MATH540: Algebraic Topology PROBLEM SET 3 STUDENT SOLUTIONS

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1 Key Problems 1. Compute π 1 of the Mobius strip. Solution (Spencer Gerhardt): MATH540: Algebraic Topology PROBLEM SET 3 STUDENT SOLUTIONS In other words, M = I I/(s, 0) (1 s, 1). Let x 0 = ( 1 2, 0). Now by horizontal compression (M, x 0 ) deformation retracts to the vertical line L = { 1 2 } I with endpoints x 0 = ( 1 2, 0) ( 1 2, 1) identified. We know an interval with endpoints identified is homotopic to S 1, so we have (M, x 0 ) (L, x 0 ) (S 1, x 0 ). As π 1 is invariant under homotopy equivalence, this implies π 1 (M, x 0 ) = π 1 (S 1, x 0 ) = Z. 2. Compute π 1 of the surface of genus two using its presentation by identifying sides of a hexagon using Seifert van-kampen theorem as we did in class with the Torus T 2. Solution (Hongying Zhao): Figure 1: X We can have two open sets U and V that cover the surface of X = genus 2, as shown in the following figures. U, V and U V are path connected. π 1 (V ) = Z Z Z Z, π 1 (U) = 1, and π 1 (U V ) = Z. Let i 1, i 2 be inclusion maps from π 1 (U V ) to π 1 (U), π 1 (V ) respectively. If γ generates π 1 (U V ), then i 1 (γ) = 1, i 2 (γ) = aba 1 b 1 cdc 1 d 1 = [a, b][c, d]. By the Van Kampen Theorem, we have that π 1 (X) = π 1 (U V ) =< a, b, c, d [a, b][c, d] >. 3. (a) Use the cell structure for the orientable surface M g of genus g given in Hatcher pg 5 to compute its fundamental group. 1

2 Figure 2: U Figure 3: V (b) Argue that M g is not homeomorphic, or even homotopy equivalent, to M h if g h. Hint: Consider the abelianization of π 1 (M g ). Solution (Christian Geske): 3. (a) Hatcher instructs us to construct the orientable surface M g of genus g as follows. Let a 1,..., a 2g be 2g copies of S 1. Take their wedge sum 2g n=1 a n with basepoint x 0 and attach a single 2-cell via the map ϕ that sends S 1 to the loop g n=1 a 2n 1a 2n ā 2n 1 ā 2n to get M g. Let N π 1 ( 2g n=1 a n, x 0 ) be the normal subgroup generated by [ϕ(s 1 )] = [ g n=1 a 2n 1a 2n ā 2n 1 ā 2n ] = g n=1 [a 2n 1][a 2n ][a 2n 1 ] 1 [a 2n ] 1. Because 2g n=1 a n is path-connected and because π 1 ( 2g n=1 a n, x 0 ) = 2g n=1 Z (the free product of 2g copies of Z), by Hatcher Proposition 1.26 we have that π 1(M g, x 0 ) = ( 2g n=1 Z)/N. Because M g is path-connected, this can be written simply as π 1 (M g ) = ( 2g n=1 Z)/N. (b) Let surface M g be given. Then π 1 (M g ) = ( 2g n=1z)/n where N is defined in part (a). Lemma 1. For all groups G, for all N G, [G, G]/([G, G] N) = [G/N, G/N] (where [, ] denotes the commutator subgroup). Proof. Let π : G G/N be the quotient map. By the properties of a homomorphism, π([g, G]) [G/N, G/N]. Because π is surjective, π([g, G]) = [G/N, G/N]. Thus the restriction of π to [G, G] is a surjective homomorphism onto [G/N, G/N] with kernel ([G, G] N). The result follows from the first isomorphism theorem. By Lemma 1 and because N [ 2g n=1z, 2g n=1 Z], [π 1(M g ), π 1 (M g )] = [ 2g n=1z, 2g n=1z]/n. Moreover, this isomorphism is the identity morphism. Then the abelianization of π 1 (M g ) is given by π 1 (M g )/[π 1 (M g ), π 1 (M g )] = π 1 (M g )/([ 2g n=1z, 2g n=1 Z]/N) = 2g n=1 Z/[ 2g n=1z, 2g n=1z] where the last isomorphism follows from the third isomorphism theorem. But this is precisely the direct sum of 2g copies of Z. If h g, then the direct sum of 2g copies of Z is not isomorphic to the direct sum of 2h copies of Z. Because the abelianization of a group does not change (up to isomorphism) under an isomorphism of the group, we conclude that π 1 (M g ) is not isomorphic to π 1 (M h ). Solution by Yueyang Ding (a) Use the cell structure for the orientable surface M g of genus g given in Hatcher pg 5 to compute its fundamental group. (b) Argue that M g is not homeomorphic, or even homotopy equivalent, to M h if g h. Hint: Consider the abelianization of π 1 (M g ). Proof. (a) Let X be the space of the 0-cell and 1-cells of M g, then X is the wedge sum of 2g circles. Thus, Π 1 (X) =< a 1, b 1, a 2, b 2,, a n, b n >. Let the 2-cell e 2 be attached to X via the map ϕ : S 1 X, then ϕ = a 1 b 1 a 1 1 b 1 1 a 2b 2 a 1 2 a 1 2 a g b g a 1 g b 1 g. By Prop 1.26, we conclude that Π 1 (M g ) = Π 1 (M g )/N, where N is the normal subgroup generated by ϕ, i.e. Π 1 (X) =< a 1, b 1,, a n, b n a 1 b 1 a 1 1 b 1 1 a g b g a 1 g b 1 g >. 2

