Assignment 9; Due Friday, March 17


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1 Assignment 9; Due Friday, March b: A icture of this set is shown below. Note that the set only contains oints on the lines; internal oints are missing. Below are choices for U and V. Notice that U, V, and U V are arcwise connected. Both U and V can be strongly deformed to a bouquet of two circles, so their fundamental grous are the free grous F (a, b) and F (c, d). Since U V can be contracted to a oint, its fundamental grou is trivial. So there are no relations and the fundamental grou of our set is the free grou F (a, b, c, d). This is not surrising since our sace is almost a bouquet of four circles. 24.4d: The sace consists of two tori joined at a oint, as below. 1
2 Choose U to be the entire first torus together with a collar about the common oint in the second torus. Choose V similarly. See the ictures below. Notice that U can be strongly deformed to a torus, so its fundamental grou is Z Z. Think of this grou as the free abelian grou on two generators, which we write F A(a, b). Each element of this grou can be written uniquely as a ower of a followed by a ower of b: a m b n. A similar statement holds for V; its fundamental grou is F A(c, d). The set U V can be deformed to a oint, so its fundamental grou is trivial and there are no relations. Thus the fundamental grou of the join of the two circles is the free roduct F A(a, b) F A(c, d). A tyical element of this grou has the form (a 3 b 2 )(c 5 d 2 )(a 5 b)(c 2 d 2 )(ab 2 )(cd) e: Below is a icture when n = 2. The sace S 1 R is a cylinder and we are to remove k disks. We get a different answer when n = 2 and when n 3. We ll first do the calculation when n = 2. Recall that a cylinder S 1 R is homeomorhic to a unctured lane R 2 {0} by using olar coordinates in the second sace. Indeed a oint in the unctured lane can be described by an angle θ S 1 and a real 0 < r <. We can ma the unctured lane to the cylinder by sending (θ, r) θ ln(r). 2
3 Hence removing k disks from S 1 R is the same as removing k + 1 disks from the lane. This sace can be strongly deformed to a bouquet of k + 1 circles, so the fundamental grou is the free grou on k + 1 generators. When n 3, we can still find a homeomorhism from S n 1 R to R n {0}, so the sace of interest is still R n minus k +1 disks. We will rove by induction on k that the fundamental grou of this object is trivial. Let U be R n minus all k + 1 disks, and let V one of these disks slightly enlarged to a bigger oen disk. Notice that V is contractible and its fundamental grou is zero. The sace U V is an annulus which can be strongly deformed to a shere S n 1. Since n 3, the fundamental grou of this shere is trivial, so π(u V) = 0. Thus there are no relations and the SeifertVan Kamen theorem states that π(u V) = π(u). But U V is R n minus k holes and U is R n minus k + 1 holes. 3
4 24.4k: The icture below shows that RP 2 {y} can be strongly deformed to a circle, so its fundamental grou equals the fundamental grou of a circle, which is Z. 25.1b: Techniques from the end of last term show that all vertices of the olygon are glued to the same oint. Let U be the interior of the entagon, and let V be the entire entagon with boundary glued together, minus the center. Note that U V is the interior of the entagon minus the center, which can be deformed to a circle. Note that U is contractible, and V can be retracted to the boundary, which is a bouquet of two circles. Hence the fundamental grou of U is trivial and the fundamental grou of V is F (a 1, a 2 ). The fundamental grou of U V is Z, and the ma π(u V) π(v) sends the generator of Z to a 2 1 a 1 2 a 1a 2 So the fundamental grou of this sace is F (a 1, a 2 )/a 1 a 2 a 2 1 = a 2. 4
5 25.1fa: Once again, techniques from last term show that all vertices are glued to the same oint. Exactly the method of the revious roblem works here. The boundary of this olygon is again a bouquet of two circles a 1 and a 2, and the generator of π(u V) = Z mas to a 2 1 a 1 2 a 2 1 a 2. Hence the required fundamental grou is F (a 1, a 2 ) with relation a 2 1 a 1 2 = a 1 2 a2 1. We can relace the generator a 2 by a 1 2 and get the relation a 2 1 a 2 = a 2 a fh: I interrete this sace as the region between the inner circle and the outer circle. This time we must be careful because it is no longer true that all vertices glue together. Indeed in both the inner and outer circle, there are two vertices and each edge goes from one to the other. Hence the inner and outer boundaries are circles. Note that a 1 