QUALIFYING EXAM, Fall Algebraic Topology and Differential Geometry

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1 QUALIFYING EXAM, Fall 2017 Algebraic Topology and Differential Geometry 1. Algebraic Topology Problem 1.1. State the Theorem which determines the homology groups Hq (S n \ S k ), where 1 k n 1. Let X S n be homeomorphic to S p S q, 1 p,q n 1. Compute the homology groups H q (S n \X). Solution. First, we state the Jordan-Brouwer Theorem: Theorem 1. Let S n 1 S n be an embedded sphere in S n. Then the complement X = S n \S n 1 has two path-connected components: X = X 1 X 2, where X 1, X 2 are open in S n. Furthermore, X 1 = X 2 = S n 1. Recall the following fact: Theorem 2. Let S k S n, 0 k n 1. Then { (1) Hq (S n \S k ) Z, if q = n k 1, = 0 if q n k 1. Denote X p := S n \S p and X q := S n \S q, where S p S q S n. Then X p X q = S n \(S p S q ) = S n \ = R n, X p X q = S n \(S p S q ) = S n \(S p S q ) = X. Then the Mayer-Vietoris exact sequence gives H j (X) = H j (X p ) H Z, if p q and j = n p 1 or j = n q 1, j (X q ) = Z Z if p = q and j = n p 1, 0 if q n k 1. Problem 1.2. Let (X,A) be a CW-pair, and Y is a CW-complex as well. Let E = C(X,Y), B = C(A,Y), and the map p : E B be defined as p : (f : X Y) (f A : A Y). Prove that the map p : E B is a Serre fiber bundle. Solution. We have to show that the map p : C(X,Y) C(A,Y) satisfies the Lifting Homotopy property. Let K be a CW-complex, and f : K C(A,Y), f : K C(X,Y) be such that the diagram f f K C(X,Y) p C(A,Y) commutes. Let F : K I C(A,Y) be a homotopy such that F K {0} = f. In particular, F C(K I,C(A,Y))

2 2 We notice that the map p : C(X,Y) C(A,Y) induces the map of the space of continuous map p # : C(K I,C(X,Y)) C(K I,C(A,Y)). Furthemore, there are natural homepmorphisms: C(K I,C(X,Y)) = C(K X I,Y), C(K I,C(A,Y)) = C(K A I,Y). We obtain a map p # : C(K X I,Y) C(K A I,Y). The homotopy F : K I C(A,Y) gives the homotopy F : K A I Y by the formula F : (k,a,t) F(k,t)(a) Y. Here F K A {0} = f, where f : (k,a) f(k)(a). Let f : K X Y be the map given as f : (k,x) f(k)(x). Since the pair (K X,K A) is CW-pair, it is a Borsuk pair. Thus any map F : K A I Y such that F K A {0} = f, where f = f A extends to a map F : K X I Y such that F K A I = F and F K X {0} = f. Problem 1.3. Define the Hopf invariant. Prove that the group π 4n 1 (S 2n ) has infinite order. Solution. Let α π 4n 1 (S 2n ), and let f : S 4n 1 S 2n be a representative of α. Let X α = S 2n f D 4n. It is easy to compute the cohomology groups of X α : { Z, q = 0,2n,4n, H q (X α ;Z) = 0, otherwise. Let a H 2n (X α ;Z), b H 4n (X α ;Z) be generators. Since a 2 = a a H 4n (X α ;Z), then a 2 = h b, where h Z. The number h(α) = h is the Hopf invariant of the element α π 4n 1 (S 2n ). We use the following fact: Lemma 3. (1) h(α) = 0 if α = 0; (2) h(ϕ 1 )+h(ϕ 2 ) = h(ϕ 1 +ϕ 2 ). We obtain a homomorphism h : π 4n 1 (S 2n ) Z. Let ι 2n π 2n (S 2n ) be a generator given by the identity map S 2n S 2n. Then [ι 2n,ι 2n ] denote the Whitehead product. Lemma 4. The Hopf invariant is not trivial, in particular, h([ι 2n,ι 2n ]) = 2. First we compute the cohomology (together with a product