Differential Topology Assignment 5 Exercise 1: Classification of smooth 1-manifolds Let M be a compact, smooth and connected 1−manifold with boundary. Show that M is either diffeomorphic to a circle or to a closed interval [a; b] ⊂ R. This then shows that every compact smooth 1−manifold is build up from connected components each diffeomorphic to a circle or a closed interval. To prove the result, show the following steps: 1. Consider a Morse function f : M ! R. Let B denote the set of boundary points of M, and C the set of critical points of f. Show that M n (B[C) consists of a finite number of connected 1−manifolds, which we will call L1;:::;LN ⊂ M. 2. Prove that fk = fjLk : Lk ! f(Lk) is a diffeomorphism from Lk to an open interval f(Lk) ⊂ R. 3. Let L ⊂ M be a subset of a smooth 1−dimensional manifold M. Show that if L is diffeomorphic to an open interval in R, then there are at most two points a, b 2 L such that a, b2 = L. Here, L is the closure of L. 4. Consider a sequence (Li1 ;:::;Lik ) 2 fL1;:::;LN g. Such a sequence is called a chain if for every j 2 f1; : : : ; k − 1g the closures Lij and Lij+1 have a common boundary point. Show that a chain of maximal length m contains all of the 1−manifolds L1;:::;LN defined above. 5. To finish the proof, we will need a technical lemma. Let g :[a; b] ! R be a smooth function with positive derivative on [a; b] n fcg for c 2 (a; b). Then there exists a smooth function g~ :[a; b] ! R with positive derivative on the whole of [a; b] such that g =g ~ on a neighborhood of a and on a neighborhood of b. 6. Finally use the previous lemma to show that if L1 and LN , contained in the maximal chain, have a common boundary point, then M is diffeomorphic to a circle. Otherwise, M is diffeo- morphic to a closed interval. Exercise 2: Brouwer's fixed point theorem In the following, we use the continous version of Brouwer's fixed-point theorem: every continuous ¯ ¯ ¯ n function f : Bn ! Bn has a fixed point. Here Bn is the closed unit ball in R . In this exercise, we show the Perron-Frobenius theorem: n×n A matrix A 2 R with Ai;j ≥ 0 for all i; j 2 f1; : : : ; ng has a non-negative eigenvalue. In the following, think of A as a linear map A : Rn ! Rn (expressed in the standard basis). 1. Show that n−1 n−1 S+ = fx = (x1; : : : ; xn) 2 S j xj ≥ 0 for all j = 1; : : : ; ng ¯ is homeomorphic to Bn−1. 2. Consider the map f(x) = Ax=kAxk to prove the theorem. Let X be a topological space and A ⊂ X a subspace. Then A is called a retract of X if there is a retraction from X to A, that is, a continuous map r : X ! A whose restriction to A is the identity map. ¯ 3. Show that if A is a retract of Bn and f : A ! A is continuous, then f has a fixed point. 1 Differential Topology Exercise 3: mod 2-degree and \one half" the fundamental theorem of algebra Let M be a smooth, compact (n+1)−dimensional manifold with boundary, Y a smooth, connected n−dimensional manifold and f : @M ! Y a smooth map. 1. Show that deg2(f) = 0 if there is a smooth extension F : M ! Y of f to M. Let M ⊂ R2 be a compact smooth submanifold with boundary. Let g : R2 ! R2 be smooth 1 g(x) with g(x) 6= 0 for all x 2 @M. Define f : @M ! S by f(x) = kg(x)k . 2. Show: if deg2(f) 6= 0, then there is an x 2 M with g(x) = 0. In the following, we will identify R2 with C, and use complex variables to specify functions. n 1 3. Compute deg2(g) for the case where g(z) = z , n 2 N and M = S = N. 4. Deduce that every complex polynomial of odd degree has a root in C. (Hint: Define a homotopy to zm, where m is the degree of the polynomial.) Exercise 4: Smooth no-retraction theorem and mod 2-degree 1. Show that the identity map id : M ! M on a compact connected smooth manifold M is not smoothly homotopic to a constant map. 2. Use this result for M = Sn in order to prove the smooth no retraction theorem for the unit ball. Exercise 5: mod 2-intersection theory Let M; N be smooth manifolds where M is compact, and Z ⊂ N a submanifold with M; N; Z boundaryless and dim M + dim Z = dim N. Assume f : M ! N smooth with f t Z. Define −1 I2(f; Z) := jf (Z)j mod 2 : 1. Show that this is well-defined. (Hint: look at exercise 4 of assignment 5.) 2. Show that I2(f; Z) is a smooth homotopy invariant among all functions f : M ! N with f t Z. You may use the following result: if f t Z and g t Z are homotopic, then there is a homotopy H : M × [0; 1] ! N with H t Z. The transversality homotopy theorem states that for any smooth map f : M ! N, there is a smooth map g : M ! N which is homotopic to f and satisfies g t Z and @g t Z. For an arbitrary smooth function f : M ! N define I2(f; Z) := I2(g; Z) ; where g : M ! N is homotopic to f and satisfies g t Z. 3. Show that this is well-defined and a homotopy invariant of functions f : M ! N. 4. Show that I2(f; fqg) does not depend on q 2 N and coincides with deg2(f). 2.