a(e, x) = x. Diagrammatically, this is encoded as the following commutative diagrams / X

Transcription

1 4. Mon, Sept. 30 Lst time, we defined the quotient topology coming from continuous surjection q : X! Y. Recll tht q is quotient mp (nd Y hs the quotient topology) if V Y is open precisely when q (V ) X is open. Exmple 4.. (Collpsing subspce) Let A X be subspce. We define reltion on X s follows: x y if both re points in A or if neither is in A nd x = y. Here, we hve one equivlence clss for the subset A, nd every point outside of A is its own equivlence clss. Stndrd nottion for the set X/ of equivlence clsses under this reltion is X/A. The universl property cn be summed up s sying tht ny mp on X which is constnt on A fctors through the quotient X/A. For exmple, we considered lst time the exmple R/(, 0] = [0, ). Exmple 4.2. I. The exponentil mp e : I! S is constnt so we get n induced continuous mp ' : I/@I! S, which is esily seen to be bijection. In fct, it is homeomorphism. Once we lern bout compctness, it will be esy to see tht this is closed mp. We show insted tht it is open. A bsis for I/@I is given by q(, b) with0<<b< nd by q([0,) [ (b, ]) with gin 0 <<b<. It is cler tht both re tken to bsis elements for the subspce topology on S. It follows tht ' is homeomorphism. Exmple 4.3. Generlizing the previous exmple, for ny closed bll D n R n+, we cn consider the quotient D n /@D n. On your homework this week, you re sked to provide continuous bijection D n /@D n! S n. Agin, we will see lter tht this must be homeomorphism D n /@D n = S n. Exmple 4.4. On S n we impose the equivlence reltion x x. The resulting quotient spce is known s n-dimensionl rel projective spce nd is denoted RP n. Consider the cse n =. We hve the hemisphere inclusion I,! S given by x 7! e ix. Then the composition I,! S RP is quotient mp tht simply identifies the to point. In other words, this is exmple 4. from bove, nd we conclude tht RP = S. However, the higher-dimensionl versions of these spces re certinly not homeomorphic. Exmple 4.5. Consider S 2n s subspce of C n. We then hve the coordinte-wise multipliction by elements of S = U() on C n. This multipliction restricts to multipliction on the subspce S 2n, nd we impose n equivlence reltion (z,...,z n ) ( z 0,..., z n ) for ll 2 S. The resulting quotient spce is the complex projective spce CP n. Exmple 4.6. On I I, we impose the reltion (0,y) (,y) nd lso the reltion (x, 0) (x, ). The resulting quotient spce is the torus T 2 = S S. We recognize this s the product of two copies of exmple 4., but bewre tht in generl product of quotient mps need not be quotient mp. torus We discussed lst time the fct tht quotient mp need not be open. Nevertheless, there is clss of open sets tht re lwys crried to open sets. Definition 4.7. Let q : X! Y be continuous surjection. We sy subset A X is sturted (with respect to q) if it is of the form q (V ) for some subset V Y. It follows tht A is sturted if nd only if q (q(a)) = A. Recll tht fiber of mp q : X! Y is the preimge of single point. Then nother description is tht A is sturted if nd only if it contins ll fibers tht it meets. Proposition 4.8. A continuous surjection q : X Y is quotient mp if nd only if it tkes sturted open sets to sturted open sets. 26