does not generate an element of any fundamental grou because it doesn t start and end at the same oint. But a 1 a 2 does define an element of a fundamental grou. Let U be the entire set minus the exterior boundary, and let V be the entire set minus the interior boundary. Note that U can be deformed to the interior boundary and V can be deformed to the exterior boundary. So π(u) is Z with generator a 1 a 2 and π(v) is Z with generator b 1 b 2. The set U V is the sace minus both boundaries. This can be deformed to a circle. The generator of this circle mas to (b 1 b 2 ) 2 in π(v) and mas to (a 1 a 2 ) 3 in π(u). So the fundamental grou is F (a 1 a 2, b 1 b 2 ) with relation (a 1 a 2 ) 3 = (b 1 b 2 ) 2. If we let C = a 1 a 2 and D = b 1 b 2, then this grou is F (C, D)/C 3 = D 2. Notice that this is exactly the grou of the trefoil knot. ) Extra Exercise: Clearly h is the identity since (e 2πi ) = 1 and (e 2πiq = 1. Suose h k (z 1, z 2 ) = (z 1, z 2 ) where 1 k <. If z 1 0, we conclude that e 2πik can only haen if divides k, which is doesn t. = 1. This If z 1 = 0, then z 2 0 and we conclude that e 2πikq = 1. This can only haen if divides kq. Since and q are relatively rime, we again conclude that divides k, which is false. 5
6 Extra Exercises Continued: We show that S 3 S 3 /Z is a covering sace as follows. Pick a oint S 3. We will show that there is an oen neighborhood U of in S 3 such that U, h(u), h 2 (U),..., h 1 (U) are disjoint. This is the key observation. Suose we succeed. I claim that π : U π(u) S 3 /Z is a homeomorhism, and the inverse image of this set is exactly the collection of sets U, h(u),..., h 1 (U), each homeomorhic to π(u). Indeed, the ma π : U π(u) is certainly onto; it is onetoone because if two oints and q ma to the same oint, then there is a k such that q = h k (), but U h k (U) =. The remaining assertions are easily checked. Next we rove that there is a U with U, h(u), h 2 (U),..., h 1 (U) disjoint. If this assertion is not correct, then for any neighborhood U of we can find unequal integers m and n between 0 and 1 such that h m (U) h n (U). Thus we can find oints q and r in U with h m (q) = h n (r). Without loss of generality suose n > m, and notice that q = h n m (r). Since does not divide n m, h n m has no fixed oints, so q and r are distinct. If no neighborhood U of works, we can find a sequence of unequal airs q n and r n such that q n and r n, and q n = h kn (r n ) for k n between 1 and 1. Since only finitely many k n are available, we can find a subsequence of n and q n such that all k n are equal to some fixed k. Then q n and r n and q n = h k (r n ). By continuity, = h k (), contradicting the assertion that h k has no fixed oints. Extra Exercises Continued: Since S 3 is simly connected, π(s 3 /Z ) is isomorhic to the deck transformation grou, and this grou is obviously the set {id, h, h 2,..., h 1 } Z since each of this is a deck transformation, and there are no other deck transformations because this set already acts transitively on π 1 () for U. When = 2 our ma is h(z 1, z 2 ) = ( z 1, z 2 ), so S 3 /Z 2 is obtained by gluing oosite oints together, and thus equals RP 3. Extra Exercise Concluded: This is the fun art, and I intend to describe some initial tries and false stes before exlaining the method ultimately used. The first goal is to ignore the grou Z and try to get the full S 3 to look like a ball B 3 with the uer hemishere glued to the lower hemishere. 6
7 Consider S 3 = {(z 1, z 2 ) z z 2 2 = 1}. Notice that a tyical z C is given in olar coordinates by giving its radius z and its angle θ. We can assume that π <= θ <= π, but we must then glue π to π. In our case it is not necessary to give both z 1 and z 2 because these quantities are related by z z 2 2 = 1. So a oint in S 3 is comletely described by giving z 1 with 0 z 1 1 and then giving a second angle θ 2. We can draw this by attaching a line segment π θ 2 π to each oint of the z 1 disk. See the icture below. However, when z 1 = 1, then z 2 = 0 and there is only one oint rather than an interval of oints. We can icture this by drawing the intervals π θ 2 π shorter and shorter near the boundary as in the icture below. We still must remember that θ 2 = π is the same oint as θ 2 = π. So to get S 3 from a ball, we must glue each oint on the uer hemishere to the oint immediately below it on the lower hemishere. This is analogous to constructing S 2 from a twodimensional disk 7
