(FALL 2011) 225A - DIFFERENTIAL TOPOLOGY FINAL < HOPF DEGREE THEOREM >

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1 (FALL 2011) 225A - DIFFERENTIAL TOPOLOGY FINAL < HOPF DEGREE THEOREM > GEUNHO GIM 1. Without loss of generality, we can assume that x = z = 0. Let A = df 0. By regularity at 0, A is a linear isomorphism. Write f(x) = Ax + u(x) where u(x) = o(). Choose a small such that Ax > u(x) on B(0, ). By using the homotopy H(t, x) = Ax + tu(x) Ax + tu(x), Ax + u(x) Ax we can conclude that the map is homotopic to the map Ax + u(x) Ax. In the last homework, we proved that GL n (R) has two (path-)connected components determined by the sign of a determinant. Thus every matrix A GL n (R) is homotopic to the identity I (if det A > 0) or a reflection R = diag( 1, 1,, 1) (if det A < 0) via a path inside GL n (R). In B(0, ), we can find a homotopy from Ax to x Ax (Case I) f preserves orientation at 0. det A = det df 0 > 0, so A is homotopic to I. Ax + u(x) W ( f, 0) = deg : B(0, ) Sk 1 Ax + u(x) x = deg : B(0, ) Sk 1 1 = deg I : B(0, ) Sk 1 = +1 since 1 I is bijective and orientation-preserving. (Case II) f reverses orientation at 0. Similarly, A is homotopic to R and W ( f, 0) = deg( 1 R) = 1 since 1 R is bijective and orientation-reversing. or Rx Rx. 1

2 2 GEUNHO GIM 2. Because z is a regular value, there are only finitely many preimages of z in int(b). Call them {x 1,, x t } and choose small balls B(x i, ) int(b) for each i. Note that u(x) = : B ( ( B i )) = (B int(b i )) S k 1 extends to B int(b i ). Since B int(b i ) is compact, we know that I( u, {y}) = 0 for a regular value y of u. Thus we can conclude that W ( f, z) = deg : B Sk 1 = deg : ( B i) S k 1 = deg : B i S k 1, i which equals the number of preimages of z counted with orientation convention(+1 if df xi is orientation-preserving, 1 otherwise) by Exercise (I guess this statement is true, but couldn t construct an extension explicitly. Instead, I proved a weaker statement which is enough for further arguments) Claim : Let B = B(0, 1) be a closed ball in R k, and let f : R k int(b) Y be a smooth map. If f : B Y is homotopic to a constant, then there exists a smooth map F : R k Y such that F = f on R k int(b(0, 2)). (Proof) Let H : B [0, 1] Y be a homotopy such that H 1 = f and H 0 is a constant map. Choose a smooth increasing function ρ : R R such that ρ(s) = 0, if s 1 3 1, if 1 2 s 3 2 s, if s 2 By using ρ, we can define a smooth map F : R k Y by f ρ() x, if 1 F (x) = x H, ρ(), if 1 Note that F is constant on the set {x 1, 1 3 }, and F = f is on Suppose f : S l R l+1 {0} has W (f, 0) = 0. Then, deg(u(x) := f(x) : S l S l ) = f(x) W (f, 0) = 0 by definition. The special case implies that u is homotopic to a constant map. On the other hand, H : S l [0, 1] R l+1 {0} gives a homotopy between f and u by H(x, t) = (1 t)f(x) + t f(x) ( 0 for any t, x). f(x) Therefore, f is homotopic to a constant map.