structure) H q (S 2n S 2n ): Z, q = 0,4n, H q (S 2n S 2n ;Z) = Z Z, q = 2n 0, otherwise. Let c 1,c 2 H 2n (S 2n S 2n ) be such generators that the homomorphisms H 2n (S1 2n ) p 1 H 2n (S1 2n S2 2n ) p 2 H 2n (S 2n p 1 induced by the projections S1 2n S 2n 1 S2 2n p 2 S 2n 2, send the generators c 1 and c 2 to the generators of the groups H 2n (S1 2n ), H 2n (S2 2n ). Let d H 4n (S 2n S 2n ) be a generator. Then c 1 c 2 = d. We also note that c 2 1 = 0 and c 2 2 = 0 since by naturality p 1(c 1 ) 2 = 0 and p 2(c 2 ) 2 = 0. So we have that the ring H (S1 2n S2 2n) is generated over Z by the elements 1, c 1, c 2 with the relations c 2 1 = 0, c 2 2 = 0. In particular, we have: (c 1 +c 2 ) 2 = c c 1c 2 +c 2 2 = 2d. Proof of Lemma 4. We consider the factor space X = S 2n S 2n /, where we identify the points (x,x 0 ) = (x 0,x), where x 0 is the base point of S 2n. We notice that 2 )

3 3 The space X = S 2n S 2n / is homeomorphic to the space S 2n f D 4n, where f is the map defining the Whitehead product [ι 2n,ι 2n ]. Indeed S 2n S 2n = (S 2n S 2n ) w D 4n, where w is the map we described above. The generator ι 2n is represented by the identical map S 2n S 2n. The composition S 4n 1 w S 2n S 2n Id Id S 2n represents the element [ι 2n,ι 2n ]. It exactly means that the identification (S 2n,x 0 ) = (x 0,S 2n ) we just did in the space S 2n S 2n is the same as to attach D 4n with the attaching map (Id Id) w. Problem 1.4. Let n 1. Consider the map g : S 4n 2 S 5 proj (S 4n 2 S 5 )/(S 4n 2 S 5 4n+3 Hopf ) = S HP n. Prove that g induces trivial homomorphism in homology and homotopy groups, however g is not homotopic to a constant map. Solution. Consider the homomorphism g : π q (S 4n 2 S 5 ) proj 4n+3 Hopf π q S π q HP n. Since π q (S 4n 2 S 5 ) = π q S 4n 2 π q S 5, the homomorphism proj factors through the projections π q S 4n+3 4n 2 proj1 π q S π q (S 4n 2 S 5 ) proj2 π q S 5 π q S 4n+3 Since the maps S 4n 2 S 4n+3 and S 5 S 4n+3 are homotopically trivial, the homomorphism g : π q (S 4n 2 S 5 ) π q S 4n+3 π q HP n is trivial as well. The homomorphism g : Hq (S 4n 2 S 5 ) proj H Hopf 4n+3 q S H q HP n is also trivial. Indeed, the group H 4n+3 S 4n+3 = Z and trivial ortherwise and H q HP n is non-trivial only if q = 4,...,4n, i.e., any homomorphism H q S 4n+3 H q HP n is trivial. Now we assume that the map g : S 4n 2 S 5 HP n is contractible. Let g t : S 4n 2 S 5 HP n be a homotopy such that g 0 = g and g 1 (S 4n 2 S 5 ) = x 0 HP n. By the Lifting Homotopy Property, there exists a homotopy f t : S 4n 2 S 5 S 4n+3 such that the diagram f t S 4n 2 S 5 g t S 4n+3 Hopf HP n commutes. Here f 0 = proj as above. Then we have that f 1 (S 4n 2 S 5 ) Hopf 1 (x 0 ) = S 3. We notice that the homomorphism proj : H 4n+3 (S 4n 2 S 5 ) H 4n+3 (S 4n+3 ) is non-trivial (it is an isomorphism). However (f 1 ) must be trivial since the map f 1 factors through S 3 : Since (f 1 ) = proj, we obtain a contradiction. S 3 S 4n+3 f 1 h 1 Hopf S 4n 2 S 5 g 1 HP n Problem 1.5. (i) Let f : RP 4n CP 2n RP 4n CP 2n be a continuous map. Prove that f always has a fixed point. (ii) Give an example of a map g : RP 2k CP k RP 2k CP k without fixed points.