3 (b) Since the abelianization of Π 1 (M g ) is the free abelian group of rank 2g, if Π 1 (M g ) = Π 1 (M h ), then h = g. Therefore, M g is not homotopy equivalent to M h if h g. 4. The space RP 2 can be thought of in several equivalent ways: as the set of lines through the origin in R 3, as the S 2 with antipodal points x and x identified, or as the quotient of D 2 with antipodal points on the boundary D 2 identified. (a) Using the second description and the quotient map S 2 RP 2, use covering space arguments to prove that π 1 (RP 2, ) is isomorphic to Z 2. (b) Using the third description above, thought of a cell complex, compute π 1 (RP 2, ) using results proved about cell complexes. Solution (Kris Joanidis) RP n is the coequaliser of the antipodal map σ and id: DIAGRAM let U + k = {x 1, x 2,...x n+1 ) S n : x k > 0} U k = {x 1, x 2,...x n+1 ) S n : x k < 0} Clearly σ(u + k ) = U k and σ(u k ) = U + k and {U + k, U k } k {1,2...n+1} is an open cover. Hence σ acts in a properly discontinuous way, so p : S n RP n is a covering map. a; Using the correspondence between fibres and cosets, we have the following isomorphisms of sets: for all n > 1. π 1 (RP n ) p (π 1 (S n )) = π 1 (RP n ) = p 1 (x 0 ) = {, } Hence π 1 (RP n ) = 2 so π 1 (RP n ) = Z 2 b; We can define the cell structure in the following way: X 0 = { } X 1 = X 0 ϕ D 1 X 2 = X 1 ψ D 2 where ϕ is the unique map S 0 X 0, and ψ : S 1 X 1 is given by ψ(e 2πiϑ ) = ϑ D 1 X 1 for ϑ [0, 1) Using Proposition 1.26 in [?] with as the basepoint, we obtain: π 1 (RP 2, ) = a a 2 = Z 2 3

4 5. The n-fold dunce cap X is a certain generalization of RP 2. Consider the quotient of D 2 by identifying each point on D 2 with its orbit under the map r : S 1 S 1 sending z e 2πı/n z. Compute π 1 (X). Solution by Nicolle Sandoval Gonzalez Solution. As defined we have X = D 2 with each point of the boundary identified with its orbit under z e 2πi/n z. So then let V = X a, for a interior of X, and U be some open neighborhood of a not intersecting the boundary. Then π 1 (U) = 0 and π 1 (U V ) = Z. Now V can be deformation retracted to the boundary where the path γ such that γ(0) = z and γ(1) = e 2πi/n z generates π 1 (V ). In particular, we can represent V as an n-sided polygon with all the edges identified. Thus, V S 1 so π 1 (V ) = Z. By Van Kampen s theorem we know π 1 (X) = Z/N where N is the least normal subgroup contaning i 1 (1)[i 2 (1)] 1 = γ n. Thus modding by N induces the relation γ n = e. Since we said γ = π 1 (V ) we obtain that π 1 (X) = γ γ n = e = Z/nZ. 6. (a) Let A be a single circle in R 3. Compute π 1 (R 3 A). (b) Let A and B be disjoint circles in R 3. Compute π 1 (R 3 (A B). (c) How does π 1 (R 3 (A B) change if the circles are linked? Solution (by Michael Hankin): Main source: Hatcher, page 47 R 3 - a circle Taking A to be the embedding of S 1 in R 3 and X = R 3 A, X is deformation retractable to the image below, which is homeomorphic to S 2 S 1 as in the image below. 4

5 We have π 1 (S 1, x 0 ) = Z, π 1 (S 2, x 0 ) = 1, and π 1 (S 1 S 2, x 0 ) = 1 Thus, by SVK fundamental group of X is π 1 (X, x 0 ) = Z R 3-2 disjoint circles Taking A and B to be the embeddings of two unlinked copies of S 1 in R 3 and X = R 3 (A B), X is deformation retractable to S 2 S 1 S 1 S 2 as in the image below. Using the same logic as in the previous problem, the fundamental group of X is π 1 (X, x 0 ) = Z Z 5