2 Proof. Exercise. A number of the exmples bove hve secretly been exmples of more generl construction, nmely the quotient under the ction of group. Definition 4.9. A topologicl group is bsed spce (G, e) with continuous multipliction m : G G! G nd inverse i : G! G stisfying ll of the usul xioms for group. Remrk 4.0. Munkres requires ll topologicl groups to stisfy the condition tht points re closed. We will not mke this restriction, though the exmples we will consider will ll stisfy this. 5. Wed, Oct. 2 Lst time, we introduced the ide of topologicl group, which is simultneously group nd spce, where the multipliction nd inverse re required to be continuous. Exmple 5.. () Any group G cn be considered s topologicl group equipped with the discrete topology. For instnce, we hve the cyclic groups Z nd C n = Z/nZ. (2) The rel line R is group under ddition, This is topologicl group becuse ddition nd multipliction by re both continuous. Note tht here Z is t the sme time both subspce nd subgroup. It is thus topologicl subgroup. (3) If we remove zero, we get the multiplictive group R = R \{0} of rel numbers. (4) Inside R, we hve the subgroup {, } of order two. (5) R n is lso topologicl group under ddition. In the cse n = 2, we often think of this s C. (6) Agin removing zero, we get the multiplictive group C = C \{0} of complex numbers. (7) Inside C we hve the subgroup of complex numbers of norm, k the circle group S = U() = SO(2). (8) This lst exmple suggests tht mtrix groups in generl re good cndidtes. For instnce, we hve the topologicl group Gl n (R). This is subspce of M n (R) = R n2. The determinnt mpping det : M n (R)! R is polynomil in the coe cients nd therefore continuous. The generl liner group is the complement of det (0). It follows tht Gl n (R) is n open subspce of R n2. (9) Inside Gl n (R), we hve the closed subgroups Sl n (R), O(n), SO(n). Let G be topologicl group nd fix some h 2 G. DefineL h : G! G by L h (g) =hg. Thisis left multipliction by h. The definition of topologicl group implies tht this is continuous, s L h is the composition G (h,id)! G G m! G. Moreover, L h is clerly inverse to L h nd continuous by the sme rgument, so we conclude tht ech L h is homeomorphism. Since L h (e) =h, we conclude tht neighborhoods round h look like neighborhoods round e. Since h ws rbitrry, we conclude tht neighborhoods round one point look like neighborhoods round ny other point. This implies tht spce like the unoin of the coordinte xes in R 2 cnnot be given the structure of topologicl group, s neighborhoods round the origin do not resemble neighborhoods round other points. The min reson for studying topologicl groups is to consider their ctions on spces. Definition 5.2. Let G be topologicl group nd X spce. A left ction of G on X is mp : G X! X which is ssocitive nd unitl. This mens tht (g, (h, x)) = (gh, x) nd 27

3 (e, x) = x. Digrmmticlly, this is encoded s the following commuttive digrms G G X id / G X X e,id / G X m id G X It is common to write g x or simply gx rther thn (g, x). / X id # Given n ction of G on spce X, we define reltion on X by x y if y = g x for some g. The equivlence clsses re known s orbits of G in X, nd the quotient of X by this reltion is typiclly written s X/G. Relly, the nottion X/G should be reserved for the quotient by right ction of G on X, nd the quotient by left ction should be G\X. Exmple 5.3. () For ny G, left multipliction gives n ction of G on itself! This is trnsitive ction, mening tht there is only one orbit, nd the quotient G/G is just point. Note tht we sw bove tht, for ech h 2 G, the mp L h : G! G is homeomorphism. This generlizes to ny ction. For ech g 2 G, the mp (g, ):X! X is homeomorphism. (2) For ny (topologicl) subgroup H pple G, left multipliction by elements of H gives left ction of H on G. Note tht n orbit here is precisely right coset Hg. The quotient is H\G, the set of right cosets of H in G. (3) Consider the subgroup Z pple R. SinceR is belin, we don t need to worry bout bout left vs. right ctions or left vs. right cosets. We then hve the quotient R/Z, which is gin topologicl group (gin, R is belin, so Z is norml). Wht is this group? Once gin, consider the exponentil mp exp : R! S given by exp(x) =e 2 ix. This is surjective, nd it is homomorphism since exp(x + y) =exp(x)exp(y). The First Isomorphism Theorem in group theory tells us tht S = R/ ker(exp), t lest s group. The kernel is precisely Z pple R, nd it follows tht S = R/Z s group. To see tht this is lso homeomorphism, we need to know tht exp : R! S is quotient mp, but this follows from our erlier verifiction tht I! S is quotient. (4) Similrly, we cn think of Z n cting on R n, nd the quotient is R n /Z n = (S ) n = T n. (5) The group Gl(n) cts on R n (just multiply mtrix with vector), but this is not terribly interesting, s there re only two orbits: the origin is closed orbit, nd the complement is n open orbit. Thus the quotient spce consists of closed point nd n open point. (6) More interesting is the ction of the subgroup O(n) on R n. Using the fct tht orthogonl mtrices preserve norms, it is not di cult to see tht the orbits re precisely the spheres round the origin. We clim tht the quotient is the spce [0, ) (thought of s subspce of R). To see this, consider the continuous surjection : R n! [0, ). By considering how this cts on open blls, you cn show tht this is n open mp nd therefore quotient. But the fibers of this mp re precisely the spheres, so it follows tht this is the quotient induced by the bove ction of O(n). 6. Fri, Oct. 4 At the end of clss lst time, we were looking t the exmple of O(n) cting on R n, nd we climed tht the quotient ws [0, ). We sw tht the reltion coming from the O(n)-ction ws the sme s tht coming from the surjection R n! [0, ). Nmely, we identify points if nd 28 X.

