabasisofV . Therefore all ··· − − ∈ − ci are zero. NowwehavetocheckthattheCi span the space V/W.Letv be any vector in V ,andlet C = v + W be the equivalence class containing v.Wecanwrite

v = d1w1 + ...+ dmwm + c1v1 + ...+ cn mvn m, − − since (w1,...,wm,v1,...,vn m)isabasisofV .Heretheci’s and the di’s are scalars. Now −

v (c1v1 + ...+ cn mvn m) W. − − − ∈ Therefore

v + W =(c1v1 + ...+ cn mvn m)+W =(c1v1 + W )+ (cn mvn m + W ). − − ··· − − This shows that v C1 + ...Cn m.  ∈ −

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