Assignment 5; Due Friday, February 10

Transcription

1 Assigmet 5; Due Friday, February b The set X is just two circles joied at a poit, ad the set X is a grid i the plae, without the iteriors of the small squares. The picture below shows that the iteriors of the circles are evely covered, ad that the juctio where the circles meet is evely covered. This eds the argumet. Let me put this exercise i cotext by tellig you a few thigs we ll prove later. Let X be a coverig space of X. The the map π : X X iduces a map π : π( X) π(x) It is ot difficult to show that this map is oe-to-oe, so we ca thik of π( X) as a subgroup of π(x). We are goig to prove that this sets up a oe-to-oe correspodece betwee subgroups of π(x) ad coverig spaces of X. The full group correspods to the trivial cover π = id : X X ad the trivial group correspods to the uique simply-coected coverig space. We will prove that the deck trasformatio group Γ of a coverig π : ( (X), x 0 ) (X, x 0 ) is trasitive o π 1 (x 0 ) if ad oly if π( X) is a ormal subgroup of π(x). I this case, we will 1

2 prove that Γ is isomorphic to the quotiet group. For example, id : X X correspods to the full group π(x) ad the quotiet group is {1}; ideed the oly deck trasformatio of X is the idetity map. Similarly, the uiversal cover correspods to the idetity subgroup, ad the quotiet is the full group π(x), which is isomorphic to the deck trasformatio group. The fudametal group of a joi of two circles is the free group o two geerators F (a, b). I claim that the fudametal group of the above grid is the group geerated by all commutators g 1 g 2 g1 1 g 1 2, ad the quotiet group is thus the free abelia group geerated by a ad b. This abeliaized group is just Z Z. Sure eough, the group of deck trasformatios is clearly Z Z because a deck trasformatio of the grid is just a traslatio (x, y) (x + m, y + ). 17.9c These problems probably are t fair because they assume some kowledge of complex aalysis. But heck, all of mathematics is itercoected. The map z is much easier to uderstad i polar coordiates. It is ( (r cos θ, r si θ) = r(cos θ + i si θ) = re iθ re iθ) = r e iθ = (r cos θ, r si θ) This map does some stretchig ad shrikig i the radial directio, which is t very importat because r : R + R + is a homeomorphism. I the agular directio it wids aroud times. Thus the map defies a -fold coverig space. I the picture below, the iverse image of U = C {egative x-axis} is covered by ope wedges. The map si(z) : C C is ot a coverig map. The easiest way to see this is to otice that the derivative of the map is cos(z), ad this derivative has a zero at, for istace π 2. So si(z) could ot be a local homeomorphism ear π 2. Oce we otice this, we ca give a easy argumet idepedet of complex aalysis. Notice that si( π 2 ) = 1 ad si( π 2 t) = si( π 2 + t). So there could ot be a ope eighborhood of π 2 carried homeomorphically to a ope eighborhood of C by si(z). The derivative of (1 z) m z is m(1 z) m 1 z + (1 z) m z 1, which equals ( ) ( ) mz + (1 z) (1 z) m 1 z 1 = (m + )z (1 z) m 1 z 1 2