6 R 3-2 linked circles Taking A and B to be the embeddings of two linked copies of S 1 in R 3 and X = R 3 (A B), X is deformation retractable to S 2 T 2 as in the image below. Using the following calculation: π 1 (S 2 T 2, x 0 ) = π 1 (S 2, x 0 ) π1({x 0},x 0) π 1 (T 2, x 0 ) = 1 1 π 1 (T 2, x 0 ) = π 1 (T 2, x 0 ) We can calculate the fundamental group of X: π 1 (X, x 0 ) = π 1 (S 2 T 2, x 0 ) = π 1 (T 2, x 0 ) = π 1 (S 1 S 1, x 0 ) = Z Z 7. (a) Define the nth dihedral group D n. (b) Define the infinite dihedral group D. (c) Let X be the Klein bottle, see Hatcher pg 51 for a cell structure of the Klein bottle. presentation of π 1 (X, ) and construct a homomorphism π 1 (X, ) D. Find a 8. Let X R 3 be the union of n lines through the origin. Compute π 1 (R 3 X). Solution (Giuseppe Martone) The sphere S 2 \ {2n points} is homotopically equivalent to R 3 \ X since the lines are all different and all passing through the origin. So we can define a retraction r : R 3 \ X S 2 \ {2n points} x l 0x S 2 where l 0x is the half line starting in 0 and passing through x. But S 2 \{2n points} = R 2 \{2n 1 points}. So we only need a formula that tell us the fundamental group of R 2 \ {m points} for all m N. We want to prove that π 1 (R 2 \ {m points}) is the free group on m generators. If m = 1 then we alrealdy know that R 2 minus a point is homotopically equivalent to S 1 (straight line homotopy). We can procede by induction on m. If we have m different points in the plane, there exists a line l that divides the plane in two halves with k points on one side and m k on the other with 0 < k < m. To see that, choose another point p in the plane such that the m points don t lie 6

7 on the same line passing through p. Since the lines passing through p are uncountable and our set of points is finite (and so it is the set of the lines passing through p and at least one of the other m points), we can find a line l with the required property. Again, since our set of points is finite, there exists an open interval I such that l I doesn t contain any of the m points. (The measure of I has to be less than the smallest distance from one of the points to l). Now take open sets U and V that contain l I and one of the two (path) connected components of R \ (X l I). Now, we can apply Seifert-van Kampen theorem and thanks to our induction hypotesis and the fact that U V = l I is contractible, we have that π 1 (R 2 \ {m points}) = π 1 (U) π1(u V ) π 1 (V ) = ( k i=1 Z i) ( m k i=1 Z i) = m i=1 Z i. 9. (a) Suppose that Y is obtained from a path-connected subspace X by attaching n-cells for a fixed n 3. Show that the inclusion X Y induces an isomorphism on π 1. (Hint: see Hatcher Proposition 1.26). (b) Use this fact to show that the complement of a discrete subspace of R n is simply-connected if n 3. Solution (Matthew Donner): Proposition 0.1. Suppose that Y is obtained from a path-connected subspace X by attaching n-cells for a fixed n 3. Show that the inclusion X Y induces a isomorphism on Π 1. Proof. We pick a basepoint x 0 in X. Let {e n α} α I denote the n-cells attached to X to form Y. Let λ α be a path in X from x 0 to the point ϕ(s 0 ), where s 0 is a basepoint of S n 1. Now, we begin by expanding Y to a space Z that deformation retracts onto Y. We obtain Z from Y by attaching rectangular strips S α with lower edge attached along λ α and right edge attached along an arc in e n α and all the left edges of the strips attached to each other. The top edges are not attached to anything, so Z deformation retracts onto Y as desired. Now, in each cell e n α we choose a point y α not in the are along which S α is attached. Let A = Z α {y α } and let B = Z X. Then A deformation retracts onto X. We show that B and A B have trivial fundamental groups. Since B is contractible, Π 1 (B) = 1. Next, we apply Van Kampen s Theorem to show that Π 1 (A B) = 1 Now, A B is covered by the open sets A α = (A B) β α e n β. Since A α deformation retracts onto S n 1, Π 1 (A α ) = 1. Thus, by Van Kampen s Theorem, Π 1 (A B) = 1. Now, since A, B, and A B are path connected and A and B are open sets, we can apply Van Kampen s Theorem: since Π 1 (A α ) = 1, Π 1 (Y ) = Π 1 (Z) = Π 1 (A) Π 1 (B) = Π 1 (X) 1 = Π 1 (X). Thus, Π 1 (Y ) = Π 1 (X). Proposition 0.2. Use this fact to show that the complement of a discrete subspace of R n is simplyconnected if n 3. Proof. Let Q be a discrete subset of R n. Then for each q Q we can find an n disk containing q in a way that none of the disks intersect. Let T denote the union of all these n-disks. We can deformation retract R n Q onto R n T. Moreover, we can attach n-disks to R n T, filling in all of the space, without chaning the fundamental group (by part a). The space that results from attaching n-disks is R n, which has a trivial fundamental group. Thus R n Q deformation retracts to a space whose fundamental group is trivial. Thus, the fundamental group of R n Q is trivial. Thus, R n Q is simply connected. 10. (a) Define the topological space RP n. (b) Write RP n as a cell complex. (c) Use the results of (??) and (9) to compute π 1 (RP n ). 7