8 by gluing the semicircle at the to to the semicircle at the bottom as shown below. Conclusion II: Already this construction looks like our revious construction of a lens sace. To ush things further, let us bring the grou Z into the icture. We wish to draw a shaded subregion of S 3 such that every oint in S 3 is equivalent under Z to a unique oint in the shaded region, excet for identifications along the boundary. Let us concentrate on the action of Z on the z 1 disk. Here the grou acts by rotation by 2π. Every oint in the disk is equivalent to a oint in a shaded region below. Once we know where the z 1 comonent goes under the ma h Z, the action on z 2 is comletely determined. So every oint in S 3 is equivalent to a unique oint in the wedge below, excet for boundary oints. Thus we can take this wedge as our shaded region. I like to think of this iece as a wedge of an orange. We must examine gluing along the boundary. We already know that the oints on the uer hemishere must be glued to corresonding oints on the lower hemishere. If we do that, our wedge folds around and becomes a sort of lens with a shar corner along the entire outside. This shar corner corresonds to the equator of the lens, and the sides of the orange have become the to and bottom of the lens. We still must examine the manner that Z forces us to glue this 8
9 to and bottom together (or equivalently, the sides of the orange together). Let us concentrate on the action of the grou Z on the equator, and thus the action on the original line over z 1 = 0. Notice that this center is a singular lace where the z 1 action of Z is trivial and the z 2 action becomes significant. This Z action is multilication by multiles of e 2πiq. To understand this multilication, we must break the equator into equal ieces; the multilication rotates each iece by q units. This begins to look like a lens sace. By continuing to the end, this method can robably be made to work. However, at this oint I ll change to a different strategy. Conclusion III: Instead of obtaining a shaded region by cutting the z 1 disk into wedges, let us obtain it by cutting the angle θ 2 into ieces. This is a little easier to do if we think of θ 2 as moving from 0 to 2π. Thus we break this into subintervals [0, 2π] = [ 0, 2π ] [ 2π, 2π 2 ]... [ 2π ( 1), 2π ] For most oints, the action of Z is comletely determined by its action on the second comonent, so we can get a shaded region by allowing any z 1, but suitably restricting θ 2. Thus our [ shaded ] region can be the set of all (z 1, θ) with z 1 in the unit disk and θ in the interval 0, 2π. There is still a restriction that θ collases at the boundary where z 1 = 1 and so z 2 = 0. Notice that this fundamental region looks like a ball B 3, so we are definitely making rogress. We need only glue boundaries of this set, which involves gluing the to to the bottom. Usually if we aly h k Z to a oint in the shaded region, we get a oint in a comletely different region, so we can ignore h k. But sometimes if we take a boundary oint of the shaded region and aly h k, we get another boundary oint of the shaded region. In that secial case, we must glue the two boundary oints together. 9
10 Indeed, notice that h k mas θ 2 = 0 to θ 2 = 2πqk. If this second angle is 2π modulo multiles of 2π, then h k glues the bottom of the ball to the to of the ball, while simultaneously rotating z 1 by k clicks. So this looks very romising. But the q is annoying. A slight modification will simlify matters. Conclusion IV; the Real Conclusion Instead of letting z 1 belong to a disk and describing z 2 by giving θ 2, let us allow z 2 to belong to a disk and describe z 1 by giving θ 1. To get a shaded region, let us divide the θ 1 interval into ieces just as we earlier divided the θ 2 interval. Our icture is exactly the same as before. We now ask how h k could ma a boundary oint of the shaded region to another boundary oint. Since h acts on θ 1 by θ 1 θ 1 + 2π, the only mas hk which send oints in the shaded region to other such oints are h and h 1 = h 1. Restrict attention to the first ma h. It mas the lower boundary θ 1 = 0 to the uer boundary θ 1 = 2π. We can ignore h 1 because it just does this same gluing in reverse. But notice that when the lower boundary is glued to the uer one, h is simultaneously rotating z 2 by 2πq. So the rule is that we glue the lower hemishere to the uer one while simultaneously rotating about the zaxis counterclockwise by q clicks. Equivalently, we glue the uer hemishere to the lower one while simultaneously rotating about the z axis clockwise by q clicks. This is exactly the descrition of the lens sace L(, q). Whew. 10
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