3 This expressio is zero if z = 0 or z = 1 or z = m+. The poits z = 0 ad z = 1 have bee omitted to form U, but m+ belogs to U uless m+ = 0 or m+ = or m+ = 1, that is, = 0 or m = or m = 0. By complex aalysis, a map is ot a local homeomorphism ear a spot where( its derivative ) is zero. So our map caot be a coverig map uless it is (1 z) or z or z 1 z. But z does t map U to C {1} uless = ±1 because it maps all th roots of uity to 1. Similar argumets show that our map is a coverig map if ad oly if it is oe of z, 1 z, 1 z, 1 1 z, z 1 z 1 z, or z. 17.9ef Notice that the map a seds (x, y) to (x, y + 1). The map b seds (x, y) to (x + 1, y + 1). 2 To show that ba = a 1 b it suffices to show that aba = b. But ba(x, y) = b(x, y + 1) = (x+ 1 2, (y+1)+1) = (x+ 1 2, y) ad therefore aba(x, y) = a(x+ 1 2, y) = (x+ 1 2, y+1) = b(x, y). It follows that wheever we fid b followed by a, we ca replace this with a 1 followed by b. So we ca get a equivalece expressio with all a s o the left ad all b s o the right: a k b l. Moreover, l is either eve or odd, as the book says i a peculiar way. I tryig to come to grips with this problem, let us figure out where (x, y)) is mapped by the various group elemets. The map a k maps (x, y) to (x + k, y). The map b maps (x, y) to (x + 1 2, y + 1) ad b2 maps it to (x + 1, ( y + 1) + 1) = (x + 1, y). So if we restrict ourselves to products of a ad b 2, we exactly get trasformatios of the iteger lattice, ad cosequetly every poit is equivalet to a poit i the uit rectagle with corers (0, 0), (0, 1), (1, 1), (1, 0). We still must come to grips with the actio of b. It maps poits o the left half of some rectagle to poits o the right half of some other rectagle. Cosequetly (x, y) is mapped to (x + 1 2, y + 1). Let us restrict to the previous uit rectagle. Here are pictures of some equivaleces iduced by the map b: 3

4 It ow follows that every poit i the plae is equivalet to a poit i the shaded rectagle below, ad poits o the boudary are equivalet as idicated. Clearly the set of equivalece classes is the Klei bottle. 17.9h We will fid a oe-to-oe correspodece betwee p 1 (x) ad p 1 (y) where x ad y are ay two poits i X. Let γ be a path from x to y i X. If q π 1 (x), there is a uique lift of γ to a path γ i X startig at q. This path eds at a elemet of p 1 (y); call this elemet ψ(q). We claim ψ : p 1 (x) p 1 (y) is oe-to-oe ad oto. It is oe-to-oe, for if ψ(q 1 ) = ψ(q 2 ), the there would be lifts γ 1 ad γ 2 of γ startig at q 1 ad q 2 respectively ad edig at the same poit. But the γ γ 1 could be lifted i two differet ways, as γ 1 γ 1 1 ad as γ 1 γ 2 1, cotradictig uiqueess of lifts. It is oto, for if q 1 p 1 (y), lift γ 1 to a path i X startig at q 1. This lift eds at a poit q p 1 (x); clearly ψ(q) = q 1 because we ca lift γ by tracig our lift of γ 1 backward. 4

5 17.9i This was doe i class. See the picture below. 17.9j Recall that RP 2 ca be obtaied from S 2 by glueig opposite poits together. I this solutio, whe p S 2 the p is this opposite poit. Let γ : (I, 0) (RP 2, x 0 ) be a parameterizatio of the simple closed curve i RP 2. Let x 1 ad x 2 = x 1 be the two poits i S 2 which project to x 0. Let γ 1 be the lift of γ to a path i S 2 startig at x 1. Notice that γ 2 = γ 1 is the the lift of γ to a path startig at x 2. The iverse image of γ is clearly the uio of the images of γ 1 ad γ 2. It may happe that γ 1 (1) = γ 1 (0). I that case γ 1 ad γ 2 are both closed curves, ad they are easily proved disjoit simple closed curves. The other possibility is that γ 1 (1) = γ 1 (0) = γ 2 (0). I that case γ 1 γ 2 is a simply closed curve ad so the iverse image is oe curve rather tha two. 17.9l This was essetially doe i class. The space RP is obtaied from S by glueig opposite poits together, so π : S RP is a two-fold cover. Sice S is simply coected, the map π(rp, x 0 ) π 1 (x 0 ) is oe-to-oe ad oto. This image set has two elemets, so π(rp, x 0 ) = Z q By defiitio, X is locally path-coected if wheever x is a poit ad U is a ope eighborhood of x, there is a ope V such that x V U ad V is path coected. Suppose X X is a coverig space ad let x X. Choose a evely covered ope eighborhood V X of π(x). The π 1 (V) = V α. Also x is i oe of these ope sets; say x V β. Now suppose U is a ope eighborhood of x i X. Replace V β by V β U ad replace V by π(v β U). This ew V is evely covered, ad this time π 1 (V) = V α ad x V β U. But X is locally arcwise coected, so we ca fid a arcwise coected ope set W 5