3 (FALL 2011) 225A - DIFFERENTIAL TOPOLOGY FINAL < HOPF DEGREE THEOREM > 3 5. Take a large ball B = B(0, r) so that it contains every preimage of 0. Then, W ( f : B R k {0}, 0) = 0 by Exercise 2. Let s assume that the special case holds when l = k 1, and that r = 1 by suitable scaling. By Exercise 4, Corollary holds for l = k 1, thus f is homotopic to a constant. By Exercise 3, we can construct g : R k R k {0} such that g = f on R k int(b(0, 2)). Note that g does not assume the value 0 because f R k int(b(0,1)) and the homotopy (between f and a constant) don t assume it. 6. Firstly, we need the following lemma. Lemma 1. Let f, g : S 1 S 1 be smooth maps. Then, f and g are homotopic if and only if deg f = deg g. Proof. ( ) Obvious. ( ) Suppose deg f = d = deg g. There exists a map h : R R such that f(e iθ ) = e ih(θ). Moreover, h(θ + 2π) h(θ) is a continuous map which takes values in a discrete set 2πZ. Thus h(θ + 2π) h(θ) = 2πm for some fixed m Z. Note that (1 t)h(θ + 2π) + tm(θ + 2π) = (1 t)h(θ) + tmθ + 2mπ. By the well-defined homotopy H(θ, t) = exp(i{(1 t)h(θ) + tmθ}), f is homotopic to the m-th power map θ m : S 1 S 1. Since m = deg θ m = deg f, m = d. Because both f and g are homotopic to θ d, they are homotopic to each other. We will use an induction on l to prove Special Case. First suppose l = 1. By the lemma, any smooth map f : S 1 S 1 with deg f = 0 is homotopic to the 0-th power map, i.e., a constant map. Assume that Special Case is true for l = 1, 2,, k 1. Let f : S k S k be a degree-0 map. Choose two distinct regular values a, b S k (possible by Sard). Let f 1 (a) = {a 1,, a s }. Take an open set U = R k such that U S k f 1 (b). By isotropy lemma, there exists a diffeomorphism h : S k S k, isotopic to the identity, such that h maps a 1,, a s into U. Since f = f id is homotopic to f h 1, we can assume that f 1 (a) U. We can view f U : U S k {b} as a map from R k to R k where a is identified with 0 in the latter case. By Exercise 2, the induction hypothesis, and the fact that U contains every preimage of a, Exercise 5 is applicable in this case. There exists a map g : U S k {a, b} such that g = f outside a compact set in U. Extend g : S k S k {a} by defining g = f on S k U. g is smoothly extended since f = g outside a compact set. We can define a homotopy H(x, t) = (1 t)f(x) + tg(x) on S k {b} = R k and extend it to S k by setting H(x, t) = f(x) when x S k U. Because g = f outside a compact set in U, and g(x) = f(x) implies H(x, t) = f(x), H is well-defined smooth homotopy between f and g. On the other hand, g is homotopic to a constant map since S k {a} is contractible. Thus, so is f.

4 4 GEUNHO GIM 7. Suppose W R M and f : W R k+1 is a smooth map. By -neighborhood theorem, there is an open set U of points in R M with distance less than from W. Choose a sufficiently small so that each point x U possesses a unique closest point w x W. Also, there exists a submersion π : U W which is the identity on W. We can extend f to U by composing g := f π : U R k+1. Take a smooth increasing function h : R R such that { 1, x /3 h(x) = 0, x 2/3 Extend g smoothly to the whole R M by setting { g(wx + (x w g(x) = x )h(d(x, W ))), x U 0, x R M U where d(x, W ) is the distance between the point x U and W. In particular, f is extendable to W. 8. Let W be a compact, connected oriented k + 1 dimensional manifold with boundary, and let f : W S k be a smooth map. If f extends to F : W S k, with F = f, then clearly deg f = 0. Conversely, suppose deg f = 0. By Exercise 7, f : W S k R k+1 extends to G : W R k+1. By taking a homotopy of F (which is invariant on W ) if necessary, we can assume that 0 is a regular value of G. By a similar argument in Exercise 6, we can take an open set U = R k+1 such that G 1 (0) U W. Take a closed ball B in U which contains G 1 (0). Note that G(x) is defined on a compact set W int(b), G(x) which has (W int(b)) = W B. Let G = G B, then G W ( G, 0) = deg : B Sk G G = deg : W Sk G = deg(f) = 0 by assumption. Since B = S k, Corollary after Exercise 4 is applicable in this case. G is homotopic to a constant map, thus by Exercise 3, G U int(b) : U int(b) R k+1 {0} enduces G : U R k+1 {0} such that G = G on U int(b ) where B U is a closed ball properly containing B. Also, it extends to W by defining G = G outside U. We get F = G G : W Sk which extends f.

5 9. (FALL 2011) 225A - DIFFERENTIAL TOPOLOGY FINAL < HOPF DEGREE THEOREM > 5 Theorem 1. (The Hopf Degree Theorem) Two maps of a compact, connected, oriented k-manifold X into S k are homotopic if and only if they have the same degree. Proof. If g, h : X S k are homotopic, they have the same degree. Conversely, suppose deg g = deg h. Let W = X [0, 1], and define f : W S k by f(x, 0) = g(x), f(x, 1) = h(x). Note that deg f = deg h deg g = 0. By Extension Theorem, f can be extended to F : W S k, which gives a homotopy between g and h.