4 4 Solution. (i) We compute the cohomology groups H (RP 4n CP 2n ;Q): { H j (RP 4n CP 2n Q, q = 0,2,,4n, ;Q) = 0, otherwise. Indeed, wehavethat H j (RP 4n ;Q) = 0. ThentheKünnethformulaimpliestheresult. Furthermore, we see that the projection RP 4n CP 2n CP 2n induces an isomorphism H (CP 2n ;Q) = H (RP 4n CP 2n ;Q) = Q[z]/z 2n+1, where z H 2 (RP 4n CP 2n ;Q). Let f : RP 4n CP 2n RP 2k CP 2n be a map. Then f (z) = λ z H 2 (RP 2k CP 2n ;Q). Then we have that f (z k ) = λ k zk. We obtain the Lefcshetz number L(f) = 1+λ+ +λ 2n. If λ = 1, then L(f) 0. If λ 1, we have L(f) = λ2n+1 1 λ 1 0. Since RP 2k CP 2n is a finite CW-complex, Lefcshetz fixed-point theorem implies that f has a fixed point. (ii) Consider CP 1 = S 2 and the intipodal map A : x x: this map does not have fixed points. By construction, the map Id A : RP 2 CP 1 RP 2 CP 1 does not have fixed points. Problem 1.6. Compute the homotopy group π q (S 2 S 2 ) for q = 1,2,3. Solution. We have that π 1 (S 2 S 2 ) = 0 since S 2 S 2 has a CW-decomposition with one zero cell and two 2-cells. Then the CAT iplies that any map S 1 S 2 S 2 is homotopic to a constant map. Then we know that π 2 (S 2 S 2 ) = π 2 (S 2 ) π 2 (S 2 ) = Z Z. We notice that the inclusion map i : S 2 S 2 S 2 S 2 induces an isomorphism π 2 (S 2 S 2 ) π 2 (S 2 S 2 ). Indeed, we have that S 2 S 2 = (S 2 S 2 ) w D 4, where w : S 3 S 2 S 2 is the Whitehead map. Now we have that the homomorphism i : π 3 (S 2 S 2 ) π 3 (S 2 S 2 ) has the kernel given by the attaching map w : S 3 S 2 S 2. Thus we have the exact sequence Thus we conclude π 3 (S 2 S 2 ) = Z Z Z. Z π 3 (S 2 S 2 ) π 3 (S 2 S 2 ) Problem 1.7. Let Mg 2 be a two-dimensional surface of genus g 1 (oriented). Compute the homotopy groups π q (Mg) 2 for all q 1. Hint: use covering spaces. Solution. The standard model for M 2 g is 4g-gone labeled as a j,b j,ā j, b j, j = 1,...,g. Here the notations ā j, b j mean that a j, b j are identified with the sides ā j, b j in the opposite direction. Let F(a j,b j,ā j, b j j = 1,...,g) denote a free group with generators a j,b j,ā j, b j j = 1,...,g. Thus the fundmental group π 1 (M 2 g) is isomorphic to the group F(a j,b j j = 1,...,g)/a 1 b 1 a 1 1 b 1 1 a g b g a 1 g b 1 g. Now we prove the following Theorem 5. π q (M 2 g) = 0 for all q 2. Proof. Let g = 1 then Mg 2 = T 2, and we have the univesral covering space p : R 2 T 2, where p : (θ 1,θ 2 ) (e iπθ1,e iπθ2 ) T 2. Then p : π q (R 2 ) π q (T 2 ) induces isomorphism for all q 2. Now, by induction, assume that π q (Mg) 2 = 0 for all q 2 for some g 1. Then we consider the following covering p : Z g Mg+1. 2 Here Z g is given by an infinite cylinder R S 1, where every coordinate (k/2, ) R S 1 is used to take a connected sum with Mg 2, see the figure below, where g = 1:

5 5 Then we have a map p : Z g M 2 g+1 which sends (x,α) (eiπx,α). It is easy to see that p is a covering. Let f : S q Z g be a map. The image of f is compact, so f(s q ) Y g Z g, where Y g has only [ N,N] S 1 with connected sums of M 2 g as before. Then Y g is nothi g but an oriented surface M 2 ḡ for some ḡ with two disks removed. Such space Y g is homotopy equivalent to one-point union of circles. Thus the map f : S q Y q Z q is homotopically trivial, i.e., this shows that π q M 2 g+1 = 0. Problem 1.8. Let T 2 = S 1 S 1 be a torus. Prove that any map f : RP 2 T 2 is homotopic to a constant map. Solution. We consider the universal covering p : R 2 T 2. Let f : RP 2 T 2 be a map, where a base point x 0 RP 2 maps to y 0 = f(x 0 ). Let ỹ 0 p 1 (y 0 ) be a point in R 2. We know that π 1 RP 2 = Z 2, and π 1 T 2 = Z Z. Thus f (π 1 RP 2 ) = 0 π 1 T 2 for any map f : RP 2 T 2. We use the following result: Theorem 6. Let p : T X be a covering space, and Z be a path-connected space, x 0 X, x 0 T, p( x 0 ) = x 0. Given a map f : (Z,z 0 ) (X,x 0 ) there exists a lifting f : (Z,z 0 ) (T, x 0 ) if and only if f (π 1 (Z,z 0 )) p (π 1 (T, x 0 )). In our case, it means that there exists a map f : RP 2 R 2 such that f(x 0 ) = ỹ 0 and f = p f. Thus the homotopy class of f is in the image of p : π 1 R 2 T 2. Hence f 0. Problem 1.9. Find H 1 (X;Z) and H 1 (X;Z) in terms of the group π, where π is a finite group. Solution. We have that H 1 (X;Z) = π/[π,π]. If π/ is finite group, then π/[π,π] is finite abelian group. The universal coefficient theorem yields H 1 (X;Z) = Ext(H 0 (X,Z),Z) Hom(H 1 (X,Z),Z) Then Ext(H 0 (X,Z),Z) = 0 since H 0 (X,Z) is free abelian, and Hom(H 1 (X,Z),Z) = 0 since H 1 (X,Z) is finite. Thus H 1 (X;Z) = 0. Problem Prove that the spaces CP S 3 and S 2 have isomorphic homotopy groups and that they are not homotopy equivalent. Solution. First, we claim that π 2 CP = Z, and π q CP = 0 for all q 0. Indeed, consider the Hopf fibre bunde S 2n+1 CP n with the fibre S 1. Then the exact sequence in homotopy groups π q (S 1 ) π q (S 2n+1 ) π q (CP n ) π q 1 (S 1 ) imlies that π 2 CP = Z and π q CP = 0 for all q 0 and q < 2n+1. Now we hve that { π q (CP S 3 ) = π q CP π q S 3 Z, q = 2, = π q S 3, q 2 The Hopf bundle S 3 S 2 gives us that π q S 2 = π q S 3 for q 3. We conclude that the homotopy groups of the spaces CP S 3 and S 2 have isomorphic homotopy groups. Now we assume that there is a homotopy equivalence f : S 2 CP S 3. Then f has to induce an isomorphism in all

6 6 homotopy groups. Then f = (f 1,f 2 ), where f 1 : S 2 CP, and f 2 : S 2 S 3. However, f 2 is homotopically trivial, and for q 3, we have that a map ϕ : S q S 2 maps to the composition which is homotopically trivial. S q ϕ S 2 f S 3