6 such that π(x) W V ad the, sice π : V β V is a homeomorphism, there is a correspodig arcwise coected ope set x W V β U. 17.9r This is almost trivial. If d 1 ad d 2 are coverig trasformatios (called deck trasformatios i class), the d 1, d 2 : X X are homeomorphisms makig a diagram commute. So d 1 d 2 is also a homeomorphism. Clearly the diagram still commutes for this compositio. Similarly d 1 1 is a homeomorphism, ad the diagram commutes for this homeomorphism. 20.7a If such a map existed, the it would iduce a cotiuous map ϕ : S /±1 S 1 /±1, where ±1 symbolizes the equivalece relatio which glues opposite poits together. Notice that S / ± 1 = RP ad S 1 / ± 1 = S 1. Thus we obtai a map ϕ : RP S 1. There would be a iduced map ϕ : Z 2 = π(rp ) π(s 1 ) = Z. All such group homomorphisms are trivial, so the ozero elemet of π(rp ) would have to map to the trivial elemet of π(s 1 ). But S 1 S 1 by z z 2 is a coverig map. which wraps the circle twice aroud itself. Cosider the diagram below. I this diagram, let the map I RP represet the ozero elemet of π(rp ). The correspodig lift I S must the be a path which starts at x 0 ad eds at x 0. So the compositio of this map with ϕ must be a path i S 1 which starts at ỹ 0 ad eds at ỹ 0. Whe this elemet is projected dow by z z 2, it becomes a closed path i S 1 ad this closed path is homotopic to the idetity because π(rp ) π(s 1 ) is the zero map. Our liftig theory the implies that the uique lift of this map back up to where it came from as a map I S 1 must be a closed loop. This cotradictio proves the exercise. 20.7b First a word about the meaig of the exercise. The sphere S 3 ca be thought of as the set of all poits i C 2 = R 4 with absolute value oe, ad thus as the set of all (z 1, z 2 ) with z z 2 2 = 1. The group Z p the acts o this space by actig o each compoet separately, where Z p acts o C as the group of rotatios of a regular p-sided polygo. Similarly Z p acts o S 1 R 2, agai as the group of rotatios of a regular p-sided polygo. 6

7 We ow wat a cotiuous map S 3 S 1. But we do t wat just ay map; we wat our map to commute with the actio of Z p. That meas that if (z 1, z 2 ) maps to a poit p S 1, ad if we the rotate z 1 ad z 2 by k polygoal clicks, the the image p should also rotate by k polygoal clicks. A special case is whe p = 2, so rotatio by oe click takes (z 1, z 2 ) to ( z 1, z 2 ) ad takes p to p. I this case there is o such map by the previous exercise. Now we are goig to prove that i geeral there is o such map. If there were such a map, the it would iduce a map S 3 / S 1 / where deotes the equivalece relatio iduced by Z p i which (z 1, z 2 ) is glued to the poit (w 1, w 2 ) obtaied by rotatig z 1 ad z 2 simultaeously by k clicks. We would get a diagram exactly as before, ad the argumet is goig to proceed exactly as it did earlier. Here is the key step. Cosider a map γ : I S 3 defiig a path which starts at x 0 ad eds at x 1 where both complex coordiates of x 1 are obtaied from the correspodig complex coordiates of x 0 by rotatig by 2π. This path iduces a loop i S3 / ad the correspodig elemet of π(s 3 / ) is ot the idetity elemet because its lift does ot begi ad ed at the same poit. This elemet has order i π(s 3 / ) because whe we lift the elemet repeated times, we get a loop i S 3 ad all loops i S 3 are homotopic to costat maps. The elemet i π(s 3 / ) must map to zero i π(s 1 ) because the latter group is isomorphic to Z ad has o ozero elemets of fiite order. O the other had, the map I S 3 S 1 obtaied by followig the top of the diagram is ot a loop, because by equivariace its ed poit is obtaied from its begiig by a rotatio of 2π. This cotradictio proves the exercise. 7