4 only if they hve the sme norm. To see tht the quotient by the O(n)-ction is homeomorphic to [0, ), it remins to show tht the norm mp R n! [0, ) is quotient mp. We know lredy tht it is continuous surjection, nd by considering bsis elements (open blls) in R n, we cn see tht it is open s well. We leve this verifiction to the industrious student! Why does the bove rgument show tht the quotient R n /O(n) is homeomorphic to [0, ). We now hve two quotient mps out of R n, nd they re defined using the sme equivlence reltion on R n. By the universl property of quotients, the two spces re homeomorphic! Let s get on with more exmples. Exmple 6.. () Let R ct on R n vi sclr multipliction. This ction preserves lines, nd within ech line there re two orbits, one of which is the origin. Note tht the only sturted open set contining 0 is R n, so the only neighborhood of 0 in the quotient is the entire spce. (2) Switching from n to n + for convenience, we cn remove tht troublesome 0 nd let R ct on X n+ = R n+ \{0}. Here the orbits re precisely the lines (with origin removed). The quotient is RP n. To see this, recll tht we defined RP n s the quotient of S n by the reltion x x. This is precisely the reltion tht rises from the ction of the subgroup C 2 = {, } ppler on S n R n+. Now notice tht the mp R n+ \{0}! S n R >0 given by x 7! x kxk, kxk is homeomorphism. Next, note tht we hve n isomorphism R = C 2 R >0. Thus the quotient (R n+ \{0})/R cn be viewed s the two step quotient (S n R >0 )/R >0 ) /C 2. But (R n R >0 )/R >0 = S n, so we re done. We cn think of RP n in yet nother wy. Consider the following digrm: D n / S n / R n+ \{0} D n / / S n /C 2 / R n+ \{0}/R The mp D n! S n is the inclusion of hemisphere. The reltion on D n is the reltion x x, but only llowed on the n. All mps on the bottom re continuous bijections, nd gin we will see lter tht they re necessrily homeomorphisms. Note tht the reltion we imposed on D n does not come from n ction of C 2 on D n. Let us write C 2 = h i. We cn try defining x x 2 Int(D x = n ) x x n ), where here the interior nd boundry re tken in S n. But this is not continuous, s the convergent sequence r r! n, 0,...,0,! (, 0,...,0) n is tken by to convergent sequence, but the new limit is not (, 0,...,0) = (, 0,...,0). (3) We hve similr story for CP n. There is n ction of C on C n+ \{0}, nd the orbits re the punctured complex lines. We clim tht the quotient is CP n. We defined CP n s quotient of n S -ction on S 2n+. We lso hve homeomorphism C n+ \{0} = S 2n+ R >0 nd n isomorphism C = S R >0. We cn then describe CPn 29

5 s the two-step quotient C n+ \{0} /C = (S 2n+ R >0 )/R >0 /S = S 2n+ /S = CP n. 30